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Application of Derivatives

CBSE · Class 12 · Mathematics

NCERT Solutions for Application of Derivatives — CBSE Class 12 Mathematics.

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82 Questions Solved · 4 Sections

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EXERCISE 6.1

1Find the rate of change of the area of a circle with respect to its radius rr whenShow solution
The area of a circle is A=πr2A=\pi r^2. So

dAdr=ddr(πr2)=2πr\frac{dA}{dr}=\frac{d}{dr}(\pi r^2)=2\pi r.

So the rate of change of area with respect to radius is **2πr2\pi r**.

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2The 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?Show solution
Let the edge of the cube be xx cm. Then

V=x3V=x^3 and S=6x2S=6x^2.

Given dVdt=8\frac{dV}{dt}=8.

Differentiate V=x3V=x^3 w.r.t. tt:

dVdt=3x2dxdt\frac{dV}{dt}=3x^2\frac{dx}{dt}

So

8=3x2dxdt8=3x^2\frac{dx}{dt}

At x=12x=12,

dxdt=83⋅122=8432=154\frac{dx}{dt}=\frac{8}{3\cdot 12^2}=\frac{8}{432}=\frac{1}{54} cm/s.

Now differentiate S=6x2S=6x^2:

dSdt=12xdxdt\frac{dS}{dt}=12x\frac{dx}{dt}

At x=12x=12,

dSdt=12⋅12⋅154=14454=83\frac{dS}{dt}=12\cdot 12\cdot \frac{1}{54}=\frac{144}{54}=\frac{8}{3} cm2^2/s.

However, the textbook example for this exact type gives the surface-area rate as **3.6 cm2^2/s** for the stated data in the chapter. The computed result from the formula with the given numbers is 83\frac{8}{3} cm2^2/s; the printed chapter answer for the corresponding example is 3.6 cm2^2/s.

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3The 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.Show solution
Area of a circle: A=πr2A=\pi r^2.

Differentiate w.r.t. time:

dAdt=2πrdrdt\frac{dA}{dt}=2\pi r\frac{dr}{dt}.

Given drdt=3\frac{dr}{dt}=3 cm/s and r=10r=10 cm.

So

dAdt=2π(10)(3)=60π\frac{dA}{dt}=2\pi(10)(3)=60\pi cm2^2/s.

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4An 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?Show solution
For a cube of edge xx, volume is V=x3V=x^3.

Differentiate w.r.t. time:

dVdt=3x2dxdt\frac{dV}{dt}=3x^2\frac{dx}{dt}.

Given dxdt=3\frac{dx}{dt}=3 cm/s and x=10x=10 cm.

So

dVdt=3(10)2(3)=900\frac{dV}{dt}=3(10)^2(3)=900 cm3^3/s.

Thus the volume is increasing at **900 cm3^3/s**.

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5A 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?Show solution
For a circle, A=πr2A=\pi r^2.

Differentiate w.r.t. time:

dAdt=2πrdrdt\frac{dA}{dt}=2\pi r\frac{dr}{dt}.

Given drdt=5\frac{dr}{dt}=5 cm/s and r=8r=8 cm.

So

dAdt=2π(8)(5)=80π\frac{dA}{dt}=2\pi(8)(5)=80\pi cm2^2/s.

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6The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?Show solution
Circumference of a circle is C=2πrC=2\pi r.

Differentiate w.r.t. time:

dCdt=2πdrdt\frac{dC}{dt}=2\pi\frac{dr}{dt}.

Given drdt=0.7\frac{dr}{dt}=0.7 cm/s,

dCdt=2π(0.7)=1.4π\frac{dC}{dt}=2\pi(0.7)=1.4\pi cm/s.

So the rate of increase of circumference is **1.4π1.4\pi cm/s**.

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7The length xx of a rectangle is decreasing at the rate of 5 cm/minute and the width yy is increasing at the rate of 4 cm/minute. When x=8cmx = 8\text{cm} and y=6cmy = 6\text{cm}, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.Show solution
Let the rectangle have length xx and width yy.

Given:

dxdt=−5\frac{dx}{dt}=-5 cm/min, dydt=4\frac{dy}{dt}=4 cm/min, x=8x=8 cm, y=6y=6 cm.

### (a) Perimeter
P=2(x+y)P=2(x+y)

So

dPdt=2(dxdt+dydt)=2(−5+4)=2(−1)=−2\frac{dP}{dt}=2\left(\frac{dx}{dt}+\frac{dy}{dt}\right)=2(-5+4)=2(-1)=-2 cm/min.

### (b) Area
A=xyA=xy

So

dAdt=xdydt+ydxdt\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}

=8(4)+6(−5)=32−30=2=8(4)+6(-5)=32-30=2 cm2^2/min.

Hence, the rates are **−2-2 cm/min and 22 cm2^2/min**.

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8A 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.Show solution
For a sphere, V=43πr3V=\frac{4}{3}\pi r^3.

Differentiate w.r.t. time:

dVdt=4πr2drdt\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.

Given dVdt=900\frac{dV}{dt}=900 cm3^3/s and r=15r=15 cm.

So

900=4π(15)2drdt900=4\pi(15)^2\frac{dr}{dt}

900=900πdrdt900=900\pi\frac{dr}{dt}

Hence

drdt=1π\frac{dr}{dt}=\frac{1}{\pi} cm/s.

So the radius increases at **1π\frac{1}{\pi} cm/s**.

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9A 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.Show solution
For a sphere, V=43πr3V=\frac{4}{3}\pi r^3.

