Miscellaneous(Exercise 6)

Question 1: Using differentials; find the approximate value of each of the following.

(a) $\left(\frac{17}{81}\right)^{\frac{1}{4}}$

(b) $(33)^{-\frac{1}{5}}$

Answer : (a) Let $\mathrm{y}=x^{\frac{1}{4}}$

Let $x=\frac{16}{81}$ and $\Delta x=\frac{1}{81}$

Then, $\Delta y=(x+\Delta x)^{\frac{1}{4}}-x^{\frac{1}{4}}$

$=\left(\frac{17}{81}\right)^{\frac{1}{4}}-\left(\frac{16}{81}\right)^{\frac{1}{4}}$

$=\left(\frac{17}{81}\right)^{\frac{1}{4}}-\frac{2}{3}$

$\therefore\left(\frac{17}{81}\right)^{\frac{1}{4}}=\frac{2}{3}+\Delta y$

Now, dy is approximately equal to $\Delta y$ and given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{1}{4}}}(\Delta y)$

$=\frac{1}{4\left(\frac{16}{81}\right)^{\frac{1}{4}}}\left(\frac{1}{81}\right)$

$=\frac{27}{4 \times 8} \times \frac{1}{81}=\frac{1}{32 \times 3}$

$=\frac{1}{96}=0.010$

Hence, the approximate value of $\left(\frac{17}{81}\right)^{\frac{1}{4}}$ is $\frac{2}{3}+0.010=0.677$

(b) Let $\mathrm{y}=x^{-\frac{1}{5}}$

Let $x=32$ and $\Delta x=1$

Then, $\Delta y=(x+\Delta x)^{-\frac{1}{5}}-x^{-\frac{1}{5}}$

$=(33)^{-\frac{1}{5}}-(32)^{-\frac{1}{5}}=(33)^{-\frac{1}{5}}-\frac{1}{2}$

$\therefore(33)^{-\frac{1}{5}}=\frac{1}{2}+\Delta y$

Now, dy is approximately equal to $\Delta y$ and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{-1}{5(x)^{\frac{6}{5}}}(\Delta x) \quad\left(\text { as } y=x^{-\frac{1}{5}}\right)$

$=-\frac{1}{5(2)^6}(1)=-\frac{1}{320}=-0.003$

Hence, the approximate value of $(33)^{-\frac{1}{5}}$ is $\frac{1}{2}+(-0.003)=0.5-0.003=0.497$

Question 2: Show that the function given by $f(x)=\frac{\log x}{x}$ has maximum at $\mathrm{x}=\mathrm{e}$ .

Answer : The given function is $f(x)=\frac{\log x}{x}$

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

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

$1-\log x=0 \Rightarrow \log x=1$

$\log x=\log e \Rightarrow x=e^{\log x}$

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

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

Now, $f^{\prime \prime}(e)=\frac{-3+2 \log e}{e^3}$

$=\frac{-3+2}{e^3}=\frac{-1}{e^3}<0$

Therefore, by second derivative test, $f$ is the maximum at $x=e$ .

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

Answer : Let $\triangle A B C$ be isosceles where $\mathrm{BC}$ is the base of fixed length b. Let the length of the two equal sides of $\triangle A B C$ be ‘a’, draw $\mathrm{AD} \perp B C$

Now, in $\triangle A D C$ , by applying the Pythagoras theorem,

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

Area of triangle $(\mathrm{A})=\frac{1}{2} b \sqrt{a^2-\frac{b^2}{4}}$

The rate of change of the area with respect to time $(\mathrm{t})$ is given by,

$\frac{d A}{d t}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^2-\frac{b^2}{4}}} \frac{d a}{d t}$

$=\frac{a b}{\sqrt{4 a^2-b^2}} \frac{d a}{d t}$

It is given that the two equal sides of the triangle are decreasing at the rate of $3 \mathrm{~cm}$ per second.

