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

Let $\triangle A B C$ be isosceles where BC is the base of fixed length $b$. Also, let the length of the two equal sides of $\triangle A B C$ be $a$. Draw $A D \perp B C$. ![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-135.jpg?height=384&width=307&top_left_y=1861&top_left_x=873) Now, in $\triangle A D C$, by applying the Pythagoras theorem, we have: $$ A D=\sqrt{a^{2}-\frac{b^{2}}{4}} $$ Area of triangle, $$ A=\frac{1}{2} b \sqrt{a^{2}-\frac{b^{2}}{4}} $$ The rate of change of the area with respect to time $(t)$ is given by, $$ \begin{aligned} \frac{d A}{d t} & =\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^{2}-\frac{b^{2}}{4}}} \frac{d a}{d t} \\ & =\frac{a b}{\sqrt{4 a^{2}-b^{2}}} \frac{d a}{d t} \end{aligned} $$ It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. Therefore, $$ \frac{d a}{d t}=-3 \mathrm{~cm} / \mathrm{s} $$ Hence, $$ \Rightarrow \frac{d A}{d t}=\frac{-3 a b}{\sqrt{4 a^{2}-b^{2}}} $$ When, $a=b$ we have: $$ \begin{aligned} \frac{d A}{d t} & =\frac{-3 b^{2}}{\sqrt{4 a^{2}-b^{2}}} \\ & =\frac{-3 b^{2}}{\sqrt{3 b^{2}}} \\ & =-\sqrt{3} b \end{aligned} $$ Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of $-\sqrt{3} b c^{2} / s$.

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

The equation of the given curve is $y^{2}=4 x$ Differentiating with respect to $x$, we have: $$ \begin{aligned} & 2 y \frac{d y}{d x}=4 \\ & \Rightarrow \frac{d y}{d x}=\frac{4}{2 y} \\ & \Rightarrow \frac{d y}{d x}=\frac{2}{y} \\ & \left.\Rightarrow \frac{d y}{d x}\right]_{(1,2)}=\frac{2}{2}=1 \end{aligned} $$ 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 $$ \begin{aligned} & \Rightarrow y-2=-x+1 \\ & \Rightarrow x+y-3=0 \end{aligned} $$

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

We have $x=a \cos \theta+a \theta \sin \theta$ Therefore, $$ \begin{aligned} \frac{d x}{d \theta} & =-a \sin \theta+a \sin \theta+a \theta \cos \theta \\ & =a \theta \cos \theta \end{aligned} $$ Also, $y=a \sin \theta-a \theta \cos \theta$ Hence, $$ \begin{aligned} \frac{d y}{d \theta} & =a \cos \theta-a \cos \theta+a \theta \cos \theta \\ & =a \theta \cos \theta \end{aligned} $$ Thus, $$ \begin{aligned} \frac{d y}{d x} & =\frac{d y}{d \theta} \cdot \frac{d \theta}{d x} \\ & =\frac{a \theta \sin \theta}{a \theta \cos \theta} \\ & =\tan \theta \end{aligned} $$ Slope of the normal at any point $\theta$ is $\frac{-1}{\tan \theta}$. The equation of the normal at a given point $(x, y)$ is given by, $$ \begin{aligned} & y-a \sin \theta+a \theta \cos \theta=\frac{-1}{\tan \theta}(x-a \cos \theta-a \theta \sin \theta) \\ & \Rightarrow y \sin \theta-a \sin ^{2} \theta+a \theta \sin \theta \cos \theta=-x \cos \theta+a \cos ^{2} \theta+a \theta \sin \theta \cos \theta \\ & \Rightarrow x \cos \theta+y \sin \theta-a\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=0 \\ & \Rightarrow x \cos \theta+y \sin \theta-a=0 \end{aligned} $$ Now, the perpendicular distance of the normal from the origin is $$ \frac{|-a|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=\frac{|-a|}{\sqrt{1}}=|-a|, \text { which is independent of } \theta $$ Hence, the perpendicular distance of the normal from the origin is constant.

Q: 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

We have $f(x)=\frac{4 \sin x-2 x-x \cos x}{2+\cos x}$ Hence, $$ \begin{aligned} f^{\prime}(x) & =\frac{(2+\cos x)(4 \cos x-2-\cos x+x \sin x)-(4 \sin x-2 x-x \cos x)(-\sin x)}{(2+\cos x)^{2}} \\ & =\frac{6 \cos x-4+2 x \sin x+3 \cos ^{2} x-2 \cos x+x \sin x \cos x+4 \sin ^{2} x-2 x \sin x-x \sin x \cos x}{(2+\cos x)^{2}} \\ & =\frac{4 \cos x-4+3 \cos ^{2} x+4 \sin ^{2} x}{(2+\cos x)^{2}} \\ & =\frac{4 \cos x-4+3 \cos ^{2} x+4-4 \cos ^{2} x}{(2+\cos x)^{2}} \\ & =\frac{4 \cos x-\cos ^{2} x}{(2+\cos x)^{2}} \end{aligned} $$ $$ f^{\prime}(x)=\frac{\cos x(4-\cos x)}{(2+\cos x)^{2}} $$ Now, $$ \begin{aligned} & f^{\prime}(x)=0 \\ & \Rightarrow \cos x=0 \text { or } \cos x=4 \end{aligned} $$ But $\cos x \neq 4$ Hence, $$ \begin{aligned} & \cos x=0 \\ & \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2} \end{aligned} $$ Now, $x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ divide $(0,2 \pi)$ into three disjoint intervals i.e., $\left(0, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$. In intervals, $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right), f^{\prime}(x)>0$ Thus, $f(x)$ is increasing for $0<x<\frac{x}{2}$ and $\frac{3 x}{2}<x<2 \pi$. In the interval $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right), f^{\prime}(x)<0$ Thus, $f(x)$ is decreasing for $\frac{\pi}{2}<x<\frac{3 \pi}{2}$.

