NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Miscellaneous Exercise

1. Show that the function given by $f(x)={\displaystyle \frac{\log x}{x}}$ has maximum at $x=e$.

Ans: $f(x)={\displaystyle \frac{\log x}{x}}$

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

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

$\Rightarrow 1-\log x=0$

$\Rightarrow \log x=1$

$\Rightarrow \log x=\log e$

$\Rightarrow x=e$

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

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

$={\displaystyle \frac{-3+2\log x}{x^3}}$

$f^{\prime \prime }(e)=={\displaystyle \frac{-3+2\log e}{e^3}}={\displaystyle \frac{-3+2}{e^3}}={\displaystyle \frac{-1}{e^3}}<0$

$f$ is the maximum at $x=e$

2. The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of $3\text{cm}$ per second. How fast is the area decreasing when the two equal sides are equal to the base? 

Ans: Let $△ABC$ be isosceles where $$BC$$is the base of fixed length $b$. 

Let the length of the two equal sides of $△ABC$ be $a$. 

Draw $AD\mathrm{\perp }BC$.

$AD=\sqrt{a^2-{\displaystyle \frac{b^2}{4}}}$

Area of triangle $={\displaystyle \frac{1}{2}}b\sqrt{a^2-{\displaystyle \frac{b^2}{4}}}$

${\displaystyle \frac{dA}{dt}}={\displaystyle \frac{1}{2}}b\cdot {\displaystyle \frac{2a}{2\sqrt{a^2-{\displaystyle \frac{b^2}{4}}}}}{\displaystyle \frac{da}{dt}}={\displaystyle \frac{ab}{\sqrt{4a^2-b^2}}}{\displaystyle \frac{da}{dt}}$

${\displaystyle \frac{da}{dt}}=-3\text{cm}/\text{s}$

$\therefore {\displaystyle \frac{da}{dt}}={\displaystyle \frac{-3ab}{\sqrt{4a^2-b^2}}}$

when $a=b$

${\displaystyle \frac{dA}{dt}}={\displaystyle \frac{-3b^2}{\sqrt{4b^2-b^2}}}={\displaystyle \frac{-3b^2}{\sqrt{3b^2}}}=-\sqrt{3b}$

3. Find the intervals in which the function $f$ given by $f(x)={\displaystyle \frac{4\sin x-2x-x\cos x}{2+\cos x}}$ Is 

(i) increasing (ii) decreasing

Ans: $f(x)={\displaystyle \frac{4\sin x-2x-x\cos x}{2+\cos x}}$

$\therefore f^{\prime }(x)={\displaystyle \frac{(2+\cos x)(4\cos x-2-\cos x+x\sin x)-(4\sin x-2x-x\cos x)(-\sin x)}{{(2+\cos x)}^2}}$

$={\displaystyle \frac{(2+\cos x)(3\cos x-2+x\sin x)+\sin x(4\sin x-2x-x\cos x)}{{(2+\cos x)}^2}}$

$={\displaystyle \frac{6\cos x-4+2x\sin x+3{\cos }^{2}x-2\cos x+x\sin x\cos x+4{\sin }^{2}x-2{\sin }^{2}x-2x\sin x-x\sin x\cos x}{{(2+\cos x)}^2}}$

$={\displaystyle \frac{4\cos x-4+3{\cos }^{2}x+4{\sin }^{2}x}{{(2+\cos x)}^2}}$

$={\displaystyle \frac{4\cos x-{\cos }^{2}x}{{(2+\cos x)}^2}}={\displaystyle \frac{\cos x(4-\cos x)}{{(2+\cos x)}^2}}$

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

$\Rightarrow \cos x=0\cos x=4$

$\cos x\ne 4$

$\cos x=0$

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

$\mathrm{In}\left(0,{\displaystyle \frac{\pi }{2}}\right)$ and $\left({\displaystyle \frac{3\pi }{2}},2\pi \right),f^{\prime }(x)>0$

$f(x)$ is increasing for $0<x<{\displaystyle \frac{x}{2}}$ and ${\displaystyle \frac{3\pi }{2}}<x<2\pi$.

$\mathrm{In}\left({\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\right),f^{\prime }(x)<0$

$f(x)$ is decreasing for ${\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{3\pi }{2}}$.

