NCERT Solutions for Class 12 Maths Chapter 6 Application Of Derivatives Ex 6.3

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

1. $f\left(x\right)={(2x-1)}^2+3$ 

2. $f\left(x\right)=9x^2+12x+2$ 

3. $f\left(x\right)=-{(x-1)}^2+10$ 

4. $g\left(x\right)=x^3+1$ 

Ans: 

i) The given function is $f\left(x\right)={(2x-1)}^2+3$.

It can be observed that ${(2x-1)}^2\ge 0$ for every $x\in R$.

Therefore, $f\left(x\right)={(2x-1)}^2+3\ge 3$ for every $x\in R$.

The minimum value of $f$ is attained when $2x-1=0$ 

$2x-1=0,x=\frac{1}{2}$ 

Minimum value of $f\left(\frac{1}{2}\right)={(2.\frac{1}{2}-1)}^2+3=3.$ 

Hence, function $f$ does not have a maximum value.

ii) The given function is $f\left(x\right)=9x^2+12x+2={(3x^2+2)}^2-2$.

It can be observed that ${(3x^2+2)}^2\ge 0$ for every $x\in R$.

Therefore, $f\left(x\right)={(3x^2+2)}^2-2\ge -2$ for every $x\in R$.

The minimum value of $f$ is attained when $3x+2=0$ 

$3x+2=0,x=-\frac{2}{3}$ 

Minimum value of $f(-\frac{2}{3})={(3\left(\frac{-2}{3}\right)+2)}^2-2=-2.$ 

Hence, function $f$ does not have a maximum value.

iii) The given function is $f\left(x\right)=-{(x-1)}^2+10.$.

It can be observed that ${(x-1)}^2\ge 0$ for every $x\in R$.

Therefore, $f\left(x\right)=-{(x-1)}^2+10\le 10$ for every $x\in R$.

The minimum value of $f$ is attained when $(x-1)=0$ 

$x-1=0,x=0$ 

Minimum value of $f=f\left(1\right)=-{(1-1)}^2+10=10$ 

Hence, function $f$ does not have a maximum value.

vi) The given function is $g\left(x\right)=x^3+1$.

Hence, function $g$ neither has a maximum value nor a minimum value.

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

i) $f\left(x\right)=|x+2|-1$ 

Ans: We know that $|x+2|\ge 0$ for every $x\in R$ 

Therefore, $f\left(x\right)=|x+2|-1\ge -1$ for every $x\in R$

The minimum value of $f$ is attained when $|x+2|=0$ 

 $\Rightarrow x=-2$ 

Minimum value of $f=f(-2)=|-2+2|-1=-1$ 

Hence, function $f$ does not have a maximum value.

ii) $g\left(x\right)=-|x+1|+3$ 

Ans: We know that $-|x+1|\le 0$ for every $x\in R$ 

Therefore, $g\left(x\right)=-|x+1|+3\le 3$ for every $x\in R$

The minimum value of $g$ is attained when $|x+1|=0$ 

 $\Rightarrow x=-1$ 

Minimum value of $g=g(-1)=-|-1+1|+3=3$ 

Hence, function $g$ does not have a maximum value.

iii) $h\left(x\right)=\sin 2x+5$ 

Ans: We know that $-1\le \sin 2x\le 1$ 

$-1+5\le \sin 2x+5\le 1+5$  

$4\le \sin 2x+5\le 6$  

Hence, the maximum and minimum values of $h$ are $6\mathrm{\&}4$ respectively.

vi) $f\left(x\right)=|\sin 4x+3|$ 

Ans: We know that $-1\le \sin 4x\le 1$ 

$2\le \sin 4x+3\le 4$  

$2\le |\sin 4x+3|\le 4$  

Hence, the maximum and minimum values of $f$ are $4\mathrm{\&}2$ respectively.

v) $h\left(x\right)=x+4,x\in (-1,1)$ 

Ans: Here, if a point $x_0$ is closest to $-1$ , then we find $\frac{x_0}{2}+1<x_0+1$ for all $x_0\in (-1,1)$.

Also, if $x_1$ is closet to $-1$ , then we find$x_1+1<\frac{x_1+1}{2}+1$ for all $x_0\in (-1,1)$.

