Differentiation

$ \begin{aligned} &\text { Let } f^{\prime}(1)=a\\ &\begin{aligned} & f^{\prime}(2)=b \\ & f^{\prime}(3)=c \\ & f(x)=x^3+a x^2+b x+c \\ & f^{\prime}(x)=3 x^2+2 a x+b \\ & f^{\prime}(x)=a=3+2 a+b \Rightarrow a+b=3 \ldots . . (1)\\ & f^{\prime \prime}(x)=6 x+2 a \\ & f^{\prime \prime}(2)=12+2 a=\frac{b}{2} \Rightarrow 4 a-b=-24 \\ & f^{\prime \prime}(x)=6 \\ & f^{\prime \prime}(3)=c=6 \\ & \Rightarrow a=\frac{-27}{5}, b=\frac{12}{5} \\ & f^{\prime}(5)=75+10 a+b \\ & =75-54+\frac{12}{5}=21+\frac{12}{5}=\frac{117}{5} \end{aligned} \end{aligned} $

To solve for $ y(x) $, we start with the determinant of the given matrix:

$ \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix} $

Using the formula for a $3 \times 3$ determinant, we expand along the first row:

$ f(x) = \sin x \cdot \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \cdot \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} $

Calculating the smaller $2 \times 2$ determinants:

$ \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} = 28 \cdot 1 - 27 \cdot 1 = 1 $

$ \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 27 \cdot 1 = 0 $

$ \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 28 \cdot 1 = -1 $

Substituting these into the determinant calculation gives:

$ f(x) = \sin x \cdot 1 - \cos x \cdot 0 + (\sin x + \cos x + 1) \cdot (-1) $

$ f(x) = \sin x - (\sin x + \cos x + 1) $

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

$ f(x) = -\cos x - 1 $

Now, calculate the derivatives:

First derivative $ f'(x) $:

$ f'(x) = \frac{d}{dx}(-\cos x - 1) = \sin x $

Second derivative $ \frac{d^2 f}{dx^2} $:

$ \frac{d^2 f}{dx^2} = \frac{d}{dx}(\sin x) = \cos x $

Finally, compute $ \frac{d^2 y}{dx^2} + y $:

$ \frac{d^2 f}{dx^2} + f(x) = \cos x - \cos x - 1 = -1 $

Thus, $ \frac{d^2 y}{dx^2} + y $ is equal to $-1$.

$\begin{aligned} & \sin (x-y) f(2 x+2 y)=f(2 x-2 y) \sin (x+y) \\ & \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)}=k(\text { say }) \\ & f(2 x+2 y)=k \sin (x+y) \\ & f(2 x)=5 \sin x \quad(\because y=0) \\ & f(x)=k \sin \frac{x}{2} \\ & f^{\prime}(x)=\frac{k}{2} \cos \frac{x}{2} \\ & f^{\prime}(0)=\frac{1}{2} \Rightarrow k=1 \\ & f(x)=\sin \frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2} \cos \frac{x}{2} \\ & f^{\prime \prime}(x)=-\frac{1}{4} \sin \frac{x}{2} \\ & 24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=-3 \end{aligned}$

$\begin{aligned} & x^2 f^{\prime}(x)-2 x f(x)=3 \\ & \left(\frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}\right)=\frac{3}{\left(x^2\right)^2} \\ & \Rightarrow \frac{d}{d x}\left(\frac{f(x)}{x^2}\right)=\frac{3}{x^4} \end{aligned}$

Integrating both sides

$\begin{aligned} & \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^2}=-\frac{1}{\mathrm{x}^3}+\mathrm{C} \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+\mathrm{Cx}^2 \\ & \text { put } \mathrm{x}=1 \\ & 4=-1+\mathrm{C} \Rightarrow \mathrm{C}=5 \\ & \mathrm{f}(\mathrm{x})=-\frac{1}{\mathrm{x}}+5 \mathrm{x}^2 \end{aligned}$

Now $2 \times f(2)=2 \times\left[-\frac{1}{2}+5 \times 2^2\right]$

$=39$

$\begin{aligned} &\log _e y=3 \sin ^{-1} x\\ &\begin{aligned} & y=e^{3 \sin ^{-1} x} \\ & \frac{d y}{d x}=e^{3 \sin ^{-1} x} \cdot \frac{3}{\sqrt{1-x^2}} \end{aligned} \end{aligned}$

$\sqrt{1-x^2} \frac{d y}{d x}=3 y$

Again differentiate

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

So value of $3 y^{\prime}\left(\sqrt{1-x^2}\right)$ at $x=\frac{1}{2}$

$\begin{aligned} & 3 \cdot \frac{3}{\sqrt{1-x^2}} e^{\sin ^{-1} x}\left(\sqrt{1-x^2}\right) \\ & =9 e^{3 \frac{\pi}{6}}=9 e^{\frac{\pi}{2}} \end{aligned}$

