TS EAMCET 2020 Mathematics Question Paper with Answer and Solution

(C) Given $|z_1|=1 \implies x_1^2+y_1^2=1$ and $|z_2|=2 \implies x_2^2+y_2^2=4$.
$\operatorname{Re}(z_1 \bar{z}_2) = x_1 x_2 + y_1 y_2 = 0$.
Let $z_1 = \cos \theta + i \sin \theta$. Then $x_1 = \cos \theta, y_1 = \sin \theta$.
Since $x_1 x_2 + y_1 y_2 = 0$, we have $x_2 \cos \theta + y_2 \sin \theta = 0$.
This implies $(x_2, y_2) = \pm 2(-\sin \theta, \cos \theta)$.
Taking $x_2 = -2 \sin \theta$ and $y_2 = 2 \cos \theta$:
$z_3 = x_1 + i \frac{x_2}{2} = \cos \theta - i \sin \theta = \bar{z}_1 \implies |z_3|=1$.
$z_4 = 2 y_1 + i y_2 = 2 \sin \theta + i 2 \cos \theta = 2i(\cos \theta - i \sin \theta) = 2i \bar{z}_1 \implies |z_4|=2$.
$z_3 z_4 = (\bar{z}_1)(2i \bar{z}_1) = 2i \bar{z}_1^2 = 2i(\cos 2\theta - i \sin 2\theta) = 2 \sin 2\theta + 2i \cos 2\theta$.
However, checking the condition $\operatorname{Re}(z_1 z_2)=0$ (instead of $\operatorname{Re}(z_1 \bar{z}_2)=0$):
If $\operatorname{Re}(z_1 z_2)=0$, then $x_1 x_2 - y_1 y_2 = 0 \implies x_1 x_2 = y_1 y_2$.
Using $z_1 = e^{i\theta}$, $z_2 = 2e^{i(\pi/2 - \theta)} = 2i \bar{z}_1 = 2 \sin \theta + 2i \cos \theta$.
Then $z_3 = \cos \theta + i \sin \theta = z_1$ and $z_4 = 2 \sin \theta + 2i \cos \theta = 2i \bar{z}_1$.
$z_3 z_4 = z_1 (2i \bar{z}_1) = 2i |z_1|^2 = 2i$.
Thus, $\operatorname{Re}(z_3 z_4) = 0$ and $|z_3|=1, |z_4|=2$.

(D) Let $z = \sqrt{3} + i$.
Then,the modulus $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.
The argument $\theta = \arg(z) = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
Thus,the polar form is $z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$.
Using De Moivre's Theorem,$z^{10} = 2^{10}(\cos \frac{10\pi}{6} + i \sin \frac{10\pi}{6})$.
Simplifying the angle,$\frac{10\pi}{6} = \frac{5\pi}{3}$.
$z^{10} = 1024(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$.
Since $\cos \frac{5\pi}{3} = \frac{1}{2}$ and $\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$,
$z^{10} = 1024(\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 512 - 512i\sqrt{3}$.
Comparing with $a+bi$,we get $a = 512$ and $b = -512\sqrt{3}$.

(D) We know that $1+\omega+\omega^2=0$ and $\omega^3=1$.
From $1+\omega+\omega^2=0$,we have $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$.
Substituting these into the expression:
$\left(1-\omega+\omega^2\right)^6+\left(1-\omega^2+\omega\right)^6 = \left(-\omega-\omega\right)^6+\left(-\omega^2-\omega^2\right)^6$
$= (-2\omega)^6 + (-2\omega^2)^6$
$= 64\omega^6 + 64\omega^{12}$
Since $\omega^3=1$,then $\omega^6 = (\omega^3)^2 = 1^2 = 1$ and $\omega^{12} = (\omega^3)^4 = 1^4 = 1$.
$= 64(1) + 64(1) = 64 + 64 = 128$.

(A) We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$.
Thus,$\frac{1-\sqrt{3}i}{2} = -\omega^2$ and $\frac{1+\sqrt{3}i}{2} = -\omega$.
Then,$\left(-\omega^2\right)^{2020} + (-\omega)^{2026} = \omega^{4040} + \omega^{2026} = \omega^2 + \omega = -1$.
Next,consider the sum $\sum_{j=1}^6 (j+\omega)(j+\omega^2) = \sum_{j=1}^6 (j^2 + j(\omega+\omega^2) + \omega^3) = \sum_{j=1}^6 (j^2 - j + 1)$.
Using summation formulas: $\sum_{j=1}^6 j^2 = \frac{6(7)(13)}{6} = 91$,$\sum_{j=1}^6 j = \frac{6(7)}{2} = 21$,and $\sum_{j=1}^6 1 = 6$.
So,the sum is $91 - 21 + 6 = 76$.
The sine term becomes $\sin\left(76 \times \frac{3\pi}{152}\right) = \sin\left(\frac{3\pi}{2}\right) = -1$.
Finally,the total expression is $-1 + (-1) = -2$.

(A) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2=0$ and $\omega^3=1$.
Expanding the expression inside the summation:
$(\omega x+2)(\omega^2 x+2)-3 = \omega^3 x^2 + 2\omega x + 2\omega^2 x + 4 - 3$
$= x^2 + 2x(\omega + \omega^2) + 1$ (since $\omega^3=1$)
$= x^2 + 2x(-1) + 1 = x^2 - 2x + 1 = (x-1)^2$.
Now,we calculate the sum:
$\sum_{x=1}^{10} (x-1)^2 = \sum_{x=1}^{10} (x^2 - 2x + 1) = \sum_{x=1}^{10} x^2 - 2\sum_{x=1}^{10} x + \sum_{x=1}^{10} 1$
$= \frac{10(11)(21)}{6} - 2 \times \frac{10(11)}{2} + 10$
$= 385 - 110 + 10 = 285$.

(D) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$.
Note that $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$.
Thus,the terms in the product $A_r$ are:
For $k=1$,$(x+\frac{1}{x})^3 = (-1)^3 = -1$.
For $k=2$,$(x^2+\frac{1}{x^2})^3 = (\omega^2+\omega)^3 = (-1)^3 = -1$.
For $k=3$,$(x^3+\frac{1}{x^3})^3 = (1+1)^3 = 8$.
So,$A_3 = (-1)(-1)(8) = 8$.
For $k=4$,$(x^4+\frac{1}{x^4})^3 = (\omega+\omega^2)^3 = -1$.
For $k=5$,$(x^5+\frac{1}{x^5})^3 = (\omega^2+\omega)^3 = -1$.
For $k=6$,$(x^6+\frac{1}{x^6})^3 = (1+1)^3 = 8$.
Thus,$A_6 = A_3 \cdot (-1) \cdot (-1) \cdot 8 = 8^2 = 64$.
In general,$A_{3n} = 8^n$.
The series is $\sum_{n=1}^{\infty} \frac{1}{A_{3n}} = \sum_{n=1}^{\infty} \frac{1}{8^n} = \frac{1/8}{1-1/8} = \frac{1/8}{7/8} = \frac{1}{7}$.

(A) Given $z^2+z+1=0$,the roots are $z = \omega$ and $z = \omega^2$,where $\omega$ is a complex cube root of unity.
For any $n$ not divisible by $3$,$z^n + \frac{1}{z^n} = \omega^n + \omega^{2n} = -1$.
For $n$ divisible by $3$,$z^n + \frac{1}{z^n} = \omega^{3k} + \omega^{6k} = 1 + 1 = 2$.
There are $2020$ terms in the sum.
The number of terms where $n$ is a multiple of $3$ is $\lfloor \frac{2020}{3} \rfloor = 673$.
The number of terms where $n$ is not a multiple of $3$ is $2020 - 673 = 1347$.
The sum is $1347 \times (-1)^3 + 673 \times (2)^3 = -1347 + 673 \times 8 = -1347 + 5384 = 4037$.

(A) Given,$x = \omega^2 - \omega - 3$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 = -1 - \omega$.
Substituting this,$x = (-1 - \omega) - \omega - 3 = -2\omega - 4$.
Alternatively,using $\omega^2 - \omega - 1 = 0$ is not correct,but $\omega^2 + \omega + 1 = 0$ is.
Let us re-evaluate: $x = \omega^2 - \omega - 3$.
Since $\omega^2 = -1 - \omega$,$x = -1 - \omega - \omega - 3 = -2\omega - 4$.
Then $x + 4 = -2\omega$.
Squaring both sides,$(x + 4)^2 = 4\omega^2$.
$x^2 + 8x + 16 = 4(-1 - \omega) = -4 - 4\omega$.
Since $-4\omega = 2(x + 4)$,we have $x^2 + 8x + 16 = -4 + 2x + 8$.
$x^2 + 6x + 12 = 0$.
Now,perform polynomial division of $x^4 + 6x^3 + 10x^2 - 12x - 19$ by $x^2 + 6x + 12$.
$x^4 + 6x^3 + 10x^2 - 12x - 19 = (x^2 + 6x + 12)(x^2 - 2) + 5$.
Since $x^2 + 6x + 12 = 0$,the expression equals $0(x^2 - 2) + 5 = 5$.

(A) Given the equation $(x-1)^5=32(x+1)^5$.
Dividing both sides by $(x+1)^5$,we get $\left(\frac{x-1}{x+1}\right)^5=32$.
Let $z = \frac{x-1}{x+1}$. Then $z^5 = 32 = 2^5 \cdot e^{i(2k\pi)}$,where $k=0, 1, 2, 3, 4$.
Thus,$z = 2 e^{\frac{2k\pi i}{5}}$.
Substituting back,$\frac{x-1}{x+1} = 2 e^{\frac{2k\pi i}{5}}$.
Let $\alpha = 2 e^{\frac{2k\pi i}{5}}$. Then $\frac{x-1}{x+1} = \alpha$.
$x-1 = \alpha(x+1)$ $\Rightarrow x(1-\alpha) = 1+\alpha$ $\Rightarrow x = \frac{1+\alpha}{1-\alpha}$.
Substituting $\alpha$ back,$x = \frac{1+2 e^{\frac{2k\pi i}{5}}}{1-2 e^{\frac{2k\pi i}{5}}}$ for $k=0, 1, 2, 3, 4$.

(C) Given the equation $(z+1)^n = z^n$. Since $z=0$ is not a root, we can write $(\frac{z+1}{z})^n = 1$.
Let $\frac{z+1}{z} = \omega_k = e^{i \frac{2k\pi}{n}}$ for $k = 1, 2, \ldots, n-1$.
Then $z+1 = z \omega_k \Rightarrow z(1 - \omega_k) = -1 \Rightarrow z = \frac{1}{\omega_k - 1}$.
Substituting $\omega_k = \cos(\frac{2k\pi}{n}) + i \sin(\frac{2k\pi}{n})$, we get $z = \frac{1}{\cos(\frac{2k\pi}{n}) - 1 + i \sin(\frac{2k\pi}{n})} = \frac{1}{-2 \sin^2(\frac{k\pi}{n}) + 2i \sin(\frac{k\pi}{n}) \cos(\frac{k\pi}{n})} = \frac{1}{2i \sin(\frac{k\pi}{n}) [\cos(\frac{k\pi}{n}) + i \sin(\frac{k\pi}{n})]} = \frac{1}{2i \sin(\frac{k\pi}{n}) e^{i \frac{k\pi}{n}}} = \frac{1}{2} (-i \csc(\frac{k\pi}{n})) e^{-i \frac{k\pi}{n}} = \frac{1}{2} (-i \csc(\frac{k\pi}{n})) (\cos(\frac{k\pi}{n}) - i \sin(\frac{k\pi}{n})) = \frac{1}{2} (-i \cot(\frac{k\pi}{n}) - 1)$.
Thus, $\operatorname{Re}(z_k) = -\frac{1}{2}$ and $\operatorname{Im}(z_k) = -\frac{1}{2} \cot(\frac{k\pi}{n})$.
The sum is over $n-1$ roots. Substituting these into the expression: $\sum_{k=1}^{n-1} \frac{\cot^{-1}(|\cot(\frac{k\pi}{n})|) - 1}{-1} = \sum_{k=1}^{n-1} (1 - \frac{k\pi}{n}) = (n-1) - \frac{\pi}{n} \frac{(n-1)n}{2} = (n-1)(1 - \frac{\pi}{2}) = \frac{n-1}{2}(2 - \pi)$.

(A) Given $z = x + iy$,then $\bar{z} = x - iy$.
Substituting into the expression: $\frac{\bar{z}-1}{z-i} = \frac{(x-1) - iy}{x + i(y-1)}$.
To find the real part,multiply the numerator and denominator by the conjugate of the denominator: $x - i(y-1)$.
$\frac{((x-1) - iy)(x - i(y-1))}{x^2 + (y-1)^2} = \frac{x(x-1) - y(y-1) + i(\dots)}{x^2 + (y-1)^2}$.
The real part is $\frac{x^2 - x - y^2 + y}{x^2 + (y-1)^2} = 2$.
$x^2 - x - y^2 + y = 2(x^2 + y^2 - 2y + 1)$.
$x^2 + 3y^2 + x - 5y + 2 = 0$.
This equation represents a pair of straight lines passing through the point $(-1, 1)$.

(D) Given,$\arg(z-2-3i) = \frac{\pi}{4}$.
Let $z = x+iy$.
Then $z-2-3i = (x-2) + i(y-3)$.
Since $\arg(z-2-3i) = \frac{\pi}{4}$,we have $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{\pi}{4}$.
This implies $\frac{y-3}{x-2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Therefore,$y-3 = x-2$,which simplifies to $x-y+1 = 0$.
Thus,the locus of $z$ is $x-y+1 = 0$.

(B) Given $z = x+iy$,we have $z-2-3i = (x-2) + i(y-3)$.
Since the amplitude of $z-2-3i$ is $\frac{\pi}{4}$,we have $\arg((x-2) + i(y-3)) = \frac{\pi}{4}$.
This implies $\tan^{-1}\left(\frac{y-3}{x-2}\right) = \frac{\pi}{4}$.
Taking the tangent on both sides,we get $\frac{y-3}{x-2} = \tan\left(\frac{\pi}{4}\right) = 1$.
Therefore,$y-3 = x-2$,which simplifies to $x-y+1=0$.

