NCERT Solutions Class 12 Maths Chapter 6

NCERT Solutions Class 12 Mathematics Chapter 6 – Application of Derivatives

Mathematics is a subject requiring a strong conceptual understanding of its applications. Hence, students are advised to study the subject from good resources to gain in-depth knowledge about  Mathematics chapters with examples, solutions, theorems as well as exercises and to score well in exams.

The Class 12 Mathematics Chapter 6 ‘Application of derivatives’ is a part of the Calculus section of Mathematics. It covers the various applications of derivatives and their role while doing calculations. The vital topics covered in Chapter 6 Class 12 Mathematics include:

  • Introduction
  • Rate of Change of Quantities
  • Increasing and Decreasing Functions
  • Tangents and Normals
  • Approximations
  • Maxima and Minima
  • Maximum and Minimum Values of a Function in a Closed Interval

The chapter’s important formulas are listed in the NCERT Solutions Class 12 Mathematics Chapter 6. Also, one can find easy ways to carry out calculations after referring to them thoroughly and consistently.

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Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 6

NCERT Solutions Class 12 Mathematics Chapter 6 is about derivatives and their applications.

This chapter is a part of Calculus and requires the study of differential calculus. One should have a good command of it to excel in this chapter.

If you have a hold on calculus, this chapter will help students to learn to carry out calculations easily. As a result, they will be able to approach problems more logically. The entire chapter is covered thoroughly in the NCERT Solutions Class 12 Mathematics Chapter 6 and is available on the Extramarks website. 

After completing Chapter 6 Mathematics Class 12, students will learn to analyse the problems with a better approach and be able to solve them easily.

Introduction

A derivative is the rate of change or the amount on which a particular function changes at one given point. 

Tangent line

In the above-given figure, the function is represented in black colour, and a tangent is represented in red colour.

Rate of Change of Quantities

The derivative ds/dt is used to show the rate of change of distance s to the time t.

Assume a particular quantity y varies with another quantity x which satisfies y = f(x), here dy/dx  or  f’(x) shows the rate of change of y w.r.t. x and [dy/dx]x = x0 or f’(x0) shows the rate of change of y w.r.t. x at x = x0.

Now, assume the two variables x and y are varying w.r.t. another variable t i.e. x = f(t) and y = g(t) 

Thus, by chain rule

dy/dx = (dy/dt)/(dx/dt) here dx/dt ≠ 0

So, the rate of change of y to x could be calculated using the rate of change of y at a given time and that of x to t.

Increasing and Decreasing Functions

The function’s derivative might determine if it is increasing or decreasing at any intervals in its domain.

Assume I will be an open interval contained in the domain of the real-valued function f. Then f is said to be 

(i) increasing on I when x1 < x2 in I => f(x1) ≤ f(x2) for all x1, x2 Є I.

(ii) strictly increasing on I if x1 < x2 in I => f(x1) < f(x2) for all x1, x2 Є I.

(iii) decreasing on I if x1 < x2 in I => f(x1) ≥ f(x2) for all x1, x2 Є I.

(iv) strictly decreasing on I if x1 < x2 in I => f(x1) > f(x2) for all x1, x2 Є I. 

However, some functions are neither increasing nor decreasing. 

The graphical representations of all these functions are given below:

Increasing and Decreasing Functions

function f will be said to be increasing at x0 when there exists an interval I = (x0 – h, x0 + h), h > 0 such that for  x1, x2 ∈  I,

x1 < x2 in I => f (x1) ≤ f (x2)

With the help of a theorem listed in the NCERT Solutions Class 12 Mathematics Chapter 6, you can test for increasing and decreasing functions for a given interval.

Tangents and Normals

For the given equation of a straight line that is passing through a given point (x0, y0) having a finite slope, m is represented as y – y0 = m(x – x0)

Tangents and Normals

Assume the given curve y = f(x), and the tangent of the curve at that point (x0, y0) will be

[dy/dx](x0, y0) or f’(x0)

Now, the equation of the tangent of the curve y = f(x) at (x0, y0) will be

y – y0 = f’(x0)(x – x0)

The slope of the normal to curve y = f(x) at  (x0, y0) is given by -1/ f’(x0) here f’(x0) ≠ 0

For the equation of normal to the curve, y = f(x) at (x0, y0) will be

Y – y0 = (-1/f’(x0))(x – x0)

(y – y0)* f’(x0) +(x – x0) = 0

Particular cases

(i) If the slope of the tangent line is zero, then tan θ = 0, i.e. θ = 0, which means a tangent line is parallel to the x-axis. For this case, the equation of the given tangent at the point (x0, y0) will be y = y0.

(ii) If θ -> π/2 then tan θ → ∞, that means the tangent line will be perpendicular to the x-axis, that is parallel to the y-axis. For this case, the equation of the tangent at (x0, y0) will be x = x0.

Approximations

An approximation is anything which is similar but not the same as something else.

Assume f : D => R, D Ì R, will be a given function and assume y = f (x). Assume ∆x denotes a small increment in terms of x.

Approximations

Now, consider the increment in y corresponding to the increment in x will be ∆y = f (x + ∆x) – f (x).

(i) The differential of x is represented as dx = ∆x.

(ii) The differential of y is represented as dy = f’(x) dx or dy = (dy/dx) * ∆x

Maxima and Minima

In the section, you will learn the different methods of calculating a function’s maximum and minimum values in a given domain. You will also learn about the absolute maximum and minimum of a function used to find the solution to many applied problems.

You will clearly understand this topic through the definitions and theorems included in the NCERT Solutions Class 12 Mathematics Chapter 6 available on the Extramarks official website.

Maximum and Minimum Values of a Function in a Closed Interval

You will learn about the two theorems to find out the absolute maximum and minimum values of a function upon a closed interval I.

Theorem: Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value, as well as f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

Theorem: Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then

(i) f′(c) = 0 when f attains its absolute maximum value at c.

(ii) f′(c) = 0 when f attains its absolute minimum value at c.

NCERT Solutions Class 12 Mathematics Chapter 6 Exercise &  Solutions

Find all the NCERT solutions to the exercises covered in the chapter in the NCERT Solutions Class 12 Mathematics Chapter 6. It is prepared by the subject experts while adhering to the NCERT book and following the CBSE guidelines and curriculum. 

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NCERT Solutions Class 12 Mathematics Chapter 6 Exercise & Answer Solutions

NCERT Solutions Class 12 Mathematics Chapter 6 Exercise and Answer Solutions are available for students to refer for free on the Extramarks website. The material offers step-by-step solutions that help students understand how to solve problems relating to the chapter. The solution also helps students solve complex questions in a simplified manner. Students may refer to NCERT Solutions Class 12 Mathematics Chapter 6 to strengthen their basics.

Students may refer to the links below to download exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 6 Application of Derivatives:

NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematics book is an excellent source for students preparing for JEE Mains, NEET, MHT-CET etc. It has questions according to the topics and concepts covered in the NCERT textbook. This aids students in solving all types of questions confidently.

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The NCERT solutions Class 12 Mathematics Chapter 6 is prepared after analysing CBSE past years’ question papers. It contains extra questions from the NCERT Class 12 Mathematics textbook. Students can refer to NCERT Exemplar for Class 12 Mathematics for more practice and face their examinations with courage. They can assure themselves that nothing remains untouched in the chapter and will score very well in all their examinations. To get good grades in exams students must refer to multiple study resources, practice a lot of questions and stick to a study schedule and follow it religiously to come out with flying colours. 

Key Features for NCERT Solutions Class 12 Mathematics Chapter 6

Regular practice is quite necessary for the students to excel. Hence, NCERT Solutions Class 12 Mathematics Chapter 6 helps students develop the habit of regular practice and to clarify their doubts then and there, take regular tests to assess their performance and get proper feedback to step up their learning. . The key features are as follows: 

  • You can find all the topics covered as well as in text to end text exercises from the chapter covered in the NCERT Solutions Class 12 Mathematics Chapter 6.
  • It helps students analyse the easy and difficult topics in the chapter.
  • After completing the NCERT Solutions Class 12 Mathematics Chapter 6, students will become experts in solving derivatives problems.

Q.1 Find the rate of change of the area of a circle with respect to its radius r when

(a) r = 3 cm (b) r = 4 cm

Ans.

The area of a circle (A) with radius (r) is given by,          A=πr2Now, the rate of change of the area with respect to its radius is given by,  dAdr=ddrπr2=2πr(a)When r=3 cm  dAdr=2π(3)=6πHence, the area of the circle is changing at the rate of 6π cm2/cm when its radius is 3 cm.(b)When r=4 cm  dAdr=2π(4)=8πHence, the area of the circle is changing at the rate of 8π cm2/cm when its radius is 8 cm.

Q.2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Ans.

Let x be the length of a side, V be the volume, and S be the surface area of the cube.Then, V = x3 and S = 6x2 where x is a function of time t.It is given that,  dVdt=8cm3/sThen, by using the chain rule, we have:    8=dVdt  =ddt(x3)        =3x2dxdt  dxdt=83x2  ...(i)Now,  dSdx=ddx(6x2)        =12xdxdt        =12x×83x2[From equation (i)]        =32xdSdx=3212[Putting x=12] dS dx = 8 3 c m 2 /s. Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasingat the rate of 8 3 cm 2 /s. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@B332@

Q.3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.

Ans.

The area of a circle (A) with radius (r) is given by,  A=πr2Now, the rate of change of area (A) with respect to time (t) is given by,dAdt=ddtπr2        =2πrdrdt[By chain rule]        =2πr×3[drdt=3cm/s]dAdt=2π(10)×3[Putting r=10 cm]        =60π  cm2/sHence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π  cm2/s.

Q.4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Ans.

Let the length of a side be x and V be the volume of the cube. Then,         V=x3Differentiating both the sides with respect to t, we getdVdt=3x2dxdt[By​​ chain rule]        =3x2(3)[dxdt=3cm/s,Given]        =9x2Since,​ edge of cube(x)=10​ cm,  sodVdt=9(10)2        =900cm3/sHence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long.

Q.5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?

Ans.

The area of a circle (A) with radius (r) is given byA=πr2Therefore, the rate of change of area (A) with respect to time (t) is given by, dAdt=ddtπr2    =2πrdrdt[By chain rule]    =2π(8)(5)[drdt=5cm/s,r=8cm, Given]    =80πHence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm2/s.

Q.6 The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Ans.

Let the circumference of a circle be C with radius (r) is given byC=2πrThen, the rate of change of circumference (C) with respectto time (t) is given by,      dCdt=ddt2πr    =2πdrdt[By chain rule]      dCdt=2π(0.7)[drdt=0.7cm/s]      dCdt=1.4πHence, the rate of increase of the circumference​​​​​ is1.4πcm/s.

Q.7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

Ans.

Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have: dxdt=5cm/minute  and  dydt=4 cm/minute(a) Perimeter of rectangle(P)=2(x+y)      Differentiating w.r.t. t, we getdPdt=2ddt(x+y)        =2(dxdt+dydt)        =2(5+4)        =2cm/minuteThus, the perimeter is decreasing at the rate of 2 cm/min.(b)Area of rectangle(A)=xy      Differentiating w.r.t. t, we get  dAdt=ddtxy=xdydt+ydxdt            [By product rule]=4x5y[dxdt=5cm/min  and  dydt=4 cm/min]=4(8)5(6)=3230=2cm2/minHence, the area of the rectangle is increasing at the rate of 2 cm2/min.

Q.8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Ans.

The volume of a sphere (V) with radius (r) is given by,V=43πr3Differentiating both sides, with respect to t, we getdvdt=ddt43πr3=43π×3r2drdt          900=4πr2drdt[dVdt=900 cm3/sec]  drdt=9004πr2=225π(15)2[r=15​ cm]=225π(225)=1πHence, the rate at which the radius of the balloon increases when the radius is 15 cmis  1πcm/sec.

Q.9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.

Ans.

Volume of sphere having radius r=43πr3V=43πr3Differentiating with respect to r, we getdVdr=ddr(43πr3)        =43πddrr3        =43π×3r2        =4πr2        =4π(10)2        =400πHence, the volu e of the balloon is increasing at the rate of 400π.

Q.10 A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Ans.

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x m away from the wall. Then in ΔABC, by Pythagoras theorem, we have: x 2 + y 2 =25 [Length of the ladder = 5 m] y= 25 x 2 Differentiating both sides with respect to t, we get dy dt = d dt 25 x 2 = 2x 2 25 x 2 dx dt [Bychain rule] = x 25-x 2 2 [ dx dt =2 cm/s] = 2 4 25 4 2 = 8 3 Hence, the height of the ladder on the wall is decreasing at the rate of 8 3 cm/s.

Q.11 A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x-coordinate.

Ans.

The equation of the curve is given as:6y=x3+2The rate of change of the position of the particle with respect to time (t) is given by,    6dydt=ddt(x3+2)    6dydt=3x2dxdtWhen the ycoordinate of the particle changes 8 times as fast as the xcoordinate i.e.,dydt=8dxdt, we have        6(8dxdt)=3x2dxdt    (16x2)dxdt=0        16x2=0          x=±4When  x=4,y=(4)3+26=11When  x=4,y=(4)3+26    =626=313Hence, the points required on the curve are(4,11) and (4,313).

Q.12 The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

Ans.

The air bubble is in the shape of a sphere.So, the volume of an air bubble (V) with radius (r) is given by,V=43πr3The rate of change of volume (V) with respect to time (t) is obtained by differentiating both sides with respect to t, so      dVdt=ddt(43πr3)      dVdt=43×3πr2drdt    =4π(1)2(12)[drdt=12cm/sand r=1cm]    =2π  cm3/sHence, the rate at which the volume of the bubble increases is 2π cm3/s. 

Q.13 A balloon, which always remains spherical, has a variable diameter 32(2x+1). Find the rate of change of its volume with respect to x.

Ans.

The volume(V) of a sphere with radius (r) is given by,V=43πr3Diameter of balloon=32(2x+1)So, the radiusof balloon=34(2x+1)V=43π34(2x+1)3    =43π×2764(2x+1)3        V=916π(2x+1)3Hence, the rate of change of volume with respect to x is given bydVdx=916πddx(2x+1)3        =916π×3(2x+1)2ddx(2x+1)                                          [By chain rule]dVdx=2716π(2x+1)2×2dVdx=278π(2x+1)2Thus, the rate of change of volume with respect to x is 278π(2x+1)2.

Q.14 Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always onesixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?

Ans.

The volume of a cone with radius (r) and height (h) is given by,    V=13πr2hGiven, height of cone=16rh=16rr=6h  V=13π(6h)2h        =12πh3Differentiating both sides with respect to time (t),we getdVdt=ddt(12πh3)        =12πddth3        =36πh2dhdt[By chain rule]12=36π(4)2dhdt[dVdt=12andh=4cm]dhdt=1236π(4)2=148πHence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 148π cm/s.

Q.15 The total cost C (x) in Rupees associated with the production of x units of an item is given
by C(x) = 0.007x3 – 0.003 x2 + 15x + 4000 Find the marginal cost when 17 units are produced

Ans.

The total cost C (x)=0.007x3 0.003 x2+ 15x + 4000Marginal cost is the rate of change of total cost with respect to output.Marginal cost (MC)=dCdx      =0.007(3x2)0.003(2x)+15When x = 17,       MC =0.021(17)20.006(17)+15        =6.0690.102+15        =20.967Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

Q.16 The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 13x2 + 26x + 15 Find the marginal revenue when x = 7.

Ans.

The total revenue in Rupees received from the sale of x units of a product is      R(x)=13x2+26x+15Marginal Revenue (MR)=ddxR(x)      =ddx(13x2+26x+15)      =26x+26      =26(7)+26[When x = 7]      =208Hence, the required marginal revenue is Rs 208.

Q.17 The rate of change of the area of a circle with respect to its radius r at r =6 cmis(A)10π (B)12π (C)8π (D)11π

Ans.

T h e a r e a ( A ) o f a c i r c l e w i t h r a d i u s ( r ) i s g i v e n b y , A = π r 2 D i f f e r e n t i a t i n g b o t h s i d e s w i t h r e s p e c t t o r , w e g e t d A d r = d d r π r 2 = 2 π r ( d A d r ) r = 6 = 2 π ( 6 ) = 12 π H e n c e , t h e r e q u i r e d r a t e o f c h a n g e o f t h e a r e a o f a c i r c l e i s 12 π . T h e c o r r e c t a n s w e r i s B .

Q.18 The total revenue in Rupees received from the sale of x units of a product is given by  Rx =3x2+36x+5.The marginal revenue, when x = 15 is:(A) 116    (B) 96    (C) 90    (D) 126

Ans.

    Marginal revenue =The rate of change of total revenue with respect to the number ofunits sold.Marginal Revenue (MR)=ddxR(x)      =ddx(3x2+36x+5)      =6x+36Putting x = 15, we get  Marginal Revenue (MR)=6(15)+36      =90+36      =126Hence, the required marginal revenue is Rs 126.The correct answer is D.

Q.19 Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Ans.

Function is given by f(x)=3x + 17Differentiating w.r.t. x, we get    f(x)=3>0in every interval of R.Thus, the function is strictly increasing on R.

Q.20 Show that the function given by f(x) = e2x is strictly increasing on R

Ans.

Let x1 and x2 be any two numbers in R.Then,we have:x1 < x22x1 < 2x2  e2x1 < e2x2f(x1)<f(x2)Hence, f is strictly increasing on R.

