CBSE Class 12 Maths Revision Notes Chapter 6

Class 12 Mathematics Chapter 6 Notes

The Mathematics subject is very demanding. It requires a strong foundation and regular practice to excel in this subject. The knowledge of basic concepts introduced in previous classes is essential to understanding the complex topics introduced in Class 12 Mathematics.

The Class 12 Mathematics Chapter 6 notes- Application of Derivatives is a continuation of Chapter 5. In the notes, students can find definitions of derivatives, increasing and decreasing functions, maxima and minima, tangents and normals, and some more concepts related to differentiation. Students will be asked to solve various problems using the first and second derivative tests. With the help of the Class 12 Mathematics Chapter 6 notes, students will learn to illustrate basic principles involved in the application of derivatives.

Extramarks, an online learning platform, provides a helping guide to gain an in-depth understanding of the concepts included in this chapter. It also aids in the preparation of class 12 boards and other national-level competitive examinations. The Class 12 Mathematics Chapter 6 notes include detailed and apt information, which ensures deep knowledge and sufficient practice.

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Key Topics Covered In Class 12 Mathematics Chapter 6 Notes

The key topics covered under Extramarks class 12 Mathematics chapter 6 notes include the following.

Rate of change of quantity:

A function y=f(x), then ddxy = f’(x) gives the rate of change of function f(x).
Let the function x= f(t) and y= f(t) be two functions such that variable x and y are varying to parameter t, then using Chain Rule, we get ddxy= ddtyddtx, ddtx 0
The rate of change of volume for decreasing side is given ddxy.

Decreasing and increasing functions:

Let a function f(x) be continuous in [p,q] and differential on (p,q), then
If f’ (x) = 0 x ∈ (p, q), then F is a constant function

If f’ (x) < 0 x ∈ (p, q), then F is a decreasing function

If f’ (x) > 0 x ∈ (p, q), then F is an increasing function

Tangent and Normal to a Curve:

A normal is a perpendicular line to the tangent. The straight-line equation passing through a point with slope m is given as y – y1= m (x – x1). If y= f (x) is a tangent to the curve at point P(x1, y1), then the slope is given as ddxy = f’(x).
Therefore, the equation becomes y – y1= ddxy (x – x1)
Another way of writing the equation of normal is y – y1= -ddyx (x – x1)

Length of tangent = y cosec = y 1+ (ddxy)2(ddxy)
Length of normal = y 1+ (ddxy)2

Slope of the curve:

Slope of normal at point P= -1Slope of tangent at point P = -1(ddxy) = -ddyx
If the normal is perpendicular to x-axis and parallel to y-axis at (x,y), then f’(x) = 0
If the normal is parallel to x-axis and perpendicular to y-axis at (x,y), then f’(x) = ∞

The angle of intersection of two curves:

If y= f1(x) and y = f2(x) are two curves that meet at point P(x1 , y1), then the angle of intersection of two curves is equal to the angle between the tangent of two curves at the point P.
The angle is given as tan = m1 – m21 + m1 m2
Slope m1 = ddxf1 and m2 = ddxf2

Maxima and Minima:

Differentiation is used to find the minima and maxima of a function. Minima is the lowest point, and Maxima is the highest point of a graph.
If f (x) ≤ f (a), when x=a, x in the domain, f(x) has absolute maximum at point a.
If f (x) ≥ f (a), when x=a, x (p, q), f (x) has a relative minimum.
If f (x) ≥ f (a), when x=a, x in the domain, f(x) has an absolute minimum at point a.
If f (x) ≤ f (a), when x=a, x (p, q), f (x) has a relative maximum.


Monotonic function: If the function f(x) is either decreasing or increasing in the domain.

Strictly increasing function: In function f(x), for every x1, x2 D. If x1> x2, then f(x1)> f(x2). i.e., when the value of x increases there is an increase in the function f(x) value.

Strictly decreasing function: In function f(x), for every x1, x2 D. If x1 x2, then f(x1) f(x2). i.e., when the value of x decreases there is a decrease in the function f(x) value.

