Application of Derivatives

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

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

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

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

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

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

So, length of rectangle portion is 125 .

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

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

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

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

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

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

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

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

Equation of the Curve:

The given equation of the curve is:

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

Differentiate the Curve:

Differentiate the curve equation with respect to $ x $:

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

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

Find Slope of the Tangent Line:

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

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

The equation of the line, therefore, is:

$ y = -x + 3 $

Set the Slopes Equal:

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

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

Simplifying gives:

$ c = (x+1)^2 $

Equation of the Curve at Tangency Point:

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

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

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

$ -x + 3 = x + 1 $

Solve for $ x $:

$ 2x = 2 $

$ x = 1 $

Calculate $ c $:

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

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

Thus, the value of $ c $ is 4.

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

The formula for the percentage error in the radius is:

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

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

$ A = \pi r^2 $

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

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

This can be expanded to:

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

Differentiating both sides with respect to the logarithmic variables:

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

Thus:

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

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

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

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

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

Simplifying gives:

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

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

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} $

Given the function:

$ f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x} $

We want to find the maximum value of this expression for $0 \leq x \leq \pi$.

Step 1: Rewrite the expression

Notice that we can rewrite the expression as:

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

Step 2: Find the first derivative

Now, let's find the derivative of this expression with respect to $x$:

$ f'(x) = 1 - 2\cos{2x} + \cos{3x} $

Step 3: Find the critical points

To find the critical points, we set $f'(x) = 0$:

$ 1 - 2\cos{2x} + \cos{3x} = 0 $

Step 4: Solve for x

This equation can be rewritten as:

$ (2\cos{x} + \sqrt{3})(2\cos{x} - \sqrt{3})(\cos{x} - 1) = 0 $

We get three possible solutions for $x$:

$ \cos{x} = \frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 1 $

Which gives us: $ x = \frac{5\pi}{6}, \frac{\pi}{6}, 0 $

Step 5: Find the second derivative

Now, let's find the second derivative of the expression:

$ f''(x) = 4\sin{2x} - 3\sin{3x} $

Step 6: Analyze the critical points Evaluate the second derivative at the critical points:

1. $f''\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \sqrt{3} < 0$, indicating a local maximum.
2. $f''\left(\frac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} > 0$, indicating a local minimum.
3. $f''(0) = 0$, inconclusive.

Step 7: Evaluate the function at the local maximum point and the boundary points Since we are looking for the maximum value of $f(x)$, we can now evaluate the function at the local maximum point and the boundary points:

1. $f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}$
2. $f(0) = 0$
3. $f(\pi) = \pi$

Step 8: Compare the values The maximum value of $f(x)$ is given by $f\left(\frac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}$.

$ \text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x} $

$ \ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right) $

$ \frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x) $

For maxima or minima, $ \frac{d y}{d x}=0 $

$ \Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0 $

$ \Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0 $

$ \Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1 $

$ \Rightarrow \frac{3 e}{4 \sin ^2 x}=e $

$ \Rightarrow \sin ^2 x=\frac{3}{4} $

$ \Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text { as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right) $

$ \Rightarrow \text { Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}} $

$ \Rightarrow \mathrm{k}^8=\mathrm{e}^{11} $

$ \therefore $ $ \left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11} $

Given, $\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]$

$ \left(x\log _e x\right) f^{\prime}(x)+f(x)\left[\log _e x+1\right] \geq 1$

$ \Rightarrow $ $ \frac{d}{d x}\left[x \log _e x f(x)\right] \geq 1$

$ \begin{aligned} &\Rightarrow \frac{d}{d x}\left[x \log _e x f(x)-x\right] \geq 0 \quad\left[\because \frac{d}{d x}(x)=1\right] \\\\ & \forall x \in[2,4] \end{aligned} $

Let $g(x)=x \log _e x f(x)-x$

As $g(x) \geq 0, \forall x \in[2,4], g(x)$ is an increasing function in $[2,4]$

$ \begin{aligned} g(2) & =2 \log _e 2 f(2)-2 \\\\ & =\log _e 2-2 \quad\left[\because f(x)=\frac{1}{2}\right] \end{aligned} $

$ \begin{aligned} & g(4)=4 \log _e 4 f(4)-4=\log _e 4-4 \\\\ &=2\left(\log _e 2-2\right)\\\\ & {\left[\therefore f(4)=\frac{1}{4}\right] } \end{aligned} $

As, $g(x)$ is an increasing function,

$ \begin{aligned} & g(2) \leq g(x) \leq g(4) \\\\ & \log _e 2-2 \leq g(x) \leq 2\left(\log _e 2-2\right) \\\\ & \log _e 2-2 \leq x \log _e x f(x)-x \leq 2\left(\log _e 2-2\right) \end{aligned} $

