MHT CET 2020 Mathematics Question Paper with Answer and Solution

(B) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.
$\cos x \cos 7 x - \cos 5 x \cos 13 x = \frac{1}{2} [2 \cos 7 x \cos x - 2 \cos 13 x \cos 5 x]$
$= \frac{1}{2} [(\cos 8 x + \cos 6 x) - (\cos 18 x + \cos 8 x)]$
$= \frac{1}{2} [\cos 6 x - \cos 18 x]$
Using the formula $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= \frac{1}{2} [-2 \sin \frac{6 x + 18 x}{2} \sin \frac{6 x - 18 x}{2}]$
$= - \sin 12 x \sin(-6 x)$
$= \sin 12 x \sin 6 x$
Since $\sin 12 x = 2 \sin 6 x \cos 6 x$,we have:
$= (2 \sin 6 x \cos 6 x) \sin 6 x = 2 \sin ^{2} 6 x \cos 6 x$.

(A) Given $A+B+C=180^{\circ}$,so $\frac{A+B+C}{2} = 90^{\circ}$.
Thus,$\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$.
Taking tangent on both sides: $\tan \left(\frac{A+B}{2}\right) = \tan \left(90^{\circ} - \frac{C}{2}\right) = \cot \left(\frac{C}{2}\right)$.
Using the formula $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$,we get:
$\frac{\tan(A/2) + \tan(B/2)}{1 - \tan(A/2) \tan(B/2)} = \frac{1}{\tan(C/2)}$.
Cross-multiplying gives: $\tan(C/2) [\tan(A/2) + \tan(B/2)] = 1 - \tan(A/2) \tan(B/2)$.
Rearranging the terms: $\tan(A/2) \tan(B/2) + \tan(B/2) \tan(C/2) + \tan(C/2) \tan(A/2) = 1$.

(B) We use the identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1 = 1 - 2 \sin^2 \frac{\theta}{2}$.
Substituting these into the expression:
$\frac{1 - (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}) + (2 \cos^2 \frac{\theta}{2} - 1)}{1 - (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}) - (1 - 2 \sin^2 \frac{\theta}{2})}$
$= \frac{2 \cos^2 \frac{\theta}{2} - 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2} - 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$
$= \frac{2 \cos \frac{\theta}{2} (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})}{2 \sin \frac{\theta}{2} (\sin \frac{\theta}{2} - \cos \frac{\theta}{2})}$
$= \frac{2 \cos \frac{\theta}{2} (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})}{-2 \sin \frac{\theta}{2} (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})}$
$= -\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = -\cot \frac{\theta}{2}$.

(A) The expression $\sqrt{3} \sin x + \cos x$ can be written in the form $R \sin(x + \alpha)$,where $R = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = 2$.
Thus,the maximum value of $\sqrt{3} \sin x + \cos x$ is $2$.
Therefore,the maximum value of the function $y = e^{5 + \sqrt{3} \sin x + \cos x}$ is $e^{5 + 2} = e^{7}$.

(D) Given $\cos 2\theta = \sin \alpha$.
We know that $\sin \alpha = \cos(\frac{\pi}{2} - \alpha)$.
So,$\cos 2\theta = \cos(\frac{\pi}{2} - \alpha)$.
The general solution for $\cos x = \cos y$ is $x = 2n\pi \pm y$,where $n \in \mathbb{Z}$.
Applying this,$2\theta = 2n\pi \pm (\frac{\pi}{2} - \alpha)$.
Dividing by $2$,we get $\theta = n\pi \pm (\frac{\pi}{4} - \frac{\alpha}{2})$,where $n \in \mathbb{Z}$.

(B) We know that $\tan \theta = \tan \alpha$ implies $\theta = n\pi + \alpha$,where $n \in Z$.
Given $\tan 3x = 1$.
Since $\tan \frac{\pi}{4} = 1$,we have $\tan 3x = \tan \frac{\pi}{4}$.
Therefore,$3x = n\pi + \frac{\pi}{4}$.
Dividing by $3$,we get $x = \frac{n\pi}{3} + \frac{\pi}{12}$,where $n \in Z$.

(C) In $\triangle ABP$ and $\triangle CDP$,$AB \parallel DC$ and $AB = DC$. Since $P$ is the mid-point of $AB$,$AP = PB = \frac{1}{2} AB = \frac{1}{2} DC$.
Consider $\triangle APR$ and $\triangle CPD$:
$\angle PAR = \angle PCD$ (alternate interior angles as $AB \parallel DC$)
$\angle APR = \angle CPD$ (vertically opposite angles)
Thus,$\triangle APR \sim \triangle CPD$ by $AA$ similarity.
Therefore,the ratio of corresponding sides is equal:
$\frac{AR}{CR} = \frac{AP}{CD} = \frac{\frac{1}{2} AB}{AB} = \frac{1}{2}$.
Hence,$R$ divides $AC$ in the ratio $1: 2$.

(A) Given the vertices of the triangle are $P(3,2,6), Q(1,4,5)$,and $R(3,5,3)$.
First,we find the direction ratios of the vectors $\vec{QP}$ and $\vec{QR}$.
The vector $\vec{QP} = (3-1, 2-4, 6-5) = (2, -2, 1)$.
The vector $\vec{QR} = (3-1, 5-4, 3-5) = (2, 1, -2)$.
Now,we calculate the dot product of $\vec{QP}$ and $\vec{QR}$:
$\vec{QP} \cdot \vec{QR} = (2)(2) + (-2)(1) + (1)(-2) = 4 - 2 - 2 = 0$.
Since the dot product is $0$,the vectors $\vec{QP}$ and $\vec{QR}$ are perpendicular to each other.
Therefore,$m \angle PQR = 90^{\circ}$.