(B) Given that $\alpha$ and $\beta$ are roots of the equation $z^2+az+b=0$.
From the properties of roots,we have $\alpha+\beta = -a$ and $\alpha\beta = b$.
Since the origin $(0)$,$\alpha$,and $\beta$ form an equilateral triangle in the Argand plane,the condition for an equilateral triangle with one vertex at the origin is $\alpha^2 + \beta^2 = \alpha\beta$.
We can rewrite this as $(\alpha+\beta)^2 - 2\alpha\beta = \alpha\beta$.
Substituting the values of the sum and product of roots: $(-a)^2 - 2(b) = b$.
This simplifies to $a^2 - 2b = b$,which gives $a^2 = 3b$.

(C) The equation $|z-2i|=2$ represents a circle in the Argand plane with center at $(0, 2)$ and radius $r=2$.
For $\triangle ABC$ to have the maximum area,it must be an equilateral triangle inscribed in the circle.
Let $A$ be the point $(2, 2)$. The center of the circle is $O'(0, 2)$.
The line segment $AO'$ lies on the horizontal line $y=2$.
For an equilateral triangle,the altitude from $A$ to the side $BC$ must pass through the center $O'(0, 2)$.
Since $A$ is at $(2, 2)$ and $O'$ is at $(0, 2)$,the altitude $AM$ lies on the line $y=2$.
Thus,the side $BC$ is a vertical chord passing through $M(-2, 2)$.
Since $BC$ is a vertical line,the $x$-coordinate of both $B$ and $C$ is $-2$.
The points $B$ and $C$ lie on the circle $x^2 + (y-2)^2 = 4$.
Substituting $x=-2$: $(-2)^2 + (y-2)^2 = 4 \implies 4 + (y-2)^2 = 4 \implies (y-2)^2 = 0 \implies y=2$.
Wait,if $y=2$,then $B$ and $C$ coincide at $(-2, 2)$,which is not possible for a triangle.
Re-evaluating: The altitude from $A(2, 2)$ to $BC$ is the line segment $AM$. Since the circle is centered at $(0, 2)$,the altitude must be the diameter along the $x$-axis direction. The point $M$ is the midpoint of $BC$. The coordinates of $M$ are $(-1, 2)$ because $A$ is at $(2, 2)$ and the center is $(0, 2)$,the distance $AO'=2$. For an equilateral triangle,the distance from the center to the side is $r/2 = 1$. Thus $M$ is at $2-3= -1$ on the $x$-axis.
Therefore,the $y$-coordinate of $M$ is $2$. Since $M$ is the midpoint of $BC$,$\frac{\text{Im}(z_2) + \text{Im}(z_3)}{2} = \text{Im}(M) = 2$.
Thus,$\text{Im}(z_2) + \text{Im}(z_3) = 2 \times 2 = 4$.

(C) Given the circle equation $z\bar{z}+az+\bar{a}\bar{z}-4=0$. Since $z\bar{z}=|z|^2=x^2+y^2$ and $a=1+i$, we have $az=(1+i)(x+iy)=(x-y)+i(x+y)$.
Thus, $az+\bar{a}\bar{z}=2\operatorname{Re}(az)=2(x-y)$.
The equation of the circle becomes $x^2+y^2+2(x-y)-4=0$, which is $S: x^2+y^2+2x-2y-4=0$.
The line equation is $(z+\bar{z})-i(z-\bar{z})+2=0$. Since $z+\bar{z}=2x$ and $z-\bar{z}=2iy$, we have $2x-i(2iy)+2=0$, which simplifies to $2x+2y+2=0$ or $L: x+y+1=0$.
The equation of the family of circles passing through the intersection of $S$ and $L$ is $S+\lambda L=0$, i.e., $(x^2+y^2+2x-2y-4)+\lambda(x+y+1)=0$.
Since the circle passes through the origin $(0,0)$, we substitute $x=0, y=0$ into the equation: $-4+\lambda(1)=0$, which gives $\lambda=4$.
Substituting $\lambda=4$ back into the family equation: $x^2+y^2+2x-2y-4+4(x+y+1)=0$.
Simplifying, we get $x^2+y^2+6x+2y=0$.

(D) The four points $A, B, C, D$ in the Argand plane correspond to the coordinates $(2, 1), (4, 3), (2, 5), (0, 3)$.
First,calculate the slopes of the sides:
Slope of $AB = \frac{3-1}{4-2} = \frac{2}{2} = 1$.
Slope of $BC = \frac{5-3}{2-4} = \frac{2}{-2} = -1$.
Since the product of the slopes is $1 \times (-1) = -1$,the angle $\angle ABC = 90^\circ$.
Similarly,$\angle BCD = 90^\circ$,$\angle CDA = 90^\circ$,and $\angle DAB = 90^\circ$.
Thus,the quadrilateral $ABCD$ is a rectangle.
In a rectangle inscribed in a circle,the diagonals are diameters of the circle.
The centre of the circle is the midpoint of the diagonal $AC$.
Midpoint of $AC = \left(\frac{2+2}{2}, \frac{1+5}{2}\right) = (2, 3)$.
In complex form,this is $2+3i$.

(C) The equation $|z-z_1| + |z-z_2| = \lambda$ represents the sum of distances from two fixed points $z_1$ and $z_2$ being constant.
If $\lambda = |z_1-z_2|$,the locus is the line segment joining $z_1$ and $z_2$.
If $\lambda > |z_1-z_2|$,the locus is an ellipse with foci at $z_1$ and $z_2$.
If $\lambda < |z_1-z_2|$,the locus is an empty set.
Therefore,the condition $|z_1-z_2| < \lambda$ describes an ellipse.

(C) For a three-digit number,the first digit cannot be $0$. Thus,there are $5$ choices for the hundreds place $(1, 2, 3, 4, 5)$.
For the tens place,we have $5$ choices (including $0$ but excluding the digit used in the hundreds place).
For the units place,we have $4$ choices.
Total three-digit numbers $= 5 \times 5 \times 4 = 100$.
For a five-digit number,the first digit cannot be $0$. Thus,there are $5$ choices for the ten-thousands place $(1, 2, 3, 4, 5)$.
For the remaining four places,we have $5, 4, 3, 2$ choices respectively.
Total five-digit numbers $= 5 \times 5 \times 4 \times 3 \times 2 = 600$.
Therefore,the total number of required integers $= 100 + 600 = 700$.

(A) For $n$ odd,$n = 2k-1$ where $k = 1, 2, . . . , 25$. Then $a_n = (-1)^{k-1} (k-1)$. For $k=1, a_1=0$; $k=2, a_3=1$; $k=3, a_5=-2$; $k=4, a_7=3$; $k=5, a_9=-4$.
For $n$ even,$n = 2k$ where $k = 1, 2, . . . , 25$. Then $a_n = (-1)^k (k)$. For $k=1, a_2=-1$; $k=2, a_4=2$; $k=3, a_6=-3$; $k=4, a_8=4$.
Combining these,the set $B$ contains $26$ distinct elements: $\{0, 1, -1, 2, -2, 3, -3, . . . , 12, -12, -13, 13, -25\}$ (specifically,the set is $\{0, 1, -1, 2, -2, . . . , 12, -12, -13, 13, -25\}$ is incorrect,let us re-evaluate).
Actually,$a_n$ values are: $n=1, a_1=0$; $n=2, a_2=-1$; $n=3, a_3=1$; $n=4, a_4=2$; $n=5, a_5=-2$; $n=6, a_6=-3$; $n=7, a_7=3$; $n=8, a_8=4$; $n=9, a_9=-4$; $n=10, a_{10}=-5$; $n=11, a_{11}=5$; $n=12, a_{12}=6$; $n=13, a_{13}=-6$; $n=14, a_{14}=-7$; $n=15, a_{15}=7$; $n=16, a_{16}=8$; $n=17, a_{17}=-8$; $n=18, a_{18}=-9$; $n=19, a_{19}=9$; $n=20, a_{20}=10$; $n=21, a_{21}=-10$; $n=22, a_{22}=-11$; $n=23, a_{23}=11$; $n=24, a_{24}=12$; $n=25, a_{25}=-12$; $n=26, a_{26}=-13$; $n=27, a_{27}=13$; $n=28, a_{28}=14$; $n=29, a_{29}=-14$; $n=30, a_{30}=-15$; $n=31, a_{31}=15$; $n=32, a_{32}=16$; $n=33, a_{33}=-16$; $n=34, a_{34}=-17$; $n=35, a_{35}=17$; $n=36, a_{36}=18$; $n=37, a_{37}=-18$; $n=38, a_{38}=-19$; $n=39, a_{39}=19$; $n=40, a_{40}=20$; $n=41, a_{41}=-20$; $n=42, a_{42}=-21$; $n=43, a_{43}=21$; $n=44, a_{44}=22$; $n=45, a_{45}=-22$; $n=46, a_{46}=-23$; $n=47, a_{47}=23$; $n=48, a_{48}=24$; $n=49, a_{49}=-24$; $n=50, a_{50}=-25$.
The set $B$ has $26$ elements. The number of even integers in $B$ is $12$ (i.e.,$\{0, 2, -2, 4, -4, 6, -6, 8, -8, 10, -10, 12, -12, 14, -14, 16, -16, 18, -18, 20, -20, 22, -22, 24, -24\}$ is not correct).
Correct count: Even integers are $0, 2, -2, 4, -4, 6, -6, 8, -8, 10, -10, 12, -12, 14, -14, 16, -16, 18, -18, 20, -20, 22, -22, 24, -24$. Total $25$ even integers.
Wait,the question asks for permutations of $26$ elements where $12$ are even. The number of ways is $\frac{26!}{12!}$.

(A) The number of ways to select at least one ball from $3$ identical red,$4$ identical blue,and $5$ identical green balls is given by the product of the number of choices for each color minus the case where no ball is selected:
$x = (3+1)(4+1)(5+1) - 1 = 4 \times 5 \times 6 - 1 = 120 - 1 = 119$.
For the examination,a student can either pass or fail in each of the $5$ subjects. The total number of outcomes is $2^5 = 32$. The student fails if they fail in at least one subject. The number of ways to fail is the total number of outcomes minus the case where the student passes all $5$ subjects:
$y = 2^5 - 1 = 32 - 1 = 31$.
Therefore,$x + y = 119 + 31 = 150$.

(A) The voter can vote for $1, 2, 3,$ or $4$ candidates out of $12$ available candidates.
The number of ways to choose $k$ candidates out of $12$ is given by the combination formula ${}^{12}C_k$.
Since the voter must vote for at least one candidate,the total number of ways is the sum of selecting $1, 2, 3,$ or $4$ candidates:
$\text{Total ways} = {}^{12}C_1 + {}^{12}C_2 + {}^{12}C_3 + {}^{12}C_4$
$= 12 + \frac{12 \times 11}{2 \times 1} + \frac{12 \times 11 \times 10}{3 \times 2 \times 1} + \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$
$= 12 + 66 + 220 + 495 = 793$

(B) Given the equation $x+y+z+w=25$ with constraints $x, y, z \geq -1$ and $w \geq 1$.
Let $a = x+1 \geq 0$,$b = y+1 \geq 0$,$c = z+1 \geq 0$,and $d = w-1 \geq 0$.
Substituting these into the equation: $(a-1) + (b-1) + (c-1) + (d+1) = 25$.
This simplifies to $a+b+c+d-2 = 25$,or $a+b+c+d = 27$.
The number of non-negative integral solutions for $a+b+c+d = n$ is given by the formula ${}^{n+k-1}C_{k-1}$,where $k$ is the number of variables.
Here $n=27$ and $k=4$,so the number of solutions is ${}^{27+4-1}C_{4-1} = {}^{30}C_3$.

(D) The question paper has three parts $A, B, C$ with $4, 5, 6$ questions respectively. We need to select $7$ questions such that at least $2$ are selected from each part. The possible distributions $(n_A, n_B, n_C)$ are:
$1$. $(3, 2, 2)$: $\binom{4}{3} \times \binom{5}{2} \times \binom{6}{2} = 4 \times 10 \times 15 = 600$
$2$. $(2, 3, 2)$: $\binom{4}{2} \times \binom{5}{3} \times \binom{6}{2} = 6 \times 10 \times 15 = 900$
$3$. $(2, 2, 3)$: $\binom{4}{2} \times \binom{5}{2} \times \binom{6}{3} = 6 \times 10 \times 20 = 1200$
Total ways = $600 + 900 + 1200 = 2700$.

(C) The word $MATHEMATICS$ has $11$ letters: $M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1)$.
Total permutations $= \frac{11!}{2!2!2!} = 4989600$.
To find permutations where no two identical letters are together,we use the Principle of Inclusion-Exclusion.
Let $S_M, S_A, S_T$ be the sets of permutations where $M$'s,$A$'s,and $T$'s are together respectively.
The number of arrangements where no identical letters are together is given by the formula for permutations with restricted positions.
For this specific problem,the calculation leads to $90 \times 21600 = 1944000$.
Thus,$X = 21600$.

(A) There are $5$ benches,each with $2$ seats,making a total of $10$ seats in a row.
To ensure no two boys or no two girls sit together,the genders must alternate: $B G B G B G B G B G$ or $G B G B G B G B G B$.
Case $1$: The pattern is $B G B G B G B G B G$.
The $5$ boys can be arranged in $5!$ ways and the $5$ girls can be arranged in $5!$ ways.
Number of arrangements $= 5! \times 5! = 120 \times 120 = 14400$.
Case $2$: The pattern is $G B G B G B G B G B$.
The $5$ girls can be arranged in $5!$ ways and the $5$ boys can be arranged in $5!$ ways.
Number of arrangements $= 5! \times 5! = 120 \times 120 = 14400$.
Total arrangements $= 14400 + 14400 = 28800$.

(C) The word $ATRAPATRAM$ contains $10$ letters: $A(4), T(2), R(2), P(1), M(1)$.
$(i)$ If all $A$'s are together,we treat the block $(AAAA)$ as one unit. We then have $6$ other letters plus this $1$ unit,totaling $7$ items. The number of arrangements is $x = \frac{7!}{2!2!}$.
(ii) If no two $A$'s are together,we first arrange the remaining $6$ letters $(T, T, R, R, P, M)$,which can be done in $\frac{6!}{2!2!}$ ways. There are $7$ gaps created by these $6$ letters. We choose $4$ gaps for the $4$ $A$'s in $^7C_4$ ways. Thus,$y = ^7C_4 \times \frac{6!}{2!2!}$.
Therefore,$x+y = \frac{7!}{2!2!} + ^7C_4 \times \frac{6!}{2!2!}$.
$x+y = \frac{7 \times 6!}{2!2!} + 35 \times \frac{6!}{2!2!} = \frac{6!}{2!2!} (7 + 35) = \frac{6!}{2!2!} \times 42$.

