Application of Derivatives

As we know that, volume of sphere

$ =v=\frac{4}{3} \pi r^3 $

and Surface area of sphere $=s=4 \pi r^2$

Now, $\frac{d \nu}{d t}=4 \pi r^2 \frac{d r}{d t}$

$ \Rightarrow \quad \frac{d r}{d t}=\frac{1}{4 \pi r^2} \times \frac{d v}{d t} $

Putting $r=6$ and $\frac{d v}{d t}=12$

( $\because$ Diameter of sphere $=12 \mathrm{~cm} \quad$ (given)

$\therefore$ Radius of sphere $=6 \mathrm{~cm}$ )

$ \frac{d r}{d t}=\frac{12}{4 \pi(6)^2}=\frac{1}{12 \pi} $

Now, $\frac{d s}{d t}=8 \pi r \cdot \frac{d r}{d t}$

$ \therefore \quad \frac{d s}{d t}=8 \pi \times 6 \times \frac{1}{12 \pi}=4 \mathrm{~cm}^2 $

(putting $r=6$ and $\frac{d r}{d t}=\frac{1}{12 \pi}$ )

Given equation of curve is

$ \begin{aligned} y & =x^2+x-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \Rightarrow \quad \frac{d y}{d x} & =2 x+1 \\ \left.\therefore \frac{d y}{d x}\right|_{x=1} & =2 \times 1+1=3=m \end{aligned} $

i.e. slope of tangent $=m=3$. and point is $(1,1)$ (by putting $x=1$ in Eq. (i))

Now, length of tangent

$ \begin{aligned} & =a=\left|\frac{y \sqrt{1+m^2}}{m}\right| \\ & =\left|\frac{1 \sqrt{1+3^2}}{3}\right|=\frac{\sqrt{10}}{3} \end{aligned} $

length of subtangent

$ =b=\left|\frac{y}{m}\right|=\left|\frac{1}{3}\right|=\frac{1}{3} $

length of normal

$ =c=\left|y \sqrt{1+m^2}\right|=|1 \times \sqrt{10}|=\sqrt{10} $

length of subnormal

$ =d=|y \cdot m|=|3|=3 $

Now, from above result, we conclude that $b

$ \begin{aligned} & \text { } y=x^4-2 x^3+x^2+5 x \\ & \Rightarrow \frac{d y}{d x}=4 x^3-6 x^2+2 x+5 \end{aligned} $

The slope of tangent line at $\left(x_1, y_1\right)$ is

$ m=4 x_1^3-6 x_1^2+2 x_1+5 $

The equation of tangent line is

$ y-y_1=\left(4 x_1^3-6 x_1^2+2 x_1+5\right)\left(x-x_1\right) $

Since, the tangent line passes through $(0,0)$

$ \begin{aligned} \Rightarrow 0-y_1 & =\left(4 x_1^3-6 x_1^2+2 x_1+5\right)\left(0-x_1\right) \\ y_1 & =\left(4 x_1^3-6 x_1^2+2 x_1+5\right) x_1 \end{aligned} $

Since, ( $x_1, y_1$ ) is on the curve

$ y_1=x_1^4-2 x_1^3+x_1^2+5 x_1 $

Substituting this into the previous equation, we get

$ \begin{aligned} & x_1^4-2 x_1^3+x_1^2+5 x_1 \\ & =\left(4 x_1^3-6 x_1^2+2 x_1+5\right) x_1 \\ \Rightarrow & x_1^4-2 x_1^3+x_1^2+5 x_1 \\ & =4 x_1^4-6 x_1^3+2 x_1^2+5 x_1 \\ \Rightarrow & 3 x_1^4-4 x_1^3+x_1^2=0 \\ \Rightarrow & x_1^2\left(3 x_1^2-4 x_1+1\right)=0 \\ \Rightarrow & x_1^2\left(3 x_1-1\right)\left(x_1-1\right)=0 \\ \Rightarrow & x_1=0, x_1=\frac{1}{3}, x_1=1 \\ & x_1=1, y_1=5 \Rightarrow x_1+y_1=1+5=6 \end{aligned} $

From the given, we have

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

Differentiating both sides, we get $\Rightarrow f^{\prime}(x)=\frac{1}{(1+x)}-1+x$

$ \begin{aligned} & \Rightarrow f^{\prime}(x)=\left\{\frac{1-1-x+x+x^2}{(1+x)}\right\} \\ & \Rightarrow f^{\prime}(x)=\frac{x^2}{(1+x)} \end{aligned} $

Hence, $f(x)$ is monotonically increasing in its domain.

