MHT CET 2023 Mathematics Question Paper with Answer and Solution

(D) Given $\tan \theta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$.
Dividing the numerator and denominator by $\cos \alpha$,we get $\tan \theta = \frac{\tan \alpha - 1}{\tan \alpha + 1}$.
We know that $\tan(\frac{\pi}{4}) = 1$,so $\tan \theta = \frac{\tan \alpha - \tan(\frac{\pi}{4})}{1 + \tan \alpha \tan(\frac{\pi}{4})}$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get $\tan \theta = \tan(\alpha - \frac{\pi}{4})$.
Thus,$\theta = \alpha - \frac{\pi}{4}$.
Then $2 \theta = 2 \alpha - \frac{\pi}{2}$.
Therefore,$\cos 2 \theta = \cos(2 \alpha - \frac{\pi}{2})$.
Using the identity $\cos(x - \frac{\pi}{2}) = \sin x$,we get $\cos 2 \theta = \sin 2 \alpha$.

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have:
$\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$
$\cos(72^{\circ}-A)\cos(72^{\circ}+A) = \cos^2 72^{\circ} - \sin^2 A$
Subtracting these:
$(\cos^2 18^{\circ} - \sin^2 A) - (\cos^2 72^{\circ} - \sin^2 A) = \cos^2 18^{\circ} - \cos^2 72^{\circ}$
Since $\cos 72^{\circ} = \sin 18^{\circ}$,this becomes $\cos^2 18^{\circ} - \sin^2 18^{\circ} = \cos(2 \times 18^{\circ}) = \cos 36^{\circ}$.
Alternatively,using $\cos(x-y)\cos(x+y) = \frac{1}{2}(\cos 2x + \cos 2y)$:
$\frac{1}{2}(\cos 36^{\circ} + \cos 2A) - \frac{1}{2}(\cos 144^{\circ} + \cos 2A) = \frac{1}{2}(\cos 36^{\circ} - \cos 144^{\circ})$
Since $\cos 144^{\circ} = -\cos 36^{\circ}$,we get $\frac{1}{2}(\cos 36^{\circ} + \cos 36^{\circ}) = \cos 36^{\circ} = \sin 54^{\circ}$.

(B) Given equation: $\sin x + \sin 5x = \sin 3x$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$2 \sin(\frac{x+5x}{2}) \cos(\frac{x-5x}{2}) = \sin 3x$
$2 \sin 3x \cos(-2x) = \sin 3x$
Since $\cos(-2x) = \cos 2x$,we have:
$2 \sin 3x \cos 2x - \sin 3x = 0$
$\sin 3x (2 \cos 2x - 1) = 0$
Case $1$: $\sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$. For $x \in (0, \frac{\pi}{2})$,$x = \frac{\pi}{3}$ is a solution.
Case $2$: $2 \cos 2x - 1 = 0 \implies \cos 2x = \frac{1}{2} = \cos \frac{\pi}{3}$.
$2x = 2n\pi \pm \frac{\pi}{3} \implies x = n\pi \pm \frac{\pi}{6}$. For $x \in (0, \frac{\pi}{2})$,$x = \frac{\pi}{6}$ is a solution.
Thus,the solutions are $x = \frac{\pi}{6}, \frac{\pi}{3}$.

(B) Given that $\sin (\theta-\alpha), \sin \theta, \sin (\theta+\alpha)$ are in $H.P.$
$\Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)}$ are in $A.P.$
$\therefore \frac{2}{\sin \theta} = \frac{1}{\sin (\theta-\alpha)} + \frac{1}{\sin (\theta+\alpha)}$
$\Rightarrow \frac{2}{\sin \theta} = \frac{\sin (\theta+\alpha) + \sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)}$
Using the formula $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$ and $\sin(A+B) \sin(A-B) = \sin^2 A - \sin^2 B$:
$\Rightarrow \frac{2}{\sin \theta} = \frac{2 \sin \theta \cos \alpha}{\sin^2 \theta - \sin^2 \alpha}$
$\Rightarrow \sin^2 \theta - \sin^2 \alpha = \sin^2 \theta \cos \alpha$
$\Rightarrow \sin^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$
Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:
$\Rightarrow \sin^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$
$\Rightarrow \sin^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
$\Rightarrow 1 - \cos^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
$\Rightarrow \cos^2 \theta = 1 - 2 \cos^2 \frac{\alpha}{2}$
Multiplying by $2$ and subtracting $1$ to get $\cos 2 \theta = 2 \cos^2 \theta - 1$:
$\Rightarrow 2 \cos^2 \theta - 1 = 2(1 - 2 \cos^2 \frac{\alpha}{2}) - 1$
$\Rightarrow \cos 2 \theta = 2 - 4 \cos^2 \frac{\alpha}{2} - 1 = 1 - 4 \cos^2 \frac{\alpha}{2}$

(A) Given the equation: $a \cos 2 \theta + b \sin 2 \theta = c$
Using the identities $\cos 2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2 \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we get:
$a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) = c$
Multiplying both sides by $(1 + \tan^2 \theta)$:
$a(1 - \tan^2 \theta) + 2b \tan \theta = c(1 + \tan^2 \theta)$
$a - a \tan^2 \theta + 2b \tan \theta = c + c \tan^2 \theta$
Rearranging the terms to form a quadratic equation in $\tan \theta$:
$(a + c) \tan^2 \theta - 2b \tan \theta + (c - a) = 0$
Since $\alpha$ and $\beta$ are the roots,$\tan \alpha$ and $\tan \beta$ are the roots of this quadratic equation.
Using the sum of roots formula for $Ax^2 + Bx + C = 0$,which is $-\frac{B}{A}$:
$\tan \alpha + \tan \beta = -\frac{-2b}{a + c} = \frac{2b}{c + a}$

(C) We use the trigonometric identity: $\cos ^2 A - \sin ^2 B = \cos(A+B) \cdot \cos(A-B)$.
Applying this to the given expression with $A = 48^{\circ}$ and $B = 12^{\circ}$:
$\cos ^2 48^{\circ} - \sin ^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cdot \cos(48^{\circ} - 12^{\circ})$
$= \cos(60^{\circ}) \cdot \cos(36^{\circ})$
Since $\cos(60^{\circ}) = \frac{1}{2}$ and $\cos(36^{\circ}) = 1 - 2\sin ^2(18^{\circ})$:
$= \frac{1}{2} \cdot (1 - 2\sin ^2 18^{\circ})$
$= \frac{1}{2} \left[ 1 - 2 \left( \frac{\sqrt{5}-1}{4} \right)^2 \right]$
$= \frac{1}{2} \left[ 1 - 2 \left( \frac{5 + 1 - 2\sqrt{5}}{16} \right) \right]$
$= \frac{1}{2} \left[ 1 - \frac{6 - 2\sqrt{5}}{8} \right] = \frac{1}{2} \left[ \frac{8 - 6 + 2\sqrt{5}}{8} \right]$
$= \frac{1}{2} \left[ \frac{2 + 2\sqrt{5}}{8} \right] = \frac{1 + \sqrt{5}}{8}$.

(C) We know that $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$.
Setting $\theta = \frac{\pi}{8}$,we have $\tan \frac{\pi}{4} = \frac{2\tan \frac{\pi}{8}}{1-\tan^2 \frac{\pi}{8}}$.
Since $\tan \frac{\pi}{4} = 1$,let $y = \tan \frac{\pi}{8}$.
Then $1 = \frac{2y}{1-y^2}$,which implies $1 - y^2 = 2y$,or $y^2 + 2y - 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$.
Since $\frac{\pi}{8}$ is in the $1^{\text{st}}$ quadrant,$\tan \frac{\pi}{8} > 0$.
Therefore,$\tan \frac{\pi}{8} = \sqrt{2} - 1$.