$ \frac{\log _e 2-2+x}{x \log _e x} \leq f(x) \leq \frac{2\left(\log _e 2-2\right)+x}{x \log _e x} $

$ \begin{aligned} & \text {Now for } x \in[2,4] \\\\ & \begin{aligned} \frac{\log _e 2-2+x}{x \log _e x} & \leq \frac{2\left(\log _e 2-2\right)+e^2}{2 \log _e 2} =1-\frac{1}{\log _e 2}<1 \end{aligned} \end{aligned} $

$ \Rightarrow \mathrm{f}(\mathrm{x}) \leq 1 \text { for } \mathrm{x} \in[2,4] $

$ \text { Also for } \mathrm{x} \in[2,4] \text { : } $

Now,

$ \begin{aligned} \frac{2\left(\log _e 2-2\right)+2}{2 \log _e 2} & \geq \frac{\log _e 2-2+4}{4 \log _e 4} =\frac{1}{8}+\frac{1}{2 \log _e 2}>\frac{1}{8} \end{aligned} $

$ \therefore f(x) \geq \frac{1}{8}, \forall x \in[2,4] $

Hence, both statements $A$ and $B$ are true.

Note : LMVT on $(\mathrm{yx}(\ln \mathrm{x}))$ not satisfied.

Hence no such function exists.

Therefore it should be bonus.

We have, $g(x)=f(x)+f(1-x)$

Differentiating both side, we get

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

As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.

Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$

So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$

$\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$.

Now,

$ \begin{aligned} & \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\ & =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi \end{aligned} $

Let the slope of tangent at any point

$(x, y)$ on a curve $y=y(x)$ is $\frac{d y}{d x}$

According to the question, $\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}(x>0)$ [Given]

$ \begin{aligned} &\text { Let } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\\\ &\Rightarrow v+x \frac{d v}{d x}=\frac{x^2+v^2 x^2}{2 v x^2}=\frac{1+v^2}{2 v} \\\\ &\Rightarrow x \frac{d v}{d x}=\frac{1+v^2}{2 v}-v=\frac{1-v^2}{2 v} \end{aligned} $

$ \begin{aligned} & \Rightarrow \int \frac{2 v d v}{1-v^2}=\int \frac{d x}{x} \text { [Taking integration on both sides] } \\\\ & \Rightarrow-\ln \left|1-v^2\right|=\ln x-\ln c \\\\ & \Rightarrow \ln \left(1-v^2\right) x=\ln c \quad(c=\text { constant }) \\\\ & \Rightarrow \left(1-y^2 / x^2\right) \cdot x=c \\\\ & \Rightarrow \left(x^2-y^2\right)=c x \end{aligned} $

Put $y(2)=0 \Rightarrow c=2$

$ \begin{array}{rlrl} & \therefore x^2-y^2 =2 x \\\\ & \therefore y^2 =x^2-2 x \\\\ & \Rightarrow y(x) = \pm \sqrt{x^2-2 x} \end{array} $

Put $x=8$, we get

$ \Rightarrow y(8)= \pm \sqrt{64-16}= \pm \sqrt{48}= \pm 4 \sqrt{3} $

Given, length of square sheet of side is $30 \mathrm{~cm}$.



Let, side of small square be $x \mathrm{~cm}$.

Since, a square piece of tin is to be made into a box without top by cutting square from each corner and folding up the flaps to from a box. Then, this shape formed a cuboidal shape.

Thus, volume of the box $(V)$

$ \begin{aligned} & =(30-2 x)(30-2 x) x \\\\ &\Rightarrow V =x(30-2 x)^2 \end{aligned} $

On differentiating both side with respect to $x$, we get

$ \begin{aligned} \frac{d V}{d x} & =(30-2 x)^2+x\{2(30-2 x) \cdot(-2)\} \\\\ & =(30-2 x)^2-4 x(30-2 x) \\\\ & =(30-2 x)\{30-6 x\} \end{aligned} $

Therefore, maximum of $V, \frac{d V}{d x}=0$

$ \begin{array}{ll} &\Rightarrow (30-2 x)(30-6 x)=0 \\\\ &\Rightarrow x=15 \text { or } x=5 \quad[x=15 \text { (not taken) }] \end{array} $

Thus, $x=5$

So, the surface area

$ \begin{array}{ll} =4 \times(30-2 x) \times x+(30-2 x)^2 & \\\\ =4 \times(30-10) \times 5+(30-10)^2 & (\text { Put } x=5) \\\\ =400+400=800 & \end{array} $

The sum of the absolute maximum and minimum values of the function $f(x) = \left|x^2 - 5x + 6\right| - 3x + 2$ in the interval $[-1, 3]$ can be found by finding the maximum and minimum values of $f(x)$ in this interval and then adding them.