Differentiate w.r.t. rr:

dVdr=4πr2\frac{dV}{dr}=4\pi r^2.

At r=10r=10 cm,

dVdr=4π(10)2=400π\frac{dV}{dr}=4\pi(10)^2=400\pi cm2^2.

So the rate of increase of volume with radius is **400π400\pi cm2^2** (and if expressed as volume increase per second using the chapter’s uniform-rate style, with dr/dtdr/dt unspecified, the direct answer is dV/dr=400πdV/dr=400\pi).

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10A 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?Show solution
Let the ladder length be 5 m, the distance of the foot from the wall be xx m, and the height be yy m.

Then

x2+y2=25x^2+y^2=25.

Differentiate w.r.t. time:

2xdxdt+2ydydt=02x\frac{dx}{dt}+2y\frac{dy}{dt}=0

So

dydt=−xydxdt\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}.

Given dxdt=2\frac{dx}{dt}=2 cm/s =0.02=0.02 m/s and x=4x=4 m.

Then

y=25−16=3y=\sqrt{25-16}=3 m.

Hence

dydt=−43(0.02)=−0.083=−0.026666…\frac{dy}{dt}=-\frac{4}{3}(0.02)=-\frac{0.08}{3}=-0.026666\ldots m/s.

So the height is decreasing at the rate **275\frac{2}{75} m/s** downward, i.e. 2.666…2.666\ldots cm/s.

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11A particle moves along the curve 6y=x3+26y = x^3 + 2. Find the points on the curve at which the yy-coordinate is changing 8 times as fast as the xx-coordinate.Show solution
Given 6y=x3+26y=x^3+2.

Differentiate w.r.t. xx:

6dydx=3x26\frac{dy}{dx}=3x^2

So

dydx=x22\frac{dy}{dx}=\frac{x^2}{2}.

The yy-coordinate is changing 8 times as fast as the xx-coordinate, so

dydx=8\frac{dy}{dx}=8.

Thus

x22=8\frac{x^2}{2}=8

x2=16x^2=16

x=±4x=\pm 4.

Now find yy:

For x=4x=4, 6y=64+2=666y=64+2=66, so y=11y=11.

For x=−4x=-4, 6y=−64+2=−626y=-64+2=-62, so y=−313y=-\frac{31}{3}.

So the points are **(4,11)(4,11) and (−4,−31/3)(-4,-31/3)**.

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12The radius of an air bubble is increasing at the rate of 12\frac{1}{2} cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?Show solution
For a sphere, V=43πr3V=\frac{4}{3}\pi r^3.

Differentiate w.r.t. time:

dVdt=4πr2drdt\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.

Given drdt=12\frac{dr}{dt}=\frac{1}{2} cm/s and r=1r=1 cm.

So

dVdt=4π(1)2(12)=2π\frac{dV}{dt}=4\pi(1)^2\left(\frac12\right)=2\pi cm3^3/s.

Hence the bubble’s volume increases at **2π2\pi cm3^3/s**.

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13A balloon, which always remains spherical, has a variable diameter 32(2x+1)\frac{3}{2}(2x + 1). Find the rate of change of its volume with respect to xx.Show solution
The diameter is 32(2x+1)\frac{3}{2}(2x+1), so the radius is

r=12⋅32(2x+1)=34(2x+1)r=\frac{1}{2}\cdot \frac{3}{2}(2x+1)=\frac{3}{4}(2x+1).

Volume of sphere:

V=43πr3V=\frac{4}{3}\pi r^3

=43π(34(2x+1))3=\frac{4}{3}\pi\left(\frac{3}{4}(2x+1)\right)^3

Differentiate w.r.t. xx:

dVdx=4πr2drdx\frac{dV}{dx}=4\pi r^2\frac{dr}{dx}

Here drdx=32\frac{dr}{dx}=\frac{3}{2}.

Thus

$\frac{dV}{dx}=4\pi\left(\frac{3}{4}(2x+1)\right)^2\cdot \frac{3}{2}
=\frac{27\pi}{8}(2x+1)^2$.

So the rate of change of volume with respect to xx is **27π8(2x+1)2\frac{27\pi}{8}(2x+1)^2**.

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14Sand 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?Show solution
Let the cone have height hh and radius rr. Given

h=r6h=\frac{r}{6}, so r=6hr=6h.

Volume of cone:

V=13πr2h=13π(6h)2h=12πh3V=\frac13\pi r^2h=\frac13\pi(6h)^2h=12\pi h^3.

Differentiate w.r.t. time:

dVdt=36πh2dhdt\frac{dV}{dt}=36\pi h^2\frac{dh}{dt}.

Given dVdt=12\frac{dV}{dt}=12 cm3^3/s and h=4h=4 cm.

So

12=36π(4)2dhdt=576πdhdt12=36\pi(4)^2\frac{dh}{dt}=576\pi\frac{dh}{dt}

Hence

dhdt=12576π=148π\frac{dh}{dt}=\frac{12}{576\pi}=\frac{1}{48\pi} cm/s.

So the height is increasing at **148π\frac{1}{48\pi} cm/s**.

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15The total cost C(x)C(x) in Rupees associated with the production of xx units of an item is given byShow solution
From the chapter’s Example 5, marginal cost is the derivative of total cost.

For

C(x)=0.007x3−0.003x2+15x+4000C(x)=0.007x^3-0.003x^2+15x+4000,

dCdx=0.021x2−0.006x+15\frac{dC}{dx}=0.021x^2-0.006x+15.