(C) We use the identity $\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$.
Here,$A = 48^{\circ}$ and $B = 12^{\circ}$.
$\cos^2 48^{\circ} - \sin^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cos(48^{\circ} - 12^{\circ})$
$= \cos 60^{\circ} \cos 36^{\circ}$
We know that $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.
Therefore,the expression equals $\frac{1}{2} \times \frac{\sqrt{5}+1}{4} = \frac{\sqrt{5}+1}{8}$.

(A) Given equation: $(1+\sqrt{1+x}) \tan x = 1+\sqrt{1-x}$
$\tan x = \frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$
Let $x = \sin \theta$.
$\tan x = \frac{1+\sqrt{1-\sin \theta}}{1+\sqrt{1+\sin \theta}} = \frac{1+\sqrt{(\cos \frac{\theta}{2} - \sin \frac{\theta}{2})^2}}{1+\sqrt{(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})^2}}$
$= \frac{1+\cos \frac{\theta}{2} - \sin \frac{\theta}{2}}{1+\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}$
$= \frac{2\cos^2 \frac{\theta}{4} - 2\sin \frac{\theta}{4}\cos \frac{\theta}{4}}{2\cos^2 \frac{\theta}{4} + 2\sin \frac{\theta}{4}\cos \frac{\theta}{4}}$
$= \frac{2\cos \frac{\theta}{4}(\cos \frac{\theta}{4} - \sin \frac{\theta}{4})}{2\cos \frac{\theta}{4}(\cos \frac{\theta}{4} + \sin \frac{\theta}{4})} = \frac{1 - \tan \frac{\theta}{4}}{1 + \tan \frac{\theta}{4}} = \tan(\frac{\pi}{4} - \frac{\theta}{4})$
Thus,$x = \frac{\pi}{4} - \frac{\theta}{4} \Rightarrow 4x = \pi - \theta$.
$\sin 4x = \sin(\pi - \theta) = \sin \theta = x$.

(A) Given $\cos 2B = \frac{\cos(A+C)}{\cos(A-C)}$.
Using the formula $\cos 2B = \frac{1-\tan^2 B}{1+\tan^2 B}$ and expanding the right side:
$\frac{1-\tan^2 B}{1+\tan^2 B} = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$.
Dividing numerator and denominator by $\cos A \cos C$:
$\frac{1-\tan^2 B}{1+\tan^2 B} = \frac{1-\tan A \tan C}{1+\tan A \tan C}$.
By componendo and dividendo or cross-multiplication:
$(1-\tan^2 B)(1+\tan A \tan C) = (1+\tan^2 B)(1-\tan A \tan C)$.
$1 + \tan A \tan C - \tan^2 B - \tan^2 B \tan A \tan C = 1 - \tan A \tan C + \tan^2 B - \tan^2 B \tan A \tan C$.
Simplifying the equation:
$2 \tan A \tan C = 2 \tan^2 B$.
$\tan^2 B = \tan A \tan C$.
Therefore,$\tan A, \tan B, \tan C$ are in Geometric Progression $(G.P.)$.

(B) We are given the expression: $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$
Expanding the terms,we get: $(a^2 - 2ab + b^2) \cos^2 \frac{C}{2} + (a^2 + 2ab + b^2) \sin^2 \frac{C}{2}$
Grouping the terms: $(a^2 + b^2) (\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab \cos^2 \frac{C}{2} + 2ab \sin^2 \frac{C}{2}$
Since $\cos^2 \theta + \sin^2 \theta = 1$,this simplifies to: $(a^2 + b^2) - 2ab (\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$
Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we have $\cos C = \cos^2 \frac{C}{2} - \sin^2 \frac{C}{2}$:
$= a^2 + b^2 - 2ab \cos C$
By the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$,so:
$= c^2$
Given $c = 4$,the value is $4^2 = 16$.

(A) We know the formula $\tan \left(\frac{A}{2}\right) = \frac{1-\cos A}{\sin A}$.
Putting $A = \frac{\pi}{4}$,we get $\frac{A}{2} = \frac{\pi}{8}$.
Therefore,$\tan \left(\frac{\pi}{8}\right) = \frac{1-\cos(\pi/4)}{\sin(\pi/4)}$.
Substituting the values $\cos(\pi/4) = \frac{1}{\sqrt{2}}$ and $\sin(\pi/4) = \frac{1}{\sqrt{2}}$,we get:
$\tan \left(\frac{\pi}{8}\right) = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2}-1$.

(D) Given equation: $16^{\sin^2 x} + 16^{\cos^2 x} = 10$
Since $\cos^2 x = 1 - \sin^2 x$,we have:
$16^{\sin^2 x} + 16^{1 - \sin^2 x} = 10$
$16^{\sin^2 x} + \frac{16}{16^{\sin^2 x}} = 10$
Let $t = 16^{\sin^2 x}$. Then $t + \frac{16}{t} = 10$,which implies $t^2 - 10t + 16 = 0$.
Solving for $t$: $(t - 8)(t - 2) = 0$,so $t = 8$ or $t = 2$.
Case $1$: $16^{\sin^2 x} = 2$ $\Rightarrow 2^{4\sin^2 x} = 2^1$ $\Rightarrow 4\sin^2 x = 1$ $\Rightarrow \sin^2 x = \frac{1}{4}$ $\Rightarrow \sin x = \pm \frac{1}{2}$.
In $[0, 2\pi]$,$\sin x = \pm \frac{1}{2}$ gives $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ ($4$ solutions).
Case $2$: $16^{\sin^2 x} = 8$ $\Rightarrow 2^{4\sin^2 x} = 2^3$ $\Rightarrow 4\sin^2 x = 3$ $\Rightarrow \sin^2 x = \frac{3}{4}$ $\Rightarrow \sin x = \pm \frac{\sqrt{3}}{2}$.
In $[0, 2\pi]$,$\sin x = \pm \frac{\sqrt{3}}{2}$ gives $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$ ($4$ solutions).
Total number of solutions = $4 + 4 = 8$.

(A) The centroid $(G)$ of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by the formula:
$G = \left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$
Substituting the given coordinates $A(-1, 2, 3)$,$B(3, -2, 1)$,$C(2, 1, 3)$,and $D(-1, -2, 4)$:
$G = \left(\frac{-1+3+2-1}{4}, \frac{2-2+1-2}{4}, \frac{3+1+3+4}{4}\right)$
$G = \left(\frac{3}{4}, \frac{-1}{4}, \frac{11}{4}\right)$
Thus,the correct option is $A$.

(C) Let the observations be $x_1, x_2, \dots, x_{20}$.
Given,variance $\sigma^2 = 5$.
We know that if each observation is multiplied by a constant $k$,the new variance $\sigma'^2$ is given by $\sigma'^2 = k^2 \times \sigma^2$.
Here,$k = 2$ and $\sigma^2 = 5$.
Therefore,the new variance $\sigma'^2 = (2)^2 \times 5 = 4 \times 5 = 20$.

(B) Given that $g(x)$ is the inverse of the function $f(x)$,we have $g(x) = f^{-1}(x)$.
This implies $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get:
$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$
Therefore,$g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$ ... $(i)$
Given $f^{\prime}(x) = \frac{1}{1+x^3}$,we substitute $x$ with $g(x)$ to get:
$f^{\prime}(g(x)) = \frac{1}{1+(g(x))^3}$ ... $(ii)$
Substituting $(ii)$ into $(i)$,we get:
$g^{\prime}(x) = \frac{1}{\frac{1}{1+(g(x))^3}} = 1+(g(x))^3$.

(D) Given,$S=5+\frac{48}{t}+t^3$.
Velocity $(V) = \frac{dS}{dt} = 0 - \frac{48}{t^2} + 3t^2$.
Setting $V=0$:
$-\frac{48}{t^2} + 3t^2 = 0
\Rightarrow 3t^2 = \frac{48}{t^2}
\Rightarrow t^4 = 16
\Rightarrow t = 2$ (since $t > 0$).
Now,acceleration $(A) = \frac{dV}{dt} = \frac{d}{dt}(-\frac{48}{t^2} + 3t^2) = \frac{96}{t^3} + 6t$.
At $t=2$,$A = \frac{96}{2^3} + 6(2) = \frac{96}{8} + 12 = 12 + 12 = 24$.

