Application of Derivatives

To find the equation of the tangent to the curve $ y = x^3 - 2x + 7 $ at the point $(1, 6)$, we need to follow these steps:

Find the derivative:

The derivative of the function $ y = x^3 - 2x + 7 $ with respect to $ x $ is given by:

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

Determine the slope of the tangent line:

Calculate the slope ($ m $) of the tangent line at the point $ (1, 6) $ using the derivative:

$ m = \left. \frac{dy}{dx} \right|_{x=1} = 3(1)^2 - 2 = 1 $

Write the equation of the tangent line:

Using the point-slope form of a line equation, $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) = (1, 6) $, we substitute $ m = 1 $:

$ y - 6 = 1(x - 1) $

Simplify the equation:

Simplifying the equation gives:

$ y - 6 = x - 1 $

$ y = x + 5 $

Thus, the equation of the tangent to the curve at the point $(1, 6)$ is $ y = x + 5 $.

The distance $ S $ travelled by a particle over time $ t $ is given by the equation:

$ S = 4t^2 + 2t + 3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

To find the velocity of the particle at $ t = 3 $ seconds, we need to differentiate equation (i) with respect to time $ t $. Velocity is defined as the derivative of distance with respect to time. Therefore, we have:

$ \text{Velocity} = \frac{\text{Distance}}{\text{Time}} $

$ \Rightarrow v = \frac{dS}{dt} = \frac{d}{dt}(4t^2 + 2t + 3) $

When we differentiate the expression, we get:

$ v = \frac{dS}{dt} = (8t + 2) \, \text{units/sec} $

Substituting $ t = 3 $ into the velocity equation, we calculate:

$ V = (8 \times 3 + 2) \, \text{units/sec} $

$ V = (24 + 2) \, \text{units/sec} $

Thus, the velocity $ V $ at $ t = 3 $ seconds is:

$ V = 26 \, \text{units/sec} $

We start with the equation:

$ a^2 x^4 + b^2 y^4 = c^6 $

This can be rearranged to express $ y^4 $:

$ y^4 = \frac{c^6 - a^2 x^4}{b^2} $

Let $ z = xy $. To find the maximum value of $ z $, we express $ z $ in terms of $ x $ and differentiate:

$ z = x \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{1/4} $

Differentiate $ z $ with respect to $ x $:

$ \frac{dz}{dx} = \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{1/4} - \frac{x}{4} \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{-3/4} \cdot 4 x^3 \frac{a^2}{b^2} $

Simplify:

$ z'(x) = \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{1/4} - \frac{x^4 a^2}{b^2} \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{-3/4} $

Set $ z'(x) = 0 $ to find critical points:

$ \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{1/4} = \frac{x^4 a^2}{b^2} \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{-1} $

$ x^4 = \frac{b^2}{a^2} \left(\frac{c^6 - a^2 x^4}{b^2}\right) $

$ x^4 = \frac{c^6 - a^2 x^4}{a^2} $

This results in:

$ x^4 = \frac{c^6}{2a^2} $

Thus, solving for $ x $:

$ x = \left(\frac{c^6}{2a^2}\right)^{1/4} = \frac{c^{6/4}}{(2^{1/4} a^{2/4})} $

Now find $ z = xy $:

$ z = \frac{c^{6/4}}{2^{1/4} a^{1/2}} \left(\frac{c^6 - a^2 x^4}{b^2}\right)^{1/4} $

Substitute $ x^4 = \frac{c^6}{2a^2} $ back into the equation:

$ z = \frac{c^{6/4}}{2^{1/4} a^{1/2}} \left(\frac{c^6}{2b^2}\right)^{1/4} $

Finally, evaluate $ z $:

$ z = \left(\frac{c^{12}}{4a^2b^2}\right)^{1/4} = \frac{c^3}{\sqrt{2ab}} $

The maximum value of $ xy $ is:

$ \frac{c^3}{\sqrt{2ab}} $

We have the function $ f(x) = x^3 - 16x^2 + 20x - 5 $.

Step 1: Find the Critical Points

First, find the derivative of $ f(x) $:

$ f'(x) = 3x^2 - 32x + 20 $

Set the derivative equal to zero to find critical points:

$ 3x^2 - 32x + 20 = 0 $

Solving this quadratic equation, we find:

$ x = \frac{32 \pm \sqrt{1024 - 240}}{6} = \frac{32 \pm \sqrt{784}}{6} $

Simplifying, we get:

$ x = \frac{32 \pm 28}{6} $

Thus, the roots are $ x = 10 $ and $ x = \frac{2}{3} $.

Step 2: Identify Intervals of Decrease

The function is strictly decreasing in the interval $(\frac{2}{3}, 10)$.

Step 3: Analyze the Number Set

The set of numbers is $\{1, 3, 5, 7, 9, \ldots, 59\}$.

Step 4: Count Favourable Outcomes

Within the decreasing interval $(\frac{2}{3}, 10)$, the numbers from the set are $\{1, 3, 5, 7, 9\}$. Hence, there are 5 favorable outcomes.

Step 5: Count Total Outcomes

The sequence $1, 3, 5, 7, \ldots, 59$ is an arithmetic progression (AP) where:

First term $a = 1$

Common difference $d = 2$

Last term $a_n = 59$

To find the number of terms $n$, use the formula:

$ a_n = a + (n-1)d $

$ 59 = 1 + (n-1) \times 2 $

$ n = 30 $

So, the total number of outcomes is 30.

Step 6: Calculate Probability

The probability $P$ is given by:

$ P = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{5}{30} = \frac{1}{6} $

Thus, the probability that a randomly drawn number from the set $\{1, 3, 5, 7, \ldots, 59\}$ lies in the interval where the function $f(x) = x^3 - 16x^2 + 20x - 5$ is strictly decreasing is $\frac{1}{6}$.

To find the equation of the normal line to the parabola given by $ y^2 = 6x $ at the point $(24, 12)$, follow these steps:

Determine the Slope of the Tangent Line:

The derivative of the parabola's equation $ y^2 = 6x $ with respect to $ x $ is:

$ 2y \frac{dy}{dx} = 6 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{3}{y} $

Substitute $ y = 12 $ to find the slope at the point $(24, 12)$:

$ \left[\frac{dy}{dx}\right]_{(24,12)} = \frac{3}{12} = \frac{1}{4} $

Find the Equation of the Normal Line:

The slope of the normal line is the negative reciprocal of the tangent line's slope. Thus, the slope of the normal line is:

$ m = -\frac{1}{\frac{1}{4}} = -4 $

Use the point-slope form of a line equation for the normal line at $(24, 12)$:

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

$ y - 12 = -4(x - 24) $

Simplify and rearrange the equation:

$ y - 12 = -4x + 96 $

$ 4x + y = 108 $

Thus, the equation of the normal line to the parabola $ y^2 = 6x $ at the point $(24, 12)$ is:

$ 4x + y = 108 $

To find the point that lies on the tangent to the curve $ x^4 e^y + 2 \sqrt{y+1} = 3 $ at the point $(1, 0)$, we begin by differentiating the equation with respect to $ x $:

$ x^4 e^y y^{\prime} + e^y 4x^3 + \frac{2 y^{\prime}}{2 \sqrt{y+1}} = 0 $

Evaluating the derivative at the point $ P(1, 0) $:

$ 1^4 e^0 y_p^{\prime} + e^0 \cdot 4 \cdot 1^3 + \frac{2 y_p^{\prime}}{2 \sqrt{0+1}} = 0 $

$ y_p^{\prime} + 4 + y_p^{\prime} = 0 $

$ \Rightarrow 2y_p^{\prime} = -4 $

$ \Rightarrow y_p^{\prime} = -2 $

Hence, the equation of the tangent at $ P(1, 0) $ is:

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

Simplifying, we get:

$ 2x + y = 2 $

Now substituting $ x = -2 $ into the tangent equation:

$ 2(-2) + y = 2 $

$ -4 + y = 2 $

$ y = 6 $

Thus, the point $ (-2, 6) $ lies on the tangent line.

