Application of Derivatives

We have,

radius of sphere $=7 \mathrm{~cm}$

Error in area $=0.08 \mathrm{~cm}^2 \quad \text { [given] }$

i.e. $\Delta A=0.08 \mathrm{~cm}^2$

We need to find

$ \begin{gathered} \Delta V=\frac{d V}{d A} \cdot \Delta A \quad \ldots \text { (1) } \\\\ \because A=4 \pi r^2 \Rightarrow \frac{d A}{d r}=8 \pi r \end{gathered} $

and $V=\frac{4}{3} \pi r^3 \Rightarrow \frac{d V}{d r}=4 \pi r^2$

So, $\frac{d V}{d A}=\frac{4 \pi r^2}{8 \pi r}=\left.\frac{r}{2} \Rightarrow \frac{d V}{d A}\right|_{r=7}=\frac{7}{2}$

Hence, $\Delta V=\frac{7}{2} \times 0.08 \quad \ldots \text { [from Eq. (i)] } $

$ =0.28 \mathrm{~cm}^3 $

Approximate error in its volume $=0.28 \mathrm{~cm}^3$.

We have, curve $y=x^3-2 x^2+3 x-4$ intersects the line

$ \begin{aligned} & y=-2 \text { at } P(h, k) \\\\ & \Rightarrow \quad k=-2 \text { and }-2=h^3-2 h^2+3 h-4 \\\\ & \Rightarrow \quad h^3-2 h^2+3 h-2=0 \\\\ & (h-1)\left(h^2-h+2\right)=0 \Rightarrow h=1 \end{aligned} $

So, point $P \equiv(1,-2)$

$ \left.\frac{d y}{d x}\right|_{\text {at } P}=3 x^2-4 x+3=2 $

Now, equation of tangent at $P(1,-2)$ is

$ \begin{aligned} & y+2=2(x-1) \\\\ \Rightarrow \quad & 2 x-y-4=0 \quad \ldots \text { (i) } \end{aligned} $

$\because$ Tangent at $P$ meets $X$-axis at $\left(x_1, y_1\right)$

$ \Rightarrow \quad y_1=0 $

From Eq. (i), we get

$ 2 x_1-0-4=0 \Rightarrow x_1=\frac{4}{2}=2 $

$ \begin{aligned} &\text { We have, }\\\\ &\begin{aligned} & f(x)=(2 x-1)(3 x+2)(4 x-3) \\\\ & \begin{aligned} & f^{\prime}(x)=2(3 x+2)(4 x-3) \\\\ &+3(2 x-1)(4 x-3) \\\\ &+4(2 x-1)(3 x+2) \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & =2\left[12 x^2-x-6\right]+3\left[8 x^2-10 x+3\right] \\\\ & \quad+4\left[6 x^2+x-2\right] \\\\ & =24 x^2-2 x-12+24 x^2-30 x+9 \\\\ & + \\\\ & +24 x^2+4 x-8 \end{aligned} $

$ \begin{aligned} &f^{\prime}(x)=72 x^2-28 x-11\\\\ &\text { For Rolle's theorem, } c \text { is such that }\\\\ &\begin{aligned} & f^{\prime}(c)=0, \text { where } c \in\left[\frac{1}{2}, \frac{3}{4}\right] \\\\ & \Rightarrow \quad 72 c^2-28 c-11=0 \\\\ & \quad c=\frac{28 \pm \sqrt{(28)^2+4 \times 11 \times 72}}{2 \times 72} \\\\ & \because \quad c=\frac{7 \pm \sqrt{247}}{36} \\\\ & \because \quad c \in\left[\frac{1}{2}, \frac{3}{4}\right] \Rightarrow c=\frac{7+\sqrt{247}}{36} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\\\ &\begin{aligned} & f(x)=\log \left(\frac{1+x}{1-x}\right)-2 x-\frac{x^3}{1-x^2} \\\\ & f^{\prime}(x)=\left(\frac{1-x}{1+x}\right)\left[\frac{1-x-(1+x)(-1)}{(1-x)^2}\right] \\\\ & -2-\left[\frac{\left(1-x^2\right) \cdot 3 x^2-x^3(-2 x)}{\left(1-x^2\right)^2}\right] \\\\ & =\left(\frac{1-x}{1+x}\right)\left[\frac{1-x+1+x}{(1-x)^2}\right] \\\\ & -2-\left[\frac{3 x^2-3 x^4+2 x^4}{\left(1-x^2\right)^2}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{2}{1-x^2}-2-\frac{3 x^2-x^4}{\left(1-x^2\right)^2} \\\\ & =\frac{2\left(1-x^2\right)-2\left(1-x^2\right)^2-3 x^2+x^4}{\left(1-x^2\right)^2} \\\\ & =\frac{2-2 x^2-2-2 x^4+4 x^2-3 x^2+x^4}{\left(1-x^2\right)^2} \\\\ & =\frac{-x^4-x^2}{\left(1-x^2\right)^2}=\frac{-x^2\left(x^2+1\right)}{(1-x)^2(1+x)^2} \end{aligned}\\\\ &\because f(x) \text { is decreasing }\\\\ &\begin{aligned} & \therefore f^{\prime}(x) \leq 0 \Rightarrow \frac{-x^2\left(x^2+1\right)}{(1-x)^2(1+x)^2} \leq 0 \\\\ & \Rightarrow \frac{x^2\left(x^2+1\right)}{(1-x)^2(1+x)^2} \geq 0 \\\\ & x \in R-\{-1,1\} \end{aligned} \end{aligned} $

We have slope of tangent

$ =6 x^2+10 x-9 $

$ \begin{aligned} & \Rightarrow \frac{d y}{d x}=6 x^2+10 x-9 \\\\ & d y=\left(6 x^2+10 x-9\right) d x \end{aligned} $

On integrating both sides, we get

$ \begin{aligned} & \int d y=\int\left(6 x^2+10 x-9\right) d x \\\\ & y=2 x^3+5 x^2-9 x+C \\\\ & \because f(2)=0 \\\\ & \Rightarrow 0=16+20-18+C \Rightarrow C=-18 \\\\ & \text { So, } y=2 x^3+5 x^2-9 x-18=f(x) \\\\ & \therefore f(-2)=2(-2)^3+5(-2)^2-9(-2)-18 \\\\ & =-16+20+18-18=4 \end{aligned} $

$ \begin{aligned} & \text { } x^2+y^2=(13)^2 \\ & \Rightarrow 2 x \cdot \frac{d x}{d t}+2 y \cdot \frac{d y}{d t}=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow 5 \times 2+12 \times \frac{d y}{d t}=0 \\ & \quad\left[\text { at } x=5, y=\sqrt{13^2-5^2}=12\right] \\ & \Rightarrow \quad \frac{d y}{d t}=\frac{-5}{6} \end{aligned}\\ &\text { The speed by which upper end falls is } \frac{5}{6} \text {. } \end{aligned} $