Given the polynomial function:

$f(x) = ax^3 + bx^2 + cx + 41$

We are provided the following conditions from the problem:

1. $f(1) = 40$

2. $f^{\prime}(1) = 2$

3. $f^{\prime \prime}(1) = 4$

First, calculate $f(1)$:

$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$

Simplifying, we get:

$a + b + c + 41 = 40$

Therefore:

$a + b + c = -1$

Next, calculate the first derivative $f^{\prime}(x)$:

$f^{\prime}(x) = 3ax^2 + 2bx + c$

Given $f^{\prime}(1) = 2$:

$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$

Simplifying, we get:

$3a + 2b + c = 2$

Next, calculate the second derivative $f^{\prime \prime}(x)$:

$f^{\prime \prime}(x) = 6ax + 2b$

Given $f^{\prime \prime}(1) = 4$:

$f^{\prime \prime}(1) = 6a(1) + 2b = 4$

Simplifying, we get:

$6a + 2b = 4$

Dividing the entire equation by 2:

$3a + b = 2$

We now have three equations:

1. $a + b + c = -1$

2. $3a + 2b + c = 2$

3. $3a + b = 2$

To solve for $a$, $b$, and $c$, follow these steps:

First, subtract the third equation from the second equation:

$(3a + 2b + c) - (3a + b) = 2 - 2$

Which simplifies to:

$b + c = 0$

So,

$c = -b$

Substitute $c = -b$ into the first equation:

$(a + b - b = -1)$

Simplifying, we get:

$a = -1$

Now substitute $a = -1$ into the third equation:

$3(-1) + b = 2$

Which simplifies to:

$-3 + b = 2$

Therefore:

$b = 5$

Next, since $c = -b$:

$c = -5$

Finally, we need to find $a^2 + b^2 + c^2$:

$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$

Simplifying, we get:

$1 + 25 + 25 = 51$

Therefore, the answer is:

Option B: 51

To determine $g^{\prime}(0)$, we start by applying the chain rule and product rule to find the derivative of the given function $g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$.

The product rule states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their product is given by:

$ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) $

Let's denote $u(x) = h(\mathrm{e}^x)$ and $v(x) = \mathrm{e}^{h(x)}$.

First, we need to find $u'(x)$ and $v'(x)$. Using the chain rule, we find:

$ u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x $

Now, the derivative of $v(x)$ is:

$ v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x) $

Using the product rule, we get the derivative of $g(x)$:

$ g'(x) = u'(x) v(x) + u(x) v'(x) $

Substituting $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the above expression, we get:

$ g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right) $

Next, we need to evaluate this at $x = 0$:

First, we know that:

$ h(0) = 0 $

$ h(1) = 1 $

$ h'(0) = 2 $

$ h'(1) = 2 $

Substituting $x = 0$ into the expressions, we get:

$ u(0) = h(\mathrm{e}^0) = h(1) = 1 $

$ v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1 $

$ u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2 $

$ v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2 $

Therefore, evaluating $g'(0)$:

$ g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right) $

$ g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right) $

$ g'(0) = 2 + 2 = 4 $

Thus, the value of $g^{\prime}(0)$ is 4, which corresponds to Option A.

The correct answer is Option A: 4.

Given the function:

$ f(x)=\left\{\begin{array}{ll} x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. $

we need to find its second derivative at specific points.

First, let’s compute the first derivative $ f^{\prime}(x) $:

$ f^{\prime}(x) = 3x^2 \sin \left( \frac{1}{x} \right) - x \cos \left( \frac{1}{x} \right) $

Next, the second derivative $ f^{\prime \prime}(x) $ is:

$ f^{\prime \prime}(x) = 6x \sin \left(\frac{1}{x}\right) - 3x \cos \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) - \frac{1}{x} \sin \left(\frac{1}{x}\right) $

Therefore, evaluating the second derivative at $ x = \frac{2}{\pi} $:

$ f^{\prime \prime} \left(\frac{2}{\pi}\right) = 6 \left( \frac{2}{\pi} \right) \sin \left(\frac{\pi}{2}\right) - 3 \left( \frac{2}{\pi} \right) \cos \left(\frac{\pi}{2}\right) - \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{2} \sin \left( \frac{\pi}{2} \right) $

Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$, this simplifies to:

$ f^{\prime \prime} \left( \frac{2}{\pi} \right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi} $

Finally, note that $ f^{\prime}(0) $ is not defined, as it involves terms like $\frac{1}{x}$ when $ x = 0 $.