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

Let $l, b$ and $h$ represent the length, breadth, and height of the tank respectively. Then, we have height, $h=2 m$ and volume of the tank, $V=8 m^{3}$ Volume of the tank $$ \begin{aligned} & V=l b h \\ & \Rightarrow 8=l \times b \times 2 \\ & \Rightarrow l b=4 \\ & \Rightarrow b=\frac{4}{l} \end{aligned} $$ Now, area of the base, $l b=4$ Area of the 4 walls, $$ \begin{aligned} A & =2 h(l+b) \\ & =4\left(l+\frac{4}{l}\right) \end{aligned} $$ Hence, $$ \frac{d A}{d l}=4\left(l-\frac{4}{l^{2}}\right) $$ Now, $$ \begin{aligned} & \frac{d A}{d l}=0 \\ & \Rightarrow\left(l-\frac{4}{l^{2}}\right)=0 \\ & \Rightarrow l^{2}=4 \\ & \Rightarrow l= \pm 2 \end{aligned} $$ However, the length cannot be negative. Therefore, we have $l=4$ Hence, $$ \begin{aligned} b & =\frac{4}{l} \\ & =\frac{4}{2} \\ & =2 \end{aligned} $$ Now, $$ \frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}} $$ When, $l=2$ Then, $$ \begin{aligned} \frac{d^{2} A}{d l^{2}} & =\frac{32}{8} \\ & =4>0 \end{aligned} $$ Thus, by second derivative test, the area is the minimum when $l=2$ We have $l=b=h=2$ Therefore, Cost of building the base in ₹ is $$ \begin{aligned} 70 \times(l b) & =70(4) \\ & =280 \end{aligned} $$ Cost of building the walls in ₹ is $$ \begin{aligned} 2 h(l+b) \times 45 & =2 \times 2(2+2) \times 45 \\ & =720 \end{aligned} $$ Required total cost is ₹ is $$ 280+720=1000 $$ Thus, the total cost of the tank will be ₹ 1000 .

Q: 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

The given function is $f(x)=(x-2)^{4}(x+1)^{3}$ Thereofore, $$ \begin{aligned} f^{\prime}(x) & =4(x-2)^{3}(x+1)^{3}+3(x+1)^{2}(x-2)^{4} \\ & =(x-2)^{3}(x+1)^{2}[4(x+1)+3(x-2)] \\ & =(x-2)^{3}(x+1)^{2}(7 x-2) \end{aligned} $$ Now, $$ \begin{aligned} & f^{\prime}(x)=0 \\ & \Rightarrow x=-1, x=\frac{2}{7}, x=2 \end{aligned} $$ 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, $x=2$ is the point of local minima. Now, as the value of $x$ varies through $-1, f^{\prime}(x)$ does not change its sign. Thus, $x=-1$ is the point of inflexion.

Question 1

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

(i) Increasing

(ii) Decreasing

Accepted Answer

We have $f(x)=x^{3}+\frac{1}{x^{3}}$ Therefore, $$ \begin{aligned} f^{\prime}(x) & =3 x^{2}-\frac{3}{x^{4}} \ & =\frac{3 x^{6}-3}{x^{4}} \end{aligned} $$ Now, $$ \begin{aligned} & f^{\prime}(x)=0 \ & \Rightarrow 3 x^{6}-3=0 \ & \Rightarrow x^{6}=1 \ & \Rightarrow x= \pm 1 \end{aligned} $$ 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

Question 2

Using differentials, find the approximate value of each of the following.

(i) $\left(\frac{17}{81}ight)^{\frac{1}{4}}$

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

Solution:

(i) $\left(\frac{17}{81}ight)^{\frac{1}{4}}$

Consider $y=(x)^{\frac{1}{4}}$

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

Then,

$$

\begin{aligned}

\Delta y & =(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}} \

& =\left(\frac{17}{81}ight)^{\frac{1}{4}}-\left(\frac{16}{81}ight)^{\frac{1}{4}} \

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

\frac{2}{3}+\Delta y & =\left(\frac{17}{81}ight)^{\frac{1}{4}}

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{4}}ight] \

& =\frac{1}{4\left(\frac{16}{81}ight)^{\frac{3}{4}}}\left(\frac{1}{81}ight) \

& =\frac{27}{4 	imes 8} 	imes \frac{1}{81} \

& =\frac{1}{32 	imes 3} \

& =\frac{1}{96} \

& =0.010

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\left(\frac{17}{81}ight)^{\frac{1}{4}} & =\frac{2}{3}+0.010 \

& =0.667+0.010 \

& =0.677

\end{aligned}

$$

Thus, the approximate value of $\left(\frac{17}{81}ight)^{\frac{1}{4}}=0.677$.

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

Consider $y=(x)^{-\frac{1}{5}}$

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

Then,

$$

\begin{aligned}

\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} \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{-1}{5(x)^{\frac{6}{5}}}(\Delta x) \quad\left[\because y=(x)^{-\frac{1}{5}}ight] \

& =-\frac{1}{5(2)^{6}}(1) \

& =-\frac{1}{320} \

& =-0.003

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(33)^{-\frac{1}{5}} & =\frac{1}{2}+(-0.003) \

& =0.5-0.003 \

& =0.497

\end{aligned}

$$

Thus, the approximate value of $(33)^{-\frac{1}{5}}=0.497$.

\section*{Question 2:}

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

Accepted Answer

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

Therefore,

$$

\begin{aligned}

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

& =\frac{1-\log x}{x^{2}}

\end{aligned}

$$

Now,

$$

\begin{aligned}

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

& \Rightarrow 1-\log x=0 \

& \Rightarrow \log x=1 \

& \Rightarrow \log x=\log e \

& \Rightarrow x=e

\end{aligned}

$$

Also,

$$

\begin{aligned}

f^{\prime \prime}(x) & =\frac{x^{2}\left(-\frac{1}{x}ight)-(1-\log x)(2 x)}{x^{4}} \

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

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

\end{aligned}

$$

Now,

$$

\begin{aligned}

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

& =\frac{-3+2}{e^{3}} \

& =\frac{-1}{e^{3}}<0

\end{aligned}

$$

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 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

Accepted Answer

Let $	riangle A B C$ be isosceles where BC is the base of fixed length $b$.

Also, let the length of the two equal sides of $	riangle A B C$ be $a$.

Draw $A D \perp B C$.