(C) Given $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$.
Differentiating with respect to $x$,we get $f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$.
Differentiating again,we get $f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$.
Substituting $x=1$ in the expression for $f^{\prime}(x)$:
$f^{\prime}(1)=3(1)^2+2(1) f^{\prime}(1)+f^{\prime \prime}(2) \Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3$ (Equation $I$).
Substituting $x=2$ in the expression for $f^{\prime \prime}(x)$:
$f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1) \Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1)$ (Equation $II$).
Substituting $f^{\prime \prime}(2)$ from Equation $II$ into Equation $I$:
$f^{\prime}(1)+12+2 f^{\prime}(1)=-3 \Rightarrow 3 f^{\prime}(1)=-15 \Rightarrow f^{\prime}(1)=-5$.
Now,find $f^{\prime \prime}(2)$ using Equation $II$:
$f^{\prime \prime}(2)=12+2(-5)=12-10=2$.
Finally,substitute $f^{\prime}(1)$ and $f^{\prime \prime}(2)$ into the original function $f(x)$ at $x=2$:
$f(2)=2^3+2^2(-5)+2(2)+6 = 8-20+4+6 = -2$.

(B) Given equation: $(2 x)^{2 y}=4 e^{2 x-2 y}$.
Taking natural logarithm on both sides:
$2 y \log(2 x) = \log(4) + \log(e^{2 x-2 y})$
$2 y \log(2 x) = 2 \log 2 + 2 x - 2 y$
Dividing by $2$:
$y \log(2 x) = \log 2 + x - y$
$y(1 + \log(2 x)) = x + \log 2$
$y = \frac{x + \log 2}{1 + \log(2 x)}$
Differentiating with respect to $x$ using the quotient rule:
$\frac{d y}{d x} = \frac{(1 + \log(2 x)) \cdot \frac{d}{d x}(x + \log 2) - (x + \log 2) \cdot \frac{d}{d x}(1 + \log(2 x))}{(1 + \log(2 x))^2}$
$\frac{d y}{d x} = \frac{(1 + \log(2 x)) \cdot 1 - (x + \log 2) \cdot \frac{1}{2 x} \cdot 2}{(1 + \log(2 x))^2}$
$\frac{d y}{d x} = \frac{1 + \log(2 x) - \frac{x + \log 2}{x}}{(1 + \log(2 x))^2}$
Multiplying by $(1 + \log(2 x))^2$:
$(1 + \log(2 x))^2 \frac{d y}{d x} = 1 + \log(2 x) - 1 - \frac{\log 2}{x}$
$= \log(2 x) - \frac{\log 2}{x} = \frac{x \log(2 x) - \log 2}{x}$.

(A) Given the rate of change of production: $\frac{dP}{dx} = 100 - 12\sqrt{x}$.
Integrating both sides with respect to $x$:
$\int dP = \int (100 - 12x^{1/2}) dx$
$P = 100x - 12 \cdot \frac{x^{3/2}}{3/2} + C$
$P = 100x - 8x^{3/2} + C$
Given that the initial production $P = 1000$ when $x = 0$:
$1000 = 100(0) - 8(0)^{3/2} + C \implies C = 1000$.
Thus,the production function is $P(x) = 100x - 8x\sqrt{x} + 1000$.
For $x = 9$ additional workers:
$P(9) = 100(9) - 8(9)\sqrt{9} + 1000$
$P(9) = 900 - 8(9)(3) + 1000$
$P(9) = 900 - 216 + 1000 = 1684$.
The new level of production is $1684$ items.

(A) The function is defined as $f(x) = \frac{x}{1+|x|}$.
We can express this piecewise as:
$f(x) = \begin{cases} \frac{x}{1-x}, & x < 0 \\ \frac{x}{1+x}, & x \geq 0 \end{cases}$
Now,we check the differentiability at $x = 0$:
Left-hand derivative $(LHD)$ at $x = 0$:
$LHD = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\frac{h}{1-h} - 0}{h} = \lim_{h \to 0^-} \frac{1}{1-h} = 1$
Right-hand derivative $(RHD)$ at $x = 0$:
$RHD = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h}{1+h} - 0}{h} = \lim_{h \to 0^+} \frac{1}{1+h} = 1$
Since $LHD = RHD = 1$,the function is differentiable at $x = 0$.
For $x \neq 0$,the function is a rational function with a non-zero denominator,so it is differentiable everywhere.
Thus,the derivative exists for all $x \in (-\infty, \infty)$.

(C) $f(x) = \sin^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is defined if $-1 \leq \frac{|x|+5}{x^2+1} \leq 1$.
Since $|x|+5 > 0$ and $x^2+1 > 0$,the left inequality $\frac{|x|+5}{x^2+1} \geq -1$ is always true.
We only need to solve $\frac{|x|+5}{x^2+1} \leq 1$.
$|x|+5 \leq x^2+1$
$x^2 - |x| - 4 \geq 0$.
Let $t = |x|$,where $t \geq 0$. Then $t^2 - t - 4 \geq 0$.
The roots of $t^2 - t - 4 = 0$ are $t = \frac{1 \pm \sqrt{1 - 4(1)(-4)}}{2} = \frac{1 \pm \sqrt{17}}{2}$.
Since $t \geq 0$,we have $t \geq \frac{1 + \sqrt{17}}{2}$.
Thus,$|x| \geq \frac{1 + \sqrt{17}}{2}$,which implies $x \in \left(-\infty, -\frac{1 + \sqrt{17}}{2}\right] \cup \left[\frac{1 + \sqrt{17}}{2}, \infty\right)$.
Comparing this with $(-\infty, -a] \cup [a, \infty)$,we get $a = \frac{1 + \sqrt{17}}{2}$.

(D) Given the equation $2^x+2^y=2$.
We can rewrite this as $2^y = 2 - 2^x$.
For $y$ to be defined,the argument of the logarithm must be positive: $2 - 2^x > 0$.
This implies $2^x < 2$.
Since the base $2 > 1$,the inequality holds when $x < 1$.
Thus,the domain is $(-\infty, 1)$ or $-\infty < x < 1$.

(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 $0$.
$2 - 2^x > 0 \Rightarrow 2^x < 2$.
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 $-\infty < x < 1$.

(C) Let $f(x) = \frac{1-x+x^2}{1+x+x^2}$.
Using the quotient rule,$f'(x) = \frac{(1+x+x^2)(-1+2x) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}$.
Expanding the numerator: $(1+x+x^2)(-1+2x) = -1+2x-x+2x^2-x^2+2x^3 = 2x^3+x^2+x-1$.
$(1-x+x^2)(1+2x) = 1+2x-x-2x^2+x^2+2x^3 = 2x^3-x^2+x+1$.
Subtracting these: $(2x^3+x^2+x-1) - (2x^3-x^2+x+1) = 2x^2-2$.
Thus,$f'(x) = \frac{2x^2-2}{(1+x+x^2)^2}$.
Setting $f'(x) = 0$ gives $2x^2-2 = 0$,so $x^2 = 1$,which means $x = 1$ or $x = -1$.
Evaluating $f(x)$ at these points:
$f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$.
$f(-1) = \frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = \frac{3}{1} = 3$.
Therefore,the minimum value of $f(x)$ is $\frac{1}{3}$.

(B) Let $y = \frac{x^2}{x^2+1}$.
$y(x^2+1) = x^2$
$yx^2 + y = x^2$
$x^2(y-1) = -y$
$x^2 = \frac{y}{1-y}$.
For $x$ to be a real number,$x^2 \geq 0$.
Therefore,$\frac{y}{1-y} \geq 0$.
This implies $y(1-y) \geq 0$ and $y \neq 1$.
Solving the inequality $y(y-1) \leq 0$,we get $0 \leq y < 1$.
Thus,the range of the function is $[0, 1)$.

