NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.5

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.5) Exercise 6.5

NCERT Maths Class 12 Exercise 6.5 Solutions deal with the Application of Derivatives. It is one of the most important chapters according to the CBSE aspect. The following article provides a vivid analysis regarding the exercise 6.5 class 12 maths along with a few other significant aspects. The article also provides an example of types of questions given in the ex 6.5 class 12 Maths NCERT solutions. You are suggested to go through the article carefully to get a firm grip on the chapter. 

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Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives part-1

Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives

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

  1. $f\left( x \right)={{\left( 2x-1 \right)}^{2}}+3$ 

  2. $f\left( x \right)=9{{x}^{2}}+12x+2$ 

  3. $f\left( x \right)=-{{\left( x-1 \right)}^{2}}+10$ 

  4. $g\left( x \right)={{x}^{3}}+1$ 

Ans: 

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

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

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

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

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

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

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

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

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

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

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

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

$x-1=0,x=0$ 

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

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) $f\left( x \right)=\left| x+2 \right|-1$ 

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

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

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

 $\Rightarrow x=-2$ 

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

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

ii) $g\left( x \right)=-\left| x+1 \right|+3$ 

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

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

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

 $\Rightarrow x=-1$ 

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

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 $6\And 4$ respectively.

vi) $f\left( x \right)=\left| \sin 4x+3 \right|$ 

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

$  2\le \sin 4x+3\le 4 $  

$  2\le \left| \sin 4x+3 \right|\le 4 $  

Hence, the maximum and minimum values of $f$ are $4\And 2$ respectively.

v) $h\left( x \right)=x+4,x\in \left( -1,1 \right)$ 

Ans: Here, if a point ${{x}_{0}}$ is closest to $-1$ , then we find $\frac{{{x}_{0}}}{2}+1<{{x}_{0}}+1$ for all ${{x}_{0}}\in \left( -1,1 \right)$.

Also, if ${{x}_{1}}$ is closet to $-1$ , then we find${{x}_{1}}+1<\frac{{{x}_{1}}+1}{2}+1$ for all ${{x}_{0}}\in \left( -1,1 \right)$.

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

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: $f\left( x \right)={{x}^{2}}$ 

$\therefore f'\left( x \right)=2x$ 

Now, $f'\left( x \right)=0\Rightarrow x=0$ 

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: $g\left( x \right)={{x}^{3}}-3x$ 

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

Now,

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

$  {g}''\left( x \right)=6x $  

$  {g}''\left( 1 \right)=6>0 $  

$  {g}''\left( -1 \right)=-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}'\left( x \right)=\cos x+\sin x $  

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

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

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

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

$  =-\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}'\left( x \right)=\cos x+\sin x $  

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

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

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

Ans: $f\left( x \right)={{x}^{3}}-6x+9x+15$ 

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

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

$  \Rightarrow 3\left( x-1 \right)\left( x-3 \right)=0 $  

$  \Rightarrow x=1,3 $  

Now,

$  {f}''\left( x \right)=6x-12=6\left( x-2 \right) $  

$  {f}''\left( 1 \right)=6\left( 1-2 \right)=-6<0 $  

$  {f}''\left( 3 \right)=6\left( 3-2 \right)=-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}'\left( x \right)=\frac{1}{2}-\frac{2}{{{x}^{2}}} $  

$  Now, $  

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

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

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

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

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

 $  {f}'\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}\left( -3 \right)-\left( 2-3x \right)\left( \frac{-1}{2\sqrt{1-x}} \right)}{1-x} \right] $  

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

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

 $  {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 $   

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: $f\left( x \right)={{e}^{x}}$ 

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

Now, if  ${f}'\left( x \right)=0$ 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 ${f}'\left( c \right)=0$.

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}'\left( x \right)=\frac{1}{x} $  

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

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

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

Now,

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

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

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

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

Now,

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

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

$  f\left( 0 \right)=0 $  

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

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

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

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

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

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

Now,

$  {f}'\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 $\left[ 0,\pi  \right]$.

  $  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 $\left[ 0,\pi  \right]$ is $\sqrt{2}$ occurring at $x=\frac{\pi }{4}$. Also, the absolute minimum value of $f$ on $\left[ 0,\pi  \right]$ is $-1$ occurring at $x=\pi $.

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

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

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

Now,

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

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

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

 $  f\left( -2 \right)=-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 $\left[ -2,\frac{9}{2} \right]$ is $8$ occurring at $x=4$. Also, the absolute minimum value of $f$ on $\left[ -2,\frac{9}{2} \right]$ is $-10$ occurring at $x=-2$.