Q.21

Show that the function given by f(x) = sin x is (a) strictly increasing in ( 0 , π 2 ) , (b) strictly decreasing in ( π 2 , π ) , (c) neither increasing nor decreasing in ( 0 , π )

Ans.

The given function is f(x) = sin x.f(x)=cosx(a)Since for each  x(0,π2),cosx>0      wehave f(x)>0.Hence, f is strictly increasing in(0,π2).(b)Since for each  x(π2,π),cosx<0      wehave f(x)<0.Hence, f is strictly decreasing in(π2,π).(c)From the results obtained in (a) and (b), it is clear that f is     neither increasing nordecreasing in (0, π).

Q.22 Find the intervals in which the function f given by f(x) = 2x2 − 3x is

(a) strictly increasing (b) strictly decreasing

Ans.

The given function is f(x)=2x2 3x.      f(x)=4x3Therefore, f(x)=04x3=0  x=34Now the point x=34 divides the real line into two disjoint intervalsnamely, (,34) and (34,).

In the interval ( , 3 4 ), f’( x )=4x3<0 Therefore, f is strictly deacreasing in this interval. Also, in the interval( 3 4 , ), f’( x )>0and so the function f is strictly increasing in this interval. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGceaqabeaacaqGjbGaaeOBaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabkhacaqG2bGaaeyyaiaabYgacaqGGaWaaeWaaeaacqGHsislcqGHEisPcaGGSaGaaGPaVpaalaaabaGaaG4maaqaaiaaisdaaaaacaGLOaGaayzkaaGaaiilaaqaaiaaxMaacaWLjaGaaeOzaiaabEcadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaaI0aGaamiEaiabgkHiTiaaiodacqGH8aapcaaIWaaabaGaamivaiaadIgacaWGLbGaamOCaiaadwgacaWGMbGaam4BaiaadkhacaWGLbGaaiilaiaabccacaqGMbGaaeiiaiaabMgacaqGZbGaaeiiaiaabohacaqG0bGaaeOCaiaabMgacaqGJbGaaeiDaiaabYgacaqG5bGaaeiiaiaabsgacaqGLbGaaeyyaiaabogacaqGYbGaaeyzaiaabggacaqGZbGaaeyAaiaab6gacaqGNbGaaeiiaiaabMgacaqGUbGaaeiiaiaabshacaqGObGaaeyAaiaabohacaqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabkhacaqG2bGaaeyyaiaabYgacaqGUaGaaeiiaiaabgeacaqGSbGaae4Caiaab+gacaqGSaGaaeiiaiaabMgacaqGUbGaaeiiaaqaaiaabshacaqGObGaaeyzaiaabccacaqGPbGaaeOBaiaabshacaqGLbGaaeOCaiaabAhacaqGHbGaaeiBaiaaykW7caaMc8UaaGPaVpaabmaabaWaaSaaaeaacaaIZaaabaGaaGinaaaacaGGSaGaeyOhIukacaGLOaGaayzkaaGaaeilaiaabccacaqGMbGaae4jamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg6da+iaaicdacaaMb8UaaGjbVlaabggacaqGUbGaaeizaiaabccacaqGZbGaae4BaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOzaiaabwhacaqGUbGaae4yaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaaeOzaiaabccacaqGPbGaae4CaiaabccacaqGZbGaaeiDaiaabkhacaqGPbGaae4yaiaabshacaqGSbGaaeyEaiaabccaaeaacaqGPbGaaeOBaiaabogacaqGYbGaaeyzaiaabggacaqGZbGaaeyAaiaab6gacaqGNbGaaeiiaiaabMgacaqGUbGaaeiiaiaabshacaqGObGaaeyAaiaabohacaqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabkhacaqG2bGaaeyyaiaabYgacaqGUaaaaaa@EC9D@

Q.23 Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing

Ans.

fx=06(x3)(x+2)=0x=2,3The points x =2 and x = 3 divide the real line into three disjoint intervals i.e.,(,2),(2,3) and (3,).

In intervals(,2) and (3,),f(x)is positive while in interval (2,3),f(x) is negative. Hence, the given function (f) is strictly increasing in intervals (,2) and (3,), while function (f) is strictly decreasing in interval 2,3 .

Q.24 Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x2 + 2x − 5 (b)10 − 6x − 2x2

(c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9x − x2

(e) (x + 1)3 (x − 3)3

Ans.

(a) We have,  f(x)=x2+ 2x5Differentiating both sides with respect to x, we getf(x)=2x+2f(x)=02x+2=0      x=1Point x =1 divides the real line into two disjoint intervals i.e.,(,1)and(1,).In interval(,1),f(x)=2x+2<0f is strictly decreasing in interval(,1).Thus, f is strictly decreasing for x < 1.In interval  (1,),f(x)=2x+2>0f is strictly increasing in interval(1,).Thus, f is strictly increasing for x > 1.(b)We have,  f(x)=106x2x2Differentiating both sides with respect to x, we getf(x)=64xf(x)=064x=0      x=32Point x =32 divides the real line into two disjoint intervalsi.e.,(,32)and(32,).In interval(,32),f(x)=64x>0f is strictly increasing in interval(,32).Thus, f is strictly increasing for x<32.In interval  (32,),f(x)=64x<0f is strictly decreasing in interval(32,).Thus, f is strictly decreasing for x >32.(c)We have,  f(x)=2x39x212x+1Differentiating both sides with respect to x, we getf(x)=6x218x12=6(x2+3x+2)f(x)=06(x2+3x+2)=0(x+1)(x+2)=0      x=1,2The points x =1 and x =2 divide the real line into three disjoint intervals i.e.,(,2),(2,1) and (1,).In intervals  (,2) and (1,),f(x)is negative. Hence, the given function (f) is strictly decreasing in intervals(,2) and (1,),i.e.,function (f) is strictly decreasing when x<2 and x>1. In intervals( 2,1 ),f( x )is positive. So, function (f) is strictly increasing in interval( 2,1 ), i.e., function (f) is strictly increasingwhen2<x<1. ( d )We have, f( x )=69x x 2 Differentiating both sides with respect to x, we get f’( x )=92x f’( x )=092x=0 x= 9 2 Point x = 9 2 divides the real line into two disjoint intervals i.e.,( , 9 2 )and( 9 2 , ). In interval( , 9 2 ),f( x )=92x>0 f is strictly increasing in interval( , 9 2 ). Thus, f is strictly increasing for x < 9 2 . In interval( 9 2 , ),f( x )=92x<0 f is strictly decreasing in interval( 9 2 , ). Thus, f is strictly decreasing for x > 9 2 . 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( e )We have, f(x)= (x + 1) 3 (x 3) 3 Differentiating both sides with respect to x, we get f’( x )= ( x + 1 ) 3 d dx ( x3 ) 3 + ( x3 ) 3 d dx ( x + 1 ) 3 [ By product rule ] f’( x )= ( x + 1 ) 3 3 ( x3 ) 2 d dx ( x3 )+ ( x3 ) 3 3 ( x + 1 ) 2 d dx ( x + 1 ) [ By chain rule ] f’( x )=3 ( x + 1 ) 3 ( x3 ) 2 ( 10 )+3 ( x3 ) 3 ( x + 1 ) 2 ( 1+0 ) f’( x )=3 ( x + 1 ) 2 ( x3 ) 2 ( x+1+x3 ) f’( x )=3 ( x + 1 ) 2 ( x3 ) 2 ( 2x2 ) =6 ( x + 1 ) 2 ( x3 ) 2 ( x1 ) Now, f’( x )=0x=1,1,3 The points x=1, x=1, and x=3 divide the real line into four disjoint intervals i.e., ( ,1 ),( 1,1 ),( 1,3 ) and ( 3, ). In intervals( ,1 )and (1, 1), 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f’( x )=6 ( x + 1 ) 2 ( x3 ) 2 ( x1 )<0 f is strictly decreasing in intervals( ,1 )and ( 1, 1 ). In intervals (1, 3) and( 3, ),f’( x )=6 ( x + 1 ) 2 ( x3 ) 2 ( x1 )>0 f is strictly increasing in intervals ( 1, 3 ) and( 3, ). 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Q.25 Showthat y=log(1+x)2x2+x,x> 1, is an increasing function of x throughout its domain.

Ans.

We have,  y=log(1+x)2x2+xDifferentiating both sides with respect to x, we get  dydx=ddxlog(1+x)ddx(2x2+x)=11+xddx(1+x)(2+x)ddx2x2xddx(2+x)(2+x)2=11+x×(0+1)2(2+x)2x(0+1)(2+x)2=11+x4(2+x)2=4+4x+x244x(1+x)(2+x)2=x2(1+x)(2+x)2Now,                  dydx=0x2(1+x)(2+x)2=0  x2=0[(1+x)0 and (2+x)20]    x=0Since x >1, point x = 0 divides the domain (1, ) in two disjoint intervals i.e.,1 <x < 0 and x > 0.When 1< x < 0, we have:x< 0x2>0x>1x+2>0(x+2)2>0dydx=x2(x+2)2>0Also, when x > 0:x>0x2>0,  (2+x2)>0dydx=x2(x+2)2>0Hence, function f is increasing throughout this domain.

Q.26 Find the values of x for whichy=[x(x2)]2is an increasingfunction.

Ans.

We have,      y=[x(x2)]2=x2(x2)2Differentiating w.r.t. x, we get  dydx=ddxx2(x2)2=x2ddx(x2)2+(x2)2ddxx2=x2.2(x2)ddx(x2)+(x2)2.2x=2x2(x2)(10)+2x(x2)2=2x(x2)(x+x2)=2x(x2)(2x2)=4x(x2)(x1)dydx=0x=0,1,2The points x = 0, x = 1, and x = 2 divide the real line into fourdisjoint intervals i.e.,(,0),(0,1),(1,2),(2,).In intervals (,0)and  (1,2),dydx<0y is strictly decreasing in intervals(,0)and  (1,2).However, in intervals (0, 1) and (2,),dydx>0y is strictly increasing in intervals (0, 1) and (2,∞).y is strictly increasing for 0 < x < 1 and x > 2.

Q.27 Prove that y=4sinθ(2+cosθ)θ is an increasing function of θ in [0,π2].

Ans.

We have,  y=4sinθ(2+cosθ)θDifferentiating w.r.t. θ, we getdy=d{4sinθ(2+cosθ)θ}        =(2+cosθ)d(4sinθ)4sinθd(2+cosθ)(2+cosθ)2dθ        =4(2+cosθ)cosθ4sinθ(0sinθ)(2+cosθ)21        =4(2+cosθ)cosθ+4sin2θ(2+cosθ)21        =8cosθ+4(cos2θ+sin2θ)(2+cosθ)21        =8cosθ+4(2+cosθ)21Now, dy=08cosθ+4(2+cosθ)21=08cosθ+4(2+cosθ)2=18cosθ+4=(2+cosθ)28cosθ+4=4+4cosθ+cos2θcos2θ4cosθ=0cosθ(cosθ4)=0cosθ=0  orcosθ=4Since cos θ4cosθ=0θ=π2Now,  dy=8cosθ+4(2+cosθ)21=8cosθ+4(2+cosθ)2(2+cosθ)2=8cosθ+444cosθcos2θ(2+cosθ)2=cosθ(4cosθ)(2+cosθ)2In interval  (0,π2),  we have cosθ > 0. Also, 4 > cosθ4cosθ> 0.cosθ(4cosθ)>0 and also (2+cosθ)2>0cosθ(4cosθ)(2+cosθ)2>0dy>0Therefore, y is strictly increasing in interval  (0,π2).Also, the given function is continuous at  x=0 and x=π2.Hence, y is increasing in interval(0,π2).

Q.28 Provethatthelogarithmicfunctionisstrictly increasingon (0, ).

Ans.

The given function if f(x)=logx      f(x)=1xIt is clear that for x > 0,f(x)=1x>0Hence, f(x) = log x is strictly increasing in interval (0,∞).

Q.29 Prove that the function f given by f(x)= x2 − x + 1 is neither strictly increasing
nor strictly decreasing on (−1, 1).

Ans.

The given function is f(x) = x2x + 1.Differentiating with respect to x, we getf’(x)=2x1Now,f(x)=02x1=0x=12The point 12divides the interval (1, 1) into two disjoint intervals i.e., (1, 12)and(12, 1).Now, for interval (1, 12),f(x)=2x1<0Therefore, f is strictly decreasing in interval(1, 12).Now, for interval (12, 1),f(x)=2x1>0So, f is strictly increasing in interval (12, 1).Hence, f is neither strictly increasing nor decreasing ininterval (1, 1).

Q.30 Whichofthefollowingfunctionsarestrictly decreasingon(0,π2)?(A)cosx (B)cos2x (C)cos3x (D)tanx

Ans.

( A )Let f 1 ( x )=cosx Differentiating w.r.t. x, we get f’ 1 ( x )= d dx cosx =sinx In interval ( 0, π 2 ), sinx is positive in first quadrant. So, f’ 1 ( x )<0 Therefore, f 1 ( x )=cosxis strictly decreasing in interval( 0, π 2 ). ( B )Let f 2 ( x )=cos2x Differentiating w.r.t. x, we get f’ 2 ( x )= d dx cos2x =2sin2x In interval ( 0, π 2 ), sin2x is positive in first quadrant. So, f’ 2 ( x )<0 Therefore, f 2 ( x )=cos2xis strictly decreasing in interval( 0, π 2 ). ( C )Let f 3 ( x )=cos3x Differentiating w.r.t. x, we get f’ 3 ( x )= d dx cos3x =3sin3x Now, f’ 3 ( x )=03sin3x=0 3x=πx= π 3 The point x= π 3 divides the interval ( 0, π 2 )into two disjoint intervals i.e.,( 0, π 3 ) and ( π 3 , π 2 ). Now, in interval( 0, π 3 ), f’ 3 ( x )=3sin3x<0 [ 0<x< π 3 0<3x<π ] f 3 is strictly decreasing in interval( 0, π 3 ). Now,ininterval( π 3 , π 2 ), π 3 <x< π 2 π<3x< 3π 2 So, sin3x is in third quadrant, where sin3x is negative. f 3 ( x )=3sin3x>0 Therefore, f 3 is strictly increasing in interval( π 3 , π 2 ). Hence, f 3 is neither increasing nor decreasing in interval( 0, π 2 ). ( D )Let f 4 ( x )=tanx Differentiating w.r.t. x, we get f’ 4 ( x )= d dx tanx = sec 2 x Ininterval ( 0, π 2 ),secx>0 sec 2 x>0 f 4 ( x )>0 f4 is strictly increasing in interval( 0, π 2 ). Therefore, functions cos x and cos 2x are strictly decreasing in( 0, π 2 ). Hence, the correct answers are A and B. 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Q.31 On which of the following intervals is the function f given byf(x)=x100 +sinx– 1strictly decreasing?(A)(0, 1)(B)(π2, π)(C)(0, π2)(D)None of these

Ans.

We  have, f(x)=x100+sinx1Differentiating w.r.t. x, we getf(x)=100x99+cosxf(x)>0[100x99>0 and cosx>0 in interval(0,1)]Thus, function f is strictly increasing in interval (0, 1).In interval (π2,π),cosx<0 and 100x99>0So,  f(x)>0Thus, function f is strictly increasing in interval (π2,π).In interval (0,π2),cosx>0 and 100x99>0100x99+cosx>0So,  f(x)>0Thus, function f is strictly increasing in interval (0,π2).Hence, function f is strictly decreasing in none of the intervals.The correct answer is D.

Q.32 Find the least value of a such that the function f given f(x) =x2+ax+1 is strictly increasing on (1, 2).

Ans.

We have,f(x)=x2+ax+1Differentiating w.r.t. x, we getf(x)=2x+aNow,function f will be increasing in(1,2),iff(x)>02x+a>0    x>a2Therefore, we have to find the least value of a such that    x>a2,  whenx(1,2)x>a2,  (when1<x<2)Thus, the least value of a for f to be increasing on(1, 2) is given by,a2=1a=2Hence, the required value of a is 2.

Q.33 Let I be any interval disjoint from (1,1).Prove that the function fgiven byf(x)=x+1x is strictly increasing onI.

Ans.

we have,f(x)=x+1xDifferentiating w.r.t. x, we getf(x)=11x2Now,f(x)=011x2=01=1x2x=±1The points x=1 and x=1 divide the real line in three disjointintervals i.e.,(,1),(1,1)  and  (1,).In interval (1, 1), it is noticed that:1<x<1x2<11<1x2,  x011x2<0,  x0f(x)=11x2<0  on (1,1)~{0}f is strictly decreasing on  (1,1)~{0}In intervals (,1)  and  (1,), it is seen thatx<1 or x>1x2>11>1x211x2>0f(x)=11x2>0 on (,1)  and  (1,).f is strictly increasing on(,1)  and  (1,).Hence, function f is strictly increasing in interval I disjoint from (1, 1).

Q.34 Provethatthefunctionfgivenbyf(x) =logsinxisstrictly increasingon (0,π2)andstrictlydecreasingon(π2,π).

Ans.

We have,f(x)=logsinxDifferentiating both sides w.r.t. x,we getf(x)=1sinxddxsinx[Bychain rule]        =1sinxcosx=cotxf(x)=cotx>0​ in interval (0,π2)f(x) is strictly increasing function in interval (0,π2).f(x)=cotx<0​ in interval (π2,π)f(x) is strictly decreasing function in interval (π2,π).

Q.35

Prove that the function f given by f(x) = logcos x is strictly decreasing on ( 0 , π 2 ) and strictly increasing on ( π 2 , π ) .