Hence, function $h\left(x\right)$ has neither maximum nor minimum value at $(-1,1).$

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

i) $f\left(x\right)=x^2$

Ans: $f\left(x\right)=x^2$ 

$\therefore f^{\prime }\left(x\right)=2x$ 

Now, $f^{\prime }\left(x\right)=0\Rightarrow x=0$ 

Thus, $x=0$ is the only critical point that could be the point of local maxima or local minima of $f.$ 

We have $f^{″}\left(0\right)=2$, which is positive.

Therefore, by second derivative test,$x=0$ is a point of local minima and local minimum value of $f$ at $x=0$ is $f\left(0\right)=0.$ 

ii) $g\left(x\right)=x^3-3x$

Ans: $g\left(x\right)=x^3-3x$ 

$\therefore g^{\prime }\left(x\right)=3x^2-3$ 

Now,

$g^{\prime }\left(x\right)=0\Rightarrow 3x^2=3\Rightarrow x=\pm 1$  

$g^{″}\left(x\right)=6x$  

$g^{″}\left(1\right)=6>0$  

$g^{″}(-1)=-6<0$  

By second derivative test, $x=1$ is a point of local minima and local minimum value of $g$ at $x=1$ is $g\left(1\right)=1^3-3=-2.$

iii) $h\left(x\right)=\sin x+\cos .0<x<\frac{\pi }{2}$

Ans: $h\left(x\right)=\sin x+\cos .0<x<\frac{\pi }{2}$ 

$\therefore h^{\prime }\left(x\right)=\cos x+\sin x$  

$h^{\prime }\left(x\right)=0\Rightarrow \sin x=\cos x\Rightarrow \tan x=1$  

$\Rightarrow x=\frac{\pi }{4}\in (0,\frac{\pi }{2})$  

$h^{″}\left(x\right)=-\sin x-\cos x$  

$h^{″}\left(\frac{\pi }{4}\right)=-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})$  

$=-\frac{1}{\sqrt{2}}=-\sqrt{2}<0$  

Therefore, by second derivative test, $x=\frac{\pi }{4}$ is a point of local maxima and the local maximum value of $h$ at $x=\frac{\pi }{4}$   

$h\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}$  

$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}.$

iv) $f\left(x\right)=\sin x-\cos x,0<x<2\pi$

Ans: $f\left(x\right)=\sin x-\cos x,0<x<2\pi$ 

$\therefore f^{\prime }\left(x\right)=\cos x+\sin x$  

$f^{\prime }\left(x\right)=0\Rightarrow \cos x=-\sin x$  

$\tan x=-1\Rightarrow x=\frac{3\pi }{4},\frac{7\pi }{4}\in (0,2\pi )$  

$f^{″}\left(x\right)=-\sin x+\cos x$  

$f^{″}\left(\frac{3\pi }{4}\right)=-\sin \frac{3\pi }{4}+\cos \frac{3\pi }{4}=-\sqrt{2}<0$  

$f^{″}\left(\frac{7\pi }{4}\right)=-\sin \frac{7\pi }{4}+\cos \frac{7\pi }{4}=\sqrt{2}>0$  

Therefore, by second derivative test, $x=\frac{3\pi }{4}$ is appoint of local maxima and the local maximum value of $f$ at $x=\frac{3\pi }{4}$ is 

$f\left(\frac{3\pi }{4}\right)=\sin \frac{3\pi }{4}\cos \frac{3\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}.$ However, $x=\frac{7\pi }{4}$ is a point of local minima and the local minimum value of $f$ at $x=\frac{7\pi }{4}$

is $f\left(\frac{7\pi }{4}\right)=\sin \frac{7\pi }{4}-\cos \frac{7\pi }{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2.}$

v) $f\left(x\right)=x^3-6x^2+9x+15$

Ans: $f\left(x\right)=x^3-6x+9x+15$ 

$\therefore f^{\prime }\left(x\right)=3x^2-12x+9$  

$f\left(x\right)=0\Rightarrow 3(x^2-4x+3)=0$  

$\Rightarrow 3(x-1)(x-3)=0$  

$\Rightarrow x=1,3$  

Now,

$f^{″}\left(x\right)=6x-12=6(x-2)$  

$f^{″}\left(1\right)=6(1-2)=-6<0$  

$f^{″}\left(3\right)=6(3-2)=-6>0$  

Therefore, by second derivative test, $x=1$ is a point of local maxima and the local maximum value of $f$ at $x=1$ is $f\left(1\right)=1-6+9+15=19$. However, $x=3$ is a point of local minima and the local minimum value of $f$ at $x=3$ is $f\left(3\right)=27-54+27+15=15.$

vi) $g\left(x\right)=\frac{x}{2}+\frac{2}{x},x>0$

Ans: $g\left(x\right)=\frac{x}{2}+\frac{2}{x},x>0$ 

$\therefore g^{\prime }\left(x\right)=\frac{1}{2}-\frac{2}{x^2}$  

$Now,$  

$g^{\prime }\left(x\right)=0gives\frac{2}{x^2}=\frac{1}{2}$  

$\Rightarrow x^2=4\Rightarrow x=\pm 2$  

Since $x>0$ we take $x=2$.