$\frac{d a}{d t}=-3 \mathrm{~cm} / \mathrm{sec}$

$\Rightarrow \frac{d A}{d t}=\frac{-3 a b}{\sqrt{4 a^2-b^2}}$

Then, when $a=b$ we have:

$\frac{d A}{d t}=\frac{-3 a b}{\sqrt{4 a^2-b^2}}=\frac{-3 b^2}{\sqrt{3 b^2}}=-\sqrt{3} b$

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of $\sqrt{3} b \mathrm{~cm} / \mathrm{sec}^2$ .

Question 4: Find the equation of the normal to curve $y^2=4 x$ at the point $(1,2)$ .

Answer : The equation of the given curve is $y^2=4 x$ ,

Differentiating with respect to $x$ , we have

$2 y \frac{d y}{d x}=4$

$\frac{d y}{d x}=\frac{4}{2 y}=\frac{2}{y}$

$\left.\therefore \frac{d y}{d x}\right]_{(1,2)}=\frac{2}{2}=1$

Now, the slope of the normal at point $(1,2)$ is $\frac{-1}{\left.\frac{d y}{d x}\right]_{(1,2)}}=\frac{-1}{1}=-1$

Equation of the normal at

$(1,2)$ is $y-2=-1(x-1) \Rightarrow y-2=-x+1$ i.e. $x+y-3=0$ .

Question 5: Show that the normal at any point $\theta$ to the curve
$x=a \cos \theta+a \theta \sin \theta, \quad y=a \sin \theta-a \theta \cos \theta$ is at a constant distance from the origin.

Answer : We have $x=a \cos \theta+a \theta \sin \theta$

$\Rightarrow \frac{d x}{d \theta}=a(- \sin \theta+ \sin \theta+ \theta \cos \theta)=a \theta \cos \theta$

Since, $y=a \sin \theta-a \theta \cos \theta$

$\Rightarrow \frac{d y}{d \theta}=a (\cos \theta- \cos \theta+\theta \sin \theta)=a \theta \sin \theta$

Now, $\frac{d y}{d x}=\frac{d y/d\theta}{dx/d \theta}$

$=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta$

Slope of the normal= $-\frac{1}{\tan \theta}$

The equation of the normal at a given point $(x, y)$ is given by,

$y-a \sin \theta+a \theta \cos \theta=-\frac{1}{\tan \theta}(x-a \cos \theta-a \theta \sin \theta)$

$\Rightarrow y \sin \theta-a \sin ^2 \theta+a \theta \sin \theta \cos \theta=-x \cos \theta+a \cos ^2 \theta+a \theta \sin \theta \cos \theta$

$\Rightarrow x \cos \theta+y \sin \theta-a\left(\sin ^2 \theta+\cos ^2 \theta\right)=0$

$\Rightarrow x \cos \theta+y \sin \theta-a=0$

Now, the perpendicular distance of the normal from the origin is constant.

$\frac{|-a|}{\sqrt{\sin ^2 \theta+\cos ^2 \theta}}=\frac{|-a|}{\sqrt{1}}=|-a|$ , which is independent of $\theta$ .

Hence, the perpendicular distance of the normal from the origin is constant.

Question 6: Find the intervals in which the function $f$ given by $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ is (i) Increasing (ii) Decreasing.

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

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

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

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

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

$\Rightarrow f^{\prime}(x)=\frac{4 \cos x-4+3 \cos ^2 x+4-4 \cos ^2 x}{(2+\cos x)^2}$

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

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

Now $f^{\prime}(x)=0 \Rightarrow \cos x=0$ or $\cos x=4$

But $\cos x \neq 4$

$\cos x=0 \text { i.e. } x=\frac{\pi}{2}, \frac{3 \pi}{2}$

Now, $x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ divides $(0,2 \pi)$ into three disjoint intervals, i.e.

$\left(0, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \text { and }\left(\frac{3 \pi}{2}, 2 \pi\right)$

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

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

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

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

Question 7: Find the intervals in which the function f given by $f(x)=x^3+\frac{1}{x^3}, x \neq 0$ is (i) increasing (ii) decreasing.

