MHT CET 2022 Mathematics Question Paper with Answer and Solution

(B) Given $y = \tan^{-1}(\sec x - \tan x)$.
We can rewrite the expression inside the inverse tangent function as:
$\sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$
Using the half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$,and $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$,we get:
$\frac{1 - \sin x}{\cos x} = \frac{(\cos\frac{x}{2} - \sin\frac{x}{2})^2}{(\cos\frac{x}{2} - \sin\frac{x}{2})(\cos\frac{x}{2} + \sin\frac{x}{2})} = \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}$
Dividing the numerator and denominator by $\cos\frac{x}{2}$,we get:
$\frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}} = \tan(\frac{\pi}{4} - \frac{x}{2})$
Thus,$y = \tan^{-1}(\tan(\frac{\pi}{4} - \frac{x}{2})) = \frac{\pi}{4} - \frac{x}{2}$
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} - \frac{x}{2}) = -\frac{1}{2}$

(C) Given $f^{\prime}(x) = \tan^{-1}(\sec x + \tan x) = \tan^{-1}\left(\frac{1+\sin x}{\cos x}\right)$.
Using trigonometric identities,$\frac{1+\sin x}{\cos x} = \frac{(\cos(x/2) + \sin(x/2))^2}{\cos^2(x/2) - \sin^2(x/2)} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} = \tan(\frac{\pi}{4} + \frac{x}{2})$.
Thus,$f^{\prime}(x) = \tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$.
Now,$f(x) = \int f^{\prime}(x) dx = \int (\frac{\pi}{4} + \frac{x}{2}) dx = \frac{\pi}{4}x + \frac{x^2}{4} + C$.
Since $f(0) = 0$,we have $0 = 0 + 0 + C$,so $C = 0$.
Therefore,$f(x) = \frac{x^2 + \pi x}{4}$.
Substituting $x = 1$,we get $f(1) = \frac{1^2 + \pi(1)}{4} = \frac{\pi+1}{4}$.

(B) Let $x = \cos \theta$,then $\theta = \cos^{-1} x$.
Substituting $x = \cos \theta$ in the expression:
$y = \sin \left(2 \tan^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)$
Using the identities $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $1-\cos \theta = 2 \sin^2 \frac{\theta}{2}$:
$y = \sin \left(2 \tan^{-1} \sqrt{\frac{2 \cos^2 \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}}\right) = \sin \left(2 \tan^{-1} \left(\cot \frac{\theta}{2}\right)\right)$
Since $\cot \frac{\theta}{2} = \tan \left(\frac{\pi}{2} - \frac{\theta}{2}\right)$:
$y = \sin \left(2 \left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) = \sin (\pi - \theta) = \sin \theta$
Since $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - x^2}$:
$y = \sqrt{1 - x^2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2 \sqrt{1 - x^2}} \times \frac{d}{dx}(1 - x^2) = \frac{1}{2 \sqrt{1 - x^2}} \times (-2x) = \frac{-x}{\sqrt{1 - x^2}}$

(B) Given $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$.
Simplify the expression inside the $\sec^{-1}$ function:
$\frac{x+x^{-1}}{x-x^{-1}} = \frac{x+\frac{1}{x}}{x-\frac{1}{x}} = \frac{x^2+1}{x^2-1} = -\frac{1+x^2}{1-x^2}$.
Let $x = \tan \theta$,then $\theta = \tan^{-1} x$.
The expression becomes $-\frac{1+\tan^2 \theta}{1-\tan^2 \theta} = -\frac{1}{\cos 2\theta} = -\sec 2\theta$.
So,$y = \sec^{-1}(-\sec 2\theta)$.
Using the identity $\sec^{-1}(-z) = \pi - \sec^{-1}(z)$,we get $y = \pi - \sec^{-1}(\sec 2\theta) = \pi - 2\theta$.
Substituting back $\theta = \tan^{-1} x$,we have $y = \pi - 2\tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\pi) - 2\frac{d}{dx}(\tan^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}\right) = -\frac{2}{1+x^2}$.

(B) Given $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$.
We can rewrite the terms inside the inverse tangent functions:
$y = \tan^{-1}\left(\frac{5x-x}{1+5x \cdot x}\right) + \tan^{-1}\left(\frac{\frac{3}{8}+x}{1-\frac{3}{8} \cdot x}\right)$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ and $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$:
$y = (\tan^{-1}(5x) - \tan^{-1}(x)) + (\tan^{-1}(\frac{3}{8}) + \tan^{-1}(x))$.
Simplifying the expression,we get:
$y = \tan^{-1}(5x) + \tan^{-1}(\frac{3}{8})$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(5x)) + \frac{d}{dx}(\tan^{-1}(\frac{3}{8}))$.
Since $\tan^{-1}(\frac{3}{8})$ is a constant,its derivative is $0$:
$\frac{dy}{dx} = \frac{1}{1+(5x)^2} \cdot \frac{d}{dx}(5x) + 0$.
$\frac{dy}{dx} = \frac{5}{1+25x^2}$.

(A) Given $y = \sin^{-1} \left( \frac{5}{13}x + \frac{12}{13}\sqrt{1-x^2} \right)$.
Let $x = \sin \theta$,then $\sqrt{1-x^2} = \cos \theta$.
Let $\cos \alpha = \frac{5}{13}$,then $\sin \alpha = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$.
Substituting these into the equation:
$y = \sin^{-1} (\sin \theta \cos \alpha + \cos \theta \sin \alpha)$
$y = \sin^{-1} (\sin(\theta + \alpha))$
$y = \theta + \alpha$
Since $\theta = \sin^{-1} x$ and $\alpha = \cos^{-1} (\frac{5}{13})$ (a constant),
$y = \sin^{-1} x + \cos^{-1} (\frac{5}{13})$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\sin^{-1} x) + \frac{d}{dx} (\cos^{-1} (\frac{5}{13}))$
$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + 0 = \frac{1}{\sqrt{1-x^2}}$.

