KCET 2011 Mathematics Question Paper with Answer and Solution

(A) Given,$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (say).
From this,we have $\log x = k(b-c)$,$\log y = k(c-a)$,and $\log z = k(a-b)$.
This implies $x = e^{k(b-c)}$,$y = e^{k(c-a)}$,and $z = e^{k(a-b)}$.
Now,consider the expression $E = x^{b+c} \cdot y^{c+a} \cdot z^{a+b}$.
Substituting the values,$E = (e^{k(b-c)})^{b+c} \cdot (e^{k(c-a)})^{c+a} \cdot (e^{k(a-b)})^{a+b}$.
Using the property $(e^m)^n = e^{mn}$,we get $E = e^{k(b^2-c^2)} \cdot e^{k(c^2-a^2)} \cdot e^{k(a^2-b^2)}$.
Combining the exponents,$E = e^{k(b^2-c^2+c^2-a^2+a^2-b^2)}$.
Since the sum in the exponent is $0$,$E = e^{k \cdot 0} = e^0 = 1$.

(B) Given the cubic equation $x^{3}-2x+1=0$ with roots $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = 0$
$\alpha\beta+\beta\gamma+\gamma\alpha = -2$
$\alpha\beta\gamma = -1$
We need to evaluate $\sum\left(\frac{1}{\alpha+\beta-\gamma}\right)$.
Since $\alpha+\beta+\gamma = 0$,we have $\alpha+\beta = -\gamma$.
Substituting this into the expression:
$\sum\left(\frac{1}{-\gamma-\gamma}\right) = \sum\left(\frac{1}{-2\gamma}\right) = -\frac{1}{2} \sum\left(\frac{1}{\gamma}\right)$
$= -\frac{1}{2} \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)$
$= -\frac{1}{2} \left(\frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\right)$
$= -\frac{1}{2} \left(\frac{-2}{-1}\right) = -\frac{1}{2} \times 2 = -1$.

(D) Let $z = \frac{1+i \sqrt{3}}{\left(1+\frac{1}{i+1}\right)^{2}}$.
First,simplify the denominator: $1+\frac{1}{i+1} = \frac{i+1+1}{i+1} = \frac{i+2}{i+1}$.
Then,$z = \frac{1+i \sqrt{3}}{\left(\frac{i+2}{i+1}\right)^{2}} = \frac{(1+i \sqrt{3})(i+1)^{2}}{(i+2)^{2}}$.
Since $(i+1)^{2} = i^{2}+1+2i = 2i$ and $(i+2)^{2} = i^{2}+4+4i = 3+4i$,we have $z = \frac{(1+i \sqrt{3})(2i)}{3+4i} = \frac{2i-2\sqrt{3}}{3+4i}$.
Now,find the modulus: $|z| = \frac{|2i-2\sqrt{3}|}{|3+4i|} = \frac{\sqrt{(-2\sqrt{3})^{2} + 2^{2}}}{\sqrt{3^{2}+4^{2}}}$.
$|z| = \frac{\sqrt{12+4}}{\sqrt{9+16}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5}$.

(D) Let $z = \frac{(1-i \sqrt{3})(1+i)}{(\sqrt{3}+i)}$.
Multiply the numerator and denominator by the conjugate of the denominator $(\sqrt{3}-i)$:
$z = \frac{(1-i \sqrt{3})(1+i)(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}$
$z = \frac{(1+i-i\sqrt{3}+\sqrt{3})(\sqrt{3}-i)}{3+1}$
$z = \frac{(1+\sqrt{3}) + i(1-\sqrt{3})}{4} \cdot (\sqrt{3}-i)$
$z = \frac{1}{4} [\sqrt{3}(1+\sqrt{3}) - i(1+\sqrt{3}) + i\sqrt{3}(1-\sqrt{3}) - i^2(1-\sqrt{3})]$
$z = \frac{1}{4} [\sqrt{3} + 3 - i - i\sqrt{3} + i\sqrt{3} - 3i + 1 - \sqrt{3}]$
$z = \frac{1}{4} [4 - 4i] = 1 - i$
The point $(1, -1)$ corresponds to $1-i$ in the Argand plane.
Since the $x$-coordinate is positive and the $y$-coordinate is negative,the point lies in the $IV^{th}$ quadrant.