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

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

Then, $f^{\prime}(x)=0$ i.e. $3 x^6-3=0$

$x^6=1,=x=\pm 1$

Now, the points $x=1$ and $x=-1$ divide the real line into three disjoint intervals i.e., $(-\infty,-1),(-1,1)$ and $(1, \infty)$

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

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

In interval $(-1,1)$ i.e., when $-1<\mathrm{x}<1, \mathrm{f}^{\prime}(\mathrm{x})<0$

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

Question 8: Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with its vertex at one end of the major axis.

Answer : The given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Let the major axis be along the $x$ -axis.

Let $A B C$ be the triangle inscribed in the ellipse where
vertex $A$ is at $(a, 0)$ .

Let the point $B(x,y)$

$y=\pm \frac{b}{a} \sqrt{a^2-x^2}$

$BC = 2y$ and $AM = a - x$

Area of $\triangle ABC=\frac{1}{2}\times 2y(a-x)$

$A = y(a-x)=\frac{b}{a} \sqrt{a^2-x^2}(a-x)--(i)$

Squaring both side

$A^2 = S(Let) = \frac{b^2}{a^2}(a^2-x^2)(a-x)^2$

Differntiate with respect to x

$\frac{dS}{dx}= \frac{b^2}{a^2}[(a^2-x^2)2(a-x)(-1)+(a-x)^2(-2x)]$

$\Rightarrow \frac{dS}{dx} = \frac{b^2}{a^2}(a-x)^2[-2(a+x)-2x]$

$\Rightarrow \frac{dS}{dx}= \frac{b^2}{a^2}(a-x)^2[-2a-4x]--(ii)$

For max and minima $\frac{dS}{dx}=0$

$\frac{b^2}{a^2}(a-x)^2[-2a-4x]=0$

$\Rightarrow a-x=0$ or $-2a -4x =0$

$\Rightarrow x = a$ or $x = -\frac{a}{2}$

Since, $x = a$ is not possible

Again differentiate of eq (ii)

$\frac{d^2S}{dx^2}=\frac{b^2}{a^2}[(a-x)^2(-4)+(-2a-4x)2(a-x)(-1)]$

$\Rightarrow \frac{d^S}{dx^2}=\frac{b^2}{a^2}(a-x)[-4(a-x)-2(-2a-4x)]$

At $x = -\frac{a}{2}$

$\frac{d^2S}{dx^2}=\frac{b^2}{a^2}[-4(a-\frac{a}{2})-2(-2a-4(-\frac{a}{2}))]$

$= \frac{b^2}{a^2}[-4(\frac{a}{2})+0]= \frac{b^2}{a^2}(-2a)<0$

Hence area of triangle is max when $x = -\frac{a}{2}$

Putting in eq (i)

Max area $= \frac{b}{a} \sqrt{a^2-(-\frac{a}{2})^2}[a-(-\frac{a}{2})]$

$= \frac{b}{a}\sqrt{\frac{3a^2}{4}}[\frac{3a}{2}]$

$= \frac{3\sqrt{3}ab}{4}$

Question 9: A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is $2 \mathrm{~m}$ and volume is $8 \mathrm{~m}^3$ . If building of tank costs ₹70 per square meters for the base and ₹45 per square metre for sides. What is the cost of least expensive tank?

Answer : Let $\mathrm{l}, \mathrm{b}$ , and $\mathrm{h}$ represent the length, breadth, and height of the tank respectively.