Non-increasing function: In function f(x), for every x1, x2 D. If x1> x2, then f(x1) f(x2). i.e., when the value of x increases, the value of function f(x) would never increase.

Non-decreasing function: In function f(x), for every x1, x2 D. If x1> x2, then f(x1) f(x2). i.e., when the value of x decreases, the value of function f(x) would never decrease.

For a function f(x) differentiable on (a, b),
f’(x)>0, if the function is increasing and f’(x)<0, if the function is decreasing

Properties of Monotonicity:

If the function f(x) is strictly increasing, then f-1 exists on the domain and is strictly increasing.
If a continuous function f(x) is strictly increasing, then f-1 is also continuous on the domain [f(a), f(b)]
If the function f(x) and g(x) is strictly increasing (decreasing), then the composite function gof(x) exists on the domain and is strictly increasing (decreasing).
If either function f(x) or g(x) is strictly increasing and the other function is strictly decreasing, then the composite function gof(x) exists on the domain and is strictly decreasing.

Critical Points:

The points where the function f(x) is not differentiative, but its derivative is equal to zero are known as critical points. The point where the maxima and minima of a function occur are critical points; however, a critical point doesn’t necessarily imply that it is the maxima and minima of a function.

Points of inflection:

In a continuous function f(x), if the first derivative may or may not be zero but the second derivative at a point must be zero, then that point is known as the point of inflection. At this point, f’’(x) can change the sign.
Case 1: y=f(x) is concave downward when the f” (x) < 0, x ∈ (a, b) Case 2: y=f(x) is concave upward in interval (a, b) when the f” (x) > 0, x ∈ (a, b)

Applications of derivatives play a significant role in the branch of Physics. It helps in the study of seismology. With the help of graphs, loss and profit occurred can also be calculated. Also, derivatives have a range of applications like temperature, distance and speed.

Class 12 Mathematics Chapter 6 notes Exercises & Answer Solutions.

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Q.1 Find the maximum area of an isosceles triangle inscribed in the ellipse


with its vertex at one end of the major axis.


A = area of isosceless Δ APPAM = aa cosθPP‘ = bsinθ+bsinθ2    = 2bsinθ2=2bsinθA= 12aacosθ2bsinθ  =ab1cosθsinθDifferentiating A w.r.t. θA‘ = ab1cosθcosθ+sinθsinθ   =abcosθcos2θ+sin2θ   =abcosθcos2θ+1+cos2θ   =abcosθ+1+2cos2θA = 0 for maxima/minimaabcosθ+12cos2θ=02cos2θ+cosθ1=02cos2θ2cosθ+cosθ1=02cosθ+1cosθ1=0θ=2π3           orθ=π2which is not possibleA<0atθ=2π3A=ab1cos2π3sin2π3orA= ab1+1232=334ab

Q.2 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.



Here AC = Hypotenuse of triangle ABCAnd AC = AP + PAC = b2+y2+a2+x2with the help of figureWe know that PD BCTherefore,ADAB=PDBCyy+a=bx+byx+b=by+ayx+yb=yb+bay=abxAC = b2+abx2+a2+x2andAC=ab22b2+abx22x3+xa2+x2AC=0for maxima/minimaab22b2+abx22x3+xa2+x2=0ab22b2+abx21x3=xa2+x2 Squaring both sidesab2b2x2+a2b2x21x6=x2a2+x2ab4b2x2+a21x6=x2a2+x2ab4b2=x6     orx=a23b13AC>0at x =a23b13AC=b2+a2b2a43b23+a2+a43b23        =b2+a23b43+a2+a43b23        =b23b23+a23+a23a23+b23        =b23+a23a23+b23        =a23+b2332

Q.3 A window is in the form of a 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.