To determine the interval where the function $ f(x) = x^x $ decreases, we first express it as:

$ f(x) = x^x = e^{x \ln x} $

To find when the function decreases, we differentiate it. Using the chain rule:

$ \frac{d}{dx}\left[e^{x \ln x}\right] = e^{x \ln x} \cdot \frac{d}{dx}[x \ln x] $

Next, we calculate the derivative of $ x \ln x $:

$ \frac{d}{dx}[x \ln x] = \ln x + 1 $

Therefore, the derivative $ f'(x) $ of the function can be written as:

$ f'(x) = e^{x \ln x} (\ln x + 1) = x^x (\ln x + 1) $

The function $ f(x) $ decreases when $ f'(x) $ is negative:

$ x^x (\ln x + 1) \leq 0 $

Since $ x^x \geq 0 $ for $ x \geq 0 $, this implies:

$ \ln x + 1 < 0 $

Solving for $ x $, we find:

$ \ln x \leq -1 $

$ x < \frac{1}{e} $

Thus, the interval where $ f(x) = x^x $ is decreasing is:

$ \left[0, \frac{1}{e}\right] $

To apply Rolle's Theorem to the function $ f(x) = x^3 + Px - 12 $ on the interval $[0,1]$, we must first ensure that the function satisfies the conditions of the theorem: $ f(x) $ must be continuous on $[0, 1]$, differentiable on $(0, 1)$, and $ f(0) = f(1) $.

Firstly, let's verify that $ f(0) = f(1) $:

$ f(0) = 0^3 + P \cdot 0 - 12 = -12 $

$ f(1) = 1^3 + P \cdot 1 - 12 = 1 + P - 12 = P - 11 $

Setting $ f(0) = f(1) $, we get:

$ -12 = P - 11 \quad \Rightarrow \quad P = -1 $

With $ P = -1 $, the function becomes:

$ f(x) = x^3 - x - 12 $

We then find $ f'(x) $ and check for points where $ f'(c) = 0 $ in the interval $(0, 1)$:

$ f'(x) = \frac{d}{dx}[x^3 - x - 12] = 3x^2 - 1 $

Setting $ f'(c) = 0 $:

$ 3c^2 - 1 = 0 \quad \Rightarrow \quad c^2 = \frac{1}{3} \quad \Rightarrow \quad c = \pm \frac{1}{\sqrt{3}} $

Since $ c $ must be in the interval $(0, 1)$, we choose the positive value:

$ c = \frac{1}{\sqrt{3}} $

Thus, for the function $ f(x) $ under the conditions of Rolle's Theorem, $ c = \frac{1}{\sqrt{3}} $ is the appropriate value within the interval $(0, 1)$.

The function given is:

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

Recall the trigonometric identity:

$ \frac{1 - \tan^2 x}{1 + \tan^2 x} = \cos 2x $

Substituting, we get:

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

To find the maximum value of this function, we first compute its derivative:

$ f'(x) = \cos x - 4\sin x \cos x = \cos x (1 - 4\sin x) $

Setting the derivative equal to zero gives us:

$\cos x = 0$

This occurs when:

$ x = \frac{\pi}{2}, \frac{3\pi}{2} $

$1 - 4\sin x = 0$

Solving for $\sin x$, we get:

$ \sin x = \frac{1}{4} $

So:

$ x = \sin^{-1} \frac{1}{4}, \quad x = \pi - \sin^{-1} \frac{1}{4} $

Thus, the critical points are:

$ x = \frac{\pi}{2}, \frac{3\pi}{2}, \sin^{-1} \frac{1}{4}, \pi - \sin^{-1} \frac{1}{4} $

Now evaluate the function at these critical points:

$ f\left(\frac{\pi}{2}\right) = 0 $

$ f\left(\frac{3\pi}{2}\right) = -2 $

For:

$ f\left( \sin^{-1} \frac{1}{4} \right) $

$ f\left( \pi - \sin^{-1} \frac{1}{4} \right) $

These values are where the function attains its maximum. The maximum value is 2.

Given the parabola:

$ y^2 = 8x $

Rewriting the equation, we have:

$ x = \frac{y^2}{8} \quad \ldots \text{(i)} $

Differentiating both sides with respect to $ y $, we find:

$ \frac{dx}{dy} = \frac{2y}{8} = \frac{y}{4} $

Thus, the slope ($ m $) of the tangent line is $ m = \frac{y}{4} $.

The equation of a tangent line passing through the point $ (1, 3) $ is:

$ (x - 1) = \frac{y}{4}(y - 3) \quad \ldots \text{(ii)} $

Combining equations (i) and (ii), we have:

$ \begin{align*} & \frac{y^2}{8} - 1 = \frac{y}{4}(y - 3) \\ \Rightarrow & y^2 - 8 = 2y^2 - 6y \\ \Rightarrow & y^2 - 6y + 8 = 0 \\ \Rightarrow & (y - 4)(y - 2) = 0 \\ \therefore & y = 4 \text{ and } y = 2 \end{align*} $

Using equation (i) to find $ x $:

For $ y = 4 $:

$ x = \frac{4^2}{8} = 2 $

For $ y = 2 $:

$ x = \frac{2^2}{8} = \frac{1}{2} $

Thus, the points are $ (2, 4) $ and $ \left(\frac{1}{2}, 2\right) $.

The required equation of the tangent line to the parabola $ y^2 = 8x $, which passes through the points $ \left(\frac{1}{2}, 2\right) $ and $ (1, 3) $, is found as follows:

$ \begin{align*} (y - 3) &= \frac{(3 - 2)}{\left(1 - \frac{1}{2}\right)}(x - 1) \\ \Rightarrow y - 3 &= \frac{1}{\frac{1}{2}}(x - 1) = 2x - 2 \\ \Rightarrow y &= 2x - 2 + 3 = 2x + 1 \end{align*} $

Thus, the equation of the tangent line is:

$ y = 2x + 1 $

Given curve

$ x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}} \quad \ldots \text { (i) } $

Let consider the point $\left(a \cos ^3 \theta, a \sin ^3 \theta\right)$

Now, slope of the tangent $=-\tan \theta$

$\therefore$ Equation of tangent is

$ y-a \sin ^3 \theta=-\tan \theta\left(x-a \cos ^3 \theta\right) $

$\Rightarrow x \sin \theta+y \cos \theta=a \sin \theta \cos \theta$

$\Rightarrow x \sin \theta+y \cos \theta=\frac{a}{2} \sin 2 \theta\quad \ldots \text { (ii) }$

Slope of normal $=\cot \theta$

$\therefore$ Equation of normal is $\left(y-a \sin ^3 \theta\right)=\cot \theta\left(x-a \cos ^3 \theta\right)$

$\Rightarrow x \cos \theta-y \sin \theta=a \cos 2 \theta \quad \ldots \text { (iii) } $

Now, $P_1=\left|\frac{-a \sin \theta \cos \theta}{\sqrt{\sin ^2 \theta+\cos ^2 \theta}}\right|=\frac{a}{2} \sin 2 \theta$

$\Rightarrow \quad 4 P_1^2=a^2 \sin ^2 2 \theta \quad \ldots \text { (iv) } $

and $P_2=\left|\frac{-a \cos 2 \theta}{\sqrt{\sin ^2 \theta+\cos ^2 \theta}}\right|=a \cos 2 \theta$