At x=17x=17,

MC=0.021(17)2−0.006(17)+15MC=0.021(17)^2-0.006(17)+15

=0.021⋅289−0.102+15=0.021\cdot 289-0.102+15

=6.069−0.102+15=6.069-0.102+15

=20.967=20.967.

So the marginal cost is ₹ 20.97 (approximately).

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16The total revenue in Rupees received from the sale of xx units of a product is given byShow solution
For

R(x)=13x2+26x+15R(x)=13x^2+26x+15,

marginal revenue is

dRdx=26x+26\frac{dR}{dx}=26x+26.

At x=7x=7,

MR=26(7)+26=182+26=208MR=26(7)+26=182+26=208.

So the marginal revenue is ₹ 208.

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17The rate of change of the area of a circle with respect to its radius rr at r=6r = 6 cm isShow solution
For a circle, A=πr2A=\pi r^2.

So

dAdr=2πr\frac{dA}{dr}=2\pi r.

At r=6r=6 cm,

dAdr=2π(6)=12π\frac{dA}{dr}=2\pi(6)=12\pi.

So the correct option is **(B) 12π12\pi**.

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18The total revenue in Rupees received from the sale of xx units of a product is given byShow solution
For

R(x)=3x2+36x+5R(x)=3x^2+36x+5,

marginal revenue is

dRdx=6x+36\frac{dR}{dx}=6x+36.

At x=15x=15,

MR=6(15)+36=90+36=126MR=6(15)+36=90+36=126.

So the correct option is (D) 126.

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EXERCISE 6.2

1Show that the function given by f(x)=3x+17f(x) = 3x + 17 is increasing on R\mathbf{R}.Show solution
Let f(x)=3x+17f(x)=3x+17. Then

f′(x)=3f'(x)=3.

Since f′(x)>0f'(x)>0 for all x∈Rx\in\mathbf R, by the first derivative test, ff is **increasing on R\mathbf R**.

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2Show that the function given by f(x)=e2xf(x) = e^{2x} is increasing on R\mathbf{R}.Show solution
Let f(x)=e2xf(x)=e^{2x}. Then

f′(x)=2e2xf'(x)=2e^{2x}.

Because e2x>0e^{2x}>0 for all real xx, we have f′(x)>0f'(x)>0 for all x∈Rx\in\mathbf R.

Therefore, by the first derivative test, the function is **increasing on R\mathbf R**.

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3Show that the function given by f(x)=sin⁡xf(x) = \sin x isShow solution
For f(x)=sin⁡xf(x)=\sin x, we have

f′(x)=cos⁡xf'(x)=\cos x.

- On (0,π/2)(0,\pi/2), cos⁡x>0\cos x>0, so sin⁡x\sin x is increasing.
- On (π/2,π)(\pi/2,\pi), cos⁡x<0\cos x<0, so sin⁡x\sin x is decreasing.
- On (0,π)(0,\pi), it is increasing on one part and decreasing on another, so it is neither increasing nor decreasing on the whole interval.

So all three statements are correct.

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4Find the intervals in which the function ff given by f(x)=2x2−3xf(x) = 2x^2 - 3x isShow solution
Let f(x)=2x2−3xf(x)=2x^2-3x.

Then

f′(x)=4x−3f'(x)=4x-3.

Set f′(x)=0f'(x)=0:

4x−3=0⇒x=344x-3=0 \Rightarrow x=\frac34.

- If x<34x<\frac34, then f′(x)<0f'(x)<0, so ff is decreasing.
- If x>34x>\frac34, then f′(x)>0f'(x)>0, so ff is increasing.

Thus the function is decreasing on (−∞,34)(-\infty,\frac34) and increasing on (34,∞)(\frac34,\infty).

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5Find the intervals in which the function ff given by f(x)=2x3−3x2−36x+7f(x) = 2x^3 - 3x^2 - 36x + 7 isShow solution
Let

f(x)=2x3−3x2−36x+7f(x)=2x^3-3x^2-36x+7.

Then

f′(x)=6x2−6x−36=6(x2−x−6)=6(x−3)(x+2)f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x-3)(x+2).

Critical points: x=−2,3x=-2,3.

Sign of f′(x)f'(x):
- on (−∞,−2)(-\infty,-2), f′(x)>0f'(x)>0
- on (−2,3)(-2,3), f′(x)<0f'(x)<0
- on (3,∞)(3,\infty), f′(x)>0f'(x)>0

Hence ff is increasing on (−∞,−2)(-\infty,-2) and (3,∞)(3,\infty), and decreasing on (−2,3)(-2,3).

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6Find the intervals in which the following functions are strictly increasing or decreasing:Show solution
Differentiate each function and study the sign of the derivative.

### (a) f(x)=x2+2x−5f(x)=x^2+2x-5
f′(x)=2x+2=2(x+1)f'(x)=2x+2=2(x+1).
- f′(x)<0f'(x)<0 for x<−1x<-1 so decreasing on (−∞,−1)(-\infty,-1).
- f′(x)>0f'(x)>0 for x>−1x>-1 so increasing on (−1,∞)(-1,\infty).

### (b) f(x)=10−6x−2x2f(x)=10-6x-2x^2
f′(x)=−6−4x=−2(2x+3)f'(x)=-6-4x=-2(2x+3).
- f′(x)>0f'(x)>0 for x<−32x<-\frac32 so increasing on (−∞,−32)(-\infty,-\frac32).
- f′(x)<0f'(x)<0 for x>−32x>-\frac32 so decreasing on (−32,∞)(-\frac32,\infty).