Step 1:Perimeter of the window when the width of the window is x and 2r is the length.2x+2r+12×2πr=10Given   2x+2r+πr=10   2x+r2+π=10        .….1For admitting the maximum light through the openingthe area of the window must be maximum.Let A = Sum of areas of rectangle and semi circle.Step2:Area of circle = πr2Area of rectangle = l ×b = 2×r×xA = 2rx + 12πr2  =r10π+2r12πr2  =10r12π+2r2For maximum aea dAdr=0andd2Adr2is negative.10π+4r=0         π+4r=10                   r=10π+4Step 3:d2Adr2=π+4      Differentiating w.r.t.ri.e.d2Adr2 is negative for r = 10π+4Aismaximum.From1we have10 = π+2r+2xPut the value of r in above equation.10 = π+2×10π+4+2x10=10π+2π+4+2x     10π+2+2xπ+4π+410π+4=10π+2+2xπ+410π+410π+2=2xπ+410π+4010π20= 2xπ+4x = 10π+4Step 4: Length of rectangle = 2r = 2 10π+4=20π+4Breath = 10π+4

Q.4 Show that semi-vertical angle of right circular cone of given surface area and maximum volume is .



Here in ΔAOC, l2=h2+r2S=πr2+πrlSπr2πr=lorl=SπrrThe total surface area of the cone isVolume of the cone is given byV = 13πr2h=13πr2l2r2V1=V2=19π2r2l2r2V1=19π2r4Sπrr2r2V1=19π2r4S2π2r22Sπ=19S2r22Sπr4V1=192S2r28Sπr3V1=0for maxima/minima192S2r8Sπr3=02Sπr3S4π=r2V1=192S224Sπr2<0Vis maximum at r2=S4π4πr2=S4πr2=πr2+πrl3πr2=πrlor rl=13i.esinAOC=13semivertical angle is sin113


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


Here, volume of the conev = 13πr2hr2=3Vπh    .………1Surface areas = πrl = πrh2+r2                         .………2Whereh= height of the cone r= radius of the cone l= Slant height of the coneS2=π2r2h2+r2by2Let S1=S2thenby1S1=3πVhh2+3Vπh=3πVh+9V2h2dS1dh=3πV+9V22h3dS1dh=0 for maxima/minima3πV+9V22h3=03πV+9V22h3=h3=6VπAs surface area is least, we havedS1dh2>0when h3=6VπTherefore curved surface area is minimum when 3πh36=VThus,πh36=13πr2hh2=2r2h=2rHence for least curved surface the altitude is 2times radius.

Q.6 Find two numbers whose sum is 24 and whose product is as large as possible.


Let one number be x. Then, the other number is (24 − x).

Let P(x) denote the product of the two numbers. Thus,

P(x) = x(24 – x) = 24 – x2

∴ P'(x) = 24 – 2x

P”(x) = -2

Now, P'(x) = 0 ⇒ x = 12

Also, P”(12)= -2 < 0

∴ By second derivative test, x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.

Q.7 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?


Let circle piece length = x m

Then square piece length = 28 – x m

For circle,
Perimeter of circle = 2


r =x ⇒ r = x/2


And for square,

Perimeter of square = 28 – x = 4a ⇒ a = (28 – x)/4

Now the sum of the areasAof circle and squareA = πr2+a2  =πx2π2+28x42  =x24π+28x42dAdx=x2π28x8For maxima/minimadAdx=0x2π28x8=0orx=28π4+πd2Adx2=0orx=28π4+πHence, A is minimum at circle piece = x = 28π4+πSquare piece = 28-x = 1124+x.

Q.8 A square piece of tin of side 18 cm is to be 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?


Let each side of the square to be cut off be x cm

Therefore, for the box:

Length = 18 – 2x

Breadth = 18 – 2x

And Height = x
Volume ‘V’ of the box: V = (18 – 2x)2 x

dVdx=182x2.1+×.2182x2     =182x186xFor maxima/minima, dVdx=0182x186x=0x=3and x = 9x = 9is not possible as the length and breadth can not be, d2ydx2=182x6+186x2at x = 3 d2ydx2<0 Volume of the box is maximum when x = 3. 