Then, we have height $(h)=2 \mathrm{~m}$

Volume of the $\operatorname{tank}=8 \mathrm{~m}^3$

Volume of the tank $=l \times b \times h$

$\therefore 8=l \times b \times 2$

$l b=4 \quad \Rightarrow b=\frac{4}{l}$

Now, area of the base $=l b=4$

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

Total cost $(C) = 70lb + 2(l+b)h\times 45$

$C = 70l\times \frac{4}{l}+90(l+\frac{4}{l})\times 2$

$\Rightarrow C = 280 + 180(l+\frac{4}{l})--(i)$

Differentiate with respect to ‘l’

$\frac{dC}{dl }= 0 +180(1-\frac{4}{l^2})$

For max and minima $\frac{dC}{dl}=0$

$180(1-\frac{4}{l^2})=0$

$1 = \frac{4}{l^2}$

$\Rightarrow l^2 = 4$

$\Rightarrow l = 2$

Again differentiate with respect to l

$\frac{d^2C}{dl^2}= 180(0-4(\frac{-2}{l^3}))$

$\Rightarrow 180(\frac{8}{l^3})$

At $l = 2$

$\frac{d^2C}{dl^2}=180(\frac{8}{(2)^3})=180>0$

Therefore cost of tank is min when $l = 2$

Putting in (i)

minimum cost $(C) = 280+180(2+\frac{4}{2})$

$= 280+180(4) = 280+720$

$= ₹ 1000$

Question 10: The sum of the perimeter of a circle and square is $\mathrm{k}$ , where $\mathrm{k}$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answer : Let $r$ be the radius of the circle and $a$ be the side of the square. Then, we have:

$2 \pi r+4 a=k($ where $k$ is constant $)$

$\Rightarrow a=\frac{k-2 \pi r}{4}$

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

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

$\frac{d A}{d r}=2 \pi r+\frac{2(k-2 \pi r)(-2 \pi)}{16}=2 \pi r-\frac{\pi(k-2 \pi r)}{4}$

Now $\frac{d A}{d r}=0$ .

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

$\Rightarrow 8 r=k-2 \pi r$

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

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

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

$\therefore$ When $r=\frac{k}{2(4+\pi)},$

$\frac{d^2 A}{d r^2}>0$ .

The sum of the area is least when $r=\frac{k}{2(4+\pi)}$

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

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

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Question 11: A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is $10 \mathrm{~m}$ . Find the dimensions of the window to admit maximum light through the whole opening

Answer: Let $x$ and $y$ be the length and breadth of the rectangular window.

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

It is given that the perimeter of the window is $10 \mathrm{~m}$ .

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

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

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

Area of the window $(A)$ is given by,

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

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

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

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

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

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

Now, $\frac{d A}{d x}=0$

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

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

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

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

Therefore, by second derivative test, the area is the maximum when length $x=\frac{20}{\pi+4} m$ .

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

$=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4} m$

Hence, the required dimensions of the window to admit maximum light is given by

Length $x=\frac{20}{\pi+4} m$ and breath $=\frac{10}{\pi+4} m$ .

Question 12: A point on 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 $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}$

Answer: Let $A B C$ be right-angled at $B$ . Let $A B=x$ and $B C=y$ .

Let $\mathrm{P}$ be a point on the hypotenuse of the triangle such that $\mathrm{P}$ is at a distance of $a$ and $b$ from the sides $\mathrm{AB}$ and $\mathrm{BC}$ respectively. Let $\angle c=\theta$

We have, $\mathrm{AC}=\sqrt{x^2+y^2}$

Now, $P C=b \operatorname{cosec} \theta$ and $A P=a \sec \theta$

$\Rightarrow \mathrm{AC}=\mathrm{AP}+\mathrm{PC} \Rightarrow A C=a \sec \theta+b \operatorname{cosec} \theta$

$\frac{d(A C)}{d \theta}=-b \operatorname{cosec} \theta \cot \theta+a \sec \theta \tan \theta$

$\therefore \frac{d(A C)}{d \theta}=0 \Rightarrow a \sec \theta \tan \theta=b \operatorname{cosec} \theta \cot \theta$

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

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

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

It can be clearly shown that $\frac{d^2(A C)}{d \theta^2}>0$ when $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$

Therefore, by second derivative test, the length of the hypotenuse is minimum when $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$ .

