Solution

Ball of salt is spherical

$\therefore \quad$ Volume of ball, $V=\frac{4}{3} \pi r^{3}$, where $r=$ radius of the ball

As per the question, $\frac{d V}{d t} \propto S$, where $S=$ surface area of the ball

$\Rightarrow \quad \frac{d}{d t}(\frac{4}{3} \pi r^{3}) \propto 4 \pi r^{2} \quad[\because S=4 \pi r^{2}]$

$\Rightarrow \quad \frac{4}{3} \pi \cdot 3 r^{2} \cdot \frac{d r}{d t} \propto 4 \pi r^{2}$

$\Rightarrow \quad 4 \pi r^{2} \cdot \frac{d r}{d t}=K \cdot 4 \pi r^{2} \quad(K=$ Constant of proportionality $)$

$\Rightarrow \quad \frac{d r}{d t}=K \cdot \frac{4 \pi r^{2}}{4 \pi r^{2}}$

$\therefore \quad \frac{d r}{d t}=K \cdot 1=K$

Hence, the radius of the ball is decreasing at constant rate.

Solution

We know that:

Area of circle, $A=\pi r^{2}$, where $r=$ radius of the circle.

and perimeter $=2 \pi r$

As per the question,

$\frac{d A}{d t} =K \text{, where } K=\text{ constant }$

$\Rightarrow \quad \frac{d}{d t}(\pi r^{2}) =K \Rightarrow \pi \cdot 2 r \cdot \frac{d r}{d t}=K$

$\therefore \quad \frac{d r}{d t} =\frac{K}{2 \pi r}\qquad$ ……(1)

$\text{ Now Perimeter } \quad c =2 \pi r$

Differentiating both sides w.r.t., $t$, we get

$\Rightarrow \quad \frac{d c}{d t}=\frac{d}{d t}(2 \pi r) \Rightarrow \frac{d c}{d t}=2 \pi \cdot \frac{d r}{d t}$

$\Rightarrow \quad \frac{d c}{d t}=2 \pi \cdot \frac{K}{2 \pi r}=\frac{K}{r}$

[From (1)]

$\Rightarrow \quad \frac{d c}{d t} \propto \frac{1}{r}$

Hence, the perimeter of the circle varies inversely as the radius of the circle.

Solution

Given that height of the kite $(h)=151.5 m$

Speed of the $k i t e(V)=10 m / s$

Let FD be the height of the kite and $A B$ be the height of the boy.

Let $AF=x m$

$\therefore \quad BG=AF=x m$

and $\frac{d x}{d t}=10 m / s$

From the figure, we get that

$ \begin{aligned} GD & =DF-GF \Rightarrow DF-AB \\ & =(151.5-1.5) m=150 m \quad[\because \quad AB=GF] \end{aligned} $

Now in $\triangle BGD$,

$ \begin{aligned} BG^{2}+GD^{2} & =BD^{2} \\ \Rightarrow x^{2}+(150)^{2} & =(250)^{2} \\ \Rightarrow x^{2}+22500 & =62500 \Rightarrow x^{2}=62500-22500 \\ \Rightarrow \quad x^{2} & =40000 \Rightarrow x=200 m \end{aligned} $

Let initially the length of the string be $y m$

$\therefore$ In $\triangle BGD$

$ BG^{2}+GD^{2}=BD^{2} \Rightarrow x^{2}+(150)^{2}=y^{2} $

Differentiating both sides w.r.t., $t$, we get

$ \begin{aligned} & \Rightarrow \quad 2 x \cdot \frac{d x}{d t}+0=2 y \cdot \frac{d y}{d t} \\ & \Rightarrow \quad 2 \times 200 \times 10=2 \times 250 \times \frac{d y}{d t} \\ & \therefore \quad \frac{d y}{d t}=\frac{2 \times 200 \times 10}{2 \times 250}=8 m / s \\ & {[\because \frac{d x}{d t}=10 m / s]} \end{aligned} $

Hence, the rate of change of the length of the string is $8 m / s$.

Solution

Let $P$ be any point at which the two roads are inclined at an angle of $45^{\circ}$.

Two men $A$ and $B$ are moving along the roads $PA$ and $PB$ respectively with the same speed ’ $V$ ‘.

Let $A$ and $B$ be their final positions such that

$AB=y$

$\angle APB=45^{\circ}$ and they move with the same speed.

$\therefore \quad \triangle APB$ is an isosceles triangle. Draw $PQ \perp AB$

$ \begin{matrix} AB=y \quad \therefore \quad AQ=\frac{y}{2} \text{ and } PA=PB=x(\text{ let }) \\ \angle APQ=\angle BPQ=\frac{45}{2}=22 \frac{1}{2} \circ \end{matrix} $

$[\because$ In an isosceles $\Delta$, the altitude drawn from the vertex, bisects the base]

Now in right $\triangle APQ$,

$ \begin{aligned} \sin 22 \frac{1}2^{\circ} & =\frac{AQ}{AP} \\ \Rightarrow \quad \sin 22 \frac{1}2^{\circ} & =\frac{\frac{y}{2}}{x}=\frac{y}{2 x} \quad \Rightarrow y=2 x \cdot \sin 22 \frac{1}2^{\circ} \end{aligned} $

Differentiating both sides w.r.t, $t$, we get

$ \begin{aligned} \frac{d y}{d t} & =2 \cdot \frac{d x}{d t} \cdot \sin 22 \frac{1}{2} \circ \\ & =2 \cdot V \cdot \frac{\sqrt{2-\sqrt{2}}}{2} \quad[\because \sin 22 \frac{1}{2} \circ=\frac{\sqrt{2-\sqrt{2}}}{2}] \\ & =\sqrt{2-\sqrt{2}} V m / s \end{aligned} $

Hence, the rate of their separation is $\sqrt{2-\sqrt{2}} V$ unit/s.

Solution

As per the given condition,

$ \begin{aligned} \frac{d \theta}{d t} & =2 \frac{d}{d t}(\sin \theta) \\ \Rightarrow \quad \frac{d \theta}{d t} & =2 \cos \theta \cdot \frac{d \theta}{d t} \Rightarrow 1=2 \cos \theta \\ \therefore \quad \cos \theta & =\frac{1}{2} \Rightarrow \quad \cos \theta=\cos \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3} \end{aligned} $

Hence, the required angle is $\frac{\pi}{3}$.

Solution

$(1.999)^{5}=(2-0.001)^{5}$

Let

$ x=2 \text{ and } \Delta x=-0.001 $

Let

$ y=x^{5} $

Differentiating both sides w.r.t, $x$, we get

$ \begin{aligned} & \frac{d y}{d x}=5 x^{4}=5(2)^{4}=80 \\ & \text{ Now } \quad \Delta y=(\frac{d y}{d x}) \cdot \Delta x=80 \cdot(-0.001)=-0.080 \\ & \therefore \quad(1.999)^{5}=y+\Delta y \\ & =x^{5}-0.080=(2)^{5}-0.080=32-0.080=31.92 \end{aligned} $

Hence, approximate value of $(1.999)^{5}$ is 31.92 .

Solution

Internal radius $r=3 cm$ and external radius $R=r+\Delta r=3.0005 cm$

$ \therefore \quad \Delta r=3.0005-3=0.0005 cm $

$$ \begin{equation*} \text{ Let } \quad y=r^{3} \Rightarrow y+\Delta y=(r+\Delta r)^{3}=R^{3}=(3.0005)^{3} \tag{i} \end{equation*} $$

Differentiating both sides w.r.t., $r$, we get

$ \begin{aligned} \frac{d y}{d r} & =3 r^{2} \\ \therefore \quad \Delta y & =\frac{d y}{d r} \times \Delta r=3 r^{2} \times 0.0005 \\ & =3 \times(3)^{2} \times 0.0005=27 \times 0.0005=0.0135 \\ \therefore \quad(3.0005)^{3} & =y+\Delta y \quad \text{ [From eq. }(i)] \\ & =(3)^{3}+0.0135=27+0.0135=27.0135 \end{aligned} $

Volume of the shell $=\frac{4}{3} \pi[R^{3}-r^{3}]$

$ \begin{aligned} & =\frac{4}{3} \pi[27.0135-27]=\frac{4}{3} \pi \times 0.0135 \\ & =4 \pi \times 0.005=4 \times 3.14 \times 0.0045=0.018 \pi cm^{3} \end{aligned} $

Hence, the approximate volume of the metal in the shell is $0.018 \pi cm^{3}$.

Solution

Let $A B$ is the height of street light post and $C D$ is the height of the man such that

$ A B=5 \frac{1}{3}=\frac{16}{3} m \text{ and } C D=2 m $

Let $BC=x$ length (the distance of the man from the lamp post) and $CE=y$ is the length of the shadow of the man at any instant. From the figure, we see that

$ \triangle ABE \sim \triangle DCE $

$\therefore \quad$ Taking ratio of their corresponding sides, we get

$ \begin{aligned} & \frac{A B}{C D}=\frac{B E}{C E} \Rightarrow \frac{A B}{C D}=\frac{B C+C E}{C E} \\ & \Rightarrow \quad \frac{16 / 3}{2}=\frac{x+y}{y} \Rightarrow \frac{8}{3}=\frac{x+y}{y} \\ & \Rightarrow \quad 8 y=3 x+3 y \Rightarrow 8 y-3 y=3 x \Rightarrow 5 y=3 x \end{aligned} $

Differentiating both sides w.r.t, $t$, we get

$ \begin{aligned} \cdot \frac{d y}{d t}=3 \cdot \frac{d x}{d t} & \\ \Rightarrow \quad \frac{d y}{d t}=\frac{3}{5} \cdot \frac{d x}{d t} & \Rightarrow \frac{d y}{d t}=\frac{3}{5} \cdot(-1 \frac{2}{3})=\frac{3}{5} \cdot(\frac{-5}{3}) \\ & {[\because \text{ man is moving in opposite direction }] } \\ & =-1 m / s \end{aligned} $

Hence, the length of shadow is decreasing at the rate of $1 m / s$. Now let $u=x+y$

$ \text{ ( } u=\text{ distance of the tip of shadow from the light post) } $

Differentiating both sides w.r.t. $t$, we get

$ \begin{aligned} \frac{d u}{d t} & =\frac{d x}{d t}+\frac{d y}{d t} \\ & =(-1 \frac{2}{3}-1)=-(\frac{5}{3}+1)=-\frac{8}{3}=-2 \frac{2}{3} m / s \end{aligned} $

Hence, the tip of the shadow is moving at the rate of $2 \frac{2}{3} m / s$ towards the light post and the length of shadow decreasing at the rate of $1 m / s$.

Solution

Given that $L=200(10-t)^{2}$ where $L$ represents the number of litres of water in the pool.

Differentiating both sides w.r.t, $t$, we get

$ \frac{d L}{d t}=200 \times 2(10-t)(-1)=-400(10-t) $

But the rate at which the water is running out

$$ \begin{equation*} =-\frac{d L}{d t}=400(10-t) \tag{1} \end{equation*} $$

Rate at which the water is running after 5 seconds

$ =400 \times(10-5)=2000 L / s \text{ (final rate) } $

For initial rate put $t=0$

$ =400(10-0)=4000 L / s $

The average rate at which the water is running out

$ =\frac{\text{ Initial rate }+ \text{ Final rate }}{2}=\frac{4000+2000}{2}=\frac{6000}{2}=3000 L / s $

Hence, the required rate $=3000 L / s$.

Solution

Let $x$ be the length of the cube

$\therefore \quad$ Volume of the cube $V=x^{3}$

Given that $\frac{d V}{d t}=K$

Differentiating Eq. (1) w.r.t. $t$, we get

$ \begin{matrix} \frac{d V}{d t} & =3 x^{2} \cdot \frac{d x}{d t}=K(\text{ constant }) \\ \therefore \quad & \frac{d x}{d t} & =\frac{K}{3 x^{2}} \end{matrix} $

Now surface area of the cube, $S=6 x^{2}$

Differentiating both sides w.r.t. $t$, we get

$ \begin{aligned} \frac{d s}{d t} & =6 \cdot 2 \cdot x \cdot \frac{d x}{d t}=12 x \cdot \frac{K}{3 x^{2}} \\ \Rightarrow \quad \frac{d s}{d t} & =\frac{4 K}{x} \Rightarrow \frac{d s}{d t} \propto \frac{1}{x} \quad(4 K=\text{ constant }) \end{aligned} $

Hence, the surface area of the cube varies inversely as the length of the side.