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

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

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

Now,

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

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

  $  f\left( 1 \right)={{\left( 1-1 \right)}^{2}}+3=3 $  

 $  f\left( -3 \right)={{\left( -3-1 \right)}^{2}}+3=19 $  

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

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

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

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

$  {p}''\left( x \right)=-36 $  

$  Now, $  

$  {p}'\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 = $p\left( -\frac{2}{3} \right)$ 

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

$  =41+16-8 $  

$  =49 $  

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

7. Find the intervals in which the function $f$ given by 

$f\left( x \right)={{x}^{3}}+\frac{1}{{{x}^{3}}},x\ne 0$ is 

i. Increasing

ii. Decreasing

Ans: We have,

$  f\left( x \right)={{x}^{3}}+\frac{1}{{{x}^{3}}} $  

 $  \therefore {f}'\left( x \right)=3{{x}^{2}}-\frac{3}{{{x}^{4}}}=\frac{3{{x}^{6}}-3}{{{x}^{4}}} $  

Then, 

 $  {f}'\left( x \right)=0\Rightarrow 3{{x}^{6}}-3=0 $  

$  \Rightarrow {{x}^{6}}=1\Rightarrow x=\pm 1 $  

Now, the points $x=1$ and $x=-1$  divided the real line into three 

disjoint intervals i.e., $\left( -\infty ,-1 \right),\left( -1,1 \right)and\left( 1,\infty  \right).$ 

In intervals $\left( -\infty ,-1 \right)and\left( 1,\infty  \right)$ i.e., when $x<-1$ and $x>1,{f}'\left( x \right)>0.$ 

Thus, when $x<-1$ and $x>1,f$ is increasing.

In intervals $\left( -1,1 \right)$ i.e., when $-1<x<1,{f}'\left( x \right)<0.$ and 

Thus, when $-1<x<1,f$ is decreasing.

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

Ans: Let 

$  f\left( x \right)=\sin 2x $  

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

$  Now, $  

$  {f}'\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 $\left[ 0,2\pi  \right]$.

$  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 

$f\left[ 0,2\pi  \right]$ 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}'\left( x \right)=\cos x-\sin x $  

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

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

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

Now, when $\left( \sin x+\cos x \right)$ 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 $x\in \left( 0,\frac{\pi }{2} \right)$.

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

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

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 $2{{x}^{3}}-24x+107$ in the interval $\left[ 1,3 \right]$ 

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

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

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

$  Now, $  

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

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

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

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

$  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 $\left[ 1,3 \right]$ is

$89$ occurring at $x=3$.

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

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

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

$  f\left( -3 \right)=2\left( -27 \right)-24\left( -3 \right)+107=125 $  

$  f\left( -1 \right)=2\left( -1 \right)-24\left( -1 \right)+107=129 $  

$  f\left( -2 \right)=2\left( -8 \right)-24\left( -2 \right)+107=139 $  

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

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

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

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

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

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

$\left[ 0,2 \right]$ at $x=1$.

$  \therefore {f}'\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 $\left[ 0,2\pi  \right].$ 

Ans: Let $f\left( x \right)=x+\sin 2x$ 

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

$  Now,{f}'\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 \left[ 0,2\pi  \right] $  

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 $\left[ 0,2\pi  \right]$.

$  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 $\left[ 0,2\pi  \right]$ is $2\pi $ occurring at $x=2\pi $ and the absolute maximum value of $f\left( x \right)$ in the interval $\left[ 0,2\pi  \right]$ 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 $\left( 24-x \right)$.

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

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

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

$  {P}''\left( x \right)=-2 $  

$  Now, $  

$  {P}'\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{{\left( 60-x \right)}^{3}} $  

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

$  ={{\left( 60+x \right)}^{3}}\left[ 60-4x \right] $  

$  And, $  

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

$  =-2\left( 60-x \right)\left( 180-6x \right) $  

$  =-12\left( 60-x \right)\left( 30-x \right) $  

$  Now,{f}'\left( x \right)=0 $  

$  \Rightarrow x=60,15 $  

When $x=60,{f}''\left( x \right)=0$ 

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

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 $y=\left( 35-x \right)$ 

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

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

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

$  =x{{\left( 35-x \right)}^{4}}\left[ 2\left( 35-x \right)-5x \right] $  

$  =7x{{\left( 35-x \right)}^{4}}\left( 10-x \right) $  

$  And, $  

$  {P}''\left( x \right)=7{{\left( 35-x \right)}^{4}}\left( 10-x \right)+7x\left[ -{{\left( 35-5 \right)}^{4}}-4{{\left( 35-x \right)}^{3}}\left( 10-x \right) \right] $  

$  =7{{\left( 35-x \right)}^{4}}\left( 10-x \right)-7x{{\left( 35-x \right)}^{4}}-28x{{\left( 35-x \right)}^{3}}\left( 10-x \right) $  

$  =7{{\left( 35-x \right)}^{3}}\left[ 350-45x+{{x}^{2}}-35x+{{x}^{2}}-40x+4{{x}^{2}} \right] $  

$  =7{{\left( 35-x \right)}^{3}}\left( 6{{x}^{2}}-120x+350 \right) $  

$  Now,{P}'\left( x \right)=0\Rightarrow x=0,x=35,x=10 $  

$  When,x=35,{f}'\left( x \right)=f\left( x \right)=0 $  

$  y=35-35=0. $  

This makes the product equal to 0.

When $x=0,y=35-0=35$ and the product ${{x}^{2}}{{y}^{5}}$ will be 0.

$x=0$ and $x=35$ cannot be the possible values of x.