Now, 

$g^{″}\left(x\right)=\frac{4}{x^3}$  

$g^{″}\left(2\right)=\frac{4}{2^3}=\frac{1}{2}>0$  

Therefore, by second derivative test,  $x=2$ is a point of local minima and the local minimum value of $g$ at $x=2$ is $g\left(2\right)=\frac{2}{2}+\frac{2}{2}=1+1=2.$

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

Ans: $g\left(x\right)=\frac{1}{x^2+2}$ 

$\therefore g^{\prime }\left(x\right)=\frac{-\left(2x\right)}{{(x^3+2)}^2}$  

$g^{\prime }\left(x\right)=0\Rightarrow \frac{-\left(2x\right)}{{(x^3+2)}^2}=0\Rightarrow x=0$  

Now, for values close to $x=0$ and to the left of $0,g^{\prime }\left(x\right)>0$. Also, for values close to $x=0$ and to the right of $g^{\prime }\left(x\right)<0.$ 

Therefore, by first derivative test $x=0$ is a point of local maxima and the local maximum value of $g\left(0\right)is\frac{1}{0+2}=\frac{1}{2}.$

viii) $f\left(x\right)=x\sqrt{1-x},x>0$

Ans: $f\left(x\right)=x\sqrt{1-x},x>0$ 

  $\therefore f^{\prime }\left(x\right)=x\sqrt{1-x}+x.\frac{1}{2\sqrt{1-x}}(-1)=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}$  

 $=\frac{2(1-x)-x}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}$  

 $f^{\prime }\left(x\right)=0\Rightarrow \frac{2-3x}{2\sqrt{1-x}}=0$  

 $\Rightarrow 2-3x=0\Rightarrow x=\frac{2}{3}$  

 $f^{″}\left(x\right)=\frac{1}{2}\left[\frac{\sqrt{1-x}(-3)-(2-3x)\left(\frac{-1}{2\sqrt{1-x}}\right)}{1-x}\right]$  

 $=\frac{\sqrt{1-x}(-3)+2(2-3x)\left(\frac{1}{2\sqrt{1-x}}\right)}{2(1-x)}$  

 $=\frac{3x-4}{4{(1-x)}^{\frac{3}{2}}}$  

 $f^{\prime }\left(\frac{2}{3}\right)=\frac{2-4}{4{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}=\frac{-1}{2{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}<0$   

Therefore, by second derivative test, $x=\frac{2}{3}$ is a point of local 

maxima and the local maximum value of $f$ at $x=\frac{2}{3}$ is $f\left(\frac{2}{3}\right)=\frac{2-4}{4{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}=\frac{-1}{2{\left(\frac{1}{3}\right)}^{\frac{3}{2}}}<0$.

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

i) $f\left(x\right)=e^x$

Ans: $f\left(x\right)=e^x$ 

$\therefore f^{\prime }\left(x\right)=e^x$ 

Now, if  $f^{\prime }\left(x\right)=0$ then $e^x=0$. But the exponential function can never assume $0$ for any value of $x$.

Therefore, there does not exist $c\in R$ such that $f^{\prime }\left(c\right)=0$.

Hence, function $f$ does not have maxima or minima.

ii) $g\left(x\right)=\log x$

Ans: We have,

$g\left(x\right)=\log x$  

$\therefore g^{\prime }\left(x\right)=\frac{1}{x}$  

Since $\log x$ is defined for a positive number $x,g^{\prime }\left(x\right)>0$ for any $x$

Therefore, there does not exist  $c\in R$ such that $g^{\prime }\left(c\right)=0$.

Hence, function $g$ does not have maxima or minima.

iii) $h\left(x\right)=x^3+x^2+x+1$

Ans: We have,

$h\left(x\right)=x^3+x^2+x+1$  

$\therefore h^{\prime }\left(x\right)=3x^2+2x+1$  

Now,

Therefore, there does not exist $c\in R$ such that $h^{\prime }\left(c\right)=0$.