Step 1: Find the Derivative and Slope of Normal

The curve is $x^2 + 3y^2 = 9$.

Differentiate both sides with respect to $x$:

$2x + 3 \cdot 2y \cdot y' = 0$

Solve for $y'$:

$2x + 6y y' = 0 \implies 6y y' = -2x \implies y' = -\frac{2x}{6y} = -\frac{x}{3y}$

The slope of the tangent is $y' = -\frac{x}{3y}$, so the slope of the normal is $-\frac{1}{y'} = \frac{3y}{x}$.

Step 2: Find the Slope of Normals at Given Points

First point: $(3\cos\theta,\, \sqrt{3}\sin\theta)$

Slope at this point:

$m_1 = \frac{3 \cdot \sqrt{3} \sin\theta}{3\cos\theta} = \sqrt{3}\tan\theta$

Second point: $(-3\sin\theta,\, \sqrt{3}\cos\theta)$

Slope at this point:

$m_2 = \frac{3 \cdot \sqrt{3} \cos\theta}{-3\sin\theta} = -\sqrt{3}\cot\theta$

Step 3: Find the Angle Between Normals ($\beta$)

Use the formula for the tangent of the angle between two lines:

$\tan\beta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

$\tan\beta = \left|\frac{\sqrt{3}\tan\theta - (-\sqrt{3}\cot\theta)}{1 + (\sqrt{3}\tan\theta)(-\sqrt{3}\cot\theta)}\right|$

$\tan\beta = \left|\frac{\sqrt{3}(\tan\theta + \cot\theta)}{1 - 3}\right|$

$\tan\beta = \left|\frac{\sqrt{3}(\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta})}{-2}\right|$

$\tan\beta = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sin\theta\cos\theta}$

Recall $2\sin\theta\cos\theta = \sin2\theta$, so:

$\tan\beta = \frac{\sqrt{3}}{\sin2\theta}$

This means $\frac{1}{\cot\beta} = \frac{\sqrt{3}}{\sin2\theta}$ or $\sqrt{3}\cot\beta = \sin2\theta$.

$ \begin{aligned} & x^2+y^2=5^2 \\ & y^2=5^2-x^2 \Rightarrow y=\sqrt{5^2-x^2} \\ & 0 \leq x \leq 5 \end{aligned} $

$ \begin{aligned} & \text { Area }=\frac{1}{2} \text { base } \times \text { height } \\ & \begin{aligned} & A=\frac{1}{2} x y=\frac{1}{2} x \sqrt{25-x^2} \\ & \frac{d A}{d x}=\frac{1}{2}\left(25-x^2\right)^{-\frac{1}{2}}\left(25-2 x^2\right) \\ &=\frac{25-2 x^2}{2 \sqrt{25-x^2}} \end{aligned} \\ & \Rightarrow \quad \frac{d A}{d x}=0 \Rightarrow \frac{25-2 x^2}{2 \sqrt{25-x^2}}=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 25-2 x^2=0 \\ & \Rightarrow \quad 25=2 x^2 \\ & \Rightarrow \quad x=\frac{5}{\sqrt{2}}, y=\frac{5}{\sqrt{2}} \\ & \Rightarrow \quad A B=\frac{5}{\sqrt{2}}, B C=\frac{5}{\sqrt{2}}, A C=5 \\ & \text { Perimeter }=A B+B C+A C \\ & =\frac{5}{\sqrt{2}}+\frac{5}{\sqrt{2}}+5=5 \sqrt{2}+5 \\ & =5(\sqrt{2}+1) \end{aligned} $

$ y=|\cos x-\sin x|+|\tan x-\cot x| $

For $^{\prime \prime} x=\frac{\pi}{3}$

$ \begin{aligned} y & =-(\cos x-\sin x)+(\tan x-\cot x) \\ & =\cos x+\sin x+\tan x-\cot x \end{aligned} $