(A) Given that $\omega^{3} = 1$ and $1+\omega+\omega^{2} = 0$.
We know that $1+\omega^{2} = -\omega$ and $1+\omega = -\omega^{2}$.
Also,$\omega^{3} = 1, \omega^{4} = \omega, \omega^{8} = \omega^{2}, \omega^{16} = \omega$.
The expression is $P = (1-\omega+\omega^{2})(1-\omega^{2}+\omega)(1-\omega+\omega^{2})(1-\omega^{2}+\omega) \ldots$ ($2n$ factors).
Substituting $1+\omega^{2} = -\omega$ and $1+\omega = -\omega^{2}$:
$P = (-\omega-\omega)(-\omega^{2}-\omega^{2})(-\omega-\omega)(-\omega^{2}-\omega^{2}) \ldots$ ($2n$ factors).
$P = (-2\omega)(-2\omega^{2})(-2\omega)(-2\omega^{2}) \ldots$ ($2n$ factors).
There are $n$ pairs of $(-2\omega)(-2\omega^{2}) = 4\omega^{3} = 4(1) = 4$.
Thus,$P = (4)^{n} = 2^{2n}$.

(D) Given,$z = x + iy$ and $\left|\frac{z-1}{z+2i}\right| = 1$.
Taking the modulus on both sides,we have $|z-1| = |z+2i|$.
Substituting $z = x + iy$,we get $|(x-1) + iy| = |x + i(y+2)|$.
Squaring both sides,we get $(x-1)^2 + y^2 = x^2 + (y+2)^2$.
Expanding the squares,$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 4y + 4$.
Simplifying the equation,$-2x + 1 = 4y + 4$,which leads to $2x + 4y + 3 = 0$.
This is the equation of a straight line.

(C) The $n$th term of the series is given by $T_{n} = \frac{\sum_{k=1}^{n} k^{2}}{\sum_{k=1}^{n} k}$.
Using the formulas $\sum k^{2} = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$,we get:
$T_{n} = \frac{n(n+1)(2n+1)}{6} \times \frac{2}{n(n+1)} = \frac{2n+1}{3}$.
The sum of the first $n$ terms is $S_{n} = \sum_{k=1}^{n} T_{k} = \sum_{k=1}^{n} \frac{2k+1}{3}$.
$S_{n} = \frac{1}{3} \left( 2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \right)$.
$S_{n} = \frac{1}{3} \left( 2 \cdot \frac{n(n+1)}{2} + n \right) = \frac{1}{3} (n^{2} + n + n) = \frac{n^{2}+2n}{3} = \frac{n(n+2)}{3}$.

(A) To find the unit's digit of $7^{171} + (177)!$,we evaluate each term separately.
First,consider $7^{171}$. The powers of $7$ follow a cycle of $4$: $7^1 = 7$,$7^2 = 49$ (ends in $9$),$7^3 = 343$ (ends in $3$),$7^4 = 2401$ (ends in $1$).
We divide the exponent $171$ by $4$: $171 = 4 \times 42 + 3$.
Thus,the unit's digit of $7^{171}$ is the same as the unit's digit of $7^3$,which is $3$.
Second,consider $(177)!$. For any $n \ge 5$,$n!$ ends in $0$ because it contains the factors $2$ and $5$.
Since $177 \ge 5$,$(177)!$ ends in $0$.
Therefore,the unit's digit of $7^{171} + (177)!$ is $3 + 0 = 3$.

(A) The prime factorization of $242$ is $2 \times 11^2$.
The divisors of $242$ are $1, 2, 11, 22, 121, 242$.
The divisors excluding $1$ and $242$ are $2, 11, 22, 121$.
The sum of these divisors is $2 + 11 + 22 + 121 = 156$.

(D) Given,$(1+x+x^{2}+x^{3})^{n}=\sum_{r=0}^{3n} a_{r} x^{r}$ and $n$ is an odd positive integer.
To find the value of $a_{0}-a_{1}+a_{2}-a_{3}+\ldots-a_{3n}$,we substitute $x = -1$ into the given expansion.
Let $f(x) = (1+x+x^{2}+x^{3})^{n} = \sum_{r=0}^{3n} a_{r} x^{r}$.
Substituting $x = -1$:
$f(-1) = a_{0} - a_{1} + a_{2} - a_{3} + \ldots - a_{3n}$.
Now,evaluate $f(-1)$ using the expression $(1+x+x^{2}+x^{3})^{n}$:
$f(-1) = (1 + (-1) + (-1)^{2} + (-1)^{3})^{n}$
$f(-1) = (1 - 1 + 1 - 1)^{n}$
$f(-1) = (0)^{n}$.
Since $n$ is an odd positive integer,$0^{n} = 0$.
Therefore,$a_{0}-a_{1}+a_{2}-a_{3}+\ldots-a_{3n} = 0$.