(B) Let $x = \tan \theta$. Since $x \in (1, \infty)$,we have $\theta \in (\frac{\pi}{4}, \frac{\pi}{2})$.
Then $2\theta \in (\frac{\pi}{2}, \pi)$.
$f(x) = \sin^{-1}\left(\frac{2\tan \theta}{1+\tan^2 \theta}\right) + \cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right)$
$f(x) = \sin^{-1}(\sin 2\theta) + \cos^{-1}(\cos 2\theta)$
Since $2\theta \in (\frac{\pi}{2}, \pi)$,$\sin^{-1}(\sin 2\theta) = \pi - 2\theta$ and $\cos^{-1}(\cos 2\theta) = 2\theta$.
Thus,$f(x) = (\pi - 2\theta) + 2\theta = \pi$.
Since $f(x) = \pi$ is a constant function,its derivative $f'(x) = 0$.

(A) Given $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$.
We can rewrite the expression inside the inverse tangent function as:
$y=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}\right)$.
Using the identity $\tan ^{-1}(A)+\tan ^{-1}(B)=\tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{d y}{d x}=\frac{d}{d x}(\tan ^{-1}(3 x+2))+\frac{d}{d x}(\tan ^{-1}(2 x-1))$.
$\frac{d y}{d x}=\frac{1}{1+(3 x+2)^2} \times 3+\frac{1}{1+(2 x-1)^2} \times 2$.
$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+4+1}+\frac{2}{4 x^2-4 x+1+1}$.
$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+5}+\frac{2}{4 x^2-4 x+2}$.
Simplifying the second term:
$\frac{d y}{d x}=\frac{3}{9 x^2+12 x+5}+\frac{1}{2 x^2-2 x+1}$.

(C) $\text{Given } f(x) = \begin{cases} \frac{1}{2}(-x-1), & x < -1 \\ \tan^{-1} x, & -1 \le x \le 1 \\ \frac{1}{2}(x-1), & x > 1 \end{cases}$
$\text{The derivative } f'(x) \text{ is defined as:}$
$f'(x) = \begin{cases} -\frac{1}{2}, & x < -1 \\ \frac{1}{1+x^2}, & -1 < x < 1 \\ \frac{1}{2}, & x > 1 \end{cases}$
$\text{At } x = -1: \text{ Left-hand derivative } = -\frac{1}{2}, \text{ Right-hand derivative } = \frac{1}{1+(-1)^2} = \frac{1}{2}. \text{ Since } -\frac{1}{2} \neq \frac{1}{2}, f(x) \text{ is not differentiable at } x = -1.$
$\text{At } x = 1: \text{ Left-hand derivative } = \frac{1}{1+1^2} = \frac{1}{2}, \text{ Right-hand derivative } = \frac{1}{2}. \text{ However, check continuity: } f(1) = \tan^{-1}(1) = \frac{\pi}{4}, \text{ but } \lim_{x \to 1^+} f(x) = \frac{1}{2}(1-1) = 0. \text{ Since } f(x) \text{ is discontinuous at } x = 1, \text{ it is not differentiable at } x = 1.$
$\text{Thus, the domain of } f'(x) \text{ is } R - \{-1, 1\}.$

(A) Let $y = \tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)$.
We can rewrite the expression inside the inverse tangent as $y = \tan ^{-1}\left(\frac{2(3 x^{3/2})}{1-(3 x^{3/2})^2}\right)$.
Using the identity $2 \tan ^{-1}(\theta) = \tan ^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,we get $y = 2 \tan ^{-1}(3 x^{3/2})$.
Now,differentiate $y$ with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(3 x^{3/2})^2} \cdot \frac{d}{dx}(3 x^{3/2})$.
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+9 x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+9 x^3} \cdot \frac{9}{2} \sqrt{x}$.
$\frac{dy}{dx} = \frac{9}{1+9 x^3} \cdot \sqrt{x}$.
Comparing this with $\sqrt{x} \cdot g(x)$,we find $g(x) = \frac{9}{1+9 x^3}$.

(D) Let $y = \sqrt{\frac{1-\tan x}{1+\tan x}}$.
Using the chain rule,$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1-\tan x}{1+\tan x}}} \cdot \frac{d}{dx}\left(\frac{1-\tan x}{1+\tan x}\right)$.
Applying the quotient rule to the inner term:
$\frac{d}{dx}\left(\frac{1-\tan x}{1+\tan x}\right) = \frac{(1+\tan x)(-\sec^2 x) - (1-\tan x)(\sec^2 x)}{(1+\tan x)^2} = \frac{-\sec^2 x - \tan x \sec^2 x - \sec^2 x + \tan x \sec^2 x}{(1+\tan x)^2} = \frac{-2\sec^2 x}{(1+\tan x)^2}$.
Now,substitute this back:
$\frac{dy}{dx} = \frac{1}{2} \sqrt{\frac{1+\tan x}{1-\tan x}} \cdot \left(\frac{-2\sec^2 x}{(1+\tan x)^2}\right) = \frac{-\sec^2 x}{\sqrt{1-\tan x} \cdot (1+\tan x)^{3/2}}$.

(B) Given $y = \cos(\sin x^2)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \frac{d}{dx}(\sin x^2)$
$\frac{dy}{dx} = -\sin(\sin x^2) \cdot \cos(x^2) \cdot 2x$
Now,substitute $x = \sqrt{\frac{\pi}{2}}$ into the derivative:
$\frac{dy}{dx} = -\sin(\sin(\frac{\pi}{2})) \cdot \cos(\frac{\pi}{2}) \cdot 2\sqrt{\frac{\pi}{2}}$
Since $\cos(\frac{\pi}{2}) = 0$,the entire expression becomes:
$\frac{dy}{dx} = -\sin(1) \cdot 0 \cdot 2\sqrt{\frac{\pi}{2}} = 0$.