Solution

Let area of the first square $A_1=x^{2}$

and area of the second square $A_2=y^{2}$

Now $A_1=x^{2}$ and $A_2=y^{2}=(x-x^{2})^{2}$

Differentiating both $A_1$ and $A_2$ w.r.t. $t$, we get

$ \begin{aligned} \frac{d A_1}{d t} & =2 x \cdot \frac{d x}{d t} \text{ and } \frac{d A_2}{d t}=2(x-x^{2})(1-2 x) \cdot \frac{d x}{d t} \\ \therefore \quad \frac{d A_2}{d A_1} & =\frac{\frac{d A_2}{d t}}{\frac{d A_1}{d t}}=\frac{2(x-x^{2})(1-2 x) \cdot \frac{d x}{d t}}{2 x \cdot \frac{d x}{d t}} \\ & =\frac{x(1-x)(1-2 x)}{x}=(1-x)(1-2 x) \\ & =1-2 x-x+2 x^{2}=2 x^{2}-3 x+1 \end{aligned} $

Hence, the rate of change of area of the second square with respect to first is $2 x^{2}-3 x+1$.

Solution

The two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is $90^{\circ}$.

Equation of the two circles are given as

$$ \begin{equation*} 2 x=y^{2} \tag{i} \end{equation*} $$

and

$$ \begin{equation*} 2 x y=k \tag{ii} \end{equation*} $$

Differentiating eq. (i) and (ii) w.r.t. $x$, we get

$ \begin{aligned} & 2.1=2 y \cdot \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{y} \Rightarrow m_1=\frac{1}{y} \\ & (m_1=\text{ slope of the tangent }) \\ & \Rightarrow \quad 2 x y=k \\ & \Rightarrow 2[x \cdot \frac{d y}{d x}+y \cdot 1]=0 \\ & \therefore \quad \frac{d y}{d x}=-\frac{y}{x} \Rightarrow m_2=-\frac{y}{x} \\ & {[m_2=\text{ slope of the other tangent }]} \end{aligned} $

If the two tangents are perpendicular to each other,

then $\quad m_1 \times m_2=-1$

$\Rightarrow \quad \frac{1}{y} \times(-\frac{y}{x})=-1 \Rightarrow \frac{1}{x}=1 \Rightarrow x=1$

Now solving $\qquad 2x=y^3 \qquad $[From(i)]

and $\qquad 2xy=k \qquad $[From(ii)]

From eq.(ii) $\quad y=\frac{k}{2x}$

Putting the value of $y$ in eq. (i)

$ \begin{aligned} 2 x & =(\frac{k}{2 x})^{2} \Rightarrow 2 x=\frac{k^{2}}{4 x^{2}} \\ \Rightarrow \quad 8 x^{3} & =k^{2} \Rightarrow 8(1)^{3}=k^{2} \Rightarrow 8=k^{2} \end{aligned} $

Hence, the required condition is $k^{2}=8$.

Solution

Given circles are $x y=4$

and

$$ \begin{equation*} x^{2}+y^{2}=8 \tag{i} \end{equation*} $$

Differentiating eq. (i) w.r.t., $x$

$$ \Rightarrow \begin{align*} x \cdot \frac{d y}{d x}+y \cdot 1 & =0 \\ \frac{d y}{d x} & =-\frac{y}{x} \Rightarrow m_1=-\frac{y}{x} \tag{iii} \end{align*} $$

where, $m_1$ is the slope of the tangent to the curve.

Differentiating eq. (ii) w.r.t. $x$

$ 2 x+2 y \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y} \Rightarrow m_2=-\frac{x}{y} $

where, $m_2$ is the slope of the tangent to the circle.

To find the point of contact of the two circles

$ m_1=m_2 \Rightarrow-\frac{y}{x}=-\frac{x}{y} \Rightarrow x^{2}=y^{2} $

Putting the value of $y^{2}$ in eq. (ii)

$ x^{2}+x^{2} =8 \Rightarrow 2 x^{2}=8 \Rightarrow x^{2}=4$

$\therefore x = \pm 2$

$\because x^{2} =y^{2} \Rightarrow y= \pm 2$

$\therefore$ The point of contact of the two circles are $(2,2)$ and $(-2,2)$.

Solution

Equation of curve is given by $\sqrt{x}+\sqrt{y}=4$ Let $(x_1, y_1)$ be the required point on the curve $\therefore \quad \sqrt{x_1}+\sqrt{y_1}=4$

Differentiating both sides w.r.t. $x_1$, we get

$ \frac{d}{d x_1} \sqrt{x_1}+\frac{d}{d x_1} \sqrt{y_1}=\frac{d}{d x_1}(4) $

$$ \begin{align*} & \Rightarrow \quad \frac{1}{2 \sqrt{x_1}}+\frac{1}{2 \sqrt{y_1}} \cdot \frac{d y_1}{d x_1}=0 \\ & \Rightarrow \quad \frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{y_1}} \cdot \frac{d y_1}{d x_1}=0 \Rightarrow \frac{d y_1}{d x_1}=-\frac{\sqrt{y_1}}{\sqrt{x_1}} \tag{i} \end{align*} $$

Since the tangent to the given curve at $(x_1, y_1)$ is equally inclined to the axes.

$\therefore \quad$ Slope of the tangent $\frac{d y_1}{d x_1}= \pm \tan \frac{\pi}{4}= \pm 1$ So, from eq. (i) we get

$ -\frac{\sqrt{y_1}}{\sqrt{x_1}}= \pm 1 $

Squaring both sides, we get

$ \frac{y_1}{x_1}=1 \Rightarrow y_1=x_1 $

Putting the value of $y_1$ in the given equation of the curve.

$ \begin{aligned} & \sqrt{x_1}+\sqrt{y_1}=4 \\ & \Rightarrow \sqrt{x_1}+\sqrt{x_1}=4 \Rightarrow 2 \sqrt{x_1}=4 \Rightarrow \sqrt{x_1}=2 \Rightarrow x_1=4 \\ & y_1=x_1 \\ & \therefore \quad y_1=4 \end{aligned} $

Hence, the required point is $(4,4)$.

Solution

We know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.

The given curves are $y=4-x^{2} \ldots$ (i) and $y=x^{2}$

Differentiating eq. ( $i$ ) and (ii) with respect to $x$, we have

$ \frac{d y}{d x}=-2 x \Rightarrow m_1=-2 x $

$m_1$ is the slope of the tangent to the curve (i).

and $\quad \frac{d y}{d x}=2 x \Rightarrow m_2=2 x$

$m_2$ is the slope of the tangent to the curve (ii).

So, $\quad m_1=-2 x$ and $m_2=2 x$

Now solving eq. (i) and (ii) we get

$\Rightarrow \quad 4-x^{2}=x^{2} \Rightarrow 2 x^{2}=4 \Rightarrow x^{2}=2 \Rightarrow x= \pm \sqrt{2}$

So, $\quad m_1=-2 x=-2 \sqrt{2}$ and $m_2=2 x=2 \sqrt{2}$

Let $\theta$ be the angle of intersection of two curves

$ \begin{aligned} & \therefore \quad \tan \theta=|\frac{m_2-m_1}{1+m_1 m_2}| \\ & \therefore \quad=|\frac{2 \sqrt{2}+2 \sqrt{2}}{1-(2 \sqrt{2})(2 \sqrt{2})}|=|\frac{4 \sqrt{2}}{1-8}|=|\frac{4 \sqrt{2}}{-7}|=\frac{4 \sqrt{2}}{7} \\ & \therefore \quad \theta=\tan ^{-1}(\frac{4 \sqrt{2}}{7}) \\ & \text{ Hence, the required angle is } \tan ^{-1}(\frac{4 \sqrt{2}}{7}) . \end{aligned} $

Solution

Given that the equation of the two curves are $y^{2}=4 x$ and

$$ \begin{equation*} x^{2}+y^{2}-6 x+1=0 \tag{i} \end{equation*} $$

Differentiating (i) w.r.t. $x$, we get $2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y}$

Slope of the tangent at $(1,2), m_1=\frac{2}{2}=1$

Differentiating (ii) w.r.t. $x \Rightarrow 2 x+2 y \cdot \frac{d y}{d x}-6=0$

$\Rightarrow \quad 2 y \cdot \frac{d y}{d x}=6-2 x \Rightarrow \frac{d y}{d x}=\frac{6-2 x}{2 y}$

$\therefore \quad$ Slope of the tangent at the same point $(1,2)$

$\Rightarrow \quad m_2=\frac{6-2 \times 1}{2 \times 2}=\frac{4}{4}=1$

We see that $m_1=m_2=1$ at the point $(1,2)$.

Hence, the given circles touch each other at the same point $(1,2)$.

Solution

We have equation of the curve $3 x^{2}-y^{2}=8$

Differentiating both sides w.r.t. $x$, we get

$\Rightarrow \quad 6 x-2 y \cdot \frac{d y}{d x}=0 \Rightarrow-2 y \frac{d y}{d x}=-6 x \Rightarrow \frac{d y}{d x}=\frac{3 x}{y}$

Slope of the tangent to the given curve $=\frac{3 x}{y}$

$\therefore \quad$ Slope of the normal to the curve $=-\frac{1}{\frac{3 x}{y}}=-\frac{y}{3 x}$.

Now differentiating both sides the given line $x+3 y=4$ $\Rightarrow \quad 1+3 \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{1}{3}$

Since the normal to the curve is parallel to the given line $x+3 y=4$.

$\therefore \quad-\frac{y}{3 x}=-\frac{1}{3} \Rightarrow y=x$

Putting the value of $y$ in $3 x^{2}-y^{2}=8$, we get

$ \begin{aligned} & 3 x^{2}-x^{2}=8 \Rightarrow 2 x^{2}=8 \Rightarrow x^{2}=4 \Rightarrow x= \pm 2 \\ & \therefore \quad y= \pm 2 \end{aligned} $

$\therefore \quad$ The points on the curve are $(2,2)$ and $(-2,-2)$.

Now equation of the normal to the curve at $(2,2)$ is

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

$\Rightarrow 3 y-6 =-x+2 \Rightarrow x+3 y=8$

$\text{ at }(-2,-2) \quad y+2 =-\frac{1}{3}(x+2) \\ \Rightarrow 3 y+6 =-x-2 \Rightarrow x+3 y=-8$

Hence, the required equations are $x+3 y=8$ and $x+3 y=-8$ or $x+3 y= \pm 8$.

Solution

Given that the equation of the curve is

$$ \begin{equation*} x^{2}+y^{2}-2 x-4 y+1=0 \tag{i} \end{equation*} $$

Differentiating both sides w.r.t. $x$, we have

$$ \begin{align*} 2 x+2 y \cdot \frac{d y}{d x}-2-4 \cdot \frac{d y}{d x} & =0 \\ \Rightarrow \quad(2 y-4) \frac{d y}{d x} & =2-2 x \Rightarrow \frac{d y}{d x}=\frac{2-2 x}{2 y-4} \tag{ii} \end{align*} $$

Since the tangent to the curve is parallel to the $y$-axis.

$\therefore \quad$ Slope $\frac{d y}{d x}=\tan \frac{\pi}{2}=\infty=\frac{1}{0}$

So, from eq. (ii) we get

$ \frac{2-2 x}{2 y-4}=\frac{1}{0} \Rightarrow 2 y-4=0 \Rightarrow y=2 $

Now putting the value of $y$ in eq. (i), we get

$\Rightarrow \quad x^{2}+(2)^{2}-2 x-8+1=0$

$ \Rightarrow \quad x^{2}-2 x+4-8+1=0 $

$ \begin{aligned} & \Rightarrow \quad x^{2}-2 x-3=0 \Rightarrow x^{2}-3 x+x-3=0 \\ & \Rightarrow \quad x(x-3)+1(x-3)=0 \Rightarrow(x-3)(x+1)=0 \end{aligned} $

$\Rightarrow \quad x=-1$ or 3

Hence, the required points are $(-1,2)$ and $(3,2)$.