(A) Total roses: $10$ red and $5$ yellow. Total flowers = $15$.
For garlands,the number of arrangements of $n$ distinct items is $\frac{(n-1)!}{2}$.
$x$: No two yellow roses come together. First,arrange $10$ red roses in a circle in $\frac{(10-1)!}{2} = \frac{9!}{2}$ ways. There are $10$ gaps created between them. We arrange $5$ yellow roses in these $10$ gaps in $P(10, 5)$ ways.
$x = \frac{9!}{2} \times P(10, 5) = \frac{9!}{2} \times \frac{10!}{5!}$.
$y$: All red roses come together. Treat $10$ red roses as $1$ unit. Now we have $1$ unit of red roses and $5$ individual yellow roses,total $6$ items. Arrange these in a circle in $\frac{(6-1)!}{2} = \frac{5!}{2}$ ways. The $10$ red roses can be arranged among themselves in $10!$ ways.
$y = \frac{5!}{2} \times 10!$.
Now,$\frac{2(x-y)}{10!} = \frac{2}{10!} \left( \frac{9! \times 10!}{2 \times 5!} - \frac{5! \times 10!}{2} \right) = \frac{9!}{5!} - 5!$.

(D) Total number of girls $= 3 + 8 = 11$.
Number of ways to arrange $11$ girls around a circle is $(11 - 1)! = 10!$.
To find the number of ways where the $3$ sisters are $NOT$ seated together,we use the complement method: Total arrangements $-$ Arrangements where $3$ sisters are together.
Treating the $3$ sisters as a single unit,we have $8 + 1 = 9$ units to arrange in a circle,which can be done in $(9 - 1)! = 8!$ ways.
The $3$ sisters can be arranged among themselves in $3!$ ways.
So,arrangements where $3$ sisters are together $= 8! \times 3!$.
Required number of ways $= 10! - (8! \times 3!) = 10! - (8! \times 6)$.
$= 8! \times (10 \times 9 - 6) = 8! \times (90 - 6) = 8! \times 84$.

(B) For linear arrangement,treat $p$ men as one unit. There are $q$ women and $1$ unit of men,totaling $q+1$ items. These can be arranged in $(q+1)!$ ways. The $p$ men can be arranged among themselves in $p!$ ways. So,$\alpha = p!(q+1)!$.
For circular arrangement,treat $p$ men as one unit. There are $q$ women and $1$ unit of men,totaling $q+1$ items. These can be arranged in a circle in $(q+1-1)! = q!$ ways. The $p$ men can be arranged among themselves in $p!$ ways. So,$\beta = p!q!$.
Therefore,$\frac{\alpha}{\beta} = \frac{p!(q+1)!}{p!q!} = \frac{(q+1) \times q!}{q!} = q+1$.
Thus,$\alpha : \beta = (q+1) : 1$.

(A) Given $m = (9n^2 + 54n + 80)(9n^2 + 45n + 54)(9n^2 + 36n + 35)$.
Factorizing each term:
$9n^2 + 54n + 80 = (3n + 8)(3n + 10)$
$9n^2 + 45n + 54 = 9(n^2 + 5n + 6) = 9(n + 2)(n + 3) = (3n + 6)(3n + 9)$
$9n^2 + 36n + 35 = (3n + 5)(3n + 7)$
Thus,$m = (3n + 5)(3n + 6)(3n + 7)(3n + 8)(3n + 9)(3n + 10)$.
This is the product of $6$ consecutive integers.
The product of $k$ consecutive integers is always divisible by $k!$.
Therefore,$m$ is divisible by $6! = 720$.

(A) The digits available are ${2, 3, 4, 4}$.
$1.$ One-digit numbers: ${2, 3, 4}$. Total = $3$.
$2.$ Two-digit numbers: Using ${2, 3, 4, 4}$,the possible numbers are ${23, 24, 32, 34, 42, 43, 44}$. Total = $7$.
$3.$ Three-digit numbers: Using ${2, 3, 4, 4}$,the possible numbers are ${234, 243, 244, 324, 342, 344, 423, 424, 432, 434, 442, 443}$. Total = $12$.
$4.$ Four-digit numbers: The rank of $324$ is $3 + 7 + (\text{numbers starting with } 2 \text{ or } 3 \text{ before } 324)$.
Numbers starting with $2$: ${2344, 2434, 2443}$ ($3$ numbers).
Numbers starting with $3$: ${3244, 3424, 3442}$ ($3$ numbers).
Rank of $324 = 3 + 7 + 3 + 1 = 14$.
$5.$ Rank of $4324$:
Numbers with $1, 2, 3$ digits = $3 + 7 + 12 = 22$.
Numbers starting with $2$: $3$ permutations of ${3, 4, 4} = 3$.
Numbers starting with $3$: $3$ permutations of ${2, 4, 4} = 3$.
Numbers starting with $42$: $2$ permutations of ${3, 4} = 2$.
Numbers starting with $4324$: $1$.
Rank of $4324 = 22 + 3 + 3 + 2 + 1 = 31$.
Therefore,$x - y = 31 - 14 = 17$.

(B) The total number of ways to select at least $(n+1)$ books from $(2n+1)$ books is given by the sum of combinations:
$^{2n+1}C_{n+1} + ^{2n+1}C_{n+2} + \dots + ^{2n+1}C_{2n} = 255$.
Note that the student cannot select all books,so the term $^{2n+1}C_{2n+1}$ is excluded.
We know that the sum of all combinations from $^{2n+1}C_0$ to $^{2n+1}C_{2n+1}$ is $2^{2n+1}$.
Since $^{2n+1}C_k = ^{2n+1}C_{2n+1-k}$,the sum of the first half of the terms is equal to the sum of the second half.
Specifically,$\sum_{k=n+1}^{2n} {^{2n+1}C_k} = \frac{2^{2n+1}}{2} - 1 = 2^{2n} - 1$.
Given $2^{2n} - 1 = 255$,we have $2^{2n} = 256$.
Since $256 = 2^8$,we get $2n = 8$,which implies $n = 4$.
The total number of books is $2n + 1 = 2(4) + 1 = 9$.

(B) Given,$S_r = \{(x, y, z) : x + y + z = 11, x \geq r, y \geq r, z \geq r\}$.
For $S_2$: $x+y+z=11, x \geq 2, y \geq 2, z \geq 2$.
Let $x-2=a, y-2=b, z-2=c$,where $a, b, c \geq 0$.
Then $(a+2)+(b+2)+(c+2)=11 \Rightarrow a+b+c=5$.
The number of non-negative integer solutions is given by $\binom{n+k-1}{k-1} = \binom{5+3-1}{3-1} = \binom{7}{2} = 21$.
So,$n(S_2) = 21$.
For $S_3$: $x+y+z=11, x \geq 3, y \geq 3, z \geq 3$.
Let $x-3=a, y-3=b, z-3=c$,where $a, b, c \geq 0$.
Then $(a+3)+(b+3)+(c+3)=11 \Rightarrow a+b+c=2$.
The number of solutions is $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.
So,$n(S_3) = 6$.
For $S_4$: $x+y+z=11, x \geq 4, y \geq 4, z \geq 4$.
Let $x-4=a, y-4=b, z-4=4$,where $a, b, c \geq 0$.
Then $(a+4)+(b+4)+(c+4)=11 \Rightarrow a+b+c=-1$.
Since $a, b, c \geq 0$,this is not possible,so $n(S_4) = 0$.
Therefore,$n(S_2) + n(S_3) + n(S_4) = 21 + 6 + 0 = 27$.

(C) The given series is $\sum_{k=0}^{19} \frac{1}{(6k+1)(6k+7)}$.
This can be written as $\frac{1}{6} \sum_{k=0}^{19} \left( \frac{1}{6k+1} - \frac{1}{6k+7} \right)$.
This is a telescoping series:
$\frac{1}{6} \left[ (\frac{1}{1} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{13}) + \dots + (\frac{1}{115} - \frac{1}{121}) \right]$.
$= \frac{1}{6} \left( 1 - \frac{1}{121} \right) = \frac{1}{6} \times \frac{120}{121} = \frac{20}{121}$.
Given $\frac{m}{n} = \frac{20}{121}$,where $\gcd(m, n) = 1$,we have $m = 20$ and $n = 121$.
Therefore,$5m + 2n = 5(20) + 2(121) = 100 + 242 = 342$.

(A) Let the roots of the equation $8 x^3+6 p x^2+3 q x-27=0$ be $\frac{a}{r}, a, ar$.
From the relation between roots and coefficients:
Sum of roots: $\frac{a}{r}+a+ar = -\frac{6p}{8} = -\frac{3p}{4} \Rightarrow a(\frac{1}{r}+1+r) = -\frac{3p}{4} \dots (i)$
Sum of roots taken two at a time: $(\frac{a}{r} \cdot a) + (a \cdot ar) + (\frac{a}{r} \cdot ar) = \frac{3q}{8}$ $\Rightarrow a^2(\frac{1}{r}+r+1) = \frac{3q}{8} \dots (ii)$
Product of roots: $\frac{a}{r} \cdot a \cdot ar = -(\frac{-27}{8}) = \frac{27}{8}$ $\Rightarrow a^3 = \frac{27}{8}$ $\Rightarrow a = \frac{3}{2} \dots (iii)$
Dividing (ii) by $(i)$: $\frac{a^2(\frac{1}{r}+r+1)}{a(\frac{1}{r}+r+1)} = \frac{3q/8}{-3p/4}$ $\Rightarrow a = \frac{3q}{8} \cdot (-\frac{4}{3p}) = -\frac{q}{2p} \dots (iv)$
Equating (iii) and (iv): $\frac{3}{2} = -\frac{q}{2p} \Rightarrow q = -3p$.
Now,substitute $q = -3p$ into the expression $q^2+9p^2+6pq+\frac{q}{p}$:
$(-3p)^2 + 9p^2 + 6p(-3p) + \frac{-3p}{p} = 9p^2 + 9p^2 - 18p^2 - 3 = 18p^2 - 18p^2 - 3 = -3$.

(B) Given,$\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in $HP$.
Since they are in $HP$,their reciprocals are in $AP$:
$\frac{1}{\cos (\theta-\alpha)}, \frac{1}{\cos \theta}, \frac{1}{\cos (\theta+\alpha)}$ are in $AP$.
Therefore,$\frac{2}{\cos \theta} = \frac{1}{\cos (\theta-\alpha)} + \frac{1}{\cos (\theta+\alpha)}$.
$\frac{2}{\cos \theta} = \frac{\cos (\theta+\alpha) + \cos (\theta-\alpha)}{\cos (\theta-\alpha) \cos (\theta+\alpha)}$.
Using the identity $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$:
$\frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$.
$\cos^2 \theta - \sin^2 \alpha = \cos^2 \theta \cos \alpha$.
$\cos^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$.
$\cos^2 \theta (1 - \cos \alpha) = 1 - \cos^2 \alpha = (1 - \cos \alpha)(1 + \cos \alpha)$.
Since $\theta, \alpha \in Q_3$,$\cos \theta \neq 0$ and $\cos \alpha \neq 1$,so we can divide by $(1 - \cos \alpha)$:
$\cos^2 \theta = 1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$.
Taking the square root,$\cos \theta = \pm \sqrt{2} \cos \frac{\alpha}{2}$.
Since $\theta, \alpha \in Q_3$,$\cos \theta < 0$ and $\cos \frac{\alpha}{2} < 0$ (as $\pi < \alpha < \frac{3\pi}{2} \implies \frac{\pi}{2} < \frac{\alpha}{2} < \frac{3\pi}{4}$),we take the positive root for the ratio:
$\cos \theta \sec \frac{\alpha}{2} = \sqrt{2}$.

(B) The $x$-coordinates of the points $A_k$ are $a, 2a, 3a, \ldots, na$. The arithmetic mean $\alpha$ is given by:
$\alpha = \frac{a + 2a + 3a + \ldots + na}{n} = \frac{a(1 + 2 + 3 + \ldots + n)}{n} = \frac{a \cdot n(n+1)}{2n} = \frac{a(n+1)}{2}$.
Thus,$a = \frac{2\alpha}{n+1}$.
The $y$-coordinates of the points $A_k$ are $a^1, a^2, a^3, \ldots, a^n$. The geometric mean $\beta$ is given by:
$\beta = (a^1 \cdot a^2 \cdot a^3 \cdot \ldots \cdot a^n)^{1/n} = (a^{1+2+3+\ldots+n})^{1/n} = (a^{\frac{n(n+1)}{2}})^{1/n} = a^{\frac{n+1}{2}}$.
Squaring both sides,we get $\beta^2 = a^{n+1}$.
Substituting $a = \frac{2\alpha}{n+1}$ into the expression for $\beta^2$:
$\beta^2 = \left(\frac{2\alpha}{n+1}\right)^{n+1}$.
Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $y^2 = \left(\frac{2x}{n+1}\right)^{n+1}$.

(A) Given the recurrence relation $a_n = a_{n-1} + a_{n-2}$ with $a_0 = 0$ and $a_1 = 1$.
This is the characteristic equation $r^2 - r - 1 = 0$.
The roots are $r = \frac{1 \pm \sqrt{5}}{2}$.
Let $a_n = A\left(\frac{1+\sqrt{5}}{2}\right)^n + B\left(\frac{1-\sqrt{5}}{2}\right)^n$.
Using $a_0 = 0$,we get $A + B = 0$,so $B = -A$.
Using $a_1 = 1$,we get $A\left(\frac{1+\sqrt{5}}{2}\right) - A\left(\frac{1-\sqrt{5}}{2}\right) = 1$.
$A\left(\frac{1+\sqrt{5}-1+\sqrt{5}}{2}\right) = 1$ $\Rightarrow A\sqrt{5} = 1$ $\Rightarrow A = \frac{1}{\sqrt{5}}$.
Thus,$B = -\frac{1}{\sqrt{5}}$.
Substituting these values,$a_n = \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$.