For $x=\frac{\pi}{6}$

$ \begin{aligned} y & =(\cos x-\sin x)-(\tan x-\cot x) \\ & =\cos x-\sin x-\tan x+\cot x \\ \frac{d y}{d x} & ]_x=\frac{\pi}{3} \\ = & \sin \frac{\pi}{3}+\cos \frac{\pi}{3}+\sec ^2 \frac{\pi}{3}+\operatorname{cosec}^2 \frac{\pi}{3} \\ = & \frac{35+3 \sqrt{3}}{6} \end{aligned} $

$ \begin{aligned} & \left.\frac{d y}{d x}\right]_{x=\frac{\pi}{6}} \\ & =-\sin \frac{\pi}{6}-\cos \frac{\pi}{6}-\sec ^2 \frac{\pi}{6}-\operatorname{cosec}^2 \frac{\pi}{6} \\ & =\frac{-35-3 \sqrt{3}}{6} \\ & \left.\left.\frac{d y}{d x}\right]_{x=\frac{\pi}{3}}+\frac{d y}{d x}\right]_{x=\frac{\pi}{6}}=0 \end{aligned} $

Given curve, $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and line $y=-\sqrt{3} x+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), $\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} y^{\prime}=0$

$ \begin{aligned} y^{\prime} l_{(\alpha, \beta)} & =-\left(\frac{\beta}{\alpha}\right)^{\frac{1}{3}} \\ -\left(\frac{\beta}{\alpha}\right)^{\frac{1}{3}} & =-\sqrt{3} \Rightarrow \frac{\beta}{\alpha}=3 \sqrt{3} \\ \beta & =3 \sqrt{3} \alpha \end{aligned} $

From Eq. (i) $\alpha^{\frac{2}{3}}+3 \alpha^{\frac{2}{3}}=4 \Rightarrow \alpha= \pm 1$

$ \begin{aligned} \beta & = \pm 3 \sqrt{3} \\ \therefore \quad \alpha^2+\beta^2 & =28 \end{aligned} $

$S=t^3-16 t^2+64 t-6$

$ \Rightarrow \frac{d S}{d t}=3 t^2-32 t+64 $

For maxima/minima $\frac{d S}{d t}=0$

$ \begin{aligned} \Rightarrow \quad & \left(3 t^2-8\right)(t-8)=0 \Rightarrow t=8, \frac{8}{3} \\ & \left.\frac{d^2 s}{d t^2}=6 t-32 \Rightarrow \frac{d^2 s}{d t^2}\right]_{t=\frac{8}{3}}<0 \end{aligned} $

So, required time, $t=\frac{8}{3}$

$ \begin{aligned} & \text { } f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x} \\ & \Rightarrow f^{\prime}(x)=\frac{-4 \cos x}{\sin ^2 x}+\frac{\cos x}{(1-\sin x)^2} \\ & f^{\prime}(x)=0 \\ & \Rightarrow \cos x\left[\frac{1}{(1-\sin x)^2}-\frac{4}{\sin ^2 x}\right]=0 \\ & \Rightarrow \cos x\left[\frac{\sin ^2 x-4(1-\sin x)^2}{\left(\sin ^2 x\right)(1-\sin x)^2}\right]=0 \\ & \Rightarrow \cos x[(\sin x+2-2 \sin x) (\sin x-2+2 \sin x)] \end{aligned} $

$ \begin{aligned} & =0 \\ \Rightarrow \quad x & =\frac{\pi}{2}, \sin x=2 \sin x=\frac{2}{3} \\ f(0) & =\text { Not possible } \\ f\left(\frac{\pi}{2}\right) & =\text { Not possible } \\ \sin x & \neq 2 \quad(\because \sin x \in[-1,1]) \end{aligned} $

$ \begin{aligned} \therefore \quad \sin x & =\frac{2}{3} \\ f(x) l_{\max } & =6+3=9=m \\ \Rightarrow \quad \cos k & =\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\sqrt{\frac{5}{m}} \end{aligned} $

Let $P(h, k)$

$ \begin{aligned} & y=x \log x \\ \Rightarrow \quad & \frac{d y}{d x}=1+\log x \\ \therefore & \text { Slope of normal }=-\frac{1}{1+\log h} \end{aligned} $