Solution

Given that $y=b \cdot e^{-x / a}$, the equation of curve and $\frac{x}{a}+\frac{y}{b}=1$, the equation of line.

Let the coordinates of the point where the curve intersects the $y$-axis be $(0, y_1)$

Now differentiating $y=b \cdot e^{-x / a}$ both sides w.r.t. $x$, we get

$ \frac{d y}{d x}=b \cdot e^{-x / a}(-\frac{1}{a})=-\frac{b}{a} \cdot e^{-x / a} $

So, the slope of the tangent, $m_1=-\frac{b}{a} e^{-x / a}$.

Differentiating $\frac{x}{a}+\frac{y}{b}=1$ both sides w.r.t. $x$, we get

$ \frac{1}{a}+\frac{1}{b} \cdot \frac{d y}{d x}=0 $

So, the slope of the line, $m_2=\frac{-b}{a}$.

If the line touches the curve, then $m_1=m_2$

$\Rightarrow \quad \frac{-b}{a} \cdot e^{-x / a}=\frac{-b}{a} \Rightarrow e^{-x / a}=1$

$\Rightarrow \quad \frac{-x}{a} \log e=\log 1 \quad$ (Taking log on both sides)

$\Rightarrow \quad \frac{-x}{a}=0 \quad \Rightarrow x=0$ Putting $x=0$ in equation $y=b \cdot e^{-x / a}$

$\Rightarrow \quad y=b \cdot e^{0}=b$

Hence, the given equation of curve intersect at $(0, b)$ i.e. on $y$-axis.

Solution

Given that $f(x)=2 x+\cot ^{-1} x+\log (\sqrt{1+x^{2}}-x)$

Differentiating both sides w.r.t. $x$, we get

$ \begin{aligned} f^{\prime}(x) & =2-\frac{1}{1+x^{2}}+\frac{1}{\sqrt{1+x^{2}}-x} \times \frac{d}{d x}(\sqrt{1+x^{2}}-x) \\ & =2-\frac{1}{1+x^{2}}+\frac{(\frac{1}{2 \sqrt{1+x^{2}}} \times(2 x-1))}{\sqrt{1+x^{2}}-x} \end{aligned} $

$ \begin{aligned} & =2-\frac{1}{1+x^{2}}+\frac{x-\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}(\sqrt{1+x^{2}-x})} \\ & =2-\frac{1}{1+x^{2}}-\frac{(\sqrt{1+x^{2}}-x)}{\sqrt{1+x^{2}}(\sqrt{1+x^{2}}-x)} \\ & =2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}} \end{aligned} $

For increasing function, $f^{\prime}(x) \geq 0$

$ \begin{aligned} & \therefore \quad 2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}} \geq 0 \\ & \Rightarrow \quad \frac{2(1+x^{2})-1+\sqrt{1+x^{2}}}{(1+x^{2})} \geq 0 \Rightarrow 2+2 x^{2}-1+\sqrt{1+x^{2}} \geq 0 \\ & \Rightarrow \quad 2 x^{2}+1+\sqrt{1+x^{2}} \geq 0 \Rightarrow 2 x^{2}+1 \geq-\sqrt{1+x^{2}} \end{aligned} $

Squaring both sides, we get $4 x^{4}+1+4 x^{2} \geq 1+x^{2}$

$\Rightarrow 4 x^{4}+4 x^{2}-x^{2} \geq 0 \Rightarrow 4 x^{4}+3 x^{2} \geq 0 \Rightarrow x^{2}(4 x^{2}+3) \geq 0$

which is true for any value of $x \in R$.

Hence, the given function is an increasing function over $R$.

Solution

Given that: $f(x)=\sqrt{3} \sin x-\cos x-2 a x+b, a \geq 1$

Differentiating both sides w.r.t. $x$, we get

$ f^{\prime}(x)=\sqrt{3} \cos x+\sin x-2 a $

For decreasing function, $f^{\prime}(x)<0$

$ \begin{matrix} \therefore & \sqrt{3} \cos x+\sin x-2 a<0 \\ \Rightarrow & 2(\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x)-2 a<0 \\ \Rightarrow & \frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x-a<0 \\ \Rightarrow & (\cos \frac{\pi}{6} \cos x+\sin \frac{\pi}{6} \sin x)-a<0 \\ \Rightarrow & \cos (x-\frac{\pi}{6})-a<0 \end{matrix} $

Since $\cos x \in[-1,1]$ and $a \geq 1$

$ \therefore \quad f^{\prime}(x)<0 $

Hence, the given function is decreasing in $R$.

Solution

Given that: $f(x)=\tan ^{-1}(\sin x+\cos x)$ in $(0, \frac{\pi}{4})$

Differentiating both sides w.r.t. $x$, we get

$ \begin{aligned} f^{\prime}(x) & =\frac{1}{1+(\sin x+\cos x)^{2}} \cdot \frac{d}{d x}(\sin x+\cos x) \\ \Rightarrow f^{\prime}(x) & =\frac{1 \times(\cos x-\sin x)}{1+(\sin x+\cos x)^{2}} \\ \Rightarrow \quad f^{\prime}(x) & =\frac{\cos x-\sin x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x} \\ \Rightarrow \quad f^{\prime}(x) & =\frac{\cos x-\sin x}{1+1+2 \sin x \cos x} \Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{2+2 \sin x \cos x} \end{aligned} $

For an increasing function $f^{\prime}(x) \geq 0$

$\therefore \quad \frac{\cos x-\sin x}{2+2 \sin x \cos x} \geq 0$

$\Rightarrow \quad \cos x-\sin x \geq 0 \quad[\because \quad(2+\sin 2 x) \geq 0.$ in $.(0, \frac{\pi}{4})]$

$\Rightarrow \cos x \geq \sin x$, which is true for $(0, \frac{\pi}{4})$

Hence, the given function $f(x)$ is an increasing function in $(0, \frac{\pi}{4})$.

Solution

Given that: $y=-x^{3}+3 x^{2}+9 x-27$

Differentiating both sides w.r.t. $x$, we get $\frac{d y}{d x}=-3 x^{2}+6 x+9$

Let slope of the cuve $\frac{d y}{d x}=Z$

$\therefore \quad z=-3 x^{2}+6 x+9$

Differentiating both sides w.r.t. $x$, we get $\frac{d z}{d x}=-6 x+6$

For local maxima and local minima, $\frac{d z}{d x}=0$

$\therefore \qquad -6x+6=0 \Rightarrow x=1\\ \qquad \Rightarrow \frac{d^2z}{dx^2}=-6<0 \quad \text{Maxima}$

Put $x=1$ in equation of the curve $y=(-1)^{3}+3(1)^{2}+9(1)-27$

$ =-1+3+9-27=-16 $

Maximum slope $=-3(1)^{2}+6(1)+9=12$

Hence, $(1,-16)$ is the point at which the slope of the given curve is maximum and maximum slope $=12$.

Solution

We have: $f(x)=\sin x+\sqrt{3} \cos x=2(\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x)$

$ \begin{aligned} & =2(\cos \frac{\pi}{3} \sin x+\sin \frac{\pi}{3} \cos x)=2 \sin (x+\frac{\pi}{3}) \\ f^{\prime}(x) & =2 \cos (x+\frac{\pi}{3}) ; f^{\prime \prime}(x)=-2 \sin (x+\frac{\pi}{3}) \\ f^{\prime \prime}(x) _{x=\frac{\pi}{6}} & =-2 \sin (\frac{\pi}{6}+\frac{\pi}{3}) \\ & =-2 \sin \frac{\pi}{2}=-2.1=-2<0 \text{ (Maxima) } \\ & =-2 \times \frac{\sqrt{3}}{2}=-\sqrt{3}<0 \text{ (Maxima) } \end{aligned} $

Maximum value of the function at $x=\frac{\pi}{6}$ is

$ \sin \frac{\pi}{6}+\sqrt{3} \cos \frac{\pi}{6}=\frac{1}{2}+\sqrt{3} \cdot \frac{\sqrt{3}}{2}=2 $

Hence, the given function has maximum value at $x=\frac{\pi}{6}$ and the maximum value is 2 .

Solution

Let $\triangle ABC$ be the right angled triangle in which $\angle B=90^{\circ}$

Let $AC=x, BC=y$

$ \begin{aligned} \therefore \quad AB & =\sqrt{x^{2}-y^{2}} \\ \angle ACB & =\theta \\ \text{ Let } \quad Z & =x+y \quad \text{ (given) } \end{aligned} $

Now area of $\triangle ABC, A=\frac{1}{2} \times AB \times BC$

$ \Rightarrow A=\frac{1}{2} y \cdot \sqrt{x^{2}-y^{2}} \Rightarrow A=\frac{1}{2} y \cdot \sqrt{(Z-y)^{2}-y^{2}} $

Squaring both sides, we get

$ \begin{aligned} A^{2} & =\frac{1}{4} y^{2}[(Z-y)^{2}-y^{2}] \Rightarrow A^{2}=\frac{1}{4} y^{2}[Z^{2}+y^{2}-2 Z y-y^{2}] \\ \Rightarrow P & =\frac{1}{4} y^{2}[Z^{2}-2 Z y] \Rightarrow P=\frac{1}{4}[y^{2} Z^{2}-2 Z y^{3}] \quad[A^{2}=P] \end{aligned} $

Differentiating both sides w.r.t. $y$ we get

$$ \begin{equation*} \frac{d P}{d y}=\frac{1}{4}[2 y Z^{2}-6 Z y^{2}] \tag{i} \end{equation*} $$

For local maxima and local minima, $\frac{d P}{d y}=0$

$\therefore \frac{1}{4}(2 y Z^{2}-6 Z y^{2})=0$

$ \begin{aligned} & \Rightarrow \quad \frac{2 y Z}{4}(Z-3 y)=0 \Rightarrow y Z(Z-3 y)=0 \\ & \Rightarrow \quad y Z \neq 0 \quad(\because y \neq 0 \text{ and } Z \neq 0) \\ & \therefore \quad Z-3 y=0 \end{aligned} $

$ \begin{matrix} \Rightarrow & y=\frac{Z}{3} \Rightarrow y=\frac{x+y}{3} \quad(\because Z=x+y) \\ \Rightarrow & 3 y=x+y \Rightarrow 3 y-y=x \Rightarrow 2 y=x \\ \Rightarrow & \frac{y}{x}=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2} \\ \therefore & \theta=\frac{\pi}{3} \end{matrix} $

Differentiating eq. (i) w.r.t. $y$, we have $\frac{d^{2} P}{d y^{2}}=\frac{1}{4}[2 Z^{2}-12 Z y]$

$ \begin{aligned} \frac{d^{2} P}{d y^{2}} \text{ at } y=\frac{Z}{3} & =\frac{1}{4}[2 Z^{2}-12 Z \cdot \frac{Z}{3}] \\ & =\frac{1}{4}[2 Z^{2}-4 Z^{2}]=\frac{-Z^{2}}{2}<0 \text{ Maxima } \end{aligned} $

Hence, the area of the given triangle is maximum when the angle between its hypotenuse and a side is $\frac{\pi}{3}$.

Solution

We have, $f(x)=x^{5}-5 x^{4}+5 x^{3}-1$

$\Rightarrow \qquad f^{\prime}(x)=5x^{4}-20 x^{3}+15 x^{2}$

For local maxima and local minima,$f^{\prime}(x)=0$

$ \begin{aligned} & \Rightarrow \quad 5 x^{4}-20 x^{3}+15 x^{2}=0 \quad \Rightarrow \quad 5 x^{2}(x^{2}-4 x+3)=0 \\ & \Rightarrow \quad 5 x^{2}(x^{2}-3 x-x+3)=0 \Rightarrow x^{2}(x-3)(x-1)=0 \\ & \therefore \quad x=0, x=1 \text{ and } x=3 \end{aligned} $

Now

$ f^{\prime \prime}(x)=20 x^{3}-60 x^{2}+30 x $

$\Rightarrow \quad f^{\prime \prime}(x) _{\text{at } x=0}=20(0)^{3}-60(0)^{2}+30(0)=0$ which is neither maxima nor minima.