We have, $x^2-y^2=4$ and $x^2+y^2=4 \sqrt{2}$

on solving, we get $x^2=2(\sqrt{2}+1)$,

$ \begin{aligned} y^2 & =2(\sqrt{2}-1) \\ m_1 & =\frac{x}{y}, m_2=-\frac{x}{y}\left\{x^2 y^2=4, x y=2\right. \\ \tan \theta & =\left|\frac{\frac{2 x}{y}}{1-\frac{x^2}{y^2}}\right|=\left|\frac{2 x y}{y^2-x^2}\right|=\left|\frac{2 x y}{-4}\right| \\ & =\left|\frac{4}{-4}\right|=1 \\ \theta & =\pi / 4 \end{aligned} $

$ \begin{aligned} & \text { }\left(\frac{h}{2}\right)^2+r^2=(12)^2 \\ & r=\sqrt{144-\frac{h^2}{4}} \end{aligned} $

Now, volume of cylinder

$ \begin{aligned} V & =\pi r^2 h \\ V & =\pi\left(144-\frac{h^2}{4}\right) \cdot h \\ \frac{d V}{d h} & =\pi\left(144-\frac{3 h^2}{4}\right)=0 \\ \Rightarrow \quad h & =8 \sqrt{3} \\ \frac{d^2 V}{d h^2} & =\pi\left(\frac{-6 h}{4}\right)<0 \end{aligned} $

So, maximum volume at $h=8 \sqrt{3}$ corresponding $r=\sqrt{144-48}=4 \sqrt{6}$

$ V=\pi \times(4 \sqrt{6})^2 \times 8 \sqrt{3}=768 \sqrt{3} \pi $

$ \begin{gathered}\therefore y=x^4-6 x^3+13 x^2-12 x+5 \\ \frac{d y}{d x}=4 x^3-18 x^2+26 x-12 \\ \because \quad 4 x^3-18 x^2+26 x-12=2 \\ \Rightarrow \quad 4 x^3-18 x^2+26 x-14=0 \end{gathered} $

No integer value of $x$ is possible in above equation but in options all are integers.

Since, the point $(1,2)$ lies on the curve $x=t^2-7 t+7, y=t^2-4 t-10$.

$ \begin{aligned}So,\,\, & t^2-7 t+7=1 \\ & t^2-4 t-10=2 \end{aligned} $

On solving Eqs. (i) and (ii), we get $t=6$

$ \begin{aligned} & \begin{array}{rlrl} x & = t^2-7 t+7 \\ \Rightarrow \quad & d x / d t =2 t-7 \\ y & =t^2-4 t-10 \\ \Rightarrow \quad \frac{d y}{d t} & =2 t-4 \\ \text { Now, }\left(\frac{d y}{d x}\right)_{t=6} & =\left(\frac{d y / d t}{d x / d t}\right)_{t=6} \\ & =\left.\frac{2 t-4}{2 t-7}\right|_{t=6}=\frac{2(6)-4}{2(6)-7}=\frac{8}{5} \\ \therefore & m \times \frac{8}{5} =-1 \Rightarrow m=-\frac{5}{8} \end{array} \end{aligned} $

$ \begin{aligned} &\text { Thus, the equation of line } L \text { be }\\ &\begin{aligned} y-2 & =-\frac{5}{8}(x-1) \\ \Rightarrow \quad 5 x+8 y-21 & =0 \\ \therefore a=5, b=8 \text { and } c & =-21 \\ \text { Now, } m(a+b+c) & =-\frac{5}{8}(5+8-21) \\ & =-\frac{5}{8}(-8)=5 \end{aligned} \end{aligned} $

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

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

For maxima or minima,

$ \begin{aligned} & & f^{\prime}(x) & =0 \\ & \Rightarrow & e^{-x}(1-x) & =0 \\ & & x & =1 \quad\left[\text { since } e^{-x} \neq 0\right] \\ & & f^{\prime}(x) & =(x-2) e^{-x} \\\Rightarrow & & f^{\prime}(1) & <0 \end{aligned} $

So, $f(x)$ ha zs a maximum value at $x=1$

Then, $x=\alpha=1$ and $\beta=\frac{1}{e}$

Hence, $(\alpha, \beta)=\left(1, \frac{1}{e}\right)$

Given, diameter of the sphere

$ =42 \mathrm{~cm} $

Then, radius $r=21 \mathrm{~cm}$

and error in measuring the diameter $=\frac{1}{77}$

$\therefore $ Error in measuring the radius,

$ \Delta r=\frac{1}{154} \mathrm{~cm} $

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

$ V^{\prime}=\frac{4}{3} \times 3 \pi r^2=4 \pi r^2 $

Let $\Delta V$ be error in the volume of sphere.

Now, $V+\Delta V=V+V^{\prime} \Delta r$

$ V-\Delta V=V-V \Delta r $

On subtracting Eq. (ii) from Eq. (i), we get

$ \begin{aligned} 2 \Delta V & =2 V^{\prime} \Delta r \\ \Delta V & =4 \pi r^2 \Delta r=4 \times \pi(21)^2 \times \frac{1}{154} \\ & =\frac{4 \times 22}{7} \times 21 \times 21 \times \frac{1}{154} \\ \Delta V & =36 \mathrm{~cm}^3 \end{aligned} $

Given, $x^2 y-x^3=8$ and $y^3-x y^2=32$

From figure, point of intersection is Y2, 4).

Now, angle between the two curves is angle between their tangents at that point

$ \begin{aligned} & \quad x^2 y-x^3=8 \Rightarrow 2 x y+x^2 \frac{d y}{d x}-3 x^2=0 \\ & \text { At }(2,4) \\ & 2 \times 2 \times 4+2^2 \frac{d y}{d x}-3 \times 2^2=0 \\ & \Rightarrow \quad 16+4 \frac{d y}{d x}-12=0 \\ & \Rightarrow \quad 4+\frac{4 d y}{d x}=0 \\ & \Rightarrow \quad \frac{d y}{d x}=-1=m_1 \text { (let) } \end{aligned} $

and

$ \begin{gathered} y^3-x y^2=32 \\ 3 y^2 \frac{d y}{d x}-x \times 2 y \frac{d y}{d x}-y^2=0 \end{gathered} $

At $(2,4)$

$ 3 \times(4)^2 \frac{d y}{d x}-2 \times 2 \times 4 \frac{d y}{d x}-(4)^2=0 $

$ \begin{array}{ll} \Rightarrow & \frac{3 d y}{d x}-\frac{d y}{d x}-1=0 \\ \Rightarrow & 2 \frac{d y}{d x}-1=0 \\ \frac{d y}{d x}=\frac{1}{2}=m_2 \text { (let) } \end{array} $

$ \text { Now, } \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\frac{\frac{1}{2}+1}{1-\frac{1}{2}}=\frac{3 / 2}{1 / 2}=3 $

Given, $f(x)=x^3-2 x^2+x-3$

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

For absolute maxima or minima put $f^{\prime}(x)=0$

$ \begin{array}{lrl} \because & 3 x^2-4 x+1 & =0 \\ \Rightarrow & 3 x^2-3 x-x+1 & =0 \\ & & (3 x-1)(x-1) =0 \\ \Rightarrow & x=\frac{1}{3} \text { or } x & =1 \end{array} $

Clearly, the critical point of $f(x)$ in $[0,2]$ is $x=\frac{1}{3}$ and 1 .