And slope of the given line $2 x-2 y=3$ is 1

$ \begin{aligned} & \Rightarrow \quad-\frac{1}{1+\log h}=1 \\ & \Rightarrow \quad 1+\ln h=-1 \\ & \Rightarrow \ln h=-2 \\ & h=e^{-2} \\ & k=h \ln h \\ & k=e^{-2} \ln e^{-2}=e^{-2}(-2)=-\frac{2}{e^2} \end{aligned} $

$\therefore$ The point $P$ is $\left(\frac{1}{e^2},-\frac{2}{e^2}\right)$

We have the curve

$ y^2=16 x \text { and } 9 x^2+\alpha y^2=25 $

On differentiating, we get slopes

$ 2 y y^{\prime}=16 \text { and } 18 x+2 \alpha y y^{\prime}=0 $

$ y^{\prime}=\frac{8}{y}, y^{\prime}=-\frac{9 x}{\alpha y} $

Clearly, $m_1=\frac{8}{y}$ and $m_2=-\frac{9 x}{\alpha y}$

$ \begin{aligned} & m_1 m_2=-1 \\ & \quad \frac{8}{y} \times\left(\frac{-9 x}{\alpha y}\right)=-1 \Rightarrow-\frac{72 x}{\alpha y^2}=-1 \\ & \quad 72 x=\alpha y^2=\alpha \cdot 16 x \\ & 16 \alpha=72 \\ & \quad \alpha=\frac{72}{16}=\frac{9}{2} \\ & \quad \therefore \quad \alpha=\frac{9}{2} \end{aligned} $

We have, $y=\sin x(1+\cos x)$

$ \begin{aligned} \frac{d y}{d x} & =\sin x(-\sin x)+(1+\cos x) \cos x \\ & =-\sin ^2 x+\cos x+\cos ^2 x \\ & =-1+\cos ^2 x+\cos x+\cos ^2 x \\ & =2 \cos ^2 x+\cos x-1 \\ \Rightarrow \frac{d y}{d x} & =(\cos x+1)(2 \cos x-1) \end{aligned} $

Given $y$ is strictly increasing

$ \begin{aligned} & \therefore \quad \frac{d y}{d x}=(\cos x+1)(2 \cos x-1)>0 \\ & \Rightarrow \cos x<-1 \text { or } \cos x>\frac{1}{2} \\ & \Rightarrow x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \text { in the interval }(-\pi, \pi) \\ & \therefore x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \end{aligned} $

Given:

Velocity $ v $ is proportional to the cube root of the displacement $ x $.

$ v \propto x^{1/3} $

This means we can write:

$ v = k x^{1/3} $

where $ k $ is a constant of proportionality.

We are asked to find how acceleration $ a $ depends on other quantities.

We know that:

$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} $

Step 1: Find $ \frac{dv}{dx} $

From $ v = k x^{1/3} $,

$ \frac{dv}{dx} = k \cdot \frac{1}{3} x^{-2/3} = \frac{k}{3} x^{-2/3} $

Step 2: Substitute in acceleration formula:

$ a = v \frac{dv}{dx} = (k x^{1/3}) \cdot \left(\frac{k}{3} x^{-2/3}\right) $

Simplify:

$ a = \frac{k^2}{3} x^{1/3 - 2/3} = \frac{k^2}{3} x^{-1/3} $

Step 3: Express in terms of velocity if desired

Since $ v = k x^{1/3} \Rightarrow x^{1/3} = \frac{v}{k} $,

$ x^{-1/3} = \frac{k}{v} $

So:

$ a = \frac{k^2}{3} \cdot \frac{k}{v} = \frac{k^3}{3v} $

Therefore:

$ a \propto \frac{1}{v} $

✅ Final Answer:

Option B: inversely proportional to its velocity

We have, $f(x)=4 x^4+4 x^3-23 x^2$$ -12 x+36=0 $

Using concept a polynomial $f(x)$ has a multiple root $r$, then $f(r)=0$ and

$ \begin{aligned} & f^{\prime}(r)=0 \\ & f^{\prime}(x)=16 x^3+12 x^2-46 x-12 \end{aligned} $

Now, using hit and trial

By taking $x=-2$

$ \begin{gathered} f(-2)=4(-2)^4+4(-2)^3-23(-2)^2 \\ -12(-2)+36 \\ =64-32-92+24+36=0 \\ \Rightarrow f^{\prime}(-2)=16(-2)^3+12(-2)^2 \\ -46(-2)-12 \\ =-128+48+92-12=0 \end{gathered} $

$\because \quad x=-2$ is a multiple root.

