Application of Derivatives 6.3 NCERT Solutions

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

(i) f(x)=(2x-1)^2+3

Solution

Given

f(x)=(2x-1)^2+3

Since a square is always non-negative, we have

(2x-1)^2\geq 0

Adding 3 on both sides, we get

f(x)=(2x-1)^2+3\geq 3

The equality part will hold true when

2x-1=0

i.e. when \displaystyle x=\frac{1}{2}

Hence, the minimum value of the function is 3 at \displaystyle x=\frac{1}{2}

The function has no maximum value.

Verification Using Derivatives

Differentiating with respect to x, we get

f'(x)=4(2x-1)

Putting f'(x)=0, we get

4(2x-1)=0

\Rightarrow \displaystyle x=\frac{1}{2}

Now, f''(x)=8

and \displaystyle f''\left(\frac{1}{2}\right)=8\gt 0

\therefore \displaystyle x=\frac{1}{2} is a point of local minima.

Also, \displaystyle f\left(\frac{1}{2}\right)=3

Hence, the minimum value of the function is 3 at \displaystyle x=\frac{1}{2}, and the function has no maximum value.

(ii) f(x)=9x^2+12x+2.

Solution

Given

f(x)=9x^2+12x+2

Taking 9 common from the first two terms, we get

\displaystyle f(x)=9\left(x^2+\frac{4}{3}x\right)+2

Adding and subtracting the square of half the coefficient of x inside the bracket, we get

\displaystyle f(x)=9\left(x^2+\frac{4}{3}x+\frac{4}{9}-\frac{4}{9}\right)+2

\displaystyle f(x)=9\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)-9\cdot\frac{4}{9}+2

Using a^2+2ab+b^2=(a+b)^2, we get

\displaystyle f(x)=9\left(x+\frac{2}{3}\right)^2-4+2

\displaystyle f(x)=9\left(x+\frac{2}{3}\right)^2-2

Since

\displaystyle 9\left(x+\frac{2}{3}\right)^2\geq 0

Therefore, Subtracting 2 from both sides, we get

\displaystyle f(x)=9\left(x+\frac{2}{3}\right)^2-2\geq -2

Equality holds true when

\displaystyle \left(x+\frac{2}{3}\right)=0

i.e. when \displaystyle x=-\frac{2}{3}

Hence, the minimum value of the function is -2 at \displaystyle x=-\frac{2}{3}.

The function has no maximum value.

Verification Using Derivatives

f'(x)=18x+12

Putting f'(x)=0 we get,

18x+12=0

\Rightarrow \displaystyle x=-\frac{2}{3}

Also f''(x)=18

and \displaystyle f''\left(-\frac{2}{3}\right)=18\gt 0

Therefore, \displaystyle x=-\frac{2}{3} is a point of local minima.

Also,

\displaystyle f\left(-\frac{2}{3}\right)=-2

Hence, the minimum value of the function is -2 at \displaystyle x=-\frac{2}{3}, and the function has no maximum value.

(iii)f(x)=-(x-1)^2+10

Solution

Since

(x-1)^2\geq 0

Therefore,

-(x-1)^2\leq 0

Adding 10 on both sides, we get

f(x)=-(x-1)^2+10\leq 10

Equality holds true when

x-1=0 i.e. when x=1

Hence, the maximum value of the function is 10 at x=1.

The function has no minimum value.

Verification Using Derivatives

f'(x)=-2(x-1)

Putting f'(x)=0, we get

x=1

Also, f''(x)=-2

and \displaystyle f''(1)=-2\lt 0

Therefore, x=1 is a point of local maxima.

Also,

f(1)=10

Hence, the maximum value of the function is 10 at x=1, and the function has no minimum value.

(iv) g(x)=x^3+1

Solution

Differentiating with respect to x, we get

g'(x)=3x^2

Putting g'(x)=0, we get

3x^2=0

\Rightarrow x=0

Again,

g''(x)=6x

At x=0,

g''(0)=0

Hence, the Second Derivative Test fails.

Using the First Derivative Test, we note that

g'(x)=3x^2\geq 0 for all x.

Thus, g'(x) does not change sign as x passes through 0.

Therefore, x=0 is neither a point of local maxima nor a point of local minima.

Hence, the function has neither a maximum value nor a minimum value.

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

(i) f(x)=|x+2|-1

Solution

Since

|x+2|\geq 0

Therefore, on subtracting 1 from both sides, we get

f(x)=|x+2|-1\geq -1

Equality holds true when

|x+2|=0

i.e. when x=-2

Hence, the minimum value of the function is -1 at x=-2.

The function has no maximum value.

Important: The function is not differentiable at x=-2, yet it attains its minimum value there. This shows that a function may have maxima or minima even at points where the derivative does not exist.

(ii) g(x)=-|x+1|+3

Solution

Since

|x+1|\geq 0

Therefore,

-|x+1|\leq 0

Hence, after adding 3 on both sides, we get

g(x)=-|x+1|+3\leq 3

Equality holds true when

|x+1|=0

i.e. whenx=-1

Hence, the maximum value of the function is 3 at x=-1.

The function has no minimum value.

Note: The function is not differentiable at x=-1, yet it attains its maximum value there.

(iii) h(x)=\sin(2x)+5

Solution

We know that,

-1\leq \sin(2x)\leq 1

Adding 5 throughout, we get

4\leq \sin(2x)+5\leq 6

That is,

4\leq h(x)\leq 6

Hence, the maximum value of the function is 6 and the minimum value is 4.

(iv) f(x)=|\sin 4x+3|

Solution

We know that,

-1\leq \sin 4x\leq 1

Adding 3 throughout, we get

2\leq \sin 4x+3\leq 4

Since \sin 4x+3 is always positive,

\therefore |\sin 4x+3|=\sin 4x+3

Thus,

2\leq f(x)\leq 4

Hence, the maximum value of the function is 4 and the minimum value is 2.

(v) h(x)=x+1,\;x\in(-1,1)

Solution

Given that

-1\lt x\lt 1

Adding 1 throughout, we get

0\lt x+1\lt 2

That is,

0\lt h(x)\lt 2

However, neither 0 nor 2 is attained because the interval (-1,1) does not include its endpoints.

The values of h(x) can be made arbitrarily close to 0 and 2, but can never equal them.

Therefore, the smallest value and the largest value of the function do not exist on the given interval.

Hence, the function has neither a maximum value nor a minimum value.

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

(i) f(x)=x^2

Solution

Given

f(x)=x^2

Differentiating with respect to x, we get

f'(x)=2x

Putting f'(x)=0, we get

2x=0

\Rightarrow x=0

Thus, x=0 is the critical point of the function.

Now, differentiating once again, we get

f''(x)=2

At x=0,

f''(0)=2\gt 0

\therefore x=0 is a point of local minima.

Also,

f(0)=0

Hence, the local minimum value of the function is 0 at x=0.

Since there is no point of local maxima, the function has no local maximum value.

(ii) g(x)=x^3-3x

Solution

Given

g(x)=x^3-3x

Differentiating with respect to x,

We get, g'(x)=3x^2-3

=3(x^2-1)

=3(x-1)(x+1)

Putting g'(x)=0, we get

3(x-1)(x+1)=0

\Rightarrow x=-1,\;1

Thus, x=-1 and x=1 are the critical points.

Using the First Derivative Test

At x=-1

Values of x	Sign of g'(x)=3(x-1)(x+1)
Close to -1 and to the left (say -1.1)	\gt 0
Close to -1 and to the right (say -0.9)	\lt 0

Thus, g'(x) changes sign from positive to negative as we move from the left of -1 to the right of -1.

\therefore x=-1 is a point of local maxima.

Also,

g(-1)=(-1)^3-3(-1)=2

Hence, the local maximum value is 2.

At x=1

Values of x	Sign of g'(x)=3(x-1)(x+1)
Close to 1 and to the left (say 0.9)	\lt 0
Close to 1 and to the right (say 1.1)	\gt 0

Thus, g'(x) changes sign from negative to positive as we move from the left of 1 to the right of 1.