When $x=10$ we have:

${P}''\left( x \right)=7{{\left( 35-10 \right)}^{3}}\left( -250 \right)<0$ 

By second derivative test, P(x) will be the maximum when $x=10$ and 

$y=35-10=25$.

Hence, the required numbers are 10 and 25.

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

Ans: Let one number be x. Then, the other number is $\left( 16-x \right)$.

Let the sum of the cubes of these numbers be denoted by S(x). Then, 

$  S\left( x \right)={{x}^{3}}+{{\left( 16-x \right)}^{3}} $  

$  \therefore {S}'\left( x \right)=3{{x}^{2}}-3{{\left( 16-x \right)}^{2}}, $  

$  {S}''\left( x \right)=6x+6\left( 16-x \right) $  

$  Now,{S}'\left( x \right)=0\Rightarrow 3{{x}^{2}}-3{{\left( 16-x \right)}^{2}}=0 $  

$  \Rightarrow {{x}^{2}}-{{\left( 16-x \right)}^{2}}=0 $  

$  \Rightarrow x=\frac{256}{32}=8 $  

$  Now, $  

$  {S}''\left( 8 \right)=6\left( 8 \right)+6\left( 16-8 \right)=96>0 $  

By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 –8= 8.

17. A square piece of tin od side 18 cm is to be made into a box without a 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?

Ans: Let the side of the square be cut off by x cm. 

Then, the length and the breath of the box will be (18 – 2x) cm each.

The box has a height of x cm. As a result, the box's volume V(x) is given by,

$  V\left( x \right)=x{{\left( 18-2x \right)}^{2}} $  

$  \therefore {V}'\left( x \right)={{\left( 18-2x \right)}^{2}}-4x\left( 18-2x \right) $  

$  =\left( 18-2x \right)\left[ 18-2x-4x \right] $  

$  =12\left( 9-x \right)\left( 3-x \right) $  

$  And, $  

$  {V}'\left( x \right)=12\left[ -\left( 9-x \right)-\left( 3-x \right) \right] $  

$  =-24\left( 6-x \right) $  

$  Now,{v}'\left( x \right)=0\Rightarrow x=9,3 $  

If x = 9, then the length and the breadth will become 0.

$  \therefore x=9 $  

$  \Rightarrow x=3 $  

$  Now,{V}''\left( 3 \right)=-24\left( 6-3 \right)=-72<0 $  

Therefore, by second derivative test, x = 3 is the point of maxima of V.

As a result, the volume of the box obtained is the biggest achievable by removing a square of side 3 cm from each corner of the square tin and building a box from the remaining sheet.

18. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without a 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 the maximum possible?

Ans: Let the side of the square be cut by x cm.

Then, the height of the box is x, the length is $45-2x$ , and the breadth is $24-2x$.

$  V\left( x \right)=x\left( 45-2x \right)\left( 24-2x \right) $  

$  =4{{x}^{3}}-138{{x}^{2}}+1080x $  

$  {V}'\left( x \right)=12{{x}^{2}}-276+1080 $  

$  =12\left( {{x}^{2}}-23x+90 \right) $  

$  =12\left( x-18 \right)\left( x-5 \right) $  

$  {V}''\left( x \right)=24x-276=12\left( 2x-23 \right) $  

$  Now,{V}'\left( x \right)=0\Rightarrow x=18,5 $  

It is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet,

Thus, x cannot be equal to 18.

$  x=5 $  

$  Now,{V}''\left( 5 \right)=12\left( 10-23 \right)=-156<0 $  

Therefore, by second derivative test x = 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5cm.

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

Ans: Let a rectangle of length l and breadth b be inscribed in the given circle of radius a.

Then, the diagonal of length 2a cm passes through the center.

(Image Will Be Updated Soon)

By Pythagoras theorem, we have:

$  {{\left( 2a \right)}^{2}}={{l}^{2}}+{{b}^{2}} $  

$  \Rightarrow {{b}^{2}}=4{{a}^{2}}-{{l}^{2}} $  

$  \Rightarrow b=\sqrt{4{{a}^{2}}-{{l}^{2}}} $  

Area of the rectangle,$A=l\sqrt{4{{a}^{2}}-{{l}^{2}}}$ 

$  \therefore \frac{dA}{dl}=\sqrt{4{{a}^{2}}-{{l}^{2}}}+l\frac{1}{2\sqrt{4{{a}^{2}}-{{l}^{2}}}}\left( -2l \right)=\sqrt{4{{a}^{2}}-{{l}^{2}}}+\frac{{{l}^{2}}}{\sqrt{4{{a}^{2}}-{{l}^{2}}}}=\frac{4{{a}^{2}}-2{{l}^{2}}}{\sqrt{4{{a}^{2}}-{{l}^{2}}}} $  

$  \frac{{{d}^{2}}A}{d{{l}^{2}}}=\frac{\sqrt{4{{a}^{2}}-{{l}^{2}}}\left( -4l \right)-\left( 4{{a}^{2}}-2{{l}^{2}} \right)\frac{\left( -2l \right)}{2\sqrt{4{{a}^{2}}-{{l}^{2}}}}}{\left( 4{{a}^{2}}-{{l}^{2}} \right)} $  