Let $f^{\prime}(1)=k$

$\Rightarrow \quad \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}=k \quad\left(\frac{0}{0}\right)$

$\begin{aligned} & \lim _\limits{x \rightarrow 0} \frac{f^{\prime}(x)}{2 x}=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)}{2}=k \\ \Rightarrow & f^{\prime \prime}(0)=2 k \end{aligned}$

Given information is not complete.

$\begin{aligned} & y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \\ & y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \\ & =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \end{aligned}$

$y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}$

$\begin{aligned} & \text { at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \\ & \therefore \quad y+y^{\prime}+y^{\prime \prime}=2 \end{aligned}$

Given that $(g \circ f)(x) = x$ for all $x \in \mathbf{R}$. This means $g(f(x)) = x$ for all $x \in \mathbf{R}$. Differentiating both sides with respect to $x$, we get:

$g'(f(x)) \cdot f'(x) = 1$

Now, we want to find the value of $8g'(2)$. To do this, we need to find a value of $x$ such that $f(x) = 2$. Let's solve for $x$:

$x^5 + 2e^{x/4} = 2$

By inspection, we see that $x = 0$ is a solution. Therefore, $f(0) = 2$. Now, we can substitute this into our differentiated equation:

$g'(f(0)) \cdot f'(0) = 1$

$g'(2) \cdot f'(0) = 1$

Let's find $f'(0)$:

$f'(x) = 5x^4 + \frac{1}{2}e^{x/4}$

$f'(0) = \frac{1}{2}$

Substituting this back into our equation:

$g'(2) \cdot \frac{1}{2} = 1$

$g'(2) = 2$

Finally, we can calculate $8g'(2)$:

$8g'(2) = 8 \cdot 2 = \boxed{16}$

$f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$

$\begin{aligned} & \mathrm{f}^{\prime}(1)=2024 \\ & \mathrm{f}(1)=1 \end{aligned}$

Partially differentiating w. r. t. x

$\begin{aligned} & \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \cdot \frac{1}{\mathrm{y}}=\frac{1}{\mathrm{f}(\mathrm{y})} \mathrm{f}^{\prime}(\mathrm{x}) \\ & \mathrm{y} \rightarrow \mathrm{x} \\ & \mathrm{f}^{\prime}(1) \cdot \frac{1}{\mathrm{x}}=\frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \\ & 2024 \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \mathrm{xf}^{\prime}(\mathrm{x})-2024 \mathrm{~f}(\mathrm{x})=0 \end{aligned}$

$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$

Also $\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$

$\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$

$\Rightarrow \operatorname{rootsin}\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$

$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$

$\begin{aligned} & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3+2 \cos ^4 x & 2 \sin ^4 x & \sin ^2 2 x \\ 2 \cos ^4 x & 3+2 \sin ^2 4 x & \sin ^2 2 x \end{array}\right| \\ & \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1, \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\ & \left|\begin{array}{ccc} 2 \cos ^4 x & 2 \sin ^4 x & 3+\sin ^2 2 x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{array}\right| \\ & \mathrm{f}(\mathrm{x})=45 \\ & \mathrm{f}^{\prime}(\mathrm{x})=0 \\ & \end{aligned}$

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

Again,

$\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$

Again

$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$

at $\mathrm{x}=\frac{1}{2}$,

$y^{\prime}-y^{\prime \prime}=\frac{736}{225}$

Thus $225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$

$\begin{aligned} & f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \\ & =\sqrt{\pi} \end{aligned}$

1. Given the equation: $3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$

2. Replace $x$ with $\frac{1}{x}$ in the original equation:
   $3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$

3. Now, we have two equations:

$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$

$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$

1. By adding the two equations, we can find $f(x)$:

$5f(x) = \frac{3}{x} - 2x - 10$

1. Now, let's differentiate both sides with respect to $x$:

$5f'(x) = -\frac{3}{x^2} - 2$

1. Now, we can find the values for $f(3)$ and $f'\left(\frac{1}{4}\right)$:

$f(3) = \frac{1}{5}(1 - 6 - 10) = -3$

$f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10$

1. Finally, calculate the expression we are interested in :

$\left|f(3) + f'\left(\frac{1}{4}\right)\right| = \left|-3 - 10\right| = 13$

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

$ \begin{aligned} & =\frac{\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x-1}{\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x} \\\\ & =\frac{\cos \left(x-\frac{\pi}{4}\right)-1}{\sin \left(x-\frac{\pi}{4}\right)} \\\\ & =\frac{-2 \sin ^2\left(\frac{x}{2}-\frac{\pi}{8}\right)}{2 \sin \left(\frac{x}{2}-\frac{\pi}{8}\right) \cos \left(\frac{x}{2}-\frac{\pi}{8}\right)} \end{aligned} $