$\Rightarrow P_2^2=a^2 \cos ^2 2 \theta$

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

$ \begin{aligned} 4 P_1^2+P_2^2 & =a^2\left(\sin ^2 2 \theta+\cos ^2 2 \theta\right) \\ & =a^2=k_1 P_1^2+k_2 P_2^2 \quad \text { [given] }\end{aligned} $

On comparing, we get

$ \begin{aligned} & \quad k_1=4, k_2=1 \\ & \therefore k_1+k_2=4+1=5 \end{aligned} $

Given the curve:

$ y = \left(\frac{x}{2024}\right)^k $

Let $ P(x_1, y_1) $ be a point on this curve. Then:

$ y_1 = \left(\frac{x_1}{2024}\right)^k $

Differentiating Eq. (i) with respect to $ x $, we have:

$ \frac{dy}{dx} = k\left(\frac{x}{2024}\right)^{k-1} \cdot \frac{1}{2024} $

Thus, the slope $ m $ at the point $ (x_1, y_1) $ is:

$ m = \frac{k}{2024}\left(\frac{x_1}{2024}\right)^{k-1} $

The length of the subnormal at point $ P $ is given by:

$ \left|y_1 m\right| = \left|\left(\frac{x_1}{2024}\right)^k \cdot \frac{k}{2024}\left(\frac{x_1}{2024}\right)^{k-1}\right| = \frac{k}{2024}\left(\frac{x_1}{2024}\right)^{2k-1} $

Since the length of the subnormal is constant, it must be independent of $ x_1 $. Thus, we have:

$ 2k - 1 = 0 $

Solving this equation gives:

$ k = \frac{1}{2} $

Given, curve $x^2+y^2=x+y \quad \ldots \text { (i) } $

and $ x^2+y^2=2 y \quad \ldots \text { (ii) } $

On differentiate Eq. (i) w.r.t. $x$, we get

$ \begin{array}{ll} & 2 x+2 y \frac{d y}{d x}=1+\frac{d y}{d x} \\ \Rightarrow \quad & (2 y-1) \frac{d y}{d x}=1-2 x \\ \Rightarrow \quad & m_1=\frac{d y}{d x}=\frac{1-2 x}{2 y-1} \end{array} $

Again, differentiate Eq. (ii) w.r.t. $x$, we get

$ \begin{array}{r} 2 x+2 y \frac{d y}{d x}=2 \frac{d y}{d x} \\ (y-1) \frac{d y}{d x}=-x \\ \Rightarrow \quad m_2=\frac{d y}{d x}=\frac{x}{1-y} \end{array} $

From Eqs. (i) and (ii), we get

$ x+y=2 y \Rightarrow x=y $

From Eq. (i), $2 x^2=2 x$

$ \begin{aligned} & \Rightarrow \quad x=0,1 \Rightarrow y=0,1 \\ & \text { At }(0,0) \end{aligned} $

$ \begin{aligned} & m_1=\frac{1-0}{0-1}=-1 \\ & m_2=\frac{0}{1-0}=0 \end{aligned} $

So, $\tan \theta=\left|\frac{m_1+m_2}{1-m_1 m_2}\right|=\left|\frac{-1+0}{1-0}\right|=1$

$\Rightarrow \quad \theta=\frac{\pi}{4}$

Given the function $ f(x) = (x-1)(x-2)(x-3) $, we know that it is a cubic polynomial. This ensures that it is continuous and differentiable in the interval $[0, 4]$.

Start by expanding the function:

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

This expands to:

$ f(x) = x^3 - 6x^2 + 11x - 6 $

Next, find the derivative $ f'(x) $:

$ f'(x) = 3x^2 - 12x + 11 $

Lagrange's Mean Value Theorem states that there exists some $ c $ in the interval $(0, 4)$ such that:

$ f'(c) = \frac{f(4) - f(0)}{4 - 0} $

Calculate $ f(4) $ and $ f(0) $:

$ f(4) = (4-1)(4-2)(4-3) = 3 \times 2 \times 1 = 6 $

$ f(0) = (0-1)(0-2)(0-3) = (-1) \times (-2) \times (-3) = -6 $

Thus,

$ f'(c) = \frac{6 - (-6)}{4} = \frac{12}{4} = 3 $

Set the derivative equal to this value:

$ 3c^2 - 12c + 11 = 3 $

Rearrange to form a quadratic equation:

$ 3c^2 - 12c + 8 = 0 $

Now, solve this quadratic equation using the quadratic formula:

$ c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

In this equation, $ a = 3 $, $ b = -12 $, and $ c = 8 $. Thus:

$ c = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 3 \cdot 8}}{6} $

$ c = \frac{12 \pm \sqrt{144 - 96}}{6} $

$ c = \frac{12 \pm \sqrt{48}}{6} $

$ c = \frac{12 \pm 4\sqrt{3}}{6} $

$ c = 2 \pm \frac{2}{\sqrt{3}} $

Since both roots $ c = 2 + \frac{2}{\sqrt{3}} $ and $ c = 2 - \frac{2}{\sqrt{3}} $ are valid within the interval $(0, 4)$, any of these can be a valid 'c' in the context of Lagrange’s Mean Value Theorem.

We start with the equation for the period $ T $:

$ T = 2 \pi \sqrt{\frac{L}{g}} $

Given that $ g $ is constant, the derivative of $ T $ with respect to $ L $ is:

$ \frac{dT}{dL} = \frac{\pi}{\sqrt{Lg}} $

The approximate error in $ T $ is:

$ dT = \frac{dT}{dL} \times \Delta L = \frac{\pi}{\sqrt{Lg}} \times \Delta L $

The relative error in $ T $ is:

$ \frac{dT}{T} = \frac{1}{2L} \Delta L $

According to the problem, the relative error in $ T $ is $ K $ times the percentage error in $ L $, which is expressed as:

$ \frac{\Delta T}{T} = K \frac{\Delta L}{L} \times 100 $

Substituting from the relative error expression, we have:

$ \frac{1}{2L} \Delta L = K \frac{\Delta L}{L} \times 100 $

Solving for $ K $, we find:

$ K = \frac{1}{200} $

Thus, the reciprocal of $ K $ is:

$ \frac{1}{K} = 200 $

We have,

$ \begin{aligned} & y^2=-2 x \text { and } x^2+y^2=8 \\ & \Rightarrow \quad x^2+2 x-8=0 \\ & x^2+4 x-2 x-8=0 \\ &(x+4)(x-2)=0 \\ & x=-4 \text { or } x=2 \end{aligned} $

If $x=2$, then $y= \pm 2$ and $x \neq-4$ because $y^2 \neq-8$

$\therefore$ Intersecting points are $(2,2)$ and $(2,-2)$

$\therefore$ Intersecting points are $(2,2)$ and $(2,-2)$

$ \begin{aligned} & 2 y \frac{d y}{d x}=2 \Rightarrow \frac{d y}{d x}=\frac{1}{y} \\ & \left.\frac{d y}{d x}\right|_{\text {at } x=(2,2)}=\frac{1}{2}=m_1(\text { let }) \end{aligned} $

Now, differentiate equation of circle

$ \begin{aligned} &\begin{aligned} & 2 x+2 y \frac{d y}{d x}=0 \\ & \frac{d y}{d x}=-\frac{x}{y} \\ &\left.\frac{d y}{d x}\right|_{\text {at }(2,2)}=-1 \text { (let) } \end{aligned}\\ &\text { Let } \theta \text { be the angle between tangent. }\\ &\begin{aligned} \tan \theta & =\left|\frac{\frac{1}{2}+1}{1-\frac{1}{2}}\right|=3 \\ \theta & =\tan ^{-1} 3 \end{aligned} \end{aligned} $

For the function $ f(x) = \sqrt{x^2 - 4} $, we need to determine the value of $ C $ such that Lagrange's Mean Value Theorem is satisfied on the interval $[2, 4]$.