(B) The general term of the expansion $(p+q)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} p^{n-k} q^{k}$.
Given that the $r$-th term and $(r+1)$-th term are equal,we have $T_{r} = T_{r+1}$.
$T_{r} = {}^{n}C_{r-1} p^{n-r+1} q^{r-1}$ and $T_{r+1} = {}^{n}C_{r} p^{n-r} q^{r}$.
Equating them: ${}^{n}C_{r-1} p^{n-r+1} q^{r-1} = {}^{n}C_{r} p^{n-r} q^{r}$.
Dividing both sides by ${}^{n}C_{r-1} p^{n-r} q^{r-1}$,we get $p = \frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} q$.
Using the property $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we have $p = \frac{n-r+1}{r} q$.
Rearranging gives $pr = (n-r+1)q = nq - rq + q = q(n+1) - rq$.
Thus,$pr + rq = q(n+1)$,which implies $r(p+q) = q(n+1)$.
Therefore,$\frac{q(n+1)}{r(p+q)} = 1$.

(D) We know that $\sin 30^{\circ} = \frac{1}{2}$.
Let $E = \sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}$.
$E = \frac{1}{2} \cdot (\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ})$.
Using the identity $\sin \theta \cdot \sin(60^{\circ}-\theta) \cdot \sin(60^{\circ}+\theta) = \frac{1}{4} \sin 3\theta$,where $\theta = 10^{\circ}$:
$\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ} = \sin 10^{\circ} \cdot \sin(60^{\circ}-10^{\circ}) \cdot \sin(60^{\circ}+10^{\circ}) = \frac{1}{4} \sin(3 \times 10^{\circ}) = \frac{1}{4} \sin 30^{\circ} = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$.
Therefore,$E = \frac{1}{2} \cdot \frac{1}{8} = \frac{1}{16}$.

(B) Given equation: $\sin 5\theta - \sin 3\theta + \sin \theta = 0$ for $\theta \in (0, \frac{\pi}{2})$.
Rearranging the terms: $(\sin 5\theta + \sin \theta) = \sin 3\theta$.
Using the sum-to-product formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$2 \sin 3\theta \cos 2\theta = \sin 3\theta$.
$2 \sin 3\theta \cos 2\theta - \sin 3\theta = 0$.
$\sin 3\theta (2 \cos 2\theta - 1) = 0$.
This gives two cases:
Case $1$: $\sin 3\theta = 0$ $\Rightarrow 3\theta = n\pi$ $\Rightarrow \theta = \frac{n\pi}{3}$. For $0 < \theta < \frac{\pi}{2}$,$\theta = \frac{\pi}{3}$.
Case $2$: $2 \cos 2\theta - 1 = 0 \Rightarrow \cos 2\theta = \frac{1}{2} = \cos \frac{\pi}{3}$.
$2\theta = 2n\pi \pm \frac{\pi}{3} \Rightarrow \theta = n\pi \pm \frac{\pi}{6}$. For $0 < \theta < \frac{\pi}{2}$,$\theta = \frac{\pi}{6}$.
Thus,the values of $\theta$ are $\frac{\pi}{6}$ and $\frac{\pi}{3}$. Comparing with the options,$\frac{\pi}{6}$ is the correct choice.

(C) Given points are $A(1, 2)$,$B(2, 4)$,and $C(4, 8)$.
To check if they are collinear,we calculate the slopes of the line segments formed by these points.
Slope of $AB = \frac{4 - 2}{2 - 1} = \frac{2}{1} = 2$.
Slope of $BC = \frac{8 - 4}{4 - 2} = \frac{4}{2} = 2$.
Since the slope of $AB$ is equal to the slope of $BC$ and they share a common point $B$,the points $A$,$B$,and $C$ lie on the same straight line.
Therefore,the points form a straight line.