$  =\frac{\left( 4{{a}^{2}}-{{l}^{2}} \right)\left( -4l \right)+1\left( 4{{a}^{2}}-2{{l}^{2}} \right)}{{{\left( 4{{a}^{2}}-{{l}^{2}} \right)}^{\frac{3}{2}}}} $  

$  =\frac{-2l\left( 6{{a}^{2}}-{{l}^{2}} \right)}{{{\left( 4{{a}^{2}}-{{l}^{2}} \right)}^{\frac{3}{2}}}} $  

$Now,\frac{dA}{dl}=0$ gives $4{{a}^{2}}=2{{l}^{2}}\Rightarrow l=\sqrt{2a}$ 

$\Rightarrow b=\sqrt{4{{a}^{2}}-2{{a}^{2}}}=\sqrt{2{{a}^{2}}}=\sqrt{2a}$ 

Now, when $l=\sqrt{2a}$ 

$\frac{{{d}^{2}}A}{d{{l}^{2}}}=\frac{-2\left( \sqrt{2a} \right)\left( -6{{a}^{2}}-2{{a}^{2}} \right)}{2\sqrt{2{{a}^{3}}}}=\frac{-8\sqrt{2{{a}^{3}}}}{2\sqrt{2{{a}^{3}}}}=-4<0$ 

By second derivative test, when $l=\sqrt{2a}$ then the area of the rectangle is the maximum.

Since $l=b=\sqrt{2a}$the rectangle is a square.

Hence, it has been proved that the square has the maximum area of all the rectangles inscribed in the given fixed circle. 

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

Ans: Let r and h denote the cylinder’s radius and height respectively.

Then, the cylinder has the following Surface Area (S):

$  S=2\pi {{r}^{2}}+2\pi rh $  

$  \Rightarrow h=\frac{S-2\pi {{r}^{2}}}{2\pi r} $  

$  =\frac{S}{2\pi }\left( \frac{1}{r} \right)-r $  

Let V be the volume of the cylinder. Then,

$  V=\pi {{r}^{2}}h=\pi {{r}^{2}}\left[ \frac{S}{2\pi }\left( \frac{1}{r} \right)-r \right]=\frac{Sr}{2} $  

$  =-\pi {{r}^{3}}=\frac{S}{2\pi }\left( \frac{1}{r} \right)-r $  

$  Then,\frac{dV}{dr}=\frac{S}{2}-3\pi {{r}^{2}},\frac{{{d}^{2}}V}{d{{r}^{2}}}=-6\pi r $  

$  Now,\frac{dV}{dr}=0\Rightarrow \frac{S}{2}=3\pi {{r}^{2}}\Rightarrow {{r}^{2}}=\frac{S}{6\pi } $  

When ${{r}^{2}}=\frac{S}{6\pi }$ then $h=\frac{6\pi {{r}^{2}}}{2\pi }\left( \frac{1}{r} \right)-r=3r-r=2r$ 

Hence, the volume is the maximum when the height is twice the radius i.e. when the height is equal to the diameter.

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

Ans: Let r and h denote the cylinder’s radius and height respectively.

Then, the cylinder has the following volume (V),

$  V=\pi {{r}^{2}}h=100 $  

$  \therefore h=\frac{100}{\pi {{r}^{2}}} $  

The cylinder has the following surface area (S),

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

$  \frac{dS}{dr}=0\Rightarrow 4\pi r=\frac{200}{{{r}^{2}}} $  

$  \Rightarrow {{r}^{3}}=\frac{200}{4\pi }=\frac{50}{\pi } $  

$  \Rightarrow r={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}} $  

Now, it is observed that when $r={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}},\frac{{{d}^{2}}S}{d{{r}^{2}}}>0.$ 

By second derivative test, when the radius of the cylinder is ${{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}$ cm, the surface area is the minimum.

When $r={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}},h=2{{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}cm.$ 

Hence the required dimensions of the can which has the minimum surface area are given as $radius={{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}cm$ and $height=2{{\left( \frac{50}{\pi } \right)}^{\frac{1}{3}}}cm.$ 

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 circle is minimum?

Ans: Let a piece of length l be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length $\left( 28-l \right)m.$ 

Let r be the circle's radius. Then $2\pi r=28-l\Rightarrow r=\frac{1}{2\pi }\left( 28-l \right).$ 

The combined areas of the square and the circle (A) is given by,

$A={{\pi }^{2}}+{{r}^{2}}$ (side of the square)

$  =\frac{{{l}^{2}}}{16}+\pi {{\left[ \frac{1}{2\pi }\left( 28-l \right) \right]}^{2}} $  

$  =\frac{{{l}^{2}}}{16}+\frac{1}{4\pi }{{\left( 28-l \right)}^{2}} $  

$  \therefore \frac{dA}{dl}=\frac{2l}{16}+\frac{2}{4\pi }\left( 28-l \right)\left( -1 \right)=\frac{l}{8}-\frac{1}{2\pi }\left( 28-l \right) $  