4. Find the intervals in which the function $f$ given by $f(x)=x^3+{\displaystyle \frac{1}{x^3}},x\ne 0$ is

(i) increasing 

(ii) decreasing

Ans: $f(x)=x^3+{\displaystyle \frac{1}{x^3}}$

$\therefore f^{\prime }(x)=3x^2-{\displaystyle \frac{3}{x^4}}={\displaystyle \frac{3x^6-3}{x^4}}$

$f^{\prime }(x)=0\Rightarrow 3x^6-3=0\Rightarrow x^6=x=\pm 1$

In $(-\mathrm{\infty },1)$ and $(1,\mathrm{\infty })$ i.e.., when $x<-1$ and $x>1,f^{\prime }(x)>0.$ when $x<-1$ and $x>1,f$ is increasing. In $(-1,1)$ i.e., when $-1<x<1,f^{\prime }(x)<0$.

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

5. Find the maximum area of an isosceles triangle inscribed in the ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ with its vertex at one end of the major axis.

Ans:

Ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$

Let $ABC$, be the triangle inscribed in the ellipse where vertex $C$ is at $(a,0)$. Since the ellipse is symmetrical with $x$-axis and $y-axis$ $y_1=\pm {\displaystyle \frac{b}{a}}\sqrt{a^2-x_1^2}.$

Coordinates of $A$ are $\left(-x_{1,}{\displaystyle \frac{b}{a}}\sqrt{a^2-x^{2}1}\right)$ and coordinates of $B$ are $\left(x_1,-{\displaystyle \frac{b}{a}}\sqrt{a^2-x_1^2}.\right)$ As the point $\left(-x_1,y_1\right)$ lies on the ellipse, the area of triangle $ABC$ is $A={\displaystyle \frac{1}{2}}\mid a\left({\displaystyle \frac{2b}{a}}\sqrt{a^2-x_1^2}\right)+\left(-x_1\right)\left(-{\displaystyle \frac{b}{a}}\sqrt{a^2-x_1^2}\right)+\left(-x_1\right)\left(-{\displaystyle \frac{b}{a}}\sqrt{a^2-x^{2}1}\right)$

$\Rightarrow A=ba\sqrt{a^2-x^{2}1}+x_1{\displaystyle \frac{b}{a}}\sqrt{a^2-x_1^2}$

$\therefore {\displaystyle \frac{dA}{d{x}_1}}={\displaystyle \frac{-2xb}{2\sqrt{a^2-x_1^2}}}+{\displaystyle \frac{b}{a}}\sqrt{a^2-x_1^2}-{\displaystyle \frac{2b{x}_1^2}{a^2\sqrt{a^2-x_1^2}}}$

$={\displaystyle \frac{b}{2\sqrt{a^2-x_1^2}}}\left[-x_{1}a+\left(a^2-x_1^2\right)-x_1^2\right]$

$={\displaystyle \frac{b\left(-2x_1^2-x_1^2+a^2\right)}{a\sqrt{a^2-x_1^2}}}$

${\displaystyle \frac{dA}{d{x}_1}}=0$

$\Rightarrow -2x_1^2-x_{1}a+a^2=0$

$\Rightarrow x_1={\displaystyle \frac{a\pm \sqrt{a^2-4(-2)\left(a^2\right)}}{2(-2)}}$

$={\displaystyle \frac{a\pm \sqrt{9a^2}}{-4}}$

$={\displaystyle \frac{a\pm 3a}{-4}}$

$\Rightarrow x_1=-a,{\displaystyle \frac{a}{2}}$

$x_1$ cannot be equal to $a$. $\therefore x_1={\displaystyle \frac{a}{2}}\Rightarrow y_1={\displaystyle \frac{b}{a}}\sqrt{a^2-{\displaystyle \frac{a^2}{4}}}={\displaystyle \frac{ba}{2a}}\sqrt{3}={\displaystyle \frac{\sqrt{3b}}{2}}$

Now,${\displaystyle \frac{d^{2}A}{d{x}^{2}1}}={\displaystyle \frac{b}{a}}\left\{{\displaystyle \frac{\sqrt{a^2-x_1^2}\left(-4x_1-a\right)-\left(-2x_1^2-x_{1}a+a^2\right){\displaystyle \frac{\left(-2x_1\right)}{2\sqrt{a^2-x_1^2}}}}{a^2-x_1^2}}\right\}$