By dividing the equation by $(x+2)^2$, we get another multiple root $x=\frac{3}{2}$

$ \begin{aligned} & \text { Given, } \alpha>\beta \\ & \therefore \quad \alpha=\frac{3}{2} \text { and } \beta=-2 \\ & \therefore 2 \alpha-\beta=3+2=5 \end{aligned} $

$ \begin{aligned} &\text { We have, the curve }\\ &x^3+y^3=2 x y \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 3 x^2+3 y^2 y^{\prime}=2 x y^{\prime}+y \cdot 2 \\ & \Rightarrow \quad y^{\prime}\left(3 y^2-2 x\right)=2 y-3 x^2 \\ & \Rightarrow \quad y^{\prime}=\frac{2 y-3 x^2}{3 y^2-2 x} \end{aligned} $

Slope of tangent $=\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{2-3}{3-2}$

$ =\frac{-1}{1}=-1 $

Slope of normal $=1$

Equation of tangent $\Rightarrow y-1=-1(x-1)$

$ \begin{aligned} & y-1=-x+1 \\ & x+y=2 \end{aligned} $

$x$-intercept of tangent $=(2,0)$

Equation of normal,

$ \begin{gathered} y-1=1(x-1) \\ y=x \end{gathered} $

$\because$ Area of $\triangle P T N=\frac{1}{2} \times N T \times P S$

$ =\frac{1}{2} \times 2 \times 1=1 \text { sq. unit } $

$f(x)=2 \sin x+\sin 2 x$

$f(x)$ is continuous and differentiable

$ f(0)=f(\pi) $

$\therefore$ Rolle's Theorem is applicable

$ \begin{aligned} & f^{\prime}(x)=2 \cos x+\cos 2 x \cdot 2 \\ & \begin{aligned} \Rightarrow f^{\prime}(c) & =2 \cos c+2\left(2 \cos ^2 c-1\right)=0 \\ & =2\left[2 \cos ^2 c+\cos c-1\right]=0 \\ & =2 \cos ^2 c+2 \cos c-\cos c-1=0 \end{aligned} \\ & \Rightarrow 2 \cos c(\cos c+1)-1(\cos c+1)=0 \\ & \Rightarrow(\cos c+1)(2 \cos c-1)=0 \\ & \Rightarrow \cos c=-1 \text { and } \cos c=\frac{1}{2} \\ & \because \quad c=\frac{\pi}{3} \end{aligned} $

$y=3 x^4-5 x^3-12 x^2+18 x+3$

Slope of tangent

$ \begin{aligned} g(x) & =\frac{d y}{d x}=12 x^3-15 x^2-24 x+18 \\ g^{\prime}(x) & =36 x^2-30 x-24 \\ & =6\left(6 x^2-5 x-4\right) \end{aligned} $

$\therefore g(x)$ is increasing         $ \left[\because g^{\prime}(x)>0\right] $

$ \begin{aligned} & x<-\frac{1}{2} \text { or } x>\frac{4}{3} \\ \therefore \quad & R-\left[-\frac{1}{2}, \frac{4}{3}\right] \end{aligned} $

$ \text { In } \triangle P O M \text {, } $

$ \begin{aligned} & \cos \theta=\frac{O M}{O P}=\frac{4}{8}=\frac{1}{2} \\ & \sin \theta=\frac{P M}{O P}=\frac{P M}{8} \end{aligned} $

On differentiating both sides w.r.t.t, we get

$ \begin{aligned} \cos \theta \frac{d \theta}{d t} & =\frac{1}{8} \frac{d(P M)}{d t} \\ \frac{4}{8} \times 6 & =\frac{1}{8} \times \frac{d(P M)}{d t} \\ \Rightarrow \quad \frac{d(P M)}{d t} & =24 \end{aligned} $