$ \begin{aligned} & \Rightarrow f(x)=-\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \\\\ & \Rightarrow f^{\prime}(x)=-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow f^{\prime \prime}(x)=-\frac{1}{2} \cdot 2 \sec \left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \sec \left(\frac{x}{2}-\frac{\pi}{8}\right)\tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \times \frac{1}{2} \\\\ & =-\frac{1}{2} \sec ^2\left(\frac{x}{2}-\frac{\pi}{8}\right) \cdot \tan \left(\frac{x}{2}-\frac{\pi}{8}\right) \end{aligned} $

$ \begin{aligned} & \text { Now, } f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right) \\\\ & =-\tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \frac{-1}{2} \sec ^2\left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \times \tan \left(\frac{7 \pi}{24}-\frac{\pi}{8}\right) \\\\ & =\frac{1}{2} \tan ^2\left(\frac{\pi}{6}\right) \times \sec ^2 \frac{\pi}{6} \\\\ & =\frac{1}{2} \times \frac{1}{3} \times \frac{4}{3}=\frac{2}{9} \end{aligned} $

Given, $2 x^y+3 y^x=20$ ..........(i)

Let $u=x^y$

On taking log both sides, we get

$\log u=y \log x$

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

$ \begin{array}{rlrl} & \frac{1}{u} \frac{d u}{d x} =y \frac{1}{x}+\log x \frac{d y}{d x} \\\\ & \Rightarrow \frac{d u}{d x} =u\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) \\\\ & \Rightarrow \frac{d u}{d x} =x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right) .........(ii) \end{array} $

Also, let $v=y^x$

On taking log both sides, we get

$\log v=x \log y$

On differentiating both sides, we get

$ \begin{array}{rlrl} & \frac{1}{v} \frac{d v}{d x} =x \frac{1}{y} \frac{d y}{d x}+\log y \cdot 1 \\\\ &\Rightarrow \frac{d v}{d x} =v\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) \\\\ &\Rightarrow \frac{d v}{d x} =y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right) ..........(iii) \end{array} $

Now, from Equation (i), $2 u+3 v=20$

$ \begin{aligned} & \Rightarrow 2 \frac{d u}{d x}+3 \frac{d v}{d x}=0 \\\\ & \Rightarrow 2 x^y\left(\frac{y}{x}+\log x \frac{d y}{d x}\right)+3 y^x\left(\frac{x}{y} \frac{d y}{d x}+\log y\right)=0 \text { [Using Eqs. (ii) and (iii)] } \end{aligned} $

On putting $x=2$ and $y=2$, we get

$ \begin{aligned} & 2(4)\left(1+\log 2 \frac{d y}{d x}\right)+3(4)\left(\frac{d y}{d x}+\log 2\right)=0 \\\\ & \Rightarrow \frac{d y}{d x}(8 \log 2+12)+(8+12 \log 2)=0 \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{2+3 \log 2}{3+2 \log 2}\right) \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{2+\log 8}{3+\log 4}\right) \end{aligned} $

$\begin{aligned} & y=x^x \\\\ & y^{\prime}=x^x(1+\ln x) \\\\ & y^{\prime \prime}=x^x(1+\ln x)^2+\frac{x^x}{x} \\\\ & f^{\prime \prime}(2)-2 f^{\prime}(2)=\left(4(1+\ln 2)^2+2\right)-(2)(4(1+\ln 2)) \\\\ & =4\left(1+(\ln 2)^2\right)+2-8 \\\\ & =4(\ln 2)^2-2 \\\\ & \end{aligned}$

$ \begin{aligned} & f^{\prime}(x)=2+\frac{1}{1+x^2}, g^{\prime}(x)=\frac{1}{\sqrt{x^2+1}} \\\\ & f^{\prime \prime}(x)=-\frac{2 x}{\left(1+x^2\right)^2}<0 \\\\ & g^{\prime \prime}(x)=-\frac{1}{2}\left(x^2+1\right)^{-3 / 2} \cdot 2 x<0 \\\\ & \left.f^{\prime}(x)\right|_{\min }=f^{\prime}(3)=2+\frac{1}{10}=\frac{21}{10} \\\\ & \left.g^{\prime}(x)\right|_{\max }=g^{\prime}(0)=1 \\\\ & \left.f^{\prime}(x)\right|_{\max }=f(3)=2+\tan ^{-1} 3 \\\\ & \left.g(x)\right|_{\max }=g(3)=\ln (3+\sqrt{10})<\ln <7<2 \end{aligned} $

$f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)$

$ \begin{aligned} & f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right) \end{aligned} $

$f^{\prime}(1)=\frac{3 \pi^{2}}{16}$

$ \begin{aligned} & f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\ &=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\ & \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0 \end{aligned} $