\therefore x=1 is a point of local minima.

Also,

g(1)=1-3=-2

Hence, the local minimum value is -2.

Verification Using the Second Derivative Test: Although the First Derivative Test already gives the answer, it is often helpful to verify the result using the Second Derivative Test, which is generally quicker and easier to apply in examinations.

Now,

g''(x)=6x

At x=-1,

g''(-1)=-6\lt 0

which confirms that x=-1 is a point of local maxima.

At x=1,

g''(1)=6\gt 0

which confirms that x=1 is a point of local minima.

(iii) h(x)=\sin x+\cos x,\quad 0\lt x\lt \frac{\pi}{2}

Solution

Recommended: Unless specifically asked to use the First Derivative Test, it is generally advisable to use the Second Derivative Test for such questions, as it is quicker and involves less computation. We shall, however, discuss both methods for better conceptual understanding.

Method 1: Using the Second Derivative Test

Given

h(x)=\sin x+\cos x

Differentiating with respect to x, we get

h'(x)=\cos x-\sin x

Putting h'(x)=0,

We get, \cos x-\sin x=0

\cos x=\sin x

\tan x=1

Since 0\lt x\lt\frac{\pi}{2}, we get

\displaystyle x=\frac{\pi}{4}

Thus, \displaystyle x=\frac{\pi}{4} is the critical point.

Now, differentiating once again, we get

h''(x)=-\sin x-\cos x

At \displaystyle x=\frac{\pi}{4},

We have, \displaystyle h''\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}

\displaystyle =-\sqrt2\lt0

\therefore x=\frac{\pi}{4} is a point of local maxima.

Also,

We have, \displaystyle h\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}

\displaystyle =\frac{1}{\sqrt2}+\frac{1}{\sqrt2}

\displaystyle =\sqrt2

Hence, the local maximum value is \sqrt2.

There is no point of local minima in the interval \left(0,\frac{\pi}{2}\right).

Method 2: Using the First Derivative Test

We have

h'(x)=\cos x-\sin x

Multiplying and dividing by \sqrt2, we get

\displaystyle h'(x)=\sqrt2\left(\frac{\cos x-\sin x}{\sqrt2}\right)

Using the identity

\displaystyle \frac{\cos x-\sin x}{\sqrt2}=\cos\left(x+\frac{\pi}{4}\right)

we get

\displaystyle h'(x)=\sqrt2\cos\left(x+\frac{\pi}{4}\right)

Let

\displaystyle \theta=x+\frac{\pi}{4}

Since 0\lt x\lt\frac{\pi}{2}, we get

\displaystyle \frac{\pi}{4}\lt\theta\lt\frac{3\pi}{4}

Putting h'(x)=0, we get

\sqrt2\cos\theta=0

\cos\theta=0

In the interval \left(\frac{\pi}{4},\frac{3\pi}{4}\right), we get

\displaystyle \theta=\frac{\pi}{2}

Thus, \displaystyle \theta=\frac{\pi}{2} is the critical point.

At \displaystyle \theta=\frac{\pi}{2}

Values of \theta	Sign of h'(x)=\sqrt2\cos\theta
Close to \frac{\pi}{2} and to the left	\gt 0 (Quadrant I ⇒ \cos\theta\gt0)
Close to \frac{\pi}{2} and to the right	\lt 0 (Quadrant II ⇒ \cos\theta\lt0)

Thus, h'(x) changes sign from positive to negative as we move from the left of \frac{\pi}{2} to the right of \frac{\pi}{2}.

\therefore \theta=\frac{\pi}{2} is a point of local maxima.

Converting back to x,

\displaystyle x=\theta-\frac{\pi}{4}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}

Hence, \displaystyle x=\frac{\pi}{4} is a point of local maxima.

Also, \displaystyle h\left(\frac{\pi}{4}\right)=\sqrt2.

Hence, the local maximum value is \sqrt2.

Note: For trigonometric functions, rewriting the derivative as a single trigonometric function often makes sign analysis much easier. This technique is particularly useful while finding increasing and decreasing intervals, as seen in NCERT Examples 12 and 13.

Alternative: The sign of \cos x-\sin x can also be checked directly by substituting suitable values on either side of the critical point. However, rewriting the derivative as a single trigonometric function often provides a clearer and more systematic approach.

(iv) f(x)=\sin x-\cos x,\quad 0\lt x\lt 2\pi

Solution

Method 1: Using the Second Derivative Test

Given

f(x)=\sin x-\cos x

Differentiating with respect to x, we get

f'(x)=\cos x+\sin x

Putting f'(x)=0,

We get, \cos x+\sin x=0

\sin x=-\cos x

\tan x=-1

Since 0\lt x\lt2\pi, we get

\displaystyle x=\frac{3\pi}{4},\frac{7\pi}{4}

Thus, \displaystyle x=\frac{3\pi}{4} and \displaystyle x=\frac{7\pi}{4} are the critical points.

Now, differentiating once again, we get

f''(x)=\cos x-\sin x

At \displaystyle x=\frac{3\pi}{4},

\displaystyle f''\left(\frac{3\pi}{4}\right)=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}

\displaystyle =-\sqrt2\lt0

\therefore x=\frac{3\pi}{4} is a point of local maxima.

Also,

We have, \displaystyle f\left(\frac{3\pi}{4}\right)=\sin\frac{3\pi}{4}-\cos\frac{3\pi}{4}

\displaystyle =\frac{1}{\sqrt2}-\left(-\frac{1}{\sqrt2}\right)

\displaystyle =\sqrt2

Hence, the local maximum value is \sqrt2.

At \displaystyle x=\frac{7\pi}{4},

We have, \displaystyle f''\left(\frac{7\pi}{4}\right)=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}

\displaystyle =\sqrt2\gt0

\therefore x=\frac{7\pi}{4} is a point of local minima.

Also, \displaystyle f\left(\frac{7\pi}{4}\right)=\sin\frac{7\pi}{4}-\cos\frac{7\pi}{4}

\displaystyle =-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}

\displaystyle =-\sqrt2

Hence, the local minimum value is -\sqrt2.

Method 2: Using the First Derivative Test

We have

f'(x)=\cos x+\sin x

Multiplying and dividing by \sqrt2, we get

\displaystyle f'(x)=\sqrt2\left(\frac{\cos x+\sin x}{\sqrt2}\right)

Using the identity

\displaystyle \frac{\cos x+\sin x}{\sqrt2}=\sin\left(x+\frac{\pi}{4}\right)

we get

\displaystyle f'(x)=\sqrt2\sin\left(x+\frac{\pi}{4}\right)

Let

\displaystyle \theta=x+\frac{\pi}{4}

Since 0\lt x\lt2\pi, we get

\displaystyle \frac{\pi}{4}\lt\theta\lt\frac{9\pi}{4}

Putting f'(x)=0, we get

\sqrt2\sin\theta=0

\sin\theta=0

In the interval \left(\frac{\pi}{4},\frac{9\pi}{4}\right), we get

\displaystyle \theta=\pi,\;2\pi

Thus, \displaystyle \theta=\pi and \displaystyle \theta=2\pi are the critical points.

At \displaystyle \theta=\pi

Values of \theta	Sign of f'(x)=\sqrt2\sin\theta
Just to the left of \theta=\pi	\gt0 (Quadrant II ⇒ \sin\theta\gt0)
Just to the right of \theta=\pi	\lt0 (Quadrant III ⇒ \sin\theta\lt0)

Thus, f'(x) changes sign from positive to negative as we move from the left of \theta=\pi to the right of \theta=\pi.

\therefore \theta=\pi is a point of local maxima.

Converting back to x,

\displaystyle x=\pi-\frac{\pi}{4}=\frac{3\pi}{4}

Hence, \displaystyle x=\frac{3\pi}{4} is a point of local maxima.

Also, \displaystyle f\left(\frac{3\pi}{4}\right)=\sin\frac{3\pi}{4}-\cos\frac{3\pi}{4}

\displaystyle =\frac{1}{\sqrt2}-\left(-\frac{1}{\sqrt2}\right)

\displaystyle =\sqrt2

Hence, the local maximum value is \sqrt2.