Now, $f(0)=-3, f\left(\frac{1}{3}\right)=-\frac{77}{27} \approx-2.851$

$ f(1)=-3, f(2)=-1 $

∴ Absolute minimum value i.e. $m=-3$ and absolute maximum value i.e. $M=-1$

Now, $M+m=-1-3=-4$

According to the question,

$ d y / d x=3 x^2-5 $

On integrating both sides, we get

$ y=\frac{3 x^3}{3}-5 x+C \Rightarrow y=x^3-5 x+C $

At $(1,2)$,

$ \begin{aligned} & & 2 & =(1)^3-5 \times 1+C \Rightarrow 2=1-5+C \\ & \Rightarrow & 2+4 & =C \Rightarrow C=6 \\ & \therefore & y & =x^3-5 x+6 \end{aligned} $

We know that, intersection satisfied the equation.

Now, from options $(-2,8)$ satisfied the curve $y=x^3-5 x+6$

                                       Let $y=\sqrt{x}$

$ \begin{aligned} \begin{aligned} \frac{d y}{d x} & =\frac{1}{2 \sqrt{x}} \Rightarrow x=1936 \\ \Delta x & =87 \\ \text { Then, } \Delta y & =\sqrt{x+\Delta x}-\sqrt{x} \\ \Delta y & =\sqrt{2023}-\sqrt{1936} \\ \Rightarrow \quad \sqrt{2023}=44+\Delta y & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} $

Approximate value of $\sqrt{2023}$ is

$ \begin{aligned} &\begin{aligned} \Delta y & =\frac{d y}{d x} \cdot \Delta x=\frac{1}{2 \sqrt{x}} \times \Delta x \\ & =\frac{1}{2 \sqrt{1936}} \times 87 \\ \Delta y & =\frac{87}{88} \approx 0.9778 \end{aligned}\\ &\text { From Eq. (i), we get }\\ &\sqrt{2023}=44+0.9778=44.9778 \end{aligned} $

$ \begin{aligned} &\text { Given, curve }\\ &\begin{aligned} & \qquad \begin{array}{c} y=x^3-10 x^2+31 x-30 \\ \frac{d y}{d x}=3 x^2-20 x+31 \end{array} \\ & \text { Slope of normal }=\frac{1}{-\frac{d y}{d x}}=-\frac{1}{14} \end{aligned} \end{aligned} $

$ \begin{array}{lr} \Rightarrow & \frac{d y}{d x}=14 \\ \Rightarrow & 3 x^2-20 x+31=14 \\ \Rightarrow & 3 x^2-20 x+17=0 \\ \Rightarrow & 3 x^2-3 x-17 x+17=0 \\ \Rightarrow & (x-1)(3 x-17)=0 \\ \Rightarrow & x=1, x=17 / 3 \end{array} $

Coordinates are integers

$ \begin{array}{ll} \therefore & x=1 \\ \Rightarrow & y=1-10+31-30=-8 \\ P(1,-8) & \end{array} $

Slope of tangent at $P=\left(\frac{d y}{d x}\right)_{(1,-8)}$

$ =3-20+31=14 $

Equation of tangent

$ \begin{aligned} & \begin{aligned} y+8 & =14(x-1) \\ 14 x-y-22 & =0 \Rightarrow \frac{x}{11 / 7}+\frac{y}{22}=1 \end{aligned} \\ & \text { intercept }=11 / 7 \end{aligned} $

Let $z=x^2 y^3$ and $x=\frac{50-3 y}{2}$

$ \begin{aligned} \therefore z & =\left(\frac{50-3 y}{2}\right)^2 y^3 \\ z & =\frac{2500 y^3+9 y^5-300 y^4}{4} \end{aligned} $

Now, $d z / d y=0$

$ \begin{array}{lr} \Rightarrow & \left(7500+45 y^2-1200 y\right) y^2=0 \\ \because & y>0 \\ \Rightarrow & 3 y^2-80 y+500=0 \\ \Rightarrow & (y-10)(3 y-50)=0 \end{array} $

$y$ is are integer.

$ \begin{array}{ll} \Rightarrow & y=10 \\ \therefore & x=10 \end{array} $

$x^2 y^3$ is maximum at $x=\alpha=10$

and   $ y=\beta=10 $

$ \therefore \quad \frac{\alpha}{2}+\frac{\beta}{5}=\frac{10}{2}+\frac{10}{5}=7 $

Given the function $ f(x) = \frac{1-x+x^2}{1+x+x^2} $, we want to find its minimum value for all real values of $ x $.

First, differentiate $ f(x) $ with respect to $ x $:

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

Simplify the expression:

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

To find the critical points, set $ f^{\prime}(x) = 0 $:

$ 2x^2 - 2 = 0 \\ x^2 = 1 \\ x = \pm 1 $

Evaluate $ f(x) $ at $ x = 1 $ and $ x = -1 $:

At $ x = 1 $:

$ f(1) = \frac{1 - 1 + 1^2}{1 + 1 + 1^2} = \frac{1}{3} $

At $ x = -1 $:

$ f(-1) = \frac{1 - (-1) + 1^2}{1 - 1 + 1^2} = 3 $

Hence, the minimum value of $ f(x) $ is $ \frac{1}{3} $.