Now, if $\tan \theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}$ , we have:

$A C=\frac{b \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{2}{3}}}+\frac{b \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{2}{3}}}$

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

Hence, the minimum length of the hypotenuses is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{2}{3}}$ .

Question 13: Find the points at which the function $f$ given by $f(x)=(x-2)^2(x+1)^3$ has

(i) local maxima

(ii) local minima

(iii) point of inflexion.

Answer: The given function is $f(x)=(x-2)^2(x+1)^3$

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

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

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

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

$\Rightarrow x=-1$ and $x=\frac{2}{7}$ or $x=2$

Now, for values of $x$ close to $\frac{2}{7}$ and to the left of $\frac{2}{7}, f^{\prime}(x)>0$ Also, for values of $x$ close to $\frac{2}{7}$ and to the right of $\frac{2}{7}, f^{\prime}(x)<0$ .

Thus, $x=\frac{2}{7}$ is the point of local maxima.

Now, for values of $\mathrm{x}$ close to 2 and to the left of $2, f^{\prime}(x)<0$ Also, for values of $\mathrm{x}$ close to 2 and to the right of $2, f^{\prime}(x)>0$ .

Thus, $x=2$ is the point of local minima.

Now, as the value of $x$ varies through $-1, f^{\prime}(x)$ does not changes its sign.

Thus, $x=-1$ is the point of inflexion.

Question 14: Find the absolute maximum and minimum values of the function $\mathrm{f}$ given by
$f(x)=\cos ^2 x+\sin x, \quad x \in[0, \pi]$

Answer: $f(x)=\cos ^2 x+\sin x$

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

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

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

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

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

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

Now, evaluating the value of $f$ at critical points $x=\frac{\pi}{2}$ and $x=\frac{\pi}{6}$ and at the end points of the interval $[0, \pi]$ (i.e., at $x=0$ and $x=\pi$ ),

we have:

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

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

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

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

Hence, the absolute maximum value of $f$ is $\frac{5}{4}$ occurring at $x=\frac{\pi}{6}$ and the absolute minimum value of $f$ is 1 occurring at $x=0, \frac{\pi}{2}$ and $\pi$ .

Question 15: Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4 r}{3}$ .

Answer : Let $O$ be the centre and $r$ be the radius of the given sphere, $O M$ $=x$

In ΔOBM, using pythagoras theorem

$OB^2=OM^2+BM^2$

$\Rightarrow r^2 = x^2 + BM^2$

$\Rightarrow BM^2 = r^2-x^2$

$AM= h = r+x$

BM is the radius of cone

Volume of cone $V = \frac{1}{3}\pi r^2 h$

$\Rightarrow V = \frac{1}{3}\pi(r^2-x^2)(r+x)--(i)$

$\Rightarrow V = \frac{1}{3}\pi(r-x)(r+x)(r+x)$

$\Rightarrow V = \frac{1}{3}\pi (r-x)(r+x)^2$

Differentiating with respect to x

$\frac{dV}{dx} = \frac{1}{3}\pi [(r-x)2(r+x)+(r+x)^2(-1)]$

$= \frac{1}{3}\pi (R+x)[2(R-x)-(R+x)]$

$= \frac{1}{3}\pi(r+x)[2r-2x-r-x]$

$=\frac{1}{3}\pi (r+x)[r-3x]$ —(ii)

For max and minima $\frac{dV}{dx}=0$

$\frac{1}{3}\pi (r+x)[r-3x]=0$

Hence $r+x=0$ or $r-3x = 0$

$x = -r$ or $x = r/3$

$x =-r$ Not possible

Agian differentiate with respect to x of eq (ii)

$\frac{d^2V}{dx^2} = \frac{1}{3}\pi[(r+x)(-2)+(r-2x).1]$

$= \frac{1}{3}\pi[-2r-2x+r-2x]$

$= \frac{1}{3}\pi[-r-4x]$

At $x = r/3$

$\frac{d^V}{dx^2}=\frac{1}{3}\pi[-r-4\times \frac{r}{3}]$

$= \frac{1}{3}[-7r/3]<0$

Hence volume of cone is max when $x=r/3$

Height of cone $= r+x$

$= r+\frac{r}{3}=\frac{4r}{3}$

Question 16: 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 $(a, b)$ .