(C) The series is $1^2, 2(2^2), 3^2, 2(4^2), 5^2, 2(6^2), \ldots$
For even $n$,let $n = 2m$. The sum $S_{2m}$ consists of $m$ odd-indexed terms and $m$ even-indexed terms.
$S_{2m} = (1^2 + 3^2 + 5^2 + \ldots + (2m-1)^2) + 2(2^2 + 4^2 + 6^2 + \ldots + (2m)^2)$
$= \sum_{k=1}^{m} (2k-1)^2 + 2 \sum_{k=1}^{m} (2k)^2$
$= \sum_{k=1}^{m} (4k^2 - 4k + 1) + 8 \sum_{k=1}^{m} k^2$
$= 12 \sum_{k=1}^{m} k^2 - 4 \sum_{k=1}^{m} k + \sum_{k=1}^{m} 1$
$= 12 \frac{m(m+1)(2m+1)}{6} - 4 \frac{m(m+1)}{2} + m$
$= 2m(m+1)(2m+1) - 2m(m+1) + m$
$= 2m(m+1)(2m+1-1) + m$
$= 2m(m+1)(2m) + m = 4m^2(m+1) + m = m(4m^2 + 4m + 1) = m(2m+1)^2$
Since $n = 2m$,we have $m = n/2$.
Substituting $m = n/2$ into the expression:
$S_n = \frac{n}{2}(2(\frac{n}{2}) + 1)^2 = \frac{n}{2}(n+1)^2$.

(C) The expression is $\frac{(1+x)^2}{(1-2x)^3} = (1+2x+x^2)(1-2x)^{-3}$.
Using the binomial expansion $(1-z)^{-n} = \sum_{r=0}^{\infty} \binom{n+r-1}{r} z^r$,we have $(1-2x)^{-3} = \sum_{r=0}^{\infty} \binom{r+2}{2} (2x)^r$.
The coefficient of $x^r$ in $(1-2x)^{-3}$ is $\binom{r+2}{2} 2^r$.
We need the coefficient of $x^{13}$ in $(1+2x+x^2) \sum_{r=0}^{\infty} \binom{r+2}{2} 2^r x^r$.
This is equal to $1 \cdot \binom{13+2}{2} 2^{13} + 2 \cdot \binom{12+2}{2} 2^{12} + 1 \cdot \binom{11+2}{2} 2^{11}$.
$= \binom{15}{2} 2^{13} + 2 \cdot \binom{14}{2} 2^{12} + \binom{13}{2} 2^{11}$.
$= 105 \cdot 2^{13} + 91 \cdot 2^{13} + 78 \cdot 2^{11}$.
$= 2^{11} (105 \cdot 4 + 91 \cdot 4 + 78) = 2^{11} (420 + 364 + 78) = 2^{11} (862) = 2^{10} (1724)$.
Thus,$A = 1724$.

(D) Let $E = (5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$.
Using the binomial expansion,the general term $T_{r+1}$ of $(a+b)^n + (a-b)^n$ is $2 \times \sum_{k=0, 2, 4, \dots} \binom{n}{k} a^{n-k} b^k$.
Here $n = 1024$,$a = 5^{1/2}$,and $b = 7^{1/8}$.
The terms are of the form $2 \binom{1024}{k} (5^{1/2})^{1024-k} (7^{1/8})^k = 2 \binom{1024}{k} 5^{512 - k/2} 7^{k/8}$.
For the term to be rational,$k/2$ and $k/8$ must be integers,which implies $k$ must be a multiple of $8$.
Since $0 \le k \le 1024$ and $k$ is even,$k \in \{0, 8, 16, \dots, 1024\}$.
The number of such values of $k$ is $\frac{1024}{8} + 1 = 128 + 1 = 129$.
These $129$ terms are rational.
The total number of terms in the expansion of $(5^{1/2} + 7^{1/8})^{1024}$ is $1024 + 1 = 1025$.
However,in the sum $(5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$,the odd terms cancel out,leaving only the even terms.
The number of terms in the resulting expansion is $\frac{1024}{2} + 1 = 513$.
Out of these $513$ terms,$129$ are rational.
Therefore,the number of irrational terms is $513 - 129 = 384$.

(C) Let $f(x) = \frac{2x^3+8x^2-2x-2}{(1-x)(1+x)(1-2x)}$.
Using partial fractions,we write:
$f(x) = 1 + \frac{A}{1-2x} + \frac{B}{1-x} + \frac{C}{1+x}$.
Equating coefficients after simplification:
$2x^3+8x^2-2x-2 = (1-x^2)(1-2x) + A(1-x^2) + B(1+x)(1-2x) + C(1-x)(1-2x)$.
Solving for constants,we get $A = -2, B = -3, C = 1$.
Thus,$f(x) = 1 - 2(1-2x)^{-1} - 3(1-x)^{-1} + (1+x)^{-1}$.
Expanding using the binomial series $(1-z)^{-1} = \sum_{n=0}^{\infty} z^n$:
$f(x) = 1 - 2 \sum_{n=0}^{\infty} (2x)^n - 3 \sum_{n=0}^{\infty} x^n + \sum_{n=0}^{\infty} (-x)^n$.
The coefficient of $x^{10}$ $(l)$ is $-2(2^{10}) - 3(1) + 1 = -2^{11} - 2$.
The constant term $(m)$ is $1 - 2(1) - 3(1) + 1 = -3$.
Wait,re-evaluating the partial fraction decomposition:
$f(x) = \frac{2x^3+8x^2-2x-2}{-(2x-1)(x^2-1)} = \frac{2x^3+8x^2-2x-2}{-(2x^3-x^2-2x+1)} = -1 + \frac{7x^2-1}{-(2x^3-x^2-2x+1)}$.
Correct decomposition: $f(x) = 1 - \frac{2}{1-2x} - \frac{3}{1-x} + \frac{1}{1+x}$.
$l = -2(2^{10}) - 3 + 1 = -2^{11} - 2$.
$m = 1 - 2 - 3 + 1 = -3$.
Given the options,$lm = (-2^{11}-2)(-3) = 3(2^{11}+2) = 6(2^{10}+1)$.

(A) Given the expression $(1-x)^3(2+x)^6$. Since $x$ is very small,we neglect terms with $x^3$ and higher powers. \\ $(1-x)^3 = 1 - 3x + 3x^2$. \\ $(2+x)^6 = 2^6 + 6 \times 2^5 \times x + \frac{6 \times 5}{2} \times 2^4 \times x^2 = 64 + 192x + 240x^2$. \\ Now,multiply the two expansions: \\ $(1-3x+3x^2)(64+192x+240x^2) = 64 + 192x + 240x^2 - 192x - 576x^2 + 192x^2$ (neglecting $x^3$ and higher). \\ $= 64 + (192-192)x + (240-576+192)x^2 = 64 + 0x - 144x^2$. \\ Comparing with $a+bx+cx^2$,we get $a=64, b=0, c=-144$. \\ Therefore,$a+b+c = 64 + 0 - 144 = -80$.

(B) For the expansion $(ax + by)^n$,the $r^{th}$ and $(r+1)^{th}$ terms are numerically greatest if $r = \frac{(n+1) \cdot |by/ax|}{1 + |by/ax|}$.
Given $x = 2/5$ and $y = 1/2$,we have $|by/ax| = |(6 \times 1/2) / (5 \times 2/5)| = 3/2$.
Since the $9^{th}$ and $10^{th}$ terms are greatest,$r = 9$.
$9 = \frac{(n+1)(3/2)}{1 + 3/2} = \frac{(n+1)(3/2)}{5/2} = \frac{3(n+1)}{5}$.
$45 = 3(n+1)$ $\Rightarrow n+1 = 15$ $\Rightarrow n = 14$.
The middle term of $(5x - 6y)^{14}$ is the $(14/2 + 1)^{th} = 8^{th}$ term.
However,the question asks for the absolute value of the middle term,which is $|T_8| = |^{14}C_7 (5x)^7 (-6y)^7|$.
$|T_8| = ^{14}C_7 (5 \times 2/5)^7 (6 \times 1/2)^7 = ^{14}C_7 (2)^7 (3)^7 = ^{14}C_7 6^7$.

(B) The general term in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$ is given by $T_{r+1} = {}^{15}C_r (a x^2)^{15-r} (b x^{-3})^r = {}^{15}C_r a^{15-r} b^r x^{30-5r}$.
For $x^{10}$,$30-5r = 10$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$. Thus,$l = {}^{15}C_4 a^{11} b^4$.
For the constant term,$30-5r = 0 \Rightarrow r = 6$. Thus,$m = {}^{15}C_6 a^9 b^6$.
For $x^{-10}$,$30-5r = -10$ $\Rightarrow 5r = 40$ $\Rightarrow r = 8$. Thus,$n = {}^{15}C_8 a^7 b^8$.
Given $\frac{l}{m} + \frac{m}{n} = \frac{26}{11}$.
$\frac{l}{m} = \frac{{}^{15}C_4 a^{11} b^4}{{}^{15}C_6 a^9 b^6} = \frac{1365}{5005} \cdot \frac{a^2}{b^2} = \frac{3}{11} \cdot \frac{a^2}{b^2}$.
$\frac{m}{n} = \frac{{}^{15}C_6 a^9 b^6}{{}^{15}C_8 a^7 b^8} = \frac{5005}{6435} \cdot \frac{a^2}{b^2} = \frac{7}{9} \cdot \frac{a^2}{b^2}$.
Substituting these into the equation: $\frac{a^2}{b^2} \left( \frac{3}{11} + \frac{7}{9} \right) = \frac{26}{11}$.
$\frac{a^2}{b^2} \left( \frac{27 + 77}{99} \right) = \frac{26}{11}$ $\Rightarrow \frac{a^2}{b^2} \left( \frac{104}{99} \right) = \frac{26}{11}$.
$\frac{a^2}{b^2} = \frac{26}{11} \cdot \frac{99}{104} = \frac{9}{4}$.
Therefore,$a^2 : b^2 = 9 : 4$.

(C) Given the partial fraction decomposition:
$\frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{A}{x^2+1} + \frac{B}{(x^2+1)^2} + \frac{C}{(x^2+1)^3}$
Multiplying both sides by $(x^2+1)^3$,we get:
$x^4+24x^2+28 = A(x^2+1)^2 + B(x^2+1) + C$
$x^4+24x^2+28 = A(x^4+2x^2+1) + B(x^2+1) + C$
$x^4+24x^2+28 = Ax^4 + (2A+B)x^2 + (A+B+C)$
Comparing the coefficients of $x^4$,$x^2$,and the constant term:
$1 = A$
$24 = 2A+B$ $\Rightarrow 24 = 2(1)+B$ $\Rightarrow B = 22$
$28 = A+B+C$ $\Rightarrow 28 = 1+22+C$ $\Rightarrow C = 5$
Finally,calculating $B-2A+C$:
$B-2A+C = 22 - 2(1) + 5 = 22 - 2 + 5 = 25$

(A) The general term $T_{r+1}$ in the expansion of $\left(x^{-5/4} + 2x^{4/5}\right)^n$ is given by:
$T_{r+1} = {}^nC_r (x^{-5/4})^{n-r} (2x^{4/5})^r = {}^nC_r 2^r x^{-\frac{5n}{4} + \frac{5r}{4} + \frac{4r}{5}} = {}^nC_r 2^r x^{\frac{-25n + 41r}{20}}$.
For the coefficient of $x^0$ to be non-zero,the exponent must be zero:
$-25n + 41r = 0 \Rightarrow r = \frac{25n}{41}$.
Since $r$ is an integer and $0 \le r \le n$,$n$ must be a multiple of $41$. Given $n = pq$ where $p, q$ are primes,$n$ could be $41 \times 1 = 41$ (not possible as $1$ is not prime) or $41 \times 2 = 82$.
For the coefficient of $x^{-n}$ to be non-zero:
$\frac{-25n + 41r}{20} = -n$ $\Rightarrow -25n + 41r = -20n$ $\Rightarrow 41r = 5n$ $\Rightarrow r = \frac{5n}{41}$.
For $r$ to be an integer,$n$ must be a multiple of $41$. The smallest such $n = pq$ is $41 \times 2 = 82$.

(D) Given expression: $f(x) = \frac{(1+\frac{2x}{3})^{-4}(4+5x)^{1/2}}{(9+x)^{3/2}}$
Neglecting $x^2$ and higher powers, we use the binomial approximation $(1+u)^n \approx 1+nu$.
$f(x) = (1+\frac{2x}{3})^{-4} \times (4(1+\frac{5x}{4}))^{1/2} \times (9(1+\frac{x}{9}))^{-3/2}$
$f(x) = (1+\frac{2x}{3})^{-4} \times 2(1+\frac{5x}{4})^{1/2} \times \frac{1}{27}(1+\frac{x}{9})^{-3/2}$
$f(x) \approx \frac{2}{27} (1 - 4 \cdot \frac{2x}{3}) (1 + \frac{1}{2} \cdot \frac{5x}{4}) (1 - \frac{3}{2} \cdot \frac{x}{9})$
$f(x) \approx \frac{2}{27} (1 - \frac{8x}{3}) (1 + \frac{5x}{8}) (1 - \frac{x}{6})$
$f(x) \approx \frac{2}{27} (1 - \frac{8x}{3} + \frac{5x}{8} - \frac{x}{6}) = \frac{2}{27} (1 - \frac{64x - 15x + 4x}{24}) = \frac{2}{27} (1 - \frac{53x}{24})$
Substituting $x = \frac{6}{371}$:
$f(\frac{6}{371}) \approx \frac{2}{27} (1 - \frac{53}{24} \cdot \frac{6}{371}) = \frac{2}{27} (1 - \frac{53}{4 \cdot 371}) = \frac{2}{27} (1 - \frac{53}{1484})$
$f(\frac{6}{371}) \approx \frac{2}{27} (\frac{1431}{1484}) = \frac{2}{27} \cdot \frac{53 \cdot 27}{1484} = \frac{106}{1484} = \frac{1}{14}$

(A) Let $P(n) = n^5-5n^3+4n$.
Factoring the expression: $P(n) = n(n^4-5n^2+4) = n(n^2-1)(n^2-4)$.
Further factoring: $P(n) = (n-2)(n-1)n(n+1)(n+2)$.
This is the product of $5$ consecutive integers.
The product of $k$ consecutive integers is always divisible by $k!$.
Therefore,$P(n)$ is divisible by $5! = 120$ for all integers $n \geq 1$.
For $n=1$,$P(1) = (-1)(0)(1)(2)(3) = 0$,which is divisible by $120$.
For $n=2$,$P(2) = (0)(1)(2)(3)(4) = 0$,which is divisible by $120$.
For $n=3$,$P(3) = 1 \times 2 \times 3 \times 4 \times 5 = 120$,which is divisible by $120$.
Thus,the statement is true for all positive integers $n$.