$f''(x) = g''(x) + 6x$

$ \Rightarrow f'(x) = g'(x) + 3{x^2} + C$

$f'(1) = g'(1) + 3 + C$

$ \Rightarrow g = 3 + 3 + C \Rightarrow C = 3$

$ \Rightarrow f'(x) = g'(x) + 3{x^2} + 3$

$ \Rightarrow f(x) = g(x) + {x^2} + 3x + C'$

$x = 2$

$f(2) = g(2) + 14 + C'$

$12 = 4 + 14 + C'$

$ \Rightarrow C' = - 6$

$ \Rightarrow f(x) = g(2) + {x^3} + 3x - 6$

$f( - 2) = g( - 2) - 8 - 6 - 6$

$g( - 2) - f( - 2) = 20$

$f'(x) - g'(x) = 3{x^2} + 3$

$x \in ( - 1,1)$

$3{x^2} + 3 \in (0,6)$

$ \Rightarrow f'(x) - g'(x) \in (0,6)$

$f(x) - g(x) = {x^3} + 3x - 6$

At $x = - 1$

$|f( - 1) - g( - 1)| = 10$

$\therefore$ Option (4) is false.

$ \begin{aligned} & y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\\\ & y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\\\ & y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\\\ & y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\\\ & y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\\\ & y^{\prime \prime}(-1)-y^{\prime}(-1)=-496 \end{aligned} $

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

Let $\mathrm{f}^{\prime}(1)=\mathrm{a}, \mathrm{f}^{\prime \prime}(2)=\mathrm{b}, \mathrm{f}^{\prime \prime \prime}(3)=\mathrm{c}$

$ \begin{aligned} & f(x)=x^3-a x^2+b x-c \\\\ & f^{\prime}(x)=3 x^2-2 a x+b \\\\ & f^{\prime \prime}(x)=6 x-2 a \\\\ & f^{\prime \prime \prime}(x)=6 \\\\ & c=6, a=3, b=6 \\\\ & f(x)=x^3-3 x^2+6 x-6 \\\\ & f(1)=-2, f(2)=2, f(3)=12, f(0)=-6 \\\\ & 2 f(0)-f(1)+f(3)=2=f(2) \end{aligned} $

$x = 2\sqrt 2 \cos t\sqrt {\sin 2t} ,\,y = 2\sqrt 2 \sin t\sqrt {\sin 2t} $

$\therefore$ ${{dx} \over {dt}} = {{2\sqrt 2 \cos 3t} \over {\sqrt {\sin 2t} }},\,{{dy} \over {dt}} = {{2\sqrt 2 \sin 3t} \over {\sqrt {\sin 2t} }}$

$\therefore$ ${{dy} \over {dx}} = \tan 3t,\,\left( {\mathrm{at}\,t = {\pi \over 4},\,{{dy} \over {dx}} = - 1} \right)$

and ${{{d^2}y} \over {d{x^2}}} = 3{\sec ^2}3t\,.\,{{dt} \over {dx}} = {{3{{\sec }^2}3t\,.\,\sqrt {\sin 2t} } \over {2\sqrt 2 \cos 3t}}$

$\left( {\mathrm{At}\,t = {\pi \over 4},\,{{{d^2}y} \over {d{x^2}}} = - 3} \right)$

$\therefore$ ${{1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \over {{{{d^2}y} \over {d{x^2}}}}} = {2 \over { - 3}} = {{ - 2} \over 3}$

Let $f(x) = {\log _{\cos x}}\cos ec\,x$

$ = {{\log \cos ec\,x} \over {\log \cos x}}$

$ \Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}$

at $x = {\pi \over 4}$

$f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}$

$\therefore$ ${\log _e}2f'(x)$ at $x = {\pi \over 4} = 4$

${\cos ^{ - 1}}\left( {{y \over 2}} \right) = {\log _e}{\left( {{x \over 5}} \right)^5}\,\,\,\,\,\,\,\,\,|y| < 2$

Differentiating on both side

$ - {1 \over {\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} }} \times {{y'} \over 2} = {5 \over {{x \over 5}}} \times {1 \over 5}$

${{ - xy'} \over 2} = 5\sqrt {1 - {{\left( {{y \over 2}} \right)}^2}} $

Square on both side

${{{x^2}y{'^2}} \over 4} = 25\left( {{{4 - {y^2}} \over 4}} \right)$

Diff on both side

$2xy{'^2} + 2y'y''{x^2} = - 25 \times 2yy'$

$xy' + y''{x^2} + 25y = 0$

$f(x) = 3{x^2} + 1$

f'(x) is bijective function

and $f(g(x)) = x \Rightarrow g(x)$ is inverse of f(x)

$g(f(x)) = x$

$g'(f(x))\,.\,f'(x) = 1$

$g'(f(x)) = {1 \over {3{x^2} + 1}}$

Put x = 4 we get

$g'(63) = {1 \over {49}}$

Let ${x^3} = \theta \Rightarrow {\theta \over 2} \in \left( {{\pi \over 4},\,{{3\pi } \over 4}} \right)$

