Application of Derivatives

$\begin{aligned} & f: R \rightarrow(0, \infty) \\ & \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1 \end{aligned}$

$\because \mathrm{f}$ is increasing

$\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x}) \\ & \because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(7 \mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 1<\lim _{\mathrm{x} \rightarrow \infty} \frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<1 \\ & \therefore\left[\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}-1\right] \\ & \Rightarrow 1-1=0 \end{aligned}$

$\begin{aligned} & f(x)=e^{x^3-3 x+1} \\ & f^{\prime}(x)=e^{x^3-3 x+1} \cdot\left(3 x^2-3\right) \\ & =e^{x^3-3 x+1} \cdot 3(x-1)(x+1) \end{aligned}$

For $\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$

$\therefore \mathrm{f}(\mathrm{x})$ is increasing function

$\begin{aligned} & \therefore \mathrm{a}=\mathrm{e}^{-\infty}=0=\mathrm{f}(-\infty) \\ & \mathrm{b}=\mathrm{e}^{-1+3+1}=\mathrm{e}^3=\mathrm{f}(-1) \\ & \mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2) \\ & \therefore \mathrm{P}\left(2 \mathrm{e}^3+4,2\right) \end{aligned}$

$\mathrm{d}=\frac{\left(2 \mathrm{e}^3+4\right)+2 \mathrm{e}^{-3}-4}{\sqrt{1+\mathrm{e}^{-6}}}=2 \sqrt{1+\mathrm{e}^6}$

$\begin{aligned} & f(0)=\left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|=12 \\ & f^{\prime}(x)=\left|\begin{array}{ccc} 3 x^2 & 4 x & 3 \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right|+ \end{aligned}$

$\begin{aligned} & \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 6 x & 2 & 3 x^2 \\ x^3-x & 4 & x^2-2 \end{array}\right|+ \\ & \left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ 3 x^2-1 & 0 & 2 x \end{array}\right| \\ & \therefore f^{\prime}(0)=\left|\begin{array}{ccc} 0 & 0 & 3 \\ 2 & 0 & 6 \\ 0 & 4 & -2 \end{array}\right|+\left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 4 & -2 \end{array}\right|+\left|\begin{array}{ccc} 0 & 1 & 1 \\ 2 & 0 & 6 \\ -1 & 0 & 0 \end{array}\right| \\ & =24-6=18 \\ & \therefore 2 f(0)+f^{\prime}(0)=42 \end{aligned}$

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \\ & =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \\ & \mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2 \end{aligned}$

$\begin{aligned} & f(-4)=-216 \\ & f(-3)=0, f(4)=49 \times 8=392 \\ & M=392, m=-216 \\ & M-m=392+216=608 \\ & \text { Ans }=\text { '3' } \end{aligned}$

Area of $\Delta$

$\begin{aligned} & =\frac{1}{2}\left|\begin{array}{ccc} 0 & 0 & 1 \\ x & y & 1 \\ -x & y & 1 \end{array}\right| \\ & \Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y| \\ & \operatorname{Area}(\Delta)=|x y|=\left|x\left(-2 x^2+54\right)\right| \\ & \frac{d(\Delta)}{d x}=\left|\left(-6 x^2+54\right)\right| \Rightarrow \frac{d \Delta}{d x}=0 \text { at } x=3 \\ & \text { Area }=3(-2 \times 9+54)=108 \end{aligned}$

$f(x)=\frac{x}{x^2-6 x-16}$

Now,

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\ & \mathrm{f}^{\prime}(\mathrm{x})<0 \end{aligned}$

Thus $f(x)$ is decreasing in

$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$

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

So, maxima (M) at x = $-$1 & minima (m) at x = 0

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x})=12 \sqrt{2} \mathrm{x}^2-3 \sqrt{2} \geq 0 \text { for }\left[\frac{1}{2}, 1\right] \\ & \mathrm{f}\left(\frac{1}{2}\right)<0 \end{aligned}$

$\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})$ is correct.