$\therefore f(x)$ has the point of inflection at $x=0$

$ \begin{aligned} f^{\prime \prime}(x) _{\text{at } x=1} & =20(1)^{3}-60(1)^{2}+30(1) \\ & =20-60+30=-10<0 \text{ Maxima } \\ f^{\prime \prime}(x) _{\text{at } x=3} & =20(3)^{3}-60(3)^{2}+30(3) \\ & =540-540+90=90>0 \text{ Minima } \end{aligned} $

The maximum value of the function at $x=1$

$ \begin{aligned} f(x) & =(1)^{5}-5(1)^{4}+5(1)^{3}-1 \\ & =1-5+5-1=0 \end{aligned} $

The minimum value at $x=3$ is

$ \begin{aligned} f(x) & =(3)^{5}-5(3)^{4}+5(3)^{3}-1 \\ & =243-405+135-1=378-406=-28 \end{aligned} $

Hence, the function has its maxima at $x=1$ and the maximum value $=0$ and it has minimum value at $x=3$ and its minimum value is -28 .

$x=0$ is the point of inflection.

Solution

Let us consider that the company increases the annual subscription by $₹ x$.

So, $x$ is the number of subscribers who discontinue the services.

$\therefore$ Total revenue, $R(x)=(500-x)(300+x)$

$ \begin{aligned} & =150000+500 x-300 x-x^{2} \\ & =-x^{2}+200 x+150000 \end{aligned} $

Differentiating both sides w.r.t. $x$, we get $R^{\prime}(x)=-2 x+200$

For local maxima and local minima, $R^{\prime}(x)=0$

$ \begin{aligned} -2 x+200 & =0 \Rightarrow x=100 \\ R^{\prime \prime}(x) & =-2<0 \text{ Maxima } \end{aligned} $

So, $R(x)$ is maximum at $x=100$

Hence, in order to get maximum profit, the company should increase its annual subscription by ₹ 100 .

Solution

The given curve is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

and the straight line $x \cos \alpha+y \sin \alpha=p$

Differentiating eq. (i) w.r.t. $x$, we get

$ \begin{aligned} & \frac{1}{a^{2}} \cdot 2 x+\frac{1}{b^{2}} \cdot 2 y \cdot \frac{d y}{d x}=0 \\ \Rightarrow \quad \frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x} & =0 \Rightarrow \frac{d y}{d x}=-\frac{b^{2}}{a^{2}} \cdot \frac{x}{y} \end{aligned} $

So the slope of the curve $=\frac{-b^{2}}{a^{2}} \cdot \frac{x}{y}$

Now differentiating eq. (ii) w.r.t. $x$, we have

$ \begin{aligned} & \cos \alpha+\sin \alpha \cdot \frac{d y}{d x} & =0 \\ \therefore & \frac{d y}{d x} & =\frac{-\cos \alpha}{\sin \alpha}=-\cot \alpha \end{aligned} $

So, the slope of the straight line $=-\cot \alpha$

If the line is the tangent to the curve, then

$ \frac{-b^{2}}{a^{2}} \cdot \frac{x}{y}=-\cot \alpha \Rightarrow \frac{x}{y}=\frac{a^{2}}{b^{2}} \cdot \cot \alpha \Rightarrow x=\frac{a^{2}}{b^{2}} \cot \alpha \cdot y $

Now from eq. (ii) we have $x \cos \alpha+y \sin \alpha=p$

$\Rightarrow \quad \frac{a^{2}}{b^{2}} \cdot \cot \alpha \cdot y \cdot \cos \alpha+y \sin \alpha=p$

$\Rightarrow \quad a^{2} \cot \alpha \cdot \cos \alpha y+b^{2} \sin \alpha y=b^{2} p$

$\Rightarrow a^{2} \frac{\cos \alpha}{\sin \alpha} \cdot \cos \alpha y+b^{2} \sin \alpha y=b^{2} p$

$\Rightarrow \quad a^{2} \cos ^{2} \alpha y+b^{2} \sin ^{2} \alpha y=b^{2} \sin \alpha p$

$\Rightarrow \quad a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=\frac{b^{2}}{y} \cdot \sin \alpha \cdot p$

$\Rightarrow \quad a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=p \cdot p \quad[\because \frac{b^{2}}{y} \sin \alpha=p]$

Hence, $\quad a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=p^{2}$

Alternate method

We know that $y=m x+c$ will touch the ellipse

$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text{ if } c^{2}=a^{2} m^{2}+b^{2} $

Here equation of straight line is $x \cos \alpha+y \sin \alpha=p$ and that of ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

$ \begin{aligned} & x \cos \alpha+y \sin \alpha=p \\ \Rightarrow \quad & y \sin \alpha=-x \cos \alpha+p \\ \Rightarrow & y=-x \frac{\cos \alpha}{\sin \alpha}+\frac{p}{\sin \alpha} \Rightarrow y=-x \cot \alpha+\frac{p}{\sin \alpha} \end{aligned} $

Comparing with $y=m x+c$, we get

$ m=-\cot \alpha \quad \text{ and } \quad c=\frac{p}{\sin \alpha} $

So, according to the condition, we get $c^{2}=a^{2} m^{2}+b^{2}$

$ \begin{aligned} & \frac{p^{2}}{\sin ^{2} \alpha}=a^{2}(-\cot \alpha)^{2}+b^{2} \\ \Rightarrow \quad & \frac{p^{2}}{\sin ^{2} \alpha}=\frac{a^{2} \cos ^{2} \alpha}{\sin ^{2} \alpha}+b^{2} \Rightarrow p^{2}=a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha \end{aligned} $

Hence, $a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=p^{2} \quad$ Hence proved.

Solution

Let $x$ be the length of the side of the square base of the cubical open box and $y$ be its height.

$\therefore \quad$ Surface area of the open box

$c^{2}=x^{2}+4 x y \Rightarrow y=\frac{c^{2}-x^{2}}{4 x}$

Now volume of the box, $V=x \times x \times y$

$\Rightarrow V=x^{2} y$

$\Rightarrow V=x^{2}(\frac{c^{2}-x^{2}}{4 x})$

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

Differentiating both sides w.r.t. $x$, we get

$$ \begin{equation*} \frac{d V}{d x}=\frac{1}{4}(c^{2}-3 x^{2}) \tag{ii} \end{equation*} $$

For local maxima and local minima, $\frac{d V}{d x}=0$

$\therefore \quad \frac{1}{4}(c^{2}-3 x^{2})=0 \Rightarrow c^{2}-3 x^{2}=0$

$\Rightarrow \quad x^{2}=\frac{c^{2}}{3}$

$\therefore \quad x=\sqrt{\frac{c^{2}}{3}}=\frac{c}{\sqrt{3}}$

Now again differentiating eq. (ii) w.r.t. $x$, we get

$ \frac{d^{2} V}{d x^{2}}=\frac{1}{4}(-6 x)=\frac{-3}{2} \cdot \frac{c}{\sqrt{3}}<0 \quad \text{ (maxima) } $

Volume of the cubical box $(V)=x^{2} y$

$ =x^{2}(\frac{c^{2}-x^{2}}{4 x})=\frac{c}{\sqrt{3}}[\frac{c^{2}-\frac{c^{2}}{3}}{4}]=\frac{c}{\sqrt{3}} \times \frac{2 c^{2}}{3 \times 4}=\frac{c^{3}}{6 \sqrt{3}} $

Hence, the maximum volume of the open box is $\frac{c^{3}}{6 \sqrt{3}}$ cubic units.

Solution

Let $x$ and $y$ be the length and breadth of a given rectangle $ABCD$ as per question, the rectangle be revolved about side AD which will make a cylinder with radius $x$ and height $y$.

$\therefore$ Volume of the cylinder $V=\pi r^{2} h$

$$ \begin{equation*} \Rightarrow \quad V=\pi x^{2} y \tag{i} \end{equation*} $$

Now perimeter of rectangle $P=2(x+y) \Rightarrow 36=2(x+y)$

$\Rightarrow \quad x+y=18 \Rightarrow y=18-x$

Putting the value of $y$ in eq. (i) we get

$ \begin{aligned} & V & =\pi x^{2}(18-x) \\ \Rightarrow & V & =\pi(18 x^{2}-x^{3}) \end{aligned} $

Differentiating both sides w.r.t. $x$, we get

$$ \begin{equation*} \frac{d V}{d x}=\pi(36 x-3 x^{2}) \tag{iii} \end{equation*} $$

For local maxima and local minima $\frac{d V}{d x}=0$

$ \begin{aligned} & \therefore \quad \pi(36 x-3 x^{2})=0 \Rightarrow 36 x-3 x^{2}=0 \\ & \Rightarrow \quad 3 x(12-x)=0 \\ & \Rightarrow \quad x \neq 0 \quad \therefore \quad 12-x=0 \Rightarrow x=12 \end{aligned} $

From eq. (ii) $y=18-12=6$

Differentiating eq. (iii) w.r.t. $x$, we get $\frac{d^{2} V}{d x^{2}}=\pi(36-6 x)$

$ \text{ at } x=12 \quad \begin{aligned} \frac{d^{2} V}{d x^{2}} & =\pi(36-6 \times 12) \\ & =\pi(36-72)=-36 \pi<0 \text{ maxima } \end{aligned} $

Now volume of the cylinder so formed $=\pi x^{2} y$

$ =\pi \times(12)^{2} \times 6=\pi \times 144 \times 6=864 \pi cm^{3} $

Hence, the required dimensions are $12 cm$ and $6 cm$ and the maximum volume is $864 \pi cm^{3}$.

Solution

Let $x$ be the edge of the cube and $r$ be the radius of the sphere. Surface area of cube $=6 x^{2}$

and surface area of the sphere $=4 \pi r^{2}$

$\therefore \quad 6 x^{2}+4 \pi r^{2}=K$ (constant) $\Rightarrow r=\sqrt{\frac{K-6 x^{2}}{4 \pi}}$

Volume of the cube $=x^{3}$ and the volume of sphere $=\frac{4}{3} \pi r^{3}$

$\therefore \quad$ Sum of their volumes $(V)=$ Volume of cube

$ \begin{matrix} \Rightarrow & V=x^{3}+\frac{4}{3} \pi r^{3} \\ \Rightarrow & V=x^{3}+\frac{4}{3} \pi \times(\frac{K-6 x^{2}}{4 \pi})^{3 / 2} \end{matrix} $

Differentiating both sides w.r.t. $x$, we get

$ \frac{d V}{d x}=3 x^{2}+\frac{4 \pi}{3} \times \frac{3}{2}(K-6 x^{2})^{1 / 2}(-12 x) \times \frac{1}{(4 \pi)^{3 / 2}} $

$$ \begin{align*} & =3 x^{2}+\frac{2 \pi}{(4 \pi)^{3 / 2}} \times(-12 x)(K-6 x^{2})^{1 / 2} \\ & =3 x^{2}+\frac{1}{4 \pi^{1 / 2}} \times(-12 x)(K-6 x^{2})^{1 / 2} \\ \therefore \quad \frac{d V}{d x} & =3 x^{2}-\frac{3 x}{\sqrt{\pi}}(K-6 x^{2})^{1 / 2} \tag{ii} \end{align*} $$

For local maxima and local minima, $\frac{d V}{d x}=0$

$\therefore \quad 3 x^{2}-\frac{3 x}{\sqrt{\pi}}(K-6 x^{2})^{1 / 2}=0$

$\Rightarrow \quad 3 x[x-\frac{(K-6 x^{2})^{1 / 2}}{\sqrt{\pi}}]=0$

$x \neq 0 \quad \therefore \quad x-\frac{(K-6 x^{2})^{1 / 2}}{\sqrt{\pi}}=0$

$\Rightarrow x=\frac{(K-6 x^{2})^{1 / 2}}{\sqrt{\pi}}$

Squaring both sides, we get

$ \begin{matrix} x^{2} =\frac{K-6 x^{2}}{\pi} \Rightarrow \pi x^{2}=K-6 x^{2} \\ \Rightarrow \pi x^{2}+6 x^{2} =K \Rightarrow x^{2}(\pi+6)=K \Rightarrow x^{2}=\frac{K}{\pi+6} \\ x =\sqrt{\frac{K}{\pi+6}} \end{matrix} $