$\therefore$ $y = {\tan ^{ - 1}}(\sec \theta - \tan \theta )$

$ = {\tan ^{ - 1}}\left( {{{1 - \sin \theta } \over {\cos \theta }}} \right)$

$\therefore$ $y = {\pi \over 4} - {\theta \over 2}$

$y = {\pi \over 4} - {{{x^3}} \over 2}$

$\therefore$ $y' = {{ - 3{x^2}} \over 2}$

$y'' = - 3x$

$\therefore$ ${x^2}y'' - 6y + {{3\pi } \over 2} = 0$

We have,

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

$ =\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|} $

$ \left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right. $

and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$

$ \begin{aligned} & =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\ & \end{aligned} $

$ =\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right) $

$ =\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right) $

$ \begin{aligned} & \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\ & \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right) \end{aligned} $

$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$

${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x $

or $f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$

$ = \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$

$f(x) = {{1 - x} \over {1 + x}}$

Now, $f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$

or $f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$

or ${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$.

Let f = ${\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)$

Put x = tan $\theta $ $ \Rightarrow $ $\theta $ = tan^–1 x

f = ${\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)$

$ \Rightarrow $ f = ${\tan ^{ - 1}}\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)$ = ${\theta \over 2}$

$ \Rightarrow $ f = ${{{{\tan }^{ - 1}}x} \over 2}$

$ \therefore $ ${{df} \over {dx}}$ = ${1 \over {2\left( {1 + {x^2}} \right)}}$ ....(1)

Let g = ${\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)$

Put x = sin $\theta $ $ \Rightarrow $ $\theta $ = sin^–1 x

$ \Rightarrow $ g = ${\tan ^{ - 1}}\left( {{{2\sin \theta \cos \theta } \over {1 - 2{{\sin }^2}\theta }}} \right)$

$ \Rightarrow $ g = tan^–1 (tan 2$\theta $) = 2$\theta $

$ \Rightarrow $ g = 2sin^-1 x

$ \Rightarrow $ ${{dg} \over {dx}}$ = ${2 \over {\sqrt {1 - {x^2}} }}$ ...(2)

Using (i) and (ii),

$ \therefore $ ${{df} \over {dg}}$ = ${1 \over {2\left( {1 + {x^2}} \right)}}{{\sqrt {1 - {x^2}} } \over 2}$

At x = ${1 \over 2}$, ${\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}}$ = ${{\sqrt 3 } \over {10}}$

$(a + \sqrt 2 b\cos x)(a - \sqrt 2 b\cos y) = {a^2} - {b^2}$

$ \Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}$

Differentiating both sides :

$0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)$

$ - 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0$

At $\left( {{\pi \over 4},{\pi \over 4}} \right)$ :

$ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0$

$ \Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}$; a, b > 0

Given y² + log_e (cos²x) = y .....(1)

Put x = 0, we get

y² + log_e (1) = y

$ \Rightarrow $ y² = y

$ \Rightarrow $ y = 0, 1

Differentiating (1) we get

2yy' + ${1 \over {\cos x}}\left( { - \sin x} \right)$ = y'

$ \Rightarrow $ 2yy' - 2tanx = y' ....(2)

From (2) when x = 0, y = 0 then y'(0) = 0

From (2) when x = 0, y = 1 then

2y' = y'

$ \Rightarrow $ y'(0) = 0

Again differentiating (2) we get

2(y')² + 2yy'' – 2sec²x = y''

from (2) when x = 0, y = 0, y’(0) = 0 then

y”(0) = -2

Also from (2) when x = 0, y = 1, y’(0) = 0 then

y”(0) = 2

$ \therefore $ |y''(0)| = 2

Given the function composition f(g(x)) is the identity function, it means f(g(x)) = x for all x.

$ \Rightarrow $ ƒ'(g(x)) g'(x) = 1

put x = a

$ \Rightarrow $ ƒ'(b) g'(a) = 1

$ \Rightarrow $ ƒ'(b) = ${1 \over 5}$

$x = 2\sin \theta - \sin 2\theta $

$ \Rightarrow $ ${{dx} \over {d\theta }}$ = $2\cos \theta - 2\cos 2\theta $

$y = 2\cos \theta - \cos 2\theta $

$ \Rightarrow $ ${{dy} \over {d\theta }}$ = –2sin$\theta $ + 2sin2$\theta $

${{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}$ = ${{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}$

This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$, leading to the expression:

$ \frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}$

This is the rate of change of y with respect to x, as a function of θ.