Hence, function $h$ does not have maxima or minima.

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

i) $f\left(x\right)=x^3,x\in [-2,2]$ 

Ans: The given function is $f\left(x\right)=x^3$.

$\therefore f^{\prime }\left(x\right)=3x^2$ 

Now,

$f^{\prime }\left(x\right)=0\Rightarrow x=0$ 

Then, we evaluate the value of $f$ at critical point $x=0$ and at endpoints of the interval $[-2,2]$.

$f\left(0\right)=0$  

 $f(-2)={(-2)}^3=-8$  

 $f\left(2\right)={\left(2\right)}^3=8$  

Hence, we can conclude that the absolute maximum value of $f$ on $[-2,2]$ is $8$ occurring at $x=-2$. Also, the absolute minimum value of $f$  on $[-2,2]$ is $-8$ occurring at $x=-2$.

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

Ans: The given function is $f\left(x\right)=\sin x+\cos x$ 

$\therefore f^{\prime }\left(x\right)=\cos x-\sin x$ 

Now,

$f^{\prime }\left(x\right)=0\Rightarrow \sin x=\cos x$  

$\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4}$  

Then, we evaluate the value of $f$ at critical point $x=\frac{\pi }{4}$ and at the endpoints of the interval $[0,\pi ]$.

  $f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$  

 $f\left(0\right)=\sin 0+\cos 0=1$  

 $f\left(\pi \right)=\sin \pi +\cos \pi =1$  

Hence, we can conclude that the absolute maximum value of $f$ on $[0,\pi ]$ is $\sqrt{2}$ occurring at $x=\frac{\pi }{4}$. Also, the absolute minimum value of $f$ on $[0,\pi ]$ is $-1$ occurring at $x=\pi$.

iii) $f\left(x\right)=4x-\frac{1}{2}{x}^2,x\in [-2,\frac{9}{2}]$ 

Ans: The given function is $f\left(x\right)=4x-\frac{1}{2}{x}^2$ 

$\therefore f^{\prime }\left(x\right)=4x-\frac{1}{2}{x}^2$

Now,

$f^{\prime }\left(x\right)=0\Rightarrow x=4$ 

Then, we evaluate the value of $f$ at critical point $x=4$ and at the endpoints of the interval $[-2,\frac{9}{2}]$.

  $f\left(4\right)=16-\frac{1}{2}\left(16\right)=8$  

 $f(-2)=-8-\frac{1}{2}\left(4\right)=-10$  

 $f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}{\left(\frac{9}{2}\right)}^2=18-\frac{81}{8}=7.875$  

Hence, we can conclude that the absolute maximum value of $f$ on $[-2,\frac{9}{2}]$ is $8$ occurring at $x=4$. Also, the absolute minimum value of $f$ on $[-2,\frac{9}{2}]$ is $-10$ occurring at $x=-2$.

iv) $f\left(x\right)={(x-1)}^2+3,x\in [-3,1]$

Ans: The given function is $f\left(x\right)={(x-1)}^2+3$ 

$\therefore f^{\prime }\left(x\right)=2(x-1)$

Now,

$f^{\prime }\left(x\right)=0\Rightarrow 2(x-1)=0,x=1$ 

Then, we evaluate the value of x at critical point $x=1$ and at the endpoints of the interval $[-3,1]$.

  $f\left(1\right)={(1-1)}^2+3=3$  

 $f(-3)={(-3-1)}^2+3=19$  

Hence, we can conclude that the absolute maximum value of f on $[-3,1]$ is $19$ taking place at $x=-3$. Also, the absolute minimum value of f on $[-3,1]$ is $3$ taking place at $x=1$.

6. Find the maximum profit that a company can make if the profit function is given by $p\left(x\right)=41-24x-18x^2$.

Ans: The profit function is given as $p\left(x\right)=41-24x-18x^2$.

$\therefore p^{\prime }\left(x\right)=-24-36x$  

$p^{″}\left(x\right)=-36$  

$Now,$  

$p^{\prime }\left(x\right)=0$  

$\Rightarrow x=\frac{-24}{36}=-\frac{2}{3}$  

$Also,$  

$p^{″}\left(\frac{-2}{3}\right)=-36<0$  

By second derivatives test, $x=-\frac{2}{3}$ is the point of local maximum of $p$

Therefore, Maximum profit = $p(-\frac{2}{3})$ 

$=41-24(-\frac{2}{3})-18{(-\frac{2}{3})}^2$  

$=41+16-8$  

$=49$  

Hence, the maximum profit that the company can make is $49$ units.