Ans.

We have,f(x)=logcosxDifferentiating both sides w.r.t. x,we getf(x)=1cosxddxcosx[Bychain rule]        =1cosx×sinx=tanxf(x)=tanx<0​ in interval (0,π2)f(x) is strictly decreasing function in interval (0,π2).f(x)=tanx>0​ in interval (π2,π)f(x) is strictly increasing function in interval (π2,π).

Q.36 Prove that the function f given byf(x) = x23x2+3x100is increasinginR.

Ans.

We have,f(x)=x33x2+3x100Differentiating w.r.t. x, we getf(x)=3x26x+3=3(x22x+1)=3(x1)2For any xR, (x1)2>0.Thus, f(x)  is always positive in R.Hence, the given function f(x) is increasing in R.

Q.37 The interval in which y=x2exis increasing is(A) (, ) (B )(2, 0) (C) (2, ) (D) (0, 2)

Ans.

We have,y=x2exDifferentiating w.r.t. x, we getdydx=x2ddxex+exddxx2      =x2ex+ex(2x)      =ex(x2+2x)Now,dydx=0ex(x2+2x)=0x=0,2[ex0]The points x=0 and x=2 divide the real line into three disjoint intervals i.e., (,0),(0,2)and  (2,).f(x)<0 in intervals (,0)and(2,).So,f(x) is decreasing in intervals (,0)and(2,).Now,f(x)>0 in interval (0,2)So,f(x) is increasing in intervals (0,2)Hence, f is strictly increasing in interval (0, 2).The correct answer is D.

Q.38 Find the slope of the tangent to the curve y = 3x4 − 4x at x = 4

Ans.

The given curve is y=3x4 4x.Differentiating w.r.t. x, we get        dydx=ddx(3x4 4x)    =12x34Then, the slope of the tangent to the given curve at x=4 is given by,(dydx)x=4=(12x34)x=4    =12(4)34    =7684    =764Thus,​ the slope of tangent to the given curve is 764.

Q.39

Find the slope of the tangent to the curve y = x 1 x 2 , x 2 a t x = 10 .

Ans.

The given curve is y=x1x2Differentiating w.r.t. x, we getdydx=ddx(x1x2)      =(x2)ddx(x1)(x1)ddx(x2)(x2)2      =(x2)(10)(x1)(10)(x2)2      =x2x+1(x2)2      =1(x2)2Thus, the slope of the tangent at x = 10 is given by,(dydx)x=10=(1(x2)2)x=10    =1(102)2    =164Thus,​ the slope of tangent to the given curve is 164.

Q.40 Find the slope of the tangent to curve y = x3−x+1 at the point whose x-coordinate is 2.

Ans.

The given curve isy=x3x+1Differentiating w.r.t.x, we getdydx=ddx(x3x+1)= 3x21The slope of tangent to the curve is(dydx)x=2=3(2)21=121=11Thus, the slope of the tangent at the point where thexcoordinate is 2 is11.

Q.41 Find the slope of the tangent to curve y = x3−3x+2 at the point whose x-coordinate
is 3.

Ans.

The given curve is      y=x33x+2Differentiating w.r.t. x, we get  dydx=ddx(x33x+2)=3x23The slope of tangent to the curve is(dydx)x=3=3(3)23    =273=24Thus, the slope of the tangent at the point where the xcoordinate is 3 is 24.

Q.42 Find the slope of the normal to the curve x =acos3θ, y=asin3θat θ=π4.

Ans.

Given that x = acos3θ and y = asin3θDifferentiating w.r.t. θ, we getdx=adcos3θ        =3acos2θdcosθ=3acos2θsinθdy=adsin3θ        =3asin2θdsinθ        =3asin2θcosθdydx=(dy)(dx)        =3asin2θcosθ3acos2θsinθ=tanθTherefore, the slope of the tangent at  θ=π4,is(dydx)(θ=π4)=(tanθ)(θ=π4)        =tanπ4        =1Hence, the slope of the normal atθ=π4,isslope of normal(M)=1slope of tangent(m)=11=1

Q.43

Find the slope of the normal to the curve x = 1 a s i n θ , y = b c o s 2 θ a t θ = π 2

Ans.

Given that:x=1asinθ and y=bcos2θDifferentiating w.r.t. θ, we get    dx=d(1asinθ)=0acosθ=acosθ  dy=d(bcos2θ)=2bcosθdcosθ=2bcosθsinθdydx=(dy)(dx)=2bcosθsinθacosθ=2basinθSlope of tangent at θ=π2,is given by    m=(dydx)θ=π2=2basinπ2=2baslope of normal    (M)=1m=1(2ba)=a2bThus, slope of normal is a2b.

Q.44 Find points at which the tangent to the curve
y = x3 − 3x2 − 9x + 7 is parallel to the x axis.

Ans.

The equation of the given curve is y= x 3 3x 2 9x + 7 Differentiating w.r.t. x, we get dy dx = d dx ( x 3 3x 2 9x + 7 ) =3 x 2 6x9 Now, the tangent is parallel to the x-axis if the slope of the tangent is zero. 3 x 2 6x9=03( x 2 2x3 )=0 3( x3 )( x+1 )=0 x=3,1 When x=3, then y= ( 3 ) 3 3 ( 3 ) 2 9( 3 ) + 7 =272727+7 =20 When x=1, then y= ( 1 ) 3 3 ( 1 ) 2 9( 1 ) + 7 =13+9+7 =12 Hence, the points at which the tangent is parallel to the x-axis are ( 3,20 ) and ( 1, 12 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@2D49@

Q.45 Find a point on the curve y = (x − 2)2 at which the tangent is parallel to the chord joining the
points (2, 0) and (4, 4).

Ans.

Given curve is y= ( x2 ) 2 Differentiating both sides, w.r.t. x, we get dy dx = d dx ( x2 ) 2 =2( x2 ) d dx ( x2 ) =2( x2 ) The slope of the chord = 40 42 [ m= y 2 y 1 x 2 x 1 ] =2 Since, the slope of the tangent = slope of the chord, we have: 2( x2 )=2 x=3 When x=3, y= ( 32 ) 2 =1 Hence, the required point is ( 3, 1 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6F0D@

Q.46 Find the point on the curve y = x3 − 11x + 5
at which the tangent is y = x − 11.

Ans.

The equation of the given curve is y= x 3 11x+5 The equation of the tangent to the given curve is y = x11 comparing it with y = mx + c, we get m=1 slope of the tangent( m )=1 Now, the slope of the tangent to the given curve at the point (x, y) is, dy dx = d dx ( x 3 11x+5 ) =3 x 2 11 3 x 2 11=1 [ Given ] x 2 = 12 3 =4 x=±2 When x=2, y= (2) 3 11(2) + 5 =822 + 5 =9. When x=2, y=(2)311 (2) + 5 =8 + 22 + 5 =19. Hence, the required points are ( 2,9 ) and ( 2, 19 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@44E6@

Q.47

Find the equation of all lines having slope1 that aretangents to the curvey=1x1,x1.

Ans.

Equation of given curve is y=1x1,x1Differentiating w.r.t. x, we getdydx=ddx1x1      =1(x1)2ddx(x1)      =1(x1)2So,m=(dydx)(x,y)      =1(x1)2According​ to question,1(x1)2=1(x1)2=1x1=±1x1=1  or  x1=1x=2,0When x=0, y=1Whenx=2,  y=1Thus, there are two tangents to the given curve having slope1.These are passing through the points (0, -1) and (2, 1). And equation of two tangents are y 1 = 1 ( x 2 ) y 1 = x + 2 x + y 3 = 0 A n d , y + 1 = 1 ( x 0 ) y + 1 = x x + y + 1 = 0

Q.48 

Find the equation of all lines having slope 2 which are tangents to the curve y = 1 x 3 , x 3 .

Ans.

Equation of given curve is y=1x3,x3Differentiating w.r.t. x, we getdydx=ddx(1x3)      =1(x3)2ddx(x3)    =1(x3)2So,m=(dydx)(x,y)      =1(x3)2According​ to question,1(x3)2=2(x3)2=2This is impossible that square of a number is negative.Hence, there is no tangent to the given curve having slope 2.

Q.49 

Find the equations of all lines having slope 0 which are tangent to the curve y = 1 x 2 2 x + 3

Ans.

The equation of the given curve isy= 1 x 2 2x+3 The slope of the tangent to the given curve at any point ( x, y ) ( dy dx ) ( x,y ) = d dx 1 x 2 2x+3 = 1 ( x 2 2x+3 ) 2 d dx ( x 2 2x+3 ) = ( 2x2 ) ( x 2 2x+3 ) 2 Since, the slope of the tangent is 0, then we have: ( 2x2 ) ( x 2 2x+3 ) 2 =0( 2x2 )=0 x=1 Whenx=1, y= 1 12+3 = 1 2 The point on the curve is ( 1, 1 2 ). The equation of the tangent through( 1, 1 2 )is y 1 2 =0( x1 ) y= 1 2 Hence, the equation of the required line isy= 1 2 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@C6CF@

Q.50

Findpointsonthecurvex29+y216 = 1atwhichthetangentsare (i) paralleltoxaxis(ii)paralleltoyaxis.

Ans.

The equation of the given curve isx29+y216=1(i)On differentiating both sides with respect to x, we have:    2x9+2y16dydx=0dydx=2x9×162y=169xy(i)When the tangent is parallel to xaxis, thenslope of tangent(m)=0dydx=0169xy=0x=0Putting​ value of x in equation (i),we ​have(0)29+y216=1y2=16  y=±4Hence, the points at which the tangents are parallel to the xaxis are (0, 4) and (0,4).(b)The tangent is parallel to the yaxis if the slope of the normal(M) is 0, i.e., M=0 So,  1169xy=0  [M=1m]yx=0y=0Putting​ value of y in equation (i),we ​havex29+(0)216=1x2=9x=±3Hence, the points at which the tangents are parallel to the yaxis are (3, 0) and (- 3, 0).

Q.51 Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x4 − 6x3+13x2−10x + 5 at (0, 5)

(ii) y = x4 − 6x3+13x2−10x + 5 at (1, 3)

(iii) y = x3 at (1, 1)

(iv) y = x2 at (0, 0)

(v) x = cos t, y = sin t at t= π 4 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGcbaacbeGaa8hEaiaa=bcacqGH9aqpcaWFGaGaa83yaiaa=9gacaWFZbGaa8hiaiaa=rhacaWFSaGaa8hiaiaa=LhacaWFGaGaeyypa0Jaa8hiaiaa=nhacaWFPbGaa8NBaiaa=bcacaWF0bGaa8hiaiaa=fgacaWF0bGaaGPaVlaaykW7caWF0bGaeyypa0ZaaSaaaeaarmWu51MyVXgaiyaacqGFapaCaeaacaWH0aaaaiaac6caaaa@5B2A@

Ans.

(i) The equation of the curve is y = x 4 6 x 3 + 13 x 2 10 x + 5 . On differentiating with respect to x, we get: d y d x = d d x ( x 4 6 x 3 + 13 x 2 10 x + 5 ) =4x318x2+26x10(dydx)(0,5)=4(0)318(0)2+26(0)10=10Thus, the slope of the tangent at (0,5)is10.The equation of the tangent is given as:y5=10(x0)y=10x+510x+y=5The slope of the normal (M)at(0,5)=110=110There fore, thee quation of the normal at (0,5) is given as:y5=110(x0)[yy1=M(xx1)]10y50=xx10y+50=0(ii) The equation of the curve is y=x46x3+13x210x+5.On differentiating with respect to x, we get:dydx=ddx(x46x3+13x210x+5)=4x318x2+26x10(dydx)(1,3)=4(1)318(1)2+26(1)10=2Thus, the slope of the tangent at(1,3)is2.The equation of the tangent is given as:

y 3 = 2 ( x 1 ) y 3 = 2 x 2 y = 2 x + 1 T h e s l o p e o f t h e n o r m a l ( M ) a t ( 1 , 3 ) = 1 2 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 1 , 3 ) i s g i v e n a s : y 3 = 1 2 ( x 1 ) [ y y 1 = M ( x x 1 ) ] x + 2 y 7 = 0 ( i i ) T h e e q u a t i o n o f t h e c u r v e i s y = x 3 O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x , w e g e t : d y d x = d d x ( x 3 ) = 3 x 2 ( d y d x ) ( 1 , 1 ) = 3 ( 1 ) 2 = 3 T h u s , t h e s l o p e o f t h e t a n g e n t a t ( 1 , 1 ) i s 3 . E q u a t i o n o f t a n g e n t a t ( 1 , 1 ) o f t h e t a n g e n t i s : y 1 = 3 ( x 1 ) y = 3 x 2 T h e s l o p e o f t h e n o r m a l a t ( 1 , 1 ) = 1 3 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 1 , 1 ) i s : y 1 = 1 3 ( x 1 ) x + 3 y 4 = 0 ( i v ) T h e e q u a t i o n o f t h e c u r v e i s y = x 2 O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x , w e g e t : d y d x = d d x ( x 2 ) = 2 x ( d y d x ) ( 0 , 0 ) = 2 ( 0 ) = 0 E q u a t i o n o f t a n g e n t a t ( 0 , 0 ) : y 0 = 0 ( x 0 ) y = 0 T h e s l o p e o f t h e n o r m a l a t ( 0 , 0 ) = 1 0 = 1 0 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 0 , 0 ) i s : y 0 = 1 0 ( x 0 ) x = 0 ( v ) T h e e q u a t i o n o f t h e c u r v e i s x = c o s t , y = s i n t . D i f f e r e n t i a t i n g w . r . t . t , w e g e t d x d t = sin t a n d d y d t = cos t d y d x = d y d t d x d t = cos t sin t = cot t ( d y d x ) x = π 4 = cot ( π 4 ) = 1 E q u a t i o n o f t a n g e n t a t ( cos π 4 , sin π 4 ) i . e . , ( 1 2 , 1 2 ) : y 1 2 = 1 ( x 1 2 ) x + y = 2 T h e s l o p e o f t h e n o r m a l a t t = π 4 : M = 1 ( d y d x ) x = π 4 = 1 1 = 1 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 1 2 , 1 2 ) : y 1 2 = 1 ( x 1 2 ) y = x

Q.52 Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is

(a) parallel to the line 2x – y + 9 = 0

(b) perpendicular to the line 5y −15x = 13.

Ans.

The equation of the given curve is       y=x22x+7 ...(i)On differentiating with respect to x, we get:  dydx=2x2slope of tangent to curve=(dydx)(x,y)=2x2(a)The equation of the line is 2xy + 9 = 0y=2x+9comparing with y=mx+c, we get    m=2When tangent is parallel to given line than,(dydx)(x,y)=m2x2=2x=2Putting x=2 in equation(i), we gety=(2)22(2)+7=7Thus, the equation of the tangent passing through (2, 7) is:  y7=2(x2)y7=2x4  y2x3=0Thus, the equation of the tangent line to the given curve (which is parallel to line 2xy + 9 = 0) is y2x3=0.(b) The equation of the line is 5y15x=13.This equation can be written asy=3x+135 which in the form of y=mx+c,So, m=3If a tangent is perpendicular to the line 5y15=13,then the slope of the tangent is1(dydx)(x,y)=132x2=13x=56Whenx=56,y=(56)22(56) + 7=21736Thus, the equation of the tangent passing through(56,21736) is    y21736=13(x56)12x+36y227=0Hence, the equation of the required tangent line is      12x+36y227=0.

Q.53 Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −2 are parallel.

Ans.

We have, y= 7x 3 + 11 Differentiating w.r.t. x, we get dy dx =21 x 2 Slope of tangent at x=2, m 1 = ( dy dx ) x=2 =21 ( 2 ) 2 =84 Slope of tangent at x=2, m 2 = ( dy dx ) x=2 =21 ( 2 ) 2 =84 Since, m 1 = m 2 Hence, the two tangents are parallel. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@FB3A@

Q.54 Find the points on the curve y = x3 at which the slope of the tangent is equal to the y coordinate of the point.

Ans.

The equation of the given curve is y = x3.Diffferentiating both sides, w.r.t. x, we get            dydx=3x2Slope of tangent at point (x, y) is given by,    (dydx)(x,y)=3x2(dydx)(x,y)=y[Given]      3x2=yBut y=x3[Given]  x3=3x2x33x2=0    x2(x3)=0x=0 and 3When x = 0, then y = 0 and when x = 3, then y = 3(3)2= 27.Hence, the required points are (0, 0) and (3, 27).

Q.55 For the curve y = 4x3 − 2x5, find all the points at which the tangents

passes through the origin.

Ans.

The equation of the given curve is     y=4x32x5...(i)Differentiating with respect to x, we getdydx=12x210x4slope of tangent at the point (x1,y1) is(dydx)(x1,y1)=12x1210x14The equation of the tangent at (x1, y1) is given byyy1=(dydx)(x1,y1)(xx1)yy1=(12x1210x14)(xx1)When the tangent passes through the origin (0, 0), then x=y=0.0y1=(12x1210x14)(0x1)  y1=(12x1210x14)x1  y1=12x1310x15​​​   ...(ii)Point (x1,y1) lies on curve (i),then      y1=4x132x15    .  (iii)Fromequation (ii) and equation(iii),wehave    12x1310x15=4x132x15      8x138x15=08x13(1x12)=0x1=0,±1When x=0, y=4 (0)32 (0)5= 0.When x=1, y=4 (1)32(1)5= 2.When x=1, y=4 (1)32(1)5=2.Hence, the required points are (0, 0), (1, 2) and (1,2).