Now putting the value of $K$ in eq. (i), we get

$6 x^{2}+4 \pi r^{2} =x^{2}(\pi+6)$

$\Rightarrow 6 x^{2}+4 \pi r^{2} =\pi x^{2}+6 x^{2}\Rightarrow 4 \pi r^{2}=\pi x^{2} \Rightarrow 4 r^{2}=x^{2}$

$\therefore 2 r =x$

$\therefore x: 2 r =1: 1$

Now differentiating eq. (ii) w.r.t $x$, we have

$ \begin{aligned} \frac{d^{2} V}{d x^{2}} & =6 x-\frac{3}{\sqrt{\pi}} \frac{d}{d x}[x(K-6 x^{2})^{1 / 2}] \\ & =6 x-\frac{3}{\sqrt{\pi}}[x \cdot \frac{1}{2 \sqrt{K-6 x^{2}}} \times(-12 x)+(K-6 x^{2})^{1 / 2} \cdot 1] \\ & =6 x-\frac{3}{\sqrt{\pi}}[\frac{-6 x^{2}}{\sqrt{K-6 x^{2}}}+\sqrt{K-6 x^{2}}] \end{aligned} $

$ \begin{aligned} & =6 x-\frac{3}{\sqrt{\pi}}[\frac{-6 x^{2}+K-6 x^{2}}{\sqrt{K-6 x^{2}}}]=6 x+\frac{3}{\sqrt{\pi}}[\frac{12 x^{2}-K}{\sqrt{K-6 x^{2}}}] \\ & \text{ Put } \quad x=\sqrt{\frac{K}{\pi+6}}=6 \sqrt{\frac{K}{\pi+6}}+\frac{3}{\sqrt{\pi}}[\frac{\frac{12 K}{\pi+6}-K}{\sqrt{K-\frac{6 K}{\pi+6}}}] \\ & =6 \sqrt{\frac{K}{\pi+6}}+\frac{3}{\sqrt{\pi}}[\frac{12 K-\pi K-6 K}{\sqrt{\frac{\pi K+6 K-6 K}{\pi+6}}}] \\ & =6 \sqrt{\frac{K}{\pi+6}}+\frac{3}{\sqrt{\pi}}[\frac{6 K-\pi K}{\sqrt{\frac{\pi K}{\pi+6}}}] \\ & =6 \sqrt{\frac{K}{\pi+6}}+\frac{3}{\pi \sqrt{K}}[(6 K-\pi K) \sqrt{\pi+6}]>0 \end{aligned} $

So it is minima.

Hence, the required ratio is $1: 1$ when the combined volume is minimum.

Solution

Let $A B$ be the diameter and $C$ be any point on the circle with radius $r$.

$\angle ACB=90^{\circ}$ [angle in the semi circle is $90^{\circ}$ ]

Let $AC=x$

$\therefore \quad BC=\sqrt{AB^{2}-AC^{2}}$

$\Rightarrow BC=\sqrt{(2 r)^{2}-x^{2}} \quad \Rightarrow BC=\sqrt{4 r^{2}-x^{2}}$

Now area of $\triangle ABC, A=\frac{1}{2} \times AC \times BC$

$ \Rightarrow \quad A=\frac{1}{2} x \cdot \sqrt{4 r^{2}-x^{2}} $

Squaring both sides, we get

Let $A^{2}=Z$

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

$\therefore \quad Z=\frac{1}{4} x^{2}(4 r^{2}-x^{2}) \quad \Rightarrow Z=\frac{1}{4}(4 x^{2} r^{2}-x^{4})$

Differentiating both sides w.r.t. $x$, we get

$$ \begin{equation*} \frac{d Z}{d x}=\frac{1}{4}[8 x r^{2}-4 x^{3}] \tag{ii} \end{equation*} $$

For local maxima and local minima $\frac{d Z}{d x}=0$

$\therefore \quad \frac{1}{4}[8 x r^{2}-4 x^{3}]=0 \Rightarrow x[2 r^{2}-x^{2}]=0$

$x \neq 0 \quad \therefore \quad 2 r^{2}-x^{2}=0$

$\Rightarrow \quad x^{2}=2 r^{2} \Rightarrow x=\sqrt{2} r=AC$

Now from eq. (i) we have

$ BC=\sqrt{4 r^{2}-2 r^{2}} \Rightarrow BC=\sqrt{2 r^{2}} \Rightarrow BC=\sqrt{2} r $

So

$ AC=BC $

Hence, $\triangle ABC$ is an isosceles triangle.

Differentiating eq. (ii) w.r.t. $x$, we get $\frac{d^{2} Z}{d x^{2}}=\frac{1}{4}[8 r^{2}-12 x^{2}]$

Put $x=\sqrt{2} r$

$ \begin{aligned} \therefore \quad \frac{d^{2} Z}{d x^{2}} & =\frac{1}{4}[8 r^{2}-12 \times 2 r^{2}]=\frac{1}{4}[8 r^{2}-24 r^{2}] \\ & =\frac{1}{4} \times(-16 r^{2})=-4 r^{2}<0 \quad \text{ maxima } \end{aligned} $

Hence, the area of $\triangle ABC$ is maximum when it is an isosceles triangle.

Solution

Let $x$ be the side of the square base and $y$ be the length of the vertical sides.

Area of the base and bottom $=2 x^{2} cm^{2}$

$\therefore$ Cost of the material required $=₹ 5 \times 2 x^{2}$

$ =₹ 10 x^{2} $

Area of the 4 sides $=4 x y cm^{2}$

$\therefore \quad$ Cost of the material for the four sides

Total cost

$$ \begin{align*} & =₹ 2.50 \times 4 x y=₹ 10 x y \\ C & =10 x^{2}+10 x y \tag{i} \end{align*} $$

New volume of the box $=x \times x \times y$

$$ \begin{matrix} \Rightarrow & 1024 =x^{2} y \\ \therefore & y =\frac{1024}{x^{2}} \tag{ii} \end{matrix} $$

Putting the value of $y$ in eq. (i) we get

$ C=10 x^{2}+10 x \times \frac{1024}{x^{2}} \Rightarrow C=10 x^{2}+\frac{10240}{x} $

Differentiating both sides w.r.t. $x$, we get

$$ \begin{equation*} \frac{d C}{d x}=20 x-\frac{10240}{x^{2}} \tag{iii} \end{equation*} $$

For local maxima and local minima $\frac{d C}{d x}=0$

$ \begin{aligned} 20-\frac{10240}{x^{2}} & =0 \\ \Rightarrow \quad 20 x^{3}-10240 & =0 \quad \Rightarrow x^{3}=512 \quad \Rightarrow \quad x=8 cm \end{aligned} $

Now from eq. (ii)

$ y=\frac{10240}{(8)^{2}}=\frac{10240}{64}=16 cm $

$\therefore$ Cost of material used $C=10 x^{2}+10 x y$

$ =10 \times 8 \times 8+10 \times 8 \times 16=640+1280=1920 $

Now differentiating eq. (iii) we get

$ \frac{d^{2} C}{d x^{2}}=20+\frac{20480}{x^{3}} $

Put $x=8$

$ =20+\frac{20480}{(8)^{3}}=20+\frac{20480}{512}=20+40=60>0 \text{ minima } $

Hence, the required cost is $₹ 1920$ which is the minimum.

Solution

Let ’ $r$ ’ be the radius of the sphere.

$\therefore \quad$ Surface area of the sphere $=4 \pi r^{2}$

Volume of the sphere $=\frac{4}{3} \pi r^{3}$

The sides of the parallelopiped are $x, 2 x$ and $\frac{x}{3}$

$\therefore \quad$ Its surface area $=2[x \times 2 x+2 x \times \frac{x}{3}+x \times \frac{x}{3}]$

$ \begin{aligned} & =2[2 x^{2}+\frac{2 x^{2}}{3}+\frac{x^{2}}{3}]=2[2 x^{2}+x^{2}] \\ & =2[3 x^{2}]=6 x^{2} \end{aligned} $

Volume of the parallelopiped $=x \times 2 x \times \frac{x}{3}=\frac{2}{3} x^{3}$

As per the conditions of the question,

Surface area of the parallelopiped

• Surface area of the sphere $=$ constant

$\Rightarrow \quad 6 x^{2}+4 \pi r^{2}=K$ (constant) $\Rightarrow 4 \pi r^{2}=K-6 x^{2}$

$\therefore \quad r^{2}=\frac{K-6 x^{2}}{4 \pi}$

Now let $\quad V=$ Volume of parallelopiped

• Volume of the sphere

$\Rightarrow \quad V=\frac{2}{3} x^{3}+\frac{4}{3} \pi r^{3}$

$\Rightarrow \quad V=\frac{2}{3} x^{3}+\frac{4}{3} \pi[\frac{K-6 x^{2}}{4 \pi}]^{3 / 2} \quad$ [from eq. (i)]

$\Rightarrow \quad V=\frac{2}{3} x^{3}+\frac{4}{3} \pi \times \frac{1}{(4)^{3 / 2} \pi^{3 / 2}}[K-6 x^{2}]^{3 / 2}$

$\Rightarrow \quad V=\frac{2}{3} x^{3}+\frac{4}{3} \pi \times \frac{1}{8 \times \pi^{3 / 2}}[K-6 x^{2}]^{3 / 2}$

$\Rightarrow \quad=\frac{2}{3} x^{3}+\frac{1}{6 \sqrt{\pi}}[K-6 x^{2}]^{3 / 2}$

Differentiating both sides w.r.t. $x$, we have

$ \begin{aligned} \frac{d V}{d x} & =\frac{2}{3} \cdot 3 x^{2}+\frac{1}{6 \sqrt{\pi}}[\frac{3}{2}(K-6 x^{2})^{1 / 2}(-12 x)] \\ & =2 x^{2}+\frac{1}{6 \sqrt{\pi}} \times \frac{3}{2} \times(-12 x)(K-6 x^{2})^{1 / 2} \\ & =2 x^{2}-\frac{3 x}{\sqrt{\pi}}[K-6 x^{2})^{1 / 2} \end{aligned} $

For local maxima and local minima, we have $\frac{d V}{d x}=0$

$ \begin{matrix} \therefore & 2 x^{2}-\frac{3 x}{\sqrt{\pi}}(K-6 x^{2})^{1 / 2}=0 \\ \Rightarrow & 2 \sqrt{\pi} x^{2}-3 x(K-6 x^{2})^{1 / 2}=0 \\ \Rightarrow & x[2 \sqrt{\pi} x-3(K-6 x^{2})^{1 / 2}]=0 \end{matrix} $

Here $x \neq 0$ and $2 \sqrt{\pi} x-3(K-6 x^{2})^{1 / 2}=0$

$ \Rightarrow \quad 2 \sqrt{\pi} x=3(K-6 x^{2})^{1 / 2} $

Squaring both sides, we get

$ 4 \pi x^{2}=9(K-6 x^{2}) \Rightarrow 4 \pi x^{2}=9 K-54 x^{2} $

$$ \begin{matrix} \Rightarrow 4 \pi x^{2}+54 x^{2} =9 K \\ K =\frac{4 \pi x^{2}+54 x^{2}}{9} \tag{ii}\\ \Rightarrow 2 x^{2}(2 \pi+27) =9 K \\ \therefore x^{2} =\frac{9 K}{2(2 \pi+27)}=3 \sqrt{\frac{K}{4 \pi+54}} \end{matrix} $$