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.9 Find both the maximum and minimum value of 3x4 – 8x + 12x2 – 48x + 25 on the interval [0, 3].


Let f(x) = 3x4 – 8x3 + 12x2 – 48x + 25, therefore,

f'(x) = 12x3 – 24x2 + 24x – 48

f'(x) = 12(x3 – 2x2 + 2x – 4)

f'(x) = 12(x2 + 2) ( x – 2)

For maxima/minima f'(x) = 0

12(x2 + 2) ( x – 2) = 0 x = 2.

Now, we have to find f(x) at x = 0, 2 and 3.

f(0) = 25, f(2) = – 39 and f(3) = 16

At x = 0, maximum value is 25

At x = 2, minimum value is – 39.


Use differential to approximate 0.037.


Let y = x, x = 0.040 and x + Δx = 0.037.Then Δx = 0.037-0.047 = -0.003for x = 0.040, y = 0.040=0.2Let dx = Δx = -0.003Now, y = xi.e., dydx=12x=10.4Therefore, dy = dydxxdxi.e.,dy=10.40.003=3100Δy=3400Hence,0.037=y+Δy = 0.2 – 3400=0.1925.

Q.11 Use differential to approximate (25)1/3.


Let y = x13and x = 27Therefore, dydx=12x213Suppose, Δx = -2Now, y+Δy= x+Δx13or2513=x13+ΔyAs Δy dyΔy=dydxΔx=13x213.2               =227=0.074      2513=3+0.074=2.926

Q.12 If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.


Let r be the radius of the sphere and Δr be the error in measuring the radius.
Then r = 9 cm and Δr = 0.03 cm. Now, the volume V of the sphere is given by

V = (4/3)

π r3

, therefore,

dV/dr =

4π r2



(9)2(0.03) = 9.72



Thus, the approximate error in calculating the volume will be 9.72




Find the equation of the tangent to the curve y = 3x2which isparallel to the line 4x-2y+5 = 0.


Let the point of contact of the tangent line parallel to the given line be x1,y1.Equation of curve is y = 3x2.…….  1and equation of given line is 4x-2y+5 = 0.Differentiating equation 1,w.r.t. x, we getdydx=ddx3x212     = 323x212dydxx1,y1=323x12Differentiating equation 2,w.r.t. x, we get     4-2dydx+0=0dydxx1,y1=2Sinceline is parallel to tangent to the curveSo,  323x12=2        3=4 3x12Squarring both sides, we get               9= 16 3x122     3x12=916x1=4148Since point x1,y1lies on curve 1,so       y1=3x12       y1=3×41482           =41162           =34Thus, the equation of tangent at 4148,34is yy1=dydxx1,y1xx1 y34=2x4148 4y34=248×414864y3=48×4124y18=48×4148x24y41+18=048x24y23=0

Q.14 Prove that the curves x = y3 and xy = k cut at right angles if 8k2 = 1.


Here x = y3 ..(1)

and xy = k ..(2)

Let (x1, y1) be the intersecting point

By solving given equations we have (x1, y1) = (k2/3, k1/3)

Slope of the equation (1) at (k2/3, k1/3) is

dy/dx = 1/2y or dy/dx at (k2/3, k1/3) = 1/ 2k1/3

And slope of the equation (2) at (k2/3, k1/3)is

dy/dx = -(y/x) or dy/dx at (k2/3, k1/3) = – 1/k1/3

Now curve cuts at right angles
⇒ (1/2)k2/3.(- 1/k1/3)= – 1
(1)3 = 8k2

⇒ 8k2 = 1.

Q.15 For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.