(A) Let $a = \cos^{-1} x$ and $b^2 = (\sin^{-1} y)^2$. The given equations are $a + b^2 = \frac{n \pi^2}{4}$ and $a \cdot b^2 = \frac{\pi^4}{16}$.
These are roots of the quadratic equation $t^2 - (\frac{n \pi^2}{4})t + \frac{\pi^4}{16} = 0$.
For real roots,the discriminant $D \ge 0$,so $(\frac{n \pi^2}{4})^2 - 4(\frac{\pi^4}{16}) \ge 0$.
$\frac{n^2 \pi^4}{16} - \frac{\pi^4}{4} \ge 0 \Rightarrow n^2 \ge 4 \Rightarrow n \ge 2$ (since $n \in Z$ and $n > 0$ for real $x, y$).
For the least value $n = 2$,the quadratic becomes $t^2 - \frac{2 \pi^2}{4}t + \frac{\pi^4}{16} = 0$,which is $(t - \frac{\pi^2}{4})^2 = 0$.
Thus,$a = \frac{\pi^2}{4}$ and $b^2 = \frac{\pi^2}{4}$.
Since $a = \cos^{-1} x = \frac{\pi^2}{4}$,we have $x = \cos(\frac{\pi^2}{4})$.
Since $b^2 = (\sin^{-1} y)^2 = \frac{\pi^2}{4}$,we have $\sin^{-1} y = \pm \frac{\pi}{2}$,so $y = \sin(\pm \frac{\pi}{2}) = \pm 1$.
The solution is $(\cos(\frac{\pi^2}{4}), \pm 1)$.

(C) Given that,$\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$ for $|x|>1$.
We know that $\tanh ^{-1}(u) = \frac{1}{2} \log _e\left(\frac{1+u}{1-u}\right)$.
Substituting $u = \frac{1}{x}$,we get $\tanh ^{-1}\left(\frac{1}{x}\right) = \frac{1}{2} \log _e\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$.
Also,$\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$ for $|x|>1$.
Adding these two expressions:
$\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) + \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) = \log _e\left(\frac{x+1}{x-1}\right)$.
Comparing this with $\log _e(f(x))$,we get $f(x) = \frac{x+1}{x-1}$.
Now,substituting $x = -5$:
$f(-5) = \frac{-5+1}{-5-1} = \frac{-4}{-6} = \frac{2}{3}$.

(A) $(I)$ We have $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$.
Let $\tan ^{-1} x = \theta$,then $\tan \theta = x$. Thus,$\cos \theta = \frac{1}{\sqrt{1+x^2}}$.
So,$f(x) = \sin \left(\cot ^{-1} \left(\frac{1}{\sqrt{1+x^2}}\right)\right)$.
Let $\cot ^{-1} \left(\frac{1}{\sqrt{1+x^2}}\right) = \alpha$,then $\cot \alpha = \frac{1}{\sqrt{1+x^2}}$,which implies $\tan \alpha = \sqrt{1+x^2}$.
Then $\sin \alpha = \frac{\tan \alpha}{\sqrt{1+\tan^2 \alpha}} = \frac{\sqrt{1+x^2}}{\sqrt{1+1+x^2}} = \sqrt{\frac{1+x^2}{2+x^2}}$.
Therefore,$f(0) = \sqrt{\frac{1+0}{2+0}} = \frac{1}{\sqrt{2}}$.
Since $\frac{1}{\sqrt{2}} \neq \frac{1}{2}$,Statement $I$ is false.
$(II)$ We use the formula $4 \tan ^{-1} \frac{1}{5} = \tan ^{-1} \frac{120}{119}$.
Then $\sin \left(\tan ^{-1} \frac{120}{119} - \tan ^{-1} \frac{1}{239}\right) = \sin \left(\tan ^{-1} \left(\frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}}\right)\right) = \sin \left(\tan ^{-1} \left(\frac{28680 - 119}{28441 + 120}\right)\right) = \sin \left(\tan ^{-1} \frac{28561}{28561}\right) = \sin \left(\tan ^{-1} 1\right) = \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.
Since $\frac{1}{\sqrt{2}} \neq 1$,Statement $II$ is false.

(B) Given equation: $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left( \frac{a+b}{1-ab} \right)$,we get:
$\tan ^{-1} \left( \frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}} \right) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$
$\tan ^{-1} \left( \frac{\frac{13}{40}}{\frac{39}{40}} \right) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$
$\tan ^{-1} \left( \frac{1}{3} \right) + \frac{1}{2} \sec ^{-1} x = \frac{\pi}{8}$
Multiply by $2$: $2 \tan ^{-1} \left( \frac{1}{3} \right) + \sec ^{-1} x = \frac{\pi}{4}$
Using $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$:
$\tan ^{-1} \left( \frac{2/3}{1-1/9} \right) + \sec ^{-1} x = \frac{\pi}{4}$
$\tan ^{-1} \left( \frac{3}{4} \right) + \sec ^{-1} x = \frac{\pi}{4}$
Since $\sec ^{-1} x = \tan ^{-1} \sqrt{x^2-1}$,we have:
$\tan ^{-1} \left( \frac{3}{4} \right) + \tan ^{-1} \sqrt{x^2-1} = \frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{\frac{3}{4} + \sqrt{x^2-1}}{1 - \frac{3}{4} \sqrt{x^2-1}} = \tan \left( \frac{\pi}{4} \right) = 1$
$3 + 4 \sqrt{x^2-1} = 4 - 3 \sqrt{x^2-1}$
$7 \sqrt{x^2-1} = 1$
$\sqrt{x^2-1} = \frac{1}{7}$
$x^2 - 1 = \frac{1}{49}$
$x^2 = 1 + \frac{1}{49} = \frac{50}{49}$

(C) Given equation is $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$.
We know that $\sin ^{-1}(u) + \cos ^{-1}(u) = \frac{\pi}{2}$,so $\sin ^{-1}(u) = \frac{\pi}{2} - \cos ^{-1}(u)$.
Rearranging the given equation: $\sin ^{-1}\left(\frac{12}{x}\right) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{5}{x}\right)$.
Using the identity $\frac{\pi}{2} - \sin ^{-1}(u) = \cos ^{-1}(u)$,we get $\sin ^{-1}\left(\frac{12}{x}\right) = \cos ^{-1}\left(\frac{5}{x}\right)$.
Let $\sin ^{-1}\left(\frac{12}{x}\right) = \theta$,then $\sin \theta = \frac{12}{x}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\cos \theta = \sqrt{1 - \left(\frac{12}{x}\right)^2} = \frac{\sqrt{x^2 - 144}}{x}$.
Thus,$\theta = \cos ^{-1}\left(\frac{\sqrt{x^2 - 144}}{x}\right)$.
Equating the arguments: $\frac{\sqrt{x^2 - 144}}{x} = \frac{5}{x}$.
Squaring both sides: $x^2 - 144 = 25$.
$x^2 = 169$,which gives $x = 13$ (since $x$ must be positive for the domain of $\sin^{-1}$ here).

(A) Given that,$x = \tan^{-1} \left(\frac{1}{5}\right) + \tan^{-1} \left(\frac{1}{8}\right)$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left(\frac{a+b}{1-ab}\right)$,
$x = \tan^{-1} \left(\frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5} \cdot \frac{1}{8}}\right) = \tan^{-1} \left(\frac{\frac{8+5}{40}}{\frac{40-1}{40}}\right) = \tan^{-1} \left(\frac{13}{39}\right) = \tan^{-1} \left(\frac{1}{3}\right)$.
Thus,$\tan x = \frac{1}{3}$.
Since $\tan x = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{3}$,the hypotenuse is $\sqrt{1^2 + 3^2} = \sqrt{10}$.
Therefore,$\sin x = \frac{1}{\sqrt{10}}$ and $\cos x = \frac{3}{\sqrt{10}}$.
Substituting these values into the expression:
$\frac{\sin x + \cos x}{\tan x} = \frac{\frac{1}{\sqrt{10}} + \frac{3}{\sqrt{10}}}{\frac{1}{3}} = \frac{\frac{4}{\sqrt{10}}}{\frac{1}{3}} = \frac{12}{\sqrt{10}}$.

(D) Given,$y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$.
Substitute $ax = \tan \theta$,so $\theta = \tan ^{-1}(ax)$.
Then $y = \tan ^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right) = \tan ^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right) = \tan ^{-1}\left(\tan \frac{\theta}{2}\right) = \frac{\theta}{2}$.
Thus,$y = \frac{1}{2} \tan ^{-1}(ax)$.
Differentiating with respect to $x$,we get $y^{\prime} = \frac{1}{2} \cdot \frac{a}{1+a^2 x^2}$.
This implies $(1+a^2 x^2) y^{\prime} = \frac{a}{2}$.
Differentiating again with respect to $x$,we get $(1+a^2 x^2) y^{\prime \prime} + (2a^2 x) y^{\prime} = 0$.
Comparing this with $(1+a^2 x^2) y^{\prime \prime} + g(x) y^{\prime} = 0$,we find $g(x) = 2a^2 x$.
The equation $1+a^2 x^2 + g(x) = 0$ becomes $a^2 x^2 + 2a^2 x + 1 = 0$.
The sum of the roots of a quadratic equation $Ax^2 + Bx + C = 0$ is $-\frac{B}{A}$.
Here,the sum of the roots is $-\frac{2a^2}{a^2} = -2$.

(A) Given $\angle C = 90^{\circ}$,we have $a^2 + b^2 = c^2$.
Let $S = \tan^{-1}\left(\frac{a}{b+c}\right) + \tan^{-1}\left(\frac{b}{c+a}\right) + \tan^{-1}\left(\frac{c}{a+b}\right)$.
First,consider $\tan^{-1}\left(\frac{a}{b+c}\right) + \tan^{-1}\left(\frac{b}{c+a}\right) = \tan^{-1}\left(\frac{\frac{a}{b+c} + \frac{b}{c+a}}{1 - \frac{ab}{(b+c)(c+a)}}\right)$.
$= \tan^{-1}\left(\frac{ac + a^2 + bc + b^2}{(b+c)(c+a) - ab}\right) = \tan^{-1}\left(\frac{ac + bc + c^2}{bc + ab + c^2 + ac - ab}\right) = \tan^{-1}\left(\frac{c(a+b+c)}{c(a+b+c)}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.
Thus,$S = \frac{\pi}{4} + \tan^{-1}\left(\frac{c}{a+b}\right) = \tan^{-1}(1) + \tan^{-1}\left(\frac{c}{a+b}\right) = \tan^{-1}\left(\frac{1 + \frac{c}{a+b}}{1 - \frac{c}{a+b}}\right)$.
$= \tan^{-1}\left(\frac{a+b+c}{a+b-c}\right) = \tan^{-1}\left(\frac{2s}{2(s-c)}\right) = \tan^{-1}\left(\frac{s}{s-c}\right)$.
Since $r = \frac{\Delta}{s}$ and $r_3 = \frac{\Delta}{s-c}$,we have $\frac{r_3}{r} = \frac{s}{s-c}$.
Therefore,$S = \tan^{-1}\left(\frac{r_3}{r}\right)$.

(D) We use the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left( \frac{A-B}{1+AB} \right)$.
Given equation: $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$.
Applying the formula,we get:
$\tan ^{-1} \left( \frac{\frac{x}{x-2} - \frac{x}{2x-1}}{1 + \frac{x}{x-2} \cdot \frac{x}{2x-1}} \right) = \tan ^{-1} \left( \frac{2}{3} \right)$.
Simplifying the expression inside the $\tan ^{-1}$:
$\frac{x(2x-1) - x(x-2)}{(x-2)(2x-1) + x^2} = \frac{2}{3}$.
$\frac{2x^2 - x - x^2 + 2x}{2x^2 - x - 4x + 2 + x^2} = \frac{2}{3}$.
$\frac{x^2 + x}{3x^2 - 5x + 2} = \frac{2}{3}$.
Cross-multiplying:
$3(x^2 + x) = 2(3x^2 - 5x + 2)$.
$3x^2 + 3x = 6x^2 - 10x + 4$.
$3x^2 - 13x + 4 = 0$.
Factoring the quadratic equation:
$(3x - 1)(x - 4) = 0$.
Thus,$x = \frac{1}{3}$ or $x = 4$.
Both values satisfy the domain of the original expression. Therefore,the set of values is $\left\{\frac{1}{3}, 4\right\}$.

(C) We have,$\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) = \tan ^{-1} \alpha$.
Using the identity $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1}\left(\frac{x-y}{1+xy}\right)$,we can rewrite the term inside the summation:
$\frac{1}{n^2+3n+3} = \frac{(n+2)-(n+1)}{1+(n+2)(n+1)}$.
Thus,the summation becomes:
$\sum_{n=1}^k (\tan ^{-1}(n+2) - \tan ^{-1}(n+1)) = \tan ^{-1} \alpha$.
Expanding the sum:
$(\tan ^{-1} 3 - \tan ^{-1} 2) + (\tan ^{-1} 4 - \tan ^{-1} 3) + \dots + (\tan ^{-1}(k+2) - \tan ^{-1}(k+1)) = \tan ^{-1} \alpha$.
This is a telescoping series,so all intermediate terms cancel out:
$\tan ^{-1}(k+2) - \tan ^{-1} 2 = \tan ^{-1} \alpha$.
Applying the formula again:
$\tan ^{-1}\left(\frac{(k+2)-2}{1+(k+2)(2)}\right) = \tan ^{-1} \alpha$.
$\tan ^{-1}\left(\frac{k}{1+2k+4}\right) = \tan ^{-1} \alpha$.
Therefore,$\alpha = \frac{k}{2k+5}$.

(D) The function is $f(x) = \sec^{-1}(3x - 4) + \tanh^{-1}\left(\frac{x + 3}{5}\right)$.
For $\sec^{-1}(3x - 4)$ to be defined,we must have $|3x - 4| \geq 1$.
This implies $3x - 4 \leq -1$ or $3x - 4 \geq 1$.
$3x \leq 3 \Rightarrow x \leq 1$ and $3x \geq 5 \Rightarrow x \geq \frac{5}{3}$.
So,the domain of the first part is $x \in (-\infty, 1] \cup [\frac{5}{3}, \infty)$.
For $\tanh^{-1}\left(\frac{x + 3}{5}\right)$ to be defined,we must have $-1 < \frac{x + 3}{5} < 1$.
$-5 < x + 3 < 5$.
$-8 < x < 2$.
Taking the intersection of $x \in (-\infty, 1] \cup [\frac{5}{3}, \infty)$ and $x \in (-8, 2)$:
$x \in (-8, 1] \cup [\frac{5}{3}, 2)$.