Values of \theta	Sign of f'(x)=\sqrt2\sin\theta
Just to the left of \theta=2\pi	\lt0 (Quadrant IV ⇒ \sin\theta\lt0)
Just to the right of \theta=2\pi	\gt0 (Quadrant I ⇒ \sin\theta\gt0)

Thus, f'(x) changes sign from negative to positive as we move from the left of \theta=2\pi to the right of \theta=2\pi.

\therefore \theta=2\pi is a point of local minima.

Converting back to x,

\displaystyle x=2\pi-\frac{\pi}{4}=\frac{7\pi}{4}

Hence, \displaystyle x=\frac{7\pi}{4} is a point of local minima.

Also,

We have, \displaystyle f\left(\frac{7\pi}{4}\right)=\sin\frac{7\pi}{4}-\cos\frac{7\pi}{4}

\displaystyle =-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}

\displaystyle =-\sqrt2

Hence, the local minimum value is -\sqrt2.

(v) f(x)=x^3-6x^2+9x+15

Solution

Method 1: Using the Second Derivative Test

Given

f(x)=x^3-6x^2+9x+15

Differentiating with respect to x,

We get, f'(x)=3x^2-12x+9

=3(x^2-4x+3)

=3(x-1)(x-3)

Putting f'(x)=0, we get

3(x-1)(x-3)=0

\Rightarrow x=1,\;3

Thus, x=1 and x=3 are the critical points.

Now, differentiating once again, we get

f''(x)=6x-12

At x=1,

f''(1)=6-12=-6\lt0

\therefore x=1 is a point of local maxima.

Also,

f(1)=1-6+9+15=19

Hence, the local maximum value is 19.

At x=3,

f''(3)=18-12=6\gt0

\therefore x=3 is a point of local minima.

Also,

f(3)=27-54+27+15=15

Hence, the local minimum value is 15.

Method 2: Using the First Derivative Test

We have

f'(x)=3(x-1)(x-3)

The critical points are x=1 and x=3.

Values of x	Sign of f'(x)=3(x-1)(x-3)
Close to 1 and to the left (say 0.9)	\gt0
Close to 1 and to the right (say 1.1)	\lt0

Thus, f'(x) changes sign from positive to negative as we move from the left of 1 to the right of 1.

\therefore x=1 is a point of local maxima.

Also, f(1)=19.

Hence, the local maximum value is 19.

Values of x	Sign of f'(x)=3(x-1)(x-3)
Close to 3 and to the left (say 2.9)	\lt0
Close to 3 and to the right (say 3.1)	\gt0

Thus, f'(x) changes sign from negative to positive as we move from the left of 3 to the right of 3.

\therefore x=3 is a point of local minima.

Also, f(3)=15.

Hence, the local minimum value is 15.

(vi) g(x)=\frac{x}{2}+\frac{2}{x},\quad x\gt0

Solution

Method 1: Using the Second Derivative Test

Given

\displaystyle g(x)=\frac{x}{2}+\frac{2}{x}

Differentiating with respect to x, we get

\displaystyle g'(x)=\frac12-\frac{2}{x^2}

Putting g'(x)=0, we get

\displaystyle \frac12-\frac{2}{x^2}=0

or \displaystyle \frac12=\frac{2}{x^2}

x^2=4

x=\pm2

Since x\gt0, we take

x=2

Thus, x=2 is the critical point.

Now, differentiating once again, we get

\displaystyle g''(x)=\frac{4}{x^3}

At x=2,

\displaystyle g''(2)=\frac{4}{8}=\frac12\gt0

\therefore x=2 is a point of local minima.

Also,

\displaystyle g(2)=\frac22+\frac22

=1+1=2

Hence, the local minimum value is 2.

There is no point of local maxima.

Method 2: Using the First Derivative Test

We have

\displaystyle g'(x)=\frac12-\frac{2}{x^2}

Putting g'(x)=0, we get the critical point x=2.

Values of x	Sign of \displaystyle g'(x)=\frac12-\frac{2}{x^2}
Close to 2 and to the left (say 1.9)	\lt0
Close to 2 and to the right (say 2.1)	\gt0

Thus, g'(x) changes sign from negative to positive as we move from the left of 2 to the right of 2.

\therefore x=2 is a point of local minima.

Also, g(2)=2.

Hence, the local minimum value is 2.

There is no point of local maxima.

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

Solution

Method 1: Using the Second Derivative Test

Given

\displaystyle g(x)=\frac{1}{x^2+2}

Differentiating with respect to x, we get

\displaystyle g'(x)=\frac{-2x}{(x^2+2)^2}

Putting g'(x)=0, we get

-2x=0

\Rightarrow x=0

Thus, x=0 is the critical point.

Now, differentiating once again, we get

\displaystyle g''(x)=\frac{6x^2-4}{(x^2+2)^3}

At x=0,

\displaystyle g''(0)=\frac{-4}{8}=-\frac12\lt0

\therefore x=0 is a point of local maxima.

Also,

\displaystyle g(0)=\frac{1}{2}

Hence, the local maximum value is \displaystyle \frac12.

There is no point of local minima.

Method 2: Using the First Derivative Test

We have

\displaystyle g'(x)=\frac{-2x}{(x^2+2)^2}

Since (x^2+2)^2\gt0 for all real values of x, the sign of g'(x) depends only on the sign of -2x.

Putting g'(x)=0, we get the critical point x=0.

Values of x	Sign of \displaystyle g'(x)=\frac{-2x}{(x^2+2)^2}
Close to 0 and to the left (say -0.1)	\gt0
Close to 0 and to the right (say 0.1)	\lt0

Thus, g'(x) changes sign from positive to negative as we move from the left of 0 to the right of 0.

\therefore x=0 is a point of local maxima.

Also,

\displaystyle g(0)=\frac12

Hence, the local maximum value is \displaystyle \frac12.

There is no point of local minima.

(viii) f(x)=x\sqrt{1-x},\quad 0\lt x\lt1

Solution

Method 1: Using the Second Derivative Test

Given

f(x)=x\sqrt{1-x}

Differentiating using the product rule,

We get, \displaystyle f'(x)=x\left(-\frac{1}{2\sqrt{1-x}}\right)+\sqrt{1-x}

\displaystyle =\frac{-x+2(1-x)}{2\sqrt{1-x}}

\displaystyle =\frac{2-3x}{2\sqrt{1-x}}

Putting f'(x)=0, we get

2-3x=0

\displaystyle x=\frac23

Thus, \displaystyle x=\frac23 is the critical point.

Now, differentiating once again, we get

\displaystyle f''(x)=\frac{3x-4}{4(1-x)^{3/2}}

At \displaystyle x=\frac23,

\displaystyle f''\left(\frac23\right)=\frac{2-4}{4\left(\frac13\right)^{3/2}}

\displaystyle =\frac{-2}{4\left(\frac13\right)^{3/2}}\lt0

\therefore x=\frac23 is a point of local maxima.

Also,

We have, \displaystyle f\left(\frac23\right)=\frac23\sqrt{\frac13}

\displaystyle =\frac{2}{3\sqrt3}

\displaystyle =\frac{2\sqrt3}{9}

Hence, the local maximum value is \displaystyle \frac{2\sqrt3}{9}.

There is no point of local minima.

Method 2: Using the First Derivative Test

We have

\displaystyle f'(x)=\frac{2-3x}{2\sqrt{1-x}}

Since 0\lt x\lt1, we have \sqrt{1-x}\gt0. Therefore, the sign of f'(x) depends only on the sign of 2-3x.

Putting f'(x)=0, we get

\displaystyle x=\frac23

Thus, \displaystyle x=\frac23 is the critical point.

Values of x	Sign of \displaystyle f'(x)=\frac{2-3x}{2\sqrt{1-x}}
Close to \frac23 and to the left	\gt0 (since 2-3x\gt0)
Close to \frac23 and to the right	\lt0 (since 2-3x\lt0)

Thus, f'(x) changes sign from positive to negative as we move from the left of \frac23 to the right of \frac23.