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-135.jpg?height=384&width=307&top_left_y=1861&top_left_x=873)

Now, in $	riangle A D C$, by applying the Pythagoras theorem, we have:

$$

A D=\sqrt{a^{2}-\frac{b^{2}}{4}}

$$

Area of triangle,

$$

A=\frac{1}{2} b \sqrt{a^{2}-\frac{b^{2}}{4}}

$$

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

$$

\begin{aligned}

\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}

\end{aligned}

$$

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second. Therefore,

$$

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

$$

Hence,

$$

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

$$

When, $a=b$ we have:

$$

\begin{aligned}

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

& =\frac{-3 b^{2}}{\sqrt{3 b^{2}}} \

& =-\sqrt{3} b

\end{aligned}

$$

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

Question 4

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

Accepted Answer

The equation of the given curve is $y^{2}=4 x$

Differentiating with respect to $x$, we have:

$$

\begin{aligned}

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

& \Rightarrow \frac{d y}{d x}=\frac{4}{2 y} \

& \Rightarrow \frac{d y}{d x}=\frac{2}{y} \

& \left.\Rightarrow \frac{d y}{d x}ight]_{(1,2)}=\frac{2}{2}=1

\end{aligned}

$$

Now, the slope of the normal at point $(1,2)$ is

$$

\frac{-1}{\left.\frac{d y}{d x}ight]_{(1,2)}}=\frac{-1}{1}=-1

$$

Equation of the normal at $(1,2)$ is

$$

\begin{aligned}

& \Rightarrow y-2=-x+1 \

& \Rightarrow x+y-3=0

\end{aligned}

$$

Question 5

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

Accepted Answer

We have $x=a \cos 	heta+a 	heta \sin 	heta$

Therefore,

$$

\begin{aligned}

\frac{d x}{d 	heta} & =-a \sin 	heta+a \sin 	heta+a 	heta \cos 	heta \

& =a 	heta \cos 	heta

\end{aligned}

$$

Also, $y=a \sin 	heta-a 	heta \cos 	heta$

Hence,

$$

\begin{aligned}

\frac{d y}{d 	heta} & =a \cos 	heta-a \cos 	heta+a 	heta \cos 	heta \

& =a 	heta \cos 	heta

\end{aligned}

$$

Thus,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d y}{d 	heta} \cdot \frac{d 	heta}{d x} \

& =\frac{a 	heta \sin 	heta}{a 	heta \cos 	heta} \

& =	an 	heta

\end{aligned}

$$

Slope of the normal at any point $	heta$ is $\frac{-1}{	an 	heta}$.

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

$$

\begin{aligned}

& y-a \sin 	heta+a 	heta \cos 	heta=\frac{-1}{	an 	heta}(x-a \cos 	heta-a 	heta \sin 	heta) \

& \Rightarrow y \sin 	heta-a \sin ^{2} 	heta+a 	heta \sin 	heta \cos 	heta=-x \cos 	heta+a \cos ^{2} 	heta+a 	heta \sin 	heta \cos 	heta \

& \Rightarrow x \cos 	heta+y \sin 	heta-a\left(\sin ^{2} 	heta+\cos ^{2} 	hetaight)=0 \

& \Rightarrow x \cos 	heta+y \sin 	heta-a=0

\end{aligned}

$$

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

$$

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

$$

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

Accepted Answer

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

Hence,

$$

\begin{aligned}

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}} \

& =\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}} \

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

& =\frac{4 \cos x-4+3 \cos ^{2} x+4-4 \cos ^{2} x}{(2+\cos x)^{2}} \

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

\end{aligned}

$$

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

$$

Now,

$$

\begin{aligned}

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

& \Rightarrow \cos x=0 	ext { or } \cos x=4

\end{aligned}

$$

But $\cos x 
eq 4$

Hence,

$$

\begin{aligned}

& \cos x=0 \

& \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2}

\end{aligned}

$$

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

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

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

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

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

Question 7

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.

Accepted Answer

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-140.jpg?height=538&width=563&top_left_y=2047&top_left_x=742)

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

Since the ellipse is symmetrical with respect to the $x$-axis and $y$-axis, we can assume the coordinates of A to be $\left(-x_{1}, y_{1}ight)$ and the coordinates of B to be $\left(-x_{1},-y_{1}ight)$

Now, we have $y_{1}= \pm \frac{b}{a} \sqrt{a^{2}-x_{1}^{2}}$

Coordinates of A are $\left(-x_{1}, \frac{b}{a} \sqrt{a^{2}-x_{1}^{2}}ight)$ and

the coordinates of B are $\left(x_{1},-\frac{b}{a} \sqrt{a^{2}-x_{1}^{2}}ight)$

As the point $\left(x_{1}, y_{1}ight)$ lies on the ellipse, the area of triangle $\mathrm{ABC}(A)$ is given by,

$$

\begin{align*}

A & =\frac{1}{2}\left|a\left(\frac{2 b}{a} \sqrt{a^{2}-x_{1}^{2}}ight)+\left(-x_{1}ight)\left(-\frac{b}{a} \sqrt{a^{2}-x_{1}^{2}}ight)+\left(-x_{1}ight)\left(-\frac{b}{a} \sqrt{a^{2}-x_{1}^{2}}ight)ight| \

& =b \sqrt{a^{2}-x_{1}^{2}}+x_{1} \frac{b}{a} \sqrt{a^{2}-x_{1}^{2}} 	ag{1}

\end{align*}

$$

Therefore,

$$

\begin{aligned}

\frac{d A}{d x_{1}} & =\frac{-2 x_{1} b}{2 \sqrt{a^{2}-x_{1}^{2}}}+\frac{b}{a} \sqrt{a^{2}-x_{1}^{2}}-\frac{-2 b x_{1}^{2}}{2 a \sqrt{a^{2}-x_{1}^{2}}} \

& =\frac{b}{a \sqrt{a^{2}-x_{1}^{2}}}\left[-x_{1} a+\left(a^{2}-x_{1}^{2}ight)-x_{1}^{2}ight] \

& =\frac{b\left(-2 x_{1}^{2}-x_{1} a+a^{2}ight)}{a \sqrt{a^{2}-x_{1}^{2}}}

\end{aligned}

$$

Now, $\frac{d A}{d x_{1}}=0$

Hence,

$$

\begin{aligned}

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

& \Rightarrow x_{1}=\frac{a \pm \sqrt{a^{2}-4(-2)\left(a^{2}ight)}}{2(-2)} \

& \Rightarrow x_{1}=\frac{a \pm \sqrt{9 a^{2}}}{-4} \

& \Rightarrow x_{1}=\frac{a \pm 3 a}{-4} \

& \Rightarrow x_{1}=-a, \frac{a}{2}

\end{aligned}

$$

But $x_{1} 
eq-a$

Therefore,

$$

\begin{aligned}

x_{1} & =\frac{a}{2} \

y_{1} & =\frac{b}{a} \sqrt{a^{2}-\frac{a^{2}}{4}} \

& =\frac{b a}{2 a} \sqrt{3} \

& =\frac{\sqrt{3} b}{2}

\end{aligned}

$$

Now,

$$

\begin{aligned}

\frac{d^{2} A}{d x_{1}^{2}} & =\frac{b}{a}\left\{\frac{\sqrt{a^{2}-x_{1}^{2}}\left(-4 x_{1}-aight)-\left(-2 x_{1}^{2}-x_{1} a+a^{2}ight) \frac{\left(-2 x_{1}ight)}{2 \sqrt{a^{2}-x_{1}^{2}}}}{a^{2}-x_{1}^{2}}ight\} \