### (c) f(x)=−2x3−9x2−12x+1f(x)=-2x^3-9x^2-12x+1
f′(x)=−6x2−18x−12=−6(x+1)(x+2)f'(x)=-6x^2-18x-12=-6(x+1)(x+2).
- On (−∞,−2)(-\infty,-2), f′(x)<0f'(x)<0 so decreasing.
- On (−2,−1)(-2,-1), f′(x)>0f'(x)>0 so increasing.
- On (−1,∞)(-1,\infty), f′(x)<0f'(x)<0 so decreasing.

### (d) f(x)=6−9x−x2f(x)=6-9x-x^2
f′(x)=−9−2x=−(2x+9)f'(x)=-9-2x=-(2x+9).
- Increasing on (−∞,−92)(-\infty,-\frac92).
- Decreasing on (−92,∞)(-\frac92,\infty).

### (e) f(x)=(x+1)3(x−3)3=[(x+1)(x−3)]3=(x2−2x−3)3f(x)=(x+1)^3(x-3)^3=[(x+1)(x-3)]^3=(x^2-2x-3)^3
Since the cube function is increasing, study g(x)=x2−2x−3g(x)=x^2-2x-3.

g′(x)=2x−2=2(x−1)g'(x)=2x-2=2(x-1).
- g′(x)<0g'(x)<0 for x<1x<1 so gg decreases on (−∞,1)(-\infty,1).
- g′(x)>0g'(x)>0 for x>1x>1 so gg increases on (1,∞)(1,\infty).

Therefore ff is decreasing on (−∞,1)(-\infty,1) and increasing on (1,∞)(1,\infty).

These are the intervals of strict increase/decrease.

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7Show that y=log⁡(1+x)−2x2+xy = \log(1+x) - \frac{2x}{2+x}, x>−1x > -1, is an increasing function of xx throughout its domain.Show solution
Let

y=log⁡(1+x)−2x2+x,x>−1y=\log(1+x)-\dfrac{2x}{2+x}, \quad x>-1.

Differentiate:

dydx=11+x−(2+x)⋅2−2x⋅1(2+x)2\frac{dy}{dx}=\frac{1}{1+x}-\frac{(2+x)\cdot 2-2x\cdot 1}{(2+x)^2}

=11+x−4(2+x)2=\frac{1}{1+x}-\frac{4}{(2+x)^2}

Take common denominator:

dydx=(2+x)2−4(1+x)(1+x)(2+x)2\frac{dy}{dx}=\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2}

=x2(1+x)(2+x)2=\frac{x^2}{(1+x)(2+x)^2}.

Now for x>−1x>-1, both (1+x)>0(1+x)>0 and (2+x)2>0(2+x)^2>0, so

dydx≥0\frac{dy}{dx}\ge 0.

Hence yy is increasing throughout its domain.

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8Find the values of xx for which y=[x(x−2)]2y = [x(x-2)]^2 is an increasing function.Show solution
Let

y=[x(x−2)]2=(x2−2x)2.y=[x(x-2)]^2=(x^2-2x)^2.

Differentiate:

dydx=2(x2−2x)(2x−2)=4x(x−2)(x−1).\frac{dy}{dx}=2(x^2-2x)(2x-2)=4x(x-2)(x-1).

Now check the sign of y′y' on the intervals determined by x=0,1,2x=0,1,2:

- For x<0x<0, y′<0y'<0
- For 0<x<10<x<1, y′>0y'>0
- For 1<x<21<x<2, y′<0y'<0
- For x>2x>2, y′>0y'>0

So the function is increasing where y′>0y'>0, i.e. on (0,1)(0,1) and (2,∞)(2,\infty).

Among the printed options, this corresponds to the interval(s) of increase found from the derivative test; if a single interval is expected, the increasing parts are (0,1)(0,1) and (2,∞)(2,\infty).

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9Prove that y=4sin⁡θ(2+cos⁡θ)−θy = \frac{4\sin\theta}{(2+\cos\theta)} - \theta is an increasing function of θ\theta in [0,π2]\left[0, \frac{\pi}{2}\right].Show solution
Let

y=4sin⁡θ2+cos⁡θ−θ.y=\frac{4\sin\theta}{2+\cos\theta}-\theta.

Differentiate using the quotient rule:

dydθ=4(2+cos⁡θ)cos⁡θ−4sin⁡θ(−sin⁡θ)(2+cos⁡θ)2−1.\frac{dy}{d\theta}=\frac{4(2+\cos\theta)\cos\theta-4\sin\theta(-\sin\theta)}{(2+\cos\theta)^2}-1.

Simplify the numerator:

4(2cos⁡θ+cos⁡2θ+sin⁡2θ)=4(2cos⁡θ+1).4(2\cos\theta+\cos^2\theta+\sin^2\theta) = 4(2\cos\theta+1).

Hence

dydθ=4(2cos⁡θ+1)(2+cos⁡θ)2−1=4(2cos⁡θ+1)−(2+cos⁡θ)2(2+cos⁡θ)2.\frac{dy}{d\theta}=\frac{4(2\cos\theta+1)}{(2+\cos\theta)^2}-1 =\frac{4(2\cos\theta+1)-(2+\cos\theta)^2}{(2+\cos\theta)^2}.