(B) Let $y = \tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$.
Substitute $x = \cos 2\theta$,which implies $\theta = \frac{1}{2} \cos ^{-1} x$.
Then $y = \tan ^{-1}\left(\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right)$.
Using trigonometric identities $1+\cos 2\theta = 2\cos^2\theta$ and $1-\cos 2\theta = 2\sin^2\theta$,we get:
$y = \tan ^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right) = \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)$.
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1-\tan\theta}{1+\tan\theta}$,we have:
$y = \tan ^{-1}\tan(\frac{\pi}{4} - \theta) = \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2} \cos ^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{2} \times \left(-\frac{1}{\sqrt{1-x^2}}\right) = \frac{1}{2 \sqrt{1-x^2}}$.

(A) We know the trigonometric identity $\cos^2 \theta - \sin^2 \theta = \cos(2\theta)$.
Given $y = \cos^2\left(\frac{5x}{2}\right) - \sin^2\left(\frac{5x}{2}\right)$,we can substitute $\theta = \frac{5x}{2}$.
Thus,$y = \cos\left(2 \times \frac{5x}{2}\right) = \cos(5x)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = -\sin(5x) \times \frac{d}{dx}(5x) = -5\sin(5x)$.
Differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2y}{dx^2} = -5 \times \cos(5x) \times \frac{d}{dx}(5x) = -5 \times 5 \cos(5x) = -25\cos(5x)$.
Since $y = \cos(5x)$,we have $\frac{d^2y}{dx^2} = -25y$.

(B) Given the function $y = e^{4x} + 2e^{-x}$.
First,find the first derivative: $\frac{dy}{dx} = 4e^{4x} - 2e^{-x}$.
Next,find the second derivative: $\frac{d^2y}{dx^2} = 16e^{4x} + 2e^{-x}$.
Substitute these into the given equation $\frac{d^2y}{dx^2} + A\frac{dy}{dx} + By = 0$:
$(16e^{4x} + 2e^{-x}) + A(4e^{4x} - 2e^{-x}) + B(e^{4x} + 2e^{-x}) = 0$.
Group the terms by $e^{4x}$ and $e^{-x}$:
$(16 + 4A + B)e^{4x} + (2 - 2A + 2B)e^{-x} = 0$.
For this to hold for all $x$,the coefficients must be zero:
$16 + 4A + B = 0$ (Equation $1$)
$2 - 2A + 2B = 0 \Rightarrow 1 - A + B = 0 \Rightarrow B = A - 1$ (Equation $2$)
Substitute $B = A - 1$ into Equation $1$:
$16 + 4A + (A - 1) = 0 \Rightarrow 5A + 15 = 0 \Rightarrow A = -3$.
Then $B = -3 - 1 = -4$.
Thus,$A = -3$ and $B = -4$.

(A) Given the equation: $2[2x - 5] - 1 = 7$ \\
Using the property $[x + n] = [x] + n$ for integer $n$,we have $[2x - 5] = [2x] - 5$ \\
Substituting this into the equation: $2([2x] - 5) - 1 = 7$ \\
$2[2x] - 10 - 1 = 7$ \\
$2[2x] - 11 = 7$ \\
$2[2x] = 18$ \\
$[2x] = 9$ \\
By the definition of the greatest integer function,$[y] = n \Rightarrow n \leq y < n + 1$ \\
So,$9 \leq 2x < 10$ \\
Dividing by $2$,we get $\frac{9}{2} \leq x < 5$ \\
Thus,$x \in \left[\frac{9}{2}, 5\right)$

(D) The function $f(x) = e^{|x| \sin x}$ is defined for all real values of $x$.
Hence,the domain of $f(x)$ is $\mathbb{R}$.
Since $|x| \sin x$ can take any real value from $-\infty$ to $\infty$,the expression $e^{|x| \sin x}$ will always be strictly greater than $0$.
Therefore,the range of $f(x)$ is $(0, \infty)$.

(B) For the function $f(x) = \frac{1}{4-x^2} + \log_{10}(x^3-x)$ to be defined:
$1$. The denominator must not be zero: $4 - x^2 \neq 0 \implies x^2 \neq 4 \implies x \neq \pm 2$.
$2$. The argument of the logarithm must be positive: $x^3 - x > 0$.
Factorizing the expression: $x(x^2 - 1) > 0 \implies x(x-1)(x+1) > 0$.
Using the wavy curve method (sign scheme),the inequality $x(x-1)(x+1) > 0$ holds for $x \in (-1, 0) \cup (1, \infty)$.
Combining the conditions $x \neq \pm 2$ and $x \in (-1, 0) \cup (1, \infty)$,we exclude $x = 2$ from the interval $(1, \infty)$.
Thus,the domain is $x \in (-1, 0) \cup (1, 2) \cup (2, \infty)$.

(C) Given $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$.
The derivative is $f'(x) = 3x^2 - 6(a - 2)x + 3a$.
Since the function changes behavior at $x = 1$,we have $f'(1) = 0$.
$3(1)^2 - 6(a - 2)(1) + 3a = 0$
$3 - 6a + 12 + 3a = 0$
$-3a + 15 = 0 \Rightarrow a = 5$.
Substituting $a = 5$ into $f(x)$,we get $f(x) = x^3 - 3(5 - 2)x^2 + 3(5)x + 7 = x^3 - 9x^2 + 15x + 7$.
Now,solve $\frac{f(x) - 14}{(x - 1)^2} = 0$ for $x \neq 1$.
$f(x) - 14 = 0 \Rightarrow x^3 - 9x^2 + 15x + 7 - 14 = 0$.
$x^3 - 9x^2 + 15x - 7 = 0$.
Since $x = 1$ is a root,we divide by $(x - 1)^2$:
$(x - 1)^2(x - 7) = 0$.
The roots are $x = 1$ and $x = 7$.
Since $x \neq 1$,the root is $x = 7$.