Q.56 Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are
parallel to the x-axis.

Ans.

The equation of the given curve is x 2 + y 2 2x3=0 ( i ) Differentiating both sides, w.r.t. x, we get 2x+2y dy dx 2=0 dy dx = 22x 2y = 1x y Slope of tangent of the given curve at point ( x 1 , y 1 )is ( dy dx ) ( x 1 , y 1 ) = 1 x 1 y 1 Itis given that ( dy dx ) ( x 1 , y 1 ) =0 1 x 1 y 1 =01 x 1 =0 x 1 =1 Since, point ( x 1 , y 1 ) lies on curve( i ), we get x 1 2 + y 1 2 2x 1 3=0 ( 1 ) 2 + y 1 2 2( 1 )3=0 y 1 2 =4 y 1 =±2 Hence, the points at which the tangents are parallel to the x-axis are ( 1, 2 ) and ( 1,2 ). 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Q.57 Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3

Ans.

The equation of the given curve is ay2=x3.On differentiating with respect to x, we have:2aydydx=3x2        dydx=3x22aySlope at the point (am2, am3)=(dydx)(am2, am3)=3(am2)22a(am3)=3a2m42a2m3=32m    Slope of normal=1(32m)=23mHence, the equation of the normal at (am2, am3) is given by,  yam3=23m(xam2)  3my3am4=2x+2am22x+3myam2(23m2)=0

Q.58 Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line
x + 14y + 4 = 0.

Ans.

The equation of the given curve is y = x3 + 2x + 6.Differentiating w.r.t. x, we getdydx=3x2+2The slope of the tangent to the given curve at any point (x1, y1)                 (dydx)(x1, y1)=3x12+2Slope of normal to the given curve at any point (x1, y1)        =1(dydx)(x1, y1)        =13x12+2The equation of the given line is           x + 14y + 4=0Differentiating w.r.t. x, we get  dydx=114Slope of lineat any point (x1, y1)        =(dydx)(x1, y1)        =114Since normal is parallel to line, so   slope of normal=slope of line        13x2+2=114  14=3x2+2    x2=123=4      x=±2When x=2, y=8+4+6=18.When x=2,y=84+6=6.Therefore, the equations of normals passing through the points (2,18) and (2,6).                          y18=114(x2)x+14y254=0and                      y+6=114(x+2)  x+14y+86=0Hence, the equations of the normals to the given curve (which are parallel to the given line) arex+14y254=0,x+14y+86=0.  

Q.59 Find the equations of the tangent and normal to the parabola y2= 4ax at the point (at2, 2at).

Ans.

The equation of parabola is y2=4axDifferentiating both sides w.r.t. x, we get    2ydydx=4adydx=4a2y=2aySlope of tangent to parabola at point (at2, 2at)  (dydx)(at2, 2at)=2a2at=1tSlope of normal to parabola at point (at2, 2at)=1(dydx)(at2, 2at)=1(1t)=tThen, the equation of the tangent at(at2, 2at)    y2at=1t(xat2)    ty=x+at2And, the equation of the normal at(at2, 2at)    y2at=t(xat2)      tx+  y=2at+at3Thus, the required equations of tangent and normal arety=x+at2 and tx+  y=2at+at3.

Q.60 Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.

Ans.

Given equations of the curves are x=y2 and xy=k.Putting x=y2 in xy=k, we get          y2.y=ky=k13Putting value of y in xy=k, we get  k13x=kx=k23Thus, the point of intersection of the given curves is(k23,k13).Differentiating the curves x=y2 and xy=k w.r.t. x, we get      ddxx=ddxy21=2ydydxdydx=12ySlope of tangent to the curvesx=y2at point (k23,k13)          m1=(dydx)(k23,k13)=12k13and ddx(xy)=ddxkxdydx+yddxx=0dydx=yxSlope of tangent to the curve xy=k at point (k23,k13)          m2=(dydx)(k23,k13)=k13k23=1k13Since,two curves intersect at right angles if     m1.m2=1      12k13×1k13=1      1=2k23Cubing both sides, we get      1=8k2.Hence, the given two curves cut at right angels if 8k2=1.

Q.61 

Find the equations of the tangent and normal to the hyperbola x 2 a 2 y 2 b 2 =1 at the point x 0 ,y 0 .

Ans.

Find the equations of the tangent and normal to the hyperbola x 2 a 2 y 2 b 2 =1 at the point x 0 ,y 0 .

Q.62 

Find the equation of the tangent to the curve y= 3x2 which is parallel to the line 4x2y + 5 = 0.

Ans.

The equation of the given curve is y=3x2Differentiating w.r.t. x, we get  dydx=ddx(3x2)12=12(3x2)12ddx(3x2)[By chain rule]=123x2×3=323x2The slope of the tangent to the given curve at any point (x1, y1),  m1=(dydx)(x1, y1)=323x12The equation of the given line is 4x2y + 5 = 0.Differentiating w.r.t. x, we get42dydx=0dydx=2The slope of the tangent to the given line at any point (x1, y1),  m2=(dydx)(x1, y1)=2Since,​ tangent is parallel to given line. So,    m1=m2323x12=2    3x12=34      3x12=9163x1=916+23x1=4116x1=4148Since, point (x1,y1) lies on given curve  y=3x2so,    y1=3x12=34Equation of the tangent passing through the point(4148,34)is                y34=2(x4148)  48x24y=23Hence, the equation of the required tangent is  48x24y=23.

Q.63 The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) – 3
(D) –1/3

Ans.

The equation of given curve is y= 2x 2 + 3sinx Differentiating w.r.t. x, we get dy dx = d dx ( 2x 2 + 3sinx ) =4x+3cosx Putting x= 0, we get m= ( dy dx ) x= 0 =4( 0 )+3cos0 =3 Slope of normal to the given curve is M= 1 m = 1 3 The correct answer is D. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@179E@

Q.64 The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, −2) (D) (−1, 2)

Ans.

The equation of the given curve is  y2= 4xDifferentiating w.r.t. x, we get2ydydx=4dydx=42y=2yTherefore, the slope of the tangent to the given curve at any point (x1, y1)=(dydx)(x1, y1)=2y1The given line is y = x + 1 which is in the form of y=mx+c,      m=1Since, tangent and line are parallel.So(dydx)(x1, y1)=m2y1=1    y1=2Since,y=x+1,so   x=y1        =21=1Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2).The correct answer is A.

Q.65 

Using differentials, find the approximate value o f e a c h o f the following up to 3 places of d e c i m a l : ( i ) 25 .3 ( i i ) 49 .5 ( i i i ) 0 .6 ( i v ) ( 0 .009 ) 1 3 ( v ) ( 0 .999 ) 1 10 ( v i ) ( 15 ) 1 4 ( v i i ) ( 26 ) 1 3 ( v i i i ) ( 255 ) 1 4 ( i x ) ( 82 ) 1 4 ( x ) ( 401 ) 1 2 ( x i ) ( 0 .0037 ) 1 2 ( x i i ) ( 26 .57 ) 1 3 ( x i i i ) ( 81 .5 ) 1 4 ( x i v ) ( 3 .968 ) 3 2 ( x v ) ( 32 .15 ) 1 5

Ans.

(i)25.3Lety=x, x=25 and Δx=0.3dydx=12xThen,      Δy=x+Δxx    =25.325    =25.3525.3=Δy+5Now, dy is approximately equal to Δy and is given by,        dy=(dydx)Δx    =12x(0.3)    =1225(0.3)    =12×5×0.3    =0.03Hence, the approximate value of25.3is 5+0.03=5.03.(ii)49.5Lety=x, x=49 and Δx=0.5dydx=12xThen,      Δy=x+Δxx    =49.549    =49.5749.5=Δy+7Now, dy is approximately equal to Δy and is given by,        dy=(dydx)Δx    =12x(0.5)    =1249(0.5)    =12×7×0.5    =0.035Hence, the approximate value of49.5is 7+ 0.035=7.035.(iii)0.6Lety=x, x=1 and Δx= 0.4dydx=12xThen,      Δy=x+Δxx    =0.61  0.6=Δy+1Now, dy is approximately equal to Δy and is given by,        dy=(dydx)Δx    =12x( 0.4)    =121( 0.4)    = 0.2Hence, the approximate value of0.6is 1 0.2=0.8(iv)(0.009)13Lety=x13, x=0.008 and Δx=0.001dydx=13x23Then,            Δy=(x+Δx)13x13          =(0.009)130.2(0.009)13=Δy+0.2Now, dy is approximately equal to Δy and is given by,      dydx=13x23(Δx)      =13(0.008)23(0.001)      =13(0.04)(0.001)=0.0010.12      =0.008Hence, the approximate value of(0.009)23is 0.2 + 0.008=0.208 ( v ) ( 0.999 ) 1 10 Let y= x 1 10 ,x=1 and Δx=0.001 dy dx = 1 10 x 9 10 Then, Δy= ( x+Δx ) 1 10 ( x ) 1 10 = ( 0.999 ) 1 10 ( 1 ) 1 10 ( 0.999 ) 1 10 =Δy+1 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 10 x 9 10 ×0.001 = 1 10 ( 1 ) 9 10 ×0.001 = 0.001 10 =0.0001 Hence, the approximate value of ( 0.999 ) 1 10 is 1+( 0.001 )=0.9999 ( vi ) ( 15 ) 1 4 Lety= x 1 4 ,x=16 and Δx=1 dy dx = 1 4 x 3 4 Then, Δy= ( x+Δx ) 1 4 x 1 4 = ( 15 ) 1 4 ( 16 ) 1 4 ( 15 ) 1 4 =Δy+2 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 4 x 3 4 ×1 = 1 4 ( 16 ) 3 4 ×1 = 1 4×8 ×1=0.03125 Hence, the approximate value of ( 15 ) 1 4 is 2+( 0.03125 )=1.96875 ( vii ) ( 26 ) 1 3 Lety= x 1 3 ,x=27 and Δx=1 dy dx = 1 3 x 2 3 Then, Δy= ( x+Δx ) 1 3 x 1 3 = ( 26 ) 1 3 ( 27 ) 1 3 ( 26 ) 1 3 =Δy+3 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 3 x 2 3 ×1 = 1 3 ( 27 ) 2 3 ×1 = 1 3×9 ×1 =0.0 370 ¯ Hence, the approximate value of ( 26 ) 1 3 is 3+( 0.0370 )=2.9629 ( viii ) ( 255 ) 1 4 Lety= x 1 4 ,x=256 and Δx=1 dy dx = 1 4 x 3 4 Then, Δy= ( x+Δx ) 1 4 x 1 4 = ( 255 ) 1 4 ( 256 ) 1 4 ( 255 ) 1 4 =Δy+4 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 4 x 3 4 ×( 1 ) = 1 4 ( 256 ) 3 4 ×( 1 ) = 1 4× 4 3 =0.0039 Hence, the approximate value of ( 255 ) 1 4 is 4+(0.0039)=3.9961 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( ix ) ( 82 ) 1 4 Lety= x 1 4 ,x=81 and Δx=1 dy dx = 1 4 x 3 4 Then, Δy= ( x+Δx ) 1 4 x 1 4 = ( 82 ) 1 4 ( 81 ) 1 4 ( 82 ) 1 4 =Δy+3 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 4 x 3 4 ×( 1 ) = 1 4 ( 81 ) 3 4 ×( 1 ) = 1 4× 3 3 =0.009 Hence, the approximate value of ( 82 ) 1 4 is 3+(0.009)=3.009 ( x ) ( 401 ) 1 2 Lety= x 1 2 ,x=400 and Δx=1 dy dx = 1 2 x Then, Δy= x+Δx x = 401 400 = 401 20 401 =Δy+20 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 2 x ( 1 ) = 1 2 400 ( 1 ) = 1 2×20 ×( 1 ) =0.025 Hence, the approximate value of 401 is 20+0.025=20.025. ( xi ) ( 0.0037 ) 1 2 Lety= x 1 2 ,x=0.0036 and Δx=0.0001 dy dx = 1 2 x Then, Δy= x+Δx x = 0.0037 0.0036 = 0.0037 0.06 0.0037 =Δy+0.06 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 2 x ( 0.0001 ) = 1 2 0.0036 ( 0.0001 ) = 1 2×0.06 ×( 0.0001 ) =0.00083 Hence, the approximate value of 0.0037 is 0.06+0.00083=0.06083 ( xii ) ( 26.57 ) 1 3 Lety= x 1 3 ,x=27 and Δx=0.43 dy dx = 1 3 x 2 3 Then, Δy= ( x+Δx ) 1 3 x 1 3 = ( 26.57 ) 1 3 ( 27 ) 1 3 ( 26.57 ) 1 3 =Δy+3 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 3 x 2 3 ×0.43 = 1 3 ( 27 ) 2 3 ×0.43 = 1 3( 9 ) ×0.43=0.015 Hence, the approximate value of ( 26.57 ) 1 3 is 30.015=2.984 ( xiii ) ( 81.5 ) 1 4 Lety= x 1 4 ,x=81 and Δx=0.5 dy dx = 1 4 x 3 4 Then, Δy= ( x+Δx ) 1 4 x 1 4 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= ( 81.5 ) 1 4 ( 81 ) 1 4 ( 81.5 ) 1 4 =Δy+3 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 4 x 3 4 ×( 0.5 ) = 1 4 ( 81 ) 3 4 ×( 0.5 ) = 0.5 4× 3 3 =0.0046 Hence, the approximate value of ( 81.5 ) 1 4 is 3+0.0046=3.0046 ( xiv ) ( 3.968 ) 3 2 Lety= x 3 2 ,x=4 and Δx=0.032 dy dx = 3 2 x 1 2 Then, Δy= ( x+Δx ) 3 2 x 3 2 = ( 3.968 ) 3 2 ( 4 ) 3 2 ( 3.968 ) 3 2 =Δy+8 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 3 2 x 1 2 ×( 0.032 ) = 3 2 ( 4 ) 1 2 ×( 0.032 ) = 3 2 ×2×( 0.032 )=0.096 Hence, the approximate value of ( 3.968 ) 3 2 is 80.096=7.904 ( xv ) ( 32.15 ) 1 5 Lety= x 1 5 ,x=32 and Δx=0.15 dy dx = 1 5 x 4 5 Then, Δy= ( x+Δx ) 1 5 x 1 5 = ( 32.15 ) 1 5 ( 32 ) 1 5 ( 32.15 ) 1 4 =Δy+2 Now, dy is approximately equal to Δy and is given by, dy=( dy dx )Δx = 1 5 x 4 5 ×( 0.15 ) = 1 5 ( 32 ) 4 5 ×( 0.15 ) = 0.15 5× 2 4 =0.00187 Hence, the approximate value of ( 32.15 ) 1 4 is 2+0.00187=2.00187 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Q.66

Find the approximate value of f 2.01 , where f x =4x 2 +5x+2

Ans.

Let x=2 and Δx=0.01. Then, we have:f(2.01)=f(x + Δx)=4(x + Δx)2+ 5(x + Δx)+ 2Since,  Δy=f(x+Δx)f(x).​ Therefore,f(x+Δx)=f(x)+Δyf(x)+f(x)Δx[dx=Δx]    f(2.01)(4x2+5x+2)+(8x+5)Δx{4(2)2+5(2)+2}+(8×2+5)(0.01)28+21×0.01=28+0.2128.21Hence, the approximate value of f (2.01) is 28.21.

Q.67 Find the approximate value of f (5.001), where f (x) = x3 – 7x2 + 15.

Ans.

Letx=5 and Δx=0.001.Then, wehave:f(5.001)=f(x+Δx)=(x+Δx)37(x+Δx)2+15Since,Δy =f(x+Δx)f(x).Therefore,f(x+Δx)=f(x)+Δyf(x)+f(x)Δx[dx=Δx]f(5.001)(x37x2+15)+(3x214x)Δx{(5)37(5)2+15}+{3(5)214(5)}(0.01)35+5×(0.01)=35+0.0534.995Hence, the approximate value of f(5.001) is 34.995.

Q.68 Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.

Ans.

Side of cube is x metres.Volume of a cube is given byV=x3dV=(dVdx)Δx=(ddxx3)Δx=(3x2)Δx=(3x2)(0.01x)[1%ofx=0.01x]=0.03x3Hence,the approximate change in the volume of thecube is 0.03x3m3.

Q.69 Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.

Ans. 

Let side of cube be x metres.Then,Surface area of cube=6x2S=6x2  dSdx=ddx(6x2)=12x  dS=(dSdx)Δx=(12x)(0.01x)[1% of x=0.01x]=0.12x2Hence, the approximate change in the surface area of the cube is 0.12x2 m2.

Q.70 If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.

Ans.

Let r be the radius of the sphere and Δr be the error in measuring the radius.Then,r = 7 m and Δr = 0.02 mNow, the volume V of the sphere is given by,  V=43πr3dVdr=ddr(43πr3)        =4πr2        dV=(dVdr)Δr        =(4πr2)Δr        =4π(7)2×0.02=3.92πm3Hence, the approximate error in calculating the volume is 3.92π m3.

Q.71 If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating ‘its surface area’

Ans.

Let r be the radius of the sphere and Δr be the error in measuring the radius.Then, r = 9 m and Δr = 0.03 mNow, the surface area of the sphere (S) is given by,S=4πr2Differentiating w.r.t. r, we get    dSdr=8πrdS=(dSdr)Δr=(8πr)×0.03  m2=(8π×9)×0.03  m2=2.16πm2Hence, the approximate error in calculating the surface area is 2.16π m2.