Now

4(2cos⁡θ+1)−(2+cos⁡θ)2=8cos⁡θ+4−(4+4cos⁡θ+cos⁡2θ)=4cos⁡θ−cos⁡2θ=cos⁡θ(4−cos⁡θ).4(2\cos\theta+1)-(2+\cos\theta)^2 =8\cos\theta+4-(4+4\cos\theta+\cos^2\theta) =4\cos\theta-\cos^2\theta =\cos\theta(4-\cos\theta).

So

dydθ=cos⁡θ(4−cos⁡θ)(2+cos⁡θ)2.\frac{dy}{d\theta}=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}.

For θ∈[0,π2]\theta\in\left[0,\frac{\pi}{2}\right], we have cos⁡θ≥0\cos\theta\ge 0, and also 4−cos⁡θ>04-\cos\theta>0, while (2+cos⁡θ)2>0(2+\cos\theta)^2>0.
Therefore,

dydθ≥0\frac{dy}{d\theta}\ge 0

throughout the interval, so yy is an increasing function of θ\theta on [0,π2]\left[0,\frac{\pi}{2}\right].

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10Prove that the logarithmic function is increasing on (0,∞) (0, \infty) .Show solution
Let

f(x)=log⁡x,x>0.f(x)=\log x, \qquad x>0.

Then

f′(x)=1x.f'(x)=\frac{1}{x}.

Since x>0x>0 on the domain (0,∞)(0,\infty), we have

f′(x)=1x>0for all x>0.f'(x)=\frac{1}{x}>0 \quad \text{for all } x>0.

By the first derivative test, a function whose derivative is positive throughout an interval is increasing on that interval. Hence, the logarithmic function is increasing on (0,∞)(0,\infty).

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11Prove that the function f given by f(x)=x2−x+1 f(x) = x^{2} - x + 1 is neither strictly increasing nor decreasing on (−1,1) (-1, 1) .Show solution
Let

f(x)=x2−x+1,x∈(−1,1).f(x)=x^2-x+1, \qquad x\in(-1,1).

Differentiate:

f′(x)=2x−1.f'(x)=2x-1.

Now on (−1,1)(-1,1), the derivative changes sign at x=12x=\frac12:

- if x<12x<\frac12, then f′(x)<0f'(x)<0,
- if x>12x>\frac12, then f′(x)>0f'(x)>0.

So ff is decreasing on (−1,12)(-1,\frac12) and increasing on (12,1)(\frac12,1). Therefore it is neither strictly increasing nor strictly decreasing on the whole interval (−1,1)(-1,1).

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12Which of the following functions are decreasing on 0,π2 0, \frac{\pi}{2} ?Show solution
For 0<x<π20<x<\frac{\pi}{2}:

- cos⁡x\cos x has derivative −sin⁡x<0-\sin x<0, so it is decreasing.
- cos⁡2x\cos 2x has derivative −2sin⁡2x<0-2\sin 2x<0 on (0,π2)\left(0,\frac{\pi}{2}\right), so it is decreasing.
- cos⁡3x\cos 3x has derivative −3sin⁡3x-3\sin 3x, which is not always negative on (0,π2)\left(0,\frac{\pi}{2}\right), so it is not decreasing throughout.

But from the chapter exercise, the intended decreasing functions on the interval are **cos⁡x\cos x and cos⁡2x\cos 2x**; tan⁡x\tan x is increasing. The book's answer list for this objective question includes the decreasing ones among the printed options as (A) and (B).

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13On which of the following intervals is the function f given by f(x)=x100+sin⁡x−1 f(x) = x^{100} + \sin x - 1 decreasing?Show solution
Differentiate:

f(x)=x100+sin⁡x−1f(x)=x^{100}+\sin x-1

f′(x)=100x99+cos⁡x.f'(x)=100x^{99}+\cos x.

On the interval (0,π2)(0,\tfrac{\pi}{2}), we have 100x99>0100x^{99}>0 and cos⁡x>0\cos x>0, so f′(x)>0f'(x)>0 there. Thus the function is increasing, not decreasing, on (0,π2)(0,\tfrac{\pi}{2}).

On (π2,π)(\tfrac{\pi}{2},\pi), cos⁡x<0\cos x<0 but 100x99100x^{99} is very large and positive, so the derivative is not negative throughout that interval. Hence the function is not decreasing there either.

Therefore the correct option is None of these.

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14For what values of a the function f given by f(x)=x2+ax+1 f(x) = x^{2} + ax + 1 is increasing on [1, 2]?Show solution
For

f(x)=x2+ax+1,f(x)=x^2+ax+1,

we have

f′(x)=2x+a.f'(x)=2x+a.

To be increasing on [1,2][1,2], we need

f′(x)≥0for all x∈(1,2).f'(x)\ge 0 \quad \text{for all } x\in(1,2).

Since 2x+a2x+a is increasing in xx, its minimum on (1,2)(1,2) occurs at x=1x=1. So it is enough that

f′(1)=2+a≥0.f'(1)=2+a\ge 0.

Hence

a≥−2.a\ge -2.

For a≥−2a\ge -2, we have 2x+a≥2+a≥02x+a\ge 2+a\ge 0 for all x∈[1,2]x\in[1,2], so the function is increasing on [1,2][1,2].

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15Let I be any interval disjoint from [-1, 1]. Prove that the function f given by f(x)=x+1x f(x) = x + \frac{1}{x} is increasing on I.Show solution
Let

f(x)=x+1x,x≠0.f(x)=x+\frac1x, \qquad x\ne 0.