(A) Given functions are $f(x) = e^{|x|}$ and $g(x) = \log x$.
The composite function $(g \circ f)(x)$ is defined as $g(f(x))$.
Substituting $f(x)$ into $g(x)$,we get:
$(g \circ f)(x) = g(e^{|x|}) = \log(e^{|x|})$.
Using the logarithmic property $\log(a^b) = b \log a$ and knowing that $\log e = 1$,we have:
$(g \circ f)(x) = |x| \log e = |x| \times 1 = |x|$.
Thus,the correct option is $A$.

(D) Given $f(x) = \frac{a-x}{a+x}$.
We are given that $(f \circ f)(x) = x$.
$f(f(x)) = f\left(\frac{a-x}{a+x}\right) = \frac{a - \left(\frac{a-x}{a+x}\right)}{a + \left(\frac{a-x}{a+x}\right)} = x$.
Simplifying the expression:
$\frac{a(a+x) - (a-x)}{a(a+x) + (a-x)} = x$
$\frac{a^2 + ax - a + x}{a^2 + ax + a - x} = x$
$a^2 + ax - a + x = x(a^2 + ax + a - x)$
$a^2 + ax - a + x = a^2x + ax^2 + ax - x^2$
Rearranging the terms:
$(a-1)x^2 + (a^2-1)x - a(a-1) = 0$
$(a-1)(x^2 + (a+1)x - a) = 0$
$(a-1)(x+a)(x-1) = 0$
Since this must hold for all $x \neq -a$,we must have $a-1 = 0$,so $a = 1$.
Thus,$f(x) = \frac{1-x}{1+x}$.
Now,calculate $f\left(-\frac{1}{2}\right)$:
$f\left(-\frac{1}{2}\right) = \frac{1 - (-1/2)}{1 + (-1/2)} = \frac{1 + 1/2}{1 - 1/2} = \frac{3/2}{1/2} = 3$.

(B) Given the function $f: R \rightarrow R$ defined by $f(x) = |x|$.
For a function to be injective (one-one),$f(x_1) = f(x_2)$ must imply $x_1 = x_2$.
Here,$f(-1) = |-1| = 1$ and $f(1) = |1| = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function is not injective.
For a function to be surjective (onto),the range must be equal to the codomain.
The codomain is $R$ (all real numbers),but the range of $f(x) = |x|$ is $[0, \infty)$.
Since the range $[0, \infty) \neq R$,the function is not surjective.
Therefore,the function is neither injective nor surjective.

(D) Given $f(x) = \frac{2x}{x-1}$.
To check for injectivity,we find the derivative: $f'(x) = \frac{(x-1)(2) - 2x(1)}{(x-1)^2} = \frac{-2}{(x-1)^2}$.
Since $f'(x) < 0$ for all $x \in A$,the function $f$ is strictly decreasing,which implies $f$ is injective.
To check for surjectivity,let $f(x) = y$. Then $y = \frac{2x}{x-1} \Rightarrow yx - y = 2x \Rightarrow x(y-2) = y \Rightarrow x = \frac{y}{y-2}$.
For $f$ to be surjective,for every $y \in R$,there must exist an $x \in A$ such that $f(x) = y$.
If $y = 2$,then $x = \frac{2}{0}$,which is undefined. Thus,$2$ has no pre-image in $A$.
Also,if $y = 4$,$x = \frac{4}{4-2} = 2$. However,$2$ is a positive integer,so $2 \notin A$.
Since there exist elements in the codomain $R$ that do not have a pre-image in $A$,$f$ is not surjective.
Therefore,$f$ is injective but not surjective.

(A) For the function $f(x)$ to be surjective,the codomain $A$ must be equal to the range of $f(x)$.
Let $y = \frac{x^2}{1-x^2}$.
Rearranging for $x^2$,we get $y(1-x^2) = x^2$,which implies $y - yx^2 = x^2$,so $y = x^2(1+y)$.
Thus,$x^2 = \frac{y}{1+y}$.
Since $x^2 \geq 0$ and $x \neq \pm 1$,we must have $\frac{y}{1+y} \geq 0$ and $x^2 \neq 1$ (which implies $\frac{y}{1+y} \neq 1$,so $y \neq 1+y$,which is always true for any finite $y$).
The inequality $\frac{y}{1+y} \geq 0$ holds when $y \in (-\infty, -1) \cup [0, \infty)$.
Therefore,the range of $f(x)$ is $(-\infty, -1) \cup [0, \infty)$,which can be written as $R - [-1, 0)$.
Hence,$A = R - [-1, 0)$.

(A) Let $y = f(x) = \frac{a^x - a^{-x}}{a^x + a^{-x}}$.
Multiply the numerator and denominator by $a^x$:
$y = \frac{a^{2x} - 1}{a^{2x} + 1}$.
Now,solve for $x$ in terms of $y$:
$y(a^{2x} + 1) = a^{2x} - 1$
$y \cdot a^{2x} + y = a^{2x} - 1$
$1 + y = a^{2x} - y \cdot a^{2x}$
$1 + y = a^{2x}(1 - y)$
$a^{2x} = \frac{1+y}{1-y}$.
Taking the logarithm base $a$ on both sides:
$2x = \log_a \left( \frac{1+y}{1-y} \right)$
$x = \frac{1}{2} \log_a \left( \frac{1+y}{1-y} \right)$.
Replacing $y$ with $x$ to find the inverse function:
$f^{-1}(x) = \frac{1}{2} \log_a \left( \frac{1+x}{1-x} \right)$.