(B) Given $f(x) = \frac{3x-2}{5x+3}$.
First,calculate $f(1)$:
$f(1) = \frac{3(1)-2}{5(1)+3} = \frac{3-2}{5+3} = \frac{1}{8}$.
Now,calculate $f(f(1)) = f\left(\frac{1}{8}\right)$:
$f\left(\frac{1}{8}\right) = \frac{3\left(\frac{1}{8}\right)-2}{5\left(\frac{1}{8}\right)+3} = \frac{\frac{3}{8}-2}{\frac{5}{8}+3}$.
Multiply numerator and denominator by $8$:
$f\left(\frac{1}{8}\right) = \frac{3-16}{5+24} = \frac{-13}{29}$.

(A) Given functions are $f(x)=2x-3$ and $g(x)=x^3+5$.
First,we find the composite function $(f \circ g)(x) = f(g(x))$.
$(f \circ g)(x) = 2(g(x)) - 3 = 2(x^3+5) - 3 = 2x^3 + 10 - 3 = 2x^3 + 7$.
Let $y = (f \circ g)(x) = 2x^3 + 7$.
To find the inverse $(f \circ g)^{-1}(y)$,we solve for $x$ in terms of $y$:
$y - 7 = 2x^3$
$x^3 = \frac{y-7}{2}$
$x = \left(\frac{y-7}{2}\right)^{1/3}$.
Thus,$(f \circ g)^{-1}(y) = \left(\frac{y-7}{2}\right)^{1/3}$.
Now,substitute $y = -9$:
$(f \circ g)^{-1}(-9) = \left(\frac{-9-7}{2}\right)^{1/3} = \left(\frac{-16}{2}\right)^{1/3} = (-8)^{1/3} = -2$.

(B) $h(x) = f(g(x))$
$h(x) = f(\sin^{-1} x) = e^{\sin^{-1} x}$
Differentiating $h(x)$ with respect to $x$ using the chain rule:
$h^{\prime}(x) = \frac{d}{dx}(e^{\sin^{-1} x}) = e^{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x)$
$h^{\prime}(x) = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}$
Now,calculating the ratio $\frac{h^{\prime}(x)}{h(x)}$:
$\frac{h^{\prime}(x)}{h(x)} = \frac{e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1-x^2}}}{e^{\sin^{-1} x}}$
$\frac{h^{\prime}(x)}{h(x)} = \frac{1}{\sqrt{1-x^2}}$

(A) Given $f(x) = e^x - x$ and $g(x) = x^2 - x$.
$h(x) = (fog)(x) = f(g(x)) = e^{x^2-x} - (x^2-x) = e^{x^2-x} - x^2 + x$.
Now,differentiate $h(x)$ with respect to $x$:
$h'(x) = e^{x^2-x}(2x-1) - 2x + 1$.
$h'(x) = (2x-1)(e^{x^2-x} - 1)$.
For $h(x)$ to be increasing,we require $h'(x) \geq 0$.
$(2x-1)(e^{x^2-x} - 1) \geq 0$.
Let $u = x^2-x$. Since $e^u - 1$ has the same sign as $u$,we have $(2x-1)(x^2-x) \geq 0$.
$(2x-1)x(x-1) \geq 0$.
Using the sign scheme for the critical points $0, \frac{1}{2}, 1$:
For $x \in [0, \frac{1}{2}]$,$(2x-1) \leq 0$ and $x(x-1) \leq 0$,so the product is $\geq 0$.
For $x \in [1, \infty)$,$(2x-1) > 0$ and $x(x-1) \geq 0$,so the product is $\geq 0$.
Thus,$h(x)$ is increasing for $x \in [0, \frac{1}{2}] \cup [1, \infty)$.

(B) Given $f(x)=\log (\sin x)$ for $0 < x < \pi$ and $g(x)=\sin ^{-1}(e^{-x})$ for $x \geq 0$.
First,we find the composition $(f \circ g)(x) = f(g(x))$.
$(f \circ g)(x) = \log(\sin(\sin^{-1}(e^{-x}))) = \log(e^{-x}) = -x$.
Now,we find the derivative $(f \circ g)^{\prime}(x) = \frac{d}{dx}(-x) = -1$.
Given $a = (f \circ g)^{\prime}(\alpha) = -1$ and $b = (f \circ g)(\alpha) = -\alpha$.
Substitute these values into the options:
For option $(B)$: $a \alpha^2 - b \alpha - a = (-1)\alpha^2 - (-\alpha)\alpha - (-1) = -\alpha^2 + \alpha^2 + 1 = 1$.
Thus,$a \alpha^2 - b \alpha - a = 1$ is correct.

(C) $f(g(x)) = f\left(\frac{7x+4}{5x-3}\right)$
$= \frac{3\left(\frac{7x+4}{5x-3}\right)+4}{5\left(\frac{7x+4}{5x-3}\right)-7}$
$= \frac{21x+12+20x-12}{35x+20-35x+21}$
$= \frac{41x}{41}$
$= x$
Similarly,$g(f(x)) = g\left(\frac{3x+4}{5x-7}\right)$
$= \frac{7\left(\frac{3x+4}{5x-7}\right)+4}{5\left(\frac{3x+4}{5x-7}\right)-3}$
$= \frac{21x+28+20x-28}{15x+20-15x+21}$
$= \frac{41x}{41}$
$= x$
Therefore,$f(g(x)) = g(f(x)) = x$. Since $x$ is not explicitly listed as an option,but $g(f(x))$ is given as option $C$,and we proved $f(g(x)) = g(f(x))$,the correct choice is $C$.

(D) Given $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2 \sqrt{x}+x$.
We can rewrite $f(g(x))$ as:
$f(g(x)) = (1 + 2\sqrt{x} + x) + 2$
$f(g(x)) = (1 + \sqrt{x})^2 + 2$
Since $g(x) = 1 + \sqrt{x}$,we substitute $g(x)$ into the equation:
$f(g(x)) = [g(x)]^2 + 2$
Therefore,the function $f(x)$ is defined as $f(x) = x^2 + 2$.
Now,we need to find $f(f(x))$:
$f(f(x)) = f(x^2 + 2)$
$f(f(x)) = (x^2 + 2)^2 + 2$
$f(f(x)) = x^4 + 4x^2 + 4 + 2$
$f(f(x)) = x^4 + 4x^2 + 6$.

(C) Given functions are $f(x)=x^2+1$ and $g(x)=\frac{1}{x}$.
First,we find the composite function $f(g(g(f(x))))$:
$f(x) = x^2+1$
$g(f(x)) = g(x^2+1) = \frac{1}{x^2+1}$
$g(g(f(x))) = g\left(\frac{1}{x^2+1}\right) = \frac{1}{\frac{1}{x^2+1}} = x^2+1$
$f(g(g(f(x)))) = f(x^2+1) = (x^2+1)^2+1$
Now,substitute $x=1$ into the expression:
$f(g(g(f(1)))) = (1^2+1)^2+1$
$= (1+1)^2+1$
$= 2^2+1$
$= 4+1 = 5$

(A) Let $f(x) = y$,which implies $x = f^{-1}(y)$.
Given $y = \frac{2x - 3}{3x - 4}$.
Multiply both sides by $(3x - 4)$:
$y(3x - 4) = 2x - 3$
$3xy - 4y = 2x - 3$
Rearrange the terms to isolate $x$:
$3xy - 2x = 4y - 3$
$x(3y - 2) = 4y - 3$
$x = \frac{4y - 3}{3y - 2}$
Since $x = f^{-1}(y)$,we have $f^{-1}(y) = \frac{4y - 3}{3y - 2}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{4x - 3}{3x - 2}$.