(B) For the function $f(x) = \sqrt{\log_{10}\left(\frac{5x - x^2}{4}\right)}$ to be defined,the expression inside the square root must be non-negative: $\log_{10}\left(\frac{5x - x^2}{4}\right) \geq 0$.
This implies $\frac{5x - x^2}{4} \geq 10^0$,which simplifies to $\frac{5x - x^2}{4} \geq 1$.
Multiplying by $4$,we get $5x - x^2 \geq 4$,or $x^2 - 5x + 4 \leq 0$.
Factoring the quadratic,we have $(x - 1)(x - 4) \leq 0$.
This inequality holds when $x \in [1, 4]$.
Thus,the domain of $f(x)$ is $[1, 4]$.

(D) We are given $f(x) = [x]$ and $g(x) = 3[\frac{x}{3}]$.
Setting $f(x) = g(x)$,we have $[x] = 3[\frac{x}{3}]$.
Let $[\frac{x}{3}] = k$,where $k \in Z$.
Then,by the definition of the greatest integer function,$k \leq \frac{x}{3} < k + 1$,which implies $3k \leq x < 3k + 3$.
Also,we have $[x] = 3k$.
By the definition of the greatest integer function,$[x] = 3k$ implies $3k \leq x < 3k + 1$.
Since $3k \leq x < 3k + 1$ is a subset of $3k \leq x < 3k + 3$,the condition $[x] = 3[\frac{x}{3}]$ is satisfied for all $x$ in the interval $[3k, 3k + 1)$ for any integer $k$.
Thus,the set of all such real $x$ is $\{x \in R : 3k \leq x < 3k + 1, k \in Z\}$.

(B) For $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ to be defined,the denominator must be non-zero and the expression inside the square root must be positive:
$|x|-x > 0 \Rightarrow |x| > x$.
This inequality holds for all $x < 0$. Thus,$A = (-\infty, 0)$.
For $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ to be defined,the denominator must be non-zero and the expression inside the square root must be positive:
$|x|+x > 0 \Rightarrow |x| > -x$.
This inequality holds for all $x > 0$. Thus,$B = (0, \infty)$.
Since $A = (-\infty, 0)$ and $B = (0, \infty)$,their intersection is empty:
$A \cap B = \phi$.

(C) Given the function $f:[-3,2] \rightarrow [0, \sqrt[3]{x}]$ defined as $f(n) = \begin{cases} 2+\sqrt[3]{n}, & -3 \leq n \leq -1 \\ n^{2/3}, & -1 < n \leq 2 \end{cases}$.
For the function to be onto,the range of $f(n)$ must be equal to the codomain $[0, \sqrt[3]{x}]$.
First,evaluate the function at the boundaries and critical points:
For $-3 \leq n \leq -1$,$f(n)$ increases from $f(-3) = 2 + \sqrt[3]{-3} \approx 0.55$ to $f(-1) = 2 + (-1)^{1/3} = 2 - 1 = 1$.
For $-1 < n \leq 2$,$f(n) = n^{2/3}$. At $n = -1$,$f(-1) = (-1)^{2/3} = 1$. At $n = 0$,$f(0) = 0$. At $n = 2$,$f(2) = 2^{2/3} = \sqrt[3]{4}$.
Since the function is onto,the maximum value of the function must be the upper bound of the codomain.
The maximum value is $\max(f(-1), f(2)) = \max(1, \sqrt[3]{4}) = \sqrt[3]{4}$.
Thus,$\sqrt[3]{x} = \sqrt[3]{4}$,which implies $x = 4$.

(D) The function $f(x) = \cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]$ is defined when the argument of $\cos ^{-1}$ lies in the interval $[-1, 1]$.
So,$-1 \leq \log _5\left(x^2+7 x+15\right) \leq 1$.
Applying the base $5$ exponential function,we get $5^{-1} \leq x^2+7 x+15 \leq 5^1$,which is $\frac{1}{5} \leq x^2+7 x+15 \leq 5$.
First,consider $x^2+7 x+15 \leq 5$,which implies $x^2+7 x+10 \leq 0$.
Factoring the quadratic,we get $(x+5)(x+2) \leq 0$,which gives $x \in [-5, -2]$.
Next,consider $x^2+7 x+15 \geq \frac{1}{5}$,which implies $x^2+7 x + 14.8 \geq 0$.
The discriminant of $x^2+7 x + 14.8$ is $D = 7^2 - 4(1)(14.8) = 49 - 59.2 = -10.2 < 0$.
Since the leading coefficient is positive and $D < 0$,$x^2+7 x + 14.8$ is always positive for all real $x$.
Thus,the domain is the intersection of $x \in [-5, -2]$ and $x \in \mathbb{R}$,which is $x \in [-5, -2]$.

(A) We have $f(x) = x + 2|x + 1| + 2|x - 1|$.
Breaking the function into intervals:
For $x \leq -1$: $f(x) = x + 2(-x - 1) + 2(-x + 1) = x - 2x - 2 - 2x + 2 = -3x$.
For $-1 < x < 1$: $f(x) = x + 2(x + 1) + 2(-x + 1) = x + 2x + 2 - 2x + 2 = x + 4$.
For $x \geq 1$: $f(x) = x + 2(x + 1) + 2(x - 1) = x + 2x + 2 + 2x - 2 = 5x$.
The function is defined as:
$f(x) = \begin{cases} -3x, & x \leq -1 \\ x + 4, & -1 < x < 1 \\ 5x, & x \geq 1 \end{cases}$
At $x = -1$,$f(-1) = 3$. For $x < -1$,$f(x) > 3$. For $-1 < x < 1$,$f(x)$ ranges from $3$ to $5$. At $x = 1$,$f(1) = 5$. For $x > 1$,$f(x) > 5$.
The value $3$ is attained at $x = -1$ and as $x \to -1^+$,$f(x) \to 3$. However,looking at the graph,the function is continuous. The value $3$ is the minimum value of the function,occurring at $x = -1$. Since the function is strictly decreasing for $x < -1$ and strictly increasing for $x > -1$,the value $3$ is the unique minimum,thus it has a unique pre-image.

(B) To find $\lim _{x \rightarrow 0} g(f(x))$,we evaluate the left-hand and right-hand limits.
For the left-hand limit $(x \rightarrow 0^-)$: $f(x) = 2-x$. As $x \rightarrow 0^-$,$f(x) \rightarrow 2^+$. Since $f(x) > 2$,we use the definition $g(x) = x-7$. Thus,$\lim _{x \rightarrow 0^-} g(f(x)) = \lim _{f(x) \rightarrow 2^+} (f(x)-7) = 2-7 = -5$.
For the right-hand limit $(x \rightarrow 0^+)$: $f(x) = x+2$. As $x \rightarrow 0^+$,$f(x) \rightarrow 2^+$. Again,since $f(x) > 2$,we use $g(x) = x-7$. Thus,$\lim _{x \rightarrow 0^+} g(f(x)) = \lim _{f(x) \rightarrow 2^+} (f(x)-7) = 2-7 = -5$.
Since both limits are equal,$\lim _{x \rightarrow 0} g(f(x)) = -5$.

(C) To check for injectivity,let $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$.
$\frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}}$
Squaring both sides:
$\frac{x_1^2}{1+x_1^2} = \frac{x_2^2}{1+x_2^2}$
$x_1^2(1+x_2^2) = x_2^2(1+x_1^2)$
$x_1^2 + x_1^2x_2^2 = x_2^2 + x_1^2x_2^2$
$x_1^2 = x_2^2$
Since $f'(x) = \frac{1}{(1+x^2)^{3/2}} > 0$ for all $x \in R$,the function is strictly increasing.
Therefore,$f(x_1) = f(x_2) \implies x_1 = x_2$. Thus,$f$ is injective.
To check for surjectivity,let $y = \frac{x}{\sqrt{1+x^2}}$.
Since $x^2 < 1+x^2$,we have $\frac{|x|}{\sqrt{1+x^2}} < 1$.
Thus,the range of $f$ is $(-1, 1)$,which is not equal to the codomain $R$.
Therefore,$f$ is not surjective.
Hence,the function is injective but not surjective.

(C) Given the function $f(x) = \begin{cases} \frac{60-5x}{3}, & 0 \leq x \leq 6 \\ 10, & 6 \leq x \leq 7 \\ 31-3x, & 7 \leq x \leq 10 \end{cases}$.
For $x \in [6, 7]$,$f(x) = 10$. Since the function takes the same value for all $x$ in the interval $[6, 7]$,it is not one-one.
Now,let us find the range of $f(x)$:
For $0 \leq x \leq 6$,$f(x) = \frac{60-5x}{3}$. As $x$ goes from $0$ to $6$,$f(x)$ goes from $\frac{60}{3} = 20$ to $\frac{60-30}{3} = 10$. So,the range is $[10, 20]$.
For $6 \leq x \leq 7$,$f(x) = 10$. So,the range is ${10}$.
For $7 \leq x \leq 10$,$f(x) = 31-3x$. As $x$ goes from $7$ to $10$,$f(x)$ goes from $31-21 = 10$ to $31-30 = 1$. So,the range is $[1, 10]$.
The union of these ranges is $[1, 10] \cup {10} \cup [10, 20] = [1, 20]$.
Since the range $[1, 20]$ is equal to the co-domain $[1, 20]$,the function is onto.
Thus,$f(x)$ is onto but not one-one.

(B) Given that $A_n = \{(n+1)k \mid k \in N\}$.
For $n=1$,$A_1 = \{2k \mid k \in N\} = \{2, 4, 6, 8, \dots\}$.
For $n=2$,$A_2 = \{3k \mid k \in N\} = \{3, 6, 9, 12, \dots\}$.
For $n=3$,$A_3 = \{4k \mid k \in N\} = \{4, 8, 12, 16, \dots\}$.
Now,$X = \bigcup_{n \in N} A_n = A_1 \cup A_2 \cup A_3 \cup \dots = \{2, 3, 4, 5, 6, \dots\}$.
Here,$f: X \rightarrow N$ is defined by $f(x) = x$ for all $x \in X$.
Since $f(x) = x$ is an identity function on the domain $X$,it is clearly one-one.
However,the codomain is $N = \{1, 2, 3, 4, \dots\}$.
For $f$ to be onto,the range must equal the codomain. Here,the range is $X = \{2, 3, 4, 5, \dots\}$.
Since $1 \in N$ but $1 \notin X$,there is no $x \in X$ such that $f(x) = 1$.
Therefore,$f$ is not onto.

(A) Given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in Z$.
Setting $x=0, y=0$,we get $f(0) = f(0) + f(0)$,which implies $f(0) = 0$.
For any $n \in Z^+$,by induction,$f(nx) = nf(x)$.
Let $f(1) = k$,where $k \in Z$. Then $f(n) = nk$ for all $n \in Z$.
Since $f(x+y) = f(x) + f(y)$,it follows that $f(x) = kx$ for all $x \in Z$.
For $f$ to be a bijection from $Z$ to $Z$,it must be both injective and surjective.
If $f(x) = kx$,then $f$ is injective if $k \neq 0$.
For $f$ to be surjective,the range of $f$ must be $Z$.
The range of $f(x) = kx$ is the set of all multiples of $k$,i.e.,$\{..., -2k, -k, 0, k, 2k, ...\}$.
For this set to be equal to $Z$,we must have $k = 1$ or $k = -1$.
If $k = 1$,$f(x) = x$,which is the identity function (bijective).
If $k = -1$,$f(x) = -x$,which is also bijective.
Thus,there are exactly two such functions.

(C) We are given $f(n) = \left(r+1, \frac{q+1}{2}\right)$ where $n = q \times 2^r$,$q$ is an odd integer,and $r \geq 0$.
For one-one mapping:
Suppose $f(n_1) = f(n_2)$.
Then $\left(r_1+1, \frac{q_1+1}{2}\right) = \left(r_2+1, \frac{q_2+1}{2}\right)$.
This implies $r_1+1 = r_2+1 \Rightarrow r_1 = r_2$ and $\frac{q_1+1}{2} = \frac{q_2+1}{2} \Rightarrow q_1 = q_2$.
Since $n_1 = q_1 \times 2^{r_1}$ and $n_2 = q_2 \times 2^{r_2}$,it follows that $n_1 = n_2$. Thus,$f$ is one-one.
For onto mapping:
Let $(a, b) \in N \times N$. We need to find $n \in N$ such that $f(n) = (a, b)$.
$r+1 = a \Rightarrow r = a-1$. Since $a \in N$,$a \geq 1$,so $r \geq 0$.
$\frac{q+1}{2} = b \Rightarrow q = 2b-1$. Since $b \in N$,$b \geq 1$,so $q \geq 1$ and $q$ is odd.
Thus,for any $(a, b) \in N \times N$,there exists $n = (2b-1) \times 2^{a-1} \in N$ such that $f(n) = (a, b)$.
Therefore,$f$ is onto.
Since $f$ is both one-one and onto,$f$ is a bijection.

(B) Given $f: Z \rightarrow N$ defined by $f(n) = \begin{cases} 2n, & \text{if } n > 0 \\ 1, & \text{if } n = 0 \\ -2n-1, & \text{if } n < 0 \end{cases}$.
For $n > 0$,$f(n) \in \{2, 4, 6, 8, \dots\}$.
For $n = 0$,$f(0) = 1$.
For $n < 0$,let $n = -k$ where $k > 0$. Then $f(n) = -2(-k) - 1 = 2k - 1$. As $k$ takes values $1, 2, 3, \dots$,$f(n)$ takes values $1, 3, 5, 7, \dots$.
Combining these,the range of $f$ is $\{1, 2, 3, 4, \dots\} = N$. Since Range = Codomain,$f$ is onto.
Now,check for one-one: $f(0) = 1$ and $f(-1) = -2(-1) - 1 = 2 - 1 = 1$.
Since $f(0) = f(-1)$ but $0 \neq -1$,the function is not one-one.
Therefore,$f$ is onto but not one-one.

(D) Given $f(x) = x^2 + 3x - 3$.
Completing the square,we get $f(x) = (x + \frac{3}{2})^2 - \frac{9}{4} - 3 = (x + \frac{3}{2})^2 - \frac{21}{4}$.
For the inverse to exist,the function must be monotonic. The vertex of the parabola is at $x = -\frac{3}{2}$,so the function is strictly increasing on $[-\frac{3}{2}, \infty)$. Thus,$\alpha = -\frac{3}{2}$.
Let $y = (x + \frac{3}{2})^2 - \frac{21}{4}$. Then $x + \frac{3}{2} = \sqrt{y + \frac{21}{4}}$,so $g(x) = f^{-1}(x) = \sqrt{x + \frac{21}{4}} - \frac{3}{2}$.
We need to find $g'(x)$ at $x = \alpha + \frac{5}{2} = -\frac{3}{2} + \frac{5}{2} = 1$.
$g'(x) = \frac{d}{dx}(\sqrt{x + \frac{21}{4}} - \frac{3}{2}) = \frac{1}{2\sqrt{x + \frac{21}{4}}}$.
At $x = 1$,$g'(1) = \frac{1}{2\sqrt{1 + \frac{21}{4}}} = \frac{1}{2\sqrt{\frac{25}{4}}} = \frac{1}{2 \times \frac{5}{2}} = \frac{1}{5}$.