\therefore x=\frac23 is a point of local maxima.

Also,

\displaystyle f\left(\frac23\right)=\frac{2\sqrt3}{9}

Hence, the local maximum value is \displaystyle \frac{2\sqrt3}{9}.

There is no point of local minima.

4. Prove that the following functions do not have maxima or minima:

(i) f(x)=e^x

Solution

Differentiating with respect to x, we get

f'(x)=e^x

Since e^x\gt0 for all x\in\mathbb R, we have

f'(x)\gt0 for all x\in\mathbb R

Therefore, f(x)=e^x is strictly increasing on \mathbb R.

Hence, the function has neither a maximum value nor a minimum value.

(ii) g(x)=\log x

Solution

The domain of \log x is (0,\infty).

Differentiating with respect to x, we get

\displaystyle g'(x)=\frac{1}{x}

Since x\gt0, we have

\displaystyle \frac1x\gt0

Therefore, g'(x)\gt0 for all x\gt0.

Hence, g(x)=\log x is strictly increasing on (0,\infty).

Therefore, the function has neither a maximum value nor a minimum value.

(iii) h(x)=x^3+x^2+x+1

Solution

Differentiating with respect to x, we get

h'(x)=3x^2+2x+1

Taking 3 common from the first two terms, we get

h'(x)=3\left(x^2+\frac{2}{3}x\right)+1

Adding and subtracting \left(\frac13\right)^2 inside the bracket,

We get, \displaystyle h'(x)=3\left(x^2+\frac23x+\frac19-\frac19\right)+1

\displaystyle =3\left(x+\frac13\right)^2-\frac13+1

\displaystyle =3\left(x+\frac13\right)^2+\frac23

Since a square is always non-negative, we have

\displaystyle 3\left(x+\frac13\right)^2\ge0

Therefore,

\displaystyle h'(x)=3\left(x+\frac13\right)^2+\frac23\gt0

for all x\in\mathbb R.

Hence, h(x)=x^3+x^2+x+1 is strictly increasing on \mathbb R.

Since the function is strictly increasing throughout its domain, it can neither attain a maximum value nor a minimum value.

5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

Important: To find the absolute maximum and absolute minimum values of a continuous function on a closed interval, evaluate the function at all critical points lying inside the interval and at the endpoints of the interval. The largest value obtained is the absolute maximum value, while the smallest value obtained is the absolute minimum value.

(i) f(x)=x^3,\quad x\in[-2,2]

Solution

Step 1: Find the critical points.

Differentiating with respect to x, we get

f'(x)=3x^2

Putting f'(x)=0, we get

3x^2=0

x=0

Thus, x=0 is the critical point.

Step 2: Take the endpoints of the interval.

The endpoints are x=-2 and x=2.

Step 3: Calculate the value of the function at all these points.

x	f(x)=x^3
-2	-8
0	0
2	8

Step 4: Identify the largest and smallest values.

The largest value is 8 and the smallest value is -8.

Hence, the absolute maximum value is 8 at x=2 and the absolute minimum value is -8 at x=-2.

(ii) f(x)=\sin x+\cos x,\quad x\in[0,\pi]

Solution

Step 1: Find the critical points.

Differentiating with respect to x, we get

f'(x)=\cos x-\sin x

Putting f'(x)=0, we get

\cos x=\sin x

\tan x=1

In the interval [0,\pi], we get

\displaystyle x=\frac{\pi}{4}

Thus, \displaystyle x=\frac{\pi}{4} is the critical point.

Step 2: Take the endpoints of the interval.

The endpoints are x=0 and x=\pi.

Step 3: Calculate the value of the function at all these points.

x	f(x)=\sin x+\cos x
0	1
\frac{\pi}{4}	\sqrt2
\pi	-1

Step 4: Identify the largest and smallest values.

The largest value is \sqrt2 and the smallest value is -1.

Hence, the absolute maximum value is \sqrt2 at \displaystyle x=\frac{\pi}{4} and the absolute minimum value is -1 at x=\pi.

(iii) f(x)=4x-\frac12x^2,\quad x\in\left[-2,\frac92\right]

Solution

Step 1: Find the critical points.

Differentiating with respect to x, we get

f'(x)=4-x

Putting f'(x)=0, we get

4-x=0

x=4

Thus, x=4 is the critical point.

Step 2: Take the endpoints of the interval.

The endpoints are x=-2 and \displaystyle x=\frac92.

Step 3: Calculate the value of the function at all these points.

x	\displaystyle f(x)=4x-\frac12x^2
-2	-10
4	8
\displaystyle \frac92	\displaystyle \frac{63}{8}

Step 4: Identify the largest and smallest values.

The largest value is 8 and the smallest value is -10.

Hence, the absolute maximum value is 8 at x=4 and the absolute minimum value is -10 at x=-2.

(iv) f(x)=(x-1)^2+3,\quad x\in[-3,1]

Solution

Step 1: Find the critical points.

Differentiating with respect to x, we get

f'(x)=2(x-1)

Putting f'(x)=0, we get

2(x-1)=0

x=1

Thus, x=1 is the critical point.

Step 2: Take the endpoints of the interval.

The endpoints are x=-3 and x=1.

Step 3: Calculate the value of the function at all these points.

x	f(x)=(x-1)^2+3
-3	19
1	3

Step 4: Identify the largest and smallest values.

The largest value is 19 and the smallest value is 3.

Hence, the absolute maximum value is 19 at x=-3 and the absolute minimum value is 3 at x=1.

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

p(x)=41-72x-18x^2

Solution

Differentiating with respect to x, we get

p'(x)=-72-36x

Putting p'(x)=0, we get

-72-36x=0

x=-2

Thus, x=-2 is the critical point.

Now, differentiating once again, we get

p''(x)=-36

Since

p''(-2)=-36\lt0

\therefore x=-2 is a point of local maxima.

Substituting x=-2 in the profit function,

We get, p(2)=41-72(-2)-18(-2)^2

=41+144-72

=113

Hence, the maximum profit that the company can make is 113.

Note: The given profit function attains its maximum value 113 at x=-2. In practical applications, x often represents the number of units produced or sold and is therefore non-negative. Interestingly, if the sign of the linear term were changed, i.e. if p(x)=41+72x-18x^2, the maximum profit would still be 113, but it would occur at x=2.

7. Find both the maximum value and the minimum value of

3x^4-8x^3+12x^2-48x+25 on the interval [0,3].

Solution

Step 1: Find the critical points.

Let

f(x)=3x^4-8x^3+12x^2-48x+25

Differentiating with respect to x,

We get, f'(x)=12x^3-24x^2+24x-48

=12(x^3-2x^2+2x-4)

=12(x-2)(x^2+2)

Putting f'(x)=0, we get

12(x-2)(x^2+2)=0

Since x^2+2\gt0 for all real values of x, we get

x=2

Thus, x=2 is the only critical point in the interval [0,3].

Step 2: Take the endpoints of the interval.

The endpoints are x=0 and x=3.

Step 3: Calculate the value of the function at all these points.

x	f(x)
0	25
2	-39
3	-20

Step 4: Identify the largest and smallest values.

The largest value is 25 and the smallest value is -39.

Hence, the maximum value of the function on [0,3] is 25 at x=0, and the minimum value is -39 at x=2.

8. At what points in the interval [0,2\pi] does the function \sin 2x attain its maximum value?

Solution

Let

f(x)=\sin 2x

Differentiating with respect to x, we get

f'(x)=2\cos 2x

Putting f'(x)=0, we get

2\cos 2x=0

\cos 2x=0

Thus, \displaystyle 2x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}

Therefore,

\displaystyle x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}

These are the critical points in the interval [0,2\pi].

Now, we calculate the value of f(x)=\sin 2x at these points.

x	\sin 2x
\frac{\pi}{4}	1
\frac{3\pi}{4}	-1
\frac{5\pi}{4}	1
\frac{7\pi}{4}	-1

The maximum value of \sin 2x is 1.