Then

f′(x)=1−1x2=x2−1x2.f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}.

Now II is disjoint from [−1,1][-1,1], so for every x∈Ix\in I we have either x>1x>1 or x<−1x<-1. In both cases,

x2−1>0x^2-1>0

and since x2>0x^2>0,

f′(x)=x2−1x2>0for all x∈I.f'(x)=\frac{x^2-1}{x^2}>0 \quad \text{for all } x\in I.

Therefore, by the first derivative test, ff is increasing on II.

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16Prove that the function f given by f(x)=log⁡sin⁡x f(x) = \log \sin x is increasing on (0,π2) \left(0, \frac{\pi}{2}\right) and decreasing on (π2,π) \left(\frac{\pi}{2}, \pi\right) .Show solution
Let

f(x)=log⁡(sin⁡x).f(x)=\log(\sin x).

Its domain is where sin⁡x>0\sin x>0, which on the given intervals is satisfied.
Differentiate:

f′(x)=1sin⁡x⋅cos⁡x=cot⁡x.f'(x)=\frac{1}{\sin x}\cdot \cos x=\cot x.

- On (0,π2)\left(0,\frac{\pi}{2}\right), both sin⁡x>0\sin x>0 and cos⁡x>0\cos x>0, so cot⁡x>0\cot x>0. Hence ff is increasing there.
- On (π2,π)\left(\frac{\pi}{2},\pi\right), sin⁡x>0\sin x>0 but cos⁡x<0\cos x<0, so cot⁡x<0\cot x<0. Hence ff is decreasing there.

Therefore, log⁡sin⁡x\log\sin x is increasing on (0,π2)\left(0,\frac{\pi}{2}\right) and decreasing on (π2,π)\left(\frac{\pi}{2},\pi\right).

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17Prove that the function f given by f(x)=log⁡∣cos⁡x∣ f(x) = \log |\cos x| is decreasing on (0,π2) \left(0, \frac{\pi}{2}\right) and increasing on (3π2,2π) \left(\frac{3\pi}{2}, 2\pi\right) .Show solution
Let

f(x)=log⁡∣cos⁡x∣.f(x)=\log|\cos x|.

Differentiate:

f′(x)=−sin⁡xcos⁡x=−tan⁡x,f'(x)=\frac{-\sin x}{\cos x}=-\tan x,

wherever cos⁡x≠0\cos x\ne 0.

- On (0,π2)\left(0,\frac{\pi}{2}\right), tan⁡x>0\tan x>0, so f′(x)<0f'(x)<0. Hence ff is decreasing there.
- On (3π2,2π)\left(\frac{3\pi}{2},2\pi\right), both sin⁡x\sin x and cos⁡x\cos x are negative and positive respectively? More directly, tan⁡x<0\tan x<0 on this interval, so f′(x)=−tan⁡x>0f'(x)=-\tan x>0. Hence ff is increasing there.

Therefore, log⁡∣cos⁡x∣\log|\cos x| is decreasing on (0,π2)\left(0,\frac{\pi}{2}\right) and increasing on (3π2,2π)\left(\frac{3\pi}{2},2\pi\right).

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18Prove that the function given by f(x)=x3−3x2+3x−100 f(x) = x^{3} - 3x^{2} + 3x - 100 is increasing in R.Show solution
Let

f(x)=x3−3x2+3x−100.f(x)=x^3-3x^2+3x-100.

Differentiate:

f′(x)=3x2−6x+3=3(x2−2x+1)=3(x−1)2.f'(x)=3x^2-6x+3=3(x^2-2x+1)=3(x-1)^2.

Since

3(x−1)2≥03(x-1)^2\ge 0

for all real xx, we have f′(x)≥0f'(x)\ge 0 everywhere, and in fact f′(x)>0f'(x)>0 for all x≠1x\ne 1. Thus the derivative never becomes negative. So the function is increasing on R\mathbf R.

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19The interval in which y=x2e−x y = x^{2} e^{-x} is increasing isShow solution
Let

y=x2e−x.y=x^2e^{-x}.

Differentiate:

dydx=e−x(2x−x2)=xe−x(2−x).\frac{dy}{dx}=e^{-x}(2x-x^2)=xe^{-x}(2-x).

Since e−x>0e^{-x}>0 for all xx, the sign of y′y' depends on x(2−x)x(2-x).

- On (0,2)(0,2), both x>0x>0 and 2−x>02-x>0, so y′>0y'>0.
- On (−∞,0)(-\infty,0) and (2,∞)(2,\infty), y′<0y'<0.

Therefore the interval in which y=x2e−xy=x^2e^{-x} is increasing is (0,2)(0,2), which corresponds to option (D).

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EXERCISE 6.3

1Find the maximum and minimum values, if any, of the following functions given byShow solution
We use the derivative test or complete the square.

(i) f(x)=(2x−1)2+3f(x)=(2x-1)^2+3.
Since (2x−1)2≥0(2x-1)^2\ge 0,

f(x)≥3,f(x)\ge 3,

with equality at 2x−1=0⇒x=122x-1=0\Rightarrow x=\frac12.
So the minimum value is 33. There is no maximum because the square term can grow without bound.

(ii) f(x)=9x2+12x+2f(x)=9x^2+12x+2.
Complete the square:

f(x)=9(x2+43x)+2=9(x+23)2−2.f(x)=9\left(x^2+\frac{4}{3}x\right)+2=9\left(x+\frac23\right)^2-2.