(A) Given the equation: $3 f(x) - f\left(\frac{1}{x}\right) = 8 \log_2 x^3 = 24 \log_2 x$ $(i)$
Replace $x$ with $\frac{1}{x}$: $3 f\left(\frac{1}{x}\right) - f(x) = 24 \log_2 \left(\frac{1}{x}\right) = -24 \log_2 x$ $(ii)$
Multiply $(i)$ by $3$: $9 f(x) - 3 f\left(\frac{1}{x}\right) = 72 \log_2 x$ $(iii)$
Add $(ii)$ and $(iii)$: $(9 f(x) - f(x)) + (3 f\left(\frac{1}{x}\right) - 3 f\left(\frac{1}{x}\right)) = 72 \log_2 x - 24 \log_2 x$
$8 f(x) = 48 \log_2 x \Rightarrow f(x) = 6 \log_2 x$
Now,calculate the values:
$f(2) = 6 \log_2 2 = 6(1) = 6$
$f(4) = 6 \log_2 4 = 6(2) = 12$
$f(8) = 6 \log_2 8 = 6(3) = 18$
Since $12 - 6 = 6$ and $18 - 12 = 6$,the common difference is constant.
Therefore,$f(2), f(4), f(8)$ are in $A$.$P$.

(C) Note that $\triangle OAB$ is a right-angled triangle. Let $OA = x \text{ ft}$ and $OB = y \text{ ft}$.
By Pythagoras theorem,$x^2 + y^2 = 13^2 = 169$.
Therefore,$y^2 = 169 - x^2$.
Differentiating with respect to time $t$,we get $2y \frac{dy}{dt} = -2x \frac{dx}{dt}$,which simplifies to $y \frac{dy}{dt} = -x \frac{dx}{dt}$.
Given that at $x = 5 \text{ ft}$,$\frac{dx}{dt} = 3 \text{ ft/sec}$.
When $x = 5$,$y = \sqrt{169 - 5^2} = \sqrt{144} = 12 \text{ ft}$.
Substituting these values: $12 \frac{dy}{dt} = -5(3) = -15$.
Thus,$\frac{dy}{dt} = -\frac{15}{12} = -\frac{5}{4} \text{ ft/sec}$.
The negative sign indicates that $B$ is moving downwards towards $O$.
Therefore,$B$ is moving at the rate of $\frac{5}{4} \text{ ft/sec}$ downwards.

(B) Let $x$ be the distance of the lower end from the wall and $y$ be the height of the top of the ladder from the floor. The length of the ladder is $L = 5 \ m$.
From the Pythagorean theorem,$x^2 + y^2 = 5^2 = 25$.
We are given that the top slides downwards at $10 \ cm/s = 0.1 \ m/s$,so $\frac{dy}{dt} = -0.1 \ m/s$.
We want to find the rate of change of the angle $\theta$ between the ladder and the floor,where $\sin \theta = \frac{y}{5}$.
Differentiating with respect to $t$: $\cos \theta \frac{d\theta}{dt} = \frac{1}{5} \frac{dy}{dt}$.
When $x = 4 \ m$,$y = \sqrt{5^2 - 4^2} = 3 \ m$.
Then $\cos \theta = \frac{x}{5} = \frac{4}{5}$.
Substituting the values: $\frac{4}{5} \frac{d\theta}{dt} = \frac{1}{5} (-0.1)$.
$\frac{d\theta}{dt} = -\frac{0.1}{4} = -0.025 \ rad/s$.
The rate of decrease is $0.025 \ rad/s$.

(D) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder from the floor. The length of the ladder is $L = 5 \ m$.
From the Pythagorean theorem,$x^2 + y^2 = 5^2 = 25$.
We are given that the top slides downwards at $10 \ cm/s = 0.1 \ m/s$,so $\frac{dy}{dt} = -0.1 \ m/s$.
We want to find the rate of change of the angle $\theta$ between the ladder and the floor,where $\sin \theta = \frac{y}{5}$.
Differentiating with respect to $t$: $\cos \theta \frac{d\theta}{dt} = \frac{1}{5} \frac{dy}{dt}$.
When $x = 4 \ m$,$y = \sqrt{25 - 4^2} = \sqrt{9} = 3 \ m$.
Then $\cos \theta = \frac{x}{5} = \frac{4}{5}$.
Substituting the values: $\frac{4}{5} \frac{d\theta}{dt} = \frac{1}{5} (-0.1)$.
$\frac{d\theta}{dt} = -\frac{0.1}{4} = -0.025 \ rad/s$.
The angle is decreasing at the rate of $0.025 \ rad/s$.

(B) Given the curve $xy = 1$,we have $y = \frac{1}{x}$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{1}{x^2}$.
The slope of the tangent at any point $(x, y)$ is $-\frac{1}{x^2}$.
The slope of the normal is the negative reciprocal of the tangent slope,which is $x^2$.
The slope of the line $ax + by + c = 0$ is $-\frac{a}{b}$.
Since the line is a normal to the curve,we equate the slopes: $x^2 = -\frac{a}{b}$.
Since $x^2$ is always non-negative for real $x$,we must have $-\frac{a}{b} > 0$,which implies $\frac{a}{b} < 0$.
This condition holds if $a$ and $b$ have opposite signs,i.e.,$(a > 0, b < 0)$ or $(a < 0, b > 0)$.

(B) Let $I = \int \frac{1}{\sin (x-a) \sin x} \,d x$.
Multiply and divide by $\sin a$:
$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin (x-a) \sin x} \,d x$.
Express $\sin a$ as $\sin (x - (x-a))$:
$I = \frac{1}{\sin a} \int \frac{\sin (x - (x-a))}{\sin (x-a) \sin x} \,d x$.
Using the identity $\sin (A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin a} \int \frac{\sin x \cos (x-a) - \cos x \sin (x-a)}{\sin (x-a) \sin x} \,d x$.
Split the integral:
$I = \frac{1}{\sin a} \left[ \int \frac{\sin x \cos (x-a)}{\sin (x-a) \sin x} \,d x - \int \frac{\cos x \sin (x-a)}{\sin (x-a) \sin x} \,d x \right]$.
$I = \frac{1}{\sin a} \left[ \int \cot (x-a) \,d x - \int \cot x \,d x \right]$.
Integrating:
$I = \frac{1}{\sin a} [\log |\sin (x-a)| - \log |\sin x|] + c$.
$I = \operatorname{cosec} a \log |\frac{\sin (x-a)}{\sin x}| + c$.

(B) Given $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx$.
Using $\sin^2 x = 1 - \cos^2 x$,we have $\sin x(1 + \sin^2 x) = \sin x(1 + 1 - \cos^2 x) = \sin x(2 - \cos^2 x)$.
So,$I = \int \frac{\sin x(2 - \cos^2 x)}{2 \cos^2 x - 1} dx$.
Let $\cos x = t$,then $-\sin x dx = dt$,so $\sin x dx = -dt$.
$I = \int \frac{t^2 - 2}{2t^2 - 1} (-dt) = \int \frac{2 - t^2}{2t^2 - 1} dt$.
$I = \frac{1}{2} \int \frac{4 - 2t^2}{2t^2 - 1} dt = \frac{1}{2} \int \frac{-(2t^2 - 1) + 3}{2t^2 - 1} dt$.
$I = \frac{1}{2} \int (-1 + \frac{3}{2t^2 - 1}) dt = \frac{1}{2} [-t + \frac{3}{2\sqrt{2}} \log |\frac{\sqrt{2}t - 1}{\sqrt{2}t + 1}|] + c$.
$I = -\frac{1}{2} \cos x + \frac{3}{4\sqrt{2}} \log |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + c$.
Comparing with $P \cos x + Q \log |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + c$,we get $P = -\frac{1}{2}$ and $Q = \frac{3}{4\sqrt{2}}$.
Note: The provided options seem to have a sign discrepancy for $P$. Based on the standard integral form,$P = -1/2$ and $Q = 3/(4\sqrt{2})$. If $P$ is $1/2$,the numerator in the integral must be $\sin x(\cos^2 x - 2)$.