& =\frac{b}{a}\left\{\frac{\left(a^{2}-x_{1}^{2}ight)\left(-4 x_{1}-aight)+x_{1}\left(-2 x_{1}^{2}-x_{1} a+a^{2}ight)}{\left(a^{2}-x_{1}^{2}ight)^{\frac{3}{2}}}ight\} \

& =\frac{b}{a}\left\{\frac{2 x^{3}-3 a^{2} x-a^{3}}{\left(a^{2}-x_{1}^{2}ight)^{\frac{3}{2}}}ight\}

\end{aligned}

$$

Also, when, $x_{1}=\frac{a}{2}$

Then,

$$

\begin{aligned}

\frac{d^{2} A}{d x_{1}^{2}} & =\frac{b}{a}\left\{\frac{2\left(\frac{a^{3}}{8}ight)-3 a^{2}\left(\frac{a}{2}ight)-a^{3}}{\left(a^{2}-\left(\frac{a}{2}ight)^{2}ight)^{\frac{3}{2}}}ight\} \

& =\frac{b}{a}\left\{\frac{\frac{a^{3}}{4}-\frac{3}{2} a^{3}-a^{3}}{\left(\frac{3 a^{2}}{4}ight)^{\frac{3}{2}}}ight\} \

& =-\frac{b}{a}\left\{\frac{\frac{9}{4} a^{3}}{\left(\frac{3 a^{2}}{4}ight)^{\frac{3}{2}}}ight\}<0

\end{aligned}

$$

Thus, the area is the maximum when $x_{1}=\frac{a}{2}$.

Hence, Maximum area of the triangle is given by,

$$

\begin{aligned}

A & =b \sqrt{a^{2}-\frac{a^{2}}{4}}+\left(\frac{a}{2}ight) \frac{b}{a} \sqrt{a^{2}-\frac{a^{2}}{4}} \

& =a b \frac{\sqrt{3}}{2}+\left(\frac{a}{2}ight) \frac{b}{a} 	imes \frac{a \sqrt{3}}{2} \

& =\frac{a b \sqrt{3}}{2}+\frac{a b \sqrt{3}}{4} \

& =\frac{3 \sqrt{3}}{4} a b

\end{aligned}

$$

Question 8

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

Accepted Answer

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

Then, we have height, $h=2 m$ and volume of the tank, $V=8 m^{3}$

Volume of the tank

$$

\begin{aligned}

& V=l b h \

& \Rightarrow 8=l 	imes b 	imes 2 \

& \Rightarrow l b=4 \

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

\end{aligned}

$$

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

Area of the 4 walls,

$$

\begin{aligned}

A & =2 h(l+b) \

& =4\left(l+\frac{4}{l}ight)

\end{aligned}

$$

Hence,

$$

\frac{d A}{d l}=4\left(l-\frac{4}{l^{2}}ight)

$$

Now,

$$

\begin{aligned}

& \frac{d A}{d l}=0 \

& \Rightarrow\left(l-\frac{4}{l^{2}}ight)=0 \

& \Rightarrow l^{2}=4 \

& \Rightarrow l= \pm 2

\end{aligned}

$$

However, the length cannot be negative.

Therefore, we have $l=4$

Hence,

$$

\begin{aligned}

b & =\frac{4}{l} \

& =\frac{4}{2} \

& =2

\end{aligned}

$$

Now,

$$

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

$$

When, $l=2$

Then,

$$

\begin{aligned}

\frac{d^{2} A}{d l^{2}} & =\frac{32}{8} \

& =4>0

\end{aligned}

$$

Thus, by second derivative test, the area is the minimum when $l=2$

We have $l=b=h=2$

Therefore,

Cost of building the base in ₹ is

$$

\begin{aligned}

70 	imes(l b) & =70(4) \

& =280

\end{aligned}

$$

Cost of building the walls in ₹ is

$$

\begin{aligned}

2 h(l+b) 	imes 45 & =2 	imes 2(2+2) 	imes 45 \

& =720

\end{aligned}

$$

Required total cost is ₹ is

$$

280+720=1000

$$

Thus, the total cost of the tank will be ₹ 1000 .

Question 9

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

Accepted Answer

Let $r$ be the radius of the circle and $a$ be the side of the square.

Then, we have:

$$

\begin{aligned}

& 2 \pi r+4 a=k \

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

\end{aligned}

$$

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

$$

\begin{aligned}

A & =\pi r^{2}+a^{2} \

& =\pi r^{2}+\frac{(k-2 \pi r)^{2}}{16}

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\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}

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \frac{d A}{d r}=0 \

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

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

\end{aligned}

$$

When, $r=\frac{k}{2(4+\pi)}, \Rightarrow \frac{d^{2} A}{d r^{2}}>0$

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

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

Then,

$$

\begin{aligned}

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

& =\frac{k(4+\pi)-\pi k}{4(4+\pi)} \

& =\frac{4 k}{4(4+\pi)} \

& =\frac{k}{4+\pi} \

& =2 r

\end{aligned}

$$

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 10

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

Accepted Answer

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

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

It is given that the perimeter of the window is $10 m$.

Therefore,

$$

\begin{aligned}

& x+2 y+\frac{\pi x}{2}=10 \

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

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

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

\end{aligned}

$$

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

$$

\begin{aligned}

A & =x y+\frac{\pi}{2}\left(\frac{x}{2}ight)^{2} \

& =x\left[5-x\left(\frac{1}{2}+\frac{\pi}{4}ight)ight]+\frac{\pi}{8} x^{2} \

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

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d A}{d x} & =5-2 x\left(\frac{1}{2}+\frac{\pi}{4}ight)+\frac{\pi}{4} x \

& =5-x\left(1+\frac{\pi}{2}ight)+\frac{\pi}{4} x \

\frac{d^{2} A}{d x^{2}} & =-\left(1+\frac{\pi}{2}ight)+\frac{\pi}{4} \

& =-1-\frac{\pi}{4}

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \frac{d A}{d x}=0 \

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

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

& \Rightarrow x\left(1+\frac{\pi}{4}ight)=5 \

& \Rightarrow x=\frac{5}{\left(1+\frac{\pi}{4}ight)} \

& \Rightarrow x=\frac{20}{\pi+4}

\end{aligned}

$$

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 is $\frac{20}{\pi+4} m$

Now,

$$

\begin{aligned}

y & =5-\frac{20}{\pi+4}\left(\frac{2+\pi}{4}ight) \

& =5-\frac{5(2+\pi)}{\pi+4} \

& =\frac{10}{\pi+4}

\end{aligned}

$$

Hence, the required dimensions of the window to admit maximum light is given by length $\frac{20}{\pi+4} m$ and breadth $\frac{10}{\pi+4} m$.