7. Find both the maximum value and the minimum value of $3x^4-8x^3+12x^2-48x+25$ on the interval [0,3].

Ans:  Let’s take f(x) = $3x^4-8x^3+12x^2-48x+25$ on [0,3]

$f^{\prime }\left(x\right)=12x^3-24x^2+24x-48$

Now $f^{\prime }\left(x\right)=0$

$12x^3-24x^2+24x-48=0$

$x^3-2x^2+2x-4=0$

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

$x=2orx=\pm \sqrt{2}$

As $x=\pm \sqrt{2}$ is imaginary, therefore this value is not considered.

At x=2, f(2) = 3(16)-8(8)+12(4)-48(2)+25 = -39

At x=0, f(0) = 3(0)-8(0)+12(0)-48(0)+25 = 25

At x=3, f(3) = 3(81)-8(27)+12(9)-48(3)+25 = 16

Therefore, the minimum value of the given function is -39 and maximum value is 25.

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

Ans: Let 

$f\left(x\right)=\sin 2x$  

$\therefore f^{\prime }\left(x\right)=2\cos 2x$  

$Now,$  

$f^{\prime }\left(x\right)=0\Rightarrow \cos 2x=0$  

$\Rightarrow 2x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}$  

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

Then, we evaluate the values of $f$ at critical points 

$x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ and at the endpoints of

the interval $[0,2\pi ]$.

$f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{2}=1,$  

$f\left(\frac{3\pi }{4}\right)=\sin \frac{3\pi }{2}=-1,$  

$f\left(\frac{5\pi }{4}\right)=\sin \frac{5\pi }{2}=1,$  

$f\left(\frac{7\pi }{4}\right)=\sin \frac{7\pi }{2}=-1,$  

$f\left(0\right)=\sin 0=0,$  

$f\left(2\pi \right)=\sin 2\pi =0$  

Hence, we can conclude that the absolute maximum value of 

$f[0,2\pi ]$ is occurring at $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}.$

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

Ans: Let $f\left(x\right)=\sin x+\cos x$ 

$\therefore f^{\prime }\left(x\right)=\cos x-\sin x$  

$f^{\prime }\left(x\right)=0\Rightarrow \sin x=\cos x$  

$\Rightarrow \tan x=1\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},...$  

$f^{\prime }\left(x\right)=-\sin x-\cos x=-(\sin x+\cos x)$  

Now, when $(\sin x+\cos x)$ is positive, i.e., 

when sin x and cos x are both positive, f” (x) will be negative.

We also know that in the first quadrant, both sin x and cos x are positive. Then, $f^{″}\left(x\right)$ will be 

negative when $x\in (0,\frac{\pi }{2})$.

As a result, we consider $x=\frac{\pi }{4}$ 

$f^{″}\left(\frac{\pi }{4}\right)=-(\sin \frac{\pi }{4}+\cos \frac{\pi }{4})=-\left(\frac{2}{\sqrt{2}}\right)=-\sqrt{2}<0$ 

By second derivative test, $f$ will be the maximum at $x=\frac{\pi }{4}$ and the 

maximum value of $f$ is  $f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}.$

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

Find the maximum value of the same function in $[-3,-1]$.

Ans: Let $f\left(x\right)=2x^3-24x-107$    

$\therefore f^{\prime }\left(x\right)=6x^2-24=6(x^2-4)$  

$Now,$  

$f^{\prime }\left(x\right)=0\Rightarrow 6(x^2-4)=0$  

$\Rightarrow x^2=4\Rightarrow x=\pm 2$  

We first consider the interval $[1,3]$.

Then, we evaluate the value of $f$ at the critical point $x\in [1,3]$ and at the endpoints of the interval $[1,3]$.

$f\left(2\right)=2\left(8\right)-24\left(2\right)+107=75$  

$f\left(1\right)=2\left(1\right)-24\left(1\right)+107=85$  

$f\left(3\right)=2\left(27\right)-24\left(3\right)+107=89$  

Hence, the absolute maximum value of $f\left(x\right)$ in the interval $[1,3]$ is

$89$ occurring at $x=3$.

Next, we consider the interval $[-3,-1]$.

Evaluate the value of $f$ at the critical point $x\in [1,3]$ and at 

the endpoints of the interval $[-3,-1]$.