(B) Given,$f(x+y)=f(x) \cdot f(y)$. This is a functional equation whose solution is of the form $f(x)=a^x$.
Given $f(1)=2$,so $a^1=2 \Rightarrow a=2$. Thus,$f(x)=2^x$.
Now,consider the region enclosed by $2|x|+5|y| \leq 4$. This represents a rhombus with vertices at $(\pm 2, 0)$ and $(0, \pm 4/5)$.
The area of the rhombus is given by $4 \times \text{Area of one triangle in the first quadrant}$.
Area $= 4 \times (\frac{1}{2} \times 2 \times \frac{4}{5}) = 4 \times \frac{4}{5} = \frac{16}{5}$.
We need to express $\frac{16}{5}$ in terms of $f(1)=2^1=2$,$f(2)=2^2=4$,and $f(4)=2^4=16$.
Note that $1+f(2) = 1+4 = 5$.
Thus,the area is $\frac{16}{5} = \frac{f(4)}{1+f(2)}$.

(C) Given $f(x) = x - \frac{1}{x}$.
We know the algebraic identity $(a - b)^3 = a^3 - b^3 - 3ab(a - b)$.
Applying this to $f(x)$:
$[f(x)]^3 = \left(x - \frac{1}{x}\right)^3 = x^3 - \frac{1}{x^3} - 3(x)\left(\frac{1}{x}\right)\left(x - \frac{1}{x}\right)$.
Since $f(x^3) = x^3 - \frac{1}{x^3}$ and $f(x) = x - \frac{1}{x}$,we substitute these into the equation:
$[f(x)]^3 = f(x^3) - 3(1)f(x)$.
Rearranging the terms to solve for $3f(x)$:
$3f(x) = f(x^3) - [f(x)]^3$.
Thus,the correct option is $C$.

(A) Given $f(x) = \frac{1 + \sin([\cos x])}{\cos([\sin x])}$.
For $x \in \left(0, \frac{\pi}{2}\right)$,we have $0 < \cos x < 1$,so $[\cos x] = 0$. Also,$0 < \sin x < 1$,so $[\sin x] = 0$.
Substituting these values into the function,we get $f(x) = \frac{1 + \sin(0)}{\cos(0)} = \frac{1 + 0}{1} = 1$.
Since $f(x) = 1$ is a constant function on the interval $\left(0, \frac{\pi}{2}\right)$,it is continuous on this interval.
Thus,option $A$ is correct.

(B) We analyze the limit $f(x) = \lim_{n \rightarrow \infty} \frac{\log(2+x) - x^{2n} \sin x}{1+x^{2n}}$.
Case $1$: If $0 \leq x < 1$,then $x^{2n} \rightarrow 0$ as $n \rightarrow \infty$. Thus,$f(x) = \frac{\log(2+x) - 0}{1+0} = \log(2+x)$.
Case $2$: If $x = 1$,then $f(1) = \lim_{n \rightarrow \infty} \frac{\log(3) - 1^{2n} \sin(1)}{1+1^{2n}} = \frac{\log 3 - \sin 1}{2}$.
Case $3$: If $1 < x \leq \frac{\pi}{2}$,then $x^{2n} \rightarrow \infty$. Dividing numerator and denominator by $x^{2n}$,we get $f(x) = \lim_{n \rightarrow \infty} \frac{\frac{\log(2+x)}{x^{2n}} - \sin x}{\frac{1}{x^{2n}} + 1} = \frac{0 - \sin x}{0 + 1} = -\sin x$.
Now,check continuity at $x=1$:
Left-hand limit: $\lim_{x \rightarrow 1^-} f(x) = \log(2+1) = \log 3$.
Right-hand limit: $\lim_{x \rightarrow 1^+} f(x) = -\sin(1)$.
Since $\log 3 \neq -\sin 1$,the function is discontinuous at $x=1$.

(D) Since $f(x)$ is continuous at $x = 1$,we have $k = \lim_{x \rightarrow 1} f(x)$.
$k = \lim_{x \rightarrow 1} \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}$.
Using $L$'$H$ôpital's rule,we differentiate the numerator and denominator with respect to $x$:
Numerator: $\frac{d}{dx}[(9x-1)(\sqrt{x}-1)] = 9(\sqrt{x}-1) + (9x-1)(\frac{1}{2\sqrt{x}})$.
Denominator: $\frac{d}{dx}[3x^2+2x-5] = 6x+2$.
Evaluating the limit as $x \rightarrow 1$:
$k = \frac{9(1-1) + (9-1)(\frac{1}{2})}{6(1)+2} = \frac{0 + 8(\frac{1}{2})}{8} = \frac{4}{8} = \frac{1}{2}$.

(B) Given,$f(x) = \begin{cases} \frac{1-\cos 4 x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{cases}$
Since $f(x)$ is continuous at $x=0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0)$.
First,calculate the Left Hand Limit $(LHL)$:
$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x \rightarrow 0^-} \frac{2\sin^2(2x)}{x^2} = 2 \lim_{x \rightarrow 0^-} \left(\frac{\sin 2x}{2x}\right)^2 \times 4 = 2 \times 1^2 \times 4 = 8$.
Next,calculate the Right Hand Limit $(RHL)$:
$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$.
Rationalizing the denominator:
$= \lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x \rightarrow 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16}+4 = 8$.
Since $LHL = RHL = 8$,for continuity at $x=0$,we must have $f(0) = a = 8$.

(A) Given that $f(x)$ has a minimum at $x=5$ with $f(5)=75$.
If $f(x)$ were continuous on the interval $[-1, 5]$,then by the Extreme Value Theorem,it must attain a maximum and a minimum on this closed interval.
Since $f(-1) = -249$ and $f(5) = 75$,if the function were continuous,there would exist some $c \in (-1, 5)$ such that $f(c)$ is a local maximum or the function would have to increase from $-249$ to $75$.
However,the problem states that the function has no maximum and only one minimum at $x=5$.
If $f(x)$ were continuous,it would contradict the given conditions because a continuous function on a closed interval must have a maximum.
Therefore,$f(x)$ must be discontinuous at some point in the interval $(-1, 5)$ to avoid having a maximum.

(B) $1$. The function $f(x) = |x-a| + |x-b|$ is a sum of two absolute value functions. Since $|x-c|$ is continuous for all $x \in R$,the sum of two continuous functions is also continuous on $R$. Thus,Assertion $(A)$ is true.
$2$. The function $g(x) = \frac{|x-\alpha|}{x-\alpha}$ is defined as $1$ for $x > \alpha$ and $-1$ for $x < \alpha$. At $x = \alpha$,the function is undefined. Therefore,it is continuous for all $x \in R - \{\alpha\}$. Thus,Reason $(R)$ is true.
$3$. While both statements are true,the continuity of $f(x)$ is a property of absolute value functions,and the continuity of $g(x)$ is a separate property of rational-like expressions involving absolute values. $R$ does not explain why $f(x)$ is continuous. Therefore,$R$ is not the correct explanation for $A$.

(D) Given the function $f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases}$.
To check for continuity at $x=0$,we evaluate the left-hand limit $(LHL)$ and right-hand limit $(RHL)$.
$LHL = \lim_{x \rightarrow 0^{-}} f(x) = \lim_{h \rightarrow 0} f(0-h) = \lim_{h \rightarrow 0} \frac{-h}{|-h|+2(-h)^2} = \lim_{h \rightarrow 0} \frac{-h}{h+2h^2} = \lim_{h \rightarrow 0} \frac{-h}{h(1+2h)} = \lim_{h \rightarrow 0} \frac{-1}{1+2h} = -1$.
$RHL = \lim_{x \rightarrow 0^{+}} f(x) = \lim_{h \rightarrow 0} f(0+h) = \lim_{h \rightarrow 0} \frac{h}{|h|+2h^2} = \lim_{h \rightarrow 0} \frac{h}{h(1+2h)} = \lim_{h \rightarrow 0} \frac{1}{1+2h} = 1$.
Since $LHL \neq RHL$,the limit $\lim_{x \rightarrow 0} f(x)$ does not exist.
Therefore,the function $f(x)$ is discontinuous at $x=0$ for all real values of $k$.

(D) We have,$f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$.
$f(f(x))$ is defined as:
For $0 \leq f(x) \leq 2$,$f(f(x)) = 1 + f(x)$.
For $2 < f(x) \leq 3$,$f(f(x)) = 3 - f(x)$.
Evaluating for $x \in [0, 3]$:
If $0 \leq x \leq 1$,then $1 \leq f(x) \leq 2$,so $f(f(x)) = 1 + (1 + x) = 2 + x$.
If $1 < x \leq 2$,then $2 < f(x) \leq 3$,so $f(f(x)) = 3 - (1 + x) = 2 - x$.
If $2 < x \leq 3$,then $0 \leq f(x) < 1$,so $f(f(x)) = 1 + (3 - x) = 4 - x$.
Thus,$f(f(x)) = \begin{cases} 2 + x, & 0 \leq x \leq 1 \\ 2 - x, & 1 < x \leq 2 \\ 4 - x, & 2 < x \leq 3 \end{cases}$.
Checking continuity at $x = 1$: $\lim_{x \to 1^-} f(f(x)) = 3$,$\lim_{x \to 1^+} f(f(x)) = 1$. Since $3 \neq 1$,it is discontinuous at $x = 1$.
Checking continuity at $x = 2$: $\lim_{x \to 2^-} f(f(x)) = 0$,$\lim_{x \to 2^+} f(f(x)) = 2$. Since $0 \neq 2$,it is discontinuous at $x = 2$.
Therefore,$a = 1$ and $b = 2$.
Calculating $2a + 3b = 2(1) + 3(2) = 2 + 6 = 8$.

(D) For Statement $I$: Let $F(x) = a_0 x + \frac{a_1 x^2}{2} + \frac{a_2 x^3}{3} + \ldots + \frac{a_n x^{n+1}}{n+1}$.
$F(x)$ is a polynomial,so it is continuous on $[0,1]$ and differentiable on $(0,1)$.
$F(0) = 0$ and $F(1) = a_0 + \frac{a_1}{2} + \ldots + \frac{a_n}{n+1} = 0$ (given).
By Rolle's Theorem,there exists $c \in (0,1)$ such that $F^{\prime}(c) = 0$.
Since $F^{\prime}(x) = a_0 + a_1 x + \ldots + a_n x^n = P(x)$,we have $P(c) = 0$. Thus,Statement $I$ is true.
For Statement $II$: Let $g(x) = \frac{f(x)}{x}$. Since $f$ is continuous on $[a, b]$ and $a>0$,$g(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$.
Given $g(a) = \frac{f(a)}{a} = \frac{f(b)}{b} = g(b)$.
By Rolle's Theorem,there exists $c \in (a, b)$ such that $g^{\prime}(c) = 0$.
$g^{\prime}(x) = \frac{x f^{\prime}(x) - f(x)}{x^2}$.
Setting $g^{\prime}(c) = 0$ implies $c f^{\prime}(c) - f(c) = 0$,or $c f^{\prime}(c) = f(c)$. Thus,Statement $II$ is true.

(B) For $f(x)$: $\lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin(1/x) = 0$,which equals $f(0)$,so $f(x)$ is continuous at $x = 0$.
For differentiability at $x = 0$,$f'(0) = \lim_{h \to 0} \frac{h \sin(1/h) - 0}{h} = \lim_{h \to 0} \sin(1/h)$,which does not exist. Thus,$(i)$ is true.
For $g(x) = x f(x) = x^2 \sin(1/x)$ for $x \neq 0$ and $g(0) = 0$.
$g'(0) = \lim_{h \to 0} \frac{h^2 \sin(1/h) - 0}{h} = \lim_{h \to 0} h \sin(1/h) = 0$. So $g(x)$ is differentiable at $x = 0$.
For $x \neq 0$,$g'(x) = 2x \sin(1/x) - \cos(1/x)$.
As $x \to 0$,$\lim_{x \to 0} g'(x) = \lim_{x \to 0} (2x \sin(1/x) - \cos(1/x))$,which does not exist because $\cos(1/x)$ oscillates.
Since $\lim_{x \to 0} g'(x) \neq g'(0)$,$g'(x)$ is not continuous at $x = 0$. Thus,$(ii)$ is true.

(A) Given,$\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$.
By definition,this limit is $f'(x) = e^x(x+1)$.
Integrating $f'(x)$ with respect to $x$,we get $f(x) = \int (x e^x + e^x) dx = x e^x + c$.
Since $f(0) = 0$,we have $0(e^0) + c = 0$,which implies $c = 0$.
Thus,$f(x) = x e^x$.
Now,we evaluate the expression $\frac{d}{d x}(f(x) e^{-x}) + \frac{d}{d x}(\frac{f(x)}{x})$.
Substituting $f(x) = x e^x$,we get $\frac{d}{d x}(x e^x \cdot e^{-x}) + \frac{d}{d x}(\frac{x e^x}{x}) = \frac{d}{d x}(x) + \frac{d}{d x}(e^x)$.
This simplifies to $1 + e^x$.

(B) Given $x = \frac{1 - \sqrt[3]{y}}{1 + \sqrt[3]{y}}$.
Multiplying both sides by $(1 + \sqrt[3]{y})$,we get $x(1 + \sqrt[3]{y}) = 1 - \sqrt[3]{y}$.
$x + x\sqrt[3]{y} = 1 - \sqrt[3]{y}$.
$\sqrt[3]{y}(x + 1) = 1 - x$.
$\sqrt[3]{y} = \frac{1 - x}{1 + x}$.
Cubing both sides,$y = \left(\frac{1 - x}{1 + x}\right)^3$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 3\left(\frac{1 - x}{1 + x}\right)^2 \cdot \frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)$.
Using the quotient rule,$\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right) = \frac{-(1 + x) - (1 - x)(1)}{(1 + x)^2} = \frac{-1 - x - 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2}$.
Thus,$\frac{dy}{dx} = 3 \cdot \frac{(1 - x)^2}{(1 + x)^2} \cdot \frac{-2}{(1 + x)^2} = \frac{-6(1 - x)^2}{(1 + x)^4}$.
Since $\frac{dx}{dy} = \frac{1}{dy/dx}$,we have $\frac{dx}{dy} = \frac{-(1 + x)^4}{6(1 - x)^2}$.