Now from eq. (i) we have

$ r^{2}=\frac{K-6 x^{2}}{4 \pi} $

$ \begin{matrix} \Rightarrow & r^{2}=\frac{\frac{4 \pi x^{2}+54 x^{2}}{9}-6 x^{2}}{4 \pi} \\ \Rightarrow & r^{2}=\frac{4 \pi x^{2}+54 x^{2}-54 x^{2}}{9 \times 4 \pi}=\frac{4 \pi x^{2}}{9 \times 4 \pi} \\ \Rightarrow \quad & r^{2}=\frac{x^{2}}{9} \quad \Rightarrow \quad r=\frac{x}{3} \quad \therefore x=3 r \end{matrix} $

Now we have $\quad \frac{d V}{d x}=2 x^{2}-\frac{3 x}{\sqrt{\pi}}(K-6 x^{2})^{1 / 2}$

Differentiating both sides w.r.t. $x$, we get

$ \begin{aligned} \frac{d^{2} V}{d x^{2}} & =4 x-\frac{3}{\sqrt{\pi}}[x \cdot \frac{d}{d x}(K-6 x^{2})^{1 / 2}+(K-6 x^{2})^{1 / 2} \cdot \frac{d}{d x} \cdot x] \\ & =4 x-\frac{3}{\sqrt{\pi}}[x \cdot \frac{1 \times(-12 x)}{2 \sqrt{K-6 x^{2}}}+(K-6 x^{2})^{1 / 2} \cdot 1] \\ & =4 x-\frac{3}{\sqrt{\pi}}[\frac{-6 x^{2}}{(K-6 x^{2})^{1 / 2}}+(K-6 x^{2})^{1 / 2}] \\ & =4 x-\frac{3}{\sqrt{\pi}}[\frac{-6 x^{2}+K-6 x^{2}}{(K-6 x^{2})^{1 / 2}}]=4 x-\frac{3}{\sqrt{\pi}}[\frac{K-12 x^{2}}{(K-6 x^{2})^{1 / 2}}] \end{aligned} $

Put $\quad x=3 \cdot \sqrt{\frac{K}{4 \pi+54}}$

$ \frac{d^{2} V}{d x^{2}}=4 \cdot 3 \sqrt{\frac{K}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}[\frac{K-12 \cdot \frac{9 K}{4 \pi+54}}{\sqrt{(K-6 \cdot \frac{9 K}{4 \pi+54})}}] $

$ \begin{aligned} & =12 \sqrt{\frac{K}{4 \pi+54}}-\frac{3}{\sqrt{\pi}} \frac{\frac{4 K \pi+54 K-108 K}{4 \pi+54}}{\sqrt{\frac{4 K \pi+54 K-54 K}{4 \pi+54}}} \\ & =12 \sqrt{\frac{K}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}[\frac{\frac{4 K \pi-54 K}{4 \pi+54}}{.\sqrt{\frac{4 K \pi}{4 \pi+54}}]}. \\ & =12 \sqrt{\frac{K}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}[\frac{4 K \pi-54 K}{\sqrt{4 K \pi} \cdot \sqrt{4 \pi+54}}] \\ & =12 \sqrt{\frac{K}{4 \pi+54}}-\frac{6 K}{\sqrt{\pi}}(\frac{2 \pi-27}{\sqrt{4 K \pi} \cdot \sqrt{4 \pi+54}}) \\ & =12 \sqrt{\frac{K}{4 \pi+54}}+\frac{6 K}{\sqrt{\pi}}[\frac{27-2 \pi}{\sqrt{4 k \pi} \cdot \sqrt{4 \pi+54}}]>0 \end{aligned} $

$[\because \quad 27-2 \pi>0]$

$\therefore \quad \frac{d^{2} V}{d x^{2}}>0 \quad$ so, it is minima.

$\quad d x^{2}$ Hence, the sum of volume is minimum for $x=3 \sqrt{\frac{K}{4 \pi+54}}$ $\therefore$ Minimum volume,

$ \text{ V at } \begin{aligned} (x=3 \sqrt{\frac{K}{4 \pi+54}}) & =\frac{2}{3} x^{3}+\frac{4}{3} \pi r^{3}=\frac{2}{3} x^{3}+\frac{4}{3} \pi \cdot(\frac{x}{3})^{3} \\ & =\frac{2}{3} x^{3}+\frac{4}{3} \pi \cdot \frac{x^{3}}{27}=\frac{2}{3} x^{3}+\frac{4}{81} \pi x^{3} \\ & =\frac{2}{3} x^{3}(1+\frac{2 \pi}{27}) \end{aligned} $

Hence, the required minimum volume is $\frac{2}{3} x^{3}(1+\frac{2 \pi}{27})$ and $x=3 r$.

Solution

Let the length of each side of the given equilateral triangle be $x cm$.

$\therefore \quad \frac{d x}{d t}=2 cm / sec$

Area of equilateral triangle $A=\frac{\sqrt{3}}{4} x^{2}$

$\therefore \quad \frac{d A}{d t}=\frac{\sqrt{3}}{4} \cdot 2 x \cdot \frac{d x}{d t}=\frac{\sqrt{3}}{2} \times 10 \times 2=10 \sqrt{3} cm^{2} / sec$

Hence, the rate of increasing of area $=10 \sqrt{3} cm^{2} / sec$.

Hence, the correct option is (c).

Solution

Length of ladder $=5 m$

Let $AB=y m$ and $BC=x m$

$\therefore$ In right $\triangle ABC$,

$ A B^{2}+BC^{2}=A C^{2} $

$\Rightarrow \quad x^{2}+y^{2}=(5)^{2} \Rightarrow x^{2}+y^{2}=25$

Differentiating both sides w.r.t $x$, we have

$ \begin{aligned} 2 x \cdot \frac{d x}{d t}+2 y \cdot \frac{d y}{d t} & =0 \\ \Rightarrow \quad x \frac{d x}{d t}+y \cdot \frac{d y}{d t} & =0 \\ \Rightarrow \quad 2 \cdot \frac{d x}{d t}+y \times(-0.1) & =0 \\ \Rightarrow \quad 2 \cdot \frac{d x}{d t}+\sqrt{25-x^{2}} \times(-0.1) & =0 \\ \Rightarrow \quad 2 \cdot \frac{d x}{d t}+\sqrt{25-4} \times(-0.1) & =0 \\ \Rightarrow \quad 2 \cdot \frac{d x}{d t}-\frac{\sqrt{21}}{10} & =0 \Rightarrow \frac{d x=2 m]}{d t}=\frac{\sqrt{21}}{20} \end{aligned} $

Now $\cos \theta=\frac{BC}{AC}$

$\Rightarrow$ $\cos \theta=\frac{x}{5}$

Differentiating both sides w.r.t. $t$, we get

$ \begin{aligned} \frac{d}{d t} \cos \theta & =\frac{1}{5} \cdot \frac{d x}{d t} \Rightarrow-\sin \theta \cdot \frac{d \theta}{d t}=\frac{1}{5} \cdot \frac{\sqrt{21}}{20} \\ \Rightarrow \quad \frac{d \theta}{d t} & =\frac{\sqrt{21}}{100} \times(-\frac{1}{\sin \theta})=\frac{\sqrt{21}}{100} \times-(\frac{1}{\frac{AB}{AC}}) \\ & =-\frac{\sqrt{21}}{100} \times \frac{AC}{AB}=-\frac{\sqrt{21}}{100} \times \frac{5}{\sqrt{21}}=-\frac{1}{20} \text{ radian/sec } \\ & {[(-) \text{ sign shows the decrease of change of angle] }} \end{aligned} $

Hence, the required rate $=\frac{1}{20} radian / sec$

Hence, the correct option is $(b)$.

Solution

Equation of curve is $y=x^{1 / 5}$

Differentiating w.r.t. $x$, we get $\frac{d y}{d x}=\frac{1}{5} x^{-4 / 5}$

$ \begin{aligned} (\text{ at } x=0) \quad \frac{d y}{d x} & =\frac{1}{5}(0)^{-4 / 5}=\frac{1}{5} \times \frac{1}{0}=\infty \\ \frac{d y}{d x} & =\infty \end{aligned} $

$\therefore \quad$ The tangent is parallel to $y$-axis.

Hence, the correct option is $(a)$.

Solution

Given equation of the curve is $3 x^{2}-y^{2}=8$

Differentiating both sides w.r.t. $x$, we get

$ 6 x-2 y \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{3 x}{y} $

$\frac{3 x}{y}$ is the slope of the tangent

$ \therefore \quad \text{ Slope of the normal }=\frac{-1}{d y / d x}=\frac{-y}{3 x} $

Now $x+3 y=8$ is parallel to the normal

Differentiating both sides w.r.t. $x$, we have

$ \begin{aligned} 1+3 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{1}{3} \\ \therefore \frac{-y}{3 x}=-\frac{1}{3} & \Rightarrow y=x \end{aligned} $

Putting $y=x$ in eq. (i) we get

$3 x^{2}-x^{2}=8 \Rightarrow 2 x^{2}=8 \Rightarrow x^{2}=4$

$\therefore x= \pm 2 \text{ and } y= \pm 2$

So the points are $(2,2)$ and $(-2,-2)$.

Equation of normal to the given curve at $(2,2)$ is

$ \begin{aligned} y-2 & =-\frac{1}{3}(x-2) \\ \Rightarrow \quad 3 y-6 & =-x+2 \Rightarrow x+3 y-8=0 \end{aligned} $

Equation of normal at $(-2,-2)$ is

$ \begin{aligned} y+2 & =-\frac{1}{3}(x+2) \\ \Rightarrow \quad 3 y+6 & =-x-2 \Rightarrow x+3 y+8=0 \end{aligned} $

$\therefore$ The equations of the normals to the curve are

$ x+3 y \pm 8=0 $

Hence, the correct option is (c).

Solution

Equation of the given curves are $a y+x^{2}=7$

and $\quad x^{3}=y$

Differentiating eq. (i) w.r.t. $x$, we have

$ \begin{aligned} & a \frac{d y}{d x}+2 x=0 \quad \Rightarrow \quad \frac{d y}{d x}=-\frac{2 x}{a} \\ & \therefore \quad m_1=-\frac{2 x}{a} \quad(m_1=\frac{d y}{d x}) \end{aligned} $

Now differentiating eq. (ii) w.r.t. $x$, we get

$ 3 x^{2}=\frac{d y}{d x} \Rightarrow m_2=3 x^{2} \quad(m_2=\frac{d y}{d x}) $

The two curves are said to be orthogonal if the angle between the tangents at the point of intersection is $90^{\circ}$.

$ \begin{aligned} & \therefore \quad m_1 \times m_2=-1 \\ & \Rightarrow \quad \frac{-2 x}{a} \times 3 x^{2}=-1 \Rightarrow \frac{-6 x^{3}}{a}=-1 \Rightarrow 6 x^{3}=a \end{aligned} $

$(1,1)$ is the point of intersection of two curves.

$ \begin{aligned} \therefore & 6(1)^{3} & =a \\ \text{ So } & a & =6 \end{aligned} $

Hence, the correct option is $(d)$.

Solution

Given that $y=x^{4}-10$

$ \begin{aligned} \frac{d y}{d x} & =4 x^{3} \\ \Delta x & =2.00-1.99=0.01 \\ \therefore \quad \Delta y & =\frac{d y}{d x} \cdot \Delta x=4 x^{3} \times \Delta x \\ & =4 \times(2)^{3} \times 0.01=32 \times 0.01=0.32 \end{aligned} $

Hence, the correct option is $(a)$.

Solution

Given that $y(1+x^{2})=2-x$

If it cuts $x$-axis, then $y$-coordinate is 0 .

$ \therefore \quad 0(1+x^{2})=2-x \Rightarrow x=2 $

Put $x=2$ in equation $(i)$

$ y(1+4)=2-2 \Rightarrow y(5)=0 \Rightarrow y=0 $

Point of contact $=(2,0)$

Differentiating eq. (i) w.r.t. $x$, we have

$ \begin{aligned} & y \times 2 x+(1+x^{2}) \frac{d y}{d x}=-1 \\ \Rightarrow & 2 x y+(1+x^{2}) \frac{d y}{d x}=-1 \Rightarrow(1+x^{2}) \frac{d y}{d x}=-1-2 x y \\ \therefore \quad & \frac{d y}{d x}=\frac{-(1+2 x y)}{(1+x^{2})} \Rightarrow \frac{d y}{d x}=\frac{-1}{(1+4)}=\frac{-1}{5} \end{aligned} $

Equation of tangent is $y-0=-\frac{1}{5}(x-2)$

$ \Rightarrow \quad 5 y=-x+2 \Rightarrow x+5 y=2 $

Hence, the correct option is (a).