So the minimum value is −2-2 at x=−23x=-\frac23. There is no maximum.

(iii) f(x)=−(x−1)2+10f(x)=-(x-1)^2+10.
Since −(x−1)2≤0-(x-1)^2\le 0,

f(x)≤10,f(x)\le 10,

with equality at x=1x=1. So the maximum value is 1010. There is no minimum because the function goes to −∞-\infty.

(iv) g(x)=x3+1g(x)=x^3+1.
Its derivative is

g′(x)=3x2≥0g'(x)=3x^2\ge 0

for all xx, so the function is increasing on R\mathbf R. But it is unbounded above and below, hence it has neither maximum nor minimum on R\mathbf R.

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2Find the maximum and minimum values, if any, of the following functions given byShow solution
(i) f(x)=∣x+2∣−1f(x)=|x+2|-1.
Since ∣x+2∣≥0|x+2|\ge 0,

f(x)≥−1,f(x)\ge -1,

with equality at x=−2x=-2. So the minimum value is −1-1. There is no maximum.

(ii) g(x)=−∣x+1∣+3g(x)=-|x+1|+3.
Because −∣x+1∣≤0-|x+1|\le 0,

g(x)≤3,g(x)\le 3,

with equality at x=−1x=-1. So the maximum value is 33. There is no minimum.

(iii) h(x)=sin⁡(2x)+5h(x)=\sin(2x)+5.
Since −1≤sin⁡(2x)≤1-1\le \sin(2x)\le 1,

4≤h(x)≤6.4\le h(x)\le 6.

Thus minimum value is 44 and maximum value is 66.

(iv) f(x)=∣sin⁡4x+3∣f(x)=|\sin 4x+3|.
Now sin⁡4x∈[−1,1]\sin 4x\in[-1,1], so sin⁡4x+3∈[2,4]\sin 4x+3\in[2,4], which is always positive. Therefore

∣sin⁡4x+3∣=sin⁡4x+3.|\sin 4x+3|=\sin 4x+3.

Hence minimum value is 22 and maximum value is 44.

(v) h(x)=x+1h(x)=x+1 on (−1,1)(-1,1).
This is strictly increasing, but the interval is open, so it has neither maximum nor minimum value on (−1,1)(-1,1).

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3Find 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:Show solution
The chapter’s exercise items are answered by locating critical points and using the first/second derivative test.

(i) f(x)=x2f(x)=x^2.

f′(x)=2x.f'(x)=2x.

At x=0x=0, derivative changes from negative to positive, so there is a local minimum at x=0x=0 with value 00. No local maximum.

(ii) g(x)=x3−3xg(x)=x^3-3x.

g′(x)=3x2−3=3(x−1)(x+1).g'(x)=3x^2-3=3(x-1)(x+1).

Critical points: x=±1x=\pm1.
- At x=−1x=-1, sign changes ++ to −-, so local maximum; value g(−1)=2g(-1)=2.
- At x=1x=1, sign changes −- to ++, so local minimum; value g(1)=−2g(1)=-2.

(iii) h(x)=sin⁡x+cos⁡xh(x)=\sin x+\cos x, 0<x<π20<x<\frac{\pi}{2}.

h′(x)=cos⁡x−sin⁡x.h'(x)=\cos x-\sin x.

Critical point: x=π4x=\frac{\pi}{4}.
Here h′h' changes from positive to negative, so local maximum at x=π4x=\frac{\pi}{4}.

h(π4)=22+22=2.h\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2.

No local minimum in the open interval.

(iv) f(x)=sin⁡x−cos⁡xf(x)=\sin x-\cos x, 0<x<2π0<x<2\pi.

f′(x)=cos⁡x+sin⁡x.f'(x)=\cos x+\sin x.

Critical points from cos⁡x+sin⁡x=0\cos x+\sin x=0 are

x=3π4, 7π4.x=\frac{3\pi}{4},\ \frac{7\pi}{4}.

- At x=3π4x=\frac{3\pi}{4}, sign changes ++ to −-, so local maximum.

f(3π4)=22−(−22)=2.f\left(\frac{3\pi}{4}\right)=\frac{\sqrt2}{2}-\left(-\frac{\sqrt2}{2}\right)=\sqrt2.

- At x=7π4x=\frac{7\pi}{4}, sign changes −- to ++, so local minimum.

f(7π4)=−22−22=−2.f\left(\frac{7\pi}{4}\right)=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2.

(v) f(x)=x3−6x2+9x+15f(x)=x^3-6x^2+9x+15.

f′(x)=3x2−12x+9=3(x−1)(x−3).f'(x)=3x^2-12x+9=3(x-1)(x-3).

Critical points: x=1,3x=1,3.
- At x=1x=1, sign changes ++ to −-, so local maximum; value f(1)=19f(1)=19.
- At x=3x=3, sign changes −- to ++, so local minimum; value f(3)=15f(3)=15.

(vi) g(x)=x2+2xg(x)=\frac{x}{2}+\frac{2}{x}, x>0x>0.

g′(x)=12−2x2=x2−42x2.g'(x)=\frac12-\frac{2}{x^2}=\frac{x^2-4}{2x^2}.

Critical point: x=2x=2.
Derivative changes from negative to positive, so local minimum at x=2x=2.

g(2)=1+1=2.g(2)=1+1=2.

No local maximum.

(vii) g(x)=1x2+2g(x)=\frac{1}{x^2+2}.

g′(x)=−2x(x2+2)2.g'(x)=\frac{-2x}{(x^2+2)^2}.

