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 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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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=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@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=