Approximate error in current

$ d I=f^{\prime}(I) \times \Delta I=\frac{\pi}{4} \times 0.01 $

$\therefore$ Percentage error in current

$ =\left(\frac{\pi}{4} \times 0.01\right) \times 100=\frac{\pi}{4} $

Given that, $y=\cos (x+y)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Differentiating on both side w.r.t. $x$, we get

$ \frac{d y}{d x}=-\sin (x+y)\left(1+\frac{d y}{d x}\right) $

Equation of tangent is $x+2 y=k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$\Rightarrow$ Slope is $-\frac{1}{2} \Rightarrow \frac{d y}{d x}=-\frac{1}{2}$

Hence, $-\frac{1}{2}=-\sin (x+y)\left(1-\frac{1}{2}\right)$

$\Rightarrow \quad-\frac{1}{2}=-\frac{1}{2} \sin (x+y)$

$\Rightarrow \sin (x+y)=1=\sin \frac{\pi}{2}$

$\Rightarrow \quad x+y=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}-x$

From Eq. (i), we get

$ \begin{aligned} \quad \frac{\pi}{2}-x & =\cos \left(x+\frac{\pi}{2}-x\right) \\ \Rightarrow \frac{\pi}{2}-x & =\cos \left(\frac{\pi}{2}\right)=0 \\ \Rightarrow \quad x & =\frac{\pi}{2} \Rightarrow y=0 \end{aligned} $

Now, from Eq. (ii), we get

$ k=\frac{\pi}{2}+0 \cdot 2=\frac{\pi}{2} $

Given that,

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

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

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

Now, $f^{\prime}(x)=0 \Rightarrow \frac{\left(e^{2 x}+2\right) e^x-2 e^{3 x}}{\left(e^{2 x}+2\right)^2}=0$

$ \begin{aligned} & \Rightarrow \quad 2 e^x-e^{3 x}=0 \\ & \Rightarrow \quad e^x\left(e^{2 x}-2\right)=0 \end{aligned} $

Since, $e^x \neq 0 e^{2 x}=2 \Rightarrow e^x=\sqrt{2}$

For maximum

$ f(x)=\frac{\sqrt{2}}{2+2}=\frac{\sqrt{2}}{4}=\frac{1}{2 \sqrt{2}} $

Thus, $0

$\Rightarrow C \in R$ such that, $f(c)=\frac{1}{3}$

So, Assertion (A) is true.

and Reason ( $R$ ) is true.

and Reason (R) is the correct explanation of Assertion (A).

Given the expression $ f(x) = x^3 + 3x^2 - 9x + \lambda $, we need to find the values of $\lambda$ such that $ f(x) $ can be expressed as $(x-\alpha)^2(x-\beta)$.

Since $(x-\alpha)$ is a repeated factor, it must also be a factor of the derivative $ f'(x) $. We compute the derivative:

$ f'(x) = 3x^2 + 6x - 9 $

Factor the derivative:

$ f'(x) = 3(x^2 + 2x - 3) = 3(x+3)(x-1) $

This shows that the potential values of $\alpha$ are the roots of $ f'(x) = 0$, which are $\alpha = 1$ and $\alpha = -3$.

Case 1: $\alpha = 1$

Substitute $\alpha = 1$ into the original expression to find $\lambda$:

$ f(1) = 1^3 + 3(1)^2 - 9(1) + \lambda = 0 $

$ 1 + 3 - 9 + \lambda = 0 $

$ \lambda = 5 $

Case 2: $\alpha = -3$

Substitute $\alpha = -3$ into the original expression to find $\lambda$:

$ f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + \lambda = 0 $

$ -27 + 27 + 27 + \lambda = 0 $

$ \lambda = -27 $

Hence, the values of $\lambda$ can be $5$ or $-27$.

The given parametric equations for the curve are $x = 2 \sin t$ and $y = 2 \cos t$.

To find the equation of the normal line at $t = \frac{\pi}{2}$, we first need to determine the slope of the tangent line. Using the derivatives:

$ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{-2 \sin t}{2 \cos t} $

This expression represents the slope of the tangent line to the curve.

Next, the slope of the normal line, which is perpendicular to the tangent line, is given by:

$ \text{Slope of normal} = -\frac{1}{\frac{-2 \sin t}{2 \cos t}} = \frac{\cos t}{\sin t} $

Substituting $t = \frac{\pi}{2}$:

$ m_s = \frac{\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}} = \frac{0}{1} = 0 $

Thus, the slope of the normal line at $t = \frac{\pi}{2}$ is 0, which indicates a horizontal line.

The point on the curve at $t = \frac{\pi}{2}$ is:

$ x = 2 \sin \frac{\pi}{2} = 2, \quad y = 2 \cos \frac{\pi}{2} = 0 $

The equation of a horizontal line through this point, where the slope ($m = 0$), is:

$ (y - 2 \cos t) = 0(x - 2 \sin t) $

Simplifying this at $t = \frac{\pi}{2}$:

$ \left(y - 2 \times \cos \frac{\pi}{2}\right) = 0 \quad \Rightarrow \quad y = 0 $

Thus, the equation of the normal line at $t = \frac{\pi}{2}$ is $y = 0$.

To find the relative maximum of the function $ f(x) = \frac{x}{5} + \frac{5}{x} $, where $ x \neq 0 $, follow these steps:

Derive the Function:

Compute the first derivative:

$ f'(x) = \frac{1}{5} - \frac{5}{x^2} $

Set the derivative equal to zero for critical points:

$ \frac{1}{5} - \frac{5}{x^2} = 0 \quad \Rightarrow \quad \frac{5}{x^2} = \frac{1}{5} $

Solve for $ x $:

$ x^2 = 25 \quad \Rightarrow \quad x = \pm 5 $

Second Derivative Test:

Compute the second derivative:

$ f''(x) = \frac{10}{x^3} $

Evaluate the second derivative at $ x = 5 $:

$ f''(5) = \frac{10}{125} = \frac{2}{25} > 0 $

Since $ f''(5) > 0 $, $ x = 5 $ is a point of relative minimum.

Evaluate the second derivative at $ x = -5 $:

$ f''(-5) = \frac{-10}{125} = -\frac{2}{25} < 0 $

Since $ f''(-5) < 0 $, $ x = -5 $ is a point of relative maximum.

Calculate the Expression:

Given that $ \alpha = -5 $, calculate:

$ \sqrt{\alpha^2 + 2\alpha - 6} = \sqrt{(-5)^2 + 2(-5) - 6} $

Simplify:

$ = \sqrt{25 - 10 - 6} = \sqrt{9} = 3 $

Therefore, $\sqrt{\alpha^2 + 2\alpha - 6} = 3$.