$={\displaystyle \frac{b}{a}}\left\{{\displaystyle \frac{\left(a^2-x_1^2\right)\left(-4x_1-a\right)+x_1\left(-2x_1^2-x_{1}a+a^2\right)}{{\left(a^2-x_1^2\right)}^{{\displaystyle \frac{2}{3}}}}}\right\}$

$={\displaystyle \frac{b}{a}}\left\{{\displaystyle \frac{2x^3-3a^{2}x-a^3}{{\left(a^2-x_1^2\right)}^{{\displaystyle \frac{3}{2}}}}}\right\}$

when $x_1={\displaystyle \frac{a}{2}}$,

${\displaystyle \frac{d^{2}A}{d{x}_1^2}}={\displaystyle \frac{b}{a}}\left\{{\displaystyle \frac{2{\displaystyle \frac{a^3}{8}}-3{\displaystyle \frac{a^3}{2}}-a^3}{{\left({\displaystyle \frac{3a^2}{4}}\right)}^{{\displaystyle \frac{3}{2}}}}}\right\}={\displaystyle \frac{b}{a}}\left\{{\displaystyle \frac{{\displaystyle \frac{a^3}{4}}-{\displaystyle \frac{3}{2}}{a}^3-a^3}{{\left({\displaystyle \frac{3a^2}{4}}\right)}^{{\displaystyle \frac{3}{2}}}}}\right\}$

$={\displaystyle \frac{b}{a}}\left\{{\displaystyle \frac{{\displaystyle \frac{9}{4}}{a}^3}{{\left({\displaystyle \frac{3a^2}{4}}\right)}^{{\displaystyle \frac{3}{2}}}}}\right\}<0$

Area is the maximum when $x_1={\displaystyle \frac{a}{2}}$. Maximum area of the triangle is $A=b\sqrt{a^2-{\displaystyle \frac{a^2}{4}}}+\left({\displaystyle \frac{a}{2}}\right){\displaystyle \frac{b}{a}}\sqrt{a^2-{\displaystyle \frac{a^2}{4}}}$

$=ab{\displaystyle \frac{\sqrt{3}}{2}}+\left({\displaystyle \frac{a}{2}}\right){\displaystyle \frac{b}{a}}\times {\displaystyle \frac{a\sqrt{3}}{2}}$

$={\displaystyle \frac{ab\sqrt{3}}{2}}+{\displaystyle \frac{ab\sqrt{3}}{4}}={\displaystyle \frac{3\sqrt{3}}{4}}ab$

6. A tank with a rectangular base and rectangular sides, open at the top is to be constructed so that its depth is $$2m$$and volume is $8m^3$. If the building of the tank costs Rs 70 per sq meter for the base and Rs 45 per sq meter for the sides. What is the cost of the least expensive tank?

Ans: Let $$l,b$$and $h$ represent the length, breadth, and height of the tank respectively.

height $(h)=2m$

Volume of the tank $=8{\text{m}}^3$ Volume of the tank $=l\times b\times \text{h}$ $8=l\times b\times 2$

$\Rightarrow lb=4\Rightarrow b={\displaystyle \frac{4}{l}}$

Area of the base $=lb=4$ 

Area of 4 walls $(A)=2h(l+b)$

$\therefore A=4\left(l+{\displaystyle \frac{4}{l}}\right)$

$\Rightarrow {\displaystyle \frac{dA}{dl}}=4\left(1-{\displaystyle \frac{4}{l^2}}\right)$

$Now,{\displaystyle \frac{dA}{dl}}=0$

$\Rightarrow l-{\displaystyle \frac{4}{l^2}}=0$

$\Rightarrow l^2=4$

$\Rightarrow l=\pm 2$

Therefore, we have $l=4$. 

$\therefore b={\displaystyle \frac{4}{l}}={\displaystyle \frac{4}{2}}=2$

${\displaystyle \frac{d^{2}A}{d{l}^2}}={\displaystyle \frac{32}{l^3}}$

$l=2,{\displaystyle \frac{d^{2}A}{d{l}^2}}={\displaystyle \frac{32}{8}}=4>0.$

Area is the minimum when $l=2$. 

We have $l=b=h=2$.