First, differentiate the function $ f(x) = \sqrt{x^2 - 4} $:

$ f'(x) = \frac{1}{2\sqrt{x^2 - 4}} \times 2x = \frac{x}{\sqrt{x^2 - 4}} $

According to Lagrange's Mean Value Theorem, there exists a point $ c $ within the interval $(2, 4)$ such that:

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

where $ a = 2 $ and $ b = 4 $. Calculating the right-hand side:

$ f(4) = \sqrt{16 - 4} = \sqrt{12} $

$ f(2) = \sqrt{4 - 4} = 0 $

Thus:

$ \frac{f(b) - f(a)}{b - a} = \frac{\sqrt{12} - 0}{4 - 2} = \frac{\sqrt{12}}{2} = \sqrt{3} $

Equating both sides of the theorem:

$ \frac{c}{\sqrt{c^2 - 4}} = \sqrt{3} $

Squaring both sides, we get:

$ \frac{c^2}{c^2 - 4} = 3 \implies c^2 = 3(c^2 - 4) $

Simplifying:

$ c^2 = 3c^2 - 12 \implies 2c^2 = 12 \implies c^2 = 6 $

Thus,

$ c = \sqrt{6} $

We reject $ c = -\sqrt{6} $ since $ c $ must be within the interval $(2, 4)$.

Therefore, the value of $ C $ is $ \sqrt{6} $.

We are given two positive integers, $x$ and $y$, with the constraint

$ x + y = 20. $

We need to maximize the expression

$ x^3 y, $

and we’re told that the maximum value is $k$, attained when $x = \alpha$ and $y = \beta$. Then, we must find

$ \frac{k}{\alpha^2 \beta^2}. $

Let’s work through the steps:

Since $x + y = 20$, write

$ y = 20 - x. $

Substitute this into the expression to be maximized:

$ f(x) = x^3 (20 - x). $

Although $x$ is required to be an integer, we can first treat it as a continuous variable and find the critical point. Differentiate $f(x)$ with respect to $x$:

$ f'(x) = \frac{d}{dx}\left(20x^3 - x^4\right) = 60x^2 - 4x^3. $

Factor the derivative:

$ f'(x) = 4x^2 (15 - x). $

Set the derivative equal to zero:

$ 4x^2 (15 - x) = 0. $

This yields two possibilities:

$x = 0$ (not acceptable since $x$ is positive)

$15 - x = 0 \quad \Longrightarrow \quad x = 15.$

For $x = 15$, compute $y$:

$ y = 20 - 15 = 5. $

So, we have $\alpha = 15$ and $\beta = 5.$

(You can verify that among the integers, this pair indeed produces the maximum value for $x^3 y$.)

Compute the maximum value $k$:

$ k = 15^3 \cdot 5. $

Now, compute $\alpha^2 \beta^2$:

$ \alpha^2 \beta^2 = 15^2 \cdot 5^2. $

Form the ratio:

$ \frac{k}{\alpha^2 \beta^2} = \frac{15^3 \cdot 5}{15^2 \cdot 5^2}. $

Simplify the ratio by canceling common factors:

$ \frac{15^3 \cdot 5}{15^2 \cdot 5^2} = \frac{15}{5} = 3. $

Lastly, check the provided options. We need an expression that equals 3 when $\alpha=15$ and $\beta=5$.

Option A: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{15}{5} + \frac{5}{15} = 3 + \frac{1}{3} = \frac{10}{3} \neq 3.$

Option B: $\frac{\alpha}{\beta} - \frac{\beta}{\alpha} = \frac{15}{5} - \frac{5}{15} = 3 - \frac{1}{3} = \frac{8}{3} \neq 3.$

Option C: $\frac{\alpha}{\beta} = \frac{15}{5} = 3.$

Option D: $\frac{\alpha + \beta}{\alpha \beta} = \frac{15 + 5}{15 \cdot 5} = \frac{20}{75} = \frac{4}{15} \neq 3.$

Thus, the correct option is Option C, which is

$ \frac{\alpha}{\beta}. $

So, the answer is: $\frac{k}{\alpha^2 \beta^2} = \frac{\alpha}{\beta}.$

If $ y = \left(1 + \alpha + \alpha^2 + \ldots\right) e^{\eta x} $, where $\alpha$ and $n$ are constants, the expression for $y$ can be simplified. The series $1 + \alpha + \alpha^2 + \ldots$ is an infinite geometric progression (GP) which sums to $\frac{1}{1 - \alpha}$, assuming $|\alpha| < 1$.

Thus, $ y = \left(\frac{1}{1-\alpha}\right) e^{nx} $.

The derivative of $y$ with respect to $x$ is:

$ \frac{d y}{d x} = \left(\frac{1}{1-\alpha}\right) e^{nx} \cdot n $

The approximate error in $y$ is given by:

$ d y = \frac{d y}{d x} \times \Delta x = \left(\frac{1}{1-\alpha}\right) e^{nx} \cdot n \cdot \Delta x $

To find the relative error in $y$, we compute:

$ \frac{d y}{y} = \frac{\left(\frac{1}{1-\alpha}\right) e^{nx} \cdot n}{\left(\frac{1}{1-\alpha}\right) e^{nx}} = n \Delta x $

This shows that the relative error in $y$ is $n \times \Delta x$.

To solve for the values of $ a $ and $ b $ given the equation of a tangent to the curve, we start with the equation:

$ y^2 = ax^3 + b $

The derivative, $ \frac{dy}{dx} $, which represents the slope of the tangent, is calculated as follows:

$ 2y \frac{dy}{dx} = 3ax^2 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{3ax^2}{2y} $

At the point $(2,3)$, the slope $\frac{dy}{dx}$ is:

$ \left. \frac{dy}{dx} \right|_{(2,3)} = \frac{3a \times 4}{2 \times 3} = 2a $

We are given that the equation of the tangent line at $(2,3)$ is:

$ y = 4x - 5 $

Since the slope of this line is 4, it follows that:

$ 2a = 4 \quad \Rightarrow \quad a = 2 $

Next, substituting the point $(2,3)$ into the original curve equation:

$ 3^2 = a \cdot 2^3 + b \quad \Rightarrow \quad 9 = 8a + b $

Substituting $ a = 2 $ into this equation:

$ 9 = 8 \times 2 + b \quad \Rightarrow \quad 9 = 16 + b \quad \Rightarrow \quad b = -7 $

Finally, we calculate $ a^2 + b^2 $:

$ a^2 + b^2 = 2^2 + (-7)^2 = 4 + 49 = 53 $

Thus, the value of $ a^2 + b^2 $ is 53.

We have $f(x)=x(x+3) e^{\frac{-x}{2}}$

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

$ \begin{aligned} & =e^{\frac{-x}{2}}\left[\frac{-x^2+x+6}{2}\right] \\ f^{\prime}(c) & =e^{\frac{-c}{2}}\left[\frac{-c^2+c+6}{2}\right]=0 \end{aligned} $

[ $\because$ Rolle's Theorem is applicable]

$ \begin{aligned} & {[\because \text { Rolle's Theorem is applicable }} \\ & \Rightarrow-c^2+c+6=0 \quad\left[\because e^{\frac{-c}{2}} \neq 0\right] \\ & \Rightarrow(3-c)(2+c)=0 \Rightarrow c=3 \text { or } c=-2 \end{aligned} $

$ \text { But } 3 \notin(-3,0) \text { so, } c=-2 $

To determine the least possible value of $ f(2024) $, given that $ f(x) $ is differentiable for all $ x \in [0, 2024] $ with $ f(0) = -2 $ and $ f'(x) \geq 5 $, we can consider the function $ f(x) = 5x - 2 $.