Next, we differentiate this function with respect to θ to find $\frac{d^2 y}{dx^2}$, yielding :

$ \frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}$

Here, $\frac{d \theta}{d x}$ is the reciprocal of $\frac{dx}{d\theta}$, so the equation becomes :

$ \frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}$

Finally, we evaluate this expression at $\theta = \pi$, yielding :

$ \frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}$

Therefore, the correct answer is (A) $\frac{3}{8}$.

Alternate Method :

First, let's find the derivatives of x and y with respect to θ :

1) $ \frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta $

2) $ \frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta $

We know that $ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} $

So, we substitute 1) and 2) into this equation :

We have

$\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}$

For simplification, let's denote the numerator as $N = \sin 2\theta - \sin \theta$ and the denominator as $D = \cos \theta - \cos 2\theta$.

We have to compute $\frac{d}{d\theta}(\frac{dy}{dx})$ which is $\frac{d}{d\theta}(\frac{N}{D})$.

We can use the quotient rule for differentiation, which states that if we have a function of the form $\frac{u}{v}$, then its derivative is given by $\frac{vu' - uv'}{v^2}$.

So here, $N' = 2\cos 2\theta - \cos \theta$ and $D' = -\sin \theta + 2\sin 2\theta$.

Applying the quotient rule :

$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}$

Substituting the expressions for $N'$, $D'$, $N$, and $D$ we get :

$\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$

Now $ \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x} $

= $\frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}$$ \times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)} $

$ \begin{aligned} \left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\ & =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8} \end{aligned} $

Given ƒ(x) = (sin(tan^–1x) + sin(cot^–1x))² – 1

= (sin(tan^–1x) + sin(${\pi \over 2}$ - tan^–1x))² – 1

= (sin(tan^–1x) + cos(tan^–1x))² – 1

= sin²(tan^–1x) + cos²(tan^–1x) + 2sin(tan^–1x)cos(tan^–1x) + 1

= 1 + sin(2tan^–1x) - 1

= sin(2tan^–1x)

Also given ${{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)$

Integrating both sides we get

y = ${1 \over 2}$ sin^-1 (f(x)) + C

= ${1 \over 2}$ sin^-1 (sin(2tan^–1x)) + C

Given $y\left( {\sqrt 3 } \right) = {\pi \over 6}$ mean x = $\sqrt 3 $ and y = ${\pi \over 6}$

$ \therefore $ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (sin(2tan^–1$\sqrt 3 $)) + C

$ \Rightarrow $ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (sin(2$ \times $${\pi \over 3}$)) + C

$ \Rightarrow $ ${\pi \over 6}$ = ${1 \over 2}$ sin^-1 (${{\sqrt 3 } \over 2}$) + C

$ \Rightarrow $ ${\pi \over 6}$ = ${1 \over 2}$ $ \times $ ${\pi \over 3}$ + C

$ \Rightarrow $ C = 0

Now y(${ - \sqrt 3 }$) means when x = ${ - \sqrt 3 }$ then find y.

y = ${1 \over 2}$ sin^-1 (sin(2tan^–1x))

= ${1 \over 2}$ sin^-1 (sin(2tan^–1(${ - \sqrt 3 }$)))

= ${1 \over 2}$ sin^-1 (sin(-2tan^–1(${ \sqrt 3 }$)))

= ${1 \over 2}$ sin^-1 (sin(-2$ \times $${\pi \over 3}$))

= ${1 \over 2}$ sin^-1 (-sin(2$ \times $${\pi \over 3}$))

= ${1 \over 2}$ sin^-1 (-${{\sqrt 3 } \over 2}$)

= ${1 \over 2}$ $ \times $ -sin^-1 (${{\sqrt 3 } \over 2}$)

= ${1 \over 2}$ $ \times $ -${\pi \over 3}$

= -${\pi \over 6}$

$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $ ....(1)

On differentiating both side of eq. (1) w.r.t. x we get,

${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$

= 0 - $\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$

Put x = ${1 \over 2}$ and y = $ - {1 \over 4}$, we get

${{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}$

= $ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}$

$ \therefore $ ${{dy} \over {dx}} = - {{\sqrt 5 } \over 2}$

x^k + y^k = a^k

$ \Rightarrow $ kx^{k - 1} + ky^{k - 1}${{{dy} \over {dx}}}$ = 0

$ \Rightarrow $ ${{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}$ = 0 ...(1)

Given ${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$ ...(2)

Comparing (1) and (2), we get

k - 1 = $ - {1 \over 3}$

$ \Rightarrow $ k = ${2 \over 3}$

$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$

= $\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}} $

= $\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} $

= $\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}} $

= ${{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}$

At $\alpha $ = ${{5\pi } \over 6}$

${\left| {\sin \alpha + \cos \alpha } \right|}$ = -(sin$\alpha $ + cos$\alpha $)

and |sin$\alpha $| = sin$\alpha $

$ \therefore $ y($\alpha $) = ${{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}$