$f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0$

Let $\cos \alpha=\mathrm{x}$,

$\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}$

$\mathrm{x}=\cos \frac{\pi}{12}$

(4) is correct.

$g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)$

$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})$ is increasing function

$\begin{aligned} & g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\ & =f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \end{aligned}$

If $\mathrm{g}$ is decreasing in $(0, \alpha)$

$\begin{aligned} & \mathrm{g}^{\prime}(\mathrm{x})<0 \\ & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\ & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\ & \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\ & \Rightarrow \mathrm{x}<\frac{9}{4} \end{aligned}$

Therefore $\alpha=\frac{9}{4}$

Then $8 \alpha=8 \times \frac{9}{4}=18$

Given, $ y=f(x)=2 x^{2}-3 x+4 $

$ \begin{array}{l} \begin{array}{l} \delta_{x}=0.02 \quad x=5 \\\\ \frac{d y}{d x}=4 x-3 \\\\ d y=(4 x-3) d x \\\\ d y=(4 \times 5-3)(0.02) \\\\ =17 \times 0.02=0.34 \\\\ \delta_{y}=f(x+\delta x)-y f(x) \\\\ \delta_{y}=f(5.02)-f(5) \\\\ =2(5.02)^{2}-3(5.02)+4-\left(2(5)^{2}-3(5)+4\right] \\\\ \left.=2[5.02)^{2}-5^{2}\right]-3(5.02-5) \\\\ =2(5.05+5)(5.02-5)-3(0.02) \\\\ =2(10.02)(0.02)-0.06 \\\\ =4.008-0.06=0.3408 \\\\ \delta y-d y=0.3408-0.34=0.0008 \end{array} \end{array} $

$ x=2(\cos 2 t+t \sin 2 t) $

$ \frac{d x}{d t}=2[-\sin 2 t \cdot 2+t \cos 2 t \cdot 2+\sin 2 t] $

$ =2[2 t \cos 2 t-\sin 2 t] $

$ y=4(\sin 2 t+t \cos 2 t) $

$ \frac{d y}{d t}=4[\cos 2 t \cdot 2+t(-\sin 2 t) 2+\cos 2 t] $

$ =4\left[3 \cos 2 t-2 t \sin ^{2} t\right] $

$ \frac{d y}{d x}=\frac{4\left[3 \cos 2 t-2 t \sin ^{2} t\right]}{2[2 t \cos 2 t-\sin 2 t]} $

$ \left(\frac{d y}{d x}\right) \text { at } t=\frac{\pi}{4}=\frac{2\left[0-2 \cdot \frac{\pi}{4}\right]}{\left[2 \frac{\pi}{4} \times 0-1\right]}=\frac{-\pi}{-1}=\pi $

$ y \text { at } t=\frac{\pi}{4}=4[1]=4 $

$ \begin{aligned} \text { Length of normal } & =\left|y_{1}\right| \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} \\\\ & =4 \sqrt{1+\pi^{2}} \end{aligned} $

To find the rate at which the water level rises in a cylindrical tank, given that water is poured at a constant rate, we start by using the formula for the volume of a cylinder:

$ V = \pi r^2 h $

Here, $ V $ is the volume, $ r $ is the radius of the base, and $ h $ is the height of the water in the tank.

Given:

The radius $ r = 3.5 \, \text{ft} $.

The rate of volume change (water being poured) $ \frac{dV}{dt} = 1 \, \text{cu ft/min} $.

We want to find $ \frac{dh}{dt} $, the rate at which the water level $ h $ is rising.