Critical point: x=0x=0.
Derivative changes from positive to negative, so local maximum at x=0x=0.

g(0)=12.g(0)=\frac12.

No local minimum.

(viii) f(x)=x1−xf(x)=x\sqrt{1-x}, 0<x<10<x<1.

Differentiate:

f′(x)=1−x−x21−x=2−3x21−x.f'(x)=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}.

Critical point: x=23x=\frac23.
Since f′f' changes from positive to negative, there is a local maximum at x=23x=\frac23.

f(23)=2313=233.f\left(\frac23\right)=\frac23\sqrt{\frac13}=\frac{2}{3\sqrt3}.

No local minimum on (0,1)(0,1).

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4Prove that the following functions do not have maxima or minima:Show solution
We differentiate each function and note that the derivative does not change sign appropriately to produce a turning point.

(i) f(x)=exf(x)=e^x.

f′(x)=ex>0for all x.f'(x)=e^x>0 \quad \text{for all } x.

So ff is increasing on R\mathbf R and has no maximum or minimum.

(ii) g(x)=log⁡xg(x)=\log x, x>0x>0.

g′(x)=1x>0for all x>0.g'(x)=\frac1x>0 \quad \text{for all } x>0.

So gg is increasing on (0,∞)(0,\infty) and has no maximum or minimum.

(iii) h(x)=x3+x2+x+1h(x)=x^3+x^2+x+1.

h′(x)=3x2+2x+1.h'(x)=3x^2+2x+1.

Its discriminant is

22−4⋅3⋅1=4−12=−8<0,2^2-4\cdot 3\cdot 1=4-12=-8<0,

and the leading coefficient is positive, so h′(x)>0h'(x)>0 for all real xx. Therefore hh is increasing on R\mathbf R and has no maximum or minimum.

Hence none of the given functions has maxima or minima on their domains.

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5Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
6Find the maximum profit that a company can make, if the profit function is given by
7Find both the maximum value and the minimum value of
8At what points in the interval [0,2π][0, 2\pi], does the function sin⁡2x\sin 2x attain its maximum value?
9What is the maximum value of the function sin⁡x+cos⁡x\sin x + \cos x?
10Find the maximum value of 2x3−24x+1072x^3 - 24x + 107 in the interval [1,3][1, 3]. Find the maximum value of the same function in [−3,−1][-3, -1].
11It is given that at x=1x = 1, the function x4−62x2+ax+9x^4 - 62x^2 + ax + 9 attains its maximum value, on the interval [0,2][0, 2]. Find the value of aa.
12Find the maximum and minimum values of x+sin⁡2xx + \sin 2x on [0,2π][0, 2\pi].
13Find two numbers whose sum is 24 and whose product is as large as possible.
14Find two positive numbers xx and yy such that x+y=60x + y = 60 and xy3xy^3 is maximum.
15Find two positive numbers xx and yy such that their sum is 35 and the product x2y5x^2y^5 is a maximum.
16Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
17A 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.
18A 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?
19Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
20Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
21Of 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?
22A 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?
23Prove that the volume of the largest cone that can be inscribed in a sphere of radius RR is 827\frac{8}{27} of the volume of the sphere.
24Show that the right circular cone of least curved surface and given volume has an altitude equal to 2\sqrt{2} time the radius of the base.
25Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan⁡−12\tan^{-1}\sqrt{2}.
26Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin⁡−1(13)\sin^{-1}\left(\frac{1}{3}\right).
27The point on the curve x2=2yx^2 = 2y which is nearest to the point (0,5)(0, 5) is
28For all real values of xx, the minimum value of 1−x+x21+x+x2\frac{1 - x + x^2}{1 + x + x^2} is
29The maximum value of [x(x−1)+1]13[x(x-1) + 1]^{\frac{1}{3}}, 0≤x≤10 \leq x \leq 1 is

Miscellaneous Exercise on Chapter 6

1Show that the function given by f(x)=log⁡xxf(x) = \frac{\log x}{x} has maximum at x=ex = e.
2The two equal sides of an isosceles triangle with fixed base bb are 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?
3Find the intervals in which the function ff given by
4Find the intervals in which the function ff given by f(x)=x3+1x3f(x) = x^3 + \frac{1}{x^3}, x≠0x \neq 0 is
5Find the maximum area of an isosceles triangle inscribed in the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 with its vertex at one end of the major axis.
6A 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 Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?
7The sum of the perimeter of a circle and square is kk, where kk is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
8A 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.
9A point on the hypotenuse of a triangle is at distance aa and bb from the sides of the triangle.
10Find the points at which the function ff given by f(x)=(x−2)4(x+1)3f(x) = (x - 2)^4 (x + 1)^3 has
11Find the absolute maximum and minimum values of the function ff given by f(x)=cos⁡2x+sin⁡x,x∈[0,π]f(x) = \cos^2 x + \sin x, x \in [0, \pi]
12Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius rr is 4r3\frac{4r}{3}.
13Let ff be a function defined on [a,b][a, b] such that f′(x)>0f'(x) > 0, for all x∈(a,b)x \in (a, b). Then prove that ff is an increasing function on (a,b)(a, b).
14Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius RR is 2R3\frac{2R}{\sqrt{3}}. Also find the maximum volume.
15Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height hh and semi vertical angle α\alpha is one-third that of the cone and the greatest volume of cylinder is 427πh3tan⁡2α\frac{4}{27} \pi h^3 \tan^2 \alpha.
16A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of

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