Answer : We have to prove that function is always increasing, i.e.

$f\left(x_2\right)>f\left(x_1\right)$ for all $x_2>x_1 . \quad\left[\right.$ Where $\left.x_1, x_2 \in[a, b]\right]$

Let $x_1$ and $x_2$ be two numbers in the interval $[a, b], i . e . x_1, x_2 \in[a, b]$ and $x_2>x_1$ .

Let the interval $\left[x_1, x_2\right]$ .

Function $f$ is continuous as well as differential in $\left[x_1, x_2\right]$ as it is continuous and differential in $[a, b]$ .

Using Mean value theorem, there exists $c \in\left[x_1, x_2\right]$ , such that

$f'(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}$

Given that $f^{\prime}(x)>0$ for all $x \in[a, b]$ .

Therefore, $f^{\prime}(c)>0$ for all $x \in\left[x_1, x_2\right]$ .

$\Rightarrow \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}>0$

$\Rightarrow f\left(x_2\right)-f\left(x_1\right)>0 \quad \Rightarrow f\left(x_2\right)>f\left(x_1\right)$

Now, for the two point $x_1, x_2 \in[a, b]$ , where $x_2>x_1$ , we have $f\left(x_2\right)>f\left(x_1\right)$ .

Hence, the functions $f$ is increasing in the interval $[a, b]$ .

Question 17: Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}}$ . Also find the maximum volume.

Answer : Consider $O$ be the centre and $R$ be the radius of the given sphere, $O M$ $=x$

In ΔOBM, using pythagoras theorem

$OB^2=OM^2+BM^2$

$\Rightarrow R^2 = x^2 + BM^2$

$\Rightarrow BM^2 = R^2-x^2$

$MN= h =2x$

BM is the radius of cone

Volume of cone $V = \pi r^2 h$

$\Rightarrow V = \pi(R^2-x^2)2x--(i)$

Differentiating with respect to x

$\frac{dV}{dx} = \pi [(R^2-x^2).2+2x(-2x)]$

$= \pi[2R^2-2x^2-4x^2]$

$\frac{dV}{dx} = \pi[2R^2-6x^2]--(ii)$

For max and minima $\frac{dV}{dx}=0$

$\pi[2R^2-6x^2]=0$

$\Rightarrow 2R^2 = 6x^2$

$\Rightarrow x^2 = \frac{R^2}{3}$

$\Rightarrow x =\frac{R}{\sqrt{3}}$

Again differentiate of eq (ii)

$\frac{d^2V}{dx^2}= \pi[0-12x]$

At $x=\frac{R}{\sqrt{3}}$

$\frac{d^2V}{dx^2}= \pi[-12(\frac{R}{\sqrt{3}})]<0$

Therefore volume of cylinde is max when $x = \frac{R}{\sqrt{3}}$

Height of cylinder $=2x= \frac{2R}{\sqrt{3}}$

Question 18: Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and having semi vertical angle $\alpha$ is one third that of the cone and the greatest volume of the cylinder is $\frac{4}{27}\pi h^2\tan^3\alpha$

Solution: Let radius of cone = r and Radius of cylinder= x

Semivertical angle of cone $=\alpha$ and height of cone $= h$

In $\triangle APQ$

$\tan\alpha = \frac{x}{AP}$

$\Rightarrow AP = x\cot\alpha$

Therefore, Height of cylinder $(QM) = AO - AP$

$= h-x\cot\alpha ---(i)$

Volume of cylinder $= \pi R^2 H$

$V = \pi x^2(h-x\cot\alpha)$

$\Rightarrow V = \pi(hx^2-x^3\cot\alpha)---(ii)$

Differentiating with respect to ‘x’