Differentiate both sides with respect to time $ t $:

$ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} $

Substitute the values:

$ 1 = \pi (3.5)^2 \frac{dh}{dt} $

Solve for $ \frac{dh}{dt} $:

$ \frac{dh}{dt} = \frac{1}{\pi \times 3.5 \times 3.5} $

Simplifying further, we perform the multiplication:

$ \pi = \text{approximately } 22/7 $

$ 3.5 = \frac{7}{2} $

So we have:

$ \frac{dh}{dt} = \frac{1}{(\frac{22}{7}) \times (\frac{7}{2}) \times (\frac{7}{2})} $

Simplify the expression:

$ \frac{dh}{dt} = \frac{7}{22} \times \frac{10}{35} \times \frac{10}{35} = \frac{2}{77} \, \text{ft/min} $

Therefore, the rate at which the water level in the tank rises is $ \frac{2}{77} \, \text{ft/min} $.

$ \begin{array}{l} y= 2 x^{3}-8 x^{2}+10 x-4 \\\\ \frac{d y}{d x}=6 x^{2}-16 x+10 \\\\ \left(\frac{d y}{d x}\right)_{(a, b)}=6 a^{2}-16 a+10=0 \\\\ 3 a^{2}-8 a+5=0 \\\\ (3 a-5)(a-1)=0 \\\\ a=\frac{5}{3} \text { or } 1 \\\\ 1 < \frac{5}{3} < 2 \\\\ a=\frac{5}{3} \end{array} $

To find the sum of the absolute minimum and maximum values of the function $ f(x) = 2x^3 + 9x^2 + 12x + 1 $ on the interval $[-3, 0]$, follow these steps:

Determine Critical Points:

Compute the derivative of $ f(x) $:

$ f'(x) = 6x^2 + 18x + 12 = 6(x^2 + 3x + 2) = 6(x+2)(x+1) $

Setting $ f'(x) = 0 $ yields critical points at $ x = -2 $ and $ x = -1 $.

Evaluate the Function at Critical Points and Endpoints:

Compute $ f(x) $ at the critical points and endpoints of the interval:

$ f(-3) = 2(-3)^3 + 9(-3)^2 + 12(-3) + 1 = -54 + 81 - 36 + 1 = -8 $

$ f(-2) = 2(-2)^3 + 9(-2)^2 + 12(-2) + 1 = -16 + 36 - 24 + 1 = -3 $

$ f(-1) = 2(-1)^3 + 9(-1)^2 + 12(-1) + 1 = -2 + 9 - 12 + 1 = -4 $

$ f(0) = 2(0)^3 + 9(0)^2 + 12(0) + 1 = 1 $

Identify the Absolute Minimum and Maximum:

Compare the function values: $ f(-3) = -8 $, $ f(-2) = -3 $, $ f(-1) = -4 $, and $ f(0) = 1 $.

The absolute minimum is $ m = -8 $.

The absolute maximum is $ M = 1 $.

Calculate $ m + M $:

$ m + M = -8 + 1 = -7 $

Therefore, $ m + M = -7 $.

To find the maximum interval where the slopes of the tangents to the curve $ y = x^4 + 5x^3 + 9x^2 + 6x + 2 $ are increasing, we need to consider the behavior of the derivative.

The slope of the tangent line is given by the first derivative, $ \frac{dy}{dx} $. We need to determine when this slope function itself is increasing, which requires the second derivative to be positive.

Let's calculate the derivatives:

First derivative (slope of the tangent):

$ \frac{dy}{dx} = 4x^3 + 15x^2 + 18x + 6 $

Second derivative:

$ \frac{d^2y}{dx^2} = 12x^2 + 30x + 18 $

To find when this second derivative is greater than zero (indicating the slope function is increasing), solve the inequality:

$ 12x^2 + 30x + 18 > 0 $

Factor the quadratic:

$ 2x^2 + 5x + 3 > 0 $

$ (x + \frac{3}{2})(x + 1) > 0 $

This inequality holds true for:

$ x \in \mathbb{R} - \left[\frac{-3}{2}, -1\right] $

Thus, the slopes of the tangents increase on the intervals outside of $\left[\frac{-3}{2}, -1\right]$.

To understand the solution, let's analyze the conditions for horizontal and vertical tangents to the curve $ y^3 - 3xy + 2 = 0 $.

