MHT CET 2024 Mathematics Question Paper with Answer and Solution

(A) Given $(\cot \alpha_1)(\cot \alpha_2) \ldots (\cot \alpha_n) = 1$.
This implies $(\cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n) = (\sin \alpha_1 \sin \alpha_2 \ldots \sin \alpha_n)$.
Let $P = \cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n$.
Then $P^2 = (\cos \alpha_1 \ldots \cos \alpha_n)(\sin \alpha_1 \ldots \sin \alpha_n)$.
$P^2 = \frac{1}{2^n} (2 \sin \alpha_1 \cos \alpha_1) (2 \sin \alpha_2 \cos \alpha_2) \ldots (2 \sin \alpha_n \cos \alpha_n)$.
$P^2 = \frac{1}{2^n} \sin(2\alpha_1) \sin(2\alpha_2) \ldots \sin(2\alpha_n)$.
Since $\sin(2\alpha_i) \leq 1$ for all $i$,we have $P^2 \leq \frac{1}{2^n}$.
Taking the square root,$P \leq \frac{1}{\sqrt{2^n}} = \frac{1}{2^{(n/2)}}$.
The maximum value is $\frac{1}{2^{(n/2)}}$.

(C) Given the equation $\sqrt{3} \sec x + 2 = 0$.
$\sqrt{3} \sec x = -2$
$\sec x = -\frac{2}{\sqrt{3}}$
$\cos x = -\frac{\sqrt{3}}{2}$
Since $\cos x$ is negative in the second and third quadrants,we find the values in $[0, 2\pi]$.
In the second quadrant: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
In the third quadrant: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
Thus,the principal solutions are $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$.

(C) Given equation: $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$
Dividing by $2$: $\frac{\sqrt{3}}{2} \sec x+\frac{1}{2} \operatorname{cosec} x=\cot x-\tan x$
Converting to $\sin x$ and $\cos x$: $\frac{\sqrt{3}}{2 \cos x}+\frac{1}{2 \sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$
Multiplying by $\sin x \cos x$: $\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos^2 x-\sin^2 x$
Using $\cos(A-B)$ and $\cos 2x$ formulas: $\cos(x-\frac{\pi}{6})=\cos 2x$
General solution: $2x = 2n\pi \pm (x-\frac{\pi}{6})$
Case $1$: $2x = 2n\pi + x - \frac{\pi}{6} \implies x = 2n\pi - \frac{\pi}{6}$
For $n=0, x=-\frac{\pi}{6} \in S$. For $n=1, x=\frac{11\pi}{6} \notin S$.
Case $2$: $2x = 2n\pi - (x - \frac{\pi}{6}) \implies 3x = 2n\pi + \frac{\pi}{6} \implies x = \frac{2n\pi}{3} + \frac{\pi}{18}$
For $n=0, x=\frac{\pi}{18} \in S$. For $n=1, x=\frac{13\pi}{18} \in S$. For $n=-1, x=-\frac{11\pi}{18} \in S$.
Sum of solutions: $-\frac{\pi}{6} + \frac{\pi}{18} + \frac{13\pi}{18} - \frac{11\pi}{18} = \frac{-3\pi + \pi + 13\pi - 11\pi}{18} = 0$.

(C) Given equation: $\tan \theta + \sec \theta = 2 \cos \theta$
$\Rightarrow \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = 2 \cos \theta$
$\Rightarrow \sin \theta + 1 = 2 \cos^2 \theta$
$\Rightarrow \sin \theta + 1 = 2(1 - \sin^2 \theta)$
$\Rightarrow 2 \sin^2 \theta + \sin \theta - 1 = 0$
$\Rightarrow (2 \sin \theta - 1)(\sin \theta + 1) = 0$
$\therefore \sin \theta = \frac{1}{2}$ or $\sin \theta = -1$
For $\theta \in (0, 2\pi)$:
If $\sin \theta = \frac{1}{2}$,then $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$.
If $\sin \theta = -1$,then $\theta = \frac{3\pi}{2}$.
However,at $\theta = \frac{3\pi}{2}$,$\tan \theta$ and $\sec \theta$ are undefined.
Thus,the valid solutions are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.
Therefore,the number of solutions is $2$.

(C) Given the equation $2^{1+|\cos x|+|\cos x|^2+\ldots} = 4$.
Since the exponent is an infinite geometric series with first term $a = 1$ and common ratio $r = |\cos x|$,where $|\cos x| < 1$,the sum is $\frac{1}{1-|\cos x|}$.
Thus,$2^{\frac{1}{1-|\cos x|}} = 2^2$.
Equating the exponents,we get $\frac{1}{1-|\cos x|} = 2$.
$1 - |\cos x| = \frac{1}{2} \Rightarrow |\cos x| = \frac{1}{2}$.
This implies $\cos x = \frac{1}{2}$ or $\cos x = -\frac{1}{2}$.
In the interval $(-\pi, \pi)$,the solutions for $\cos x = \frac{1}{2}$ are $x = \frac{\pi}{3}, -\frac{\pi}{3}$.
The solutions for $\cos x = -\frac{1}{2}$ are $x = \frac{2\pi}{3}, -\frac{2\pi}{3}$.
Therefore,the total number of solutions is $4$.

(A) The given equation is defined for $x \neq \frac{\pi}{2}, \frac{3 \pi}{2}$.
Given equation: $\tan x + \sec x = 2 \cos x$
$\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$
$\Rightarrow \frac{\sin x + 1}{\cos x} = 2 \cos x$
$\Rightarrow \sin x + 1 = 2 \cos^2 x$
$\Rightarrow \sin x + 1 = 2(1 - \sin^2 x)$
$\Rightarrow \sin x + 1 = 2(1 - \sin x)(1 + \sin x)$
$\Rightarrow (1 + \sin x)[1 - 2(1 - \sin x)] = 0$
$\Rightarrow (1 + \sin x)(2 \sin x - 1) = 0$
Case $1$: $1 + \sin x = 0 \Rightarrow \sin x = -1$. This implies $x = \frac{3 \pi}{2}$,but at $x = \frac{3 \pi}{2}$,$\tan x$ and $\sec x$ are undefined. So,this is not a solution.
Case $2$: $2 \sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2}$.
In the interval $[0, 2 \pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}$.
Thus,the number of solutions is $2$.