Initial Condition: $ f(0) = -2 $.

Derivative Condition: We require $ f'(x) \geq 5 $ for all $ x $. The simplest function meeting this requirement is $ f(x) = 5x - 2 $, where the derivative $ f'(x) = 5 $ is a constant function satisfying the derivative condition.

Evaluating at $ x = 2024 $:

$ f(2024) = 5 \times 2024 - 2 = 10120 - 2 = 10118 $

Thus, the least possible value of $ f(2024) $ is $ 10118 $.

We are given that a point is moving along the curve described by the equation $ y = x^3 - 3x^2 + 2x - 1 $ and that the $ y $-coordinate of the point is increasing at a rate of 6 units per second. We need to determine the rate of change of the $ x $-coordinate of the point when the point is at $(2, -1)$.

Steps to Find the Rate of Change of the $ x $-coordinate:

Differentiating the Curve Equation:

Given $ y = x^3 - 3x^2 + 2x - 1 $, differentiate $ y $ with respect to $ t $ to express $\frac{dy}{dt}$:

$ \frac{dy}{dt} = \frac{d}{dt}(x^3 - 3x^2 + 2x - 1) = (3x^2 - 6x + 2) \frac{dx}{dt} $

Using Given Information:

We know $\frac{dy}{dt} = 6$. Substitute this value into the differentiated equation:

$ (3x^2 - 6x + 2) \frac{dx}{dt} = 6 $

Solving for $\frac{dx}{dt}$:

Rearrange the equation to solve for $\frac{dx}{dt}$:

$ \frac{dx}{dt} = \frac{6}{3x^2 - 6x + 2} $

Substitute the Point $(2, -1)$:

Substitute $ x = 2 $ into the equation to find $\frac{dx}{dt}$ at this specific point:

$ \frac{dx}{dt} = \frac{6}{3(2)^2 - 6(2) + 2} = \frac{6}{3 \times 4 - 6 \times 2 + 2} $

Simplify the Expression:

$ \frac{dx}{dt} = \frac{6}{12 - 12 + 2} = \frac{6}{2} = 3 $

Therefore, the rate of change of the $x$-coordinate of the point when at $(2, -1)$ is $3$ units per second.

We have.

$ \begin{aligned} & f(x)=x^3+2 a x^2+3(a+1) x+5 \\ & f'(x)=3 x^2-4 a x+3 a+3 \end{aligned} $

As the function is increasing in its entire demain

$ \begin{aligned} & f(x)>0 \\ & 3 x^2-4 x+3 x+3>0 \end{aligned} $

As cellicient of $x^2>0$, then

$ \begin{aligned} & D<0 \\ & 16 a^2-36 a-36<0 \\ & 4 a^2-9 a-9<0 \\ & (4 a+3)(a-3)<0 \\ & \Rightarrow \quad a \in \left(\frac{-3}{4}, 3\right) \end{aligned} $

,$3 f(\cos x)+2 f(\sin x)=5 x\quad \text{... (i)}$

$3 f(\sin x)+2 f(\cos x)=5\left(\frac{\pi}{2}-x\right)\quad \text{... (ii)}$

Solving Eqs. (i) and (ii) simultaneously, we get

$\begin{aligned} & f(\cos x)=5 x-\pi \\ \Rightarrow & -\sin x f^{\prime}(\cos x)=5 \\ \Rightarrow & f^{\prime}(\cos x)=\frac{-5}{\sin x} \end{aligned}$

Similarly,

$\begin{aligned} f^{\prime}(\sin x) & =\frac{-5}{\cos x} \\ & =f^{\prime}(\cos x)+f^{\prime}(\sin x) \\ & =\frac{-5}{\sin x}-\frac{5}{\cos x} \end{aligned}$

Given curve $3 y=6 x-5 x^3$

$\Rightarrow \quad \frac{d y}{d x}=2-5 x^2$

Slope of normal $=\frac{-1}{2-5 x^2}$

Equation of normal with slope $\frac{-1}{2-5 x^2}$ passing through origin $(0,0)$.

$\frac{\left(y-y_1\right)}{x-x_1}=m$

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

Since, $(h, k)$ lies on normal

$\begin{aligned} & \frac{k}{h}=\frac{-1}{2-5 h^2} \\ \Rightarrow & \frac{k}{h}=\frac{1}{5 h^2-2} \quad \text{... (i)} \end{aligned}$

Since, $(h, k)$ lies on curve

$\begin{aligned} 3 k & =6 h-5 h^3 \\ \Rightarrow \quad 3 k & =h\left(6-5 h^2\right) \\ \Rightarrow \quad \frac{k}{h} & =\frac{6-5 h^2}{3}\quad \ \text{... (ii)} \end{aligned}$

On solving Eqs. (i) and (ii),

$\begin{array}{ll} & \frac{1}{5 h^2-2}=\frac{6-5 h^2}{3} \\ \Rightarrow & 3=\left(6-5 h^2\right)\left(5 h^2-2\right) \\ \Rightarrow & 3=30 h^2-12-25 h^4+10 h^2 \\ \Rightarrow & 15=-25 h^4+40 h^2 \\ \Rightarrow & 5 h^4-8 h^2+3=0 \end{array}$

For simplicity, let $h^2=z$

The equation becomes $5 z^2-8 z+3=0$

$(5 z-3)(z-1)=0$

Now, $z=\frac{3}{5}, 1$

Putting $h^2=z$

$\Rightarrow \quad h^2=\frac{3}{5} \Rightarrow h= \pm \sqrt{\frac{3}{5}}$

It is not in option.

or $(z-1)=0$

$z=1$

Putting $h^2=z$

$\Rightarrow \quad \begin{aligned} h^2 & =1 \\ h & = \pm 1 \end{aligned}$

But $h=-1$ is not in option.

$\therefore h=1$ is correct

Slope of line joining the points $(0,3)$ and $(5,-2) \text { is } \frac{3+2}{0-5}=-1$

This is equal to the slope of tangent on the curve and that is given by

$\begin{aligned} & \frac{d y}{d x}=\frac{-c}{(x+1)^2} \\ \Rightarrow & \frac{d y}{d x}=-1 \\ \Rightarrow & c=(x+1)^2\quad \text{... (i)} \end{aligned}$

Equation of line joining the points $(0,3)$ and $(5,2)$ is given by $(y-3)=-1(x-0)$.

Solving equation of tangent and the curve for point of intersection

$\frac{c}{x+1}+x=3\quad \text{... (ii)}$

Solving Eqs. (i) and (ii),

$x=1$

On putting this in Eq. (ii), we get $c=4$

Given, $y=\frac{b^2}{a-x}+\frac{a^2}{x}$

$\begin{aligned} & \therefore \quad \frac{d y}{d x}=\frac{-b^2(-1)}{(a-x)^2}+\left(\frac{-a^2}{x^2}\right) \Rightarrow \frac{d y}{d x}=\frac{b^2}{(a-x)^2}-\frac{a^2}{x^2} \\ & \because \quad \frac{d y}{d x}=0, x=\frac{a^2}{a \pm b} \\ & \therefore \frac{d^2 y}{d x^2}=\frac{2 a^2}{x^3}+\frac{2 b^2}{(a-x)^3} \end{aligned}$

So, $\left.\frac{d^2 y}{d x^2}\right|_{x=\frac{a^2}{a+b}}>0$

Therefore, $y$ is minimum at $x=\frac{a^2}{a+b}$

$\begin{aligned} y_{\min } & =\frac{b^2}{a-\left(\frac{a^2}{a+b}\right)}+\frac{a^2}{(a)^2}(a+b) \\ & =\frac{b^2(a+b)}{a^2+a b-a^2}+(a+b)=(a+b)\left[\frac{b}{a}+1\right] \\ & =\frac{(a+b)^2}{a} \end{aligned}$