Q.72 If f (x) = 3x2 + 15x + 5, then the approximate value of f (3.02) is
A. 47.66 B. 57.66 C. 67.66 D. 77.66

Ans.

Let x=3 and Δx=0.02. Then, we have:f(3.02)=f(x + Δx)=3(x + Δx)2+ 15(x + Δx)+ 5Since,  Δy=f(x+Δx)f(x).​ Therefore,f(x+Δx)=f(x)+Δyf(x)+f(x)Δx[dx=Δx]    f(3.02)(3x2 +15x + 5)+(6x+15)Δx{3(3)2+15(3)+5}+(6×3+15)(0.01)77+33×0.02=77+0.6677.66Hence, the approximate value of f (3.02) is 77.66.The correct answer is D.

Q.73 The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is

A. 0.06 x3 m3 B. 0.6 x3 m3

C. 0.09 x3 m3 D. 0.9 x3 m3

Ans.

Let side of cube is x metres. SoThe Volume of a cube is given by V=x3      dV=(dVdx)Δx    =(ddxx3)Δx  =(3x2)Δx  =(3x2)(0.03x)[3% ofx=0.03x]  =0.09x3Hence, the approximate change in the Volume of the cube is 0.09x3 m3.The correct option is C.

Q.74 Find the maximum and minimum values, if any, of the following functions given by

(i) f(x) = (2x − 1)2 + 3 (ii) f(x)= 9x2+12x+2

(iii) f(x) = −(x − 1)2 + 10 (iv) g(x)= x3 + 1

Ans.

(i) The given function isf(x)=(2x1)2+ 3It is clear that (2x1)2 0forxRTherefore,f(x)=(2x1)2+ 33forxRThe minimum value of f is attained when 2x1=0x=12Minimum value of f=f(12)      =(2×121)2+ 3      =3Hence, function f does not have a maximum value.(ii) The given function isf(x)=9x2+12x+2=(3x + 2)22.It is clear that (3x + 2)2 0forxRTherefore,f(x)=(3x + 2)222forxRThe minimum value of f is attainedwhen 3x + 2=0x=23Minimum value of f=f(23)      =(2×23+2)22      =2Hence, function f does not have a maximum value.(iii) The given function is f(x)=(x1)2+10It is clear that (x1)20 for every xRTherefore, f(x)=(x1)2+1010 for every xRThe maximum value of f is attained when (x1)=0x=1Maximum value of f=f(1)=(11)2+10=10Hence, function f does not have a minimum value.(iv)The given function is g(x) = x3+1Here,function g has neithera maximum value nor a minimum value.

Q.75

Find the maximum and minimum values, if any, of the following functions given by:ix+21                  iigx=x+1+3iiihx=sin2x+5        ivfx=sin4x+3vhx=x+1,  x1,1

Ans.

(i) f(x)=|x+2|1We know that  |x+2|0xRTherefore, f(x)=|x+2|11xRThe minimum value of f is attained when|x+2|=0x=2Minimum value of f=f(2)      =|2+2|1      =1Thus, function f does not have a maximum value.(ii)g(x)=|x+1|+3Since, |x+1|0 xRTherefore, g(x)=|x+1|+33 xRThe maximum value of g is attained when|x+1|=0x=1Maximum value of g=g(1)    =|1+1|+3    =3Hence, function g does not have a minimum value.(iii)f(x)=|sin4x+3|      Since,1sin4x12sin4x+34Thus, the maximum and minimum values of f are 4 and 2 respectively.(v) h(x)=x+1, x(1,1)Here, if a point x0 is closest to1, then we findx0+12<x0+1 for all x0(1,1)Also, if x1 is closest to 1, then x1+1<x1+12+1x(1,1)Hence, function h(x) has neither maximum nor minimum value in (1,1).

Q.76 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

( i ) f( x ) = x 2 ( ii ) g( x ) = x 3 3x ( iii ) h( x )= sinx + cosx, 0<x< π 2 ( iv ) f( x )= sinxcosx, 0<x< π 2 ( v ) f( x ) = x 3 6 x 2 + 9x + 15 ( vi ) g( x )= x 2 + 2 x , x> 0 ( vii )g( x )= 1 x 2 +2 ( viii )f( x )=x 1x ,x>0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGceaqabeaacaaMc8UaaGPaVpaabmaabaGaaCyAaaGaayjkaiaawMcaaiaabccacaWHMbWaaeWaaeaacaWH4baacaGLOaGaayzkaaGaaeiiaiabg2da9iaabccacaWH4bWaaWbaaSqabeaacaWHYaaaaOGaaiiOaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaacckadaqadaqaaiaahMgacaWHPbaacaGLOaGaayzkaaGaaeiiaiaahEgadaqadaqaaiaahIhaaiaawIcacaGLPaaacaqGGaGaeyypa0JaaeiiaiaahIhadaahaaWcbeqaaiaahodaaaGccqGHsislcaWHZaGaaCiEaaqaamaabmaabaGaaCyAaiaahMgacaWHPbaacaGLOaGaayzkaaGaaeiiaiaahIgadaqadaqaaiaahIhaaiaawIcacaGLPaaacqGH9aqpcaqGGaGaaC4CaiaahMgacaWHUbGaaCiEaiaabccacqGHRaWkcaqGGaGaaC4yaiaah+gacaWHZbGaaCiEaiaacYcacaqGGaGaaCimaiabgYda8iaahIhacqGH8aapdaWcaaqaaeXatLxBI9gBaGGbaiab=b8aWbqaaiaaikdaaaaabaWaaeWaaeaacaWHPbGaaCODaaGaayjkaiaawMcaaiaabccacaWHMbWaaeWaaeaacaWH4baacaGLOaGaayzkaaGaeyypa0JaaeiiaiaahohacaWHPbGaaCOBaiaahIhacqGHsislcaWHJbGaaC4BaiaahohacaWH4bGaaiilaiaacckacaWLjaGaaCimaiabgYda8iaahIhacqGH8aapdaWcaaqaaiab=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@00BA@

Ans.

(i)f(x)=x2  ...(i)Differentiating w.r.t. x, we getf(x)=2x  ...(ii)For maxima or minima,   f(x)=02x=0x=0Thus, x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.differentiatingequation(ii)w.r.t. x, we getf(x)=2, which is positive.Therefore, by second derivative test, x=0 is a point of local minima and local minimum value of f at x=0 is f(0)=0.(ii)g(x)=x33x  ...(i)Differentiating w.r.t. x, we get      g(x)=3x23  ...(ii)For maxima or minima, we have      g(x)=03x23=0  x=±1Differentiating equation (ii)w.r.t. x, we get      g(x)=6xPutting x=1, we get      g(1)=6(1)=6>0

So, g( x ) is minimum at x=1 and minimum value is g( 1 )= ( 1 ) 3 3( 1 )=2 Putting x=1, we get g( 1 )=6( 1 )=6<0 So, g( x ) is maximum at x=1 and maximum value is g( 1 )= ( 1 ) 3 3( 1 )=2. ( iii )h( x )=sinx + cosx, 0 <x< π 2 ( i ) Differentiating w.r.t. x, we get h’( x )=cosxsinx( ii ) For maxima or minima, we have h’( x )=0cosxsinx=0 cosx=sinx cosx sinx =1 cotx=cot45°x=45° Differentiating equation ( ii )w.r.t. x, we get h”( x )=sinxcosx putting x=45°,we get h”( 45° )=sin45°cos45° =( 1 2 + 1 2 )<0 Thus, h( x ) is maximum at x=45°, then maximum value is h( 45° )=sin45°+cos45° = 1 2 + 1 2 = 2 2 = 2 . ( iv )f( x )=sinxcosx,0<x<2π Differentiating w.r.t. x, we get f’( x )=cox+sinx For maxima or minima, f’( x )=0cox+sinx=0 sinx=cosx tanx=1x= 3π 4 , 7π 4 ( 0,2π ) andf( x )=sinx+cosx f( 3π 4 )=sin 3π 4 +cos 3π 4 = 1 2 1 2 = 2 <0 Therefore, by second derivative test,x= 3π 4 is a point of local maxima and the local maximum value of f atx= 3π 4 is f( 3π 4 )=sin 3π 4 cos 3π 4 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= 1 2 + 1 2 = 2 and f( 7π 4 )=sin 7π 4 +cos 7π 4 = 1 2 + 1 2 = 2 >0 Therefore, by second derivative test,x= 7π 4 is a point of local minima and the local minimum value of f atx= 7π 4 is f( 7π 4 )=sin 7π 4 cos 7π 4 = 1 2 1 2 = 2 ( v )f( x )= x 3 6x 2 + 9x + 15 Differentiating w.r.t. x, we get f’( x )=3 x 2 12x + 9 For maxima or minima, we have f’( x )=03( x 2 4x + 3 )=0 ( x3 )( x1 )=0 x=3,1 Now, f( x )=6x12 Putting x=3 in f”( x ), we get f( 3 )=6( 3 )12=6>0 So, f( x )islocalminimum at x=3, then local minimum value=f( 3 ) = ( 3 ) 3 6 ( 3 ) 2 + 9( 3 )+15 =15 Putting x=1 in f”( x ), we get f( 1 )=6( 1 )12=6<0 So, f( x )islocalmaximum at x=1, thenlocal maximum value=f( 1 ) = ( 1 ) 3 6 ( 1 ) 2 + 9( 1 )+15 =19 ( vi )g( x )= x 2 + 2 x Differentiating w.r.t. x, we get g’( x )= 1 2 2 x 2 andg( x )= 4 x 3 Formaxima or minima, we have g’( x )=0 1 2 2 x 2 =0 x 2 =4x=±2 Since, x>0 so, x=2 then, g”( 2 )= 4 2 3 = 1 2 >0 Therefore, by second derivative test, x = 2 is a point of local minima and the local minimum value of g at x = 2 is 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g(2)= 2 2 + 2 2 =2 ( vii )g( x )= 1 x 2 +2 Differentiating w.r.t. x, we get g’( x )= 1 ( x 2 +2 ) 2 d dx ( x 2 +2 ) = 2x ( x 2 +2 ) 2 For maxima or minima, we have g’( x )=0 2x ( x 2 +2 ) 2 =0 x=0 Now, for values close to x=0 and to the left of 0, g( x )>0.Also, for values close to x=0 and to the right of 0, g( x )<0. Therefore, by first derivative test, x = 0 is a point of local maxima and the localmaximum value ofg( 0 )= 1 0+2 = 1 2 ( viii )f( x )=x 1x ,x>0 Differentiating w.r.t. x, we get f( x )= d dx ( x 1x ) =x d dx 1x + 1x d dx x =x( 1 2 1x )+ 1x = x+22x 2 1x = 23x 2 1x For maximum or minimum, we have f( x )=0 23x 2 1x =0 23x=0x= 2 3 f( x )= d dx ( 23x 2 1x ) = 2 1x d dx ( 23x )( 23x ) d dx 2 1x ( 2 1x ) 2 = 2 1x ( 03 )( 23x )( 2 2 1x ) ( 2 1x ) 2 = 6 1x + ( 23x ) 1x 4( 1x ) = 6( 1x )+23x 1x 4( 1x ) = 3x4 4 ( 1x ) 3 2 Puttingx= 2 3 in f”( x ), we get f”( 2 3 )= 3( 2 3 )4 4 ( 1 2 3 ) 3 2 = 24 4 ( 1 3 ) 3 2 = 2 ( 4 3 3 ) = 3 3 2 <0 So,f( x ) is maximum at x= 2 3 and maximum value is f( 2 3 )= 2 3 1 2 3 = 2 3 × 1 3 = 2 3 9 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Q.77 Prove that the following functions do not have maxima or minima:
(i) f(x) = ex (ii) g(x) = logx

(iii) h(x) = x3 + x2 + x + 1

Ans.

(i) We have,f(x)=exDifferentiating w.r.t. x, we get        f(x)=exFor maxima or minima,we have        f(x)=0ex=0Which is impossible, because the exponential function can never assume 0 for any value of x.Therefore, there is no number c for that f(c)=0.Thus,function f is neither maxima nor minima.(ii)We have,g(x)=log xDifferentiating w.r.t. x, we get        g(x)=1xFor maxima or minima,        g(x)=01x=0    x=10Thus, there is no point for maxima or minima.Hence, function g(x) does not have maxima or minima.(iii)We have,h(x)=x3 + x2 + x + 1Differentiating w.r.t. x, we geth(x)=3x2+2x+1For maxima or minima, we haveh(x)=03x2+2x+1=0x=b±b24ac2a=2±224(3)(1)2(3)=2±86=1±i23=RHence, function h does not have maxima or minima.

Q.78

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:ifx=x3,  x2,2  iifx=sinx+cosx,  x,πiiifx=4x12x2,x2,92  ivfx=x12+3,x3,1

Ans.

(i) The given function is f(x) = x3Differentiating w.r.t. x, we getf(x)=3x2For maxima or minima, we havef(x)=03x2=0x=0Then, we evaluate the value of f at critical point x = 0 and at end points of the interval [2, 2].f(0) = 0f(2)=(2)3= 8  f(2)=(2)3=8Hence, we can conclude that the absolute maximum value of f on [2, 2] is 8 occurring at x=2. Also, the absolute minimum value of f on [2, 2] is 8 occurring at x=2.(ii)The given function is f(x)=sin x + cos x.Differentiating w.r.t. x, we getf(x)=cos xsinxFor maxima or minima, we havef(x)=0cos xsinx=0tanx=1x=π4Then, we evaluate the value of f at critical pointx=π4and at the end points of the interval[0,π].f(π4)=sinπ4 + cosπ4=12+12=22=2f(0)=sin0 + cos0=0+1=1f(π)=sinπ + cosπ=01=1Hence, we can conclude that the absolute maximum value of f on [0,π] is 2 occurring atx=π4 and the absolute minimum value of f on [0, π] is1 occurring at x = π.(iii)The given function is f(x)=4x12x2Differentiating w.r.t. x, we getf(x)=4xFormaxima or minima, we havef(x)=04x=0x=4Then, we evaluate the value of f at critical point x=4 and at the end points of the interval  [2,92].f(4)=4(4)12(4)2=168=8f(2)=4(2)12(2)2=82=8f(92)=4(92)12(92)2=18818=634Hence, we can conclude that the absolute maximum value of f on [2,92]is 8 occurring at x=4 and the absolute minimum value of f on[2,92] is 10 occurring at x=2.(iv) The given function is  f(x)=(x1)2+3Differentiating w.r.t. x, we getf(x)=2(x1)For maximum and minimum, we havef(x)=02(x1)=0      x=1Then, we evaluate the value of f at critical point x = 1 and at the end points of the interval [3, 1].  f(1)=(11)2+3=3f(3)=(31)2+3=19Hence, we can conclude that the absolute maximum value of f on [3, 1] is 19 occurring at x=3 and the minimum value of f on [3, 1] is 3 occurring at x=1.

Q.79 Find the maximum profit that a company can make, if the profit function is given by

p(x) = 41 − 24x − 18x2

Ans.

We have,p(x)=4124x18x2Differentiating w.r.t. x, we getp(x)=2436xandp(x)=36For​​ maxima or minima, we havep(x)=02436x=0x=2436=23So,p(23)=36<0By second derivative test,x=23,is the point of local maxima of p.maximum profit=4124(23)18(23)2        =41+168        =49Hence, the maximum profit that the company can make is 49 units.

Q.80 

Find both the maximum value and the minimum value of 3x 4 8x 3 +12x 2 48x+25 on the interval 0,3 .

Ans.

Let  f(x)=3x48x3+12x248x+25Differentiating w.r.t. x, we get  f(x)=12x324x2+24x48f(x)=36x248x+24Formaxima or minima, we havef(x)=012x324x2+24x48=0(x2)(x2+2)=0x=2,22 is not real, so x=2.Then, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3].f(0)=3(0)48(0)3+12(0)248(0)+25=25f(2)=3(2)48(2)3+12(2)248(2)+25=39f(3)=3(3)48(3)3+12(3)248(3)+25=16So, the functin f(x) is minimum at x=2 and minimum value is 39. The function is maximum at x=0 and maximum valueis 25.

Q.81 At what point in the interval [0, 2π], does the function sin 2x attain its maximum value?

Ans.

Letf(x)=sin2xDifferentiating w.r.t.x, we getf(x)=2cos2xFor maxima or minima, we havef(x)=02cos2x=02x=π2,3π2,5π2,7π2x=π4,3π4,5π4,7π4Then, we evaluate the values of fat critical pointsx=π4,3π4,5π4,7π4and at the end point so f the interval [0, 2 π].f(π4)=sin2(π4)=sin(π2)=1,f(3π4)=sin2(3π4)=sin(3π2)=1,f(5π4)=sin2(5π4)=sin(5π2)=1,f(7π4)=sin2(7π4)=sin(7π2)=1,f(0)=sin2(0)=sin(0)=0,f(2π)=sin2(2π)=sin4π=0,Hence,we can conclude that the ab solute maximum value of fon[0, 2π] is o ccurring at x=π4andatx=5π4.

Q.82 What is the maximum value of the function sin x + cos x?

Ans.

Letf(x)=sinx+cosxDifferentiating w.r.t. x, we getf(x)=cosxsinxf(x)=sinxcosxFor maxima or minima, we havef(x)=0cosxsinx=0tanx=1x=π4,5π4,...So,f(π4)=sinπ4cosπ4  =(12+12)=2<0By second derivative test, f will be the maximum atx=π4,So maximum value=f(π4)  =sinπ4+cosπ4  =12+12=2

Q.83 Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3]. Find the maximum
value of the same function in [- 3, -1].