Given that the length of the sub-tangent at any point $ P $ on a curve is proportional to the abscissa $ x $ of the point $ P $, we start by expressing this relationship mathematically:

$ \frac{y}{\frac{dy}{dx}} \propto x $

This implies:

$ \frac{y}{\frac{dy}{dx}} = kx $

where $ k $ is the constant of proportionality. By rearranging, we get:

$ \frac{dy}{y} = \frac{dx}{kx} $

Integrating both sides, we obtain:

$ \int \frac{dy}{y} = \int \frac{dx}{kx} $

which results in:

$ \log y = \frac{1}{k} \log x + \log c $

Rearranging this:

$ \log y = \log x^{1/k} + \log c $

Therefore, we have:

$ \log y = \log(x^{1/k} c) $

Exponentiating both sides gives us:

$ y = x^{1/k} c $

Thus, the equation of the curve is:

$ y = x^{1/k} c $

Given that

Semi-vertical angle of a right circular cone $=45^{\circ}$

Radius of base of the cone $=14 \mathrm{~cm}$ with an error $(\sqrt{2}-1) / 11$

$ \text { Total surface area }=\pi r^2+\pi r s $

$ \begin{aligned} & \text { Error }=\frac{d r}{r}=\frac{\sqrt{2}-1}{11} \\ & \Rightarrow \quad d r=r \times \frac{\sqrt{2}-1}{11}=\frac{14}{11}(\sqrt{2}-1) \end{aligned} $

From $\triangle A B C, \tan 45^{\circ}=\frac{r}{h} \Rightarrow r=h$

$ \begin{array}{ll} \Rightarrow & h=14 \mathrm{~cm} \Rightarrow s^2=h^2+r^2=2 r^2 \\ \Rightarrow & s=\sqrt{2} \cdot r=14 \sqrt{2} \end{array} $

Since, error in total surface area

$ d s=\frac{\partial s}{\partial r} d r+\frac{\partial s}{\partial h} \cdot d h $

After simplifying

$ d s=8 $

$ \begin{aligned} &\triangle A B E \text { and } \triangle C D E \text { are similar }\\ &\begin{aligned} & \frac{A B}{C D}=\frac{B E}{D E} \\ \Rightarrow \quad & \frac{6}{1 \cdot 8}=\frac{x+y}{y}=6 y=1 \cdot 8 x+1 \cdot 8 y \\ \Rightarrow \quad & 7 y=3 x \end{aligned} \end{aligned} $

On differentiating both sides w.r to $t$, we get

$ 3 \frac{d x}{d t}=7 \frac{d y}{d t} $

$\therefore$ Main velocity

$ \Rightarrow \quad \frac{d y}{d t}=\frac{3 \times 7}{7}=3 $

Given that

$ C_1: 2 x^2+k y^2=30 $

and $C_2: 3 y^2=28 x$ cut each other orthogonaly.

$ \begin{array}{lr} \Rightarrow & 2 x^2+k y^2=30 \\ \Rightarrow & 4 x+2 k y \cdot \frac{d y}{d x}=0 \\ \Rightarrow & \frac{d y}{d x}=\frac{-4 x}{2 k y}=-\frac{2 x}{k y} \\ \therefore & m_1=-\frac{2 x}{k y} \end{array} $

$ \begin{aligned} &\begin{aligned} & \text { } \begin{aligned}Now,\,\, 3 y^2 & =28 x \\ 6 y \cdot \frac{d y}{d x} & =28 \Rightarrow \frac{d y}{d x}=\frac{28}{6 y}=\frac{14}{3 y} \\ \therefore \quad m_2 & =\frac{14}{3 y} \end{aligned} \end{aligned}\\ &\text { For orthogonally }\\ &m_1 m_2=-1 \end{aligned} $

$ \Rightarrow \quad-\frac{2 x}{k y} \cdot \frac{14}{3 y}=-1, \quad k=\frac{3 y^2}{28 x} $

Now, find intersection point of both curve

$ \begin{aligned} \frac{y^2}{x} & =\frac{28}{3} \quad\left[\text { from curve } C_2\right] \\ \Rightarrow \quad K & =\frac{3}{28} \times \frac{28}{3}=1 \\ K & =1 \end{aligned} $

We have, $f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}$

Interval in which $f(x)$ strictly increasing $f^{\prime}(x)>0$.