$  \frac{{{d}^{2}}A}{d{{l}^{2}}}=\frac{l}{8}+\frac{1}{2\pi }>0 $  

$  Now,\frac{dA}{dl}=0\Rightarrow \frac{l}{8}-\frac{1}{2\pi }\left( 28-l \right)=0 $  

$  \Rightarrow \left( \pi +4 \right)l-112=0 $  

$  \Rightarrow l=\frac{112}{\pi +4} $  

Thus, when $l=\frac{112}{\pi +4},\frac{{{d}^{2}}A}{d{{l}^{2}}}>0$ 

By second derivative test, area (A) is the minimum when $l=\frac{112}{\pi +4}$ . Hence, when the length of the wire is$\frac{112}{\pi +4}$ cm in making the square the combined area is the minimum while the length of the wire in making the  circle is $28-\frac{112}{\pi +4}=\frac{28\pi }{\pi +4}cm.$

23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.

Ans: Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

(Image Will Be Updated Soon)

Let V be the volume of the cone.

Then, $V=\frac{1}{3}\pi {{r}^{2}}h$ 

Height of the cone is given by,

$h=R+AB=R+\sqrt{{{R}^{2}}-{{r}^{2}}}$                                                  (ABC is a right triangle)

$  V=\frac{1}{3}\pi {{r}^{2}}\left( R+\sqrt{{{R}^{2}}-{{r}^{2}}} \right) $  

$  =\frac{1}{3}\pi {{r}^{2}}R+\frac{1}{3}\pi {{r}^{2}}\sqrt{{{R}^{2}}-{{r}^{2}}} $  

$  \frac{dV}{dr}=\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{{{R}^{2}}-{{r}^{2}}}+\frac{1}{3}\pi {{r}^{2}}.\frac{\left( -2r \right)}{2\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  =\frac{2}{3}\pi rR+\frac{2}{3}\pi r\sqrt{{{R}^{2}}-{{r}^{2}}}-\frac{1}{3}\pi .\frac{{{r}^{3}}}{\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  =\frac{2}{3}\pi rR+\frac{2\pi {{R}^{2}}-3\pi {{r}^{3}}}{3\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  \frac{{{d}^{2}}V}{d{{r}^{2}}}=\frac{2}{3}\pi rR+\frac{9\left( {{R}^{2}}-{{r}^{2}} \right)\left( 2\pi {{R}^{2}}-9\pi {{r}^{2}} \right)+2\pi r{{R}^{2}}-3\pi {{r}^{4}}}{27{{\left( {{R}^{2}}-{{r}^{2}} \right)}^{\frac{3}{2}}}} $  

$  Now,\frac{dV}{dr}=0\Rightarrow \pi \frac{2}{3}rR=\frac{3\pi {{r}^{3}}-2\pi {{R}^{2}}}{3\sqrt{{{R}^{2}}-{{r}^{2}}}} $  

$  \Rightarrow 2R=2R\sqrt{{{R}^{2}}-{{r}^{2}}}=3{{r}^{2}}-2{{R}^{2}} $  

$  \Rightarrow 4{{R}^{4}}-4{{R}^{2}}{{r}^{2}}=9{{r}^{4}}+4R $  

$  \Rightarrow 9{{r}^{4}}=8{{R}^{2}}{{r}^{2}} $  

$  \Rightarrow {{r}^{2}}=\frac{8}{9}{{R}^{2}} $  

When ${{r}^{2}}=\frac{8}{9}{{R}^{2}}$ then $\frac{{{d}^{2}}V}{d{{r}^{2}}}<0$ 

By second derivative test, the volume of the cone is the maximum when   

${{r}^{2}}=\frac{8}{9}{{R}^{2}}$.

When ${{r}^{2}}=\frac{8}{9}{{R}^{2}},h=R+\sqrt{{{R}^{2}}-\frac{8}{9}{{R}^{2}}}=R+\frac{R}{3}=\frac{4}{3}R$ 

Therefore, 

$  =\frac{1}{3}\pi \left( \frac{8}{9}{{R}^{2}} \right)\left( \frac{4}{3}R \right) $  

$  =\frac{8}{27}\left( \frac{4}{3}\pi {{R}^{3}} \right) $  

$  =\frac{8}{27}\times \left( Volume \right) $  

Hence, the volume of the largest cone that can be inscribed in the sphere is $\frac{8}{27}$ the volume of the sphere.

24. Show that the right circular cone of the least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.

Ans: Let r and h be the radius and the altitude of the cone respectively.