Let (x1, y1) be the required point on the given curve y = 4x3 – 2x5.
Slope of the curve dy/dx = 12x2 – 10x4

Slope of the curve at (x1, y1) = 12x12 – 10x14

Thus, the equation of tangent is

y – y1 = (12x12 – 10x14)(x – x1)

As tangent passes through the origin

y1 = (12x12 – 10x14)(x1)

⇒ 4x13 – 2x15 = 12x13 – 10x15 or 8x15 – 8x13 =0

⇒ 8x13 ( x12 – 1) =0

⇒ x1 = 0, – 1, 1

at x1 = 0 ⇒ y1 = 0

at x1 = – 1 ⇒ y1 = – 2, and

at x1 = 1 ⇒ y1 = 2

Hence, the required points are (0, 0), (-1, -2) and (1, 2).

Q.16 Find the equation of tangent to the curve given by x =asin3t, y = bcos3t at a point where t =



Here, x =asin3t, y = bcos3t …(1)
Differentiating (1) w.r.t.t, we get

dx/dt = 3asin2tcost and dydt = – 3bcos2t sint

Therefore, dy/dx = -(b/a) cot t

Hence, slope of the tangent at t =


is given by

dy/dx = – (b/a) cot


= 0.

Hence, the equation of tangent is given by

y – 0 = 0(xa), i.e. y = 0.

Q.17 Let I be any interval disjoint from (-1, 1). Prove that the function

given by f(x) = x + (1/x) is strictly increasing on I.


We have f x=x+1xfx= 11x2=x21x2Now x I x 1,1     x1orx1 x21     x210x21x20     fx0Thus fx 0 for all x IHence fxis strictly increasing on I.


Show that y = log1+x2x2+x, x >-1 is an increasing function of x throughout its domain.


fx= log 1+x2x2+x,x>1fx=11+x2x+2xx+22=11+x4x+22=x2x+1x+22for to be increasing fx>0x2x+1x+22>01x+1>0x>1Hence, y = logx2xx+2is an increasing function of x for all values of x > -1.

Q.19 Find the intervals in which the function f given by f(x) = sinx + cosx, 0 ≤ x ≤ 2


is strictly increasing or strictly decreasing.


We have,

f(x) = sinx + cosx

or, f'(x) = cosx – sinx, for turning point f'(x) = 0 it gives sinx = cosx

i.e., x= 14,5π4

three disjoint intervals 0,π4)π4,5π4and (5π4,2πfx>0for all x 0,π4) and (5π4,2πfx<0for all x π4, 5π4fx is strictly increasing on x 0,π4) and (5π4,2πand fxis strictly decreasing on π4,5π4.

Q.20 Find the intervals in which the function f given by f(x) = 4x3 – 6x2 – 72x + 30 is (a) strictly increasing (b) strictly decreasing.


We have f(x) = 4x3 – 6x2 – 72x + 30 or f'(x) = 12x2 – 12x – 72

= 12(x2 – x -6) = 12(x – 3) (x + 2).

Therefore, f'(x) = 0 gives x = – 2 , 3.

There are three disjoint intervals, (– , – 2), (– 2, 3) and (3, ).

f'(x) > 0 for all x ∈ (– ∞, – 2) and (3, ∞),

f'(x) < 0 for all x(– 2, 3)

Q.21 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 one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?


Let height of the cone = h = 4 cm, and radius of the cone = r
Given h = (1/6)r or 6h = r

Sand is pouring from a pipe at the rate of 12 cm/s

i.e., dv/dt = 12, where v = volume of the cone.

We know that the volume of cone

v = (1/3)

π r2

h = (1/3)


(6h)2h = 12



On differentiating w.r.t.t, we get

dv/dt = 12 x 3



⇒dh/dt = 1/(


x 3 x 4 x 4) = 1/48



Q.22 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.


Here 6y = x3 + 2 ...(1)

Differentiating w.r.t t. we get

6 dy/dt = 3x2 dx/dt …(2)

the y-coordinate is changing 8 times as fast as the x-coordinate.

i.e., dy/dt = 8 dx/dt, putting in (2), we get

48 dx/dt = 3x2 dx/dt

x2 = 16, or

x = ± 4

by (1), we get

y = 11 at x = 4 and

y = – 31/3 at x = – 4.

Q.23 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 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?