$ \begin{aligned} & \frac{d}{d x}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)>0 \\ \Rightarrow & \frac{1}{2 \sqrt{x}}-\frac{1}{2 x^{3 / 2}}>0 \Rightarrow \frac{1}{\sqrt{x}}>\frac{1}{x \cdot x^{1 / 2}} \\ \Rightarrow \quad & x \sqrt{x}>\sqrt{x} \Rightarrow \sqrt{x}(x-1)>0 \\ \Rightarrow \quad & \sqrt{x}(x-1)>0 \end{aligned} $

$ \therefore \quad x \in(1, \infty) $

Let's address the value of Lagrange's Mean Value Theorem for the function $ f(x) = e^x + 24 $ over the interval $[0, 1]$.

According to the theorem, if a function $ f $ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one point $ c $ such that:

$ f'(c) = \frac{f(b) - f(a)}{b - a} $

For this specific function:

$ f(x) = e^x + 24 $

Interval: $[0, 1]$

Calculate $ f(0) $ and $ f(1) $:

$ f(0) = e^0 + 24 = 1 + 24 = 25 $

$ f(1) = e^1 + 24 = e + 24 $

The derivative, $ f'(x) $, is:

$ f'(x) = e^x $

Next, let's find the average rate of change over the interval $[0, 1]$:

$ \frac{f(1) - f(0)}{1 - 0} = \frac{(e + 24) - 25}{1} = e - 1 $

According to Lagrange's Theorem, set:

$ f'(c) = \text{Average rate of change} $

Thus, we have:

$ e^c = e - 1 $

To find $ c $, solve for $ c $:

$ c = \log(e - 1) $

This yields the point $ c = \log(e - 1) $ as the value of the theorem for the function over the given interval.

Given the curve $ y = x^2 + x $ and the point $(1, 2)$, we need to find the equation of the normal to the curve at this point.

First, find the derivative of the curve to get the slope of the tangent:

$ \frac{dy}{dx} = 2x + 1 $

Calculate the slope at the point $(1, 2)$:

$ \left.\frac{dy}{dx}\right|_{(1, 2)} = 2 \times 1 + 1 = 3 $

The slope of the tangent line at $(1, 2)$ is 3. The slope of the normal line is the negative reciprocal of this, which is:

$ -\frac{1}{3} $

Using the point-slope form of the equation of a line, the equation of the normal line is:

$ y - y_1 = m(x - x_1) $

Here, $x_1 = 1$, $y_1 = 2$, and the slope $m = -\frac{1}{3}$.

Substitute the values into the equation:

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

Simplify:

$ 3(y - 2) = -(x - 1) $

$ 3y - 6 = -x + 1 $

Rearrange to standard form:

$ x + 3y - 7 = 0 $

Thus, the equation of the normal to the curve at the point $(1, 2)$ is:

$ x + 3y - 7 = 0 $

The position of a particle, given by its displacement $ s $ at time $ t $, is expressed as $ s = 2t^3 - 9t $.

To find velocity $ v $, we differentiate the displacement with respect to time $ t $:

$ v = \frac{d}{dt}(2t^3 - 9t) = 6t^2 - 9 $

The velocity vanishes (i.e., $ v = 0 $) at:

$ 6t^2 - 9 = 0 $

Solving for $ t $, we get:

$ t^2 = \frac{9}{6} = \frac{3}{2} $

$ t = \sqrt{\frac{3}{2}} $

Next, to find the acceleration $ a $, we differentiate the velocity:

$ a = \frac{d}{dt}(6t^2 - 9) = 12t $

Substituting $ t = \sqrt{\frac{3}{2}} $ into the acceleration equation, we find:

$ a = 12 \times \sqrt{\frac{3}{2}} $

Simplifying further, we get:

$ a = 12 \times \frac{\sqrt{6}}{\sqrt{2}} $

Thus:

$ a = 6\sqrt{6} $

Therefore, the acceleration at the time when the velocity vanishes is $ 6\sqrt{6} $.