Hence, the function \sin 2x attains its maximum value at

\displaystyle x=\frac{\pi}{4}\text{ and }\frac{5\pi}{4}.

9. What is the maximum value of the function \sin x+\cos x?

Solution

Let

f(x)=\sin x+\cos x

Differentiating with respect to x, we get

f'(x)=\cos x-\sin x

Putting f'(x)=0,

We \cos x=\sin x

\tan x=1

\displaystyle x=\frac{\pi}{4}+n\pi,\quad n\in\mathbb Z

Now, differentiating once again, we get

f''(x)=-\sin x-\cos x

At \displaystyle x=\frac{\pi}{4},

\displaystyle f''\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}=-\sqrt2\lt0

\therefore x=\frac{\pi}{4} is a point of local maxima.

Also, \displaystyle f\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}

\displaystyle =\frac{1}{\sqrt2}+\frac{1}{\sqrt2}

\displaystyle =\sqrt2

Hence, the maximum value of the function \sin x+\cos x is

\boxed{\sqrt2}

Alternative Method: Using the identity \sin x+\cos x=\sqrt2\sin\left(x+\frac{\pi}{4}\right), the maximum value is obtained when \sin\left(x+\frac{\pi}{4}\right)=1. Therefore, the maximum value is \sqrt2.

10. Find the maximum value of 2x^3-24x+107 in the interval [1,3]. Find the maximum value of the same function in [-3,-1].

Solution

Let

f(x)=2x^3-24x+107

Step 1: Find the critical points.

Differentiating with respect to x,

We get, f'(x)=6x^2-24

=6(x^2-4)

=6(x-2)(x+2)

Putting f'(x)=0, we get

6(x-2)(x+2)=0

x=2,\,-2

Thus, the critical points are x=2 and x=-2.

On the interval [1,3]

The critical point lying in this interval is x=2.

The endpoints are x=1 and x=3.

x	f(x)=2x^3-24x+107
1	85
2	75
3	89

The largest value is 89.

Hence, the maximum value of the function on [1,3] is 89, attained at x=3.

On the interval [-3,-1]

The critical point lying in this interval is x=-2.

The endpoints are x=-3 and x=-1.

x	f(x)=2x^3-24x+107
-3	125
-2	139
-1	129

The largest value is 139.

Hence, the maximum value of the function on [-3,-1] is 139, attained at x=-2.

11. It is given that at x=1, the function f(x)=x^4-62x^2+ax+9 attains its maximum value on the interval [0,2]. Find the value of a.

Solution

Given

f(x)=x^4-62x^2+ax+9

Since the function attains its maximum value at x=1, and x=1 lies in the interior of the interval [0,2], it must be a critical point.

Therefore,

f'(1)=0

Differentiating with respect to x, we get

f'(x)=4x^3-124x+a

Putting x=1, we get

f'(1)=4-124+a

=a-120

Since f'(1)=0, we have

a-120=0

a=120

To verify that x=1 is indeed a point of local maxima, we find the second derivative.

f''(x)=12x^2-124

f''(1)=12-124=-112\lt0

Hence, x=1 is a point of local maxima.

Therefore, the required value of a is

\boxed{120}

12. Find the maximum and minimum values of x+\sin 2x on [0,2\pi].

Solution

Let

f(x)=x+\sin 2x

Step 1: Find the critical points.

Differentiating with respect to x, we get

f'(x)=1+2\cos 2x

Putting f'(x)=0, we get

1+2\cos 2x=0

\cos 2x=-\frac12

In the interval 0\le x\le 2\pi, we get

\displaystyle 2x=\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\frac{10\pi}{3}

Therefore,

\displaystyle x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}

Thus, the critical points are \displaystyle \frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}.

Step 2: Take the endpoints of the interval.

The endpoints are x=0 and x=2\pi.

Step 3: Calculate the value of the function at all these points.

x	f(x)=x+\sin 2x
0	0
\displaystyle \frac{\pi}{3}	\displaystyle \frac{\pi}{3}+\frac{\sqrt3}{2}
\displaystyle \frac{2\pi}{3}	\displaystyle \frac{2\pi}{3}-\frac{\sqrt3}{2}
\displaystyle \frac{4\pi}{3}	\displaystyle \frac{4\pi}{3}+\frac{\sqrt3}{2}
\displaystyle \frac{5\pi}{3}	\displaystyle \frac{5\pi}{3}-\frac{\sqrt3}{2}
2\pi	2\pi

Step 4: Identify the largest and smallest values.

From the table above, the largest value is

2\pi

and the smallest value is

0.

Hence, the maximum value of the function on [0,2\pi] is

2\pi

attained at x=2\pi.

The minimum value of the function on [0,2\pi] is

0

attained at x=0.

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

Solution

Let one of the numbers be x.

Since their sum is 24, the other number is

24-x

Therefore, the product of the two numbers is

P=x(24-x)

P=24x-x^2

Thus, the product is a function of x.

To find the maximum product, we differentiate P with respect to x.

\displaystyle \frac{dP}{dx}=24-2x

Putting \displaystyle \frac{dP}{dx}=0, we get

24-2x=0

x=12

Thus, the critical point is x=12.

Now,

\displaystyle \frac{d^2P}{dx^2}=-2

Since

\displaystyle \frac{d^2P}{dx^2}=-2\lt0,

\therefore x=12 is a point of local maxima.

Hence, the first number is

12

and the second number is

24-12=12.

Therefore, the required numbers are 12 and 12.

The maximum product is

12\times12=144.

14. Find two positive numbers x and y such that x+y=60 and xy^3 is maximum.

Solution

We are given that

x+y=60

Therefore,

x=60-y

The quantity to be maximised is

P=xy^3

Substituting x=60-y, we get

P=(60-y)y^3

P=60y^3-y^4

Thus, P is a function of y.

To find the maximum value of P, we differentiate with respect to y.

\displaystyle \frac{dP}{dy}=180y^2-4y^3

=4y^2(45-y)

Putting \displaystyle \frac{dP}{dy}=0, we get

4y^2(45-y)=0

y=0\text{ or }45

Since the numbers are positive, we take

y=45

Now,

\displaystyle \frac{d^2P}{dy^2}=360y-12y^2

At y=45,

\displaystyle \frac{d^2P}{dy^2}=360(45)-12(45)^2

=16200-24300

Simplifying

=-8100\lt0

\therefore y=45 gives the maximum value of P.

Substituting in x+y=60, we get

x=60-45=15

Hence, the required positive numbers are

\boxed{x=15,\;y=45}

The maximum value of xy^3 is

15\times45^3=1366875.

15. Find two positive numbers x and y such that their sum is 35 and the product x^2y^5 is a maximum.

Solution

We are given that

x+y=35

Therefore,

x=35-y

The quantity to be maximised is

P=x^2y^5

Substituting x=35-y, we get

P=(35-y)^2y^5

Taking logarithms on both sides, we get

\log P=2\log(35-y)+5\log y

Differentiating with respect to y, we get

\displaystyle \frac{1}{P}\frac{dP}{dy}=-\frac{2}{35-y}+\frac{5}{y}

Putting \displaystyle \frac{dP}{dy}=0, we get

-\frac{2}{35-y}+\frac{5}{y}=0

\displaystyle \frac{5}{y}=\frac{2}{35-y}

or 5(35-y)=2y

175-5y=2y

7y=175

Thus, y=25

Therefore,

x=35-25=10

Now,

\displaystyle \frac{1}{P}\frac{dP}{dy}=-\frac{2}{35-y}+\frac{5}{y}

Differentiating once again, we get

\displaystyle \frac{d}{dy}\left(\frac{1}{P}\frac{dP}{dy}\right)=-\frac{2}{(35-y)^2}-\frac{5}{y^2}

At y=25, this expression is negative.

Hence, P=x^2y^5 attains its maximum value when

x=10,\quad y=25

Therefore, the required positive numbers are

\boxed{x=10,\;y=25}

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

Solution

Let the two positive numbers be x and y.