(B) Let $u = \cot ^{-1} x$ and $v = \log (1+x^{2})$.
First,we find the derivative of $u$ with respect to $x$:
$\frac{du}{dx} = -\frac{1}{1+x^{2}}$.
Next,we find the derivative of $v$ with respect to $x$ using the chain rule:
$\frac{dv}{dx} = \frac{1}{1+x^{2}} \times \frac{d}{dx}(1+x^{2}) = \frac{2x}{1+x^{2}}$.
Now,we find the derivative of $u$ with respect to $v$:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-1/(1+x^{2})}{2x/(1+x^{2})} = -\frac{1}{2x}$.
Thus,the correct option is $B$.

(C) Given $y = \tan^{-1} \left[ \frac{x - \sqrt{1 - x^2}}{x + \sqrt{1 - x^2}} \right]$.
Let $x = \cos \theta$,then $\theta = \cos^{-1} x$.
Substituting $x = \cos \theta$ into the expression:
$y = \tan^{-1} \left[ \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \right]$
Divide numerator and denominator by $\cos \theta$:
$y = \tan^{-1} \left[ \frac{1 - \tan \theta}{1 + \tan \theta} \right]$
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$:
$y = \tan^{-1} \left[ \tan \left( \frac{\pi}{4} - \theta \right) \right] = \frac{\pi}{4} - \theta$
Substitute back $\theta = \cos^{-1} x$:
$y = \frac{\pi}{4} - \cos^{-1} x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \cos^{-1} x \right) = 0 - \left( -\frac{1}{\sqrt{1 - x^2}} \right) = \frac{1}{\sqrt{1 - x^2}}$.

(A) Given $f(x) = \sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)$.
Method $1$: Using the chain rule:
$f^{\prime}(x) = \frac{d}{dx}\left[\sin^{-1}\left(\sqrt{\frac{1-x}{2}}\right)\right] = \frac{1}{\sqrt{1-\left(\sqrt{\frac{1-x}{2}}\right)^{2}}} \cdot \frac{d}{dx}\left(\sqrt{\frac{1-x}{2}}\right)$
$= \frac{1}{\sqrt{1-\frac{1-x}{2}}} \cdot \frac{1}{2\sqrt{\frac{1-x}{2}}} \cdot \left(-\frac{1}{2}\right)$
$= \frac{1}{\sqrt{\frac{1+x}{2}}} \cdot \frac{1}{\sqrt{2}\sqrt{1-x}} \cdot \left(-\frac{1}{2}\right)$
$= \frac{\sqrt{2}}{\sqrt{1+x}} \cdot \frac{1}{\sqrt{2}\sqrt{1-x}} \cdot \left(-\frac{1}{2}\right) = \frac{-1}{2\sqrt{1-x^{2}}}$.
Method $2$: Using substitution:
Let $x = \cos \theta$,then $\sqrt{\frac{1-x}{2}} = \sqrt{\frac{1-\cos \theta}{2}} = \sqrt{\frac{2\sin^{2}(\theta/2)}{2}} = \sin(\theta/2)$.
So,$f(x) = \sin^{-1}(\sin(\theta/2)) = \theta/2 = \frac{1}{2}\cos^{-1}(x)$.
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^{2}}}\right) = \frac{-1}{2\sqrt{1-x^{2}}}$.
Thus,the correct option is $A$.

(A) Let $x = \cos(2\theta)$,so $2\theta = \cos^{-1}(x)$ and $\theta = \frac{1}{2} \cos^{-1}(x)$.
Then $\sqrt{1+x} = \sqrt{1+\cos(2\theta)} = \sqrt{2\cos^2(\theta)} = \sqrt{2} \cos(\theta)$ and $\sqrt{1-x} = \sqrt{1-\cos(2\theta)} = \sqrt{2\sin^2(\theta)} = \sqrt{2} \sin(\theta)$.
Substituting these into the expression for $y$:
$y = \sin^{-1} \left[ \frac{\sqrt{2}\cos(\theta) + \sqrt{2}\sin(\theta)}{2} \right] = \sin^{-1} \left[ \frac{1}{\sqrt{2}} \cos(\theta) + \frac{1}{\sqrt{2}} \sin(\theta) \right]$.
Using the identity $\sin(\frac{\pi}{4} + \theta) = \sin(\frac{\pi}{4}) \cos(\theta) + \cos(\frac{\pi}{4}) \sin(\theta) = \frac{1}{\sqrt{2}} \cos(\theta) + \frac{1}{\sqrt{2}} \sin(\theta)$.
So,$y = \sin^{-1} [\sin(\frac{\pi}{4} + \theta)] = \frac{\pi}{4} + \theta = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 + \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{1}{2\sqrt{1-x^2}}$.

(D) Given $y = \sec(\tan^{-1} x)$.
Using the chain rule,$\frac{dy}{dx} = \sec(\tan^{-1} x) \tan(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x)$.
Since $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$,we have $\frac{dy}{dx} = \sec(\tan^{-1} x) \cdot x \cdot \frac{1}{1+x^2}$.
At $x = 1$,$\tan^{-1}(1) = \frac{\pi}{4}$.
So,$\frac{dy}{dx} = \sec(\frac{\pi}{4}) \cdot 1 \cdot \frac{1}{1+1^2} = \sqrt{2} \cdot \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Alternatively,let $\tan^{-1} x = \theta$,then $\tan \theta = x$. Since $\sec^2 \theta = 1 + \tan^2 \theta$,we have $\sec \theta = \sqrt{1+x^2}$.
Thus,$y = \sqrt{1+x^2}$.
Then $\frac{dy}{dx} = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$.
At $x = 1$,$\frac{dy}{dx} = \frac{1}{\sqrt{1+1^2}} = \frac{1}{\sqrt{2}}$.