(D) Let $y = \cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)$.
Simplifying the expression inside the inverse cosine function:
$y = \cos ^{-1}\left(\frac{x^2-1}{x^2+1}\right) = \cos ^{-1}\left(-\left(\frac{1-x^2}{1+x^2}\right)\right)$.
Using the property $\cos ^{-1}(-z) = \pi - \cos ^{-1}(z)$,we get:
$y = \pi - \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.
Substitute $x = \tan \theta$,which implies $\theta = \tan ^{-1} x$:
$y = \pi - \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)$.
Since $\cos 2\theta = \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}$,we have:
$y = \pi - \cos ^{-1}(\cos 2\theta) = \pi - 2\theta = \pi - 2\tan ^{-1} x$.
Differentiating with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(\pi - 2\tan ^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}\right) = \frac{-2}{1+x^2}$.

(D) Let $y = \tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$.
We can rewrite the argument of $\tan^{-1}$ as follows:
$y = \tan^{-1}\left(\frac{2(3x\sqrt{x})}{1-(3x\sqrt{x})^2}\right)$.
Using the formula $2\tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,where $\theta = 3x\sqrt{x} = 3x^{3/2}$,we get:
$y = 2\tan^{-1}(3x^{3/2})$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(3x^{3/2})^2} \cdot \frac{d}{dx}(3x^{3/2})$.
$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.
$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot \frac{9}{2} \sqrt{x} = \frac{9\sqrt{x}}{1+9x^3}$.
Given $\frac{dy}{dx} = \sqrt{x} \cdot g(x)$,we have $\sqrt{x} \cdot g(x) = \frac{9\sqrt{x}}{1+9x^3}$.
Therefore,$g(x) = \frac{9}{1+9x^3}$.

(B) Given $y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}\right)$.
We can rewrite the expression as $y = \sin^{-1}\left(3\left(\frac{x}{2}\right) - 4\left(\frac{x}{2}\right)^3\right)$.
Let $\frac{x}{2} = \sin \theta$,which implies $\theta = \sin^{-1}\left(\frac{x}{2}\right)$.
Substituting this into the equation,we get $y = \sin^{-1}(3\sin \theta - 4\sin^3 \theta)$.
Using the trigonometric identity $\sin(3\theta) = 3\sin \theta - 4\sin^3 \theta$,we have $y = \sin^{-1}(\sin 3\theta) = 3\theta$.
Substituting $\theta$ back,$y = 3\sin^{-1}\left(\frac{x}{2}\right)$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1 - (x/2)^2}} \cdot \frac{1}{2}$.
$\frac{dy}{dx} = \frac{3}{2 \cdot \sqrt{\frac{4-x^2}{4}}} = \frac{3}{2 \cdot \frac{\sqrt{4-x^2}}{2}} = \frac{3}{\sqrt{4-x^2}}$.

(D) Let $y = \tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ and $z = \cos ^{-1}(4x^3 - 3x)$.
For $y$,substitute $x = \cos 2\theta$. Then $\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} = \sqrt{\frac{2\sin^2 \theta}{2\cos^2 \theta}} = \tan \theta$.
So,$y = \tan^{-1}(\tan \theta) = \theta = \frac{1}{2} \cos^{-1} x$.
For $z$,substitute $x = \cos \theta$. Then $z = \cos^{-1}(\cos 3\theta) = 3\theta = 3 \cos^{-1} x$.
Now,differentiate $y$ and $z$ with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left( \frac{-1}{\sqrt{1-x^2}} \right) = \frac{-1}{2\sqrt{1-x^2}}$.
$\frac{dz}{dx} = 3 \left( \frac{-1}{\sqrt{1-x^2}} \right) = \frac{-3}{\sqrt{1-x^2}}$.
Finally,$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-1/(2\sqrt{1-x^2})}{-3/\sqrt{1-x^2}} = \frac{1}{6}$.

(A) Given $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$.
Let $\theta=\cot ^{-1} \sqrt{\frac{1+x}{1-x}}$,then $\cot \theta = \sqrt{\frac{1+x}{1-x}}$.
Squaring both sides,we get $\cot^2 \theta = \frac{1+x}{1-x}$.
Using the identity $1+\cot^2 \theta = \csc^2 \theta$,we have $\csc^2 \theta = 1 + \frac{1+x}{1-x} = \frac{1-x+1+x}{1-x} = \frac{2}{1-x}$.
Since $\sin^2 \theta = \frac{1}{\csc^2 \theta}$,we get $\sin^2 \theta = \frac{1-x}{2}$.
Substituting this back into the expression for $y$,we get $y = \sin^2 \theta = \frac{1-x}{2}$.
Now,differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx} \left(\frac{1}{2} - \frac{x}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2}$.

(D) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Given equation is $x^2 y^2 = \frac{\pi}{2}$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(\frac{\pi}{2})$
$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$
Divide by $2x$ (assuming $x \neq 0$):
$y^2 + xy \frac{dy}{dx} = 0$
Substitute $x = 1$ and $y = 2$:
$(2)^2 + (1)(2) \frac{dy}{dx} = 0$
$4 + 2 \frac{dy}{dx} = 0$
$2 \frac{dy}{dx} = -4$
$\frac{dy}{dx} = -2$

(A) Given $y = \sin^{-1}\left(\frac{\log x^2}{1+(\log x)^2}\right) = \sin^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right)$.
Let $\log x = \tan \theta$,then $y = \sin^{-1}\left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta$.
Substituting back,$y = 2 \tan^{-1}(\log x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \times \frac{1}{1+(\log x)^2} \times \frac{1}{x}$.
At $x = 1$,$\log 1 = 0$,so $\left(\frac{dy}{dx}\right)_{x=1} = 2 \times \frac{1}{1+0^2} \times \frac{1}{1} = 2 \times 1 \times 1 = 2$.

(B) Given $y = \sec(\tan^{-1} x)$.
Let $\tan^{-1} x = \theta$,then $\tan \theta = x$.
We know that $\sec^2 \theta = 1 + \tan^2 \theta = 1 + x^2$,so $\sec \theta = \sqrt{1 + x^2}$.
Thus,$y = \sqrt{1 + x^2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}$.
At $x = 1$:
$\left(\frac{dy}{dx}\right)_{x=1} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}$.