$x^2+y-7=4 x$

$ y=-x^2+4 x+7 $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & y^{\prime}=-2 x+4 \\ & m=-2 x+4 \end{aligned} $

At $(1,10)$

$ m=-2 \times 1+4=2 $

Equation of tangent,

$ \begin{aligned} & & y-10 & =2(x-1) \\ \Rightarrow & & y & =2 x+8 \end{aligned} $

Curve I $x^2-y^2=4$

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & 2 x-2 y y^{\prime}=0 \Rightarrow x=y y^{\prime} \\ \Rightarrow & y^{\prime}=\frac{x}{y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \Rightarrow & m_1=\frac{x}{y} \end{aligned} $

Curve II $y^2=3 x$

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & & 2 y y^{\prime} & =3 \times 1 \\ \Rightarrow & & y^{\prime} & =\frac{3}{2 y} \\ \Rightarrow & & m_2 & =\frac{3}{2 y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

To get intersection point,

$ x^2-4=y^2, y^2=3 x $

On putting $x^2-4=3 x$,

$ x^2-3 x-4=0 $

$ \begin{aligned} &\begin{aligned} \Rightarrow & & x^2-4 x+x-4 & =0 \\ \Rightarrow & & x(x-4)+1(x-4) & =0 \\ \Rightarrow & & (x-4)(x+1) & =0 \\ & & x & =4,-1 \end{aligned}\\ &\text { From the graph it is clear that } x=4 \end{aligned} $

$ \therefore \quad y^2=3 \times 4 \Rightarrow y= \pm 2 \sqrt{3} $

At point $(4,2 \sqrt{3})$ from Eqs. (i) and (ii), we get

$ \begin{aligned} m_1 & =\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}} \\ m_2 & =\frac{3}{2(2 \sqrt{3})}=\frac{\sqrt{3}}{4} \\ \tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \tan \theta & =\left|\frac{\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{4}}{1+\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{4}}\right| \\ & =\left|\frac{\frac{8-3}{4 \sqrt{3}}}{\frac{4 \sqrt{3}+2 \sqrt{3}}{4 \sqrt{3}}}\right|=\frac{5}{6 \sqrt{3}} \end{aligned} $

$f(x)=2 x^3-3 x^2-36 x+9$

$ \begin{aligned} & x \in[-3,3] \\ & f(x)=2 x^3-3 x^2-36 x+9 \end{aligned} $

On differentiating w.r.t. $x$, we get

$ f^{\prime}(x)=6 x^2-6 x-36 $

On putting $f^{\prime}(x)=0$ for critical numbers.

$ \begin{aligned} & \Rightarrow \quad 6 x^2-6 x-36=0 \\ & \Rightarrow \quad x^2-x-6=0 \\ & \Rightarrow \quad x^2-3 x+2 x-6=0 \\ & \Rightarrow \quad x(x-3)+2(x-3)=0 \\ & \Rightarrow \quad(x-3)(x+2)=0 \\ & \quad x=3,-2 \\ & f(3)=2 \times 27-3 \times 9-36 \times 3+9 \\ & =54-27-108+9=-72 \,\,\,\,\text { (min.) }\\ & f(-2)=2(-8)-3(4)-36(-2)+9 \\ & =-16-12+72+9=53 \,\,\,\text { (max.) }\\ & f(-3)=2(-27)-3(9)-36(-3)+9 \\ & =-54-27+108+9=36 \end{aligned} $

The absolute maximum value is 53 .

To find approximate value of

$ (28)^{1 / 3}=(27+1)^{1 / 3} $

Let $x=27, \Delta x=1, y=x^{1 / 3}$

$ \frac{d y}{d x}=\frac{1}{3} x^{-2 / 3}=\frac{1}{3}(27)^{-2 / 3}=\frac{1}{3} 3^{-2}=\frac{1}{27} $

Approximate value of $3 \sqrt{28}$

$ \begin{aligned} & =y+\frac{d y}{d x} \cdot \Delta x=27^{\frac{1}{3}}+\frac{1}{27} \times 1 \\ & =3+0.037=3.037 \end{aligned} $

Given, curve is $y=x^2$

If the curve is dragged along the positive $X$-axis to a distance of ' $a$ ' units, then new curve will be $y=(x-a)^2$

So, point of intersection

$ \begin{aligned} x^2 & =(x-a)^2 \\ \Rightarrow \quad x-a & = \pm x \Rightarrow x=a / 2 \end{aligned} $

At $x=\frac{a}{2}, y=\frac{a^2}{4}$

Point of intersection $\left(\frac{a}{2}, \frac{a^2}{4}\right)$

Curve $1 y=x^2$

$ \left(\frac{d y}{d x}\right)_1=2 x=2 \times \frac{a}{2}=a $

Curve II $y=(x-a)^2$

$ \begin{aligned} \left(\frac{d y}{d x}\right)_2 & =2(x-a)=2\left(\frac{a}{2}-a\right) \\ & =2 \times-\frac{a}{2}=-a \end{aligned} $

$ \begin{aligned} &\text { Angle between two curve is } \theta \text {. }\\ &\begin{aligned} \therefore \quad \tan \theta & =\left|\frac{\left(\frac{d y}{d x}\right)_1-\left(\frac{d y}{d x}\right)_2}{1+\left(\frac{d y}{d x}\right)_1\left(\frac{d y}{d x}\right)_2}\right| \\ & =\left|\frac{a-(-a)}{1+(a)(-a)}\right|=\frac{2|a|}{\left|1-a^2\right|} \end{aligned} \end{aligned} $

Here, $x$ and $y$ are two positive integers, then

$ \begin{array}{rlrl} & & x+2 y & =10 \\ \Rightarrow & y & =\frac{10-x}{2} \end{array} $

Let $P=x^2 y^3$

$ \begin{aligned} \Rightarrow P & =x^2\left(\frac{10-x}{2}\right)^3=\frac{1}{8} x^2(10-x)^3 \\ \Rightarrow \frac{d P}{d x} & =\frac{1}{8}\left\{2 x(10-x)^3-3 x^2(10-x)^2\right\} \\ & =\frac{x(10-x)^2}{8}(2(10-x)-3 x) \end{aligned} $

$ \frac{d P}{d x}=\frac{x(10-x)^2(20-5 x)}{8} $

For $P$ to be maximum, $\frac{d P}{d x}=0$

$ \begin{aligned} & \Rightarrow \quad \frac{x(10-x)^2(20-5 x)}{8}=0 \\ & \Rightarrow x=4, x \neq 0,10 \\ & \begin{aligned} \frac{d^2 P}{d x^2}= & \frac{1}{8}\left[(10-x)^2(20-5 x)+x[ \right. \\ & \left.-2(10-x)(20-5 x)-(10-x)^2 \cdot 5\right] \end{aligned} \\ & \text { At } x=4, \frac{d^2 P}{d x^2}<0 \end{aligned} $