Analyzing Horizontal Tangents:

For a tangent at a point $ P(\alpha, \beta) $ to be horizontal, the derivative $ y' $ must be zero. We find the derivative as follows:

$ 3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0 $

Solving for $ y' $:

$ y' = \frac{y}{y^2 - x} $

For the tangent to be horizontal at $ (\alpha, \beta) $, we set:

$ \left. y' \right|_{\alpha, \beta} = \frac{\beta}{\beta^2 - \alpha} = 0 $

Which implies $ \beta = 0 $. Substituting $ \beta = 0 $ into the original curve equation:

$ 0^3 - 3\alpha \cdot 0 + 2 = 0 \quad \Rightarrow \quad 2 = 0 $

This is a contradiction, so there are no points where the tangent is horizontal. Thus, $ n(A) = 0 $.

Analyzing Vertical Tangents:

For a tangent at a point $ Q(a, b) $ to be vertical, the denominator of $ y' $ must be zero, leading to:

$ b^2 - a = 0 \quad \Rightarrow \quad a = b^2 $

Substituting $ a = b^2 $ back into the original curve equation:

$ b^3 - 3b \cdot b^2 + 2 = 0 $

$ b^3 - 3b^3 + 2 = 0 $

$ -2b^3 + 2 = 0 $

$ b^3 = 1 $

$ b = 1 $

Thus, $ a = b^2 = 1^2 = 1 $. Therefore, the point is $ Q(1, 1) $, implying $ n(B) = 1 $.

Conclusion:

The number of points in sets $ A $ and $ B $ are:

$ n(A) = 0 $

$ n(B) = 1 $

Thus, the total $ n(A) + n(B) = 1 $.

Given the curves $ y=f(x) $ and $ x=g(y) $, and a common point $ P(x, y) $ on both curves, we are tasked with evaluating their tangents and how they relate at this point.

For the curve $ y = f(x) $, by differentiating both sides with respect to $ x $, we have:

$ \frac{dy}{dx} = f'(x) = Q(x) $

This is the slope of the tangent to the curve $ y = f(x) $ at the point $ P $.

Next, considering the curve $ x = g(y) $, we differentiate with respect to $ y $:

$ \frac{dx}{dy} = g'(y) = -Q(x) $

This represents the slope of the tangent to the curve $ x = g(y) $ at the same point $ P $.

Now, see that

$ \left(\frac{dy}{dx}\right) \times \left(\frac{dx}{dy}\right) = Q(x) \times (-Q(x)) = -1 $

The product of the slopes being $-1$ indicates that the tangents are perpendicular. Therefore, the tangent line to one curve at point $ P $ serves as the normal line to the other curve at the same point. This confirms that the tangent of one curve is normal to the other at point $ P $.

Given, $ 7+6 x-3 x^{2} $ or $ -3 x^{2}+6 x+1 $

As coefficient of $ x^{2} $ is less than 0 , then $ f\left(\frac{-b}{2 a}\right) $ is minima or maxima.

$ \frac{-b}{2 a}=\frac{-6}{2 \times(-3)}=1 $

$ f(\mathrm{l})=-3+6+7=10 $

Here, $ \alpha=1, \beta=10 $

Let $ m $ and $ n $ be roots of $ x^{2}+x-10=0 $

$ \begin{array}{l} m+n=-1 \text { and } m n=-10 \\\\ m^{2}+n^{2}=(m+n)^{2}-2 m n=21 \end{array} $

We have, $ y^{3}=4 x^{5} $

$ 3 y^{2} \frac{d y}{d x}=20 x^{4} $

$ \left[\frac{d y}{d x}\right]_{(4,16)}=\frac{20 \times 4^{4}}{3 \times(16)^{2}}=\frac{20}{3} $