$f(-3)=2(-27)-24(-3)+107=125$  

$f(-1)=2(-1)-24(-1)+107=129$  

$f(-2)=2(-8)-24(-2)+107=139$  

Hence, the absolute maximum value of $f\left(x\right)$ in the interval $[-3,-1]$ is

$139$ occurring at $x=-2$.

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

Ans: Let $f\left(x\right)=x^4-62x^2+ax+9$ 

$\therefore f^{\prime }\left(x\right)=4x^2-124x+a$ 

It is given that function $f$ attains its maximum value on the interval 

$[0,2]$ at $x=1$.

$\therefore f^{\prime }\left(1\right)=0$  

$\Rightarrow 4-124+a=0$  

$\Rightarrow a=120$  

Hence, the value of $a$ is $120.$

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

Ans: Let $f\left(x\right)=x+\sin 2x$ 

$\therefore f^{\prime }\left(x\right)=1+2\cos 2x$  

$Now,f^{\prime }\left(x\right)=0\Rightarrow \cos 2x=-\frac{1}{2}=\cos \frac{2\pi }{3}$  

$2x=2\pi \pm \frac{2\pi }{3},n\in Z$  

$\Rightarrow x=\pi \pm \frac{\pi }{3},n\in Z$  

$\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}\in [0,2\pi ]$  

Then, we evaluate the value of at $f$critical points $\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$ and the end points of the interval $[0,2\pi ]$.

$f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}+\sin \frac{2\pi }{3}=\frac{\pi }{3}+\frac{\sqrt{3}}{2}$  

$f\left(\frac{2\pi }{3}\right)=\frac{2\pi }{3}+\sin \frac{4\pi }{3}=\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$  

$f\left(\frac{4\pi }{3}\right)=\frac{4\pi }{3}+\sin \frac{8\pi }{3}=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}$  

$f\left(\frac{5\pi }{3}\right)=\frac{5\pi }{3}+\sin \frac{10\pi }{3}=\frac{5\pi }{3}-\frac{\sqrt{3}}{2}$  

$f\left(0\right)=0+\sin 0=0$  

$f\left(2\pi \right)=2\pi +\sin 4\pi =2\pi$  

Hence, we can conclude that the absolute maximum value of $f\left(x\right)$ in 

the interval $[0,2\pi ]$ is $2\pi$ occurring at $x=2\pi$ and the absolute maximum value of $f\left(x\right)$ in the interval $[0,2\pi ]$ is $0$ occurring at $x=0.$

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

Ans: Let one number be $x$. Then, the other number is $(24-x)$.

Let $p\left(x\right)$ denote the product of the two numbers. Thus, we have:

$P\left(x\right)=x(24-x)=24x-x^2$  

$\therefore P^{\prime }\left(x\right)=24-2x$  

$P^{″}\left(x\right)=-2$  

$Now,$  

$P^{\prime }\left(x\right)=0\Rightarrow x=12$  

$Also,$  

$P^{″}\left(12\right)=-2<0$  

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 $24-12=12.$ 

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

Ans: It is assumed that the two numbers are $x$ and $y$ such that $x+y=60$.

$y=60-x$ 

Let $f\left(x\right)=x{y}^3$ 

$\Rightarrow f\left(x\right)=x{(60-x)}^3$  

$\therefore f^{\prime }\left(x\right)={(60+x)}^3-3x{(60-x)}^2$  

$={(60+x)}^3[60-4x]$  

$And,$  

$f^{″}\left(x\right)=-2(60-x)(60-4x)-4{(60-x)}^2$  

$=-2(60-x)(180-6x)$  

$=-12(60-x)(30-x)$  

$Now,f^{\prime }\left(x\right)=0$  

$\Rightarrow x=60,15$  

When $x=60,f^{″}\left(x\right)=0$ 

When $x=15,f^{″}\left(x\right)=-12(60-15)(30-15)<0.$ 

By second derivative test, $x=15$ is a point of local maxima of f.

Thus, function $x{y}^3$ maximum when $x=15$ and $y=60-15=45$.

Hence, the required numbers are 15 and 45.

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

Ans: Let one number be x. Then, the other number is $y=(35-x)$ 

Let $p\left(x\right)=x^2{y}^5$. Then, we have:

$P\left(x\right)=x^2{(35-x)}^5$  

$\therefore P^{\prime }\left(x\right)=2x{(35-x)}^5-5x^2{(35-x)}^4$