Then, the cone has the following volume (V):

$V=\frac{1}{3\pi }\pi {{r}^{2}}h\Rightarrow h=\frac{3V}{{{r}^{2}}}$ 

The cone has the following surface area (S),

$S=\pi rl$  (l = slant height)

$  =\pi r\sqrt{{{r}^{2}}+{{h}^{2}}} $  

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

$  =\frac{1}{r}\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}} $  

$  \frac{dS}{dr}=\frac{r.\frac{6{{\pi }^{2}}{{r}^{5}}}{2\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}}}-\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}}}{{{r}^{2}}} $  

$  =\frac{2{{\pi }^{2}}{{r}^{6}}-9{{V}^{2}}}{{{r}^{2}}\sqrt{{{\pi }^{2}}{{r}^{6}}+9{{V}^{2}}}} $  

$Now,\frac{dS}{dr}=0\Rightarrow 2{{\pi }^{2}}{{r}^{6}}=9{{V}^{2}}\Rightarrow {{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}}$ 

Thus, it can be easily verified that when ${{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}},\frac{{{d}^{2}}S}{d{{r}^{2}}}<0$ 

By second derivative test, the surface area of the cone is the least when ${{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}}$

When, ${{r}^{6}}=\frac{9{{V}^{2}}}{2{{\pi }^{2}}},h=\frac{3V}{\pi {{r}^{2}}}{{\left( \frac{2{{\pi }^{2}}{{r}^{6}}}{9} \right)}^{\frac{1}{2}}}=\frac{3}{\pi {{r}^{2}}}\frac{\sqrt{2\pi {{r}^{3}}}}{3}=\sqrt{2}r$

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to $\sqrt{2}$ times the radius of the base.

25. Show that the semi-vertical angle of the cone of the maximum volume and of the given slant height is tan-1$\sqrt{2}$.

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

It is clear that $\theta \in \left[ 0,\frac{\pi }{2} \right]$ 

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

The slant height of the cone is given as constant.

(Image Will Be Updated Soon)

Now r = l $\sin \theta $ and h = l $\cos \theta $ 

The volume (V) of the cone is given by,

$  V=\frac{1}{3}\pi {{r}^{2}}h $  

$  =\frac{1}{3}\pi \left( {{l}^{2}}{{\sin }^{2}}\theta  \right)\left( l\cos \theta  \right) $  

$  =\frac{1}{3}\pi {{l}^{3}}{{\sin }^{2}}\theta \cos \theta  $  

$  \therefore \frac{dV}{d\theta }=\frac{{{l}^{3}}\pi }{3}\left[ {{\sin }^{2}}\theta \left( -\sin \theta  \right)+\cos \theta \left( 2\sin \theta \cos \theta  \right) \right] $  

$  =\frac{{{l}^{3}}\pi }{3}\left[ -{{\sin }^{3}}\theta +2\sin \theta {{\cos }^{2}}\theta  \right] $  

$  \frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{{{l}^{3}}\pi }{3}\left[ -3{{\sin }^{2}}\theta \cos \theta +2{{\cos }^{3}}\theta -4{{\sin }^{2}}\theta \cos \theta  \right] $  

$  =\frac{{{l}^{3}}\pi }{3}\left[ 2{{\cos }^{2}}\theta -7{{\sin }^{2}}\theta \cos \theta  \right] $  

$  Now,\frac{dV}{d\theta }=0 $  

$  \Rightarrow {{\sin }^{3}}\theta =2\sin \theta {{\cos }^{2}}\theta  $  

$  \Rightarrow \tan \theta =2 $  

$  \Rightarrow \theta ={{\tan }^{-1}}\sqrt{2} $  

Now, when $\theta ={{\tan }^{-1}}\sqrt{2},$then ${{\tan }^{2}}\theta =2$ or ${{\sin }^{2}}\theta =2{{\cos }^{2}}\theta $ 

Then,

$\frac{{{d}^{2}}V}{d{{\theta }^{2}}}=\frac{{{l}^{3}}\pi }{3}\left[ 2{{\cos }^{3}}\theta -14{{\cos }^{3}}\theta  \right]=-4\pi {{l}^{3}}{{\cos }^{3}}\theta <0$     (For $\theta \in \left[ 0,\frac{\pi }{2} \right]$)

By second derivative test, the volume (V) is the maximum when $\theta ={{\tan }^{-1}}\sqrt{2}$. Hence, the semi-vertical angle of the cone of the maximum volume is  ${{\tan }^{-1}}\sqrt{2}$ for a given slant height.

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

(A) ($2\sqrt{2}$,4)           

(B) ($2\sqrt{2}$,0) 

(C) (0, 0)

(D) (2, 2)

Ans: The given curve is ${{x}^{2}}=2y$ 

For each value of x, y will be $\frac{{{x}^{2}}}{2}$ the position of the point will be $\left( x,\frac{{{x}^{2}}}{2} \right)$

The distance d(x) between the points $\left( x,\frac{{{x}^{2}}}{2} \right)$ and $\left( 0,5 \right)$ is given by

$  d\left( x \right)=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( \frac{{{x}^{2}}}{2}-5 \right)}^{2}}}=\sqrt{\frac{{{x}^{4}}}{4}-4{{x}^{2}}+25} $  

$  \therefore {d}'\left( x \right)=\frac{\left( {{x}^{3}}-8x \right)}{\sqrt{{{x}^{4}}-16{{x}^{2}}+100}} $  

$  Now,{d}'\left( x \right)=0\Rightarrow {{x}^{3}}-8x=0 $  

$  \Rightarrow x=0,\pm 2\sqrt{2} $  

And ${d}''\left( x \right)=\frac{\left( {{x}^{4}}-16{{x}^{2}}+100 \right)\left( 3{{x}^{2}}-8 \right)-2{{\left( {{x}^{3}}-8x \right)}^{2}}}{{{\left( {{x}^{4}}-16{{x}^{2}}+100 \right)}^{\frac{3}{2}}}}$ 

When $x=0$ then ${d}''\left( x \right)=\frac{36\left( -8 \right)}{{{6}^{3}}}<0$ 

When $x=\pm 2\sqrt{2}$ then $d\left( x \right)>0$ 

By second derivative test, d(x) is the minimum at $x=\pm 2\sqrt{2}$.