Cost of building base $$=$$ Rs $70\text{x}(lb)=$ Rs$70(4)=$ Rs 280

Cost of building walls $=Rs2h(l+h)\times 45=Rs90(2)(2+2)=$ 

Rs $8(90)=$ Rs 720

Required total cost $=Rs(280+720)=$ Rs 1000

7. The sum of the perimeter of a circle and square is $k$, where $k$ is some constant. Prove that the sum of their area is least when the side of the square is double the radius of the circle. 

Ans: $2\pi r+4a=k$ (where $k$ is constant) $\Rightarrow a={\displaystyle \frac{k-2\pi r}{4}}$

the sum of the areas of the circle and the square $(A)$ is given by,

$A=\pi r^2+a^2=\pi r^2+{\displaystyle \frac{{(k=2\pi r)}^2}{16}}$

$\therefore {\displaystyle \frac{dA}{dr}}=2\pi r+{\displaystyle \frac{2(k-2\pi r)(2\pi )}{16}}=2\pi r$

$=-{\displaystyle \frac{\pi (k-2\pi r)}{4}}$

Now, ${\displaystyle \frac{dA}{dr}}=0$

$\Rightarrow 2\pi r={\displaystyle \frac{\pi (k-2\pi r)}{4}}$

$8r=k-2\pi r$

$\Rightarrow (8+2\pi )r=k$

$\Rightarrow r={\displaystyle \frac{k}{8+2\pi }}={\displaystyle \frac{k}{2(4+\pi )}}$

Now, ${\displaystyle \frac{d^{2}A}{d{r}^2}}=2\pi +{\displaystyle \frac{{\pi }^2}{2}}>0$

$\therefore$ where $r={\displaystyle \frac{k}{2(4+\pi )}},{\displaystyle \frac{d^{2}A}{d{r}^2}}>0.$

area is least when $r={\displaystyle \frac{k}{2(4+\pi )}}$ where $r={\displaystyle \frac{k}{2(4+\pi )}},$

$a={\displaystyle \frac{k-2\pi \left[{\displaystyle \frac{k}{2(4+\pi )}}\right]}{4}}={\displaystyle \frac{8k+2\pi k-2\pi k}{2(4+\pi )\times 4}}={\displaystyle \frac{k}{4+\pi }}=2r$

8. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10\text{m}$. Find the dimensions of the window to admit maximum light through the whole opening. 

Ans: $x$ and $y$ are the length and breadth of the rectangular window.

Radius of semicircular opening $={\displaystyle \frac{x}{2}}$

$\therefore x+2y+{\displaystyle \frac{\pi \text{x}}{2}}=10$

$\Rightarrow x\left(1+{\displaystyle \frac{\pi }{2}}\right)+2y=10$

$\Rightarrow 2y=10-x\left(1+{\displaystyle \frac{\pi }{2}}\right)$

$\Rightarrow y=5-x\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\pi }{4}}\right)$

$A=xy+{\displaystyle \frac{\pi }{2}}{\left({\displaystyle \frac{x}{2}}\right)}^2$

$=x\left[5-x\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\pi }{4}}\right)\right]+{\displaystyle \frac{\pi }{8}}\text{x}{x}^2$

$=5x-x^2\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\pi }{4}}\right)+{\displaystyle \frac{\pi }{8}}\text{x}{x}^2$

$\therefore {\displaystyle \frac{dA}{dx}}=5-2x\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\pi }{4}}\right)+{\displaystyle \frac{\pi }{4}}x$

${\displaystyle \frac{d^{2}A}{d{x}^2}}=-\left(1-{\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{\pi }{4}}=-1-{\displaystyle \frac{\pi }{4}}$

${\displaystyle \frac{dA}{dx}}=0$

$\Rightarrow 5-x\left(1+{\displaystyle \frac{\pi }{2}}\right)+{\displaystyle \frac{\pi }{4}}x=0$

$\Rightarrow 5-x-{\displaystyle \frac{\pi }{4}}x=0$

$\Rightarrow x\left(1+{\displaystyle \frac{\pi }{4}}\right)=5$

$\Rightarrow x={\displaystyle \frac{5}{\left(1+{\displaystyle \frac{\pi }{4}}\right)}}={\displaystyle \frac{20}{\pi +4}}$

$x={\displaystyle \frac{20}{\pi +4}},{\displaystyle \frac{d^{2}A}{d{x}^2}}<0.$

area is maximum when length $x={\displaystyle \frac{20}{\pi +4}}m$. 