Let AC be the ladder = 5m and BC is distance the distance between the foot of the ladder and wall = 4m. BC is increasing at the rate of 2cm/s dydx=2m/sNow in ΔABCx2+y2=25          .….1x=3     y = 4 givendiff. equation 1w.r.t.t,we get2xdxdt+2ydydt=06dxdt+8×2=0dxdt= –166=83m/sheight is decreasing at the rate of 83m/s.

Q.24 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cm3 of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.


As volume of the baloon with radius r which always remains spherical is V = 43nrr3VolumeVof the baloon is increasing at the rate of 900 cm3sdVdt=900cm3sdVdt=4πr2drdt900 = 4π152drdtdrdt=9004π152=9004×15×15π=1πcm/s

Q.25 The length


of a rectangle is decreasing at the rate of 3 cm/minute and the width yis increasing at the rate of 2 cm/minute. When x = 10 cm and y = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.


Since the length x is decreasing and the width y is increasing with respect to time, we have

dx/dt = – 3 cm/min and dy/dt = 2 cm/min

(a) the perimeter of rectangle is P = 2 (x + y)

Therefore, rate of change of perimeter and dy/dt = 2 cm/min

dP/dt = 2{dx/dt + dy/dt} = 2( – 3 + 2) = – 2 cm/min
Thus, perimeter is decreasing at the rate of 2 cm/min.

(b) The area A of the rectangle is given by A = xy.

Therefore, the rate of change of area A with respect to time t is

dA/dt = x dy/dt + ydx/dt = 10 x 2 + 6 x ( – 3) = 20 -18cm2/min

Thus, area is increasing at the rate of 2 cm/min.

Q.26 A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?


Area A of a circle with radius r is given by A =



Therefore, the rate of change of area A with respect to time t is

dA/dt = 2


r dr/dt.

It is given that dr/dt = 4 cm/s

Therefore, when r = 10 cm,

dA/dt = 2


x 10 x 4 = 80



Thus, the enclosed area is increasing at the rate of 80


cm2/s, when r = 10 cm.

Q.27 Find the rate of change of the area of a circle with respect to its radius r when r = 5cm.


The area A of a circle with radius r is given by A =

π r2


Therefore, the rate of change of the area A with respect to its radius r is given by

dA/dr = 2


r = 10


(Since, r = 5 cm given)

Thus, the area of the circle is changing at the rate of 10



Q.28 Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 2%.


V = x3ordV=dVdxΔx=3x2Δx         =3x20.02x=0.06x3m3

Q.29 Show that the function given by f(x) = 5x – 2 is strictly increasing on R.


Let x1andx1be any two numbers in R. Thenx1<x25x15<x25x12<5x22fx1<fx2Thus, f is strictly increasing on R.

Q.30 Find the slope of the tangent to the curve y = x3x at x = 2.


The slope of the tangent atx= 2is given bydydxx-2= 3x21x-2=11

Q.31 If x = at, y = at2, then give the equation of tangent at t = 1.


dydx=dydtdxdtdydx=2,att=1x=a=1,y=aequation of tangent at t = 1ya = 2xa2xy=a

Q.32 Find the rate of change of the area of a circle with respect to its radius r when r = 1 cm.


The area of a circle A =



dAdr=2πr=2πr = 1cmgiven

Thus, the area of the circle is changing at the rate of 2


cm2/s when its radius is 1 cm.

Q.33 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?


Volume of cube = a3.

dVdt=3a2dadt=900dadt= 2cm/sgiven

Thus, the volume of cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long.

Q.34 Find the rate of change of volume of a spherical body with respect to radius, when the radius is 10 cm.


As valume of sphere V = 43πr3dVdr=4πr2=4π102=400πHence, the valume of the spherical body is increasing at the rate of 400π cm3/s

Q.35 Show that the function f given by f(x) = x3 – 3x2 + 4x, x ∈ R is strictly increasing on R.


f‘(x) = 3x2 – 6x + 4
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0 in every inetrval of R

Therefore, the function f is strictly increasing on R.