$ \text { Area of rectangle point }=2 x x $

$ \begin{aligned} \text { track } & =500 \\ \therefore 500 & =2 x+2 \pi r \\ 250 & =x+\pi r \end{aligned} $

So, area of rectangle point $f(r)$

$ =2 r(250-\pi r) $

Area to be maximum $f^{\prime}(r)=0$

$ \begin{aligned} & \Rightarrow 500-4 \pi r \\ & \Rightarrow r=\frac{125}{\pi} \end{aligned} $

So, length of rectangle portion is 125 .

Given that $x$ is real and $\alpha, \beta$ are the maximum and minimum values of $\frac{x^2-x+1}{x^2+x+1}$, respectively.

$Let\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, y=\frac{x^2-x+1}{x^2+x+1} $

$ \Rightarrow y x^2+y x+y=x^2-x+1 $

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

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(x \in R)$

$ \begin{aligned} & \text { Consider } b^2-4 a c \geq 0 \\ & \Rightarrow \quad(y+1)^2-4(y-1)^2 \geq 0 \\ & \Rightarrow(y+1+2 y-2)(y+1-2 y+2) \geq 0 \\ & \Rightarrow \quad(3 y-1)(y-3) \leq 0 \end{aligned} $

$ \begin{aligned} &\text { Range }=\left[\frac{1}{3}, 3\right]\\ &\begin{aligned} & \because \text { Minimum value }=\frac{1}{3}=\beta \\ & \text { Maximum value }=3=\alpha \\ & \alpha+\beta=3+\frac{1}{3}=\frac{10}{3} \end{aligned} \end{aligned} $

To find the value of $ c $ such that the line joining the points $(0, 3)$ and $(5, -2)$ is tangent to the curve $ y = \frac{c}{x+1} $, follow these steps:

Equation of the Curve:

The given equation of the curve is:

$ y = \frac{c}{x+1} $

Differentiate the Curve:

Differentiate the curve equation with respect to $ x $:

$ \frac{dy}{dx} = \frac{-c}{(x+1)^2} $

This derivative gives the slope of the tangent to the curve.

Find Slope of the Tangent Line:

Calculate the slope of the line joining the points $(0, 3)$ and $(5, -2)$:

$ \text{Slope} = \frac{-2 - 3}{5 - 0} = -1 $

The equation of the line, therefore, is:

$ y = -x + 3 $

Set the Slopes Equal:

Since the line is tangent to the curve, set the slopes equal:

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

Simplifying gives:

$ c = (x+1)^2 $

Equation of the Curve at Tangency Point:

For the tangency condition, equate the line and the curve:

$ -x + 3 = \frac{c}{x+1} $

Substitute $ c = (x+1)^2 $ into this equation:

$ -x + 3 = x + 1 $

Solve for $ x $:

$ 2x = 2 $

$ x = 1 $

Calculate $ c $:

Substitute $ x = 1 $ back into $ c = (x+1)^2 $:

$ c = (1+1)^2 = 4 $

Thus, the value of $ c $ is 4.

If the percentage error in the radius of a circle is 3%, we can calculate the percentage error in its area as follows:

The formula for the percentage error in the radius is:

$ \frac{\delta r}{r} \times 100 = 3 $

The area $ A $ of a circle is given by:

$ A = \pi r^2 $

Taking the logarithm of both sides of the area formula gives:

$ \log A = \log(\pi r^2) $

This can be expanded to:

$ \log A = \log \pi + 2 \log r $

Differentiating both sides with respect to the logarithmic variables:

$ \frac{1}{A} \cdot \delta A = 0 + 2 \cdot \frac{1}{r} \delta r $

Thus:

$ \frac{\delta A}{A} = \frac{2 \delta r}{r} $

Multiplying both sides by 100 to find the percentage error in the area:

$ 100 \times \frac{\delta A}{A} = 2 \times \frac{\delta r}{r} \times 100 $

Substituting the known percentage error of the radius from Equation (i):

$ 100 \times \frac{\delta A}{A} = 2 \times 3 $

Simplifying gives:

$ 100 \times \frac{\delta A}{A} = 6 $

Therefore, the percentage error in the area is 6%.