Ans.

Let fx=2x3 24x + 107Differentiating w.r.t. x, we get  fx=6x224fx=12xFor maxima or minima, we getfx=06x224=06x24=0x=±2We first consider the interval [1, 3].Then, we evaluate the value of f at the critical point x=2[1, 3] and at the end points of the interval [1, 3].f(2) =2(8)24(2)+107=1648+107=75f(1) =2(1)24(1)+107=224+107=85f(3) =2(27)24(3)+107=5472+107=89Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x=3.Next, we consider the interval [3,1].Now, we find the value of f(x) at the critical point x=2 [3, 1] and at the end points ofthe interval [3, 1].f(3)=2 (27)24(3) + 107    =54 + 72 + 107    =125f(1)=2(1)24 (1) + 107    =2 + 24 + 107     =129f(2)=2(8)24 (2) + 107    =16 + 48 + 107    =139Hence, the absolute maximum value of f(x) in the interval [3,1] is 139 occurring at x=2.

Q.84 It is given that x = 1, the function x4 – 62 x2 + ax + 9 attains its maximum value, on the interval [0, 2].

Find the value of a.

Ans.

Let f(x)=x462x2+ax+9Differentiating w.r.t. x, we getf(x)=4x3124x+aSince, the given that function f attains its maximum value on the interval [0, 2] at x=1.f(1)=04(1)3124(1)+a=0        4124+a=0            120+a=0a=120Hence, the value of a is 120.

Q.85 Find the maximum and minimum values of x + sin2x on [0, 2π].

Ans.

Let f(x)=x + sin 2x.Differentiating w.r.t. x, we getf(x)=1+2cos2xf(x)=4sin2xNow, for maxima or minima, we havef(x)=01+2cos2x=0cos2x=12=cosπ3        =cos(ππ3)cos2x=cos2π3  2x=2±2π3,    nZ    x=±π3,        nZ    x=π3,2π3,4π3,5π3(0,2π)Then, we evaluate the value of f at critical pointsx=π3,2π3,4π3,5π3and at the end points of the interval [0, 2π].f(π3)=π3 + sin (2π3)=π3 + sin (ππ3)=π3 + sin (π3)=π3 + 32f(π3)=π3 + sin (2π3)=2π3+32f(2π3)=2π3 + sin (4π3)=2π332f(4π3)=4π3 + sin (8π3)=4π3+32f(5π3)=5π3 + sin (10π3)=5π332  f(0)=0+sin0=0  f(2π)=2π+sin2π=0Hence, we can conclude that the absolute maximum value of f(x) in the interval [0, 2π] is 2π occurring at x = 2π and the absolute minimum value of f(x) in the interval [0, 2π]is 0 occurring at x = 0.

Q.86 Find two numbers whose sum is 24 and whose product is as large as possible.

Ans.

Let two numbers be x and (24x)andP(x)  denotes the product of two numbers, thenP(x)=x(24x)=24xx2Differentiating w.r.t.x, we getdP(x)dx=242xP(x)=2Formaxima or minima, we haveP(x)=0242x=0x=12So,P(12)=2<0By second derivative test, x=12 is the point of local maximaof P. Hence, the product of the numbers is the maximum when the numbers are 12 and 2412=12.

Q.87 Find two positive numbers x and y such that x + 6 = 60 and xy3 is maximum.

Ans.

The two numbers are x and y such that x + y = 60. y=60x Let f( x )=x y 3 f( x )=x ( 60x ) 3 Differentiating f( x ) w.r.t. x, we get f’( x )= d dx { x ( 60x ) 3 } =x d dx ( 60x ) 3 + ( 60x ) 3 d dx x =x.3 ( 60x ) 2 ( 01 )+ ( 60x ) 3 ×1 = ( 60x ) 3 3x ( 60x ) 2 = ( 60x ) 2 ( 60x3x ) = ( 60x ) 2 ( 604x ) f( x )= ( 60x ) 2 d dx ( 604x )+( 604x ) d dx ( 60x ) 2 = ( 60x ) 2 ( 04 )+( 604x ).2( 60x )( 01 ) =4 ( 60x ) 2 2( 604x )( 60x ) =2( 60x )( 1202x+604x ) =2( 60x )( 1806x ) =12( 60x )( 30x ) For maxima or minima, we have f’( x )=0 ( 60x ) 2 ( 604x )=0 x=60, 60 4 =60,16 When, x=60 f”( 60 )=12( 6060 )( 3060 )=0 Thus, at x=60, f( x ) is neither maximum nor minimum. When, x=15 f”( 60 )=12( 6015 )( 3015 )<0 So,f( x ) is maximum when x=15 and y=6015=45 Hence, the required numbers are 15 and 45. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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Q.88 Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.

Ans.

Let one number be x. Then, the other number is y=(35x). Let P(x)= x 2 y 5 . Then, we have: P x = x 2 35x 5 Differentiating w.r.t. x, we get P’ x = x 2 d dx 35x 5 + 35x 5 d dx x 2 = x 2 ×5 35x 4 d dx 35x + 35x 5 ×2x =5 x 2 35x 4 01 +2x 35x 5 =x 35x 4 5x+702x = 35x 4 70x7 x 2 P x = 35x 4 d dx 70x7 x 2 + d dx 35x 4 70x7 x 2 P x = 35x 4 7014x + 4 35x 3 d dx 35x 70x7 x 2 P x = 35x 4 7014x +4 35x 3 01 70x7 x 2 =7 35x 3 35x 102x 4 10x x 2 =7 35x 3 6 x 2 120x+350 For maximum or minimum, we have

P’ x =0 35x 4 70x7 x 2 =0 7x 35x 4 10x =0x=0,10,35 Atx=0, P 0 =7 350 3 6× 0 2 120×0+350 =7 35 3 350 >0 So, function is minimum at x=0. Atx=35, y=3535=0 So,P x =35×0=0 Thus,x=35 will not be possible value of x. Atx=10, P 10 =7 3510 3 6× 10 2 120×10+350 =7 25 3 6001200+350 <0 So, function is maximum at x=10. Therefore, the two positive numbers are x=10 and y=3510=25 Hence, the required numbers are 10 and 25.

Q.89 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Ans.

Let one number be x. Then, the other number is y=(16x).Let S(x)=x3+y3. Then, we have:        S(x)=x3+(16x)3Differentiating w.r.t. x, we get    S(x)=ddxx3+ddx(16x)3    =3x2+3(16x)2ddx(16x)    =3x2+3(16x)2(01)    =3x23(16x)2  S(x)=ddx3x2ddx3(16x)2    =6x6(16x)ddx(16x)    =6x6(16x)(01)  =6x+6(16x)=96For maxima or minima,f(x)=03x23(16x)2=0    3x23(25632x+x2)=0      3(x2256+32xx2)=0      3(32x256)=0x=25632=8Then,S(8)=96>0By second derivative test, x = 8 is the point of local minima of S.Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 168=8.

Q.90 A square piece of tin of side 18 cm is to made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?

Ans.

Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (182x) cm each and the height of the box is x cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@BDFE@

MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@AAB1@

Therefore, the volume V(x) of the box is given by, V( x )=x(18 2x) 2 Differentiatingw.r.t. x, we get V’( x )=x d dx ( 182x ) 2 + ( 182x ) 2 d dx x MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGceaqabeaacaqGubGaaeiAaiaabwgacaqGYbGaaeyzaiaabAgacaqGVbGaaeOCaiaabwgacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG2bGaae4BaiaabYgacaqG1bGaaeyBaiaabwgacaqGGaGaaeOvaiaabIcacaqG4bGaaeykaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOyaiaab+gacaqG4bGaaeiiaiaabMgacaqGZbGaaeiiaiaabEgacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabkgacaqG5bGaaeilaaqaaiaabAfadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaqG4bGaaeikaiaabgdacaqG4aGaeyOeI0IaaeOmaiaabIhacaqGPaWaaWbaaSqabeaacaqGYaaaaaGcbaGaamiraiaadMgacaWGMbGaamOzaiaadwgacaWGYbGaamyzaiaad6gacaWG0bGaamyAaiaadggacaWG0bGaamyAaiaad6gacaWGNbGaaGjbVlaabEhacaqGUaGaaeOCaiaab6cacaqG0bGaaeOlaiaabccacaqG4bGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqGNbGaaeyzaiaabshaaeaacaqGwbGaae4jamaabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadIhadaWcaaqaaiaadsgaaeaacaWGKbGaamiEaaaadaqadaqaaiaabgdacaqG4aGaeyOeI0IaaeOmaiaabIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaabkdaaaGccqGHRaWkdaqadaqaaiaabgdacaqG4aGaeyOeI0IaaeOmaiaabIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaabkdaaaGcdaWcaaqaaiaadsgaaeaacaWGKbGaamiEaaaacaWG4baaaaa@AB15@

=x.2(182x)ddx(182x)+(182x)2×1=2x(182x)(02)+(182x)2=(182x)(4x+182x)=2(9x)×6(3x)=12(9x)(3x)V(x)=12{(9x)ddx(3x)+(3x)ddx(9x)}=12{(9x)(01)+(3x)(01)}=12(9+x3+x)=24(x6)For maxima or minima, we havev(x)=012(9x)(3x)=0x=3,9If x = 9, then the length and the breadth will become 0.x9.When x=3,V(3)=24(36)=72<0By second derivative test, x = 3 is the point of maxima of V.Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

Q.91 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the ‘volume of the box is maximum’ ?

Ans.

Let the side of the square to be cut off be x cm. Then, the height of the box is x, the length is ( 452x ), and the breadth is ( 242x ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@AE05@

Therefore, the volume V(x) of the box is given by,V(x)=x(452x)(242x)=x(108090x48x+4x2)=1080x138x2+4x3Differentiating w.r.t. x, we getV(x)=ddx(1080x138x2+4x3)

=1080276x+12 x 2 =12( x 2 23x+90 ) V( x )=12( 2x23 ) For maxima or minima, we have V’( x )=012( x 2 23x+90 )=0 ( x18 )( x5 )=0x=5,18 It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. Thus, x cannot be equal to 18. When x=5, V( 5 )=12( 2×523 ) =12( 1023 )=156<0 By second derivative test, x=5 is the point of maxima. Hence, the side of the square to be cut off to make the volume of the box maximum is 5 cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@C091@

Q.92 Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Ans.

Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.Then, the diagonal passes through the centre and is of length 2a cm. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@C32A@

Now, by applying the Pythagoras theorem, we have: ( 2a ) 2 = l 2 + b 2 b 2 =4 a 2 l 2 b= 4 a 2 l 2 Area of the rectangle, A=lb =l 4 a 2 l 2 Differentiating w.r.t. l, we get dA dl = d dl ( l 4 a 2 l 2 ) = 4 a 2 l 2 d dl l+l d dl 4 a 2 l 2 = 4 a 2 l 2 +l 1 2 4 a 2 l 2 ( 2l ) = 4 a 2 l 2 l 2 4 a 2 l 2 = 4 a 2 l 2 l 2 4 a 2 l 2 = 4 a 2 2 l 2 4 a 2 l 2 d 2 A d l 2 = d dl ( 4 a 2 2 l 2 4 a 2 l 2 ) = 4 a 2 l 2 d dl ( 4 a 2 2 l 2 )( 4 a 2 2 l 2 ) d dl 4 a 2 l 2 ( 4 a 2 l 2 ) 2 = 4 a 2 l 2 ( 04l )( 4 a 2 2 l 2 )× 1 2 4 a 2 l 2 ( 2l ) 4 a 2 l 2 = 4l 4 a 2 l 2 + 4 a 2 l2 l 3 4 a 2 l 2 ( 4 a 2 l 2 ) = 4l( 4 a 2 l 2 )+4 a 2 l2 l 3 ( 4 a 2 l 2 ) 3 2 = 16 a 2 l+4 l 3 +4 a 2 l2 l 3 ( 4 a 2 l 2 ) 3 2 = 2 l 3 12 a 2 l ( 4 a 2 l 2 ) 3 2 For maxima or minima, we have dA dl =0 4 a 2 2 l 2 4 a 2 l 2 =0 4 a 2 2 l 2 =0 l 2 = 4 a 2 2 =2 a 2 l= 2 aandb= 4 a 2 2 a 2 = 2 a When l= 2 a, d 2 A d l 2 = 2 2 a( 2 a 2 6 a 2 ) ( 4 a 2 2 a 2 ) 3 2 = 2 2 a( 4 a 2 ) ( 2 a 2 ) 3 2 <0 By the second derivative test, when l= 2 a, then the area of the rectangle is the maximum. Since l=b= 2 a, the rectangle is a square. Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area. 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Q.93 Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.

Ans.

Let r and h be the radius and height of the cylinder respectively.Then, the surface area (S) of the cylinder is given by,        S=2πr2+2πrhh=S2πr22πr=S2π(1r)rLet V be the volume of the cylinder. Then,      V=2πr2h=2πr2{S2π(1r)r}=Sr2πr3Differentiating w.r.t. r,we getdVdr=S23πr2 and d2Vdr2=6πrFor maxima or minima, we havedVdr=0S23πr2=0r2=S6πr=±S6π=S6π,S6π(Neglect)When  r=S6πd2Vdr2=6π(S6π)<0By second derivative test, the volume is the maximum whenr=S6πS=6πr2. h=6πr22πr22πr=4πr22πr=2rHence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

Q.94 Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions
of the can which has the minimum surface area?

Ans.

Let r and h be the radius and height of the cylinder respectively.Then, volume (V) of the cylinder is given by,V=πr2h=100h=100πr2Surface area (S) of the cylinder is given by,    S=2πr2+2πrh=2πr2+2πr(100πr2)    =2πr2+(200r)Differentiating w.r.t. x, we get      dSdr=ddr(2πr2)+ddr(200r)    =4πr200r2      d2Sdr2=ddr4πrddr200r2    =4π+400r3For maxima or minima, we have      dSdr=04πr200r2=0    4πr=200r2r3=2004π=50πWhen  r=(50π)13So,d2Sdr2>0  By second derivative test, the surface area is the minimum when the radius of the cylinder is  (50π)13cm.When  r=(50π)13cm, h=100π(50π)23=2(50π)13cmHence, the required dimensions of the can which has the minimum surface area is given  by radius=(50π)13 andheight=2(50π)13cm.

Q.95 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Ans.

Letcirclepiecelength=xmThensquarepiecelength=(28x)mForcircle,Perimeterofcircle =2πr=xr=x2πAndforsquare,Perimeterofsquare=28x=4aa=28x4Nowthesumoftheareas AofcircleandsquareA=πr2+a2=π(x2π)2+(28x4)2=x24π+(28x4)2dAdx=x2π28x8d2Adx2=12π+18For maxima orminima,wehavedAdx=0x2π28x8=0 or x=28π4+πd2Adx2>0atx=28π4+πAisminimumatcirclepiece=x=28π4+πSquarepiece =28x=1124+π

Q.96 

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8 27 of the volume of the sphere.

Ans.

DAB is the cone inscribed in a sphere of radius R.Volume of cone V=13πr2hWhere r is the base radius of cone and h its heightHeight of cone h=OC+OD =R+OC=R+x Radius of Cone=r where r2=R2OD2=R2x2      V=13π(R2x2)(R+x)=13π[R3+R2xx2Rx3]  dVdx=13π[R22xR3x2] =13π[R23xRxR3x2] =13π(R3x)(R+x)        d2Vdx2=13π{(R3x)ddx(R+x)+(R+x)ddx(R3x)} =13π(R3x3R3x) =13π(2R6x) =23π(R+3x)dVdx=0x=R3  or  Rd2Vdx2<0  at  x=R3Hence  x=R3  gives maximum volume.    V=13π(R2R29)(R+R3)=13π×8R29×4R3=827[43πR3]Hence the volume of cone =827(Volume of sphere) is the largest cone that could be inscribed in a given sphere .

Q.97 

Show that the right circular cone of least curved surface and given volume has analtitude equal to 2 timethe radius of the base.

Ans.

Here, Volume of the coneisV. V= 1 3 π r 2 h r 2 = 3V πh and surface area be Sof cone. S=πrl=πr h 2 + r 2 Whereh=height of the cone r=radius of the cone l=slant height of the cone S 2 = π 2 r 2 ( h 2 + r 2 ) Let S 1 = S 2 then S 1 = π 2 r 2 ( h 2 + r 2 ) S 1 = 3πV h ( h 2 + 3V πh )=3πVh+ 9 V 2 h 2 [ r 2 = 3V πh ] d S 1 dh =3πV+9 V 2 ( 2 h 3 ) d 2 S 1 d h 2 = 54 V 2 h 4 d S 1 dh =0 for maxima/minima 3πV+9 V 2 ( 2 h 3 )=0 3πV=9 V 2 ( 2 h 3 ) h 3 = 6V π d 2 S 1 d h 2 >0 at h 3 = 6V π Therefore curved surface area is minimum at π h 3 6 =V. Thus, π h 3 6 = 1 3 π r 2 h h 2 =2 r 2 h= 2 r Hence for least curved surface the altitude is 2 times radius. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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b8aWjaadIgadaahaaWcbeqaaiaaiodaaaaakeaacaaI2aaaaiabg2da9maalaaabaGaaGymaaqaaiaaiodaaaGae8hWdaNa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Q.98 

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan 1 2 .

Ans.