We are given that

x+y=16

Therefore,

y=16-x

The quantity to be minimised is

S=x^3+y^3

Substituting y=16-x, we get

S=x^3+(16-x)^3

Expanding, we get

S=x^3+4096-768x+48x^2-x^3

S=48x^2-768x+4096

Thus, S is a function of x.

Differentiating with respect to x, we get

\displaystyle \frac{dS}{dx}=96x-768

Putting \displaystyle \frac{dS}{dx}=0, we get

96x-768=0

x=8

Therefore,

y=16-8=8

Now,

\displaystyle \frac{d^2S}{dx^2}=96

Since

\displaystyle \frac{d^2S}{dx^2}=96\gt0,

\therefore x=8 gives the minimum value of S.

Hence, the required numbers are

\boxed{8\text{ and }8}.

The minimum value of the sum of the cubes is

8^3+8^3=1024.

17. 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?

Solution

Let the side of each square cut from the corners be x cm.

Since a square of side x cm is removed from each corner, the length of the base is reduced by x cm from both ends.

Therefore, the length of the box becomes

18-2x cm.

Similarly, the breadth of the box becomes

18-2x cm.

When the sides are folded upward, the height of the box is equal to the side of the square removed.

Therefore, the height of the box is

x cm.

Hence, the volume of the box is

V=(18-2x)(18-2x)(x)

V=x(18-2x)^2

Expanding, we get

V=x(324-72x+4x^2)

V=324x-72x^2+4x^3

Thus, V is a function of x.

To find the maximum volume, we differentiate with respect to x.

So, \frac{dV}{dx}=324-144x+12x^2

=12(x^2-12x+27)

=12(x-3)(x-9)

Putting \frac{dV}{dx}=0, we get

12(x-3)(x-9)=0

x=3\text{ or }9

Now, x=9 is not possible because then

18-2x=18-18=0

which would give a box with zero base area and hence zero volume.

Therefore, the only feasible critical point is

x=3

Now,

\frac{d^2V}{dx^2}=24x-144

At x=3,

We have, \frac{d^2V}{dx^2}=72-144

=-72\lt0

\therefore x=3 gives the maximum value of the volume.

Hence, the side of the square to be cut off from each corner should be

3\text{ cm}.

18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares 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?

Solution

Let the side of each square cut from the corners be x cm.

Since a square of side x cm is removed from each corner, the length of the box is reduced by x cm from both ends.

Therefore, the length of the box becomes

45-2x cm.

Similarly, the breadth of the box becomes

24-2x cm.

Since the breadth of the box must remain positive, we require 24-2x\gt0, i.e. x\lt12.

When the sides are folded upward, the height of the box is equal to the side of the square removed.

Therefore, the height of the box is

x cm.

Hence, the volume of the box is

V=x(45-2x)(24-2x)

Expanding, we get

V=x(1080-138x+4x^2)

V=1080x-138x^2+4x^3

Thus, V is a function of x.

To find the maximum volume, we differentiate with respect to x.

So, \frac{dV}{dx}=1080-276x+12x^2

=12(x^2-23x+90)

=12(x-5)(x-18)

Putting \frac{dV}{dx}=0, we get

12(x-5)(x-18)=0

x=5\text{ or }18

Now, x=18 is not possible because

24-2x=24-36=-12

which would give a negative breadth. Hence, this value is not feasible.

Therefore, the only feasible critical point is

x=5

Now,

\frac{d^2V}{dx^2}=24x-276

At x=5,

\frac{d^2V}{dx^2}=120-276

=-156\lt0

Therefore, x=5 gives the maximum value of the volume.

Hence, the side of the square to be cut off from each corner should be

5\text{ cm}.

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

Solution

Let the radius of the given circle be r.

Consider a rectangle inscribed in the circle with centre at the origin.

Let one vertex of the rectangle in the first quadrant be (x,y).

Since the vertex lies on the circle of radius r, we have

x^2+y^2=r^2

Therefore,

y=\sqrt{r^2-x^2}

The length and breadth of the rectangle are

2x and 2y respectively.

Hence, the area of the rectangle is

A=(2x)(2y)=4xy

Substituting y=\sqrt{r^2-x^2}, we get

A=4x\sqrt{r^2-x^2}

To find the maximum area, we differentiate with respect to x.

\displaystyle \frac{dA}{dx}=4\left[\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}}\right]

\displaystyle =\frac{4(r^2-2x^2)}{\sqrt{r^2-x^2}}

Putting \displaystyle \frac{dA}{dx}=0,

We get,r^2-2x^2=0

\displaystyle x^2=\frac{r^2}{2}

\displaystyle x=\frac{r}{\sqrt2}

Substituting in x^2+y^2=r^2, we get

\displaystyle y^2=r^2-\frac{r^2}{2}=\frac{r^2}{2}

\displaystyle y=\frac{r}{\sqrt2}

Thus,

x=y

Therefore, the length and breadth of the rectangle are equal.

Hence, the rectangle is a square.

Thus, of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

20. Show that the right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.

Solution

Let the radius of the base of the cylinder be r and its height be h.

Let the total surface area of the cylinder be S, which is constant.

Then

2\pi rh+2\pi r^2=S

Dividing both sides by 2\pi r, we get

\displaystyle h+r=\frac{S}{2\pi r}

Therefore,

\displaystyle h=\frac{S}{2\pi r}-r

The volume of the cylinder is

V=\pi r^2h

Substituting the value of h, we get

\displaystyle V=\pi r^2\left(\frac{S}{2\pi r}-r\right)

\displaystyle V=\frac{Sr}{2}-\pi r^3

Thus, V is a function of r.

To find the maximum volume, we differentiate with respect to r.

\displaystyle \frac{dV}{dr}=\frac{S}{2}-3\pi r^2

Putting \displaystyle \frac{dV}{dr}=0, we get

\displaystyle \frac{S}{2}-3\pi r^2=0

\displaystyle S=6\pi r^2

Now,

\displaystyle \frac{d^2V}{dr^2}=-6\pi r

Since r\gt0, we have

\displaystyle \frac{d^2V}{dr^2}\lt0

Therefore, the volume is maximum at the critical point obtained above.

Substituting this value of S in

2\pi rh+2\pi r^2=S

we get

2\pi rh+2\pi r^2=6\pi r^2

2\pi rh=4\pi r^2

h=2r

Thus, the height of the cylinder is equal to the diameter of its base.

Hence, the right circular cylinder of given surface area and maximum volume is such that

\boxed{h=2r}

i.e. the height is equal to the diameter of the base.

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

Solution

Let the radius of the base of the cylindrical can be r cm and its height be h cm.

Since the volume of the cylinder is 100 cubic centimetres, we have

\pi r^2h=100

Therefore,

\displaystyle h=\frac{100}{\pi r^2}

The total surface area of the closed cylindrical can is

S=2\pi rh+2\pi r^2

Substituting the value of h, we get

\displaystyle S=2\pi r\left(\frac{100}{\pi r^2}\right)+2\pi r^2

\displaystyle S=\frac{200}{r}+2\pi r^2

Thus, S is a function of r.

To find the minimum surface area, we differentiate with respect to r.

\displaystyle \frac{dS}{dr}=-\frac{200}{r^2}+4\pi r

Putting \displaystyle \frac{dS}{dr}=0, we get

-\frac{200}{r^2}+4\pi r=0

4\pi r=\frac{200}{r^2}

4\pi r^3=200

or \pi r^3=50

\displaystyle r=\sqrt[3]{\frac{50}{\pi}}

Now,

\displaystyle \frac{d^2S}{dr^2}=\frac{400}{r^3}+4\pi

Since r\gt0, we have

\displaystyle \frac{d^2S}{dr^2}\gt0

Therefore, the surface area is minimum at the critical point obtained above.

Substituting \pi r^3=50 in

\displaystyle h=\frac{100}{\pi r^2}

we get

\displaystyle h=\frac{100r}{\pi r^3}

\displaystyle h=\frac{100r}{50}

i.e. h=2r

Hence, the dimensions of the can of minimum surface area are

\displaystyle r=\sqrt[3]{\frac{50}{\pi}}\text{ cm}

and

\displaystyle h=2\sqrt[3]{\frac{50}{\pi}}\text{ cm}.