Now, $y=5-{\displaystyle \frac{20}{\pi +4}}\left({\displaystyle \frac{2+\pi }{4}}\right)=5-{\displaystyle \frac{5(2+\pi )}{\pi +4}}={\displaystyle \frac{10}{\pi +4}}m$

the required dimensions length $={\displaystyle \frac{20}{\pi +4}}m$ and breadth $={\displaystyle \frac{10}{\pi +4}}m$.

9. A point of the hypotenuse of a triangle is at distance $a$ and $b$ from the sides of the triangle. Show that the minimum length of the hypotenuse is ${\left(a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}\right)}^{{\displaystyle \frac{3}{2}}}$ 

Ans: $△ABC$ right-angled at $B$. $AB=x$ and $BC=y$.

$P$ be a point on hypotenuse such that $\text{P}$ is at a distance of a and $\text{b}$ from the sides $$AB$$and $$BC$$respectively.

$\mathrm{\angle }c=\theta$

$AC=\sqrt{x^2+y^2}$

$PC=b\mathrm{cosec}\theta$

$AP=a\sec \theta$

$AC=AP+PC$

$AC=b\mathrm{cosec}\theta +a\sec \theta \dots ..(1)$

$\therefore {\displaystyle \frac{d(AC)}{d\theta }}=-b\mathrm{cosec}\theta \cot \theta +a\sec \theta \tan \theta$

$\therefore {\displaystyle \frac{d(AC)}{d\theta }}=0$

$\Rightarrow a\sec \theta \tan \theta =b\cos \mathrm{ec}\theta \cot \theta$

$\Rightarrow {\displaystyle \frac{a}{\cos \theta }}\cdot {\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{b}{\sin \theta }}{\displaystyle \frac{\cos \theta }{\sin \theta }}$

$\Rightarrow a{\sin }^3\theta =b{\cos }^3\theta$

$\Rightarrow (a{)}^{{\displaystyle \frac{1}{3}}}\sin \theta =(b{)}^{{\displaystyle \frac{1}{3}}}\cos \theta$

$\Rightarrow \tan \theta ={\left({\displaystyle \frac{b}{a}}\right)}^{{\displaystyle \frac{1}{3}}}$

$\therefore \sin \theta ={\displaystyle \frac{{(b)}^{{\displaystyle \frac{1}{3}}}}{\sqrt{a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}}}}$ and $\cos \theta ={\displaystyle \frac{{(a)}^{{\displaystyle \frac{1}{3}}}}{\sqrt{a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}}}}\dots ..(2)$

clearly ${\displaystyle \frac{d^2(AC)}{d{\theta }^2}}<0$ when $\tan \theta ={\left({\displaystyle \frac{b}{a}}\right)}^{{\displaystyle \frac{1}{3}}}$.

the length of the hypotenuse is the maximum when $\tan \theta ={\left({\displaystyle \frac{b}{a}}\right)}^{{\displaystyle \frac{1}{3}}}$. 

Now, when $\tan \theta ={\left({\displaystyle \frac{b}{a}}\right)}^{{\displaystyle \frac{1}{3}}}$

$\tan \theta ={\left({\displaystyle \frac{b}{a}}\right)}^{{\displaystyle \frac{1}{3}}}$

$AC={\displaystyle \frac{b\sqrt{a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}}}{b^{{\displaystyle \frac{1}{3}}}}}+{\displaystyle \frac{a\sqrt{a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}}}{a^{{\displaystyle \frac{1}{3}}}}}$

$=a\sqrt{a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}}\left(b^{{\displaystyle \frac{2}{3}}}+a^{{\displaystyle \frac{2}{3}}}\right)$

$={\left(a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}\right)}^{{\displaystyle \frac{3}{2}}}$

maximum length of the hypotenuse is $={\left(a^{{\displaystyle \frac{2}{3}}}+b^{{\displaystyle \frac{2}{3}}}\right)}^{{\displaystyle \frac{3}{2}}}$.