When $x=\pm 2\sqrt{2},y=4.$

Hence, the point on the curve ${{x}^{2}}=2y$ which is nearest to the point $\left( 0,5 \right)$ 

is $\left( \pm 2\sqrt{2},4 \right).$ 

Therefore, the correct option is A.

27. For all real values of x, the minimum value of $\frac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}}$ is

(A) 0   

(B) 1  

(C) 3   

(D) $\frac{1}{3}$ .

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

$  {f}'\left( x \right)=\frac{\left( 1+x+{{x}^{2}} \right)\left( -1-2x \right)-\left( 1-x+{{x}^{2}} \right)\left( 1+2x \right)}{{{\left( 1+x+{{x}^{2}} \right)}^{2}}} $  

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

$  \therefore {f}'\left( x \right)=0\Rightarrow {{x}^{2}}=1\Rightarrow x=\pm 1 $  

$  Now,{f}''\left( x \right)=4\frac{\left( 1+3x-{{x}^{3}} \right)}{{{\left( 1+x+{{x}^{2}} \right)}^{3}}} $  

$  And,{f}''\left( 1 \right)=\frac{4\left( 1+3-1 \right)}{{{\left( 1+1+1 \right)}^{3}}}=\frac{4}{9}>0 $  

$  And,{f}''\left( -1 \right)=\frac{4\left( 1-3+1 \right)}{{{\left( 1+1+1 \right)}^{3}}}=\frac{-4}{27}<0 $  

By second derivative test, f is the minimum at $x=1$ and the minimum 

value is given by $f\left( 1 \right)=\frac{1}{3}$ 

The correct option is D.

28. The maximum value of ${{\left[ x\left( x+1 \right)+1 \right]}^{\frac{1}{3}}},0\le x\le 1$ is

(A) ${{\left( \frac{1}{3} \right)}^{\frac{1}{3}}}$ 

(B) $\frac{1}{2}$ 

(C) 1 

(D) 0

Ans: Let $f\left( x \right)={{\left[ x\left( x+1 \right)+1 \right]}^{\frac{1}{3}}}.$ 

$  \therefore {f}'\left( x \right)=\frac{2x+1}{3{{\left[ x\left( x+1 \right)+1 \right]}^{\frac{2}{3}}}} $  

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

But, x = -1/2 is not part of the interval [0,1]

Now, evaluate the value of f at the endpoints of the interval [0, 1] {i.e., 

at x = 0 and x =1}.

$  f\left( 0 \right)={{\left[ 0\left( 0-1 \right)+1 \right]}^{\frac{1}{3}}}=1 $  

$  f\left( 0 \right)={{\left[ 1\left( 1-1 \right)+1 \right]}^{\frac{1}{3}}}=1 $  

Hence, we can conclude that the maximum value of  f is 1 in the 

interval [0, 1].

The correct option is 13.

Class 12 Maths Chapter 6 Exercise 6.5

What Is The Significance Of Ex 6.5 Class 12 NCERT Solutions?

Application of Derivatives Exercise 6.5 is a very important chapter in terms of weightage of marks in the CBSE board exam. The application of entire Differential Calculus is based on the Application of Derivatives exercise 6.5. NCERT solutions for class 12 maths chapter 6 exercise 6.5 provided by Vedantu is the perfect material which will help you not only to understand the chapter but also to identify your lacunae. The significances of class 12 Maths NCERT solutions chapter 6 exercise 6.5 are as follows:

  • Practising exercise 6.5 class 12 Maths NCERT solutions regularly will certainly boost up your preparation.

  • The questions asked in exercise 6.5 Maths class 12 are prepared entirely according to the CBSE guidelines.

  • You can identify your loopholes by practising class 12 Maths ex 6.5 solutions.

  • Calculus is the most scoring segment of class 12 Maths and you score full marks in the class 12 maths ch 6 ex 6.5. 

  • After completing the usual textbook problems NCERT solutions class 12 maths chapter 6 exercise 6.5 are the best materials to enhance your level of preparation.

  • You cannot solve all the problems yourself. But you should try to solve them on your own in the first attempt. Then you can verify them with Maths NCERT solutions class 12 chapter 6 exercises 6.5.

  • If you can complete ex 6.5 Maths class 12 in a time-bound manner then you should go for harder problems. The level of difficulty is gradually increased in the NCERT Maths class 12 exercise 6.5 solutions.