= -1 - cot$\alpha $

$ \therefore $ ${{dy} \over {d\alpha }}$ = cosec²$\alpha $

So ${{{dy} \over {d\alpha }}}$ at $\alpha $ = ${{5\pi } \over 6}$,

= cosec²${{5\pi } \over 6}$ = 4

$y = {\tan ^{ - 1}}\left( {{{\tan x - 1} \over {\tan x + 1}}} \right)$

$ \Rightarrow $ $ - {\tan ^{ - 1}}\left( {{{1 - \tan x} \over {1 + \tan x}}} \right)$

$ \Rightarrow $ $ - {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)$

$ \Rightarrow $ $ - \left( {{\pi \over 4} - x} \right)$

$ \Rightarrow $ ${{dy} \over {dx}} = 1$

Now if differentiation of ${x \over 2}$ w.r.t is ${1 \over 2}$

$ \Rightarrow $ So differentiation of y w.r.t ${x \over 2}$ is $1 \over {1 \over 2}$ = 2

y = 1 $ \Rightarrow $ x = 0

${e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0$

$ \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}$

$ \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0$

x = 0, y = 1

$ \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0$

$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}$

f(x) = ln(sin x), g(x) = sin^–1 (e^–x)

f(g(x)) = ln(sin(sin^–1 e^–x)) = -x

f(g($\alpha $)) = – $\alpha $ = b

As f(g(x)) = – x

$ \therefore $ (f(g(x)))' = – 1

$ \Rightarrow $ (f(g($\alpha $)))' = – 1 = a

$ \therefore $ b = – $\alpha $, a = – 1

$ \therefore $ a$\alpha $² - b$\alpha $ - a = - $\alpha $² + $\alpha $² + 1 = 1

Given ƒ(1) = 1, ƒ'(1) = 3

Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x))²

On differentiating both sides with respect to x we get,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x)

Now at x = 1,

${{dy} \over {dx}}$ = ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1)

= ƒ'(ƒ(1)).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= ƒ'(1).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1)

= 3$ \times $3$ \times $3 + 2$ \times $3

= 33

$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$

As x $ \in $ $\left( {0,{\pi \over 2}} \right)$ then

${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$ = ${\left( {{\pi \over 3} + x} \right)}$

$ \Rightarrow $ 2y = ${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$

$ \Rightarrow $ 2y = ${\left( {{\pi \over 6} - x} \right)^2}$

$ \therefore $ $2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$

$ \Rightarrow $ ${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$

(2x)^2y = 4e^2x-2y

2y$\ell $n2x = $\ell $n4 + 2x $-$ 2y

y = ${{x + \ell n2} \over {1 + \ell n2x}}$

y ' = ${{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}$

y '${\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]$

Differentiating with respect to x,

$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$

at   $x = e$  we get

$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$

$ \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,$

as   $y(e) = \sqrt {4 + {e^2}} $

f(x) = x³ + x²f '(1) + xf ''(2) + f '''(3)

$ \Rightarrow $  f '(x) = 3x² + 2xf '(1) + f ''(x)     . . . . . (1)

$ \Rightarrow $  f ''(x) = 6x + 2f '(1)     . . . . . . (2)

$ \Rightarrow $  f '''(x) = 6      . . . . . .(3)

put x = 1 in equation (1) :

f '(1) = 3 + 2f '(1) + f ''(2)     . . . . .(4)

put x = 2 in equation (2) :

f ''(2) = 12 + 2f '(1)     . . . . .(5)

from equation (4) & (5) :

$-$3 $-$ f '(1) = 12 + 2f'(1)

$ \Rightarrow $  3f '(1) = $-$ 15

$ \Rightarrow $  f '(1) = $-$ 5 $ \Rightarrow $  f ''(2) = 2      . . . . .(2)

put x = 3 in equation (3) :

f ''' (3) = 6

$ \therefore $  f(x) = x³ $-$ 5x² + 2x + 6

f(2) = 8 $-$ 20 + 4 + 6 = $-$ 2

x = 3 tan t and y = 3 sec t

So that ${{dx} \over {dt}}$ = 3sec²t and ${{dy} \over {dt}}$ = 3 sec t tan t

${{dy} \over {dx}}$ = ${{dy/dt} \over {dx/dt}}$ = sin t

${{{d^2}y} \over {d{x^2}}}$ = (cos t)$.{{dt} \over {dx}}$

${{{d^2}y} \over {d{x^2}}} = \left( {\cos t} \right).{1 \over {3{{\sec }^2}t}}$

${{{d^2}y} \over {d{x^2}}}$ = ${1 \over 3}$(cos³ t)

${\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{t = \pi /4}}$ = ${1 \over 3} \times {\left( {{1 \over {\sqrt 2 }}} \right)^3} = {1 \over {6\sqrt 2 }}$