(C) Given $f(x) = \cos(\log x)$.
We need to evaluate the expression $E = f(x) \cdot f(y) - \frac{1}{2} \left( f\left(\frac{x}{y}\right) + f(xy) \right)$.
Substitute the function definition:
$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( \cos(\log(x/y)) + \cos(\log(xy)) \right)$.
Using logarithmic properties $\log(x/y) = \log x - \log y$ and $\log(xy) = \log x + \log y$:
$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( \cos(\log x - \log y) + \cos(\log x + \log y) \right)$.
Apply the trigonometric identity $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$ where $A = \log x$ and $B = \log y$:
$E = \cos(\log x) \cdot \cos(\log y) - \frac{1}{2} \left( 2 \cos(\log x) \cos(\log y) \right)$.
$E = \cos(\log x) \cdot \cos(\log y) - \cos(\log x) \cdot \cos(\log y) = 0$.

(C) Given the functional equation $f(x+y)=f(x) \cdot f^{\prime}(y)+f^{\prime}(x) \cdot f(y)$.
Putting $x=0$ and $y=0$,we get $f(0)=f(0) \cdot f^{\prime}(0)+f^{\prime}(0) \cdot f(0) = 2 f(0) \cdot f^{\prime}(0)$.
Since $f(0)=1$,we have $1 = 2(1) \cdot f^{\prime}(0)$,which implies $f^{\prime}(0)=\frac{1}{2}$.
Now,keeping $x$ as a variable and setting $y=0$,we get $f(x+0)=f(x) \cdot f^{\prime}(0)+f^{\prime}(x) \cdot f(0)$.
Substituting $f(0)=1$ and $f^{\prime}(0)=\frac{1}{2}$,we obtain $f(x) = f(x) \cdot \frac{1}{2} + f^{\prime}(x) \cdot 1$.
This simplifies to $f^{\prime}(x) = \frac{1}{2} f(x)$,or $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{2}$.
Integrating both sides with respect to $x$,we get $\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{2} dx$,which results in $\log(f(x)) = \frac{1}{2}x + C$.
Using $f(0)=1$,we have $\log(1) = 0 + C$,so $C=0$.
Thus,$\log(f(x)) = \frac{1}{2}x$.
For $x=4$,$\log(f(4)) = \frac{1}{2} \times 4 = 2$.

(C) We have $I = \int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we write $\frac{5x}{2} = 2x + \frac{x}{2}$.
$I = \int \frac{\sin(2x + \frac{x}{2})}{\sin \frac{x}{2}} dx = \int \frac{\sin 2x \cos \frac{x}{2} + \cos 2x \sin \frac{x}{2}}{\sin \frac{x}{2}} dx$.
$I = \int (\sin 2x \cot \frac{x}{2} + \cos 2x) dx$.
Alternatively,using the sum-to-product identity or expansion:
$\frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} = \frac{\sin(2x + \frac{x}{2})}{\sin \frac{x}{2}} = \frac{\sin 2x \cos \frac{x}{2} + \cos 2x \sin \frac{x}{2}}{\sin \frac{x}{2}} = \cos 2x + \sin 2x \cot \frac{x}{2}$.
Using $\sin 2x = 2 \sin x \cos x$ and $\cot \frac{x}{2} = \frac{1+\cos x}{\sin x}$:
$\sin 2x \cot \frac{x}{2} = 2 \sin x \cos x \cdot \frac{1+\cos x}{\sin x} = 2 \cos x(1+\cos x) = 2 \cos x + 2 \cos^2 x$.
Since $2 \cos^2 x = 1 + \cos 2x$,we have:
$\cos 2x + 2 \cos x + 1 + \cos 2x = 1 + 2 \cos x + 2 \cos 2x$.
Integrating: $\int (1 + 2 \cos x + 2 \cos 2x) dx = x + 2 \sin x + \sin 2x + C$.

(A) We know that the algebraic identity for the difference of two $n$-th powers is given by:
$(x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1}) = x^n - a^n$.
Substituting this into the integral,we get:
$\int(x^n - a^n)dx$.
Integrating term by term with respect to $x$:
$\int x^n dx - \int a^n dx = \frac{x^{n+1}}{n+1} - a^n x + C$.

(C) Let $I = \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} dx$.
Substitute $\log x = t$,so $x = e^t$ and $dx = e^t dt$.
Then $I = \int \frac{(t-1)^2}{(1+t^2)^2} e^t dt = \int e^t \left( \frac{t^2-2t+1}{(1+t^2)^2} \right) dt$.
We can rewrite the integrand as $\int e^t \left( \frac{t^2+1-2t}{(1+t^2)^2} \right) dt = \int e^t \left( \frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2} \right) dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = \frac{1}{1+t^2}$ and $f'(t) = -\frac{2t}{(1+t^2)^2}$.
Thus,$I = e^t \left( \frac{1}{1+t^2} \right) + C$.
Substituting back $t = \log x$,we get $I = \frac{e^{\log x}}{1+(\log x)^2} + C = \frac{x}{1+(\log x)^2} + C$.

(B) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$.
Multiply the numerator and denominator by $e^x$:
$I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} \,d x$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1+t)^2}$.
Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$.
Solving for constants,we get $A=1, B=-1, C=-1$.
So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
Integrating term by term:
$I = \log |t| - \log |1+t| + \frac{1}{1+t} + C$.
$I = \log \left| \frac{t}{1+t} \right| + \frac{1}{1+t} + C$.
Substituting $t = x e^x$ back:
$I = \log \left( \frac{x e^x}{1+x e^x} \right) + \frac{1}{1+x e^x} + C$.