So, $x=4$ is point of maximum

$ y=\frac{10-x}{2}=\frac{10-4}{2}=3 $

So, $x^2+2 y^3=16+2 \times 27=70$

Given, equation of curve

$ \sin y=\sqrt{3} x \sin \left(\frac{\pi}{6}+y\right) $

When $y=0$,

$ \begin{array}{rlrl} \sin 0 & =\sqrt{3} x \sin \left(\frac{\pi}{6}+0\right) \\ 0 & =\sqrt{3} x \times \frac{\sqrt{3}}{2} \\ \Rightarrow \quad x & =0 \end{array} $

Now, differentiate equation of curve w.r.t. $y$,

we get, $\cos y=\sqrt{3} x \cos \left(\frac{\pi}{6}+y\right)$

$ \begin{array}{r} +\sqrt{3} \sin \left(\frac{\pi}{6}+y\right) \frac{d x}{d y} \\ \Rightarrow \quad-\frac{d x}{d y}=\frac{\sqrt{3} x \cos \left(\frac{\pi}{6}+y\right)-\cos y}{\sqrt{3} \sin \left(\frac{\pi}{6}+y\right)} \end{array} $

Slope of normal at $(0,0)$

$ =-\frac{\cos 0}{\sqrt{3} \sin \frac{\pi}{6}}=-\frac{2}{\sqrt{3}} $

Equation of normal is

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

Assertion : Given, two curve $y^2=4 x$ and $x^2=-2 y$. Solving the two equation to determine the intersecting points $=\left(-\frac{x^2}{2}\right)^2=4 x$

$ \Rightarrow \quad x^4-16 x=0 $

Either $x=0$

or $\quad x^3=16$

or $\quad x=2(2)^{1 / 3}$

If $x=0, y=0$

If $x=2(2)^{\frac{1}{3}}$, then

$ \begin{aligned} y & =-\frac{x^2}{2} \\ & =-\frac{4(2)^{2 / 3}}{2}=-2(2)^{2 / 3} \end{aligned} $

∴ The two curve does not intersect at the point

$(1,2)$.

Reason : The curve are said to intersect orthogonally if the product of the slopes of the tangents drawn to two curves at their point of intersection is -1 .

Hence, Assertion is false but Reason is true.

$ \begin{aligned} & f(x)=\left\{\begin{array}{cl} 1+6 x-3 x^2, & x \leq 1 \\ x+\log _2\left(b^2+7\right), & x>1 \end{array}\right. \\ & f^{\prime}(x)=\left\{\begin{array}{cl} 6-6 x, & x \leq 1 \\ 1, & x>1 \end{array}\right. \end{aligned} $

$f(1)$ is the maximum value of $f(x)$,

$ \begin{aligned} & & f^{\prime}(x) & \geq 0 \\ \Rightarrow & & 6-6 x & \geq 0 \Rightarrow 6 x \leq 6 \\ \Rightarrow & & x & \leq 1 \\ \therefore& & & x \in[-1,1] \end{aligned} $

$ \begin{aligned} &\text {Given, equation of curves, }\\ &\begin{aligned} & & x^2+y^2=4 \text { and } y^2 & =3 x \\ & & x^2+3 x-4 & =0 \\ &\Rightarrow & x^2+4 x-x-4 & =0 \\ & \Rightarrow & x(x+4)-1(x+4) & =0 \\ & \Rightarrow & (x+4)(x-1) & =0 \\ &\Rightarrow & x & =1\,\,\,\,[x \neq-4] \\ & & y & = \pm \sqrt{3} \end{aligned} \end{aligned} $

The intersecting points are

$ \begin{aligned} & (1, \sqrt{3}) \text { and }(1,-\sqrt{3}) \\ \therefore \quad & x^2+y^2=4 \end{aligned} $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & & 2 x+2 y \frac{d y}{d x} & =0 \\ & \Rightarrow & \frac{d y}{d x} & =\frac{-x}{y} \end{aligned} $

$\begin{aligned} \frac{d y}{d x} \text { at }(1, \sqrt{3}) & =\frac{-1}{\sqrt{3}} \quad\left(\text { let } m_1\right) \\ \frac{d y}{d x} \text { at }(1,-\sqrt{3}) & =\frac{1}{\sqrt{3}} \quad\left(\text { let } m_2\right) \\ y^2=3 x \Rightarrow \frac{d y}{d x} & =\frac{3}{2 y} \\ \frac{d y}{d x} \text { at } x=(1, \sqrt{3}) & =\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \quad\left(\text { let } m_3\right) \\ \frac{d y}{d x} \text { at } x=(1,-\sqrt{3}) & =\frac{-3}{2 \sqrt{3}}=\frac{-\sqrt{3}}{2}\left(\text { let } m_4\right) \\ \tan \theta & =\left|\frac{m_1-m_3}{1+m_1 m_3}\right|\end{aligned}$

$ \begin{aligned} &=\left|\frac{-\frac{1}{2 \sqrt{3}}-\sqrt{3}}{1-\frac{1}{2}}\right|=\left|\frac{-5}{\sqrt{3}}\right|=\frac{5}{\sqrt{3}}\\ &\text { Also, }\\ &\tan \theta=\left|\frac{m_2-m_4}{1+m_2 m_4}\right|=\left|\frac{\frac{1}{2 \sqrt{3}}-\sqrt{3}}{1-\frac{1}{2}}\right|=\frac{5}{\sqrt{3}} \end{aligned} $

$ \text { In } \triangle A B C \text {, } $

$ \begin{aligned} \tan \pi / 3 & =\frac{\sqrt{3}}{H} \Rightarrow \sqrt{3}=\frac{\sqrt{3}}{H} \\ H & =1 \text { unit } \end{aligned} $

In $\triangle A D E$

$ \begin{aligned} \tan \pi / 3 & =r / h \Rightarrow \sqrt{3}=r / h \\ r & =\sqrt{3} h \Rightarrow r=\pi r^2(H-h) \\ r & =\pi 3 h^2(1-h)=3 \pi\left(h^2-h^3\right) \\ \frac{d v}{d h} & =3 \pi\left(2 h-3 h^2\right) \Rightarrow \frac{d v}{d h}=0 \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\, h & =2 / 3 \quad \quad[\because h \neq 0] \\ \frac{d^2 v}{d h^2} & =3 \pi(2-6 h] \\ \frac{d^2 v}{d h^2} \text { at } h & =\frac{2}{3}=3 \pi\left(2-6\left(\frac{2}{3}\right)\right) \\ & =3 \pi[2-4]<0 \end{aligned} $

Volume is maximum at $h=2 / 3$

Height of cylinder $=H-h=1-\frac{2}{3}=\frac{1}{3}$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } A=4 \pi r^2 \\ & \Rightarrow \quad d A=8 \pi r d r \Rightarrow 0.02=(8 \pi)(10)(d r) \\ & \Rightarrow \quad d r=\frac{0.02}{80 \pi} \Rightarrow V=\frac{4}{3} \pi r^3 \\ & \Rightarrow d V=4 \pi r^2 d r=4 \pi \times 100 \times \frac{0.02}{80 \pi} \\ & \quad=0.1 \mathrm{~cm}^3 \end{aligned} \end{aligned} $

$ \begin{aligned} & y^2=4 x, x^2+y^2=5 \\ & \frac{d y}{d x}=\frac{2}{y}, \frac{d y}{d x}=\frac{-x}{y} \end{aligned} $

For point of intersection,

$ \begin{aligned} & x^2+4 x-5=0 \\ & \Rightarrow \quad x=-5 \text { or } x=1 \end{aligned} $

∴ Points of intersection are $(1,2)$, (1, -2)

$\therefore$ Slopes of tangents to two curves at points of intersection are $1, \frac{-1}{2}$ or $-1, \frac{1}{2}$.