First, let's find the critical points of $f(x)$. To do this, we will find the zeros of the expression inside the absolute value:

$x^2 - 5x + 6 = 0$

Solving this quadratic equation, we find that the zeros are $x = 2$ and $x = 3$. These are the critical points of $f(x)$.

Next, we evaluate $f(x)$ at these critical points and at the endpoints of the interval $[-1, 3]$:

$f(-1) = \left|(-1)^2 - 5(-1) + 6\right| - 3(-1) + 2 = 17$

$f(2) = \left|(2)^2 - 5(2) + 6\right| - 3(2) + 2 = -4$

$f(3) = \left|(3)^2 - 5(3) + 6\right| - 3(3) + 2 = -7$

So, the minimum value of $f(x)$ is $-7$ and the maximum value is $17$.

So, the sum of the absolute maximum and minimum values of the function is $-7 + 17 = 10$.

Note : The absolute maximum and minimum values of a function are the largest and smallest values that the function takes on a given interval, respectively. These values are also called the "extrema" of the function. The absolute maximum value is the highest point on the graph of the function, and the absolute minimum value is the lowest point.

$ \ell_{1}+\ell_{2}=20 \Rightarrow \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=-1$

$\mathrm{A}_{1}=\left(\frac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\frac{\ell_{2}}{2 \pi}\right)^{2}$

Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\frac{\ell_{1}^{2}}{8}+\frac{3 \ell_{2}^{2}}{4 \pi}$

$\frac{\mathrm{ds}}{\mathrm{d} \ell_1}=0 \Rightarrow \frac{2 \ell_{1}}{8}+\frac{6 \ell_{2}}{4 \pi} \cdot \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=0$

$\Rightarrow \frac{\ell_{1}}{4}=\frac{6 \ell_{2}}{4 \pi} $

$\Rightarrow \frac{\pi \ell_{1}}{\ell_{2}}=6$

$f'(x)=x^2+2b+ax$

$g'(x)=x^2+a+2bx$

$\Rightarrow x=1$ is common root

$a+2b+1=0$

$y'=270x^4-540x^3-210x^2+360x+210$

Slope of normal $=-\frac{1}{90}$

$\therefore$ Slope of tangent = 90

$\therefore$ Number of normal will be number of solutions of

$270x^4-540x^3-210x^2+360x+210=90$

$\Rightarrow 9x^4-18x^3-7x^2+12x+4=0$

$\therefore x=1,2,-\frac{1}{3},-\frac{2}{3}$ are roots

$f^{\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9)$

$x_{1}<0, x_{2}>0$

$\Rightarrow f^{\prime}(0)<0$

$\Rightarrow p<\frac{9}{2}$

$p \in \left( { - \infty ,{9 \over 2}} \right)$

$ \begin{aligned} & f(x)=8 x^3-36 x+8 \\\\ & =4\left(2 x^3-9 x+2\right) \\\\ & =4(x-2)\left(2 x^2+4 x-1\right) \\\\ & =4(x-2)\left(x-\frac{-2+\sqrt{6}}{2}\right)\left(x-\frac{-2 \sqrt{6}}{2}\right) \end{aligned} $

Local maxima occurs at $x=\frac{-2+\sqrt{6}}{2}=x_0$

$ f\left(x_0\right)=12 \sqrt{6}-\frac{33}{2} $

$g(x)=f(-x)-f(x)$

$ \begin{aligned} & =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\ & =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\ & =\frac{1+e^{x}}{1-e^{x}} \\\\ g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\ & =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0 \end{aligned} $

So both statements are correct

$\mathrm{DR}^{\prime} \mathrm{S}$ of $\mathrm{OG}=1,1,1$

DR'S of AF $=-1,1,1$

DR'S of CE = 1, 1, -1

$\mathrm{DR}^{\prime} \mathrm{S}$ of $\mathrm{BD}=1,-1,1$

Equation of $\mathrm{OG} \Rightarrow \frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{1}$

Equation of $\mathrm{AB} \Rightarrow \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{0}$

Normal to both the line's

$ =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{array}\right|=\hat{i}+\hat{j}-2 \hat{k} $

$\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}$

S.D. $=\frac{|\hat{\mathrm{i}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})|}{|\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}|}=\frac{1}{\sqrt{6}}$

$ \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} $