(B) Given $y=\tan ^{-1}\left(\frac{2+3 x}{3-2 x}\right)+\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)$
We can rewrite the first term as $\tan ^{-1}\left(\frac{\frac{2}{3}+x}{1-\frac{2}{3} x}\right) = \tan ^{-1}\left(\frac{2}{3}\right) + \tan ^{-1}(x)$
We can rewrite the second term as $\tan ^{-1}\left(\frac{5 x-x}{1+(5 x)(x)}\right) = \tan ^{-1}(5 x) - \tan ^{-1}(x)$
Adding these together,we get $y = \tan ^{-1}\left(\frac{2}{3}\right) + \tan ^{-1}(x) + \tan ^{-1}(5 x) - \tan ^{-1}(x)$
Simplifying,$y = \tan ^{-1}\left(\frac{2}{3}\right) + \tan ^{-1}(5 x)$
Now,differentiating with respect to $x$:
$\frac{d y}{d x} = 0 + \frac{1}{1+(5 x)^2} \cdot \frac{d}{d x}(5 x)$
$\frac{d y}{d x} = \frac{5}{1+25 x^2}$

(B) Given $f(x) = \frac{x^2-x}{x^2+2x}$. For $x \neq 0$,we can simplify this as $f(x) = \frac{x-1}{x+2}$.
Let $y = \frac{x-1}{x+2}$.
To find $f^{-1}(x)$,we solve for $x$ in terms of $y$:
$y(x+2) = x-1$
$yx + 2y = x - 1$
$yx - x = -1 - 2y$
$x(y-1) = -(1+2y)$
$x = \frac{2y+1}{1-y}$.
Thus,$f^{-1}(x) = \frac{2x+1}{1-x}$.
Now,we differentiate $f^{-1}(x)$ with respect to $x$ using the quotient rule:
$\frac{d}{dx}(f^{-1}(x)) = \frac{d}{dx}\left(\frac{2x+1}{1-x}\right) = \frac{(1-x)(2) - (2x+1)(-1)}{(1-x)^2}$
$= \frac{2 - 2x + 2x + 1}{(1-x)^2} = \frac{3}{(1-x)^2}$.
Evaluating at $x = 2$:
$\frac{d}{dx}(f^{-1}(x))|_{x=2} = \frac{3}{(1-2)^2} = \frac{3}{(-1)^2} = 3$.

(C) Given the differential equation $\frac{dy}{dx} = y + 3$.
Separating the variables,we get $\frac{dy}{y + 3} = dx$.
Integrating both sides,we have $\int \frac{dy}{y + 3} = \int dx + C$.
This gives $\log(y + 3) = x + C$ ... $(i)$.
Given $y(0) = 2$,substitute $x = 0$ and $y = 2$ into $(i)$:
$\log(2 + 3) = 0 + C \Rightarrow C = \log 5$.
Substituting $C$ back into $(i)$,we get $\log(y + 3) = x + \log 5$.
Taking the exponential of both sides,$y + 3 = e^{x + \log 5} = 5e^x$.
Thus,$y = 5e^x - 3$.
Now,calculate $y(\log 2)$:
$y(\log 2) = 5e^{\log 2} - 3 = 5(2) - 3 = 10 - 3 = 7$.

(A) Given $y = A \cos nx + B \sin nx$ ...$(i)$
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = -An \sin nx + Bn \cos nx$
Again differentiating with respect to $x$,we get:
$\frac{d^2 y}{dx^2} = -An^2 \cos nx - Bn^2 \sin nx$
Factoring out $-n^2$:
$\frac{d^2 y}{dx^2} = -n^2 (A \cos nx + B \sin nx)$
Substituting $y$ from equation $(i)$:
$\frac{d^2 y}{dx^2} = -n^2 y$

(A) Given $y = a x^{n+1} + b x^{-n}$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = a(n+1) x^n - bn x^{-n-1}$.
Now,differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2 y}{dx^2} = a(n+1)n x^{n-1} - bn(-n-1) x^{-n-2}$.
$\frac{d^2 y}{dx^2} = a n(n+1) x^{n-1} + b n(n+1) x^{-n-2}$.
Factor out $n(n+1)$:
$\frac{d^2 y}{dx^2} = n(n+1) [a x^{n-1} + b x^{-n-2}]$.
Multiply both sides by $x^2$:
$x^2 \frac{d^2 y}{dx^2} = n(n+1) [a x^{n+1} + b x^{-n}]$.
Since $y = a x^{n+1} + b x^{-n}$,we substitute $y$ into the equation:
$x^2 \frac{d^2 y}{dx^2} = n(n+1) y$.

(D) Given: $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$.
Squaring the first equation: $(x^2+y^2)^2 = (t+\frac{1}{t})^2$.
$x^4+y^4+2x^2y^2 = t^2+\frac{1}{t^2}+2$.
Substituting $x^4+y^4 = t^2+\frac{1}{t^2}$ into the equation:
$(t^2+\frac{1}{t^2}) + 2x^2y^2 = t^2+\frac{1}{t^2}+2$.
$2x^2y^2 = 2 \implies x^2y^2 = 1$.
Thus,$y^2 = \frac{1}{x^2}$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = -2x^{-3} = -\frac{2}{x^3}$.
$\frac{dy}{dx} = -\frac{1}{x^3y}$.

(A) Given $F(x) = \left(f\left(\frac{x}{2}\right)\right)^2 + \left(g\left(\frac{x}{2}\right)\right)^2$.
Differentiating with respect to $x$,we get:
$F^{\prime}(x) = 2 f\left(\frac{x}{2}\right) \cdot f^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2} + 2 g\left(\frac{x}{2}\right) \cdot g^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}$
$F^{\prime}(x) = f\left(\frac{x}{2}\right) \cdot f^{\prime}\left(\frac{x}{2}\right) + g\left(\frac{x}{2}\right) \cdot g^{\prime}\left(\frac{x}{2}\right)$
Since $g(x) = f^{\prime}(x)$,we have $g^{\prime}(x) = f^{\prime \prime}(x) = -f(x)$.
Substituting these into the derivative expression:
$F^{\prime}(x) = f\left(\frac{x}{2}\right) \cdot g\left(\frac{x}{2}\right) + g\left(\frac{x}{2}\right) \cdot (-f\left(\frac{x}{2}\right))$
$F^{\prime}(x) = f\left(\frac{x}{2}\right) \cdot g\left(\frac{x}{2}\right) - g\left(\frac{x}{2}\right) \cdot f\left(\frac{x}{2}\right) = 0$
Since $F^{\prime}(x) = 0$,$F(x)$ is a constant function.
Given $F(5) = 5$,it follows that $F(x) = 5$ for all $x$.
Therefore,$F(10) = 5$.