Solution

Given that $y=x^{3}-12 x+18$

Differentiating both sides w.r.t. $x$, we have

$\Rightarrow \quad \frac{d y}{d x}=3 x^{2}-12$

Since the tangents are parallel to $x$-axis, then $\frac{d y}{d x}=0$

$ \begin{matrix} \therefore \qquad 3 x^{2}-12 =0 \Rightarrow x= \pm 2 d x \\ \therefore \qquad y _{x=2} =(2)^{3}-12(2)+18=8-24+18=2 \\ y _{x=-2} =(-2)^{3}-12(-2)+18=-8+24+18=34 \end{matrix} $

$\therefore \quad$ Points are $(2,2)$ and $(-2,34)$

Hence, the correct option is (d).

Solution

Equation of the curve is $y=e^{2 x}$

Slope of the tangent $\frac{d y}{d x}=2 e^{2 x} \Rightarrow \frac{d y}{d x _{(0,1)}}=2 \cdot e^{0}=2$

$\therefore \quad$ Equation of tangent to the curve at $(0,1)$ is

$ y-1=2(x-0) $

$ \Rightarrow \quad y-1=2 x \Rightarrow y-2 x=1 $

Since the tangent meets $x$-axis where $y=0$

$ \therefore \quad 0-2 x=1 \quad \Rightarrow \quad x=\frac{-1}{2} $

So the point is $(-\frac{1}{2}, 0)$

Hence, the correct option is (b).

Solution

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

$ \begin{matrix} \frac{d x}{d t}=2 t+3 \text{ and } \frac{d y}{d t}=4 t-2 \\ \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{4 t-2}{2 t+3} \end{matrix} $

Now $(2,-1)$ lies on the curve

$ \begin{matrix} \therefore 2=t^{2}+3 t-8 \Rightarrow t^{2}+3 t-10=0 \\ \Rightarrow t^{2}+5 t-2 t-10=0 \\ \Rightarrow t(t+5)-2(t+5)=0 \\ \Rightarrow \quad(t+5)(t-2)=0 \\ \therefore t=2, t=-5 \text{ and }-1=2 t^{2}-2 t-5 \\ \Rightarrow 2 t^{2}-2 t-4=0 \\ \Rightarrow t^{2}-t-2 =0 \Rightarrow t^{2}-2 t+t-2=0 \\ \Rightarrow t(t-2)+1(t-2) =0 \Rightarrow(t+1)(t-2)=0 \\ \Rightarrow t =-1 \quad \text{ and } t=2 \end{matrix} $

So $t=2$ is common value

$ \therefore \quad \text{ Slope } \frac{d y}{d x _{x=2}}=\frac{4 \times 2-2}{2 \times 2+3}=\frac{6}{7} $

Hence, the correct option is (b).

Solution

The given curves are $x^{3}-3 x y^{2}+2=0$ and

$$ \begin{equation*} 3 x^{2} y-y^{3}-2=0 \tag{i} \end{equation*} $$

Differentiating eq. (i) w.r.t. $x$, we get

$ \begin{aligned} & 3 x^{2}-3(x \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 1)=0 \\ & \Rightarrow \quad x^{2}-2 x y \frac{d y}{d x}-y^{2}=0 \Rightarrow 2 x y \frac{d y}{d x}=x^{2}-y^{2} \\ & \therefore \quad \frac{d y}{d x}=\frac{x^{2}-y^{2}}{2 x y} \\ & m_1=\frac{x^{2}-y^{2}}{2 x y} \end{aligned} $

Differentiating eq. (ii) w.r.t. $x$, we get

$ \begin{aligned} 3[x^{2} \frac{d y}{d x}+y \cdot 2 x]-3 y^{2} \cdot \frac{d y}{d x} & =0 \\ x^{2} \frac{d y}{d x}+2 x y-y^{2} \frac{d y}{d x} & =0 \Rightarrow(x^{2}-y^{2}) \frac{d y}{d x}=-2 x y \\ \frac{d y}{d x} & =\frac{-2 x y}{x^{2}-y^{2}} \end{aligned} $

So the slope of the curve $m_2=\frac{-2 x y}{x^{2}-y^{2}}$

Now

$ m_1 \times m_2=\frac{x^{2}-y^{2}}{2 x y} \times \frac{-2 x y}{x^{2}-y^{2}}=-1 $

So the angle between the curves is $\frac{\pi}{2}$.

Hence, the correct option is (c).

Solution

The given function is $f(x)=2 x^{3}+9 x^{2}+12 x-1$

$ f^{\prime}(x)=6 x^{2}+18 x+12 $

For increasing and decreasing $f^{\prime}(x)=0$

$ \begin{matrix} \therefore x^{2}+18 x+12 =0 \\ \Rightarrow x^{2}+3 x+2 =0 \Rightarrow x^{2}+2 x+x+2=0 \\ \Rightarrow x(x+2)+1(x+2) =0 \Rightarrow(x+2)(x+1)=0 \\ \Rightarrow x =-2, x=-1 \end{matrix} $

The possible intervals are $(-\infty,-2),(-2,-1),(-1, \infty)$

Now

$ f^{\prime}(x)=(x+2)(x+1) $

$ \begin{matrix} \Rightarrow & f^{\prime}(x) _{(-\infty,-2)}=(-)(-)=(+) \text{ increasing } \\ \Rightarrow & f^{\prime}(x) _{(-2,-1)}=(+)(-)=(-) \text{ decreasing } \\ \Rightarrow & f^{\prime}(x) _{(-1, \infty)}=(+)(+)=(+) \text{ increasing } \end{matrix} $

Hence, the correct option is $(b)$.

Solution

Given that

$ \begin{aligned} f(x) & =2 x+\cos x \\ f^{\prime}(x) & =2-\sin x \\ f^{\prime}(x) & >0 \forall x \end{aligned} $

Since

So $f(x)$ is an increasing function.

Hence, the correct option is $(d)$.

Solution

Here $y=x(x-3)^{2}$

$ \frac{d y}{d x}=x \cdot 2(x-3)+(x-3)^{2} \cdot 1 \Rightarrow \frac{d y}{d x}=2 x(x-3)+(x-3)^{2} $

$ \begin{aligned} & \text{ For increasing and decreasing } \frac{d y}{d x}=0 \\ & \therefore \quad 2 x(x-3)+(x-3)^{2}=0 \Rightarrow(x-3)(2 x+x-3)=0 \\ & \Rightarrow \quad(x-3)(3 x-3)=0 \Rightarrow 3(x-3)(x-1)=0 \\ & \therefore \quad x=1,3 \\ & \frac{d y}{d x}=(x-3)(x-1) \\ & \text{ For }(-\infty, 1)=(-)(-)=(+) \text{ increasing } \\ & \text{ For }(1,3)=(-)(+)=(-) \text{ decreasing } \\ & \text{ For }(3, \infty)=(+)(+)=(+) \text{ increasing } \end{aligned} $

So the function decreases in $(1,3)$ or $1<x<3$

Hence, the correct option is $(a)$.

Solution

Here,

$ \begin{aligned} f(x) & =4 \sin ^{3} x-6 \sin ^{2} x+12 \sin x+100 \\ f^{\prime}(x) & =12 \sin ^{2} x \cdot \cos x-12 \sin x \cos x+12 \cos x \\ & =12 \cos x[\sin ^{2} x-\sin x+1] \\ & =12 \cos x[\sin ^{2} x+(1-\sin x)] \end{aligned} $

$\because \quad 1-\sin x \geq 0$ and $\sin ^{2} x \geq 0$

$\therefore \sin ^{2} x+1-\sin x \geq 0 \quad$ (when $\cos x>0$ )

Hence, $f^{\prime}(x)>0$, when $\cos x>0$ i.e., $x \in(\frac{-\pi}{2}, \frac{\pi}{2})$

So, $f(x)$ is increasing where $x \in(\frac{-\pi}{2}, \frac{\pi}{2})$ and $f^{\prime}(x)<0$

when $\cos x<0$ i.e. $x \in(\frac{\pi}{2}, \frac{3 \pi}{2})$

Hence, $f(x)$ is decreasing when $x \in(\frac{\pi}{2}, \frac{3 \pi}{2})$

As $(\frac{\pi}{2}, \pi) \in(\frac{\pi}{2}, \frac{3 \pi}{2})$

So $f(x)$ is decreasing in $(\frac{\pi}{2}, \pi)$

Hence, the correct option is (b).

Solution

Here, Let $\quad f(x)=\cos x$; So, $f^{\prime}(x)=-\sin x$

$ f^{\prime}(x)<0 \text{ in }(0, \frac{\pi}{2}) $

So

Hence, the correct option is (c).

$ f(x)=\cos x \text{ is decreasing in }(0, \frac{\pi}{2}) $

Solution

Here,

$ \begin{aligned} f(x) & =\tan x-x \quad \text{ So, } f^{\prime}(x)=\sec ^{2} x-1 \\ f^{\prime}(x) & >0 \forall x \in R \end{aligned} $

So $f(x)$ is always increasing

Hence, the correct option is $(a)$.

Solution

Let

$ \begin{aligned} f(x) & =x^{2}-8 x+17 \\ f^{\prime}(x) & =2 x-8 \end{aligned} $

For local maxima and local minima, $f^{\prime}(x)=0$

$ \therefore \quad 2 x-8=0 \Rightarrow x=4 $

So, $x=4$ is the point of local maxima and local minima.

$ \begin{aligned} f^{\prime \prime}(x) & =2>0 \text{ minima at } x=4 \\ \therefore \quad f(x) _{x=4} & =(4)^{2}-8(4)+17 \\ & =16-32+17=33-32=1 \end{aligned} $

So the minimum value of the function is 1

Hence, the correct option is (c).

Solution

Let

$ f(x)=x^{3}-18 x^{2}+96 x ; \text{ So, } f^{\prime}(x)=3 x^{2}-36 x+96 $

For local maxima and local minima $f^{\prime}(x)=0$

$\therefore \quad 3 x^{2}-36 x+96=0$

$\Rightarrow \quad x^{2}-12 x+32=0 \Rightarrow x^{2}-8 x-4 x+32=0$

$\Rightarrow x(x-8)-4(x-8)=0 \Rightarrow(x-8)(x-4)=0$

$\therefore \quad x=8,4 \in[0,9]$

So, $x=4,8$ are the points of local maxima and local minima.

Now we will calculate the absolute maxima or absolute minima at $x=0,4,8,9$

$ \begin{aligned} \therefore \quad f(x) & =x^{3}-18 x^{2}+96 x \\ f(x) _{x=0} & =0-0+0=0 \end{aligned} $

$ \begin{aligned} f(x) _{x=4} & =(4)^{3}-18(4)^{2}+96(4) \\ & =64-288+384=448-288=160 \\ f(x) _{x=8} & =(8)^{3}-18(8)^{2}+96(8) \\ & =512-1152+768=1280-1152=128 \\ f(x) _{x=9} & =(9)^{3}-18(9)^{2}+96(9) \\ & =729-1458+864=1593-1458=135 \end{aligned} $

So, the absolute minimum value of $f$ is 0 at $x=0$

Hence, the correct option is $(b)$.

Solution

We have

$ \begin{aligned} f(x) & =2 x^{3}-3 x^{2}-12 x+4 \\ f^{\prime}(x) & =6 x^{2}-6 x-12 \end{aligned} $

For local maxima and local minima $f^{\prime}(x)=0$

$\therefore \quad 6 x^{2}-6 x-12=0$

$\Rightarrow \quad x^{2}-x-2=0 \Rightarrow x^{2}-2 x+x-2=0$

$\Rightarrow \quad x(x-2)+1(x-2)=0 \Rightarrow(x+1)(x-2)=0$

$\Rightarrow x=-1,2$ are the points of local maxima and local minima Now $\quad f^{\prime \prime}(x)=12 x-6$

$ \begin{aligned} f^{\prime \prime}(x) _{x=-1} & =12(-1)-6=-12-6=-18<0, \text{ maxima } \\ f^{\prime \prime}(x) _{x=2} & =12(2)-6=24-6=18>0 \text{ minima } \end{aligned} $

So, the function is maximum at $x=-1$ and minimum at $x=2$ Hence, the correct option is (c).