Let $(x, y)$ be on the parabola $y=x^2+4 x+3$ which is closet to the line $y=3 x+2$

Perpendicular distance between a point $\left(x_1, y_1\right)$ and a line $(a x+b y+c=0)$

$D=\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|$

$\left[\therefore\right.$ line $3 x-y+2=0$ at point $\left.\left(x_1, y_1\right)\right]$

$\begin{aligned} D & =\left|\frac{3 x-y+2}{\sqrt{(3)^2+(-1)^2}}\right|=\frac{\left|3 x-\left(x^2+4 x+3\right)+2\right|}{\sqrt{9+1}} \\ \Rightarrow \quad D & =\frac{\left|-x^2-x-1\right|}{\sqrt{10}} \\ \Rightarrow \quad D & =\frac{x^2+x+1}{\sqrt{10}}=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{10}} \end{aligned}$

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

$\begin{aligned} & \frac{d D}{d x}=\frac{\left(x+\frac{1}{2}\right)^2}{\sqrt{10}}+\frac{\frac{3}{4}}{\sqrt{10}} \\ \Rightarrow & \frac{d D}{d x}=0, x=\frac{-1}{2} \Rightarrow y=x^2+4 x+3 \\ \Rightarrow & y=\left(\frac{-1}{2}\right)^2+(4)\left(\frac{-1}{2}\right)+3 \Rightarrow y=\frac{1}{4}+3-2=0 \\ \Rightarrow & y=\frac{1+12-8}{4} \Rightarrow y=\frac{5}{4} \\ & \frac{d D}{d x}=\frac{2\left(x+\frac{1}{2}\right)}{\sqrt{10}} \Rightarrow \frac{d^2 D}{d x^2}>0 \end{aligned}$

Hence, minimum at $\left(\frac{-1}{2}, \frac{5}{4}\right)$.

Given, equation of cuve $y^2-2 x^3-4 y+8=0$

$\Rightarrow 2 y y^{\prime}-6 x^2-4 y^{\prime}=0 \Rightarrow y^{\prime}=\frac{3 x^2}{y-2}$

Since, $p(x, y)$ is a point on this curve.

$\therefore \quad \frac{y-2}{x-1}=\frac{3 x^2}{y-2}$

(for tangent lines passing through $(1,2)$)

$\begin{aligned} & \Rightarrow \quad(y-2)^2=3 x^2(x-1) \\ & \therefore \quad y^2-2 x^3-4 y+8=0 \\ & \Rightarrow \quad(y-2)^2-2 x^3+4=0 \\ & \Rightarrow \quad 3 x^2(x-1)-2 x^3+4=0 \\ & \Rightarrow \quad 3 x^3-3 x^2-2 x^3+4=0 \\ & \Rightarrow \quad x^3-3 x^2+4=0 \\ & \Rightarrow \quad(x+1)(x-2)=0 \\ & \therefore \quad x=-1,2 \end{aligned}$

For $x=-1$, there is no real value of $y$.

For $x=2, y=2 \pm 2 \sqrt{3}$

$\therefore \quad y^{\prime}=\frac{3 x^2}{y-2}= \pm 2 \sqrt{3}$

Thus, tangent lines drawn from the point $(1,2)$ is $y=2 \pm 2 \sqrt{3}(x-1)$.

$\therefore$ Two tangents are possible.

The given straight line is $x \cos \alpha+y \sin \alpha=p$ and given curve is $\left(\frac{x}{a}\right)^n+\left(\frac{y}{b}\right)^n=2$

On differentiating, $\frac{n x^{n-1}}{a^n}+\frac{n y^{n-1}}{b^n} y^{\prime}=0$

At $(a, b)$, then $y^1=-\frac{b}{a}$

Thus, equation of tangent at $(a, b)$ is

$y-b=-\frac{b}{a}(x-a)$

$\begin{aligned} & \Rightarrow \quad a y-a b+b x=a b \quad \text{... (i)}\\ & \Rightarrow \quad b x+a y=2 a b \quad \text{... (ii)} \end{aligned}$

Comparing Eqs. (i) and (ii) on both sides, we get

$\begin{aligned} & \quad \frac{b}{\cos \alpha}=\frac{a}{\sin \alpha}=\frac{2 a b}{p} \\ & \Rightarrow \quad \cos \alpha=\frac{p}{2 a} \text { and } \sin \alpha=\frac{p}{2 b} \\ & \therefore \text { Thus, } \sin ^2 \alpha+\cos ^2 \alpha=1 \\ & \Rightarrow \quad \frac{p^2}{4 a^2}+\frac{p^2}{4 b^2}=1 \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{4}{p^2}=\frac{k}{p^2} \end{aligned}$

Hence, the value of $k$ is 4.

Given, $y^2=4 a x$ and $x y=c^2$ cut orthogonally.

Let they intersect at $\left(x_1, y_1\right)$.

$\begin{aligned} \therefore \quad & y^2=4 a x \Rightarrow 2 y \frac{d y}{d x}=4 a \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{2 a}{y} \end{aligned}$

$\therefore \quad \frac{d y}{d x}\left(x_1, y_1\right)=\frac{2 a}{y_1} \quad \text{... (i)}$

$\begin{aligned} \text { and } & & x y & =c^2 \\ \Rightarrow & & y+x \frac{d y}{d x} & =0 \\ \Rightarrow & & \frac{d y}{d x}\left(x_1, y_1\right) & =-\frac{y_1}{x_1} \quad \text{... (ii)} \end{aligned}$

From Eqs (i) and (ii), we get

$\begin{aligned} \frac{2 a}{y_1} \times-\frac{y_1}{x_1} & =-1 \\ \Rightarrow \quad x_1 & =2 a \end{aligned}$

From $y^2=4 a x$

$\begin{aligned} \Rightarrow \quad & y_1 =\sqrt{4 a \cdot 2 a}=2 \sqrt{2} a \Rightarrow x_1 y_1=c^2 \\ \Rightarrow \quad & 2 a(2 \sqrt{2} a) =c^2 \Rightarrow \frac{c^2}{a^2}=4 \sqrt{2} \\ \Rightarrow \quad & c^4 =32 a^4 \end{aligned}$

Let $h$ be the height and $r$ be the radius. The volume of cylinder is $\pi r^2 h$.

The surface area is $2 \pi r^2+2 \pi r h$

$\begin{aligned} & =2 \pi\left(r^2+\frac{V}{r \pi}\right) \\ & \Rightarrow \frac{d}{d r}(\text { surface })=2 \pi\left(2 r-\frac{V}{\pi r^2}\right) \end{aligned}$

$\text { Also } \frac{d}{d r}(\text { surface })=0$

$ \begin{aligned} & \Rightarrow \quad 2 \pi\left(2 r-\frac{V}{\pi r^2}\right)=0 \\ & \Rightarrow \quad r^3=\frac{V}{2 \pi} \Rightarrow r=\left(\frac{V}{2 \pi}\right)^{1 / 3} \\ & \therefore \quad h=(2)^{2 / 3} \cdot\left(\frac{V}{\pi}\right)^{1 / 3} \Rightarrow h: r=2: 1 \end{aligned}$

Given that, two particles $P$ and $Q$ located at the points with coordinate $P\left(t, t^3-16 t-3\right), Q\left(t+1, t^3-6 t-6\right)$ are moving in a plane. Thus,

$\begin{aligned} P Q & =\sqrt{(t+1-t)^2+\left(t^3-6 t-6-t^3+16 t+3\right)^2} \\ & =\sqrt{1+(10 t-3)^2} \end{aligned}$

$\therefore(P Q)^2=1+(10 t-3)^2 \geq 1$

Thus, minimum value of $P Q$ is 1.