22. 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?

Solution

Let x metres be the length of the wire used to form the square.

Then the length of the wire used to form the circle is

28-x metres.

Area of the square

The perimeter of the square is x.

Therefore, its side is

\displaystyle \frac{x}{4}

Hence, the area of the square is

\displaystyle \left(\frac{x}{4}\right)^2=\frac{x^2}{16}

Area of the circle

The circumference of the circle is 28-x.

Therefore,

2\pi r=28-x

\displaystyle r=\frac{28-x}{2\pi}

Hence, the area of the circle is

\displaystyle \pi r^2=\pi\left(\frac{28-x}{2\pi}\right)^2

\displaystyle =\frac{(28-x)^2}{4\pi}

Combined Area

Therefore, the combined area is

\displaystyle A=\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}

To minimise the combined area, we differentiate with respect to x.

\displaystyle \frac{dA}{dx}=\frac{x}{8}-\frac{28-x}{2\pi}

Putting \displaystyle \frac{dA}{dx}=0, we get

\displaystyle \frac{x}{8}=\frac{28-x}{2\pi}

2\pi x=8(28-x)

\pi x=112-4x

or x(\pi+4)=112

\displaystyle x=\frac{112}{\pi+4}

Thus, the critical point is

\displaystyle x=\frac{112}{\pi+4}

Now,

\displaystyle \frac{d^2A}{dx^2}=\frac18+\frac{1}{2\pi}

Since

\displaystyle \frac18+\frac{1}{2\pi}\gt0

for all values of x,

\therefore the combined area is minimum when

\displaystyle x=\frac{112}{\pi+4}

Hence, the wire should be cut into two pieces of lengths

\displaystyle \frac{112}{\pi+4}\text{ m}

and

\displaystyle 28-\frac{112}{\pi+4}=\frac{28\pi}{\pi+4}\text{ m}.

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

Solution

Let a right circular cone be inscribed in a sphere of radius R.

Also, let h be the height of the cone and r be the radius of its base.

Consider the axial cross-section of the cone and the sphere as shown in the diagram:

So, The centre of the sphere lies on the axis of the cone.

Let the vertex of the cone be at the top of the sphere.

Then the distance of the base of the cone from the centre of the sphere is

h-R.

From the right triangle OMB,

We get, r^2+(h-R)^2=R^2

\therefore r^2=R^2-(h-R)^2

=R^2-(R^2-2Rh+h^2)

So, r^2=2Rh-h^2

Volume of the cone

We know, \displaystyle V=\frac13\pi r^2 h

\displaystyle =\frac13\pi h(2Rh-h^2)

\displaystyle =\frac13\pi(2Rh^2-h^3)

To find the maximum volume, differentiate with respect to h.

\displaystyle \frac{dV}{dh}=\frac13\pi(4Rh-3h^2)

\displaystyle =\frac13\pi h(4R-3h)

Putting \displaystyle \frac{dV}{dh}=0, we get

h=0 or 4R-3h=0

\therefore h=\frac{4R}{3}

Thus, the non-zero critical point is

\displaystyle h=\frac{4R}{3}

Now,

\displaystyle \frac{d^2V}{dh^2}=\frac13\pi(4R-6h)

At \displaystyle h=\frac{4R}{3},

\displaystyle \frac{d^2V}{dh^2}=\frac13\pi\left(4R-8R\right)

\displaystyle =-\frac{4\pi R}{3}\lt0

Hence, the volume is maximum when

\displaystyle h=\frac{4R}{3}.

Substituting \displaystyle h=\frac{4R}{3} in V=\frac13\pi(2Rh^2-h^3),

We get, \displaystyle V_{\max}=\frac13\pi\left[2R\left(\frac{4R}{3}\right)^2-\left(\frac{4R}{3}\right)^3\right]

\displaystyle =\frac13\pi\left(\frac{32R^3}{9}-\frac{64R^3}{27}\right)

Thus, \displaystyle =\frac13\pi\cdot\frac{32R^3}{27}

Thus, \displaystyle V_{\max}=\frac{32\pi R^3}{81}

The volume of the sphere is

\displaystyle \frac43\pi R^3.

Therefore,

\displaystyle \frac{V_{\max}}{\text{Volume of sphere}}=\frac{\frac{32\pi R^3}{81}}{\frac43\pi R^3}

\displaystyle =\frac{32}{81}\times\frac34=\frac{8}{27}

or \displaystyle V_{\max}=\frac{8}{27}\times\text{(Volume of Sphere)}

\therefore the volume of the largest cone that can be inscribed in a sphere is \displaystyle \frac{8}{27} of the volume of the sphere.

24. Show that the right circular cone of least curved surface and given volume has an altitude equal to \sqrt2 times the radius of the base.

Solution

Let the radius of the base of the cone be r and its altitude be h.

Let the volume of the cone be V, which is constant.

Then

\displaystyle \frac13\pi r^2h=V

Therefore,

\displaystyle h=\frac{3V}{\pi r^2}

The curved surface area of the cone is

S=\pi rl

where l is the slant height.

Since

l=\sqrt{r^2+h^2}

we get

\displaystyle S=\pi r\sqrt{r^2+h^2}

Substituting \displaystyle h=\frac{3V}{\pi r^2}, we obtain

\displaystyle S=\pi r\sqrt{r^2+\frac{9V^2}{\pi^2r^4}}

To simplify the differentiation, let us minimise S^2 instead of S. Since S\gt0, both attain their minimum value at the same point.

Therefore,

\displaystyle S^2=\pi^2r^2\left(r^2+\frac{9V^2}{\pi^2r^4}\right)

\displaystyle S^2=\pi^2r^4+\frac{9V^2}{r^2}

Let

\displaystyle F(r)=\pi^2r^4+\frac{9V^2}{r^2}

To minimise the curved surface area, we differentiate F(r) with respect to r.

\displaystyle F'(r)=4\pi^2r^3-\frac{18V^2}{r^3}

Putting F'(r)=0,

We get, \displaystyle 4\pi^2r^3=\frac{18V^2}{r^3}

\displaystyle 4\pi^2r^6=18V^2

\displaystyle 2\pi^2r^6=9V^2

Also, from

\displaystyle V=\frac13\pi r^2h

we have

\displaystyle 9V^2=\pi^2r^4h^2

Substituting in the previous equation,

We get, \displaystyle 2\pi^2r^6=\pi^2r^4h^2

2r^2=h^2

h=\sqrt2\,r

Now,

\displaystyle F''(r)=12\pi^2r^2+\frac{54V^2}{r^4}

Since r\gt0, we have

\displaystyle F''(r)\gt0

Therefore, F(r) and hence the curved surface area is minimum at the critical point obtained above.

Hence, the right circular cone of least curved surface and given volume satisfies

\boxed{h=\sqrt2\,r}

i.e. the altitude is \sqrt2 times the radius of the base.

25. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is \tan^{-1}(\sqrt2).

Solution

Let the slant height of the cone be l, which is constant.

Let \theta be the semi-vertical angle of the cone.

Then, from the right triangle formed by the radius, altitude and slant height of the cone, we have

r=l\sin\theta

and

h=l\cos\theta

The volume of the cone is

\displaystyle V=\frac13\pi r^2h

Substituting the values of r and h, we get

\displaystyle V=\frac13\pi(l\sin\theta)^2(l\cos\theta)

\displaystyle V=\frac{\pi l^3}{3}\sin^2\theta\cos\theta

Since \displaystyle \frac{\pi l^3}{3} is constant, it is sufficient to maximise

f(\theta)=\sin^2\theta\cos\theta

Differentiating with respect to \theta, we get

\displaystyle f'(\theta)=2\sin\theta\cos^2\theta-\sin^3\theta

\displaystyle =\sin\theta\left(2\cos^2\theta-\sin^2\theta\right)

Putting f'(\theta)=0, we get

\displaystyle \sin\theta\left(2\cos^2\theta-\sin^2\theta\right)=0

Ignoring the trivial solution \sin\theta=0,

We get,2\cos^2\theta=\sin^2\theta

\tan^2\theta=2

\tan\theta=\sqrt2

Therefore,

\displaystyle \theta=\tan^{-1}(\sqrt2)

Now, differentiating once again, we get

\displaystyle f''(\theta)=2\cos^3\theta-7\sin^2\theta\cos\theta

Using \tan^2\theta=2, we have

\sin^2\theta=2\cos^2\theta

Therefore,

\displaystyle f''(\theta)=2\cos^3\theta-14\cos^3\theta

\displaystyle =-12\cos^3\theta\lt0

Hence, \theta=\tan^{-1}(\sqrt2) gives the maximum value of f(\theta) and therefore the maximum volume of the cone.