(A) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$.
Substitute $x = \tan t$,so $dx = \sec^2 t dt$.
Since $x > 0$,$t = \tan^{-1} x \in (0, \pi/2)$.
Note that $\sec^{-1} \sqrt{1+x^2} = \sec^{-1} \sqrt{1+\tan^2 t} = \sec^{-1} \sec t = t$.
Also,$\cos^{-1} \left(\frac{1-x^2}{1+x^2}\right) = \cos^{-1} \cos 2t = 2t$.
Substituting these into the integral:
$I = \int \frac{e^t}{1+\tan^2 t} [t^2 + 2t] \sec^2 t dt = \int e^t (t^2 + 2t) dt$.
Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = t^2$ and $f'(t) = 2t$:
$I = e^t t^2 + c = e^{\tan^{-1} x} (\tan^{-1} x)^2 + c$.

(D) Let $I = \int \frac{x+1}{x(1+x e^x)^2} dx$.
Multiply numerator and denominator by $e^x$:
$I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} dx$.
Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting this 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: $1 = A(1+t)^2 + Bt(1+t) + Ct$.
For $t=0$,$A=1$. For $t=-1$,$C=-1$. Comparing coefficients of $t^2$,$A+B=0 \Rightarrow B=-1$.
So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$.
$I = \log |t| - \log |1+t| + \frac{1}{1+t} + c$.
Substituting $t = x e^x$:
$I = \log |x e^x| - \log |1+x e^x| + \frac{1}{1+x e^x} + c$.
$I = \log \left| \frac{x e^x}{1+x e^x} \right| + \frac{1}{1+x e^x} + c$.

(B) Let $I = \int \frac{\tan x + \tan \alpha}{\tan x - \tan \alpha} dx$.
$= \int \frac{\frac{\sin x}{\cos x} + \frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x} - \frac{\sin \alpha}{\cos \alpha}} dx$
$= \int \frac{\sin x \cos \alpha + \sin \alpha \cos x}{\sin x \cos \alpha - \sin \alpha \cos x} dx$
$= \int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} dx$.
Let $t = x - \alpha$,so $x = t + \alpha$ and $dx = dt$.
$I = \int \frac{\sin (t + 2\alpha)}{\sin t} dt$
$= \int \frac{\sin t \cos 2\alpha + \cos t \sin 2\alpha}{\sin t} dt$
$= \cos 2\alpha \int 1 dt + \sin 2\alpha \int \cot t dt$
$= t \cos 2\alpha + \sin 2\alpha \log |\sin t| + c$
$= (x - \alpha) \cos 2\alpha + \log |\sin (x - \alpha)| \sin 2\alpha + c$.
Comparing this with $A(x) \cos 2\alpha + B(x) \sin 2\alpha + c$,we get $A(x) = x - \alpha$ and $B(x) = \log |\sin (x - \alpha)|$.

(A) Let $I = \int(\sqrt{\tan x} + \sqrt{\cot x}) dx$.
We can write this as $I = \int \left(\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}\right) dx$.
Simplifying the integrand,we get $I = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
Squaring both sides of $t = \sin x - \cos x$,we get $t^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - 2 \sin x \cos x$.
Thus,$2 \sin x \cos x = 1 - t^2$,which implies $\sin x \cos x = \frac{1 - t^2}{2}$.
Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{\frac{1 - t^2}{2}}} = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}}$.
Integrating,we obtain $I = \sqrt{2} \sin^{-1}(t) + c$.
Substituting $t = \sin x - \cos x$ back,we get $I = \sqrt{2} \sin^{-1}(\sin x - \cos x) + c$.

(D) Let $I = \int \frac{\sin x}{3+4 \cos ^2 x} \,dx$.
Substitute $\cos x = t$, then $-\sin x \,dx = dt$, which implies $\sin x \,dx = -dt$.
Substituting these into the integral, we get $I = \int \frac{-dt}{3+4 t^2} = -\int \frac{dt}{(\sqrt{3})^2 + (2t)^2}$.
Using the standard integral formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$, we have:
$I = -\frac{1}{\sqrt{3}} \cdot \frac{1}{2} \tan^{-1}(\frac{2t}{\sqrt{3}}) + C = -\frac{1}{2\sqrt{3}} \tan^{-1}(\frac{2 \cos x}{\sqrt{3}}) + C$.
Comparing this with $A \tan^{-1}(B \cos x) + C$, we identify $A = -\frac{1}{2\sqrt{3}}$ and $B = \frac{2}{\sqrt{3}}$.
Therefore, $A + B = -\frac{1}{2\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{-1 + 4}{2\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.

(C) 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:
$I = \int -\frac{1}{4} t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + c = -t^{1/4} + c$.
Substituting back $t = 1 + \frac{1}{x^4} = \frac{x^4+1}{x^4}$:
$I = -(1 + \frac{1}{x^4})^{1/4} + c = -\left(\frac{x^4+1}{x^4}\right)^{1/4} + c = -\frac{(x^4+1)^{1/4}}{x} + c$.

(D) Let $I = \int \frac{dx}{\sin x + \cos x}$.
Using the substitution $\tan \frac{x}{2} = t$,we have $dx = \frac{2}{1+t^2} dt$,$\sin x = \frac{2t}{1+t^2}$,and $\cos x = \frac{1-t^2}{1+t^2}$.
Substituting these into the integral:
$I = \int \frac{\frac{2}{1+t^2}}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}} dt = \int \frac{2}{2t + 1 - t^2} dt = -2 \int \frac{1}{t^2 - 2t - 1} dt$.
Completing the square in the denominator:
$t^2 - 2t - 1 = (t-1)^2 - 2 = (t-1)^2 - (\sqrt{2})^2$.
Using the formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + c$:
$I = -2 \times \frac{1}{2\sqrt{2}} \log \left| \frac{t-1-\sqrt{2}}{t-1+\sqrt{2}} \right| + c = -\frac{1}{\sqrt{2}} \log \left| \frac{\tan \frac{x}{2} - 1 - \sqrt{2}}{\tan \frac{x}{2} - 1 + \sqrt{2}} \right| + c$.
Simplifying the expression:
$I = -\frac{1}{\sqrt{2}} \log \left| \frac{\tan \frac{x}{2} - (\sqrt{2} + 1)}{\tan \frac{x}{2} + \sqrt{2} - 1} \right| + c$.

(C) Let $I = \int \frac{(x^2-1) dx}{x^3 \sqrt{2x^4-2x^2+1}}$.
Divide the numerator and denominator inside the square root by $x^4$:
$I = \int \frac{(x^2-1) dx}{x^3 \cdot x^2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} = \int \frac{(\frac{1}{x^3} - \frac{1}{x^5}) dx}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}$.
Let $t = 2-\frac{2}{x^2}+\frac{1}{x^4}$.
Then $dt = (\frac{4}{x^3} - \frac{4}{x^5}) dx$,which implies $(\frac{1}{x^3} - \frac{1}{x^5}) dx = \frac{dt}{4}$.
Substituting these into the integral:
$I = \int \frac{dt/4}{\sqrt{t}} = \frac{1}{4} \int t^{-1/2} dt = \frac{1}{4} \cdot \frac{t^{1/2}}{1/2} + c = \frac{1}{2} \sqrt{t} + c$.
Substituting $t$ back,we get $I = \frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}} + c$.

(C) Let $I = \int \frac{5 \tan x}{\tan x - 2} \, dx = \int \frac{5 \sin x}{\sin x - 2 \cos x} \, dx$.
We express the numerator as $5 \sin x = A(\sin x - 2 \cos x) + B \frac{d}{dx}(\sin x - 2 \cos x)$.
$5 \sin x = A(\sin x - 2 \cos x) + B(\cos x + 2 \sin x)$.
Equating the coefficients of $\sin x$ and $\cos x$:
$A + 2B = 5$ and $-2A + B = 0$.
From the second equation,$B = 2A$. Substituting into the first: $A + 2(2A) = 5 \implies 5A = 5 \implies A = 1$.
Then $B = 2(1) = 2$.
Thus,$I = \int \left( 1 + 2 \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} \right) \, dx$.
$I = \int 1 \, dx + 2 \int \frac{d(\sin x - 2 \cos x)}{\sin x - 2 \cos x} = x + 2 \log |\sin x - 2 \cos x| + c$.
Comparing this with $x + a \log |\sin x - 2 \cos x| + c$,we get $a = 2$.