Let θ be the semi-vertical angle of the cone. It is clear thatθ[ 0, π 2 ]. Let r, h, and l be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@F5BB@

Now, r=l sin θ and h=l cos θThe volume (V) of the cone is given by,V=13πr2h        =13π(l sin θ)2(lcos θ)        =13πl3sin2θcosθDifferentiating w.r.t. θ, we get      dV=13πl3(sin2θdcosθ+cosθdsin2θ)        =13πl3(sin3θ+2sinθcos2θ)&  d2V2=13πl3(3sin2θcosθ+3cos3θ2sin2θcosθ)For maxima or minima, we have      dV=013πl3(sin3θ+2sinθcos2θ)=0sin3θ=2sinθcos2θtan2θ=2tanθ=2    θ=tan12When  θ=tan12sin2θ=2cos2θ  d2V2=13πl3(3×2cos2θ.cosθ+3cos3θ2×2cos2θ.cosθ)  =13πl3(6cos3θ+3cos3θ4cos3θ)  =13πl3(4cos3θ)<0By second derivative test, the volume (V) is the maximum when  θ=tan12.

Q.99 

Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin -1 1 3 .

Ans.

Here in  ΔAOC ,  l2=h2+r2The total surface area of the cone isS=πr2+πrlSπr2πr=lor  l=SπrrVolume of the cone is given by      V=13πr2h=13πr2l2r2Let    V1=V2=19π2r4(l2r2)V1=19π2r4([Sπrr]2r2)V1=19π2r4(S2π2r22Sπ)=19(S2r22Sπr4)Differentiatingw.r.t. r, we get    V1=19(2S2r8Sπr3)V1=19(2S224Sπr2)For maxima  or  minima,V1=019(2S2r8Sπr3)=02S2r=8Sπr3    S4π=r2V1=19(2S224Sπr2)<0 at r2=S4πV is maximum at r2=S4π  4πr2=S4πr2=πr2+πrl3πr2=πrlor rl=13i.e sinAOC=13[InΔAOC,sin(OAC)=rl] semivertical angle is sin1(13).

Q.100

The point on the curve x2=2y which is nearest to the point 0, 5 isA22,4  B22,0  C0,0  D2,2

Ans.

The given curve is x2=2y.Let P(x,y) be a point on x2=2y and A(0,5) be the given point.Then, AP2=(x0)2+(y5)2  =x2+(x225)2[x2=2yy=x22]Let  Z=AP2. Then, Z is maximum or minimum according as AP is maximum or minimum.Now, Z=x2+(x225)2Differentiating w.r.t. x, we get        dZdx=ddx{x2+(x225)2}      =2x+2(x225)ddx(x225)      =2x+2(x225)(x0)      =2x+x310x      =x38x      d2Zdx2=3x28For maxima or minima, we have        dZdx=0x38x=0x(x28)=0x=0,±22When x=0,      d2Zdx2=3(0)28=8<0So,Z is maximum at x=0 i.e., point P is not nearest to curve.When x=±22    d2Zdx2=3(±22)28    =248=16>0So,Z is minimum at x=±22 i.e., point P is nearest to curve.Putting x=±22 in x2=2y, we get      y=x22=(±22)22=82=4Hence, the point (±22,4) on x2=2y is the nearest to the point (0,5).Thus,the option A is correct.

Q.101

For all real values of x, the minimum value of  1x+x21+x+x2  isA0   B1   C3   D13

Ans.

Let f(x)=1x+x21+x+x2Differentiating w.r.t. x, we getf(x)=ddx(1x+x21+x+x2)=(1+x+x2)ddx(1x+x2)(1x+x2)ddx(1+x+x2)(1+x+x2)2=(1+x+x2)(01+2x)(1x+x2)(0+1+2x)(1+x+x2)2=1xx2+2x+2x2+2x3(1x+x2+2x2x2+2x3)(1+x+x2)2=1xx2+2x+2x2+2x31+xx22x+2x22x3(1+x+x2)2=2+2x2(1+x+x2)2=2(x21)(1+x+x2)2f(x)=(1+x+x2)2ddx2(x21)2(x21)ddx(1+x+x2)2(1+x+x2)4=(1+x+x2)22(2x0)2(x21)2(1+x+x2)ddx(1+x+x2)(1+x+x2)4=4x(1+x+x2)24(x21)(1+x+x2)(0+1+2x)(1+x+x2)4=4(1+x+x2){(1+x+x2)x(x21)(1+2x)}(1+x+x2)4=4{x+x2+x3x2+12x3+2x}(1+x+x2)3=4(x3+3x+1)(1+x+x2)3For maxima or minima, we havef(x)=02(x21)(1+x+x2)2=0x21=0x=±1Whenx=1,f(1)=4(13+3×1+1)(1+1+12)3=4(3)27=49>0f(1)=4{(1)3+3(1)+1}{1+(1)+(1)2}3=4(1)1=4<0By second derivative test, f is the minimum at x = 1 and the minimum value is given byf(1)=11+11+1+1=13Thus,the correct answer is D.

Q.102 

The maximum value of x x-1 +1 1 3 ,0£x£1is A 1 3 1 3 B 1 2 C 1 D 0

Ans.

Let f(x)=[x(x1)+1]13,0x1Differentiating w.r.t. x, we getf(x)=ddx[x(x1)+1]13=13[x(x1)+1]23ddx[x(x1)+1]=13[x(x1)+1]23(ddxx2ddxx+ddx1)=13[x(x1)+1]23(2x1)=(2x1)3[x(x1)+1]23For maxima or minima, we havef(x)=0(2x1)3[x(x1)+1]23=02x1=0x=12Then, we evaluate the value of f at critical pointx=12and at the end points of the interval [0, 1].f(0)=[(0)(01)+1]13=0f(12)=[(12)(121)+1]13=(34)13f(1)=[(1)(11)+1]13=1Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1.The correct answer is C.

Q.103 Using differentials, find the approximate value of each of the following:

( a ) ( 17 81 ) 1 4 ( b ) ( 33 ) 1 5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGcbaWaaeWaaeaaieqacaWFHbaacaGLOaGaayzkaaGaaGPaVlaaykW7daqadaqaamaalaaabaGaa8xmaiaa=DdaaeaacaWF4aGaa8xmaaaaaiaawIcacaGLPaaadaahaaWcbeqaamaalaaabaGaa8xmaaqaaiaa=rdaaaaaaOGaaCzcaiaaxMaacaWLjaGaaCzcamaabmaabaGaa8NyaaGaayjkaiaawMcaaiaaykW7caaMc8+aaeWaaeaacaWFZaGaa83maaGaayjkaiaawMcaamaaCaaaleqabaGaa8xlamaalaaabaGaa8xmaaqaaiaa=vdaaaaaaaaa@562C@

Ans.

(a)  Let    y=x14where x=1681 and Δx=181            dydx=14x34Then,Δy=(x+Δx)14x14=(1781)14(1681)14=(1781)1423      (1781)14=Δy+23Now, dy is approximately equal to Ny and is given by,            dy=(dydx)Δx=(14x34)Δx={14(1681)34}181=274×8×181=196=0.010Hence, the approximate value of  (1781)14=0.010+23=0.010+0.667=0.677.(b)  Let  y=x15 and x=32​​ and Δx=1(33)15=12+ΔyNow, dy is approximately equal to Δy and is given by,    dy=(dydx)Δx=15x65Δx=15(32)65×1=15(2)6=0.003Hence, the approximate value of(33)15=120.003=0.50.003=0.497

Q.104

Show that the function given byfx = logxx has maximum at x = e.

Ans.

Given: fx=logxxDifferentiating w.r.t. x, we getfx=ddxlogxx=xddxlogxlogxddxxx2=x.1xlogx.1x2=1logxx2fx=ddx1logxx2=x2ddx1logx1logxddxx2x22=x201x1logx×2xx4=x2x+2xlogxx4fx=3+2logxx3For maxima or minima, we havefx=01logxx2=01logx=0logx=1        logx=loge              x=eNow,fx=3+2logee3=3+2e3=1e3<0Therefore, fxis maximum at x=e.

Q.105 The two equal sides of an isosceles triangle with fixed base b and decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Ans.

Let ΔABC be isosceles where BC is the base of fixed length b. Let the length of the two equal sides of ΔABC be a. DrawADBC MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@A8F2@

Now, in ΔADC, by applying the Pythagoras theorem, we have: AD= a 2 b 2 4 = 1 2 4 a 2 b 2 Area of triangle( A )= 1 2 b× 1 2 4 a 2 b 2 The rate of change of the area with respect to time( t ) is given by, dA dt = 1 4 b× 8a 2 4 a 2 b 2 da dt It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second, i.e., da dt =3cm/s dA dt =b× a 4 a 2 b 2 ×3= 3ab 4 a 2 b 2 When a=b, we have dA dt = 3 b 2 4 b 2 b 2 = 3 b 2 3 b = 3 b Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of 3 bc m 2 /s. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@1B83@

Q.106 Find the equation of the normal to curve y2 = 4x at the point (1, 2).

Ans.

The given curve​​is y2= 4xDifferenetiating w.r.t. x,we get2ydydx=4    dydx=42y=2y      m=(dydx)(1,2)=22=1Slope of Normal is      M=1m=11=1Equation of normal at the point (1,2)is      y2=(1)(x1)y2=x+1x+y=3

Q.107

Show that the normal at any point θ to the curvex =acosθ +aθsinθ,  y =asinθaθcosθis at a constant distance from the origin.

Ans.

Wehave,x=acosθ+aθsinθ  and    y=asinθaθcosθThen differentiating w.r.t. x, we get  dx=adcosθ+ad(θsinθ)=asinθ+a(θcosθ+sinθ)=aθcosθ  dy=adsinθad(θcosθ)=acosθa(θsinθ+cosθ)=aθsinθdydx=(dy)(dx)=aθsinθaθcosθ=tanθSlope of normal=1tanθThe equation of the normal at a given point (x, y) is given by,          yasinθ+aθcosθ=1tanθ(xacosθaθsinθ)        yasinθ+aθcosθ=cosθsinθ(xacosθaθsinθ)ysinθasin2θ+aθsinθcosθ=xcosθ+acos2θ+aθsinθcosθ      ysinθasin2θ=xcosθ+acos2θxcosθ+ysinθa(sin2θ+cos2θ)=0xcosθ+ysinθa=0[sin2θ+cos2θ=1]Now, the perpendicular distance of the normal from the origin:p=|0.cosθ+0.sinθa|cos2θ+sin2θ=a,which is independent of θ.Hence, the perpendicular distance of the normal from the origin is constant.

Q.108

Find the intervals in which the function f given byfx =4sinx2xxcosx2+cosxis (i) increasing (ii) decreasing.

Ans.

The function is f(x)=4sinx2xxcosx2+cosxDifferentiating w.r.t. x, we getf(x)={(2+cosx)ddx(4sinx2xxcosx)(4sinx2xxcosx)ddx(2+cosx)(2+cosx)2}f(x)={(2+cosx)(4cosx2+xsinxcosx)(4sinx2xxcosx)(0sinx)(2+cosx)2}f(x)={(2+cosx)(3cosx2+xsinx)+sinx(4sinx2xxcosx)(2+cosx)2}f(x)=6cosx+3cos2x42cosx+2xsinx+xsinxcosx                                        +4sin2x2xsinxxsinxcosx(2+cosx)2f(x)=4cosx+3cos2x4+4sin2x(2+cosx)2f(x)=4cosx+3cos2x4+4(1cos2x)(2+cosx)2    =4cosx+3cos2x4+44cos2x(2+cosx)2    =4cosxcos2x(2+cosx)2    =cosx(4cosx)(2+cosx)2Now,f(x)=0cosx(4cosx)=0cosx=0  or4cosx=0cosx=0butcosx4x=π2,3π2, which divides interval (0,2π) into three disjoint intervals,i.e., (0,π2),(π2,3π2)  and  (3π2,2π).In  intervals,(0,π2),(3π2,2π),  f(x)>0Thus, f(x) is increasing for  0<x<π2​​  and  3π2<x<2π.In the interval(π2,3π2),  f(x)<0.Thus, f(x) is decreasing forπ2<x<3π2.

Q.109 

Find the intervals in which the function f given by f x =x 3 + 1 x 3 ,x¹0is (i) increasing (ii) decreasing.

Ans.

The function is f(x)=x3+1x3Differentiating w.r.t. x, we getf(x)=ddx(x3+1x3)=3x23x4Then, f(x)=03x23x4=03(x61)x4=0x61=0x=±1Now, the points x=1 and x=1 divide the real line into three disjoint intervalsi.e., (,1),(1,1)​ and(1,).In  intervals(,1)and(1,) i.e., x<1 and x>1,f(x)>0Thus, when x <1 and x > 1, f is increasing.In interval (1, 1) i.e., when  1 < x < 1,F(x)<0Thus, when 1<x<1, f is decreasing.

Q.110

Find the maximum area of an isosceles triangle inscribed in the ellipse x2a2+y2b2=1withits vertex at one end of the major axis.

Ans.

The given ellipse is x2a2+y2b2=1.Let the major axis be along the xaxis.Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0). Since the ellipse is symmetrical with respect to the xaxis and yaxis, we can assume the coordinates of A to be(x1, y1) and the coordinates of B to be (x1,y1).Now, we have  y1=±baa2x12Coordinates of A are  (x1,baa2x12), the coordinates of B are  (x1,baa2x12)andthe coordinates of C are(a,0).As the point (x1, y1) lies on the ellipse, the area(A) of ΔABC:A=12|a(baa2x12baa2x12)+(x1)(baa2x120)+(x1)(0baa2x12)|A=12|a(2baa2x12)x1(baa2x12)x1(baa2x12)|A=ba2x12+x1baa2x12  ...(i)Differentiating w.r.t. x1, we get  dAdx1=ddx1(ba2x12+x1baa2x12)=bx1a2x12+baa2x12bx12a2x12=baa2x12(ax1+a2x12x12)=b(2x12ax1+a2)aa2x12Now,  dAdx1=02x12ax1+a2=0x1=a±a24(2)a22(2)=a±9a24x1=a±3a4=a,  a2But, x1 cannot be equal to a.x1=a2y1=baa2a24=ba2a3=3b2Now,d2Adx12=ba{a2x12(4x1a)(2x12x1a+a2)(2x1)2a2x12(a2x12)}=ba{(a2x12)(4x1a)+x1(2x12x1a+a2)(a2x12)32}=ba{2x133a2xa3(a2x12)32}Also,  whenx1=a2, thend2Adx12=ba{2(a2)33a2(a2)a3(a2a222)32}=ba×a343a32a3(3a222)32=ba{94a3(3a222)32}<0Thus, the area is the maximum whenx1=a2.Maximum area of the triangle is:A=ba2a24+(a2)baa2a24    =ab32+ab43=334ab

Q.111 A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m3. If building of tank costs ₹ 70 per sq meters for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?

Ans.

Let l, b, and h represent the length, breadth, and height of the tank respectively.Then, we have height (h) = 2 mVolume of the tank = 8m3Volume of the tank = l × b × h                                  8 = l × b × 2      lb=4b=4lNow, area of the base=lb=4Area of the 4 walls (A)=2h (l + b)      A=2×2l+4lDifferentiating w.r.t. x, we getdAdl=414l2For maxima or minima, dAdl=0414l2=0  l2=4l=±2However, the length cannot be negative.Therefore, we have l = 4.b=4l=42=2Now,d2Adl2=32l3When l=2,d2Adl2=3223=4>0Thus, by second derivative test, the area is the minimum when l = 2.We have l = b = h = 2.Cost of building the base= 70×(lb)=  704=  280.Cost of building the walls= 2h (l+b)×45       = 902(2+2)       = 720  Required total cost= (280+720)      =  1000Hence, the total cost of the tank will be 1000.

Q.112 The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the
side of square is double the radius of the circle.

Ans.

Let r be the radius of the circle and a be the side of the square.Then, we have: 2πr+4a=ka= k2πr 4 The sum of the areas of the circle and the square=A A=π r 2 + a 2 =π r 2 + ( k2πr 4 ) 2 =π r 2 + ( k2πr ) 2 16 dA dr = d dr π r 2 + d dr ( k2πr ) 2 16 =2πr+ 2( k2πr ) 16 ( 02π ) dA dr =2πr π( k2πr ) 4 For maximum or minimum, we have dA dr =02πr π( k2πr ) 4 =0 2πr= π( k2πr ) 4 8πr=π( k2πr ) r= k 2( 4+π ) Now, d 2 A d r 2 =2π+ π 2 2 >0 Whenr= k 2( 4+π ) , d 2 A d r 2 =2π+ π 2 2 >0 The sum of the areas is least whenr= k 2( 4+π ) . Putting value of r in a= k2πr 4 ,we get a= k2π( k 2( 4+π ) ) 4 = 4k+πkπk 4( 4+π ) = k ( 4+π ) =2r Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle. 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Q.113 A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find
the dimensions of the window to admit maximum light through the whole opening.

Ans.