Question 11

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.

Accepted Answer

A sphere of fixed radius $(R)$ is given.

Let $r$ and $h$ be the radius and the height of the cylinder respectively.

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-156.jpg?height=513&width=501&top_left_y=1798&top_left_x=770)

From the given figure, we have $h=2 \sqrt{R^{2}-r^{2}}$

The volume $(V)$ of the cylinder is given by,

$$

\begin{aligned}

V & =\pi r^{2} h \

& =2 \pi r^{2} \sqrt{R^{2}-r^{2}}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

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

\frac{d V}{d r} & =4 \pi r \sqrt{R^{2}-r^{2}}+\frac{2 \pi r^{2}(-2 r)}{2 \sqrt{R^{2}-r^{2}}} \

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

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

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

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \frac{d V}{d r}=0 \

& \Rightarrow \frac{4 \pi r R^{2}-6 \pi r^{3}}{\sqrt{R^{2}-r^{2}}}=0 \

& \Rightarrow 4 \pi r R^{2}=6 \pi r^{3} \

& \Rightarrow r^{2}=\frac{2 R^{2}}{3}

\end{aligned}

$$

Also,

$$

\begin{aligned}

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

& =\frac{\left(R^{2}-r^{2}ight)\left(4 \pi R^{2}-18 \pi r^{2}ight)+r\left(4 \pi r R^{2}-6 \pi r^{3}ight)}{\left(R^{2}-r^{2}ight)^{\frac{3}{2}}} \

& =\frac{\left(4 \pi R^{4}-22 \pi r^{2} R^{2}+12 \pi r^{4}+4 \pi r^{2} R^{2}ight)}{\left(R^{2}-r^{2}ight)^{\frac{3}{2}}}

\end{aligned}

$$

Now, it can be observed that at $r^{2}=\frac{2 R^{2}}{3}, \Rightarrow \frac{d^{2} V}{d r^{2}}<0$

Thus, the volume is the maximum when $r^{2}=\frac{2 R^{2}}{3}$.

When, $r^{2}=\frac{2 R^{2}}{3}$

Then, the height of the cylinder is

$$

\begin{aligned}

2 \sqrt{R^{2}-\frac{2 R^{2}}{3}} & =2 \sqrt{\frac{R^{2}}{3}} \

& =\frac{2 R}{\sqrt{3}}

\end{aligned}

$$

Hence, the volume of the cylinder is the maximum when the height of the cylinder is $\frac{2 R}{\sqrt{3}}$.

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

Accepted Answer

Let $	riangle A B C$ be right-angled at $\mathrm{B}, A B=x, B C=y$ and $\angle C=	heta$.

Also, let P be a point on the hypotenuse of the triangle such that P is at a distance of $a$ and $b$ from the sides AB and BC respectively.

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-149.jpg?height=427&width=435&top_left_y=1210&top_left_x=813)

We have, $A C=\sqrt{x^{2}+y^{2}}$

Now,

$$

\begin{aligned}

& P C=b \operatorname{cosec} 	heta \

& A P=a \sec 	heta

\end{aligned}

$$

Hence,

$$

\begin{align*}

& A C=A P+P C \

& A C=b \operatorname{cosec} 	heta+a \sec 	heta 	ag{1}

\end{align*}

$$

Therefore,

$$

\frac{d(A C)}{d 	heta}=-b \operatorname{cosec} 	heta \cot 	heta+a \sec 	heta 	an 	heta

$$

Now,

$$

\begin{aligned}

& \frac{d(A C)}{d 	heta}=0 \

& \Rightarrow-b \operatorname{cosec} 	heta \cot 	heta+a \sec 	heta 	an 	heta=0 \

& \Rightarrow a \sec 	heta 	an 	heta=b \operatorname{cosec} 	heta \cot 	heta \

& \Rightarrow \frac{a}{\cos 	heta} \cdot \frac{\sin 	heta}{\cos 	heta}=\frac{b}{\sin 	heta} \cdot \frac{\cos 	heta}{\sin 	heta} \

& \Rightarrow a \sin ^{3} 	heta=b \cos ^{3} 	heta \

& \Rightarrow(a)^{\frac{1}{3}} \sin 	heta=(b)^{\frac{1}{3}} \cos 	heta \

& \Rightarrow 	an 	heta=\left(\frac{b}{a}ight)^{\frac{1}{3}}

\end{aligned}

$$

Thus,

$$

\begin{equation*}

\sin 	heta=\frac{(b)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}} 	ext { and } \cos 	heta=\frac{(a)^{\frac{1}{3}}}{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}} 	ag{2}

\end{equation*}

$$

It can be clearly shown that $\frac{d^{2}(A C)}{d 	heta^{2}}<0$, when $	an 	heta=\left(\frac{b}{a}ight)^{\frac{1}{3}}$.

By second derivative test the length of the hypotenuse is the maximum when $	an 	heta=\left(\frac{b}{a}ight)^{\frac{1}{3}}$ When, $	an 	heta=\left(\frac{b}{a}ight)^{\frac{1}{3}}$, we have:

$$

\begin{aligned}

A C & =\frac{b \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{(b)^{\frac{1}{3}}}+\frac{a \sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{(a)^{\frac{1}{3}}} \quad[	ext { Using (1) and (2) }] \

& =\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\left(b^{\frac{2}{3}}+a^{\frac{2}{3}}ight) \

& =\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}ight)^{\frac{3}{2}}

\end{aligned}

$$

Thus, the maximum length of the hypotenuses is $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}ight)^{\frac{3}{2}}$ proved.