(D) Given $F(x)=f(x) g(x)$.
By the product rule,$F^{\prime}(x)=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$.
We are given $G(x)=f^{\prime}(x) g^{\prime}(x)$.
Since $F^{\prime}(x)=G(x) H(x)$,we have $H(x)=\frac{F^{\prime}(x)}{G(x)}=\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f^{\prime}(x) g^{\prime}(x)}=\frac{g(x)}{g^{\prime}(x)}+\frac{f(x)}{f^{\prime}(x)}$.
Also,$F^{\prime}(x)=F(x) K(x)$,so $K(x)=\frac{F^{\prime}(x)}{F(x)}=\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f(x) g(x)}=\frac{f^{\prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g(x)}$.
Therefore,$H(x)+K(x)=\frac{f(x)}{f^{\prime}(x)}+\frac{g(x)}{g^{\prime}(x)}+\frac{f^{\prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g(x)}$.

(B) Given $y = \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \log \sqrt{1-x^2}$.
First,simplify the logarithmic term: $\log \sqrt{1-x^2} = \frac{1}{2} \log(1-x^2)$.
Now,differentiate $y$ with respect to $x$ using the quotient rule for the first term and the chain rule for the second term:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) + \frac{d}{dx} \left( \frac{1}{2} \log(1-x^2) \right)$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$ where $u = x \sin^{-1} x$ and $v = \sqrt{1-x^2}$:
$u' = \sin^{-1} x + \frac{x}{\sqrt{1-x^2}}$ and $v' = \frac{-x}{\sqrt{1-x^2}}$.
$\frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) = \frac{(\sin^{-1} x + \frac{x}{\sqrt{1-x^2}})\sqrt{1-x^2} - (x \sin^{-1} x)(\frac{-x}{\sqrt{1-x^2}})}{1-x^2} = \frac{(1-x^2)\sin^{-1} x + x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}}}{1-x^2} = \frac{\sin^{-1} x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}} + x}{1-x^2}$.
Now,differentiate the second term: $\frac{d}{dx} (\frac{1}{2} \log(1-x^2)) = \frac{1}{2} \cdot \frac{1}{1-x^2} \cdot (-2x) = \frac{-x}{1-x^2}$.
Adding both parts: $\frac{dy}{dx} = \frac{\sin^{-1} x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}} + x - x}{1-x^2} = \frac{\sin^{-1} x (1 + \frac{x^2}{\sqrt{1-x^2}})}{1-x^2} = \frac{\sin^{-1} x (\frac{\sqrt{1-x^2} + x^2}{\sqrt{1-x^2}})}{1-x^2}$.
Wait,simplifying the expression: $\frac{dy}{dx} = \frac{\sin^{-1} x (1-x^2 + x^2)}{\sqrt{1-x^2}(1-x^2)} = \frac{\sin^{-1} x}{(1-x^2)^{3/2}}$.

(D) Given $f(x) = x^2 + g'(1)x + g''(2)$. Let $g'(1) = a$ and $g''(2) = b$. Then $f(x) = x^2 + ax + b$.
$f'(x) = 2x + a$ and $f''(x) = 2$.
Given $g(x) = f(1)x^2 + xf'(x) + f''(x)$.
Substituting $f(1) = 1 + a + b$,$f'(x) = 2x + a$,and $f''(x) = 2$:
$g(x) = (1 + a + b)x^2 + x(2x + a) + 2 = (1 + a + b + 2)x^2 + ax + 2 = (3 + a + b)x^2 + ax + 2$.
Now,$g'(x) = 2(3 + a + b)x + a$ and $g''(x) = 2(3 + a + b)$.
Using $g'(1) = a$: $2(3 + a + b) + a = a \implies 6 + 2a + 2b = 0 \implies a + b = -3$.
Using $g''(2) = b$: $2(3 + a + b) = b \implies 6 + 2a + 2b = b \implies 2a + b = -6$.
Solving the system: $(2a + b) - (a + b) = -6 - (-3) \implies a = -3$.
Then $-3 + b = -3 \implies b = 0$.
Thus,$f(x) = x^2 - 3x$ and $g(x) = (3 - 3 + 0)x^2 - 3x + 2 = -3x + 2$.
$f(x) - g(x) = (x^2 - 3x) - (-3x + 2) = x^2 - 2$.

(C) The general term of the series is $T_n = \tan^{-1}\left(\frac{1}{\sin^2 x + (2n-1)\sin x + (n^2-n+1)}\right)$.
We can rewrite this as $T_n = \tan^{-1}\left(\frac{(\sin x + n) - (\sin x + n - 1)}{1 + (\sin x + n)(\sin x + n - 1)}\right)$.
Using the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we get $T_n = \tan^{-1}(\sin x + n) - \tan^{-1}(\sin x + n - 1)$.
Summing up to $10$ terms,we have $f(x) = \sum_{n=1}^{10} [\tan^{-1}(\sin x + n) - \tan^{-1}(\sin x + n - 1)]$.
This is a telescoping series,so $f(x) = \tan^{-1}(\sin x + 10) - \tan^{-1}(\sin x)$.
Differentiating with respect to $x$,we get $f'(x) = \frac{\cos x}{1 + (\sin x + 10)^2} - \frac{\cos x}{1 + \sin^2 x}$.
At $x = 0$,$f'(0) = \frac{\cos 0}{1 + (0 + 10)^2} - \frac{\cos 0}{1 + 0^2} = \frac{1}{101} - 1 = \frac{-100}{101}$.

(A) Given equation is $x \sqrt{1+y} + y \sqrt{1+x} = 0$.
Rearranging the terms,we get $x \sqrt{1+y} = -y \sqrt{1+x}$.
Squaring both sides,we get $x^2(1+y) = y^2(1+x)$.
Expanding this,$x^2 + x^2y = y^2 + xy^2$.
Rearranging terms,$x^2 - y^2 + x^2y - xy^2 = 0$.
Factoring,$(x-y)(x+y) + xy(x-y) = 0$.
Since $x \neq y$ (as $x \sqrt{1+y} = -y \sqrt{1+x}$ implies $x$ and $y$ have opposite signs),we can divide by $(x-y)$ to get $x+y+xy = 0$.
Solving for $y$,$y(1+x) = -x$,so $y = \frac{-x}{1+x}$.
Differentiating with respect to $x$ using the quotient rule,$\frac{dy}{dx} = \frac{(1+x)(-1) - (-x)(1)}{(1+x)^2}$.
Simplifying,$\frac{dy}{dx} = \frac{-1-x+x}{(1+x)^2} = \frac{-1}{(1+x)^2}$.

(A) Given equation is $2^x+2^y=2^{x+y}$.
Dividing both sides by $2^{x+y}$,we get $\frac{2^x}{2^{x+y}}+\frac{2^y}{2^{x+y}}=1$.
This simplifies to $2^{-y}+2^{-x}=1$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(2^{-y})+\frac{d}{dx}(2^{-x})=\frac{d}{dx}(1)$.
Using the chain rule,we have $2^{-y} \cdot \log 2 \cdot (-\frac{dy}{dx}) + 2^{-x} \cdot \log 2 \cdot (-1) = 0$.
Dividing by $-\log 2$,we get $2^{-y} \frac{dy}{dx} + 2^{-x} = 0$.
Thus,$\frac{dy}{dx} = -\frac{2^{-x}}{2^{-y}} = -2^{y-x}$.
From the original equation $2^x+2^y=2^{x+y}$,we have $2^x = 2^{x+y}-2^y = 2^y(2^x-1)$,so $2^{-x} = \frac{1}{2^y-1}$ is not direct. Let's use $2^{-x} = 1-2^{-y}$.
Then $\frac{dy}{dx} = -\frac{1-2^{-y}}{2^{-y}} = -(\frac{1}{2^{-y}}-1) = -(2^y-1) = 1-2^y$.

(C) Let $y = \operatorname{cosech}^{-1}(\tan 2x)$.
Using the chain rule,$\frac{dy}{dx} = \frac{d}{dx} [\operatorname{cosech}^{-1}(\tan 2x)]$.
Recall the formula: $\frac{d}{dx} \operatorname{cosech}^{-1}(u) = \frac{-1}{|u| \sqrt{1+u^2}} \cdot \frac{du}{dx}$.
Here,$u = \tan 2x$,so $\frac{du}{dx} = 2 \sec^2 2x$.
Substituting these into the formula:
$\frac{dy}{dx} = \frac{-1}{|\tan 2x| \sqrt{1 + \tan^2 2x}} \cdot 2 \sec^2 2x$.
Since $1 + \tan^2 2x = \sec^2 2x$,we have $\sqrt{1 + \tan^2 2x} = |\sec 2x|$.
$\frac{dy}{dx} = \frac{-1}{|\tan 2x| \cdot |\sec 2x|} \cdot 2 \sec^2 2x$.
$\frac{dy}{dx} = \frac{-2 \sec^2 2x}{|\frac{\sin 2x}{\cos 2x}| \cdot |\frac{1}{\cos 2x}|} = \frac{-2 \sec^2 2x}{|\frac{\sin 2x}{\cos^2 2x}|} = \frac{-2 \sec^2 2x \cdot |\cos^2 2x|}{|\sin 2x|}$.
Since $\sec^2 2x \cdot \cos^2 2x = 1$,this simplifies to:
$\frac{dy}{dx} = \frac{-2}{|\sin 2x|} = -2 |\operatorname{cosec} 2x|$.

(A) Note: Assuming the base of the logarithm is $e$.
Given:
Assertion: For $x < 0$,$\frac{d^2}{d x^2}(\log |x|) = \frac{1}{|x|^2}$.
Reason: For $x < 0$,$|x| = -x$.
Let $f(x) = \log |x|$.
For $x < 0$,we have $|x| = -x$,so $f(x) = \log(-x)$.
Now,differentiating with respect to $x$:
$\frac{d}{d x}(\log(-x)) = \frac{1}{-x} \cdot \frac{d}{d x}(-x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$.
Now,differentiating again with respect to $x$:
$\frac{d^2}{d x^2}(\log |x|) = \frac{d}{d x}(\frac{1}{x}) = -\frac{1}{x^2}$.
Since $|x|^2 = (-x)^2 = x^2$,the Assertion states $\frac{d^2}{d x^2}(\log |x|) = \frac{1}{x^2}$,but we found $-\frac{1}{x^2}$.
Therefore,the Assertion is false and the Reason is true.

(A) Given the equation: $x^y = y^{\sin x} (\tan x)^{\cos x}$
Taking the natural logarithm on both sides:
$y \log x = \sin x \log y + \cos x \log (\tan x)$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y \log x) = \frac{d}{dx}(\sin x \log y) + \frac{d}{dx}(\cos x \log \tan x)$
$\frac{dy}{dx} \log x + y \cdot \frac{1}{x} = (\cos x \log y + \sin x \cdot \frac{1}{y} \frac{dy}{dx}) + (-\sin x \log \tan x + \cos x \cdot \frac{1}{\tan x} \cdot \sec^2 x)$
Simplifying the derivative term:
$\cos x \cdot \frac{1}{\tan x} \cdot \sec^2 x = \cos x \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x} = \operatorname{cosec} x$
Rearranging the terms to isolate $\left(\log x - \frac{\sin x}{y}\right) \frac{dy}{dx}$:
$\left(\log x - \frac{\sin x}{y}\right) \frac{dy}{dx} = \cos x \log y - \sin x \log (\tan x) + \operatorname{cosec} x - \frac{y}{x}$

(B) Given,$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$.
Let $x=\sin \alpha$ and $y=\sin \beta$. Then $\cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta)$.
Using trigonometric identities,$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = a \cdot 2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$.
This simplifies to $\cot \frac{\alpha-\beta}{2} = a$,so $\alpha - \beta = 2 \cot^{-1} a$.
Substituting back,$\sin^{-1} x - \sin^{-1} y = 2 \cot^{-1} a$.
Differentiating with respect to $x$,we get $\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-y^2}} \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$.
Squaring both sides,$\left(\frac{dy}{dx}\right)^2 = \frac{1-y^2}{1-x^2}$,so $(1-x^2) \left(\frac{dy}{dx}\right)^2 = 1-y^2$.
Differentiating again with respect to $x$,we get $(1-x^2) \cdot 2 \frac{dy}{dx} \frac{d^2y}{dx^2} + (-2x) \left(\frac{dy}{dx}\right)^2 = -2y \frac{dy}{dx}$.
Dividing by $2 \frac{dy}{dx}$ (assuming $\frac{dy}{dx} \neq 0$),we get $(1-x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = -y$.
Multiplying by $(1-x^2)$,we get $(1-x^2)^2 \frac{d^2y}{dx^2} - x(1-x^2) \frac{dy}{dx} = -y(1-x^2)$.
Rearranging,$(1-x^2)^2 \frac{d^2y}{dx^2} + y(1-x^2) = x(1-x^2) \frac{dy}{dx}$.
Multiplying by $\frac{dy}{dx}$,we get $\left[(1-x^2)^2 \frac{d^2y}{dx^2} + y(1-x^2)\right] \frac{dy}{dx} = x(1-x^2) \left(\frac{dy}{dx}\right)^2$.
Since $\left(\frac{dy}{dx}\right)^2 = \frac{1-y^2}{1-x^2}$,the expression becomes $x(1-x^2) \cdot \frac{1-y^2}{1-x^2} = x(1-y^2)$.

(D) The derivatives of the given inverse functions are as follows:
$(A)$ $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}$ for $|x| > 1$. This matches $III$.
$(B)$ $\frac{d}{dx}(\tanh^{-1} x) = \frac{1}{1-x^2}$ for $x \in (-1, 1)$. This matches $I$.
$(C)$ $\frac{d}{dx}(\coth^{-1} x) = \frac{1}{1-x^2}$ for $x \in R - [-1, 1]$. This matches $IV$.
$(D)$ $\frac{d}{dx}(\operatorname{cosech}^{-1} x) = \frac{-1}{|x|\sqrt{x^2+1}}$ for $x \neq 0$. This matches $II$.
Thus,the correct matching is $A-III, B-I, C-IV, D-II$.