Let x and y be the length and breadth of the rectangular window.Radius of the semicircular opening x 2 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@948F@

It is given that the perimeter of the window is 10 m. Perimeterof window=x+2y+ πx 2 x+2y+ πx 2 =10 x( 1+ π 2 )+2y=10 y=5 1 2 x( 1+ π 2 ) Now, area( A )of window=xy+ 1 2 π ( x 2 ) 2 A=xy+ 1 2 π ( x 2 ) 2 A=x{ 5 1 2 x( 1+ π 2 ) }+ 1 8 π x 2 =5x 1 2 x 2 ( 1+ π 2 )+ 1 8 π x 2 Differentiating w.r.t. x, we get dA dx =5x( 1+ π 2 )+ 1 4 πx and d 2 A d x 2 =( 1+ π 2 )+ 1 4 π =1 1 2 π For maxima or minima, we have dA dx =0 5x( 1+ π 2 )+ 1 4 πx=0 x= 20 π+4 Thus, whenx= 20 π+4 , d 2 A d x 2 <0 Therefore, by second derivative test, the area is the maximum when lengthx= 20 π+4 m. So, y=5 20 π+4 ( 2+π 4 ) =5 5( 2+π ) π+4 = 5π+20105π π+4 = 10 π+4 m Hence, the required dimensions of the window to admit maximum light is givenlength= 20 π+4 m and breadth= 10 π+4 m. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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b8aWjaadIhadaahaaWcbeqaaiaaikdaaaaakeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpcaaI1aGaamiEaiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaGaamiEamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaaGymaiabgUcaRmaalaaabaGae8hWdahabaGaaGOmaaaaaiaawIcacaGLPaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI4aaaaiab=b8aWjaadIhadaahaaWcbeqaaiaaikdaaaaakeaacaWGebGaamyAaiaadAgacaWGMbGaamyzaiaadkhacaWGLbGaamOBaiaadshacaWGPbGaamyyaiaadshacaWGPbGaamOBaiaadEgacaqGGaGaae4Daiaab6cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiaiaabIhacaqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiabgkDiElaaxMaacaWLjaGaaCzcaiaaxMaacaaMc8+aaSaaaeaacaWGKbGaamyqaaqaaiaadsgacaWG4baaaiabg2da9iaaiwdacqGHsislcaWG4bWaaeWaaeaacaaIXaGaey4kaSYaaSaaaeaacqWFapaCaeaacaaIYaaaaaGaayjkaiaawMcaaiabgUcaRmaalaaabaGaaGymaaqaaiaaisdaaaGae8hWdaNaamiEaaqaaiaadggacaWGUbGaamizaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamyqaaqaaiaadsgacaWG4bWaaWbaaSqabeaacaaIYaaaaaaakiabg2da9iabgkHiTmaabmaabaGaaGymaiabgUcaRmaalaaabaGae8hWdahabaGaaGOmaaaaaiaawIcacaGLPaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI0aaaaiab=b8aWbqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9iabgkHiTiaaigdacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaiab=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Q.114 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.Show that the minimum length of the
hypotenuse is (a2/3+ b2/3) .

Ans.

Let ΔABC be right-angled at B. Let AB=x and BC=y.Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D053@

L e t C = θ I n Δ A B C , A C = x 2 + y 2 N o w , P C = b c o s e c θ A n d , A P = a s e c θ A C = A P + P C A C = b c o s e c θ + a s e c θ . . . ( 1 ) D i f f e r e n t i a t i n g w . r . t . θ , w e g e t d d θ ( A C ) = b cosec θ cot θ + a sec θ tan θ a n d d 2 d θ 2 ( A C ) = b ( cosec 3 θ cosec θ cot 2 θ ) + a ( sec 3 θ + sec θ tan 2 θ ) = b ( cosec 3 θ + cosec θ cot 2 θ ) + a ( sec 3 θ + sec θ tan 2 θ )

F o r m a x i m a o r m i n i m a , w e h a v e d d θ ( A C ) = 0 b cosec θ cot θ + a sec θ tan θ = 0 b cosec θ cot θ = a sec θ tan θ b sin θ × cos θ sin θ = a cos θ × sin θ cos θ a sin 3 θ = b cos 3 θ tan 3 θ = b a tan θ = ( b a ) 1 3 sin θ = b 1 3 a 2 3 + b 2 3 a n d c o s θ = a 1 3 a 2 3 + b 2 3 . . . ( i i ) S o , d 2 d θ 2 ( A C ) > 0 f o r tan θ = ( b a ) 1 3 T h e r e f o r e , b y s e c o n d d e r i v a t i v e t e s t , t h e l e n g t h o f t h e h y p o t e n u s e i s t h e m i n i m u m w h e n tan θ = ( b a ) 1 3 . S o , A C = b c o s e c θ + a s e c θ = b sin θ + a cos θ = b a 2 3 + b 2 3 b 1 3 + a a 2 3 + b 2 3 a 1 3

= b 2 3 a 2 3 + b 2 3 + a 2 3 a 2 3 + b 2 3 = a 2 3 + b 2 3 ( b 2 3 + a 2 3 ) = ( a 2 3 + b 2 3 ) 3 2 H e n c e , t h e m i n i m u m l e n g t h o f t h e h y p o t e n u s e i s ( a 2 3 + b 2 3 ) 3 2 .

Q.115

Find the points at which the function f given byfx =x-24 x+13has(i) local maxima (ii) local minima(iii) point of inflexion

Ans.

The given function is f( x )= ( x2 ) 4 ( x+1 ) 3 Differentiatingw.r.t. x, we get f’( x )= ( x2 ) 4 .3 ( x+1 ) 2 +4 ( x2 ) 3 ( x+1 ) 3 = ( x2 ) 3 ( x+1 ) 2 ( 3x6+4x+4 ) = ( x2 ) 3 ( x+1 ) 2 ( 7x2 ) Now, f’( x )=0 ( x2 ) 3 ( x+1 ) 2 ( 7x2 )=0 x=2,1and 2 7 Now, for values of x close to 2 7 and to the left of 2 7 ,f( x )>0. Also, for values of x close to 2 7 and to the right of 2 7 ,f( x )<0. Thus, x= 7 2 is the point of local maxima. Now, for values of x close to 2 and to the left of2,f( x )<0. Also, for values of x close to 2 and to the right of 2,f( x )>0. Thus, x=2is the point of local minima. Now, as the value of x varies through1,f’( x )does not changes its sign. Thus, x = -1 is the point of inflexion. 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Q.116

Find the absolute maximum and minimum values of the function f given by fx=cos2x+sinx,x0,π

Ans.

The given function is f( x )= cos 2 x+sinx Differentiating w.r.t. x, we get f’( x )=2cosxsinx+cosx Now,f( x )=0 2cosxsinx+cosx=0 cosx( 2sinx+1 )=0 cosx=0 and 2sinx+1=0 x= π 2 [ 0,π ] and sinx= 1 2 x= π 6 [ 0,π ] Now, evaluating the value of f at critical pointsx= π 2 and x= π 6 and at the end points of the interval[ 0,π ],wehave f( π 6 )= cos 2 π 6 +sin π 6 = 5 4 f( 0 )= cos 2 0+sin0=1 f( π )= cos 2 π+sinπ=1 f( π 2 )= cos 2 π 2 +sin π 2 =1 Hence, the absolute maximum value of f is 5 4 occurring atx= π 6 and absolute minimum value of f is 1 occurring atx=0, π 2 andπ. 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Q.117 

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is4r3.

Ans.

A sphere of fixed radius (r) is given. Let R and h be the radius

and the height of the cone respectively.

The volume (V) of the cone= 1 3 π R 2 h Now, from the right triangle BCD, we have: BC= r 2 R 2 h= r 2 R 2 AC=r+ r 2 R 2 V= 1 3 π R 2 ( r+ r 2 R 2 ) = 1 3 πr R 2 + 1 3 π R 2 r 2 R 2 Differentiating w.r.t. R, we get dV dR = 2 3 πrR+ 2 3 πR r 2 R 2 + 1 3 π R 2 × 1 2 r 2 R 2 ×( 02R ) dV dR = 2 3 πrR+ 2 3 πR r 2 R 2 1 3 r 2 R 2 π R 3 = 2 3 πrR+ 2πR( r 2 R 2 )π R 3 3 r 2 R 2 = 2 3 πrR+ 2πR r 2 3π R 3 3 r 2 R 2 d 2 V d R 2 = d dR ( 2 3 πrR )+ d dR ( 2πR r 2 3π R 3 3 r 2 R 2 ) = 2 3 πr + 3 r 2 R 2 d dR ( 2πR r 2 3π R 3 )( 2πR r 2 3π R 3 ) d dR 3 r 2 R 2 ( 3 r 2 R 2 ) 2 = 2 3 πr+ 3 r 2 R 2 ( 2π r 2 9π R 2 )( 2πR r 2 3π R 3 ) 3( 02R ) 2 r 2 R 2 9( r 2 R 2 ) = 2 3 πr+ 6( r 2 R 2 )( 2π r 2 9π R 2 )+6R( 2πR r 2 3π R 3 ) 18 ( r 2 R 2 ) 3 2 For maxima or minima, we have dV dR =0 2 3 πrR+ 2πR r 2 3π R 3 3 r 2 R 2 =0 2 3 πrR= 3π R 3 2πR r 2 3 r 2 R 2 2r r 2 R 2 =3 R 2 2 r 2 Squarring both sides, we get 4 r 2 ( r 2 R 2 )= ( 3 R 2 2 r 2 ) 2 R 2 = 8 r 2 9 When R 2 = 8 r 2 9 , d 2 V d R 2 <0 The volume is the maximum when R 2 = 8 r 2 9 . When R 2 = 8 r 2 9 , the height of the cone =r+ r 2 8 r 2 9 =r+ r 3 = 4r 3 Hence, it can be seen that the altitude of the right circular cone of maximum volumethat can be inscribed in a sphere of radius r is 4r 3 . 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Q.118

Let f be a function defined on a, b such that fx>0,a,b.Then prove that f is an increasing function on a,b.

Ans.

Let (x1, x2)(a,b) such that x1<x2. Consider the subinterval[x1, x2].Since f(x) is differentiable on (a,b) and [x1,x2](a, b).Therefore, f(x) is continuous on [x1, x2] and differentiable at(x1, x2). By the Lagranges mean value theorem, there existsc(x1, x2) such that f(c)=f(x2)f(x1)x2x1...(i)Since, f(x)>0 for all x(a, b), so in particular, f(c)>0Now,      f(c)>0f(x2)f(x1)x2x1>0f(x2)f(x1)>0      f(x2)>f(x1)f(x1)<f(x2)Since, x1, x2 are arbitrary points in (a, b).Therefore,x1<x2f(x1)<f(x2) for all x1, x2(a, b)Hence, f(x) is increasing on (a, b).

Q.119 

Show that the height of the cylinder of maximum volume that can be inscribed in asphere of radius R is 2R 3 .Also find the maximum volume.

Ans.

A sphere of fixed radius (R) is given.Let r and h be the radius and the height of the cylinder respectively.

From the given figure, we have  h=2R2r2    The volume (V) of the cylinder=πr2hV=2πr2R2r2Differentiating w.r.t. r, we get  dVdr=2πddr(r2R2r2)=2π(r2ddrR2r2+R2r2ddrr2)=2π(r2(02r)2R2r2+2rR2r2)=2π(2r(R2r2)r3(R2r2))=4πrR26πr3R2r2Now,d2Vdr2=ddr4πrR26πr3R2r2        =R2r2ddr(4πrR26πr3)(4πrR26πr3)ddrR2r2(R2r2)2        =R2r2(4πR218πr2)(4πrR26πr3)2r2R2r2R2r2        =(R2r2)(4πR218πr2)+r(4πrR26πr3)(R2r2)32For maxima or minima, we havedVdr=04πrR26πr3R2r2=04πrR26πr3=0r2=2R23Whenr2=2R23,d2Vdr2<0The volume is the maximum when  r2=2R23.

When  r2=2R23,the height of the cylinder=2R22R23      =2R3Hence, the volume of the cylinder is the maximum when the height of the cylinder is  2R3.Maximum Volume = πR22R314.4R23.2R3 = πR22R3113 = 4πR333

Q.120

Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle θ is onethird that of the cone and the greatest volume of cylinder is 427πh3tan2α.

Ans.

In given right circular cone of fixed height is h and semivertical angle be α.Then a cylinder of radius R and height H is inscribed in the cone. Then GAO=α,OG=r,OA=h,OE=R​ and  CE=H.We have, r=htanαNow, since ΔAOG is similar to ΔCEG, we have:AOOG=CEEGhr=HrR[EG=OGOE

H= h r ( rR ) = h htanα ( htanαR ) = 1 tanα ( htanαR ) Now, the volume (V) of the cylinder V=π R 2 h = π R 2 tanα ( htanαR ) =π R 2 h π R 3 tanα Differentiating w.r.t. R, we get dV dR =2πRh 3π R 2 tanα and d 2 V d R 2 =2πh 6πR tanα For maxima or minima, we have dV dr =02πRh 3π R 2 tanα =0 πR( 2h 3R tanα )=0R= 2htanα 3 So, d 2 V d R 2 <0forR= 2htanα 3 By second derivative test, the volume of the cylinder is the greatest whenR= 2htanα 3 . Then,H= 1 tanα ( htanαR ) = 1 tanα ( htanα 2htanα 3 ) = h 3 Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest. Now, the maximum volume of the cylinder V=π R 2 H =π ( 2htanα 3 ) 2 ( h 3 ) V= 4 27 π h 3 tan 2 α 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Q.121 A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic mere per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m3/h (B) 0.1 m3/h
(C) 1.1 m3/h (D) 0.5 m3/h

Ans.

Let r be the radius of the cylinder.Then, volume (V) ofthe cylinder=πr2hPutting r=10m, we getV=π(10)2h=100πhDifferentiating with respect to time t, we get:dVdt=ddt100πh=100πdhdtThe tank is being filled with wheat at the rate of 314 cubicmetres per hour.dVdt=314m3/hThus,wehave:314=100πdhdtdhdt=314100×3.14=1Hence, the depth of wheat is increasing at the rate of 1m/h.The correct answer is A.

Q.122 The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, -1) is
(A) 22/7 (B)6/7 (C) 7/6 (D) – 6/7

Ans.

The given curve is x=t2+3t8 and y=2t22t5Differentiating w.r.t. x, we getdxdt=2t+3 and dydt=4t2dydx=dydtdxdt=4t22t+3Given is point (2,1).When x=2,2=t2+3t8t2+3t10=0(t+5)(t2)=0t=5,2Aty=1,1=2t22t52t22t4=0t2t2=0(t2)(t+1)=0t=2,1The common value of t is 2, soslope=(dydx)t=2=4(2)22(2)+3  =824+3=67Thus,the correct answer is B.

Q.123 The line y = mx + 1 is a tangent to the curve y2 = 4x if the value of m is
(A) 1 (B) 2
(C) 3 (D) 1/2

Ans.

The equation of the tangent to the given curve is y = mx + 1.Substituting y=mx + 1 in y2= 4x, we get:    (mx + 1)2= 4xm2x2+2mx+1=4xm2x2+2(m2)x+1=0  ...(i)Since a tangent touches the curve at one point, the roots of equation (i) must be equal.Therefore, we have:D=0b24ac=0{2(m2)}24(m2)1=0      4(m2)2=4m2m24m+4=m2m=1Hence, the required value of m is 1.The correct answer is A.

Q.124 The normal at the point (1, 1) on the curve 2y + x2 = 3 is
(A) x + y = 0 (B) x – y = 0
(C) x + y + 1= 0 (D) x – y = 0

Ans.

The equation of the given curve is 2y + x 2 = 3. Differentiating with respect to x, we have: 2 dy dx +2x=0 dy dx =x dy dx 1,1 =1 The slope of the normal to the given curve at point (1, 1) = 1 dy dx 1,1 = 1 1 =1 Hence, the equation of the normal to the given curve at (1, 1) is y1=1 x1 y1=x1 xy=0 Thus,the correct answer is B.

Q.125 The normal to the curve x2 = 4y passing (1, 2) is

(A) x + y = 3 (B) x – y = 3
(C) x + y = 1 (D) x – y = 1

Ans.

The equation of the given curve is x2= 4y.Differentiating with respect to x, we have:2x=4dydxdydx=x2The slope of the normal to the given curve at point (h, k) is:(dydx)(h,k)=2hEquation of the normal at point (h, k) is:yk=2h(xh)Now, it is given that the normal passes through the point (1, 2).Therefore, we have:2k=2h(1h)k=2+2h(1h)  ...  (i)Since (h, k) lies on the curve x2=4y, we have h2=4k.k=h24From equation (i), we have:    h24=2+2h(1h)h3=8h+88hh=2k=224=1Hence, the equation of the normal is given as:y1=22(x2)y1=x+2x+y=3Thus,the correct answer is A.

Q.126

The points on the curve 9y2=x3, where the  normal to the curve makes equal intercepts  with theaxes areA 483B  4,83C438    D438

Ans.

The equation of the given curve is 9y2=x3.Differentiating with respect to x, we have:9(2y)dydx=3x2dydx=3x218y=x26yThe slope of the tangent to the given curve at point (h,k)is(dydx)(h,k)=h26kThe slope of the normal to the given curve at point (h,k)is1(dydx)(h,k)=6kh2The equation of the normal to the curve at(h,k)is        yk=6kh2(xh)h2yh2k=6kx+ykhh2y+6kx=h2k+ykhxh2k+ykh6k+yh2k+ykhh2=1Since, the normal makes equal intercepts with the axes.Therefore, We have:h2k+ykh6k=h2k+ykhh2    h(h+y)6=(h+y)kh      h6=kh      h2=6k  ...(ii)Also, the point (h,k)  lies on the curve9y2=x3, so we have  9k2=h3...(ii)From equation(i)and(ii), we have   9(h26)2=h3h44=h3    h=4From equation (ii),​ we have  9k2=43k2=649k=±83Hence, the required points are(4,±83).Thus, the correct answer is A.

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