Show that the volume of the greatest cylinder that can be inscribedin a cone of heighthand semvertical angleα is 427πh3tan2α.


Let x and y be the radius and height of the cylinder which is inscribed in the cone of give heighthLet r and α be the radius and semivertical angle of the cone.Then from the figure, we havePO = h, OA = r, LM = ON = x, LO = y, PL = POLO = hy.From similar triangles PLM and POA, we have PLPO=LMOAhyh=xr  x=r)hyh      .…….IIf V denotes the volume of the cylinder, then      V = πx2y       .………II   V = πr2hy2h2y   V = πr2h2h2+y22hyy   V = πr2h2h2y+y32hy2Therefore,    dVdy=πr2h2h2+3y22hyand d2Vdy2=πr2h26y4hPut  dVdy=0,we get     πr2h2h2+3y24hy=0        h2+3y24hy=0        h23hyhy+3y2=0                 h3yhy=0Either h-3y = 0 or hy = 0 Either h = 3y or h = yor             y = h3              .……IIINow,     d2Vdy2=πr2h2613h4h                     =πr2h22h==2πr2h<0.Thus, the volume is maximum when y = 13h.In triangle POA, we have           tanα= rh             r = htanα         .……IVWhen y = 13h,then    x = rh13hh=23r                            x=23htanα    .……VFrom II,III and V,we obtainVolume of the greatest cylinder                             =π23htanα213h                             =427πh3tan2α.


A car start from a point M at time t = 0 second and stopsat point N. The distance x, in metres, covered by it, in secondsis given by x = t22t3. Find the time taken by it to reach Nand also find distance between M and N.


We have,   x = t22t3x=2t2t33dxdt= 4tt2This gives velocity of the car at any time t.Suppose the car stops at N after time t1. At t = t1,dxdt=0dxdttt1=04t1t12=0t14t1=0     t1=4t1=0 is for point MThus, the car takes 4 seconds to reach at N.The distance between M and N is the value of x at t = t1i.e. at t = 4.MN = Value of x at t = 4=2×42433=32643=32m.


Find points onthe curve x29y216=1at which the tangentsare parallel to the ixaxis   iiyaxis


Let Px1,y1 be a point on the curve x29y216=1.Then,  x29y216=1The equation of the curve is  x29y216=1Differentiating both sides with respect to x, we get2x92y16dydx=0  dydx=16x9ydydxx1,y1=16x19y1iIf the tangent at Px1,y1is parallel to xaxis, thendydxx1,y1=016x19y1=016x1=0x1=0.Putting x1 = 0 in i, we get y12=16, Which is impossible as y1is real.Hence, there is no point on the curve where tangent is parallel to xaxis.iiIf the tangent at Px1,y1is parallel to yaxis, thendydxx1,y1=09y116x1=0y1=0.Putting y1=0in i,we get x1=±3.Hence, required points are 3,0and 3,0.


Find the local maximum or local minimum, if any, of the functionfx=sin4x+cos4x,0<x<n2. Using the first derivative test.


We have , y = fx=sin4+cos4xdydx=4sin3xcosx4cos3xsinxdydx=4sinxcosxcos2xsin2xdydx=2sin2xcos2x= sin4x For a local maximum or a local minimum, we have   dydx=0sin4x=0    sin4x=04x = π 0<x<π20<4x<2πx = π4 In the left n bd of x = π4, we have x<π44x<πsin4x>0sin4x<0dydx<0In the right nbd of x = π4,  we havex > π44x>πsin4x<04sin4x>0dydx>0Thus, dydx changes sign from negative to positives as x increases through π4.So, x = π4  is a point of local minimum.The local maximum value of fx at x isfπ4=sinπ44+cosπ44=12.

Q.40 Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.