$x^3-2 x^2 y^2+5 x+y-5=0$

On differentiating w.r.t. $x$, we get $3 x^2-4 x y^2-4 x^2 y \cdot y^{\prime}+5+y^{\prime}-0=0\quad \text{.... (i)}$

On putting $x=1, y=1$,

$\begin{aligned} & 3-4-4 y^{\prime}+5+y^{\prime}=0 \\ & \Rightarrow \quad y^{\prime}=\frac{4}{3} \\ & \text { or } \quad y^{\prime}(1)=\frac{4}{3} \\ & \end{aligned}$

On differentiating Eq. (i) w.r.t. $x$,

$6 x-4 y^2-8 x y y^{\prime}-8 x y y^{\prime}-4 x^2 y^{\prime} \cdot y^{\prime}-4 x^2 y y^{\prime \prime}+y^{\prime \prime}=0$

On putting $x=1, y=1$ and $y^{\prime}=\frac{4}{3}$ to obtain $y^{\prime \prime}(1)$.

$\begin{array}{ll} \Rightarrow & 6-4-8 \times \frac{4}{3}-8 \times \frac{4}{3}-4 \times \frac{4}{3} \times \frac{4}{3}-4 \times y^{\prime \prime}+y^{\prime \prime}=0 \\ \Rightarrow & 2-\frac{64}{3}-\frac{64}{9}-3 y^{\prime \prime}=0 \\ \Rightarrow & \frac{-238}{9}=3 y^{\prime \prime} \\ \Rightarrow & y^{\prime \prime}=-\frac{238}{27} \\ \text { or } & y^{\prime \prime}(1)=-\frac{238}{27} \end{array}$

$\begin{aligned} y & =x^3-3 x^2-8 x-4 \\ y & =3 x^2+7 x+4 \end{aligned}$

$\because$ Curves touch each other.

$\begin{aligned} & \therefore \quad x^3-3 x^2-8 x-4=3 x^2+7 x+4 \\ & \Rightarrow \quad x^3-6 x^2-15 x-8=0 \end{aligned}$

On putting $x=-1$

$\begin{aligned} \mathrm{LHS} & =-1-6+15-8 \\ & =-15+15=0=\text { RHS } \end{aligned}$

On putting $x=-1$ in any curve equation,

$y=-1-3+8-4=0$

$\therefore(-1,0)$ is the intersection point.

Equation of tangent,

$\begin{aligned} & y-y_1=m\left(x-x_1\right) \quad \text{... (i)}\\ & y=x^3-3 x^2-8 x-4 \\ & y^{\prime}=3 x^2-6 x-8 \end{aligned}$

$y^{\prime} |_{(-1,0)}=3+6-8=1 \Rightarrow m=1$

Put the values into Eq. (i),

$\begin{array}{rlrl} & y-0=1(x+1) \\ \Rightarrow & y=x+1 \\ \Rightarrow & x-y+1=0 \end{array}$

$\begin{aligned} & f(x)=\frac{x}{1+4 x+x^2} \\ & f^{\prime}(x)=\frac{\left(1+4 x+x^2\right)(1)-x(4+2 x)}{\left(1+4 x+x^2\right)^2}=\frac{1-x^2}{\left(1+4 x+x^2\right)^2} \\ & \text { Put } f^{\prime}(x)=0 \\ & \Rightarrow \quad x= \pm 1 \end{aligned}$

At $x=-1$, it gives minimum value.

At $x=1$, it gives maximum value.

Put $x=1$ in $f(x)$, we get

$f(1)=\frac{1}{1+4+1}=\frac{1}{6}$

$\begin{aligned} f(x) & =x+\frac{4}{x+2} \\ f^{\prime}(x) & =1+4\left(-\frac{1}{(x+2)^2}\right) \\ f^{\prime}(x) & =\frac{(x+2)^2-4}{(x+2)^2} \end{aligned}$

$\begin{array}{lr} \text { Put } & f^{\prime}(x)=0 \\ \Rightarrow & (x+2)^2-4=0 \\ \Rightarrow & (x+2)^2=4 \\ \Rightarrow & x+2= \pm 2 \end{array}$

For positive value, $x+2=2 \Rightarrow x=0$

For negative value, $x+2=-2 \Rightarrow x=-4$

$f(0)=0+\frac{4}{0+2}=\frac{4}{2}=2$

and $f(-4)=-4+\frac{4}{-4+2}=-4-2=-6$

As per the option given,

Minimum value $=2$

$\begin{aligned} & f(x)=a x^3+b x^2+c x+d \\ & f^{\prime}(x)=3 a x^2+2 b x+c \end{aligned}$

For no extrema, the roots of $f^{\prime}(x)$ will not be real i.e. $f^{\prime}(x)$ has imaginary roots.

$\begin{aligned} & \Rightarrow\quad(2 b)^2-4(3 a)(c)<0 \\ & \Rightarrow \quad 4 b^2<4(3 a c) \\ & \Rightarrow \quad b^2<3 a c \end{aligned} $

Length of subnormal $=y m$

$\begin{aligned} & \Rightarrow \quad y m=x-1 \\ & \Rightarrow \quad y \frac{d y}{d x}=x-1 \Rightarrow y d y=(x-1) d x \\ & \Rightarrow \int y d y=\int(x-1) d x \\ & \Rightarrow \quad \frac{y^2}{2}=\frac{(x-1)^2}{2}+c \quad \text{... (i)} \end{aligned}$

$\because$ Curve passes through $(1,2)$.

$\therefore$ On putting $x=1$ and $y=2$ into Eq. (i), we get

$\frac{4}{2}=\frac{0}{2}+c \Rightarrow c=2$

From Eq. (i), $y^2=(x-1)^2+4$

$\begin{array}{ll} \Rightarrow & \frac{y^2}{1}-\frac{(x-1)^2}{1}=4 \\ \Rightarrow & \frac{y^2}{2^2}-\frac{(x-1)^2}{2^2}=1 \\ \Rightarrow & \frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1 \end{array}$

where, $b=a$

Vertices are $(0, \pm b) \equiv(0, \pm 2)$.

Radius of ball = 10 cm

Given, $\frac{dV}{dt}=50$ cm$^3$/min

Let a be the thickness of ice.

Image

$\begin{aligned} & \text { Volume of ice part }=\frac{4}{3} \pi\left(R^3-r^3\right) \\ & V=\frac{4}{3} \pi\left[(10+a)^3-10^3\right] \\ & =\frac{4}{3} \pi\left[a^3+30 a^2+300 a\right] \\ & \frac{d V}{d t}=\frac{4}{3} \pi \frac{d}{d t}\left(a^3+30 a^2+300 a\right) \\ & 50=\frac{4}{3} \pi\left(3 a^2 \frac{d a}{d t}+60 a \frac{d a}{d t}+300 \frac{d a}{d t}\right) \\ & \Rightarrow \quad 50 \times \frac{3}{4 \pi}=\frac{d a}{d t}\left(3 a^2+60 a+300\right) \\ & \therefore \quad \frac{d a}{d t}=\frac{75}{2 \pi}\left(3 a^2+60 a+300\right) \\ \end{aligned}$

When, $\quad a=15$

$d a / d t=1 / 50 \pi$

Let $f(x, y)=2 x+3 y$,

Given,

$\begin{aligned} \text { Given, } \quad x y & =6 \\ \Rightarrow \quad y & =\frac{6}{x} \Rightarrow f(x)=2 x+3 \times \frac{6}{x} \\ & =\frac{2 x^2+18}{x}=2 x+\frac{18}{x} \end{aligned}$

Now, $f^{\prime}(x)=2-\frac{18}{x^2}$

Equate $f^{\prime}(x)=0$

$\begin{aligned} & \Rightarrow \quad 2-\frac{18}{x^2}=0 \Rightarrow \frac{18}{x^2}=2 \\ & \Rightarrow \quad x^2=9 \Rightarrow x= \pm 3 \\ & \Rightarrow \quad f^{\prime \prime}(x)=\frac{2 \times 18}{x^3}=\frac{36}{x^3} \Rightarrow f^{\prime \prime}(3) > 0 \\ \end{aligned}$

$\Rightarrow \quad x=3$ is minima.