$ \therefore \quad|\tan \theta|=\left|\frac{1+1 / 2}{1-1 / 2}\right|=3 $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } f(x)=-(x-2)^3(x+2)^2 \\ & f^{\prime}(x)=-\left[(x-2)^3 \cdot 2(x+2)\right. \left.+(x+2)^2 \cdot 3(x-2)^2\right] \\ & =-(x-2)^2(x+2)[2(x-2)+3(x+2)] \\ & =-(x-2)^2(x+2)[5 x+2] \\ & \text { on putting } f^{\prime}(x)=0 \\ & \Rightarrow \quad x=-2, \frac{-2}{5}, 2 \end{aligned} \end{aligned} $

$f(x)$ has local maxima, at $x=\frac{-2}{5}$

∴ Local maximum value

$ \begin{aligned} & =-\left(-\frac{2}{5}-2\right)^3\left(\frac{-2}{5}+2\right)^2 \\ & =\left(\frac{12}{5}\right)^3 \cdot\left(\frac{8}{5}\right)^2=\frac{(12)^3 8^2}{5^5} \end{aligned} $

$ \begin{aligned} &\text { Given curves, } y^2=4 x\\ &2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y} \end{aligned} $

$ \text { At }(1,2) \frac{d y}{d x}=\frac{2}{2}=1 $

Equation of tangent

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

Slope of normal $=-1$

Equation of normal

$ y-2=-1(x-1) \quad x+y=3 \Rightarrow N(0,3) $

$ \begin{aligned}Area \,\,of \,\, \Delta T P N & =\frac{1}{2}\left|\begin{array}{lll} 0 & 1 & 1 \\ 0 & 3 & 1 \\ 1 & 2 & 1 \end{array}\right| \\ & =\frac{1}{2}|1-3|=1 \end{aligned} $

Given,

$ C_1: y^2=4 a x, C_2: x^2+\frac{y^2}{2}=c^2 $

Let point of intersection of two curves is

$ \begin{aligned} & \left(x_1, y_1\right) . C_1: y^2=4 a x \\ & \Rightarrow \quad 2 y\left(\frac{d y}{d x}\right)_1=4 a \Rightarrow\left(\frac{d y}{d x}\right)_1=\frac{2 a}{y_1} \end{aligned} $

Let angle between two curves be $\theta$.

$ \begin{aligned} & C_2: x^2+\frac{y^2}{2}=c^2 \\ \Rightarrow & 2 x+\frac{2 y}{2}\left(\frac{d y}{d x}\right)_2=0 \\ \Rightarrow & \quad\left(\frac{d y}{d x}\right)_2=-\frac{2 x_1}{y_1} \\ \because & \tan \theta=\left|\frac{\left(\frac{d y}{d x}\right)_1-\left(\frac{d y}{d x}\right)_2}{1+\left(\frac{d y}{d x}\right)_1\left(\frac{d y}{d x}\right)_2}\right| \end{aligned} $

$ \begin{aligned} & =\left|\frac{\frac{2 a}{y_1}+\frac{2 x_1}{y_1}}{1-\frac{4 a x_1}{y_1^2}}\right|=\left|\frac{\frac{2 a+2 x_1}{y_1}}{1-\frac{y_1^2}{y_1^2}}\right| \\ \tan \theta & =\left|\frac{\frac{2 a+2 x_1}{y_1}}{0}\right|=\infty \\ \theta & =\frac{\pi}{2} \end{aligned} $

If $c \in(a, b)$

$ f(c-\delta) < f(c) \text { and } f(c+\delta) < f(c) $

So, $f(c)$ is point of local maxima Also $[f(\alpha-\delta)-f(\alpha)] f(\alpha+\delta)<0$

$\Rightarrow \quad f(\alpha-\delta)-f(\alpha)<0$ and $f(\alpha+\delta)>0$ or $\quad f(\alpha-\delta)-f(\alpha)>0$ and $f(\alpha+\delta)<0$

i.e. No maxima and minima at $\alpha, \alpha \in(a, b)$

Given that, at any instant of time Rate of change in volume w.r.t. time $=$ rate of change in surface area w.r.t. time

i.e.,

$ \begin{equation*} \frac{d v}{d t}=\frac{d s}{d t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Volume of sphere of radius $(r), V=\frac{4}{3} \pi r^3$

Differentiating w.r.t. ' $t$ ', we get

$ \begin{align*} & \frac{d v}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi \frac{d}{d t}\left(r^3\right) \\ & \frac{d v}{d t}=\frac{4}{3} \pi\left(3 r^2\right) \frac{d r}{d t} \text { or } \frac{d v}{d t}=4 \pi r^2 \frac{d r}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

Surface area of sphere of radius $(r)$,

$ S=4 \pi r^2 $

Differentiating w.r.t. ' $t$ ', we get

$ \begin{align*} \frac{d s}{d t} & =\frac{d}{d t}\left(4 \pi r^2\right)=4 \pi \frac{d}{d t}\left(r^2\right) \\ \Rightarrow \quad \frac{d s}{d t} & =8 \pi r \frac{d r}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{align*} $

Putting the values from Eqs. (ii) and (iii) in Eq. (i), we get

$ 4 \pi r^2 \frac{d r}{d t}=8 \pi r \frac{d r}{d t} \text { or } r=2 $

Given that,

Rate of increase in, $\frac{d V}{d t}=4 \pi \mathrm{cc} / \mathrm{sec}$

Volume, $V=288 \pi \mathrm{cc}$

As we know, volume of a sphere of radius $(r)$,

$ \begin{equation*} V=\frac{4}{3} \pi r^3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Differentiating above expression w.r.t. ' $t^{\prime}$ on both sides, we get

$ \begin{align*} & \frac{d V}{d t}=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi 3 r^2 \frac{d r}{d t} \\ & \frac{d V}{d t}=4 \pi r^2 \frac{d r}{d t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{align*} $

From Eq. (i) $(288 \pi)=\frac{4}{3} \pi r^3$

or  $ r=6 \mathrm{~cm} $

Putting this value in Eq. (ii), we get

$ 4 \pi=4 \pi(6)^2 \frac{d r}{d t} \text { or } \frac{d r}{d t}=\frac{1}{36} $

Given function,

$ f(x)=x-\log \left(\frac{1+x}{x}\right), x>0 $

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

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

[from chain rule]

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

Since $x>0$, therefore there is no possibility of $f^{\prime}(x)$ to be zero at any value of $x$.