Let ABC be a triangle inscribed in a given circle with centre Oand radius r. The area of the triangle will be maximum if itsvertex A opposite to the base BC ia at a maximum distancefrom the base BC. This is possible only when A lies on the diameter perpendicular to BC. Thus AD BC.So, triangle ABC must be an isosceles triangle.Let OD = x.In right triangle ODB, we haveOB2=OD2+BD2r2=x2+BD2BD=r2x2 BC = 2BD = 2 r2x2. AD = AO +OD = r +x.Let A denote the area of ΔABC. Then,    A = 12BC×ADA = 12×2r2x2×r+xA = r+xr2x2dAdx= r2x2xr+xr2x2dAdx= r2rx2x2r2x2For maximum or minimum values of A, we must have     dAdx=0r2rx2x2r2x2=0r2xr+x=0            r2x = 0          x=r2  r+x0Now,  dAdx=r2rx2x2r2x2     d2Adx2=r4xr2x2+r2rx2x2xr2x232    d2Adx2x=r2Thus, A is maximum when x = r2.BD=r2x2BD = 3r2 In Δ ODB, we havetanθ = BDODtanθ=3r2r2=3θ=60οBAC = θ = 60οBut, AB = AC. Therefore, B = CThus, we have A =B = C = 60οHence, A is maximum when Δ ABC is equilateral.


Water is running into a conical vessel, 15 cm deep and 5cm inradius, at the rate of 0.1cm3/sec.When the water is 6 cm deep,find at what rate isithe water level rising?iithe watersurface area increasing?iii the wetted surface of the vessel increasing?


Let v be the volume of the water in the cone i.e. the volume of the watercone VABat any time t. Let VO‘ = h, OA‘ = r and VA‘ = l. Let αbe the semivertical angle of the cone, Then,   A = πr2A=πh29     3r=hdAdt=2πh9dhdtWhen h = 6, dhdt=140n We have         dAdt=2π×69×140π=130cm2/sec.Thus, the watersurface area is increasing at the rate of 130cm2/sec.iiiLet S be the wetted surface area of the vessel at any time t.     Then S = πrl.   From figure, we have l2=VA2=VO2+OA2l2=h2+r2l2=h2+h29       3r=h      l=10h3   S=πrlS=πh310h3S=π910h2dsdt=2π10h9dhdtSince,h = 6, dhdt=140n.Therefore,dsdt=2π10h9×6×140π         =10h30cm2/sec. Thus. the wetted surface area of the vessel is increasingat the rate of 1030cm2/sec.


Find the intervals in which fx=x-13x-22is increasing or decreasing.


fx=3x12x22+2x13x2fx=x12x23x6+2x2           =x12x25x8For fx to be increasing, we must have fx>0.x12x25x8>0x2fx-8>0and x 1     x12>0 for all x 15x85x2>0and x 1x85x2>0and x 15>0x<85orx>2 and x 1x, 11,85orx2, So, fx is increasing on , 11,852,.

For fxto bedecreasing, we must have     fx<0.x12x2fx-8<0x25x8<0 and x 1 x12>0for all x 15x85x2<0 and x 1x85x2<0 and x 1   5>0x 585,2and x 1x 585,2 So, fx is decreasing on 85,2


The function f given by f( x ) = x 3 6x 2 + 14x, xR is ________on R.


The given function, f(x)=x36x2+14xDifferentiating w.r.t. x, we getf(x)=3x212x+14=3x212x+12+2=3(x24x+4)+2=3(x2)2+2>0 in every interval of R.Therefore, the function f is strictly increasing¯ on R.

Q.44 If x = 4at and y = 3at2, then the equation of tangent at t = 2 is _______________.


Given, x=4at and y=3at2Differentiating w.r.t. x, we getdxdt=4a and dydt=6atdydx=(dydt)(dxdt)      =6at4a      =3t2(dydx)t=2=3(2)2      =3At t=2,          x=4a(2)and y=3a(2)2      =8a    =12aThe equation of tangent is,  yy1=(dydx)t=2(xx1)y12a=3(x8a)y12a=3x24a3xy=12aTherefore, the required equation of tangent is 3xy=12a¯.

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