(C) Given the equation $x^{2}+y^{2}=1$.
Differentiating both sides with respect to $y$,we get:
$2x \frac{dx}{dy} + 2y = 0$
$2x \frac{dx}{dy} = -2y$
$\frac{dx}{dy} = -\frac{y}{x}$
Now,differentiating $\frac{dx}{dy}$ with respect to $y$ using the quotient rule:
$\frac{d^{2}x}{dy^{2}} = \frac{d}{dy} \left( -\frac{y}{x} \right) = -\left[ \frac{x(1) - y(\frac{dx}{dy})}{x^{2}} \right]$
Substitute $\frac{dx}{dy} = -\frac{y}{x}$ into the expression:
$\frac{d^{2}x}{dy^{2}} = -\left[ \frac{x - y(-\frac{y}{x})}{x^{2}} \right]$
$\frac{d^{2}x}{dy^{2}} = -\left[ \frac{x + \frac{y^{2}}{x}}{x^{2}} \right] = -\left[ \frac{x^{2} + y^{2}}{x^{3}} \right]$
Since $x^{2}+y^{2}=1$,we have:
$\frac{d^{2}x}{dy^{2}} = -\frac{1}{x^{3}}$

(A) Given $y = e^{4x} \cos 5x$.
Applying the product rule,$\frac{dy}{dx} = e^{4x}(-5 \sin 5x) + \cos 5x(4e^{4x}) = e^{4x}(4 \cos 5x - 5 \sin 5x)$.
Now,differentiating again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = e^{4x}(-20 \sin 5x - 25 \cos 5x) + (4 \cos 5x - 5 \sin 5x)(4e^{4x})$.
Evaluating at $x=0$:
$\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0} = e^{0}(-20 \sin 0 - 25 \cos 0) + (4 \cos 0 - 5 \sin 0)(4e^{0})$.
Since $\sin 0 = 0$ and $\cos 0 = 1$:
$\left(\frac{d^{2}y}{dx^{2}}\right)_{x=0} = 1(0 - 25) + (4 - 0)(4) = -25 + 16 = -9$.

(A) Given $\sqrt{x+y}+\sqrt{y-x}=5$.
Rearranging the terms,we get $\sqrt{y-x}=5-\sqrt{x+y}$.
Squaring both sides,we have $y-x = 25 + (x+y) - 10\sqrt{x+y}$.
Simplifying,$-2x = 25 - 10\sqrt{x+y}$,which gives $10\sqrt{x+y} = 2x + 25$.
Differentiating both sides with respect to $x$:
$10 \times \frac{1}{2\sqrt{x+y}} \times (1 + \frac{dy}{dx}) = 2$.
$\frac{5}{\sqrt{x+y}} \times (1 + \frac{dy}{dx}) = 2$.
$1 + \frac{dy}{dx} = \frac{2\sqrt{x+y}}{5}$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{2}{5} \times \frac{1}{2\sqrt{x+y}} \times (1 + \frac{dy}{dx})$.
Substituting the value of $(1 + \frac{dy}{dx})$ from the previous step:
$\frac{d^2y}{dx^2} = \frac{1}{5\sqrt{x+y}} \times \frac{2\sqrt{x+y}}{5} = \frac{2}{25}$.

(B) Given $x = a(1 - \cos \theta)$ and $y = a(\theta - \sin \theta)$.
First,find $\frac{dx}{d\theta} = a \sin \theta$ and $\frac{dy}{d\theta} = a(1 - \cos \theta)$.
Then,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1 - \cos \theta)}{a \sin \theta} = \frac{2 \sin^{2}(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \tan(\theta/2)$.
Now,differentiate with respect to $x$: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\tan(\theta/2)) = \sec^{2}(\theta/2) \cdot \frac{1}{2} \cdot \frac{d\theta}{dx}$.
Since $\frac{dx}{d\theta} = a \sin \theta$,we have $\frac{d\theta}{dx} = \frac{1}{a \sin \theta}$.
Substituting this: $\frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sec^{2}(\theta/2) \cdot \frac{1}{a(2 \sin(\theta/2) \cos(\theta/2))} = \frac{1}{4a \sin(\theta/2) \cos^{3}(\theta/2)} = \frac{1}{4a \sin(\theta/2) \cos(\theta/2) \cos^{2}(\theta/2)} = \frac{1}{2a \sin \theta \cos^{2}(\theta/2)}$.
Alternatively,$\frac{d^{2}y}{dx^{2}} = \frac{1}{2} \sec^{2}(\theta/2) \cdot \frac{1}{a(2 \sin(\theta/2) \cos(\theta/2))} = \frac{1}{4a \sin(\theta/2) \cos^{3}(\theta/2)} = \frac{\sec^{4}(\theta/2)}{4a}$.

(D) We are given the function $y = \cos^{2}\left(\frac{5x}{2}\right) - \sin^{2}\left(\frac{5x}{2}\right)$.
Using the trigonometric identity $\cos(2\theta) = \cos^{2}(\theta) - \sin^{2}(\theta)$,we can simplify the expression by setting $\theta = \frac{5x}{2}$.
Thus,$y = \cos\left(2 \times \frac{5x}{2}\right) = \cos(5x)$.
Now,we find the first derivative with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\cos(5x)) = -5 \sin(5x)$.
Next,we find the second derivative with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(-5 \sin(5x)) = -5 \times 5 \cos(5x) = -25 \cos(5x)$.
Since $y = \cos(5x)$,we substitute $y$ back into the expression:
$\frac{d^{2}y}{dx^{2}} = -25y$.

(B) Given the function $f(x) = e^{x} g(x)$.
Applying the product rule for differentiation,we have $f^{\prime}(x) = \frac{d}{dx}(e^{x}) \cdot g(x) + e^{x} \cdot \frac{d}{dx}(g(x))$.
This simplifies to $f^{\prime}(x) = e^{x} g(x) + e^{x} g^{\prime}(x) = e^{x} (g(x) + g^{\prime}(x))$.
Now,substitute $x = 0$ into the derivative expression:
$f^{\prime}(0) = e^{0} (g(0) + g^{\prime}(0))$.
Given $g(0) = 4$ and $g^{\prime}(0) = 2$,and knowing $e^{0} = 1$:
$f^{\prime}(0) = 1 \cdot (4 + 2) = 6$.