(D) Given $h(x) = [f(x)]^2 + [g(x)]^2$.
Taking the derivative with respect to $x$:
$h^{\prime}(x) = 2f(x)f^{\prime}(x) + 2g(x)g^{\prime}(x)$.
Since $f^{\prime}(x) = g(x)$,we have $g^{\prime}(x) = f^{\prime \prime}(x)$.
Given $f^{\prime \prime}(x) = -f(x)$,we substitute these into the derivative:
$h^{\prime}(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$.
Since $h^{\prime}(x) = 0$,$h(x)$ is a constant function.
Therefore,$h(5) = h(10) = 1$.

(D) Given $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5 \dots (i)$
Replacing $x$ with $\frac{1}{x}$,we get $8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5 \dots (ii)$
Multiply $(i)$ by $4$ and $(ii)$ by $3$:
$32 f(x)+24 f\left(\frac{1}{x}\right)=4x+20 \dots (iii)$
$18 f(x)+24 f\left(\frac{1}{x}\right)=\frac{3}{x}+15 \dots (iv)$
Subtracting $(iv)$ from $(iii)$:
$14 f(x)=4x-\frac{3}{x}+5 \implies f(x)=\frac{1}{14}\left(4x-\frac{3}{x}+5\right)$
Then $f'(x)=\frac{1}{14}\left(4+\frac{3}{x^2}\right)$
At $x=-1$:
$f(-1)=\frac{1}{14}(-4+3+5)=\frac{4}{14}=\frac{2}{7}$
$f'(-1)=\frac{1}{14}(4+3)=\frac{7}{14}=\frac{1}{2}$
Given $y=x^2 f(x)$,differentiating w.r.t $x$:
$\frac{dy}{dx}=2x f(x)+x^2 f'(x)$
At $x=-1$:
$\left.\frac{dy}{dx}\right|_{x=-1}=2(-1)f(-1)+(-1)^2 f'(-1)$
$= -2\left(\frac{2}{7}\right) + 1\left(\frac{1}{2}\right) = -\frac{4}{7} + \frac{1}{2} = \frac{-8+7}{14} = -\frac{1}{14}$

(B) Given $h(x) = [f(x)]^2 + [g(x)]^2$.
Differentiating $h(x)$ with respect to $x$,we get:
$h^{\prime}(x) = 2f(x)f^{\prime}(x) + 2g(x)g^{\prime}(x)$.
Since $f^{\prime}(x) = g(x)$,it follows that $f^{\prime \prime}(x) = g^{\prime}(x)$.
Given $f^{\prime \prime}(x) = -f(x)$,we have $g^{\prime}(x) = -f(x)$.
Substituting these into the derivative expression:
$h^{\prime}(x) = 2f(x)g(x) + 2g(x)(-f(x)) = 2f(x)g(x) - 2f(x)g(x) = 0$.
Since $h^{\prime}(x) = 0$,$h(x)$ is a constant function.
Therefore,$h(5) = h(10) = 1$.

(D) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule,the derivative is:
$\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x)f^{\prime}(x)$.
Evaluating at $x=1$:
$\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1)f^{\prime}(1)$.
Given $f(1)=1$ and $f^{\prime}(1)=3$,we have:
$f(f(1)) = f(1) = 1$ and $f(f(f(1))) = f(1) = 1$.
Substituting these values:
$\left. \frac{dy}{dx} \right|_{x=1} = f^{\prime}(1) \cdot f^{\prime}(1) \cdot f^{\prime}(1) + 2(1)(3) = 3 \cdot 3 \cdot 3 + 6 = 27 + 6 = 33$.

(C) Let $y = f(f(f(x))) + (f(x))^2$.
Applying the chain rule,the derivative is:
$\frac{dy}{dx} = f^{\prime}(f(f(x))) \cdot f^{\prime}(f(x)) \cdot f^{\prime}(x) + 2f(x) \cdot f^{\prime}(x)$.
At $x=1$,we have:
$\left(\frac{dy}{dx}\right)_{x=1} = f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot f^{\prime}(1) + 2f(1) \cdot f^{\prime}(1)$.
Given $f(1)=1$ and $f^{\prime}(1)=5$,we substitute these values:
$= f^{\prime}(f(f(1))) \cdot f^{\prime}(f(1)) \cdot 5 + 2(1)(5)$.
Since $f(1)=1$,then $f(f(1)) = f(1) = 1$.
$= f^{\prime}(1) \cdot f^{\prime}(1) \cdot 5 + 10$.
$= 5 \cdot 5 \cdot 5 + 10 = 125 + 10 = 135$.

(B) Given $x = \sin t$,we have $\frac{dx}{dt} = \cos t$.
Given $y = a e^{t \sqrt{2}} + b e^{-t \sqrt{2}}$.
Then $\frac{dy}{dt} = a \sqrt{2} e^{t \sqrt{2}} - b \sqrt{2} e^{-t \sqrt{2}} = \sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})}{\cos t}$.
Let $y' = \frac{dy}{dx}$. Then $y' \cos t = \sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}})$.
Differentiating both sides with respect to $t$:
$y'' \cos t - y' \sin t = \sqrt{2} \cdot \sqrt{2}(a e^{t \sqrt{2}} + b e^{-t \sqrt{2}}) = 2y$.
Since $\cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2}$,we have $y'' \sqrt{1 - x^2} - y' \frac{dx}{dt} \cdot \frac{1}{\cos t} \cdot \sin t = 2y$.
Actually,using $y' = \frac{dy}{dx}$,we have $\frac{d}{dx}(y' \cos t) = \frac{d}{dx}(\sqrt{2}(a e^{t \sqrt{2}} - b e^{-t \sqrt{2}}))$.
$y'' \cos t + y'(-\sin t) \frac{dt}{dx} = 2y \frac{dt}{dx}$.
Since $\frac{dt}{dx} = \frac{1}{\cos t}$,we get $y'' \cos t - y' \frac{\sin t}{\cos t} = 2y \frac{1}{\cos t}$.
Multiplying by $\cos t$: $y'' \cos^2 t - y' \sin t = 2y$.
Substituting $\cos^2 t = 1 - x^2$ and $\sin t = x$:
$(1 - x^2) y'' - x y' = 2y$.
Comparing with $(1 - x^2) y'' - x y' = ky$,we get $k = 2$.