Solution

We have

$ f(x)=\sin x \cos x $

$ \begin{matrix} \Rightarrow \quad & f(x)=\frac{1}{2} \cdot 2 \sin x \cos x=\frac{1}{2} \sin 2 x \\ & f^{\prime}(x)=\frac{1}{2} \cdot 2 \cos 2 x \\ \Rightarrow \quad & f^{\prime}(x)=\cos 2 x \end{matrix} $

Now for local maxima and local minima $f^{\prime}(x)=0$

$ \begin{aligned} \therefore \quad \cos 2 x & =0 \\ 2 x & =(2 n+1) \frac{\pi}{2}, \quad n \in I \\ \Rightarrow \quad x & =(2 n+1) \frac{\pi}{4} \end{aligned} $

$ \begin{aligned} x & =\frac{\pi}{4}, \frac{3 \pi}{4} \ldots \\ f^{\prime \prime}(x) & =-2 \sin 2 x \\ f^{\prime \prime}(x) _{x=\frac{\pi}{4}} & =-2 \sin 2 \cdot \frac{\pi}{4}=-2 \sin \frac{\pi}{2}=-2<0 \text{ maxima } \\ f^{\prime \prime}(x) _{x=\frac{3 \pi}{4}} & =-2 \sin 2 \cdot \frac{3 \pi}{4}=-2 \sin \frac{3 \pi}{2}=2>0 \text{ minima } \end{aligned} $

So $f(x)$ is maximum at $x=\frac{\pi}{4}$

$\therefore \quad$ Maximum value of $f(x)=\sin \frac{\pi}{4} \cdot \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2}$

Hence, the correct option is (b).

Solution

We have $f(x)=2 \sin 3 x+3 \cos 3 x$

$ \begin{aligned} f^{\prime}(x) & =2 \cos 3 x \cdot 3-3 \sin 3 x \cdot 3=6 \cos 3 x-9 \sin 3 x \\ f^{\prime \prime}(x) & =-6 \sin 3 x \cdot 3-9 \cos 3 x \cdot 3 \\ & =-18 \sin 3 x-27 \cos 3 x \\ f^{\prime \prime}(\frac{5 \pi}{6}) & =-18 \sin 3(\frac{5 \pi}{6})-27 \cos 3(\frac{5 \pi}{6}) \\ & =-18 \sin (\frac{5 \pi}{2})-27 \cos (\frac{5 \pi}{2}) \\ & =-18 \sin (2 \pi+\frac{\pi}{2})-27 \cos (2 \pi+\frac{\pi}{2}) \\ & =-18 \sin \frac{\pi}{2}-27 \cos \frac{\pi}{2}=-18 \cdot 1-27 \cdot 0 \\ & =-18<0 \text{ maxima } \end{aligned} $

Maximum value of $f(x)$ at $x=\frac{5 \pi}{6}$

$ \begin{aligned} f(\frac{5 \pi}{6}) & =2 \sin 3(\frac{5 \pi}{6})+3 \cos 3(\frac{5 \pi}{6})=2 \sin \frac{5 \pi}{2}+3 \cos \frac{5 \pi}{2} \\ & =2 \sin (2 \pi+\frac{\pi}{2})+3 \cos (2 \pi+\frac{\pi}{2})=2 \sin \frac{\pi}{2}+3 \cos \frac{\pi}{2}=2 \end{aligned} $

Hence, the correct option is (a).

Solution

Given that $\qquad y=-x^{3}+3 x^{2}+9 x-27$

$ \frac{d y}{d x}=-3 x^{2}+6 x+9 $

$\therefore \quad$ Slope of the given curve,

$ \begin{aligned} m & =-3 x^{2}+6 x+9 \\ \frac{d m}{d x} & =-6 x+6 \end{aligned} $

$ (\frac{d y}{d x}=m) $

For local maxima and local minima, $\frac{d m}{d x}=0$

$ \therefore \quad-6 x+6=0 \Rightarrow x=1 $

Now

$ \frac{d^{2} m}{d x^{2}}=-6<0 \quad \text{ maxima } $

$\therefore \quad$ Maximum value of the slope at $x=1$ is

$ m _{x=1}=-3(1)^{2}+6(1)+9=-3+6+9=12 $

Hence, the correct option is (b).

Solution

We have

$ f(x)=x^{x} $

Taking $\log$ of both sides, we have

$ \log f(x)=x \log x $

Differentiating both sides w.r.t. $x$, we get

$ \begin{aligned} \frac{1}{f(x)} \cdot f^{\prime}(x) & =x \cdot \frac{1}{x}+\log x \cdot 1 \\ \Rightarrow \quad f^{\prime}(x) & =f(x)[1+\log x]=x^{x}[1+\log x] \end{aligned} $

To find stationary point, $f^{\prime}(x)=0$

$ \begin{aligned} & \therefore \quad x^{x}[1+\log x]=0 \\ & x^{x} \neq 0 \quad \therefore \quad 1+\log x=0 \\ & \Rightarrow \quad \log x=-1 \Rightarrow x=e^{-1} \Rightarrow x=\frac{1}{e} \end{aligned} $

Hence, the correct option is $(b)$.

Solution

Let

$ f(x)=(\frac{1}{x})^{x} $

Taking $\log$ on both sides, we get

$ \begin{aligned} \log [f(x)] & =x \log \frac{1}{x} \\ \Rightarrow \quad \log [f(x)] & =x \log x^{-1} \Rightarrow \log [f(x)]=-[x \log x] \end{aligned} $

Differentiating both sides w.r.t. $x$, we get

$ \begin{aligned} \frac{1}{f(x)} \cdot f^{\prime}(x) =-[x \cdot \frac{1}{x}+\log x \cdot 1]=-f(x)[1+\log x] \\ \Rightarrow f^{\prime}(x) =-(\frac{1}{x})^{x}[1+\log x] \end{aligned} $

For local maxima and local minima $f^{\prime}(x)=0$

$ \begin{aligned} \quad-(\frac{1}{x})^{x}[1+\log x]=0 \Rightarrow(\frac{1}{x})^{x}[1+\log x]=0 \\ (\frac{1}{x})^{x} \neq 0 \\ \quad \therefore 1+\log x=0 \Rightarrow \log x=-1 \Rightarrow x=e^{-1} \end{aligned} $

So, $x=\frac{1}{e}$ is the stationary point.

Now $\quad f^{\prime}(x)=-(\frac{1}{x})^{x}[1+\log x]$

$ \begin{aligned} & f^{\prime \prime}(x)=-[(\frac{1}{x})^{x}(\frac{1}{x})+(1+\log x) \cdot \frac{d}{d x}(x)^{x}] \\ & f^{\prime \prime}(x)=-[(e)^{1 / e}(e)+(1+\log \frac{1}{e}) \frac{d}{dx}(\frac{1}{e})^{1 / e}] \\ & x=\frac{1}{e}=-e^{\frac{1}{e}}<0 \text{ maxima } \end{aligned} $

$\therefore \quad$ Maximum value of the function at $x=\frac{1}{e}$ is

$ f(\frac{1}{e})=(\frac{1}{1 / e})^{1 / e}=e^{1 / e} $

Hence, the correct option is (c).

Solution

We have

$$ \begin{align*} & y=4 x^{2}+2 x-8 \tag{i}\\ & y=x^{3}-x+13 \tag{ii} \end{align*} $$

Differentiating eq. (i) w.r.t. $x$, we have

$ \frac{d y}{d x}=8 x+2 \Rightarrow m_1=8 x+2 $

[ $m$ is the slope of curve $(i)]$

Differentiating eq. (ii) w.r.t. $x$, we get

$\frac{dy}{dx}=3x^2-1 \Rightarrow m_2=3x^2-1 \quad$[$m_2$ is the slope of curve (ii)]

If the two curves touch each other, then $m_1=m_2$

$ \begin{aligned} & \therefore \quad 8 x+2=3 x^{2}-1 \\ & \Rightarrow \quad 3 x^{2}-8 x-3=0 \Rightarrow 3 x^{2}-9 x+x-3=0 \\ & \Rightarrow \quad 3 x(x-3)+1(x-3)=0 \Rightarrow(x-3)(3 x+1)=0 \\ & \therefore \quad x=3, \frac{-1}{3} \end{aligned} $

Putting $x=3$ in eq. (i), we get

$ y=4(3)^{2}+2(3)-8=36+6-8=34 $

So, the required point is $(3,34)$

Now for $x=-\frac{1}{3}$

$ \begin{aligned} y & =4(\frac{-1}{3})^{2}+2(\frac{-1}{3})-8=4 \times \frac{1}{9}-\frac{2}{3}-8 \\ & =\frac{4}{9}-\frac{2}{3}-8=\frac{4-6-72}{9}=\frac{-74}{9} \end{aligned} $

$\therefore \quad$ Other required point is $(-\frac{1}{3}, \frac{-74}{9})$.

Hence, the required points are $(3,34)$ and $(-\frac{1}{3}, \frac{-74}{9})$.

Solution

We have $y=\tan x$. So, $\frac{d y}{d x}=\sec ^{2} x$

$\therefore \quad$ Slope of the normal $=\frac{-1}{\sec ^{2} x}=-\cos ^{2} x$

at the point $(0,0)$ the slope $=-\cos ^{2}(0)=-1$

So the equation of normal at $(0,0)$ is $y-0=-1(x-0)$

$\Rightarrow \quad y=-x \Rightarrow y+x=0$

Hence, the required equation is $y+x=0$.

Solution

We have $f(x)=\sin x-a x+b \quad \Rightarrow f^{\prime}(x)=\cos x-a$

For increasing the function $f^{\prime}(x)>0$

$\therefore \quad \cos x-a>0$

Since $\quad \cos x \in[-1,1]$

$ \therefore \quad a<-1 \Rightarrow a \in(-\infty,-1) $

Hence, the value of $a$ is $(-\infty,-1)$.

Solution

We have $f(x)=\frac{2 x^{2}-1}{x^{4}}$

$ \begin{aligned} f^{\prime}(x) & =\frac{x^{4}(4 x)-(2 x^{2}-1) \cdot 4 x^{3}}{x^{8}} \\ \Rightarrow f^{\prime}(x) & =\frac{4 x^{5}-(2 x^{2}-1) \cdot 4 x^{3}}{x^{8}}=\frac{4 x^{3}[x^{2}-2 x^{2}+1]}{x^{8}}=\frac{4(-x^{2}+1)}{x^{5}} \end{aligned} $

For decreasing the function $f^{\prime}(x)<0$

$ \begin{aligned} & \therefore \quad \frac{4(-x^{2}+1)}{x^{5}}<0 \Rightarrow-x^{2}+1<0 \quad \Rightarrow \quad x^{2}>1 \\ & \therefore \quad x> \pm 1 \Rightarrow x \in(1, \infty) \end{aligned} $

Hence, the required interval is $(1, \infty)$.

Solution

Here,

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

For maximum and minimum value $f^{\prime}(x)=0$

$ \therefore \quad a-\frac{b}{x^{2}}=0 \Rightarrow x^{2}=\frac{b}{a} \Rightarrow x= \pm \sqrt{\frac{b}{a}} $

Now

$ f^{\prime \prime}(x)=\frac{2 b}{x^{3}} $

$ f^{\prime \prime}(x) _{x=\sqrt{\frac{b}{a}}}=\frac{2 b}{(\frac{b}{a})^{3 / 2}}=2 \frac{a^{3 / 2}}{b^{1 / 2}}>0 \quad(\because a, b>0) $

Hence, minima

So the least value of the function at $x=\sqrt{\frac{b}{a}}$ is

$ f(\sqrt{\frac{b}{a}})=a \cdot \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b} $

Hence, least value is $2 \sqrt{a b}$.