$\frac{dV}{dx}=\pi(h2x-3x^2\cot\alpha)$

$\Rightarrow \frac{dV}{dx}=\pi(2xh-3x^2\cot\alpha)$

For max and minima $\frac{dV}{dx}=0$

$\pi(2xh-3x^2\cot\alpha)=0$

$\Rightarrow 2xh = 3x^2\cot\alpha$

$\Rightarrow 2h=3x\cot\alpha$

$\Rightarrow x = \frac{2}{3}\tan\alpha$

Again differentiate with respect to x

$\frac{d^2V}{dx^2}= \pi(2h-6x\cot\alpha)$

At $x = \frac{2}{3}h\tan\alpha$

$\frac{d^2V}{dx^2} = \pi(2h-6(\frac{2}{3}h\tan\alpha)\cot\alpha)$

$= \pi(2h-4h)=\pi\times -2h<0$

Therefore the volume of cylinder is max when $x=\frac{2}{3}h\tan\alpha$

Now from eq (i)

Height of cylinder $= h-\frac{2}{3}h\tan\alpha\cot\alpha$

$=h-\frac{2}{3}=\frac{1}{3}h$

Volume of cylinder $=\pi x^2(h-x\cot\alpha)$

$= \pi (\frac{2}{3})\tan\alpha)^2\times \frac{h}{3}$

$= \pi\frac{4}{9}h^2\tan^2\alpha\frac{h}{3}$

Volume of cylinder $= \frac{4}{27}\pi h^3\tan^2\alpha$

Question 19: A cylindrical tank of radius $10 \mathrm{~m}$ is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of wheat is increasing at the rate of:

(A) $1 \mathrm{~m} / \mathrm{h}$

(B) $0.1 \mathrm{~m} / \mathrm{h}$

(D) $0.5 \mathrm{~m} / \mathrm{h}$

Solution: Option (A) is correct.

Consider $y$ be the depth of the wheat in the cylindrical tank whose radius is $10 \mathrm{~m}$ at time t.

$\mathrm{V}=$ Volume of wheat in cylindrical tank at time $t$ ,

So, ${t=\pi(10)^2 y=100 \pi y}$ cubic meter

We are given that $\frac{d \mathrm{~V}}{d t}=314$ cubic meter $/ \mathrm{hr}$

So, $\frac{d}{d t} 100 \pi y=314$

$100 \pi y=314$

$100(3.14) y=314$

Therefore, $y=1 \mathrm{~m} / \mathrm{h}$

Question 20: The slope of the tangent to the curve $x=t^2+3 t-8, y=2 t^2-2 t-5$ at the point (2, $\left.-1\right)$ is:

(A) $\frac{22}{7}$

(B) $\frac{6}{7}$

(D) $\frac{-6}{7}$

Solution: Option (B) is correct.

Solution: Equation of the curves are $x=t^2+3 t-8 \ldots(1)$ and $y=2$ $\frac{d x}{d t}=2 t+3$ and $\frac{d y}{d t}=4 t-2$ Slope of the tangent to the given curve at point $(\mathrm{x}, \mathrm{y})$ is

$(x, y)=\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{4 t-2}{2 t+3}$

At the given point $(2,-1)=x=2$ and $y=-1$

At $x=2$ , from equation $(1), 2=t^2+3 t-8$

$t^2+3 t-10=0$

$(t+5)(t-2)=0$

$t=-5, t=2$

At $y=-1$ , From eq(2) $-1=2t^2-2t-5$

$2t^2-2t-4=0$

$\Rightarrow t^2 - t -2 =0$

$(t-2)(t+1)=0$

$t=2, t=-1$

Here, common value of $t$ in the two sets of values is 2 .
Again, from equation (3),

Slope of the tangent to the given curve at point $(2,-1)=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}$

Question 21: The line $y=m x+1$ is a tangent to the curve $y^2=4 x$ if the value of $m$ is:

(A) 1

(B) 2

(D) $1 / 2$

Solution: Option (A) is correct.