(D) Given,$f(x) = \frac{\log_2(x+3)}{x^2+3x+2}$.
For the function to be defined,the denominator must not be zero:
$x^2 + 3x + 2 \neq 0$
$(x+1)(x+2) \neq 0$
$x \neq -1$ and $x \neq -2$.
Also,the argument of the logarithm must be positive:
$x + 3 > 0$
$x > -3$.
Combining these conditions,the domain is $x \in (-3, \infty) - \{-1, -2\}$.

(A) For the function $f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}}$ to be defined,the expression inside the square root must be non-negative:
$\sin^{-1}(2x) + \frac{\pi}{6} \geq 0$
$\sin^{-1}(2x) \geq -\frac{\pi}{6}$
Also,the domain of $\sin^{-1}(\theta)$ is $[-1, 1]$,so $-1 \leq 2x \leq 1$,which implies $-0.5 \leq x \leq 0.5$.
The range of $\sin^{-1}(2x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Combining these,we have $-\frac{\pi}{6} \leq \sin^{-1}(2x) \leq \frac{\pi}{2}$.
Taking the sine of all parts:
$\sin(-\frac{\pi}{6}) \leq 2x \leq \sin(\frac{\pi}{2})$
$-\frac{1}{2} \leq 2x \leq 1$
Dividing by $2$:
$-\frac{1}{4} \leq x \leq \frac{1}{2}$
Thus,the domain is $\left[-\frac{1}{4}, \frac{1}{2}\right]$.

(D) Given the equation $2^x + 2^y = 2$.
Rearranging for $2^y$,we get $2^y = 2 - 2^x$.
Taking the logarithm base $2$ on both sides,$y = \log_2(2 - 2^x)$.
For the function to be defined,the argument of the logarithm must be strictly positive: $2 - 2^x > 0$.
This implies $2^x < 2$.
Since $2 = 2^1$,we have $2^x < 2^1$.
Since the base $2 > 1$,the inequality holds for $x < 1$.
Thus,the domain is $(-\infty, 1)$,which is written as $-\infty < x < 1$.

(D) Given the equation $2^x + 2^y = 2$.
Rearranging for $2^y$,we get $2^y = 2 - 2^x$.
Since $2^y > 0$ for all real $y$,the expression $2 - 2^x$ must be strictly greater than $0$.
$2 - 2^x > 0$
$2^x < 2$
$2^x < 2^1$
Since the base $2 > 1$,the inequality holds for the exponents:
$x < 1$.
Thus,the domain of the function is $(-\infty, 1)$,which is written as $-\infty < x < 1$.

(D) Given equation: $[x]^2-5[x]+6=0$
Let $[x] = a$.
Then the equation becomes $a^2-5a+6=0$.
Factoring the quadratic equation: $(a-2)(a-3)=0$.
This gives $a=2$ or $a=3$.
Substituting back $[x]=a$,we get $[x]=2$ or $[x]=3$.
For $[x]=2$,the range of $x$ is $x \in [2,3)$.
For $[x]=3$,the range of $x$ is $x \in [3,4)$.
Combining these intervals,we get $x \in [2,3) \cup [3,4) = [2,4)$.

(B) For the function $f(x)$ to be defined,the expression inside the square root must be strictly positive and the argument of $\sin^{-1}$ must be in the interval $[-1, 1]$.
$1$. For the denominator: $9 - x^2 > 0$ $\Rightarrow x^2 < 9$ $\Rightarrow -3 < x < 3 \dots (i)$.
$2$. For the numerator: $-1 \leq x - 3 \leq 1 \Rightarrow 2 \leq x \leq 4 \dots (ii)$.
$3$. Taking the intersection of $(i)$ and $(ii)$,we get $2 \leq x < 3$.
Thus,the domain is $[2, 3)$.

(C) Let $y = \frac{x^2+x+2}{x^2+x+1}$.
$y(x^2+x+1) = x^2+x+2$
$(y-1)x^2 + (y-1)x + (y-2) = 0$.
For $x \in R$,the discriminant $D \geq 0$.
$D = (y-1)^2 - 4(y-1)(y-2) \geq 0$.
$(y-1)[(y-1) - 4(y-2)] \geq 0$.
$(y-1)(-3y+7) \geq 0$.
$(y-1)(3y-7) \leq 0$.
Thus,$1 < y \leq \frac{7}{3}$.
Since $x^2+x+1 = (x+1/2)^2 + 3/4 > 0$,$y$ can never be $1$ because the numerator $x^2+x+2$ is always greater than the denominator $x^2+x+1$.
Therefore,the range is $\left(1, \frac{7}{3}\right]$.

(D) Given $f(x) = \log_{e}\left(\frac{1-x}{1+x}\right)$.
We need to evaluate $f\left(\frac{2x}{1+x^2}\right)$.
Substituting $\frac{2x}{1+x^2}$ for $x$ in $f(x)$:
$f\left(\frac{2x}{1+x^2}\right) = \log_{e}\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)$
$= \log_{e}\left(\frac{\frac{1+x^2-2x}{1+x^2}}{\frac{1+x^2+2x}{1+x^2}}\right)$
$= \log_{e}\left(\frac{(1-x)^2}{(1+x)^2}\right)$
$= \log_{e}\left(\left(\frac{1-x}{1+x}\right)^2\right)$
$= 2\log_{e}\left(\frac{1-x}{1+x}\right)$
$= 2f(x)$.

(D) Given that $g(x)=x^2+x-1$ and $(g \circ f)(x)=4 x^2-10 x+5$.
By definition,$(g \circ f)(x) = g(f(x))$.
So,$(f(x))^2 + f(x) - 1 = 4x^2 - 10x + 5$.
To find $f(2)$,we substitute $x=2$ into the equation:
$(f(2))^2 + f(2) - 1 = 4(2)^2 - 10(2) + 5$.
$(f(2))^2 + f(2) - 1 = 4(4) - 20 + 5$.
$(f(2))^2 + f(2) - 1 = 16 - 20 + 5$.
$(f(2))^2 + f(2) - 1 = 1$.
$(f(2))^2 + f(2) - 2 = 0$.
Let $y = f(2)$,then $y^2 + y - 2 = 0$.
Factoring the quadratic equation: $(y+2)(y-1) = 0$.
Thus,$y = -2$ or $y = 1$.
Checking the options,$-2$ is provided as option $D$.