Slope of normal $ =-\frac{3}{20} $

Equation $ 3 x+20 y=K $

$ 12+320=K \Rightarrow 332=K $

Frence, equation of normal is $ 3 x+20 y=332 $

Given, $ x^{3} y^{4}=2^{7} $

$ \begin{array}{l} 3 x^{2} y^{4} \frac{d x}{d t}+4 x^{3} y^{3} \frac{d y}{d t}=0 \\\\ \Rightarrow \frac{d y}{d t}=\frac{-3 x^{2} y^{4}}{4 x^{3} y^{3}} \frac{d x}{d t} \quad\left\{\because \frac{d x}{d t}=-8\right\} \\\\ \Rightarrow \frac{d y}{d t}=-\frac{3}{4} \times \frac{2^{2} \times 2^{4}}{2^{3} \times 2^{3}} \times-8 \\\\ \Rightarrow \frac{d y}{d t}=+6 \end{array} $

If $ f(x) $ satisfies the condition of

Rolle's theorem

$ \begin{array}{ll} & f(-5)=f(4)=0 \\\\ \Rightarrow & -125+25 a-5 b=64+16 a+4 b \\\\ \Rightarrow & 9 a-9 b=189 \Rightarrow a-b=21 \\\\ \therefore & f(4)=0 \\\\ \Rightarrow & 64+16 a+4 b+40=0 \\\\ \Rightarrow & 4 a+b=-26 \\\\ \Rightarrow & a-b=21 \end{array} $

On adding Eq. (i) and (ii), we get

$ \begin{aligned} 5 a & =-5 \Rightarrow a=-1 \text { and } b=-22 \\\\ \therefore \quad f(x) & =x^{3}-x^{2}-22 x+40 \\\\ f^{\prime}(x) & =3 x^{2}-2 x-22=0 \\\\ x & =\frac{2 \pm \sqrt{4+264}}{6} \Rightarrow x=\frac{1 \pm \sqrt{67}}{3} \end{aligned} $

Given that $ x $ and $ y $ are 2 positive number.

Also, $ x+y=24 \Rightarrow x=24-y $

Let $ s=x^{3} y^{5}=(24-y)^{3} y^{5} $

Now,

$ \begin{array}{l} \frac{d s}{d y}=3(24-y)^{2} \times(-1) \times y^{5}+(24-y)^{3} \times 5 y^{4} \\\\ =(24-y)^{2} \cdot y^{4}[-3 y+(24-y) 5] \\\\ =8(24-y)^{2} \cdot y^{4} \cdot(15-y) \\\\ \frac{d s}{d y}=0 \text { at } y=0,15,24 \end{array} $

Now, applying first derivative test on 15 only, because $ x $ and $ y $ both are positive.

$ \therefore x $ and $ y $ cannot be 0 so, $ y $ cannot be 0 and 24.

$ \therefore $ Sign of $ \frac{d s}{d y} $ around 15 is

$ \therefore $ We have, S at maximum, when $ y=15 $

So, $ x=24-15=9 $

$ \therefore x^{2}+y^{2}=9^{2}+15^{2}=81+225=306 $

We have,

$4+3 x-7 x^2$ attains its maximum value $M$ at $x=\alpha$

$ \therefore \quad \alpha=\frac{-(3)}{2(-7)}=\frac{3}{14} $

Now, $M=4+3\left(\frac{3}{14}\right)-7\left(\frac{3}{14}\right)^2=\frac{121}{28}$

Also, we have $5 x^2-2 x+1$ attains its minimum value $M$ at $x=\beta$

$ \therefore \quad \beta=\frac{-(-2)}{2(5)}=\frac{2}{10}=\frac{1}{5} $

Now, $m=5\left(\frac{1}{5}\right)^2-2\left(\frac{1}{5}\right)+1=\frac{4}{5}$

$ \begin{aligned} \therefore \quad \frac{28(M-\alpha)}{5(m+\beta)} & =\frac{28\left(\frac{121}{28}-\frac{3}{14}\right)}{5\left(\frac{4}{5}+\frac{1}{5}\right)} \\\\ & =\frac{115}{5}=23 \end{aligned} $