(D) Given the differential equation $f^{\prime}(x) = 2 f(x)$.
Rearranging the terms,we get $\frac{f^{\prime}(x)}{f(x)} = 2$.
Integrating both sides with respect to $x$:
$\int \frac{f^{\prime}(x)}{f(x)} dx = \int 2 dx$
$\ln |f(x)| = 2x + C$.
Using the initial condition $f(0) = 3$:
$\ln |f(0)| = 2(0) + C \Rightarrow \ln 3 = C$.
Substituting $C$ back into the equation:
$\ln |f(x)| = 2x + \ln 3$.
To find $f(2)$,substitute $x = 2$:
$\ln |f(2)| = 2(2) + \ln 3 = 4 + \ln 3$.
Taking the exponential on both sides:
$f(2) = e^{4 + \ln 3} = e^{4} \cdot e^{\ln 3} = 3 e^{4}$.

(B) The function $f(x) = \frac{x+2}{9-x^{2}}$ is defined for all real values of $x$ except where the denominator is zero.
Setting the denominator to zero: $9 - x^{2} = 0$.
This implies $x^{2} = 9$,so $x = \pm 3$.
Thus,the function is undefined at $x = 3$ and $x = -3$.
Therefore,the domain of the function is the set of all real numbers except $\{-3, 3\}$,which is written as $R - \{-3, 3\}$.

(C) For the function $f(y) = \frac{\cos^{-1}(y-5)}{\sqrt{25-y^2}}$ to be defined,the following conditions must be satisfied:
$1$. The argument of $\cos^{-1}$ must be in the interval $[-1, 1]$,so $-1 \leq y-5 \leq 1$. Adding $5$ to all sides gives $4 \leq y \leq 6$.
$2$. The denominator must be non-zero and the expression inside the square root must be positive,so $25-y^2 > 0$,which implies $y^2 < 25$,or $-5 < y < 5$.
$3$. Taking the intersection of the two conditions: $y \in [4, 6]$ and $y \in (-5, 5)$.
$4$. The intersection is $4 \leq y < 5$,which can be written as the interval $[4, 5)$.
Therefore,the domain is $[4, 5)$.

(C) The greatest integer function $[x]$ is defined as the greatest integer less than or equal to $x$.
If $x$ is an integer,then $[x] = x$,which implies $[x] + 1 = x + 1 > x$.
If $x$ is not an integer,then $[x] < x < [x] + 1$.
In both cases,we have $[x] + 1 > x$.

(C) For $f(x) = \sqrt{x}$ to be defined in the set of real numbers,the expression under the square root must be non-negative.
Therefore,$x \geq 0$.
The domain is the set of all non-negative real numbers,which is represented as $[0, \infty)$ or $R^{+} \cup \{0\}$.
Thus,the correct option is $C$.

(D) Given the relation $R = \{(x, y) : y = x + \frac{6}{x}, x, y \in N, x < 6\}$.
We test each value of $x \in \{1, 2, 3, 4, 5\}$ to see if $y$ is a natural number $(N)$:
For $x = 1, y = 1 + \frac{6}{1} = 7 \in N$.
For $x = 2, y = 2 + \frac{6}{2} = 2 + 3 = 5 \in N$.
For $x = 3, y = 3 + \frac{6}{3} = 3 + 2 = 5 \in N$.
For $x = 4, y = 4 + \frac{6}{4} = 4 + 1.5 = 5.5 \notin N$.
For $x = 5, y = 5 + \frac{6}{5} = 5 + 1.2 = 6.2 \notin N$.
Thus,the relation $R = \{(1, 7), (2, 5), (3, 5)\}$.
The domain is the set of first elements: $\{1, 2, 3\}$.
The range is the set of second elements: $\{5, 7\}$.

(B) Given $f(x) = \frac{2x+3}{3x-2}$.
We need to find $(f \circ f)(x) = f(f(x))$.
$f(f(x)) = \frac{2 \left( \frac{2x+3}{3x-2} \right) + 3}{3 \left( \frac{2x+3}{3x-2} \right) - 2}$
Multiplying the numerator and denominator by $(3x-2)$:
$= \frac{2(2x+3) + 3(3x-2)}{3(2x+3) - 2(3x-2)}$
$= \frac{4x + 6 + 9x - 6}{6x + 9 - 6x + 4}$
$= \frac{13x}{13} = x$
Since $(f \circ f)(x) = x$,the function is an identity function.

(D) Let $y = \frac{x-3}{5-x}$.
$y(5-x) = x-3$
$5y - xy = x - 3$
$5y + 3 = x + xy$
$5y + 3 = x(1+y)$
$x = \frac{5y+3}{1+y}$.
For $x$ to be defined,the denominator $1+y \neq 0$,which means $y \neq -1$.
Thus,the range of the function is $R - \{-1\}$.
Therefore,the correct option is $D$.

(C) Given function is $f(x) = \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}}$.
To check if the function is even or odd,we evaluate $f(-x)$:
$f(-x) = \frac{e^{-x} + e^{-(-x)}}{e^{-x} - e^{-(-x)}}$
$f(-x) = \frac{e^{-x} + e^{x}}{e^{-x} - e^{x}}$
We can rewrite the denominator by factoring out a negative sign:
$f(-x) = \frac{e^{x} + e^{-x}}{-(e^{x} - e^{-x})}$
$f(-x) = -\left( \frac{e^{x} + e^{-x}}{e^{x} - e^{-x}} \right)$
$f(-x) = -f(x)$
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.