(C) Given $f(x) = x^3 + e^{\frac{x}{2}}$.
First,find the value of $x$ such that $f(x) = 1$.
$x^3 + e^{\frac{x}{2}} = 1$.
By inspection,if $x = 0$,then $f(0) = 0^3 + e^0 = 0 + 1 = 1$.
Thus,$f(0) = 1$,which implies $g(1) = 0$.
We know that $g(f(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule:
$g^{\prime}(f(x)) \cdot f^{\prime}(x) = 1$.
To find $g^{\prime}(1)$,we set $x = 0$:
$g^{\prime}(f(0)) \cdot f^{\prime}(0) = 1 \Rightarrow g^{\prime}(1) \cdot f^{\prime}(0) = 1$.
Now,calculate $f^{\prime}(x)$:
$f^{\prime}(x) = 3x^2 + \frac{1}{2} e^{\frac{x}{2}}$.
$f^{\prime}(0) = 3(0)^2 + \frac{1}{2} e^0 = 0 + \frac{1}{2} = \frac{1}{2}$.
Therefore,$g^{\prime}(1) = \frac{1}{f^{\prime}(0)} = \frac{1}{1/2} = 2$.

(D) Given $f(x) = \frac{x}{2-x}$ and $g(x) = \frac{x+1}{x+2}$.
First,find $(g \circ f)(x) = g(f(x)) = \frac{f(x)+1}{f(x)+2} = \frac{\frac{x}{2-x} + 1}{\frac{x}{2-x} + 2} = \frac{\frac{x+2-x}{2-x}}{\frac{x+4-2x}{2-x}} = \frac{2}{4-x}$.
Now,find $(g \circ g \circ f)(x) = g((g \circ f)(x)) = g\left(\frac{2}{4-x}\right)$.
Substitute $x = \frac{2}{4-x}$ into $g(x) = \frac{x+1}{x+2}$:
$(g \circ g \circ f)(x) = \frac{\frac{2}{4-x} + 1}{\frac{2}{4-x} + 2} = \frac{\frac{2 + 4 - x}{4-x}}{\frac{2 + 8 - 2x}{4-x}} = \frac{6-x}{10-2x}$.

(B) Given the function $f(x) = \frac{\alpha x}{x+1}$.
We need to find $f(f(x))$:
$f(f(x)) = f\left(\frac{\alpha x}{x+1}\right) = \frac{\alpha \left(\frac{\alpha x}{x+1}\right)}{\frac{\alpha x}{x+1} + 1}$
Simplify the expression:
$f(f(x)) = \frac{\frac{\alpha^2 x}{x+1}}{\frac{\alpha x + x + 1}{x+1}} = \frac{\alpha^2 x}{(\alpha + 1)x + 1}$
Given $f(f(x)) = x$,so:
$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$
Assuming $x \neq 0$,we divide by $x$:
$\frac{\alpha^2}{(\alpha + 1)x + 1} = 1$
$\alpha^2 = (\alpha + 1)x + 1$
For this to hold for all $x$ in the domain,the coefficient of $x$ must be $0$ and the constant term must be $1$:
$\alpha + 1 = 0 \implies \alpha = -1$
Checking the constant term: $(-1)^2 = 1$,which is true.
Thus,$\alpha = -1$.

(A) Given $f(x) = x + \frac{1}{x}$.
Let $y = f(x) = x + \frac{1}{x}$.
Multiplying by $x$,we get $xy = x^2 + 1$,which implies $x^2 - xy + 1 = 0$.
Using the quadratic formula to solve for $x$:
$x = \frac{y \pm \sqrt{y^2 - 4}}{2}$.
Since the domain is $x \in [1, \infty)$,we must have $x \ge 1$.
If we take $x = \frac{y - \sqrt{y^2 - 4}}{2}$,then for $y=2$,$x=1$,but for $y > 2$,this value would be less than $1$.
Thus,we choose the positive root: $x = \frac{y + \sqrt{y^2 - 4}}{2}$.
Therefore,$f^{-1}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

(A) Given: $f(x)=\frac{a-x}{a+x}$.
Since $(f \circ f)(x)=x$,we have:
$f(f(x)) = \frac{a-f(x)}{a+f(x)} = x$
Substituting $f(x) = \frac{a-x}{a+x}$:
$\frac{a-\frac{a-x}{a+x}}{a+\frac{a-x}{a+x}} = x$
$\frac{a(a+x)-(a-x)}{a(a+x)+(a-x)} = x$
$\frac{a^2+ax-a+x}{a^2+ax+a-x} = x$
$a^2+ax-a+x = x(a^2+ax+a-x)$
$a^2+ax-a+x = a^2x+ax^2+ax-x^2$
$(a-1)x^2 + (a^2-1)x - (a^2-a) = 0$
$(a-1)x^2 + (a-1)(a+1)x - a(a-1) = 0$
For this to hold for all $x$,we must have $a-1=0$,so $a=1$.
Thus,$f(x) = \frac{1-x}{1+x}$.
Now,$f\left(-\frac{1}{5}\right) = \frac{1-(-1/5)}{1+(-1/5)} = \frac{1+1/5}{1-1/5} = \frac{6/5}{4/5} = \frac{6}{4} = 1.5$.

(B) Given $f(x) = (x+1)^2 - 1$ for $x \geqslant -1$.
Since $f(x)$ is an increasing function for $x \geqslant -1$,the solutions to $f(x) = f^{-1}(x)$ are the same as the solutions to $f(x) = x$.
Setting $f(x) = x$:
$(x+1)^2 - 1 = x$
$x^2 + 2x + 1 - 1 = x$
$x^2 + x = 0$
$x(x+1) = 0$
This gives $x = 0$ or $x = -1$.
Both values satisfy the condition $x \geqslant -1$.
Thus,the set is $\{0, -1\}$.

(C) Given equation: $[x]^2-5[x]+6=0$
Let $[x] = a$.
Then the equation becomes $a^2 - 5a + 6 = 0$.
Factoring the quadratic equation: $(a-2)(a-3) = 0$.
This gives $a = 2$ or $a = 3$.
Substituting back $[x] = a$,we get $[x] = 2$ or $[x] = 3$.
By the definition of the greatest integer function:
If $[x] = 2$,then $x \in [2, 3)$.
If $[x] = 3$,then $x \in [3, 4)$.
Combining these intervals,we get $x \in [2, 3) \cup [3, 4) = [2, 4)$.
Thus,the correct option is $C$.