Question 13

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

Accepted Answer

The given function is $f(x)=(x-2)^{4}(x+1)^{3}$ Thereofore, $$ \begin{aligned} f^{\prime}(x) & =4(x-2)^{3}(x+1)^{3}+3(x+1)^{2}(x-2)^{4} \ & =(x-2)^{3}(x+1)^{2}[4(x+1)+3(x-2)] \ & =(x-2)^{3}(x+1)^{2}(7 x-2) \end{aligned} $$ Now, $$ \begin{aligned} & f^{\prime}(x)=0 \ & \Rightarrow x=-1, x=\frac{2}{7}, x=2 \end{aligned} $$ 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 $x$ close to 2 and to the left of $2, f^{\prime}(x)<0$ Also, for values of $x$ 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 change its sign. Thus, $x=-1$ is the point of inflexion.

Question 14

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

Accepted Answer

We have $f(x)=\cos ^{2} x+\sin x$

Therefore,

$$

\begin{aligned}

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

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

\end{aligned}

$$

Now,

$$

\begin{aligned}

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

& \Rightarrow-2 \sin x \cos x+\cos x=0 \

& \Rightarrow \cos x=2 \sin x \cos x \

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

& \Rightarrow \sin x=\frac{1}{2} 	ext { or } \cos x=0 \

& \Rightarrow x=\frac{\pi}{6} 	ext { or } \frac{\pi}{2} \quad \because x \in[0, \pi]

\end{aligned}

$$

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

$$

\begin{aligned}

f\left(\frac{\pi}{6}ight) & =\cos ^{2}\left(\frac{\pi}{6}ight)+\sin \left(\frac{\pi}{6}ight) \

& =\left(\frac{\sqrt{3}}{2}ight)^{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}ight) & =\cos ^{2}\left(\frac{\pi}{2}ight)+\sin \left(\frac{\pi}{2}ight) \

& =0+1 \

& =1

\end{aligned}

$$

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 $x=0, \frac{\pi}{2}, \pi$.

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 $\frac{4 r}{3}$.

Accepted Answer

A sphere of fixed radius $(r)$ is given.

Let $R$ and $h$ be the radius and the height of the cone respectively.

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-153.jpg?height=324&width=301&top_left_y=1130&top_left_x=879)

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

$$

V=\frac{1}{3} \pi R^{2} h

$$

Now, from the right $	riangle B C D$, we have:

$$

\begin{aligned}

& B C=\sqrt{r^{2}-R^{2}} \

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

\end{aligned}

$$

Hence,

$$

\begin{aligned}

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

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

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d V}{d R} & =\frac{2}{3} \pi R r+\frac{2}{3} \pi R \sqrt{r^{2}-R^{2}}+\frac{\pi R^{2}}{3} \cdot \frac{(-2 R)}{2 \sqrt{r^{2}-R^{2}}} \

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

& =\frac{2}{3} \pi R r+\frac{2 \pi R\left(r^{2}-R^{2}ight)-\pi R^{3}}{3 \sqrt{r^{2}-R^{2}}} \

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

\end{aligned}

$$

Now,

$$

\begin{aligned}

& \frac{d V}{d R^{2}}=0 \

& \Rightarrow \frac{2}{3} \pi R r+\frac{2 \pi R r^{2}-3 \pi R^{3}}{3 \sqrt{r^{2}-R^{2}}}=0 \

& \Rightarrow \frac{2 \pi R r}{3}=\frac{3 \pi R^{3}-2 \pi R r^{2}}{3 \sqrt{r^{2}-R^{2}}} \

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

& \Rightarrow 4 r^{2}\left(r^{2}-R^{2}ight)=\left(3 R^{2}-2 r^{2}ight)^{2} \

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

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

& \Rightarrow 9 R^{2}=8 r^{2} \

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

\end{aligned}

$$

Also,

$$

\begin{aligned}

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

& =\frac{2 \pi r}{3}+\frac{3 \sqrt{r^{2}-R^{2}}\left(2 \pi r^{2}-9 \pi R^{2}ight)+\left(2 \pi R r^{2}-3 \pi R^{3}ight)(3 R) \frac{1}{\sqrt{r^{2}-R^{2}}}}{9\left(r^{2}-R^{2}ight)}

\end{aligned}

$$

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

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

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

Then, height of the cone

$$

\begin{aligned}

h & =r+\sqrt{r^{2}-\frac{8 r^{2}}{9}} \

& =r+\sqrt{\frac{r^{2}}{9}} \

& =r+\frac{r}{3} \

& =\frac{4 r}{3}

\end{aligned}

$$

Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $\frac{4 r}{3}$.

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 $(a, b)$.

Accepted Answer

Let $x_{1}, x_{2} \in(a, b)$ such that $x_{1}>x_{2}$ Consider the sub-interval $\left[x_{1}, x_{2} ight]$ Since $f(x)$ is differentiable on $(a, b)$ and $\left[x_{1}, x_{2} ight] \subset(a, b)$. Therefore, $f(x)$ is continuous on $\left[x_{1}, x_{2} ight]$ and differentiable on $\left(x_{1}, x_{2} ight)$. By the Lagrange's mean value theorem, there exists $c \in\left(x_{1}, x_{2} ight)$ such that $$ \begin{equation*} f(c)=\frac{f\left(x_{2} ight)-f\left(x_{1} ight)}{x_{1}-x_{2}} ag{1} \end{equation*} $$ Since, $f^{\prime}(x)>0$ for all $x \in(a, b)$, so in particular, $$ \begin{aligned} & f^{\prime}(c)>0 \ & \Rightarrow \frac{f\left(x_{2} ight)-f\left(x_{1} ight)}{x_{1}-x_{2}}>0 \quad[ ext { Using }(1)] \ & \Rightarrow f\left(x_{2} ight)-f\left(x_{1} ight)>0 \ & \Rightarrow f\left(x_{2} ight)>f\left(x_{1} ight) \ & \Rightarrow f\left(x_{1} ight)

Question 17

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

Accepted Answer

The given right circular cone of fixed height $h$ and semi-vertical angle $\alpha$ can be drawn as:

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-158.jpg?height=432&width=435&top_left_y=1467&top_left_x=813)

Here, a cylinder of radius $R$ and height $H$ is inscribed in the cone.

Then, $\angle G A O=\alpha, O G=r, O A=h, O E=R$ and $C E=H$.