Given, $x=\cos 2 t+\log (\tan t)$

$ \begin{aligned} & \text { and } \quad y=2 t+\cot 2 t \\\\ & \therefore \quad \frac{d x}{d t}=(-\sin 2 t)(2) \\\\ & +\frac{1}{\tan t} \sec ^2 t=-2 \sin 2 t+\frac{1}{\sin t \cos t} \\\\ & \text { and } \frac{d y}{d x}=2-\left(\operatorname{cosec}^2 2 t(2)\right. \\\\ & =2\left(1-\operatorname{cosec}^2 2 t\right) \\\\ & \therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2\left(1-\operatorname{cosec}^2 2 t\right)}{-2 \sin 2 t+\frac{1}{\sin t \cos t}} \\\\ & =\frac{2\left(-\cot ^2 2 t\right)}{-2 \sin 2 t+\frac{2}{\sin 2 t}} \\\\ & =\frac{\frac{\cos ^2 2 t}{\sin ^2 2 t}}{\sin 2 t-\frac{1}{\sin 2 t}} \end{aligned} $

$ \begin{aligned} & =\frac{\cos ^2 2 t}{\left(\sin ^2 2 t-1\right)(\sin 2 t)} \\\\ & =\frac{\cos ^2 2 t}{-\cos ^2 2 t \sin 2 t}=-\operatorname{cosec} 2 t \end{aligned} $

Let $f(x)=\sqrt[3]{x}$

$ \therefore \quad f^{\prime}(x)=\frac{1}{3}(x)^{-2 / 3} $

We know that, $f(x+\Delta x) \approx f(x)+f^{\prime}(x) \Delta x$

$ \begin{array}{ll} \therefore \quad & \sqrt[3]{730}=\sqrt[3]{729+1} \\\\ & \approx \sqrt[3]{729}+\frac{1}{3}(729)^{-2 / 3} \\\\ & =9+\frac{1}{3 \times 81} \approx 9+0.0041 \\\\ & =9.0041 \end{array} $

Hence, the approximate value of $\sqrt[3]{730}$ is 9.0041

Given curves are $y^2=x$ and

$ x^2+y^2=2 $

The points of intersection of these curves are $(1,1)$ and $(1,-1)$

Now,

$ \begin{array}{l|l} y^2=x & x^2+y^2=2 \\\\ 2 y \frac{d y}{d x}=1 & 2 x+2 y \frac{d y}{d x}=0 \\\\ \frac{d y}{d x}=\frac{1}{2 y} & \frac{d y}{d x}=-\frac{x}{y} \\\\ \left(\frac{d y}{d x}\right)_{(1,1)}=\frac{1}{2(1)}=\frac{1}{2} & \left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{1}{1}=-1 \\\\ m_1=\frac{1}{2} & m_2=-1 \end{array} $

$\begin{aligned} & \therefore \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \because \theta \text { is acute } \\\\ & =\left|\frac{\frac{1}{2}-(-1)}{1+\left(\frac{1}{2}\right)(-1)}\right|=\left|\frac{\frac{3}{2}}{\frac{1}{2}}\right|=3\end{aligned}$

Semi-vertical angle $=\frac{60^{\circ}}{2}=30^{\circ}$

$ \tan 30^{\circ}=\frac{r}{h} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{r}{h} \\\\ \Rightarrow & h=\sqrt{3} r \end{array} $

When $h=3 m$, then $r=\frac{3}{\sqrt{3}}=\sqrt{3} m$

$ \begin{aligned} & \text { Volume, } V=\frac{1}{3} \pi r^2 h \\\\ & \Rightarrow \quad 3 V=\pi r^2(\sqrt{3} r) \\\\ & \Rightarrow \quad \sqrt{3} V=\pi r^3 \end{aligned} $