(D) Let $I = \int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}$.
Using $\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}$ and $\sec 2\theta = \frac{1}{\cos 2\theta}$,we have:
$I = \int \frac{\sec ^2 \theta d \theta}{\frac{\sin 2 \theta+1}{\cos 2 \theta}} = \int \frac{\cos 2 \theta \sec ^2 \theta d \theta}{1+\sin 2 \theta}$.
Since $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ and $1 + \sin 2\theta = (\cos \theta + \sin \theta)^2$,we get:
$I = \int \frac{(\cos \theta - \sin \theta)(\cos \theta + \sin \theta) \sec^2 \theta d \theta}{(\cos \theta + \sin \theta)^2} = \int \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \sec^2 \theta d \theta$.
Dividing the numerator and denominator by $\cos \theta$,we get:
$I = \int \frac{1-\tan \theta}{1+\tan \theta} \sec^2 \theta d \theta$.
Let $t = \tan \theta$,then $dt = \sec^2 \theta d\theta$:
$I = \int \frac{1-t}{1+t} dt = \int \left(-1 + \frac{2}{1+t}\right) dt = -t + 2 \log |1+t| + C$.
Substituting $t = \tan \theta$ back,we get:
$I = -\tan \theta + 2 \log |1+\tan \theta| + C$.
Comparing this with $\lambda \tan \theta + 2 \log |f(\theta)| + C$,we find $\lambda = -1$ and $f(\theta) = 1+\tan \theta$.

(B) Let $I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} \, dx$.
Multiply numerator and denominator by $\sqrt{1-\sqrt{x}}$:
$I = \int \sqrt{\frac{(1-\sqrt{x})^2}{1-x}} \, dx = \int \frac{1-\sqrt{x}}{\sqrt{1-x}} \, dx$.
Split the integral:
$I = \int \frac{1}{\sqrt{1-x}} \, dx - \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx$.
The first part is $-2\sqrt{1-x}$.
For the second part,let $x = \cos^2 \theta$,so $dx = -2 \cos \theta \sin \theta \, d\theta$.
$\int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx = \int \frac{\cos \theta}{\sin \theta} (-2 \cos \theta \sin \theta) \, d\theta = -2 \int \cos^2 \theta \, d\theta = -\int (1 + \cos 2\theta) \, d\theta = -(\theta + \frac{\sin 2\theta}{2}) = -\theta - \sin \theta \cos \theta$.
Substituting back: $I = -2\sqrt{1-x} - (-\theta - \sin \theta \cos \theta) + C = -2\sqrt{1-x} + \theta + \sin \theta \cos \theta + C$.
Since $x = \cos^2 \theta$,$\theta = \cos^{-1} \sqrt{x}$,$\cos \theta = \sqrt{x}$,and $\sin \theta = \sqrt{1-x}$.
Thus,$I = -2\sqrt{1-x} + \cos^{-1} \sqrt{x} + \sqrt{x(1-x)} + C$.

(A) To evaluate the integral $I = \int \sqrt{\frac{1+x}{1-x}} \, dx$,we multiply the numerator and denominator inside the square root by $(1+x)$:
$I = \int \sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}} \, dx = \int \sqrt{\frac{(1+x)^2}{1-x^2}} \, dx$
$I = \int \frac{1+x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{1-x^2}} \, dx + \int \frac{x}{\sqrt{1-x^2}} \, dx$
For the first part,$\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x$.
For the second part,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$:
$\int \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int u^{-1/2} \, du = -\frac{1}{2} \cdot 2u^{1/2} = -\sqrt{1-x^2}$.
Combining these,we get $I = \sin^{-1} x - \sqrt{1-x^2} + C$.

(D) Given $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$.
We can rewrite the numerator as $x^2 + 1 - 1 + \sin^2 x = (1 + x^2) - (1 - \sin^2 x) = (1 + x^2) - \cos^2 x$.
Substituting this into the integral:
$f(x) = \int \frac{(1 + x^2) - \cos^2 x}{1 + x^2} \cdot \sec^2 x \, dx$
$f(x) = \int \left( 1 - \frac{\cos^2 x}{1 + x^2} \right) \sec^2 x \, dx$
$f(x) = \int \sec^2 x \, dx - \int \frac{\cos^2 x \cdot \sec^2 x}{1 + x^2} \, dx$
Since $\cos^2 x \cdot \sec^2 x = 1$,we have:
$f(x) = \int \sec^2 x \, dx - \int \frac{1}{1 + x^2} \, dx$
$f(x) = \tan x - \tan^{-1} x + C$.
Given $f(0) = 0$:
$0 = \tan(0) - \tan^{-1}(0) + C \implies 0 = 0 - 0 + C \implies C = 0$.
Thus,$f(x) = \tan x - \tan^{-1} x$.
Therefore,$f(1) = \tan(1) - \tan^{-1}(1) = \tan(1) - \frac{\pi}{4}$.

(B) We are given the integral $I = \int \frac{x e^{2x}}{(1+2x)^2} dx$.
Let $2x = t$,then $2 dx = dt$,which implies $dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{(t/2) e^t}{(1+t)^2} \cdot \frac{dt}{2} = \frac{1}{4} \int \frac{t e^t}{(1+t)^2} dt$.
We can rewrite the numerator $t$ as $(t+1-1)$:
$I = \frac{1}{4} \int e^t \left( \frac{t+1-1}{(1+t)^2} \right) dt = \frac{1}{4} \int e^t \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = \frac{1}{1+t}$ and $f'(t) = -\frac{1}{(1+t)^2}$:
$I = \frac{1}{4} e^t \left( \frac{1}{1+t} \right) + C$.
Substituting $t = 2x$ back:
$I = \frac{e^{2x}}{4(1+2x)} + C$.

(A) Let $I = \int \frac{dx}{\cos x \sqrt{\cos 2x}}$.
We know that $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$.
So,$I = \int \frac{dx}{\cos x \sqrt{\frac{1-\tan^2 x}{1+\tan^2 x}}} = \int \frac{dx}{\cos x \frac{\sqrt{1-\tan^2 x}}{\sec x}}$.
Since $\frac{1}{\cos x} = \sec x$,we have $I = \int \frac{\sec x \cdot \sec x}{\sqrt{1-\tan^2 x}} \, dx = \int \frac{\sec^2 x}{\sqrt{1-\tan^2 x}} \, dx$.
Let $\tan x = t$,then $\sec^2 x \, dx = dt$.
Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{1-t^2}}$.
Using the standard integral formula $\int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1}(t) + C$,we get $I = \sin^{-1}(\tan x) + C$.