(D) Given $f(x) = \frac{2x+3}{3x-2}$.
We need to find $(f \circ f)(x) = f(f(x))$.
$(f \circ f)(x) = f\left(\frac{2x+3}{3x-2}\right) = \frac{2\left(\frac{2x+3}{3x-2}\right) + 3}{3\left(\frac{2x+3}{3x-2}\right) - 2}$.
Multiplying the numerator and denominator by $(3x-2)$,we get:
$(f \circ f)(x) = \frac{2(2x+3) + 3(3x-2)}{3(2x+3) - 2(3x-2)}$.
$(f \circ f)(x) = \frac{4x + 6 + 9x - 6}{6x + 9 - 6x + 4} = \frac{13x}{13} = x$.
Since $(f \circ f)(x) = x$,and $g(x) = x$ is an odd function,the result is an odd function.

(A) Given $f(x) = \frac{4x+7}{7x-4}$.
First,calculate $f(2)$:
$f(2) = \frac{4(2)+7}{7(2)-4} = \frac{8+7}{14-4} = \frac{15}{10} = \frac{3}{2}$.
Next,calculate $f[f(2)] = f(\frac{3}{2})$:
$f(\frac{3}{2}) = \frac{4(\frac{3}{2})+7}{7(\frac{3}{2})-4} = \frac{6+7}{\frac{21}{2}-4} = \frac{13}{\frac{21-8}{2}} = \frac{13 \times 2}{13} = 2$.
Finally,calculate $f\{f[f(2)]\} = f(2)$:
$f(2) = \frac{3}{2}$.
Alternatively,observe that $f(f(x)) = x$,which implies $f(f(f(x))) = f(x)$.
Thus,$f(f(f(2))) = f(2) = \frac{3}{2}$.

(B) $(f \circ g)(x) = f[g(x)] = f(2x + 1) = (2x + 1)^{2} - 3(2x + 1) + 4$
$= 4x^{2} + 4x + 1 - 6x - 3 + 4 = 4x^{2} - 2x + 2$
Given $f(x) = (f \circ g)(x)$
$\therefore x^{2} - 3x + 4 = 4x^{2} - 2x + 2$
$3x^{2} + x - 2 = 0$
$3x^{2} + 3x - 2x - 2 = 0$
$3x(x + 1) - 2(x + 1) = 0$
$(x + 1)(3x - 2) = 0$
$\therefore x = -1, \frac{2}{3}$

(C) Given $f(x) = \frac{3x+4}{5x-7}$ and $g(x) = \frac{7x+4}{5x-3}$.
We need to find $(g \circ f)(x) = g(f(x))$.
$g(f(x)) = \frac{7(f(x)) + 4}{5(f(x)) - 3}$
Substitute $f(x) = \frac{3x+4}{5x-7}$:
$g(f(x)) = \frac{7(\frac{3x+4}{5x-7}) + 4}{5(\frac{3x+4}{5x-7}) - 3}$
Multiply numerator and denominator by $(5x-7)$:
$g(f(x)) = \frac{7(3x+4) + 4(5x-7)}{5(3x+4) - 3(5x-7)}$
$g(f(x)) = \frac{21x + 28 + 20x - 28}{15x + 20 - 15x + 21}$
$g(f(x)) = \frac{41x}{41} = x$
Therefore,$(g \circ f)(3) = 3$.

(D) Given $f(x) = 2x^{2} + bx + c$.
Since $f(0) = 3$,we have $2(0)^{2} + b(0) + c = 3$,which gives $c = 3$.
Now,$f(x) = 2x^{2} + bx + 3$.
Given $f(2) = 1$,we have $2(2)^{2} + b(2) + 3 = 1$.
$8 + 2b + 3 = 1 \Rightarrow 2b + 11 = 1 \Rightarrow 2b = -10 \Rightarrow b = -5$.
Thus,$f(x) = 2x^{2} - 5x + 3$.
First,calculate $f(1) = 2(1)^{2} - 5(1) + 3 = 2 - 5 + 3 = 0$.
Now,$(f \circ f)(1) = f(f(1)) = f(0)$.
Since $f(0) = 3$,we get $(f \circ f)(1) = 3$.

(B) Given $f(x)=2x-3$ and $g(x)=x^{3}+5$.
First,find the composite function $(fog)(x)$:
$(fog)(x) = f(g(x)) = f(x^{3}+5) = 2(x^{3}+5)-3 = 2x^{3}+10-3 = 2x^{3}+7$.
Let $y = (fog)(x) = 2x^{3}+7$.
To find the inverse,solve for $x$ in terms of $y$:
$y-7 = 2x^{3} \Rightarrow x^{3} = \frac{y-7}{2} \Rightarrow x = \left(\frac{y-7}{2}\right)^{\frac{1}{3}}$.
Replacing $y$ with $x$,we get $(fog)^{-1}(x) = \left(\frac{x-7}{2}\right)^{\frac{1}{3}}$.

(C) Given $f(x) = 7x + 8 = y$.
To find the inverse function $f^{-1}(y)$,we solve for $x$ in terms of $y$:
$7x = y - 8$
$x = \frac{y - 8}{7}$
Thus,$f^{-1}(y) = \frac{y - 8}{7}$,which implies $f^{-1}(x) = \frac{x - 8}{7}$.
Given $f^{-1}(12) = \frac{k}{7}$,we substitute $x = 12$ into the inverse function:
$f^{-1}(12) = \frac{12 - 8}{7} = \frac{4}{7}$.
Comparing $\frac{4}{7}$ with $\frac{k}{7}$,we get $k = 4$.

(B) To find the inverse function $f^{-1}(x)$,we set $f(x) = y$.
$y = \frac{4x}{5} + 3$
Subtract $3$ from both sides:
$y - 3 = \frac{4x}{5}$
Multiply both sides by $5$:
$5(y - 3) = 4x$
Divide by $4$:
$x = \frac{5(y - 3)}{4}$
Since $x = f^{-1}(y)$,we have $f^{-1}(y) = \frac{5(y - 3)}{4}$.
Replacing $y$ with $x$,we get:
$f^{-1}(x) = \frac{5(x - 3)}{4}$
Thus,the correct option is $B$.