10. Find the points at which the function $f$ given by $f(x)=(x-2{)}^4(x+1{)}^3$ has

(i) local maxima

(ii) local minima

(iii) point of inflexion

Ans: $f(x)=(x-2{)}^4(x+1{)}^3$

$\therefore f^{\prime }(x)=4(x-2{)}^3(x+1{)}^3+3(x+1{)}^2(x-2{)}^4$

$=(x-2{)}^3(x+1{)}^2[4(x+1)+3(x-2)]$

$=(x-2{)}^3(x+1{)}^2(7x-2)$

$f^{\prime }(x)=0\Rightarrow x=-1$ and $x={\displaystyle \frac{2}{7}}$ or $x=2$

for $x$ close to ${\displaystyle \frac{2}{7}}$ and to left of ${\displaystyle \frac{2}{7}},f^{\prime }(x)>0$.

for $x$ close to ${\displaystyle \frac{2}{7}}$ and to right of ${\displaystyle \frac{2}{7}},f^{\prime }(x)>0$. $x={\displaystyle \frac{2}{7}}$ is point of local minima.

as the value of $x$ varies $f^{\prime }(x)$ does not changes its sign.

$x=-1$ is point of inflexion.

11. Find the absolute maximum and minimum values of the function $f$ given by

$f(x)={\cos }^{2}x+\sin x,x\in [0,\pi ]$

Ans: $f(x)={\cos }^{2}x+\sin x$

$f^{\prime }(x)=2\cos x(-\sin x)+\cos x$

$=-2\sin x\cos x+\cos x$

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

$\Rightarrow 2\sin x\cos x=\cos x\Rightarrow \cos x(2\sin x-1)=0$

$\Rightarrow \sin x={\displaystyle \frac{1}{2}}$ or $\cos x=0$

$\Rightarrow x={\displaystyle \frac{\pi }{6}}$, or ${\displaystyle \frac{\pi }{2}}$ as $x\in [0,\pi ]$

$f\left({\displaystyle \frac{\pi }{6}}\right)={\cos }^2{\displaystyle \frac{\pi }{6}}+\sin {\displaystyle \frac{\pi }{6}}={\left({\displaystyle \frac{\sqrt{3}}{2}}\right)}^2+{\displaystyle \frac{1}{2}}={\displaystyle \frac{5}{4}}$

$f(0)={\cos }^{2}0+\sin 0=1+0=1$

$f(\pi )={\cos }^2\pi +\sin \pi =(-1{)}^2+0=1$

$f\left({\displaystyle \frac{\pi }{2}}\right)={\cos }^2{\displaystyle \frac{\pi }{2}}+\sin {\displaystyle \frac{\pi }{2}}=0+1=1$

absolute maximum value of $f$ is ${\displaystyle \frac{5}{4}}$ at $x={\displaystyle \frac{\pi }{6}}$

The absolute minimum value of $f$ is 1 at $x=0,x={\displaystyle \frac{\pi }{2}}$, and $\pi$.

12. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $\mathrm{r}$ is ${\displaystyle \frac{4r}{3}}$.

Ans: $V={\displaystyle \frac{1}{3}}\pi R^{2}h$

$BC=\sqrt{r^2-R^2}$

$h=r+\sqrt{r^2-R^2}$

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

$\therefore {\displaystyle \frac{dV}{dR}}={\displaystyle \frac{2}{3}}\pi Rr+{\displaystyle \frac{2\pi }{3}}\pi R\sqrt{r^2-R^2}+{\displaystyle \frac{R^2}{3}}\cdot {\displaystyle \frac{(-2R)}{2\sqrt{r^2-R^2}}}$

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

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

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

${\displaystyle \frac{dV}{d{R}^2}}=0$

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

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

$\Rightarrow 4r^2\left(r^2-R^2\right)={\left(3R^2-2r^2\right)}^2$

$\Rightarrow 14r^4-4r^2{R}^2=9R^4+4r^4-12R^2{r}^2$

$\Rightarrow 9R^4-8r^2{R}^2=0$

$\Rightarrow 9R^2=8r^2$

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

${\displaystyle \frac{d^{2}V}{d{R}^2}}={\displaystyle \frac{2\pi r}{3}}+{\displaystyle \frac{3\sqrt{r^2-R^2}\left(2\pi r^2-9\pi R^2\right)-\left(2\pi R^3-3\pi R^3\right)(-6R){\displaystyle \frac{1}{2{\sqrt{r^2-R}}^2}}}{9\left(r^2-R^2\right)}}$