(D) Let $I = \int \frac{3 x^{13}+2 x^{11}}{\left(2 x^4+3 x^2+1\right)^4} d x$.
Divide the numerator and denominator by $x^{16}$ inside the integral:
$I = \int \frac{3 x^{-3}+2 x^{-5}}{\left(2+3 x^{-2}+x^{-4}\right)^4} d x$.
Let $u = 2+3 x^{-2}+x^{-4}$.
Then $du = (-6 x^{-3}-4 x^{-5}) d x = -2(3 x^{-3}+2 x^{-5}) d x$.
So,$(3 x^{-3}+2 x^{-5}) d x = -\frac{1}{2} du$.
Substituting these into the integral:
$I = \int \frac{-1/2}{u^4} du = -\frac{1}{2} \int u^{-4} du = -\frac{1}{2} \left( \frac{u^{-3}}{-3} \right) + C = \frac{1}{6 u^3} + C$.
Substituting $u$ back:
$I = \frac{1}{6(2+3 x^{-2}+x^{-4})^3} + C = \frac{1}{6(\frac{2x^4+3x^2+1}{x^4})^3} + C = \frac{x^{12}}{6(2 x^4+3 x^2+1)^3} + C$.

(B) Given $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} dx$.
Divide the numerator and denominator by $x^{14}$ inside the integral:
$f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^2} dx$.
Let $t = x^{-5} + x^{-7} + 2$. Then $dt = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = dt$.
Substituting this into the integral:
$f(x) = -\int \frac{dt}{t^2} = \frac{1}{t} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C$.
Simplifying the expression:
$f(x) = \frac{x^7}{1 + x^2 + 2x^7} + C$.
Given $f(0) = 0$,we have $0 = \frac{0}{1} + C$,so $C = 0$.
Thus,$f(x) = \frac{x^7}{x^2 + 1 + 2x^7}$.
Evaluating at $x = 1$:
$f(1) = \frac{1^7}{1^2 + 1 + 2(1)^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.

(A) Given $\int e^{x^2} \cdot x^3 \, dx = e^{x^2} f(x) + C$.
Let $I = \int e^{x^2} \cdot x^2 \cdot x \, dx$.
Substitute $t = x^2$,then $dt = 2x \, dx$,so $x \, dx = \frac{1}{2} dt$.
$I = \frac{1}{2} \int t e^t \, dt$.
Using integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = t$ and $dv = e^t \, dt$,then $du = dt$ and $v = e^t$.
$I = \frac{1}{2} [t e^t - \int e^t \, dt] = \frac{1}{2} [t e^t - e^t] + C = \frac{1}{2} e^t (t - 1) + C$.
Substituting $t = x^2$ back: $I = \frac{1}{2} e^{x^2} (x^2 - 1) + C$.
Comparing with $e^{x^2} f(x) + C$,we get $f(x) = \frac{1}{2} (x^2 - 1)$.
Check $f(1) = \frac{1}{2} (1^2 - 1) = 0$,which is consistent.
Therefore,$f(2) = \frac{1}{2} (2^2 - 1) = \frac{1}{2} (4 - 1) = \frac{3}{2}$.

(A) Given $f(x) = \sqrt{\tan x}$ and $g(x) = \sin x \cdot \cos x$.
We need to evaluate the integral $I = \int \frac{f(x)}{g(x)} dx$.
$I = \int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} dx$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sqrt{\tan x}}{\frac{\sin x \cdot \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} dx = \int \frac{\sqrt{\tan x}}{\tan x} \cdot \sec^2 x dx$.
$I = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$.
Let $u = \tan x$,then $du = \sec^2 x dx$.
$I = \int \frac{1}{\sqrt{u}} du = \int u^{-1/2} du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$.
Substituting back $u = \tan x$,we get $I = 2\sqrt{\tan x} + C$.

(C) Let $t = e^x$,then $dt = e^x dx$.
Substituting this into the integral,we get:
$\int \frac{dt}{(t+2)(t+1)}$.
Using partial fractions:
$\frac{1}{(t+2)(t+1)} = \frac{A}{t+2} + \frac{B}{t+1}$.
$1 = A(t+1) + B(t+2)$.
For $t = -1$,$B = 1$.
For $t = -2$,$A = -1$.
So,$\int (\frac{-1}{t+2} + \frac{1}{t+1}) dt = -\log|t+2| + \log|t+1| + C$.
$= \log|\frac{t+1}{t+2}| + C$.
Substituting $t = e^x$ back,we get:
$\log \left(\frac{e^x+1}{e^x+2}\right) + C$.

(B) We have the integral $I = \int \frac{5(x^6+1)}{x^2+1} \, dx$.
Using the algebraic identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$,where $a=x^2$ and $b=1$,we can write $x^6+1 = (x^2)^3+1^3 = (x^2+1)(x^4-x^2+1)$.
Substituting this into the integral:
$I = \int \frac{5(x^2+1)(x^4-x^2+1)}{x^2+1} \, dx$
$I = 5 \int (x^4-x^2+1) \, dx$
Integrating term by term:
$I = 5 \left( \frac{x^5}{5} - \frac{x^3}{3} + x \right) + C$
$I = x^5 - \frac{5x^3}{3} + 5x + C$.

(C) Let $I = \int_2^3 \frac{\log x}{x} d x$.
Substitute $u = \log x$,then $du = \frac{1}{x} dx$.
When $x = 2$,$u = \log 2$.
When $x = 3$,$u = \log 3$.
Thus,$I = \int_{\log 2}^{\log 3} u du = \left[ \frac{u^2}{2} \right]_{\log 2}^{\log 3}$.
$I = \frac{1}{2} \left( (\log 3)^2 - (\log 2)^2 \right)$.
Using the identity $a^2 - b^2 = (a+b)(a-b)$,we get:
$I = \frac{1}{2} (\log 3 + \log 2)(\log 3 - \log 2)$.
Since $\log a + \log b = \log(ab)$ and $\log a - \log b = \log(\frac{a}{b})$,
$I = \frac{1}{2} \log(3 \times 2) \log(\frac{3}{2}) = \frac{1}{2} \log 6 \log \frac{3}{2}$.