Equation of the curve is $y^2=4 x$

$2 y \frac{d y}{d x}=4.1$

$\frac{d y}{d x}=\frac{2}{y}$

Slope of the tangent to the given curve at point $(x, y)$

$\frac{d y}{d x}=\frac{2}{y}$

$\frac{2}{y}= m$

$y=\frac{2}{m} \ldots(ii)$

Now $y=m x+1$

$\frac{2}{m}=m x+1$

$m x=\frac{2}{m}-1$

$x=\frac{2-m}{m} .$

Putting the values of $x$ and $y$ in equation (1), $\frac{4}{m^2}=\frac{4(2-m)}{m^2}$

$2-m=1 \Rightarrow m=1$

Question 22: The normal at the point $(1,1)$ on the curve $2 y+x^2=3$ is:

(A) $x+y=0$

(B) $x-y=0$

(D) $x-y=1$

Solution: Option (B) is correct.

Equation of the given curve is $2 y+x^2=3$

$2 \frac{d y}{d x}+2 x=0$

$\frac{d y}{d x}=-x$

Slope of the tangent to the given curve at point $(1,1)$ is

$\frac{d y}{d x}=-x=-1=m$

$\text { (say) }$

$\text { Slope of the normal }=\frac{-1}{m}=\frac{-1}{-1}=1$

Equation of the normal at $(1,1)$ is $y-1=1(x-1)$

$y-1=x-1$

$x-y=0$

Question 23: The normal to the curve $x^2=4 y$ passing through $(1,2)$ is:

(A) $x+y=3$

(B) $x-y=3$

(D) $x-y=1$

Solution: Option $(A)$ is correct.

Equation of the curve is $x^2=4 y$

$2 x=4 \frac{d y}{d x}$

$\frac{d y}{d x}=\frac{x}{2}--(i)$

Slope of the normal at $(x, y)$ is

$-\frac{d x}{d y}=\frac{-2}{x}--(ii)$

Again slope of normal at given point $(1,2)=\frac{y-2}{x-1}--(iii)$

From equation (2) and (3), we have $\frac{-2}{x}=\frac{y-2}{x-1}$

$-2 x+2=x y-2 x$

$x y=2$

$y=\frac{2}{x}$

From equation (1),

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

$x^3=8$

$x=2$

$y=\frac{2}{x}=\frac{2}{2}=1$

Now, at point $(2,1)$ , slope of the normal from equation $(2)=\frac{-2}{x}=\frac{-2}{2}=-1$

Equation of the normal is $y-1=-1(x-2)$

$y-1=-x+2$

or $x+y=3$

Question 24: The points on the curve $9 y^2=x^3=$ where the normal to the curve make equal intercepts with axes are:

(A) $\left(4, \pm \frac{8}{3}\right)$

(B) $\left(4,-\frac{8}{3}\right)$

(D) $\left(\pm 4, \frac{3}{8}\right)$

Solution: Option (A) is correct.

Equation of the curve is $9 y^2=x^3--(i)$

$18 y \frac{d y}{d x}=3 x^2$

$\frac{d y}{d x}=\frac{3 x^2}{18 y}=\frac{x^2}{6 y}$

Slope of the tangent to curve (1) at any point $(x, y)$ is

$\frac{dy}{dx} =\frac{x^2}{6y}$

Slope of the normal $=-\frac{1}{dy/dx}$ .

Slopes of lines with equal intercepts on the axis are $=\pm1$

So $-6y = \pm x^2$

if $-6y = x^2$

$y=-\frac{x^2}{6}--(ii)$

From (i) and (ii) we have

$x=4$ and $y=-8/3$

If $-6y = -x^2$

$y =\frac{x^2}{6}--(iii)$

we have, $x=4$ and $y = 8/3$

Required points are $= (4, \pm 8/3)$

class 12 maths chapter 6 Miscellaneous ncert solution

Miscellaneous(Exercise 6)

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