What is the Procedure of Practising with Ex 6.5 Class 12 Maths NCERT Solutions?

You should remember the following steps at the time of solving the problems of Application of Derivatives exercise 6.5:

  1. If it’s a straight numerical problem, then focus on its construct and if it is a problem sum try to figure out the logic and what you have been asked to find. 

  2. Don’t look at the solution first.

  3. Try to recollect the formula(s) that you need to apply to solve the application of derivative. With repeated practice, you should be able to do that within a few seconds. 

  4. Solve the problem carefully and check the answer with the solution.

  5. If your answer is correct, try to speed up next time. 

  6. If your answer is wrong, look at the solution and solve it again- this time following the steps as detailed in the solution. 

  7. Close the solution PDF and practise the sum once again.

Why Should You Choose Vedantu?

If you want to solve the Maths NCERT solutions class 12 chapter 6 exercise 6.5 scientifically and appropriately, then Vedantu is the best option for you. The reasons in support of choosing Vedantu are furnished below:

  • Ex 6.5 class 12 NCERT solutions are well-crafted and prepared by the veteran teachers of Vedantu.

  • The problems and solutions which are given in NCERT solutions class 12 maths chapter 6 exercise 6.5 are entirely based on the latest CBSE guidelines.

  • The difficulty level has been increased gradually so that the students can enhance their level of confidence in a rhythmic manner.

  • Class 12 Maths ex 6.5 solutions are the perfect combination of theory and practical knowledge.

  • You can access the NCERT solutions for class 12 maths chapter 6 exercise 6.5 in PDF format entirely free of cost from the official website of Vedantu.

  • Hence, you are just one click away from the perfect material for the class 12 CBSE Maths exam.

FAQs (Frequently Asked Questions)

Q1. What is the solution to the following problem as found in NCERT solutions for class 12 maths chapter 6 exercise 6.5?

Ans: Prove that the function h (x) = x3 + x2 + x + 1 has no maxima or minima

Maths NCERT solutions class 12 chapter 6 exercise 6.5 has this problem solved for you in the easiest way. The solution for this is in the following:

 It is given that, 

h (x) = x3 + x2 + x + 1

So, h (x) = 3x2 + 2x + 1

Therefore, cЄR suh that h (c) = 0

Hence, it can be said that the function h does not have a maxima or minima. 

Q2. What are the topics contained in the Application of Derivatives exercise 6.5?

Ans: The following topics are covered in the Application of Derivatives class 12:

6.1: Introduction

6.2: Rate of change of quantities

6.3: Increasing and Deceasing Functions

6.4: Tangents and Normals

6.5: Approximations

6.6: Maxima and Minima

6.7: Maximum and Minimum Values of a Function in a Closed Interval

6.8: Summary

Q3. How many units are there in Class 12 Maths?

Ans: In the NCERT Class 12 Maths textbook, there are 6=six units in total. These units are: 

  • Unit 1 - Relations and Functions

  • Unit 2 - Algebra

  • Unit 3 - Calculus

  • Unit 4 - Vectors and Three-Dimensional Geometry

  • Unit 5 - Linear Programming

  • Unit 6 - Probability

You can get complete solutions to all the units at the Vedantu website (evdantu.com) free of cost. Check out the NCERT Solutions Class 12 Maths and practice all the questions to get full marks! 

Q4. Which Chapter is the Application Of Derivatives?

Ans: Chapter 6 is the Application of derivatives in the NCERT Class 12 Maths textbook. You can get a thorough explanation of the chapter in easy language with full solutions to the exercises and other related questions by downloading the NCERT Solutions Class 12 Maths Chapter 6 on the Vedantu website (vedantu.com) and app. Practice the NCERT Solutions available in PDF to increase your chances of scoring full marks on questions based on this chapter in your class test and even board exams.  

Q5. What are the easy chapters in NCERT Class 12 Maths?

Ans: Some of the easy but very important chapters in NCERT Class 12 Maths are - Matrices Determinants, Integrals, Three-Dimensional Geometry, Applications of Derivatives, Relations and Functions - Inverse Trigonometric Functions, Probability, and Continuity and Differentiability. These chapters are comparatively easy but require equal amounts of practice and hard work like other chapters. You can get full NCERT Solutions Class 12 Maths at the page NCERT Solutions Class 12 Maths Chapter 6 on Vedantu’s website. 

Q6. Is Class 12 Maths difficult?

Ans: Many students fear Maths and find Class 12 Maths very tough. However, this may not be the case. Everything depends on you and your strategy. If you have a negative mindset, you will not be able to score well. You need to practice Class 12 Maths daily, and the best companion that can make things easy for you is Vedantu’s NCERT Solutions Class 12 Maths.  

Q7. How can I understand NCERT Class 12 Maths Chapter 6?

Ans: For many students, Maths can be a tough subject. But if you choose the right material, things can get very easy for you. The best way to understand NCERT Class 12 Maths Chapter 6 is to get NCERT Solutions Class 12 Maths Chapter 3 available on Vedantu. The NCERT Maths Solutions for Class 12 has full answers to all the questions and an easy explanation of concepts and theories. 

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