We have, $r=h 	an \alpha$

Now, since $	riangle A O G$ is similar to $	riangle C E G$, we have:

$$

\begin{aligned}

& \frac{A O}{O G}=\frac{C E}{E G} \

& \Rightarrow \frac{h}{r}=\frac{H}{r-R} \quad \quad[\because E G=O G-O E] \

& \Rightarrow H=\frac{h}{r}(r-R) \

& \Rightarrow H=\frac{h}{h 	an \alpha}(h 	an \alpha-R) \

& \Rightarrow H=\frac{1}{	an \alpha}(h 	an \alpha-R)

\end{aligned}

$$

Now, the volume $(V)$ of the cylinder is given by,

$$

\begin{aligned}

V & =\pi R^{2} H \

& =\frac{\pi R^{2}}{	an \alpha}(h 	an \alpha-R) \

& =\pi R^{2} h-\frac{\pi R^{3}}{	an \alpha}

\end{aligned}

$$

Therefore,

$$

\frac{d V}{d R}=2 \pi R h-\frac{3 \pi R^{2}}{	an \alpha}

$$

Now,

$$

\begin{aligned}

& \frac{d V}{d R}=0 \

& \Rightarrow 2 \pi R h-\frac{3 \pi R^{2}}{	an \alpha}=0 \

& \Rightarrow 2 \pi R h=\frac{3 \pi R^{2}}{	an \alpha} \

& \Rightarrow 2 h 	an \alpha=3 R \

& \Rightarrow R=\frac{2 h}{3} 	an \alpha

\end{aligned}

$$

Also

$$

\frac{d^{2} V}{d R^{2}}=2 \pi h-\frac{6 \pi R}{	an \alpha}

$$

For $R=\frac{2 h}{3} 	an \alpha$, we have:

$$

\begin{aligned}

\frac{d^{2} V}{d R^{2}} & =2 \pi h-\frac{6 \pi}{	an \alpha}\left(\frac{2 h}{3} 	an \alphaight) \

& =2 \pi h-4 \pi h \

& =-2 \pi h<0

\end{aligned}

$$

By second derivative test, the volume of the cylinder is the greatest when $R=\frac{2 h}{3} 	an \alpha$. When, $R=\frac{2 h}{3} 	an \alpha$

Then,

$$

\begin{aligned}

H & =\frac{1}{	an \alpha}\left(h 	an \alpha-\frac{2 h}{3} 	an \alphaight) \

& =\frac{1}{	an \alpha}\left(\frac{h 	an \alpha}{3}ight) \

& =\frac{h}{3}

\end{aligned}

$$

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.

Thus, the maximum volume of the cylinder can be obtained as:

$$

\begin{aligned}

\pi\left(\frac{2 h}{3} 	an \alphaight)^{2}\left(\frac{h}{3}ight) & =\pi\left(\frac{4 h^{2}}{9} 	an ^{2} \alphaight)\left(\frac{h}{3}ight) \

& =\frac{4}{27} \pi h^{3} 	an ^{2} \alpha

\end{aligned}

$$

Hence, the given result is proved.

Question 18

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of

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

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

(C) $1.1 \mathrm{~m} / \mathrm{h}$

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

Accepted Answer

Let $r$ be the radius of the cylinder.

Then, volume $(V)$ of the cylinder is given by,

$$

\begin{aligned}

V & =\pi r^{2} h \

& =\pi(10)^{2} h \

& =100 \pi h

\end{aligned}

$$

Differentiating with respect to time $(t)$, we have:

$$

\frac{d V}{d t}=100 \pi \frac{d h}{d t}

$$

The tank is being filled with wheat at the rate of $314 m^{3} / h$

$$

\frac{d V}{d t}=314 m^{3} / h

$$

Thus, we have:

$$

\begin{aligned}

100 \pi \frac{d h}{d t} & =314 \

\frac{d h}{d t} & =\frac{314}{100(3.14)} \

& =1

\end{aligned}

$$

Hence, the depth of wheat is increasing at the rate of $1 m / h$.

Thus, the correct option is $\mathbf{A}$.

Question 19

The slope of the tangent to the curve $x=t^{2}+3 t-8, y=2 t^{2}-2 t-5$ at the point $(2,-1)$ is

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

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

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

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

Accepted Answer

The given curve is $x=t^{2}+3 t-8$ and $y=2 t^{2}-2 t-5$.

Therefore,

$$

\begin{aligned}

& \frac{d x}{d t}=2 t+3 \

& \frac{d y}{d t}=4 t-2

\end{aligned}

$$

Hence,

$$

\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{4 t-2}{2 t+3}

$$

The given point is $(2,-1)$

At $x=2$, we have:

$$

\begin{aligned}

& t^{2}+3 t-8=2 \

& \Rightarrow t^{2}+3 t-10=0 \

& \Rightarrow(t-2)(t+5)=0 \

& \Rightarrow t=2 	ext { or } t=-5

\end{aligned}

$$

At $y=-1$,we have:

$$

\begin{aligned}

& 2 t^{2}-2 t-5=-1 \

& \Rightarrow 2 t^{2}-2 t-4=0 \

& \Rightarrow 2\left(t^{2}-t-2ight)=0 \

& \Rightarrow(t-2)(t+1)=0 \

& \Rightarrow t=2 	ext { or } t=-1

\end{aligned}

$$

The common value is $t=2$

Hence, the slope of the tangent to the given curve at point $(2,-1)$ is

$$

\begin{aligned}

\left.\frac{d y}{d x}ight]_{t=2} & =\frac{4(2)-2}{2(2)+3} \

& =\frac{8-2}{4+3} \

& =\frac{6}{7}

\end{aligned}

$$

Thus, the correct option is B.

Question 20

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

(C) 3

(D) $\frac{1}{2}$

Accepted Answer

The equation of the tangent to the given curve is $y=m x+1$

Now, substituting $y=m x+1$ in $y^{2}=4 x$, we get:

$$

\begin{align*}

& (m x+1)^{2}=4 x \

& \Rightarrow m^{2} x^{2}+1+2 m x-4 x=0 \

& \Rightarrow m^{2} x^{2}+(2 m-4) x+1=0 	ag{1}

\end{align*}

$$

Since a tangent touches the curve at one point, the roots of equation (1) must be equal.

Therefore, we have:

$$

\begin{aligned}

& 	ext { Discriminant }=0 \

& \Rightarrow(2 m-4)^{2}-4\left(m^{2}ight)(1)=0 \

& \Rightarrow 4 m^{2}+16-16 m-4 m^{2}=0 \

& \Rightarrow 16-16 m=0 \

& \Rightarrow m=1

\end{aligned}

$$

Hence, the required value of $m$ is 1 .