$={\displaystyle \frac{2\pi r}{3}}+{\displaystyle \frac{3\sqrt{r^2-R^2}\left(2\pi r^2-9\pi R^2\right)-\left(2\pi R^3-3\pi R^3\right)(3R){\displaystyle \frac{1}{2{\sqrt{r^2-R}}^2}}}{9\left(r^2-R^2\right)}}$

when $R^2={\displaystyle \frac{8r^2}{9}},{\displaystyle \frac{d^{2}V}{d{R}^2}}<0$.

volume is the maximum when $R^2={\displaystyle \frac{8r^2}{9}}$. $R^2={\displaystyle \frac{8r^2}{9}}$, 

height of the cone $=r+\sqrt{r^2-{\displaystyle \frac{8R^2}{9}}}=r+\sqrt{{\displaystyle \frac{r^2}{9}}}=r+{\displaystyle \frac{r}{3}}={\displaystyle \frac{4r}{3}}$

13. Let f be a function defined on $[a,b]$such that $f^{{}^{\prime }}(x)>0$ for all $x\in (a,b)$. Then prove that f is an increasing function on (a, b).

Ans: Consider I be the interval (a, b)

Given that $f^{{}^{\prime }}(x)>0$for all x in an interval I. Consider $x_1,x_1,\in$I with $x_1<x_2$.

By Lagrange’s Mean Value Theorem, we have, 

$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f^{{}^{\prime }}(c)$where $x_1<c<x_2$

$\Rightarrow f(x_2)-f(x_1)=(x_2-x_1)f^{{}^{\prime }}(c)$where $x_1<c<x_2$

Now $x_1<x_2$

$\Rightarrow x_2-x_1>0$……….(1)

 Also, $f^{{}^{\prime }}(x)>0$for all $x$ in an interval I

$\Rightarrow f^{{}^{\prime }}(c)>0$

From equation (1), $f(x_2)-f(x_1)>0$

$\Rightarrow f(x_1)<f(x_2)$

Thus, for every pair of points $x_1,x_1,\in$I with $x_1<x_2$

$\Rightarrow f(x_1)<f(x_2)$

Therefore, f(x) is strictly increasing in I.

14. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is ${\displaystyle \frac{2R}{\sqrt{3}}}$, also find the maximum volume.

Ans: $h=2\sqrt{R^2-r^2}$

$V=\pi r^{2}h=2\pi r^2\sqrt{R^2-r^2}$

$\therefore {\displaystyle \frac{dV}{dr}}=4\pi r\sqrt{R^2-r^2}+{\displaystyle \frac{2\pi r^2(-2r)}{2\sqrt{R^2-r^2}}}$

$=4\pi r\sqrt{R^2-r^2}-{\displaystyle \frac{2\pi r^3}{\sqrt{R^2-r^2}}}$

$={\displaystyle \frac{4\pi r\left(R^2-r^2\right)-2\pi r^3}{\sqrt{R^2-r^2}}}$

$={\displaystyle \frac{4\pi r{R}^2-6\pi r^3}{\sqrt{R^2-r^2}}}$

Now, ${\displaystyle \frac{dV}{dr}}=0\Rightarrow 4\pi r{R}^2-6\pi r^3=0$

$\Rightarrow r^2={\displaystyle \frac{2R^2}{3}}$

${\displaystyle \frac{d^{2}V}{d{r}^2}}={\displaystyle \frac{\sqrt{R^2-r^2}\left(4\pi R^2-18\pi r^2\right)-\left(4\pi r{R}^2-6\pi r^3\right){\displaystyle \frac{(-2r)}{2\sqrt{R^2-r^2}}}}{\left(R^2-r^2\right)}}$

$={\displaystyle \frac{\left(R^2-r^2\right)\left(4\pi R^2-18\pi r^2\right)+r\left(4\pi r{R}^2-6\pi r^3\right)}{{\left(R^2-r^2\right)}^{{\displaystyle \frac{3}{2}}}}}$

$={\displaystyle \frac{4\pi R^4-22\pi r^2{R}^2+12\pi r^4+4\pi r^2{R}^2}{{\left(R^2-r^2\right)}^{{\displaystyle \frac{3}{2}}}}}$