(D) Let $I = \int \frac{dx}{x^2(x^4+1)^{\frac{3}{4}}}$.
We can rewrite the integral as $I = \int \frac{dx}{x^2 \cdot x^3 (1 + \frac{1}{x^4})^{\frac{3}{4}}} = \int \frac{dx}{x^5 (1 + \frac{1}{x^4})^{\frac{3}{4}}}$.
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,we get:
$I = -\frac{1}{4} \int t^{-\frac{3}{4}} dt$.
Integrating with respect to $t$,we have:
$I = -\frac{1}{4} \cdot \frac{t^{\frac{1}{4}}}{\frac{1}{4}} + c = -t^{\frac{1}{4}} + c$.
Substituting back $t = 1 + \frac{1}{x^4}$,we get:
$I = -(1 + \frac{1}{x^4})^{\frac{1}{4}} + c = -\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}} + c$.

(D) Let $I = \int \frac{\log (\cot x)}{\sin 2 x} \,d x$.
Put $\log (\cot x) = t$.
Differentiating both sides with respect to $x$:
$\frac{1}{\cot x} \cdot (-\csc^2 x) \,d x = dt$.
$\Rightarrow \frac{\sin x}{\cos x} \cdot (-\frac{1}{\sin^2 x}) \,d x = dt$.
$\Rightarrow \frac{-1}{\sin x \cos x} \,d x = dt$.
Since $\sin 2x = 2 \sin x \cos x$, we have $\sin x \cos x = \frac{\sin 2x}{2}$.
Substituting this: $\frac{-1}{\frac{\sin 2x}{2}} \,d x = dt \Rightarrow \frac{-2}{\sin 2x} \,d x = dt \Rightarrow \frac{d x}{\sin 2x} = \frac{-dt}{2}$.
Now, substitute these into the integral:
$I = \int t \cdot (\frac{-dt}{2}) = -\frac{1}{2} \int t \,dt$.
$I = -\frac{1}{2} \cdot \frac{t^2}{2} + c = -\frac{1}{4} t^2 + c$.
Substituting $t = \log (\cot x)$ back:
$I = -\frac{1}{4} [\log (\cot x)]^2 + c$.

(C) Let $I = \int \frac{dx}{x \sqrt{1-x^3}}$.
Multiply the numerator and denominator by $x^2$:
$I = \int \frac{x^2 dx}{x^3 \sqrt{1-x^3}}$.
Let $1-x^3 = t^2$,then $-3x^2 dx = 2t dt$,so $x^2 dx = -\frac{2}{3} t dt$.
Also,$x^3 = 1-t^2$.
Substituting these into the integral:
$I = \int \frac{-\frac{2}{3} t dt}{(1-t^2) t} = -\frac{2}{3} \int \frac{dt}{1-t^2} = \frac{2}{3} \int \frac{dt}{t^2-1}$.
Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right| + c$:
$I = \frac{2}{3} \times \frac{1}{2} \log \left|\frac{t-1}{t+1}\right| + c = \frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right| + c$.
Comparing this with the given expression,we find $k = \frac{1}{3}$.

(B) Let $I = \int \frac{e^x(1+x)}{\cos^2(e^x \cdot x)} dx$.
Substitute $t = e^x \cdot x$.
Then,by the product rule,$dt = (e^x \cdot x + e^x \cdot 1) dx = e^x(x+1) dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{\cos^2 t} = \int \sec^2 t dt$.
The integral of $\sec^2 t$ is $\tan t + c$.
Therefore,$I = \tan(x \cdot e^x) + c$.

(B) Let $I = \int (1 - \cos x) \operatorname{cosec}^2 x \, dx$
Using the identity $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have:
$I = \int \frac{2 \sin^2 \frac{x}{2}}{\sin^2 x} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{(2 \sin \frac{x}{2} \cos \frac{x}{2})^2} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \frac{1}{\cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$
Integrating $\sec^2 \frac{x}{2}$,we get:
$I = \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{1/2} + c$
$I = \tan \frac{x}{2} + c$

(D) Given integral $I = \int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} \,d x$.
Using the identity $1+\cos \theta = 2 \cos^2 \frac{\theta}{2}$, we have $\cos 8x + 1 = 2 \cos^2 4x$.
The denominator is $\frac{\cos 2x}{\sin 2x} - \frac{\sin 2x}{\cos 2x} = \frac{\cos^2 2x - \sin^2 2x}{\sin 2x \cos 2x} = \frac{\cos 4x}{\frac{1}{2} \sin 4x} = \frac{2 \cos 4x}{\sin 4x}$.
Substituting these into the integral:
$I = \int \frac{2 \cos^2 4x}{\frac{2 \cos 4x}{\sin 4x}} \,d x = \int \cos 4x \sin 4x \,d x$.
Multiply and divide by $2$:
$I = \frac{1}{2} \int 2 \sin 4x \cos 4x \,d x = \frac{1}{2} \int \sin 8x \,d x$.
Integrating $\sin 8x$ gives $\frac{-\cos 8x}{8}$.
So, $I = \frac{1}{2} \left( \frac{-\cos 8x}{8} \right) + c = \frac{-\cos 8x}{16} + c$.
Comparing with $A \cos 8x + c$, we get $A = \frac{-1}{16}$.

(A) Let $I = \int x \sqrt{\frac{2 \sin \left(x^2+1\right)-\sin 2\left(x^2+1\right)}{2 \sin \left(x^2+1\right)+\sin 2\left(x^2+1\right)}} \, dx$
Using $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$I = \int x \sqrt{\frac{2 \sin \left(x^2+1\right) - 2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}{2 \sin \left(x^2+1\right) + 2 \sin \left(x^2+1\right) \cos \left(x^2+1\right)}} \, dx$
$I = \int x \sqrt{\frac{1 - \cos \left(x^2+1\right)}{1 + \cos \left(x^2+1\right)}} \, dx$
Using $1 - \cos \theta = 2 \sin^2(\theta/2)$ and $1 + \cos \theta = 2 \cos^2(\theta/2)$:
$I = \int x \sqrt{\frac{2 \sin^2 \left(\frac{x^2+1}{2}\right)}{2 \cos^2 \left(\frac{x^2+1}{2}\right)}} \, dx$
$I = \int x \tan \left(\frac{x^2+1}{2}\right) \, dx$
Let $t = \frac{x^2+1}{2}$,then $dt = x \, dx$.
$I = \int \tan t \, dt = \log |\sec t| + c$
$I = \log \left| \sec \left(\frac{x^2+1}{2}\right) \right| + c$

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^5 x + \cos ^3 x \sin ^2 x + \sin ^3 x \cos ^2 x + \cos ^5 x)^2} \,dx$.
Factor the denominator: $\sin ^5 x + \sin ^3 x \cos ^2 x + \cos ^3 x \sin ^2 x + \cos ^5 x = \sin ^3 x(\sin ^2 x + \cos ^2 x) + \cos ^3 x(\sin ^2 x + \cos ^2 x) = \sin ^3 x + \cos ^3 x$.
Thus, $I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^3 x + \cos ^3 x)^2} \,dx$.
Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{\tan ^2 x \sec ^2 x}{(1 + \tan ^3 x)^2} \,dx$.
Let $t = 1 + \tan ^3 x$. Then $dt = 3 \tan ^2 x \sec ^2 x \,dx$, which implies $\tan ^2 x \sec ^2 x \,dx = \frac{1}{3} dt$.
Substituting these into the integral: $I = \frac{1}{3} \int \frac{1}{t^2} \,dt = \frac{1}{3} (-t^{-1}) + c = -\frac{1}{3t} + c$.
Substituting $t$ back: $I = \frac{-1}{3(1 + \tan ^3 x)} + c$.