On differentiating Both sides w.r.t. $t$, we get

$ \begin{aligned} & \sqrt{3} \frac{d V}{d t}=\pi\left(3 r^2\right) \frac{d r}{d t} \\\\ \Rightarrow & \sqrt{3} \times \frac{1}{\sqrt{3}}=3 r^2 \pi \frac{d r}{d t} \\\\ \Rightarrow & \left.\frac{d r}{d t}\right|_{h=3 m}=\frac{1}{3(\sqrt{3})^2 \pi}=\frac{1}{9 \pi} \mathrm{~m} / \mathrm{min} \end{aligned} $

Semi-vertical angle $=\frac{60^{\circ}}{2}=30^{\circ}$

$ \tan 30^{\circ}=\frac{r}{h} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{r}{h} \\\\ \Rightarrow & h=\sqrt{3} r \end{array} $

When $h=3 m$, then $r=\frac{3}{\sqrt{3}}=\sqrt{3} m$

$ \begin{aligned} & \text { Volume, } V=\frac{1}{3} \pi r^2 h \\\\ & \Rightarrow \quad 3 V=\pi r^2(\sqrt{3} r) \\\\ & \Rightarrow \quad \sqrt{3} V=\pi r^3 \end{aligned} $

On differentiating Both sides w.r.t. $t$, we get

$ \begin{aligned} & \sqrt{3} \frac{d V}{d t}=\pi\left(3 r^2\right) \frac{d r}{d t} \\\\ \Rightarrow & \sqrt{3} \times \frac{1}{\sqrt{3}}=3 r^2 \pi \frac{d r}{d t} \\\\ \Rightarrow & \left.\frac{d r}{d t}\right|_{h=3 m}=\frac{1}{3(\sqrt{3})^2 \pi}=\frac{1}{9 \pi} \mathrm{~m} / \mathrm{min} \end{aligned} $

Let $r$ and $h$ be radius and height of the cone.

Let the semi-vertical angle be $\theta$.

Here, $\theta=\tan ^{-1}\left(\frac{r}{h}\right)$

in right-angle triangle, $O D C$

$ \begin{array}{rlrl} 3^2 & =(h-3)^2+r^2 \\\\ \Rightarrow r^2 & =9-\left(h^2+9-6 h\right) \\\\ \Rightarrow r^2 & =6 h-h^2 \end{array} $

Volume of cone, $V=\frac{1}{3} \pi\left(6 h-h^2\right) h$

$ \begin{array}{ll} \quad V=\frac{1}{3} \pi\left(6 h^2-h^3\right) \\\\ \therefore \frac{d V}{d h}=\frac{1}{3} \pi\left(12 h-3 h^2\right) \\\\ \text { and } \quad \frac{d^2 V}{d h^2}=\frac{1}{3} \pi(12-6 h) \end{array} $

On setting $\frac{d V}{d h}=0$, we get $h=4$

For $h=4, \frac{d^2 V}{d h^2}<0$

$\therefore$ Maximum volume will be occurs at $h=4$

Now, $r^2=6 h-h^2=6(4)-4^2=8$

$ \Rightarrow \quad r=2 \sqrt{2} $

Hence, $\theta=\tan ^{-1}\left(\frac{r}{h}\right)$

$ =\tan ^{-1}\left(\frac{2 \sqrt{2}}{4}\right)=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) $

Given,

$ f(x)=k x^3-3 x^2-12 x+8 \text { is strictly } $

decreasing, $\forall x \in R$

$\therefore f^{\prime}(x)=0$ will not have any real solution.

Now, $f^{\prime}(x)=3 k x^2-6 x-12$

$\because 3 k x^2-6 x-12=0$ does not have any real solution.

$ \begin{array}{ll} \therefore & D<0 \\ \Rightarrow & (-6)^2-4(3 k)(-12)<0 \\ \Rightarrow & 36+36(4 k)<0 \Rightarrow 1+4 k<0 \\ \Rightarrow & k<-\frac{1}{4} \end{array} $

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} $

$ \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} $.