(D) The equation of the hyperbola is $x^2-y^2=9$,which can be written as $\frac{x^2}{3^2}-\frac{y^2}{3^2}=1$.
Here,$a=3$ and $b=3$.
The eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{9}{9}} = \sqrt{2}$.
The latus rectum is at $x=ae = 3\sqrt{2}$.
The area bounded by the hyperbola and its latus rectum in the first quadrant is $\int_{a}^{ae} y \, dx = \int_{3}^{3\sqrt{2}} \sqrt{x^2-9} \, dx$.
Since the hyperbola is symmetric about both axes,the total area is $4 \int_{3}^{3\sqrt{2}} \sqrt{x^2-9} \, dx$.
Using the formula $\int \sqrt{x^2-a^2} \, dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|$,we get:
Area $= 4 \left[ \frac{x}{2}\sqrt{x^2-9} - \frac{9}{2}\log|x+\sqrt{x^2-9}| \right]_{3}^{3\sqrt{2}}$.
$= 4 \left[ (\frac{3\sqrt{2}}{2}\sqrt{18-9} - \frac{9}{2}\log|3\sqrt{2}+\sqrt{18-9}|) - (0 - \frac{9}{2}\log|3+0|) \right]$.
$= 4 \left[ \frac{3\sqrt{2}}{2}(3) - \frac{9}{2}\log(3\sqrt{2}+3) + \frac{9}{2}\log(3) \right]$.
$= 4 \left[ \frac{9\sqrt{2}}{2} - \frac{9}{2}\log(\frac{3\sqrt{2}+3}{3}) \right]$.
$= 4 \left[ \frac{9\sqrt{2}}{2} - \frac{9}{2}\log(\sqrt{2}+1) \right]$.
$= 18[\sqrt{2}-\log(\sqrt{2}+1)]$ sq. units.

(A) We need to evaluate the integral $I = \int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$.
Using trigonometric identities,$1 - \sin x = 1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$.
Substituting these into the integral:
$I = \int e^x \left( \frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2 \sin^2 \frac{x}{2}} - \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx$
$I = \int e^x \left( \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2} - \cot \frac{x}{2} \right) dx$.
Let $f(x) = -\cot \frac{x}{2}$. Then $f'(x) = -(-\operatorname{cosec}^2 \frac{x}{2}) \cdot \frac{1}{2} = \frac{1}{2} \operatorname{cosec}^2 \frac{x}{2}$.
Since the integral is of the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$,
$I = e^x (-\cot \frac{x}{2}) + c = -e^x \cot \frac{x}{2} + c$.

(B) Let $I = \int \frac{x^2+1}{(x+1)^2} dx$.
We can rewrite the numerator as $x^2+1 = (x^2+2x+1) - 2x = (x+1)^2 - 2x$.
So,$I = \int \frac{(x+1)^2 - 2x}{(x+1)^2} dx = \int 1 dx - \int \frac{2x}{(x+1)^2} dx$.
For the second integral,write $2x = 2(x+1) - 2$.
$I = x - \int \frac{2(x+1)-2}{(x+1)^2} dx = x - 2 \int \frac{1}{x+1} dx + 2 \int \frac{1}{(x+1)^2} dx$.
Integrating these terms,we get $I = x - 2 \log |x+1| + 2 \left( -\frac{1}{x+1} \right) + c$.
$I = x - 2 \log |x+1| - \frac{2}{x+1} + c$.

(A) Let $I = \int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x$
Using the identities $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$:
$I = \int \tan ^{-1}\left(\sqrt{\frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}}\right) d x$
$I = \int \tan ^{-1}\left(\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}\right) d x$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$I = \int \tan ^{-1}\left(\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}\right) d x$
Since $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$:
$I = \int \tan ^{-1}\left(\tan \left(\frac{\pi}{4} - \frac{x}{2}\right)\right) d x$
$I = \int \left(\frac{\pi}{4} - \frac{x}{2}\right) d x$
$I = \frac{\pi}{4} x - \frac{x^2}{4} + c$

(B) We are given the integral $I = \int \frac{4 e^x-25}{2 e^x-5} dx$.
To solve this,we rewrite the numerator in terms of the denominator:
$4 e^x - 25 = 2(2 e^x - 5) - 15$.
Wait,let's adjust the split:
$4 e^x - 25 = 2(2 e^x - 5) - 15$.
Actually,a better split is:
$4 e^x - 25 = 2(2 e^x - 5) - 15$.
Let's re-evaluate:
$4 e^x - 25 = 2(2 e^x - 5) - 15$.
Wait,$2(2 e^x - 5) = 4 e^x - 10$.
So,$4 e^x - 25 = (4 e^x - 10) - 15 = 2(2 e^x - 5) - 15$.
Thus,$\int \frac{4 e^x-25}{2 e^x-5} dx = \int \left( 2 - \frac{15}{2 e^x-5} \right) dx$.
This does not match the form $Ax + B \log(2 e^x - 5)$.
Let's try another approach:
Multiply numerator and denominator by $e^{-x}$:
$\int \frac{4 - 25 e^{-x}}{2 - 5 e^{-x}} dx$.
Let $u = 2 - 5 e^{-x}$,then $du = 5 e^{-x} dx$,so $dx = \frac{du}{5 e^{-x}} = \frac{du}{5(2-u)/5} = \frac{du}{2-u}$.
This is complex. Let's use the method:
$4 e^x - 25 = A(2 e^x - 5) + B(2 e^x)$.
$4 e^x - 25 = (2A + 2B) e^x - 5A$.
Comparing coefficients:
$-5A = -25 \implies A = 5$.
$2A + 2B = 4 \implies 10 + 2B = 4 \implies 2B = -6 \implies B = -3$.
So,$\int \frac{4 e^x-25}{2 e^x-5} dx = \int \left( 5 - 3 \frac{2 e^x}{2 e^x-5} \right) dx = 5x - 3 \log|2 e^x - 5| + c$.
Therefore,$A=5$ and $B=-3$.