(A) Let $I = \int \frac{\operatorname{cosec} x \, dx}{\cos^2(1 + \log \tan \frac{x}{2})}$.
Let $t = 1 + \log(\tan \frac{x}{2})$.
Differentiating both sides with respect to $x$,we get:
$dt = \frac{1}{\tan \frac{x}{2}} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} \, dx$.
Using the identity $\tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}$ and $\sec^2 \frac{x}{2} = \frac{1}{\cos^2 \frac{x}{2}}$,we have:
$dt = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \cdot \frac{1}{\cos^2 \frac{x}{2}} \cdot \frac{1}{2} \, dx = \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \, dx$.
Since $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we get $dt = \frac{1}{\sin x} \, dx = \operatorname{cosec} x \, dx$.
Substituting these into the integral:
$I = \int \frac{1}{\cos^2 t} \, dt = \int \sec^2 t \, dt$.
Integrating,we get $I = \tan t + c$.
Substituting back the value of $t$,we get $I = \tan(1 + \log(\tan \frac{x}{2})) + c$.

(B) Let $I = \int \left( \frac{\tan \left( \frac{1}{x} \right)}{x} \right)^2 \, dx$.
Substitute $t = \frac{1}{x}$,then $dt = -\frac{1}{x^2} \, dx$,which implies $dx = -x^2 \, dt = -\frac{1}{t^2} \, dt$.
Substituting these into the integral:
$I = \int \tan^2(t) \cdot \frac{1}{x^2} \cdot (-x^2 \, dt) = -\int \tan^2(t) \, dt$.
Using the identity $\tan^2(t) = \sec^2(t) - 1$:
$I = -\int (\sec^2(t) - 1) \, dt = \int (1 - \sec^2(t)) \, dt$.
Integrating term by term:
$I = t - \tan(t) + c$.
Substituting back $t = \frac{1}{x}$:
$I = \frac{1}{x} - \tan \left( \frac{1}{x} \right) + c$.

(A) Let $I = \int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \,d x$.
Since $\sin 2x = 2 \sin x \cos x$,we have $I = \int \frac{1}{\cos ^3 x \sqrt{2 \sin x \cos x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{1}{\cos ^3 x \sqrt{\sin x} \sqrt{\cos x}} \,d x$.
This simplifies to $I = \frac{1}{\sqrt{2}} \int \frac{1}{\cos^{3.5} x \sqrt{\sin x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{\sec^{3.5} x}{\sqrt{\tan x}} \,d x$.
Wait,let us rewrite the integrand as $\frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 \sin x \cos x}} = \frac{\sec^2 x \cdot \sec^2 x}{\sqrt{2 \tan x \cos^2 x}} = \frac{\sec^4 x}{\sqrt{2 \tan x}}$.
So,$I = \frac{1}{\sqrt{2}} \int \frac{\sec^4 x}{\sqrt{\tan x}} \,d x = \frac{1}{\sqrt{2}} \int \frac{(1+\tan^2 x) \sec^2 x}{\sqrt{\tan x}} \,d x$.
Let $\tan x = t$,then $\sec^2 x \,d x = dt$.
$I = \frac{1}{\sqrt{2}} \int \frac{1+t^2}{\sqrt{t}} \,dt = \frac{1}{\sqrt{2}} \int (t^{-1/2} + t^{3/2}) \,dt$.
$I = \frac{1}{\sqrt{2}} \left( \frac{t^{1/2}}{1/2} + \frac{t^{5/2}}{5/2} \right) + c = \frac{1}{\sqrt{2}} \left( 2\sqrt{t} + \frac{2}{5} t^{5/2} \right) + c$.
$I = \sqrt{2} \sqrt{t} + \frac{\sqrt{2}}{5} t^{5/2} + c = \sqrt{2} \left( \sqrt{\tan x} + \frac{1}{5} (\tan x)^{5/2} \right) + c$.

(B) Substitute $x = \tan \theta$,then $dx = \sec^2 \theta \, d\theta$.
Then,$f(x) = \int \frac{\tan^2 \theta \sec^2 \theta \, d\theta}{\sec^2 \theta(1+\sec \theta)} = \int \frac{\tan^2 \theta \, d\theta}{1+\sec \theta}$.
Since $\tan^2 \theta = \sec^2 \theta - 1$,we have $\int \frac{\sec^2 \theta - 1}{1+\sec \theta} \, d\theta = \int (\sec \theta - 1) \, d\theta$.
Integrating gives $f(x) = \log |\sec \theta + \tan \theta| - \theta + c$.
Substituting back $x = \tan \theta$ and $\sec \theta = \sqrt{1+x^2}$,we get $f(x) = \log |x + \sqrt{1+x^2}| - \tan^{-1} x + c$.
Given $f(0) = 0$,we have $\log |0 + 1| - \tan^{-1}(0) + c = 0$,which implies $c = 0$.
Thus,$f(x) = \log |x + \sqrt{1+x^2}| - \tan^{-1} x$.
Evaluating at $x=1$,$f(1) = \log |1 + \sqrt{2}| - \tan^{-1}(1) = \log (1+\sqrt{2}) - \frac{\pi}{4}$.

(B) Let $I = \int x^5 e^{-4 x^3} \,d x$.
Substitute $x^3 = t$,then $3x^2 \,d x = dt$,which implies $x^2 \,d x = \frac{1}{3} dt$.
Since $x^5 = x^3 \cdot x^2$,the integral becomes $I = \frac{1}{3} \int t e^{-4 t} dt$.
Using integration by parts,$\int u v \,dt = u \int v \,dt - \int (u' \int v \,dt) dt$,where $u = t$ and $v = e^{-4 t}$.
$I = \frac{1}{3} \left( t \cdot \frac{e^{-4 t}}{-4} - \int 1 \cdot \frac{e^{-4 t}}{-4} dt \right)$.
$I = \frac{1}{3} \left( -\frac{t e^{-4 t}}{4} + \frac{1}{4} \int e^{-4 t} dt \right)$.
$I = \frac{1}{3} \left( -\frac{t e^{-4 t}}{4} - \frac{e^{-4 t}}{16} \right) + c$.
$I = -\frac{t e^{-4 t}}{12} - \frac{e^{-4 t}}{48} + c$.
$I = \frac{e^{-4 t}}{48} (-4 t - 1) + c$.
Substituting $t = x^3$ back,we get $I = \frac{1}{48} e^{-4 x^3} (-4 x^3 - 1) + c$.
Comparing this with the given form,$f(x) = -4 x^3 - 1$.

(B) Let $I = \int \frac{\log \left(x^2+a^2\right)}{x^2} \,d x$.
Using integration by parts,$\int u v \,d x = u \int v \,d x - \int \left( \frac{du}{dx} \int v \,d x \right) d x$.
Let $u = \log \left(x^2+a^2\right)$ and $v = x^{-2}$.
Then $\frac{du}{dx} = \frac{2x}{x^2+a^2}$ and $\int v \,d x = -\frac{1}{x}$.
$I = \log \left(x^2+a^2\right) \cdot \left(-\frac{1}{x}\right) - \int \left( \frac{2x}{x^2+a^2} \cdot \left(-\frac{1}{x}\right) \right) d x$.
$I = -\frac{\log \left(x^2+a^2\right)}{x} + 2 \int \frac{1}{x^2+a^2} \,d x$.
Using the standard integral $\int \frac{1}{x^2+a^2} \,d x = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + c$.
$I = -\frac{\log \left(x^2+a^2\right)}{x} + \frac{2}{a} \tan^{-1} \left(\frac{x}{a}\right) + c$.