$\therefore$ Minimum value $f(3)=2(3)+\frac{18}{3}=6+6=12$

Volume $=\frac{4}{3} \pi r^3$, given $\frac{d V}{d t}=30 \mathrm{~cm}^3 / \mathrm{min}$.

$\Rightarrow \quad \frac{d V}{d t}=\frac{4}{3} \pi \cdot 3 r^2 \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t} \Rightarrow 30=4 \pi r^2 \frac{d r}{d t}$/p>

or $\quad \frac{d r}{d t}=\frac{15}{2 \pi r^2}$ Surface Area $=4 \pi r^2$ ..... (i)

Then, $\frac{d S}{d t}=4 \pi \cdot 2 r \frac{d r}{d t}=8 \pi r \frac{d r}{d t}$

Use Eq. (i), $\frac{d S}{d t}=8 \pi r \cdot\left(\frac{15}{2 \pi r^2}\right)=\frac{60}{r}$

When, $r=6 \mathrm{~cm}$

$\left(\frac{d S}{d t}\right)_{r=6}=\frac{60}{6}=10 \mathrm{~cm}^2 / \mathrm{min} .$

$\begin{aligned} & g(x)=\frac{1}{6} f\left(3 x^2-1\right)+\frac{1}{2} f\left(1-x^2\right) \\ & g^{\prime}(x)=\frac{1}{6} f^{\prime}\left(3 x^2-1\right)(6 x)+\frac{1}{2} f^{\prime}(1-x)^2(-2 x) \\ & =x\left(f^{\prime}\left(3 x^2 \quad 1\right)+(-1) f^{\prime}\left(1-x^2\right)\right) \end{aligned}$

$g(x)$ is increasing, when $g^{\prime}(x) > 0$

Case 1 $f^{\prime}\left(3 x^2-1\right)-f^{\prime}\left(1-x^2\right) > 0$ and $x > 0$ ....... (a)

$\Rightarrow \quad f^{\prime}\left(3 x^2-1\right)>f^{\prime}\left(1-x^2\right)$ ....... (i)

Given that, $f^{\prime \prime}(x) > 0$

$\begin{array}{ll} \Rightarrow & f(x) \text { is increasing. } \\ \Rightarrow & f^{\prime}(x) > 0\end{array}$

From Eq. (i), $3 x^2-1 > 1-x^2$

$\begin{gathered} 4 x^2 > 2 \Rightarrow x^2>\frac{1}{2} \\ \Rightarrow \quad x < -\frac{1}{\sqrt{2}} \text { and } x > \frac{1}{\sqrt{2}} \\ \therefore \quad x < \frac{-1}{\sqrt{2}} \text { and } x > \frac{1}{\sqrt{2}} \\ \Rightarrow x \in\left(-\infty,-\frac{1}{\sqrt{2}}\right) \cup\left(\frac{1}{\sqrt{2}}, \infty\right) \quad \text{.......(b)} \end{gathered}$

From (a) and (b)

$x \in\left(\frac{1}{\sqrt{2}}, \infty\right) $

Case 2 When $x < 0$

$\begin{aligned} & \text { and } f^{\prime}\left(3 x^2-1\right) < f^{\prime}\left(1-x^2\right) \\ & \Rightarrow 3 x^2-1 < 1-x^2 \\ & \qquad x^2<\frac{1}{2} \Rightarrow x \in\left(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \\ & \because \quad x < 0 \Rightarrow x \in\left(-\frac{1}{\sqrt{2}}, 0\right)\end{aligned}$

Including case- 1 and case-2.

$x \in\left(-\frac{1}{\sqrt{2}}, 0\right) \cup\left(\frac{1}{\sqrt{2}}, \infty\right)$

Given,

$\begin{aligned} & f(x)=2 x^3-9 a x^2+12 a^2 x+1 \\ & f^{\prime}(x)=6 x^2-18 a x+12 a^2 \end{aligned}$

Equate $f^{\prime}(x)=0$

$\begin{aligned} & \Rightarrow \quad 6 x^2-18 a x+12 a^2=0 \\ & \text { or } \quad x^2-3 a x+2 a^2=0 \\ & \Rightarrow \quad(x-a)(x-2 a)=0 \\ & \Rightarrow \quad x=a, 2 a \\ & \text { Now, } \quad f^{\prime \prime}(x)=12 x-18 a \\ & \Rightarrow \quad f^{\prime \prime}(a)=12 a-18 a < 0 \\ & \Rightarrow \quad f^{\prime \prime}(2 a)=24-18 a > 0 \\ \end{aligned}$

$\therefore$ Minimum value attained at $x=2 a$

Maximum value attained at $x=a$

$\begin{aligned} & \therefore \quad p=a \text { and } q=2 a \Rightarrow p^2=q \text { gives, } a^2=2 a \\ & \Rightarrow \quad a(a-2)=0 \Rightarrow a=0 \text { and } a=2 \\ & \text { Since, } a \neq 0, a=2 .\end{aligned}$

$\begin{aligned} 2 y \frac{d y}{d x} & =4 a x^3 \\ (d y / d x)_{(3,6)} & =(2 a)\left(x^3 / y\right)_{3,6}=(2 a)(27 / 6)=9 a \quad {[\because \text { slope of } y=4 x-6 \text { is 4] }}\\ 9 a & =4 \\ \quad & \\ \Rightarrow \quad a & =\frac{4}{9} \Rightarrow 36=\frac{4}{9}(81)+b \Rightarrow b=0\end{aligned}$

$2\alpha+\beta=8$

$f'(x)=6x^2-18ax+12a^2$

Critical points

$f'(x)=0\Rightarrow 6x^2-18ax+12a^2=0$

$\Rightarrow (x-a)(x-2a)=0\Rightarrow x=a,x=2a$

Now, $f''(x)=12x-18a$

at $x=a$

$f''(a)=-6a < 0$,

as $a > 0$

$x=a$ is point of maxima

at $x=2a$

$f''(2a)=6a > 0$

$x=2a$ is point of minima

$\therefore \alpha=a,\beta=2a$

$\because 2\alpha+\beta=8$

$2a+2a=8$

$\Rightarrow a=2$

$A=4\pi r^2$

$dA=8\pi r\,dr=8\pi9 \,. (0.03) =2.16 \pi$ cm$^2$

$\begin{aligned} V & =\frac{\pi}{3}\left(r^2 h\right) \\ d V & =\frac{\pi}{3}\left(2 r h d r+r^2 d h\right) \quad[Q=10 \mathrm{~cm} \text { and } h=20 \mathrm{~cm}] \\ 0 & =200 \cdot 1+25 d h \Rightarrow d h=-8 \mathrm{~cm} / \mathrm{s}\end{aligned}$

Let $x=2, x+\Delta x=1.99$

$\Rightarrow \quad \Delta x=-0.01$

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

$\begin{aligned} & \frac{d y}{d x}=3 x^2-4 \\ & \left(\frac{d y}{d x}\right)_{x=2}=3.2^2-4=8 \\ & \text { Now, } \Delta y=\left(\frac{d y}{d x}\right)_{x=2} \Delta x \\ & =8 \times(-0.01)=-0.08 \\ \end{aligned}$

Approximate change in $y=-0.08$