(A) Let $I = \int \frac{2 e^x+3 e^{-x}}{3 e^x+4 e^{-x}} d x$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{2 e^{2x}+3}{3 e^{2x}+4} d x$.
We want to express the numerator in terms of the denominator and its derivative.
Let $2 e^{2x} + 3 = A_1(3 e^{2x} + 4) + B_1 \frac{d}{dx}(3 e^{2x} + 4)$.
$2 e^{2x} + 3 = 3 A_1 e^{2x} + 4 A_1 + 6 B_1 e^{2x}$.
Comparing coefficients:
$3 A_1 + 6 B_1 = 2$ and $4 A_1 = 3 \implies A_1 = \frac{3}{4}$.
Substituting $A_1$: $3(\frac{3}{4}) + 6 B_1 = 2 \implies \frac{9}{4} + 6 B_1 = 2 \implies 6 B_1 = 2 - \frac{9}{4} = -\frac{1}{4} \implies B_1 = -\frac{1}{24}$.
Thus,$I = \int \left( \frac{3}{4} - \frac{1}{24} \frac{6 e^{2x}}{3 e^{2x} + 4} \right) d x$.
$I = \frac{3}{4} x - \frac{1}{24} \log(3 e^{2x} + 4) + C$.
Comparing with $Ax + B \log(3 e^{2x} + 4) + C$,we get $A = \frac{3}{4}$ and $B = -\frac{1}{24}$.

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \cos x}$.
Using the substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int_0^{\frac{\pi}{2}} \frac{d x}{5 + 4 \left( \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} \right)} = \int_0^{\frac{\pi}{2}} \frac{\sec^2(x/2) dx}{5(1+\tan^2(x/2)) + 4(1-\tan^2(x/2))}$
$I = \int_0^{\frac{\pi}{2}} \frac{\sec^2(x/2) dx}{9 + \tan^2(x/2)}$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,so $2 dt = \sec^2(x/2) dx$.
When $x=0, t=0$. When $x=\frac{\pi}{2}, t=\tan(\frac{\pi}{4})=1$.
$I = \int_0^1 \frac{2 dt}{9 + t^2} = 2 \int_0^1 \frac{dt}{3^2 + t^2} = 2 \left[ \frac{1}{3} \tan^{-1} \left( \frac{t}{3} \right) \right]_0^1$
$I = \frac{2}{3} \left( \tan^{-1} \left( \frac{1}{3} \right) - \tan^{-1}(0) \right) = \frac{2}{3} \tan^{-1} \left( \frac{1}{3} \right)$.

(A) Let $I = \int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} dx$.
Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$,we can factor the numerator:
$\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x) = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x)$.
Since $\sin^2 x + \cos^2 x = 1$,this simplifies to $(\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)$.
We know that $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x$.
Substituting this back into the integral:
$I = \int \frac{(\sin^2 x - \cos^2 x)(1 - 2 \sin^2 x \cos^2 x)}{1 - 2 \sin^2 x \cos^2 x} dx$.
$I = \int (\sin^2 x - \cos^2 x) dx$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we have $\sin^2 x - \cos^2 x = -\cos 2x$.
$I = \int -\cos 2x dx = -\frac{1}{2} \sin 2x + C$.

(C) Let $I = \int(3-x) \sqrt{4-x} \, dx$.
Substitute $u = 4-x$,then $du = -dx$,which means $dx = -du$.
Also,$x = 4-u$,so $3-x = 3-(4-u) = u-1$.
Substituting these into the integral:
$I = \int (u-1) \sqrt{u} (-du) = \int (1-u) \sqrt{u} \, du$
$I = \int (u^{1/2} - u^{3/2}) \, du$
$I = \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2} + C$
$I = \frac{2}{3} u^{3/2} - \frac{2}{5} u^{5/2} + C$
Substituting $u = 4-x$ back:
$I = \frac{2}{3} (4-x)^{3/2} - \frac{2}{5} (4-x)^{5/2} + C$.

(D) To solve the integral $I = \int \frac{dx}{2+\cos x}$,we use the substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$.
Substituting this into the integral,we get:
$I = \int \frac{dx}{2 + \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}} = \int \frac{1+\tan^2(x/2)}{2(1+\tan^2(x/2)) + 1-\tan^2(x/2)} dx$
$I = \int \frac{\sec^2(x/2)}{3+\tan^2(x/2)} dx$.
Let $u = \tan(x/2)$,then $du = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 du$.
Substituting these into the integral:
$I = \int \frac{2 du}{3+u^2} = 2 \int \frac{du}{(\sqrt{3})^2 + u^2}$.
Using the standard formula $\int \frac{dx}{a^2+x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = 2 \cdot \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{u}{\sqrt{3}}\right) + C = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{\tan(x/2)}{\sqrt{3}}\right) + C$.

(A) Let $I = \int \frac{x}{\sqrt{1-2 x^4}} \, dx$.
Substitute $u = \sqrt{2} x^2$.
Then $du = 2 \sqrt{2} x \, dx$,which implies $x \, dx = \frac{du}{2 \sqrt{2}}$.
Substituting these into the integral:
$I = \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{2 \sqrt{2}} = \frac{1}{2 \sqrt{2}} \int \frac{du}{\sqrt{1-u^2}}$.
Using the standard integral formula $\int \frac{1}{\sqrt{1-u^2}} \, du = \sin^{-1}(u) + C$,we get:
$I = \frac{1}{2 \sqrt{2}} \sin^{-1}(u) + C = \frac{1}{2 \sqrt{2}} \sin^{-1}(\sqrt{2} x^2) + C$.