Thus, the semi-vertical angle of the cone of maximum volume and given slant height is

\boxed{\tan^{-1}(\sqrt2)}

Observation: Since \tan\theta=\sqrt2, we have r:h=\sqrt2:1. Thus, for a cone of maximum volume and fixed slant height, the radius is \sqrt2 times its altitude.

26. Show that the semi-vertical angle of a right circular cone of given surface area and maximum volume is \sin^{-1}\left(\frac13\right).

Solution

Let r be the radius of the base, h the height, and l the slant height of the cone.

Let \theta be the semi-vertical angle of the cone.

Then, from the right triangle formed by the radius, height and slant height, we have

r=l\sin\theta

and

h=l\cos\theta

Let the total surface area of the cone be S, which is constant.

Then

S=\pi rl+\pi r^2

Substituting r=l\sin\theta, we get

S=\pi l^2\sin\theta+\pi l^2\sin^2\theta

S=\pi l^2\sin\theta(1+\sin\theta)

Therefore,

\displaystyle l^2=\frac{S}{\pi\sin\theta(1+\sin\theta)}

The volume of the cone is

\displaystyle V=\frac13\pi r^2h

Substituting r=l\sin\theta and h=l\cos\theta, we get

\displaystyle V=\frac13\pi l^3\sin^2\theta\cos\theta

Using

\displaystyle l^2=\frac{S}{\pi\sin\theta(1+\sin\theta)},

we obtain

\displaystyle V=\frac13\pi\left(\frac{S}{\pi\sin\theta(1+\sin\theta)}\right)^{3/2}\sin^2\theta\cos\theta

Ignoring constant factors, it is sufficient to maximise

\displaystyle f(\theta)=\frac{\sqrt{\sin\theta}\,\cos\theta}{(1+\sin\theta)^{3/2}}

To simplify the differentiation, let

\displaystyle F(\theta)=f(\theta)^2=\frac{\sin\theta\,\cos^2\theta}{(1+\sin\theta)^3}

Since f(\theta)\gt0, both f(\theta) and F(\theta) attain their maximum value at the same point.

Let

u=\sin\theta

Then, \displaystyle F(u)=\frac{u(1-u^2)}{(1+u)^3}

\displaystyle =\frac{u(1-u)(1+u)}{(1+u)^3}

\displaystyle =\frac{u(1-u)}{(1+u)^2}

Differentiating with respect to u, we get

\displaystyle F'(u)=\frac{(1-2u)(1+u)^2-2u(1-u)(1+u)}{(1+u)^4}

\displaystyle =\frac{1-3u}{(1+u)^3}

Putting F'(u)=0, we get

1-3u=0

\displaystyle u=\frac13

Since u=\sin\theta, we obtain

\displaystyle \sin\theta=\frac13

Therefore,

\displaystyle \theta=\sin^{-1}\left(\frac13\right)

Now,

\displaystyle F''(u)=\frac{-2(2-3u)}{(1+u)^4}

At \displaystyle u=\frac13,

\displaystyle F''\left(\frac13\right)\lt0

Therefore, F(u) and hence the volume is maximum when

\displaystyle \sin\theta=\frac13.

Hence, the semi-vertical angle of the cone of given surface area and maximum volume is

\boxed{\sin^{-1}\left(\frac13\right)}.

27. The point on the curve x^2=2y which is nearest to the point (0,5) is

• (A) (2\sqrt2,4)
• (B) (2\sqrt2,0)
• (C) (0,0)
• (D) (2,2)

Solution

Let (x,y) be any point on the curve.

Since

x^2=2y

we have

\displaystyle y=\frac{x^2}{2}

The square of the distance of the point (x,y) from (0,5) is

D^2=x^2+(y-5)^2

Substituting \displaystyle y=\frac{x^2}{2}

We get, \displaystyle D^2=x^2+\left(\frac{x^2}{2}-5\right)^2

\displaystyle =x^2+\frac{x^4}{4}-5x^2+25

\displaystyle =\frac{x^4}{4}-4x^2+25

Let

\displaystyle f(x)=\frac{x^4}{4}-4x^2+25

To find the nearest point, we minimise D^2.

Differentiating with respect to x, we get

f'(x)=x^3-8x

=x(x^2-8)

Putting f'(x)=0, we get

x=0,\;\pm2\sqrt2

Now,

f''(x)=3x^2-8

At x=\pm2\sqrt2,

f''(\pm2\sqrt2)=16\gt0

\therefore x=\pm2\sqrt2 gives the minimum value of D^2.

Substituting x=\pm2\sqrt2 in \displaystyle y=\frac{x^2}{2}, we get

\displaystyle y=\frac{8}{2}=4

Hence, the nearest points on the curve are

(2\sqrt2,4) and (-2\sqrt2,4).

✅️ Among the given options, the correct answer is (A) (2\sqrt2,4).

28. For all real values of x, the minimum value of

\displaystyle \frac{1-x+x^2}{1+x+x^2}

is

• (A) 0
• (B) 1
• (C) 3
• (D) \displaystyle \frac13

Solution

Let

\displaystyle y=\frac{1-x+x^2}{1+x+x^2}

Since

1-x+x^2=(1+x+x^2)-2x

we get

\displaystyle y=1-\frac{2x}{1+x+x^2}

Therefore, the minimum value of y will occur when

\displaystyle \frac{2x}{1+x+x^2}

is maximum.

Let

\displaystyle f(x)=\frac{2x}{1+x+x^2}

Differentiating with respect to x,

We get, \displaystyle f'(x)=\frac{2(1+x+x^2)-2x(1+2x)}{(1+x+x^2)^2}

\displaystyle =\frac{2-2x^2}{(1+x+x^2)^2}

\displaystyle =\frac{2(1-x^2)}{(1+x+x^2)^2}

Putting f'(x)=0, we get

1-x^2=0

x=\pm1

Now,

\displaystyle f(1)=\frac23

and

\displaystyle f(-1)=-2

Since \displaystyle \frac23>-2, the maximum value of f(x) is attained at

x=1.

Therefore,

\displaystyle y_{\min}=1-\frac23

\displaystyle =\frac13

✅️ Hence, the minimum value is (D) \displaystyle \frac13.

29. The maximum value of \left[x(x-1)+1\right]^{1/3}, for 0\le x\le1 is

• (A) \left(\frac13\right)^{1/3}
• (B) \frac12
• (C) 1
• (D) 0

Solution

Let

\displaystyle f(x)=\left[x(x-1)+1\right]^{1/3}

Since the cube root function is strictly increasing, f(x) will attain its maximum value whenever

g(x)=x(x-1)+1

attains its maximum value.

Now,

g(x)=x^2-x+1

Differentiating with respect to x, we get

g'(x)=2x-1

Putting g'(x)=0, we get

2x-1=0

\displaystyle x=\frac12

Now,

g''(x)=2\gt0

Therefore, \displaystyle x=\frac12 gives a minimum value of g(x).

Hence, to find the maximum value on the interval [0,1], we check the endpoints.

g(0)=1

and

g(1)=1

Thus, the maximum value of g(x) on [0,1] is 1.

Therefore,

\displaystyle f_{\max}=\sqrt[3]{1}=1

✅️ Hence, the correct answer is (C) 1.