(A) Let $y = f(x) = \frac{3x+2}{5x-3}$.
Then $y(5x - 3) = 3x + 2$.
Expanding this,we get $5xy - 3y = 3x + 2$.
Rearranging the terms to solve for $x$,we have $5xy - 3x = 3y + 2$.
Factoring out $x$,we get $x(5y - 3) = 3y + 2$.
Thus,$x = \frac{3y+2}{5y-3}$.
By definition,$f^{-1}(y) = x = \frac{3y+2}{5y-3}$.
Replacing $y$ with $x$,we obtain $f^{-1}(x) = \frac{3x+2}{5x-3}$.
Since $f^{-1}(x) = f(x)$,the correct option is $A$.

(C) Given the equation: $[x]^{2} - 5[x] + 6 = 0$.
Let $t = [x]$. Then the equation becomes $t^{2} - 5t + 6 = 0$.
Factoring the quadratic equation: $(t - 3)(t - 2) = 0$.
This gives $t = 2$ or $t = 3$.
Thus,$[x] = 2$ 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 $x \in [2, 3) \cup [3, 4) = [2, 4)$.
Therefore,the correct option is $C$.

(A) Given $f(x) = x^{2} - 3x + 4$.
First,we find $f(2x + 1)$:
$f(2x + 1) = (2x + 1)^{2} - 3(2x + 1) + 4$
$= (4x^{2} + 4x + 1) - 6x - 3 + 4$
$= 4x^{2} - 2x + 2$.
Since $f(x) = f(2x + 1)$,we equate the two expressions:
$x^{2} - 3x + 4 = 4x^{2} - 2x + 2$
$3x^{2} + x - 2 = 0$
Factoring the quadratic equation:
$3x^{2} + 3x - 2x - 2 = 0$
$3x(x + 1) - 2(x + 1) = 0$
$(x + 1)(3x - 2) = 0$
Thus,$x = -1$ or $x = \frac{2}{3}$.

(A) To evaluate the integral $I = \int \frac{dx}{\sqrt{5+4x-x^2}}$,we first complete the square for the quadratic expression inside the square root.
$5+4x-x^2 = -(x^2-4x-5) = -((x-2)^2 - 4 - 5) = -( (x-2)^2 - 9 ) = 9 - (x-2)^2$.
Substituting this back into the integral,we get:
$I = \int \frac{dx}{\sqrt{3^2 - (x-2)^2}}$.
Using the standard integration formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we obtain:
$I = \sin^{-1}\left(\frac{x-2}{3}\right) + c$.

(A) Given $f^{\prime}(x)=k(\cos x+\sin x)$.
Integrating both sides with respect to $x$,we get:
$f(x) = \int k(\cos x + \sin x) dx = k(\sin x - \cos x) + C$.
Using the condition $f(0) = 9$:
$f(0) = k(\sin 0 - \cos 0) + C = k(0 - 1) + C = -k + C = 9$ ...$(1)$.
Using the condition $f\left(\frac{\pi}{2}\right) = 15$:
$f\left(\frac{\pi}{2}\right) = k\left(\sin \frac{\pi}{2} - \cos \frac{\pi}{2}\right) + C = k(1 - 0) + C = k + C = 15$ ...$(2)$.
Adding equations $(1)$ and $(2)$:
$(-k + C) + (k + C) = 9 + 15 \Rightarrow 2C = 24 \Rightarrow C = 12$.
Substituting $C = 12$ into equation $(2)$:
$k + 12 = 15 \Rightarrow k = 3$.
Thus,$f(x) = 3(\sin x - \cos x) + 12$.

(C) Multiply numerator and denominator by $e^x$:
$I = \int \frac{4 e^{2x} + 6}{9 e^{2x} - 4} dx$
Let $4 e^{2x} + 6 = A(18 e^{2x}) + B(9 e^{2x} - 4)$
Comparing coefficients of $e^{2x}$ and constants:
$18A + 9B = 4$ and $-4B = 6$
From $-4B = 6$,we get $B = -\frac{3}{2}$
Substituting $B$ in $18A + 9B = 4$:
$18A + 9(-\frac{3}{2}) = 4 \Rightarrow 18A - \frac{27}{2} = 4 \Rightarrow 18A = \frac{35}{2} \Rightarrow A = \frac{35}{36}$
Now,$I = \int \left[ \frac{\frac{35}{36}(18 e^{2x})}{9 e^{2x} - 4} - \frac{\frac{3}{2}(9 e^{2x} - 4)}{9 e^{2x} - 4} \right] dx$
$I = \frac{35}{36} \log |9 e^{2x} - 4| - \frac{3}{2} x + c$
Comparing with $Ax + B \log |9 e^{2x} - 4| + c$,we get $A = -\frac{3}{2}$ and $B = \frac{35}{36}$.

(B) We are given the integral $I = \int \frac{dx}{\cos 2x + \sin^2 x}$.
Using the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$,we substitute this into the denominator:
$I = \int \frac{dx}{1 - 2\sin^2 x + \sin^2 x}$
$I = \int \frac{dx}{1 - \sin^2 x}$
Using the identity $1 - \sin^2 x = \cos^2 x$,we get:
$I = \int \frac{dx}{\cos^2 x}$
$I = \int \sec^2 x \, dx$
The integral of $\sec^2 x$ is $\tan x + c$.
Therefore,$I = \tan x + c$.

(D) We are given the integral $I = \int \frac{dx}{\cos 2x - \cos^2 x}$.
Using the trigonometric identity $\cos 2x = 2\cos^2 x - 1$,we substitute this into the denominator:
$I = \int \frac{dx}{(2\cos^2 x - 1) - \cos^2 x} = \int \frac{dx}{\cos^2 x - 1}$.
We know that $\cos^2 x - 1 = - (1 - \cos^2 x) = -\sin^2 x$.
Substituting this back into the integral:
$I = \int \frac{dx}{-\sin^2 x} = -\int \csc^2 x \, dx$.
Since the derivative of $\cot x$ is $-\csc^2 x$,the integral of $-\csc^2 x$ is $\cot x + c$.
Therefore,$I = \cot x + c$.