So, for the given function $f^{\prime}(x) \neq 0$ Also, for $\forall x \in \mathbf{R} f^{\prime}(x)>0$ i.e., function $f(x)$ is strictly increasing its defined domain.

So, reason correctly explains given assertion.

We have,

$ x^5-8 x^4+25 x^3-38 x^2+28 x-8=0 $

Let

$ \begin{aligned} f(x)= & x^5-8 x^4+25 x^3-38 x^2+28 x-8 \\ & f(2)=32-128+200-152+56-8 \\ & f(2)=0 \\ \Rightarrow & f^{\prime}(x)=5 x^4-32 x^3+75 x^2-76 x+28 \\ & f^{\prime}(2)=80-256+300-152+28 \\ \Rightarrow & f^{\prime}(2)=0 \\ & f^{\prime \prime}(x)=20 x^3-96 x^2+150 x-76 \\ & f^{\prime \prime}(2)=160-384+300-76 \\ \Rightarrow & f^{\prime \prime}(2)=0 \\ & f^{\prime \prime \prime}(x)=60 x^2-192 x+150 \\ & f^{\prime \prime \prime}(2)=240-384+150=6 \\ \Rightarrow & f^{\prime \prime \prime}(2) \neq 0 \end{aligned} $

∴ Given, $\alpha$ is a root of multiplicity 3

$ \begin{array}{lrl} \therefore & \alpha & =2 \\ \Rightarrow & \alpha^2-5 \alpha+6 & =4-10+6=0 \end{array} $

Given, $d \theta=9 \mathrm{~min}$

In radian $d \theta=\frac{9}{60} \times \frac{\pi}{180}=\frac{\pi}{1200}$

$ \begin{aligned} A & =\frac{1}{2} b c \sin \theta \Rightarrow \frac{d A}{d \theta}=\frac{1}{2} b c \cos \theta \\ \Rightarrow \quad d A & =\frac{1}{2} b c \cos \theta d \theta \\ \frac{d A}{A} \times 100 & =\frac{\frac{1}{2} b c \cos \theta d \theta}{\frac{1}{2} b c \sin \theta} \times 100 \\ & =\cot \theta d \theta \times 100 \\ & =\cot \frac{3 \pi}{8} \times \frac{\pi}{1200} \times 100 \\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, {\left[\because \theta=67^{\circ}=\frac{3 \pi}{8}\right] } & \\ & =(\sqrt{2}-1) \frac{\pi}{12}\left[\because \cos \frac{3 \pi}{8}=\sqrt{2}-1\right] \end{aligned} $

Given, $y=f(x)$

Equation of tangent of curve $y=f(x)$ at $P(\alpha, 1)$ is

$ y-1=f(\alpha)(x-\alpha) $

Equation of normal of curve at $P(\alpha, 1)$ is

$ y-1=-\frac{1}{f^{\prime}(\alpha)}(x-a) $

∴ Tangent cut of $X$-axis at $A$

$ \therefore \quad A=\left(\alpha-\frac{1}{f^{\prime} \alpha}, 0\right) $

∴ Normal cut of $X$-axis at $B$

$ \therefore \quad B=\left(\alpha+f^{\prime}(\alpha), 0\right) $

Point $C=(\alpha, 0)$

$ \begin{aligned} & A C+B C=\alpha-\alpha+\frac{1}{f^{\prime}(\alpha)}+\alpha+f^{\prime}(\alpha)-\alpha \\ & A C-B C=\frac{1}{f^{\prime}(\alpha)}+f^{\prime}(\alpha) \end{aligned} $

$A C+B C$ is minimum

When, $f^{\prime}(\alpha)=1$

∴ Equation of tangent of curve

$ y-1=1(x-\alpha) \Rightarrow x-y=\alpha+1 $

∴ Equation of tangent of curve is parallel to $x-y=0$.

$ \text { } \begin{aligned}Given,\,\, y & =3 x^5+15 x-8 \\ \frac{d x}{d t} & =\frac{1}{5} \mathrm{unit} / \mathrm{sec} \\ \Rightarrow \quad \frac{d y}{d t} & =6 \mathrm{unit} / \mathrm{sec} \\ \frac{d y}{d t} & =\left(15 x^4+15\right) \frac{d x}{d t} \end{aligned} $

$ \Rightarrow \quad \,\,\,\,\,\,\,6=15\left(x^4+1\right) \times \frac{1}{5} $

$ \begin{array}{lc} \Rightarrow \mathrm{M} x^4+1=2 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x^4=1 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\quad x= \pm 1 \\ \Rightarrow\,\,\,\,\,\,\,\,\,\,\, y(1)=3+15-8=10 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, y(-1)=-3-15-8=-26 \\ \therefore A(1,10), B(-1,-26) \\ \text { Slope of } A B=\frac{-26-10}{-1-1}=\frac{-36}{-2}=18 \end{array} $

$ \text { Given, } a+b \text { is constant } $

Let    $ a+b=k $

Area of $\triangle A B C$ say $A$

$ \begin{aligned} A & =\frac{1}{2} \times a \times \sqrt{b^2-a^2} \\ \Rightarrow \quad A^2 & =\frac{1}{4} a^2\left(b^2-a^2\right) \\ A^2 & =\frac{a^2}{4}\left((k-a)^2-c^2\right) \\ & =\frac{k^2 a^2-2 k a^3}{4} \end{aligned} $

Differentiating w.r.t. ' $a$ ',

$ 2 A \frac{d A}{d a}=\frac{2 k^2 a-6 k a^2}{4} $

For maximum or minimum $\frac{d A}{d a}=0$

$ \begin{aligned} k^2 a-3 k a^2 & =0 \Rightarrow a=k / 3 \\ \Rightarrow \quad \frac{d^2 A}{d a^2} & =\frac{-k^2}{4 A}<0 \end{aligned} $

$A^{\prime}$ s maximum value, $a=k / 3$

Now, $a=\frac{k}{3}, b=2 \frac{k}{3}$

$ \Rightarrow \quad \cos \theta=\frac{a}{b}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} $