Thus, the correct option is $\mathbf{A}$.

Question 21

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

(A) $x+y=0$

(B) $x-y=0$

(C) $x+y+1=0$

(D) $x-y=1$

Accepted Answer

The equation of the given curve is $2 y+x^{2}=3$

Differentiating with respect to $x$, we have:

$$

\begin{aligned}

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

& \Rightarrow \frac{d y}{d x}=-x \

& \left.\Rightarrow \frac{d y}{d x}ight]_{(1,1)}=-1

\end{aligned}

$$

The slope of the normal to the given curve at point $(1,1)$ is

$$

\frac{-1}{\left.\frac{d y}{d x}ight]_{(1,1)}}=1

$$

Hence, the equation of the normal to the given curve at $(1,1)$ is given as:

$$

\begin{aligned}

& \Rightarrow y-1=1(x-1) \

& \Rightarrow y-1=x-1 \

& \Rightarrow x-y=0

\end{aligned}

$$

Thus, the correct option is $\mathbf{B}$.

Question 22

The normal to the curve $x^{2}=4 y$ passing $(1,2)$ is

(A) $x+y=3$

(B) $x-y=3$

(C) $x+y=1$

(D) $x-y=1$

Accepted Answer

The equation of the given curve is $x^{2}=4 y$

Differentiating with respect to $x$, we have:

$$

\begin{aligned}

& 2 x=4 \cdot \frac{d y}{d x} \

& \Rightarrow \frac{d y}{d x}=\frac{x}{2}

\end{aligned}

$$

The slope of the normal to the given curve at point $(h, k)$ is

$$

\frac{-1}{\left.\frac{d y}{d x}ight]_{(h, k)}}=-\frac{2}{h}

$$

Hence, the equation of the normal to the given curve at $(h, k)$ is given as:

$$

y-k=-\frac{2}{h}(x-h)

$$

Now, it is given that the normal passes through the point $(1,2)$

Therefore, we have:

$$

\begin{align*}

& 2-k=-\frac{2}{h}(1-h) \

& \Rightarrow k=2+\frac{2}{h}(1-h) 	ag{1}

\end{align*}

$$

Since $(h, k)$ lies on the curve $x^{2}=4 y$, we have $h^{2}=4 k$

$$

\Rightarrow k=\frac{h^{2}}{4}

$$

From equation (1), we have:

$$

\begin{aligned}

& \frac{h^{2}}{4}=2+\frac{2}{h}(1-h) \

& \Rightarrow \frac{h^{3}}{4}=2 h+2-2 h \

& \Rightarrow \frac{h^{3}}{4}=2 \

& \Rightarrow h^{3}=8 \

& \Rightarrow h=2

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

& k=\frac{h^{2}}{4} \

& \Rightarrow k=1

\end{aligned}

$$

Hence, the equation of the normal is given as:

$$

\begin{aligned}

& \Rightarrow y-1=\frac{-2}{2}(x-2) \

& \Rightarrow y-1=-x+2 \

& \Rightarrow x+y=3

\end{aligned}

$$

Thus, the correct option is $\mathbf{A}$.

Question 23

The points on the curve $9 y^{2}=x^{3}$, where the normal to the curve makes equal intercepts with the axes are

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

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

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

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

Accepted Answer

The equation of the given curve is $9 y^{2}=x^{3}$

Differentiating with respect to $x$, we have:

$$

\begin{aligned}

& 9(2 y) \frac{d y}{d x}=3 x^{2} \

& \Rightarrow \frac{d y}{d x}=\frac{x^{2}}{6 y}

\end{aligned}

$$

The slope of the normal to the given curve at point $\left(x_{1}, y_{1}ight)$ is

$$

\frac{-1}{\left.\frac{d y}{d x}ight]_{\left(x_{1}, y_{1}ight)}}=-\frac{6 y_{1}}{x_{1}^{2}}

$$

The equation of the normal to the curve at $\left(x_{1}, y_{1}ight)$ is

$$

\begin{aligned}

& y-y_{1}=-\frac{6 y_{1}}{x_{1}^{2}}\left(x-x_{1}ight) \

& \Rightarrow x_{1}^{2} y-x_{1}^{2} y_{1}=-6 x y_{1}+6 x_{1} y_{1} \

& \Rightarrow 6 x y_{1}+x_{1}^{2} y=6 x_{1} y_{1}+x_{1}^{2} y_{1} \

& \Rightarrow \frac{6 x y_{1}+x_{1}^{2} y}{6 x_{1} y_{1}+x_{1}^{2} y_{1}}=1 \

& \Rightarrow \frac{6 x y_{1}}{6 x_{1} y_{1}+x_{1}^{2} y_{1}}+\frac{x_{1}^{2} y}{6 x_{1} y_{1}+x_{1}^{2} y_{1}}=1 \

& \Rightarrow \frac{x}{\frac{x_{1}\left(6+x_{1}ight)}{6}}+\frac{y}{\frac{y_{1}\left(6+x_{1}ight)}{x_{1}}}=1

\end{aligned}

$$

It is given that the normal makes equal intercepts with the axes.

Therefore, we have:

$$

\begin{align*}

& \Rightarrow \frac{x_{1}\left(6+x_{1}ight)}{6}=\frac{y_{1}\left(6+x_{1}ight)}{x_{1}} \

& \Rightarrow \frac{x_{1}}{6}=\frac{y_{1}}{x_{1}} \

& \Rightarrow x_{1}^{2}=6 y_{1} 	ag{1}

\end{align*}

$$

Also, the point $\left(x_{1}, y_{1}ight)$ lies on the curve, so we have

$$

\begin{equation*}

9 y_{1}^{2}=x_{1}^{3} 	ag{2}

\end{equation*}

$$

From (1) and (2), we have:

$$

\begin{aligned}

& \Rightarrow 9\left(\frac{x_{1}^{2}}{6}ight)^{2}=x_{1}^{3} \

& \Rightarrow \frac{x_{1}^{4}}{4}=x_{1}^{3} \

& \Rightarrow x_{1}=4

\end{aligned}

$$

From (2), we have:

$$

\begin{aligned}

& \Rightarrow 9 y_{1}^{2}=4^{3} \

& \Rightarrow y_{1}^{2}=\frac{64}{9} \

& \Rightarrow y_{1}= \pm \frac{8}{3}

\end{aligned}

$$

Hence, the required points are $\left(4, \pm \frac{8}{3}ight)$

Thus, the correct option is $\mathbf{A}$.