(A) Given $f^{\prime}(x)=k(\cos x-\sin x)$.
Substitute $x=0$ into the derivative: $f^{\prime}(0)=k(\cos 0-\sin 0)=k(1-0)=k$.
Since $f^{\prime}(0)=3$,we have $k=3$.
Now,integrate $f^{\prime}(x)$ to find $f(x)$:
$f(x)=\int 3(\cos x-\sin x) \, dx = 3(\sin x+\cos x)+C$.
Use the condition $f\left(\frac{\pi}{2}\right)=15$ to find $C$:
$f\left(\frac{\pi}{2}\right)=3(\sin \frac{\pi}{2}+\cos \frac{\pi}{2})+C = 3(1+0)+C = 3+C$.
Setting $3+C=15$,we get $C=12$.
Therefore,$f(x)=3(\sin x+\cos x)+12$.

(A) To evaluate the integral $\int \frac{d x}{x^{2}+4 x+13}$,we complete the square in the denominator:
$x^{2}+4 x+13 = (x^{2}+4 x+4) + 9 = (x+2)^{2} + 3^{2}$.
Now,the integral becomes $\int \frac{d x}{(x+2)^{2} + 3^{2}}$.
Using the standard formula $\int \frac{d x}{x^{2}+a^{2}} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c$,where $a=3$ and the variable is $(x+2)$:
$\int \frac{d x}{(x+2)^{2} + 3^{2}} = \frac{1}{3} \tan^{-1}\left(\frac{x+2}{3}\right) + c$.
Comparing this with the given options,the correct option is $A$.

(A) Let $I = \int \frac{d x}{\cos x \sqrt{\cos 2 x}}$.
Divide the numerator and denominator by $\cos x$ inside the square root,or rewrite the expression:
$I = \int \frac{d x}{\cos x \sqrt{\cos^2 x - \sin^2 x}} = \int \frac{d x}{\cos x \cdot \cos x \sqrt{1 - \tan^2 x}} = \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} d x$.
Substitute $t = \tan x$,then $dt = \sec^2 x d x$.
Thus,$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$.
Substituting back $t = \tan x$,we get $I = \sin^{-1}(\tan x) + c$.

(B) Let $\cos ^{-1} x = t$. Then $x = \cos t$ and $dx = -\sin t \ dt$. Also,$\frac{dx}{\sqrt{1-x^2}} = -dt$.
Substituting these into the integral:
$I = \int e^t \left[ \frac{\cos t - \sin t}{\sin t} \right] (-dt) = \int e^t \left[ \frac{\sin t - \cos t}{\sin t} \right] dt$
Wait,let us simplify the expression directly:
$I = \int e^{\cos ^{-1} x} \left( \frac{x}{\sqrt{1-x^2}} - 1 \right) dx$.
Let $t = \cos ^{-1} x$,so $x = \cos t$ and $dx = -\sin t \ dt$.
$I = \int e^t \left( \frac{\cos t}{\sin t} - 1 \right) (-\sin t \ dt) = \int e^t (\sin t - \cos t) dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$,we get:
$I = e^t \sin t + c$.
Since $t = \cos ^{-1} x$,$\sin t = \sqrt{1 - \cos^2 t} = \sqrt{1 - x^2}$.
This does not match the options. Let us re-evaluate:
$I = \int e^t (\sin t - \cos t) dt = -\int e^t (\cos t - \sin t) dt = -e^t \cos t + c$.
Substituting back $t = \cos ^{-1} x$:
$I = -e^{\cos ^{-1} x} \cdot x + c = -x e^{\cos ^{-1} x} + c$.

(B) Let $I = \int \frac{5^{x}}{\sqrt{5^{-2x}-5^{2x}}} dx$.
We can rewrite the denominator as:
$\sqrt{5^{-2x}-5^{2x}} = \sqrt{\frac{1}{5^{2x}} - 5^{2x}} = \sqrt{\frac{1 - (5^{2x})^2}{5^{2x}}} = \frac{\sqrt{1 - (5^{2x})^2}}{5^x}$.
Substituting this into the integral:
$I = \int \frac{5^x}{\frac{\sqrt{1 - (5^{2x})^2}}{5^x}} dx = \int \frac{5^{2x}}{\sqrt{1 - (5^{2x})^2}} dx$.
Let $t = 5^{2x}$. Then $dt = 5^{2x} \cdot \ln(5) \cdot 2 dx = 2 \ln(5) \cdot 5^{2x} dx$.
So,$5^{2x} dx = \frac{dt}{2 \ln(5)} = \frac{dt}{\ln(25)}$.
Substituting into the integral:
$I = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{dt}{\ln(25)} = \frac{1}{\ln(25)} \int \frac{dt}{\sqrt{1 - t^2}}$.
$I = \frac{1}{\ln(25)} \sin^{-1}(t) + c = \frac{\sin^{-1}(5^{2x})}{\ln(25)} + c$.

(B) Let $I = \int \frac{d x}{(x+2) \sqrt{x+1}}$.
Substitute $\sqrt{x+1} = t$,which implies $x+1 = t^2$,so $x = t^2 - 1$ and $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int \frac{2t \, dt}{(t^2 - 1 + 2) \cdot t}$
$I = \int \frac{2t \, dt}{(t^2 + 1) \cdot t}$
$I = 2 \int \frac{dt}{t^2 + 1}$
Using the standard integral formula $\int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + c$:
$I = 2 \tan^{-1}(t) + c$
Substituting back $t = \sqrt{x+1}$:
$I = 2 \tan^{-1}(\sqrt{x+1}) + c$.