(B) We know that $x^4+x^2+1 = (x^2+1)^2 - x^2 = (x^2+x+1)(x^2-x+1)$.
Therefore,the integral becomes:
$I = \int \frac{(x^2+x+1)(x^2-x+1)}{x^2-x+1} \,d x$
$I = \int (x^2+x+1) \,d x$
Integrating term by term:
$I = \frac{x^3}{3} + \frac{x^2}{2} + x + c$
Thus,the correct option is $B$.

(B) Given that $f^{\prime}(x)=4 x^3-3 x^{-4}$.
Integrating both sides with respect to $x$:
$f(x) = \int (4x^3 - 3x^{-4}) dx$
$f(x) = 4 \cdot \frac{x^4}{4} - 3 \cdot \frac{x^{-3}}{-3} + c$
$f(x) = x^4 + \frac{1}{x^3} + c$
Given that $f(2) = 0$,substitute $x = 2$:
$0 = (2)^4 + \frac{1}{2^3} + c$
$0 = 16 + \frac{1}{8} + c$
$0 = \frac{128+1}{8} + c$
$c = -\frac{129}{8}$
Therefore,$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$.

(C) First, perform polynomial long division on the integrand $\frac{x^3-7x+6}{x^2+3x}$.
Dividing $x^3-7x+6$ by $x^2+3x$ gives a quotient of $(x-3)$ and a remainder of $(2x+6)$.
Thus, $\frac{x^3-7x+6}{x^2+3x} = x-3 + \frac{2x+6}{x^2+3x}$.
Simplify the remainder term: $\frac{2x+6}{x^2+3x} = \frac{2(x+3)}{x(x+3)} = \frac{2}{x}$ for $x \neq -3$.
Now, integrate the expression: $\int (x-3+\frac{2}{x}) dx$.
$= \int x dx - \int 3 dx + \int \frac{2}{x} dx$.
$= \frac{x^2}{2} - 3x + 2 \log |x| + c$.

(C) Let $I = \int 3^{3^x} \cdot 3^x \, dx$.
Substitute $t = 3^x$.
Then,$dt = 3^x \log 3 \, dx$,which implies $3^x \, dx = \frac{1}{\log 3} \, dt$.
Substituting these into the integral:
$I = \int 3^t \cdot \frac{1}{\log 3} \, dt$.
$I = \frac{1}{\log 3} \int 3^t \, dt$.
Using the formula $\int a^x \, dx = \frac{a^x}{\log a} + c$:
$I = \frac{1}{\log 3} \cdot \frac{3^t}{\log 3} + c$.
$I = \frac{3^t}{(\log 3)^2} + c$.
Substituting $t = 3^x$ back:
$I = \frac{3^{3^x}}{(\log 3)^2} + c$.

(D) Let $I = \int \frac{\log \sqrt{x}}{3 x} \,d x$.
We know that $\log \sqrt{x} = \log (x^{1/2}) = \frac{1}{2} \log x$.
Substituting this into the integral:
$I = \int \frac{\frac{1}{2} \log x}{3 x} \,d x = \frac{1}{6} \int \frac{\log x}{x} \,d x$.
Let $u = \log x$,then $du = \frac{1}{x} \,d x$.
The integral becomes $I = \frac{1}{6} \int u \,du = \frac{1}{6} \cdot \frac{u^2}{2} + c = \frac{u^2}{12} + c$.
Substituting $u = \log x$ back,we get $I = \frac{1}{12} (\log x)^2 + c$.

(D) Given $I = \int \frac{dx}{x^2(x^4+1)^{3/4}}$.
We can rewrite the integral by taking $x^4$ common from the bracket:
$I = \int \frac{dx}{x^2 \left(x^4(1 + \frac{1}{x^4})\right)^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{3/4}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{3/4}}$.
Let $t = 1 + \frac{1}{x^4}$.
Then $dt = -\frac{4}{x^5} dx$,which implies $\frac{dx}{x^5} = -\frac{1}{4} dt$.
Substituting these into the integral:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \left( \frac{t^{1/4}}{1/4} \right) + c = -t^{1/4} + c$.
Substituting back $t = 1 + \frac{1}{x^4}$:
$I = -\left(1 + \frac{1}{x^4}\right)^{1/4} + c$.

(B) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$.
Substitute $2x-1 = t^2$,which implies $x = \frac{t^2+1}{2}$ and $dx = t \, dt$.
Then $x+1 = \frac{t^2+1}{2} + 1 = \frac{t^2+3}{2}$.
Substituting these into the integral:
$I = \int \frac{(\frac{t^2+3}{2}) t \, dt}{t} = \int \frac{t^2+3}{2} \, dt$.
$I = \frac{1}{2} (\frac{t^3}{3} + 3t) + c = \frac{t^3}{6} + \frac{3t}{2} + c$.
Factor out $\frac{t}{6}$:
$I = \frac{t}{6} (t^2 + 9) + c$.
Substitute $t = \sqrt{2x-1}$ back:
$I = \frac{\sqrt{2x-1}}{6} (2x-1+9) + c = \frac{\sqrt{2x-1}}{6} (2x+8) + c$.
$I = \sqrt{2x-1} \cdot \frac{2(x+4)}{6} + c = \sqrt{2x-1} \cdot \frac{x+4}{3} + c$.
Comparing this with $f(x) \sqrt{2x-1} + c$,we get $f(x) = \frac{x+4}{3}$.

(B) Let $I = \int(2x+4)\sqrt{x-1}dx$.
Substitute $t = x-1$,so $x = t+1$ and $dx = dt$.
Then $I = \int(2(t+1)+4)\sqrt{t}dt = \int(2t+6)\sqrt{t}dt$.
$I = \int(2t^{3/2} + 6t^{1/2})dt$.
Integrating term by term:
$I = 2 \cdot \frac{t^{5/2}}{5/2} + 6 \cdot \frac{t^{3/2}}{3/2} + c$.
$I = \frac{4}{5}t^{5/2} + 4t^{3/2} + c$.
Substituting back $t = x-1$:
$I = \frac{4}{5}(x-1)^{5/2} + 4(x-1)^{3/2} + c$.
Comparing this with $a(x-1)^{5/2} + b(x-1)^{3/2} + c$,we get $a = \frac{4}{5}$ and $b = 4$.
Therefore,$2a+b = 2(\frac{4}{5}) + 4 = \frac{8}{5} + 4 = \frac{8+20}{5} = \frac{28}{5}$.