(D) Let $I = \int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} dx$.
Substitute $t = \log (\sec x + \tan x)$.
Differentiating both sides with respect to $x$,we get:
$dt = \frac{1}{\sec x + \tan x} (\sec x \tan x + \sec^2 x) dx$.
$dt = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} dx$.
$dt = \sec x dx$.
Now,the integral becomes:
$I = \int \frac{dt}{\sqrt{t}} = \int t^{-1/2} dt$.
Integrating with respect to $t$:
$I = \frac{t^{1/2}}{1/2} + c = 2\sqrt{t} + c$.
Substituting back $t = \log (\sec x + \tan x)$:
$I = 2\sqrt{\log (\sec x + \tan x)} + c$.

(B) We need to evaluate the integral $I = \int \frac{dx}{\sqrt{(x-1)(x-2)}}$.
First,expand the denominator: $(x-1)(x-2) = x^2 - 3x + 2$.
Now,complete the square for the quadratic expression: $x^2 - 3x + 2 = (x^2 - 3x + \frac{9}{4}) - \frac{9}{4} + 2 = (x - \frac{3}{2})^2 - \frac{1}{4} = (x - \frac{3}{2})^2 - (\frac{1}{2})^2$.
The integral becomes $I = \int \frac{dx}{\sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + c$,we get:
$I = \log |(x - \frac{3}{2}) + \sqrt{(x - \frac{3}{2})^2 - (\frac{1}{2})^2}| + c$.
Simplifying the expression inside the square root back to the original form,we get:
$I = \log |(x - \frac{3}{2}) + \sqrt{x^2 - 3x + 2}| + c$.

(D) Let $I = \int \frac{1+\log x}{\cos^{2}(x \log x)} dx$.
Substitute $t = x \log x$.
Differentiating with respect to $x$ using the product rule: $\frac{dt}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x$.
Therefore,$(1 + \log x) dx = dt$.
Substituting these into the integral: $I = \int \frac{1}{\cos^{2} t} dt = \int \sec^{2} t dt$.
The integral of $\sec^{2} t$ is $\tan t + c$.
Substituting back $t = x \log x$,we get $I = \tan(x \log x) + c$.

(A) Let $I = \int x^{x}(1+\log x) dx$.
Consider the substitution $u = x^{x}$.
Taking the natural logarithm on both sides,$\log u = x \log x$.
Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$\frac{du}{dx} = u(1 + \log x) = x^{x}(1 + \log x)$.
Therefore,$du = x^{x}(1 + \log x) dx$.
Substituting this into the integral,$I = \int du = u + c$.
Replacing $u$ with $x^{x}$,we get $I = x^{x} + c$.
Comparing this with the given expression $k x^{x} + c$,we find $k = 1$.
Since $\log_{e} e = 1$,the correct option is $A$.

(B) Let $I = \int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx$.
Substitute $\sqrt{x} = t$.
Then,differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} \, dx = dt$,which implies $\frac{1}{\sqrt{x}} \, dx = 2 \, dt$.
Substituting these into the integral,we get $I = \int \cos(t) \cdot 2 \, dt$.
$I = 2 \int \cos(t) \, dt$.
Integrating $\cos(t)$,we get $I = 2 \sin(t) + c$.
Substituting back $t = \sqrt{x}$,we obtain $I = 2 \sin \sqrt{x} + c$.

(A) Let $I = \int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} dx$.
Let $u = 7^{x}$. Then $du = 7^{x} \log 7 dx$,so $7^{x} dx = \frac{du}{\log 7}$.
The integral becomes $I = \int 7^{7^{u}} 7^{u} \frac{du}{\log 7} = \frac{1}{\log 7} \int 7^{7^{u}} 7^{u} du$.
Let $v = 7^{u}$. Then $dv = 7^{u} \log 7 du$,so $7^{u} du = \frac{dv}{\log 7}$.
Substituting this into the integral,we get $I = \frac{1}{\log 7} \int 7^{v} \frac{dv}{\log 7} = \frac{1}{(\log 7)^{2}} \int 7^{v} dv$.
Since $\int 7^{v} dv = \frac{7^{v}}{\log 7} + C$,we have $I = \frac{7^{v}}{(\log 7)^{3}} + C$.
Substituting back $v = 7^{u} = 7^{7^{x}}$,we get $I = \frac{7^{7^{7^{x}}}}{(\log 7)^{3}} + C$.

(B) Let $I = \int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x$.
Substitute $u = x - \frac{1}{x}$.
Then $du = (1 + \frac{1}{x^2}) dx = \frac{x^2+1}{x^2} dx$.
Substituting these into the integral,we get:
$I = \int \sqrt{u} du = \int u^{1/2} du$.
Integrating,we get:
$I = \frac{u^{3/2}}{3/2} + c = \frac{2}{3} u^{3/2} + c$.
Substituting back $u = x - \frac{1}{x}$:
$I = \frac{2}{3} (x - \frac{1}{x})^{3/2} + c$.
Comparing this with the given expression $\frac{2}{3}(x-\frac{1}{x})^k + c$,we find $k = \frac{3}{2}$.

(C) Let $I = \int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x$.
We can rewrite the numerator as $(1-2e^{-x}) + 4e^{-x}$.
So,$I = \int \frac{(1-2e^{-x}) + 4e^{-x}}{1-2e^{-x}} d x$.
$I = \int \left( 1 + \frac{4e^{-x}}{1-2e^{-x}} \right) d x$.
$I = \int 1 d x + 2 \int \frac{2e^{-x}}{1-2e^{-x}} d x$.
Let $u = 1-2e^{-x}$,then $du = 2e^{-x} d x$.
$I = x + 2 \int \frac{1}{u} du = x + 2 \ln|u| + c$.
Substituting back $u$,we get $I = x + 2 \ln|1-2e^{-x}| + c$.