(B) The given lines are:
$x+3y-6=0$
$2x+y-4=0$
$kx-3y+1=0$
These lines are concurrent if the determinant of their coefficients is zero:
$\left|\begin{array}{ccc} 1 & 3 & -6 \\ 2 & 1 & -4 \\ k & -3 & 1 \end{array}\right| = 0$
Expanding along the first row $(R_1)$:
$1(1(1) - (-3)(-4)) - 3(2(1) - k(-4)) - 6(2(-3) - k(1)) = 0$
$1(1-12) - 3(2+4k) - 6(-6-k) = 0$
$-11 - 6 - 12k + 36 + 6k = 0$
$-6k + 19 = 0$
$6k = 19$
$k = \frac{19}{6}$

(D) Let the required circle be $S: x^{2}+y^{2}+2gx+2fy+c=0$.
For a circle $x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$,the condition for orthogonality is $2gg_{i}+2ff_{i}=c+c_{i}$.
For the given circles:
$S_{1}: x^{2}+y^{2}+6x+0y-1=0 \implies 2g(3)+2f(0)=c-1 \implies 6g=c-1$ $(i)$
$S_{2}: x^{2}+y^{2}+0x-3y+2=0 \implies 2g(0)+2f(-3/2)=c+2 \implies -3f=c+2$ (ii)
$S_{3}: x^{2}+y^{2}+x+y-3=0 \implies 2g(1/2)+2f(1/2)=c-3 \implies g+f=c-3$ (iii)
Subtracting (ii) from $(i)$: $6g+3f=-3 \implies 2g+f=-1$ (iv)
Subtracting (iii) from $(i)$: $5g-f=2$ $(v)$
Adding (iv) and $(v)$: $7g=1 \implies g=1/7$.
Substituting $g$ in (iv): $2(1/7)+f=-1 \implies f=-1-2/7 = -9/7$.
The centre of the circle is $(-g, -f) = (-1/7, 9/7)$.

(C) The given equations of the circles are:
$C_1: x^2 + y^2 - 6x - 8y + 9 = 0$
$C_2: x^2 + y^2 = 1$
For $C_1$,the center is $(3, 4)$ and the radius $R_1 = \sqrt{3^2 + 4^2 - 9} = \sqrt{9 + 16 - 9} = \sqrt{16} = 4$.
For $C_2$,the center is $(0, 0)$ and the radius $R_2 = 1$.
The distance between the centers $C_1(3, 4)$ and $C_2(0, 0)$ is $d = \sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = 5$.
We observe that $R_1 + R_2 = 4 + 1 = 5$.
Since the distance between the centers $d = R_1 + R_2$,the two circles touch each other externally.
When two circles touch each other externally,they have $3$ common tangents (two direct common tangents and one transverse common tangent).

(B) Given equation of the conic is $3x^{2} - 4y + 6x - 3 = 0$.
Rearranging the terms,we get $3x^{2} + 6x = 4y + 3$.
Taking $3$ as a common factor,$3(x^{2} + 2x) = 4y + 3$.
Completing the square inside the bracket,$3(x^{2} + 2x + 1 - 1) = 4y + 3$.
$3((x + 1)^{2} - 1) = 4y + 3$.
$3(x + 1)^{2} - 3 = 4y + 3$.
$3(x + 1)^{2} = 4y + 6$.
$(x + 1)^{2} = \frac{4}{3}(y + \frac{6}{4}) = \frac{4}{3}(y + \frac{3}{2})$.
Comparing this with the standard form of a parabola $X^{2} = 4aY$,where $X = x + 1$ and $Y = y + \frac{3}{2}$,we have $4a = \frac{4}{3}$.
The length of the latus rectum is $4a = \frac{4}{3}$.

(A) Let the coordinates of points $P$ and $Q$ on the parabola $y^{2} = 4ax$ be $(at_{1}^{2}, 2at_{1})$ and $(at_{2}^{2}, 2at_{2})$ respectively.
Since $PQ$ is a focal chord,$t_{1}t_{2} = -1$.
The focal distances of points $P$ and $Q$ are $r_{1} = a(1 + t_{1}^{2})$ and $r_{2} = a(1 + t_{2}^{2})$.
The sum of the reciprocals is $\frac{1}{r_{1}} + \frac{1}{r_{2}} = \frac{1}{a(1 + t_{1}^{2})} + \frac{1}{a(1 + t_{2}^{2})}$.
Substituting $t_{2} = -\frac{1}{t_{1}}$,we get $\frac{1}{a(1 + t_{1}^{2})} + \frac{1}{a(1 + \frac{1}{t_{1}^{2}})} = \frac{1}{a(1 + t_{1}^{2})} + \frac{t_{1}^{2}}{a(t_{1}^{2} + 1)}$.
$= \frac{1 + t_{1}^{2}}{a(1 + t_{1}^{2})} = \frac{1}{a}$.

(D) The given curve is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. The point $P$ is $\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$.
Taking the derivative,$\frac{2x}{a^{2}} + \frac{2y}{b^{2}}\frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{b^{2}x}{a^{2}y}$.
At $P$,the slope of the tangent $m_{t} = -\frac{b^{2}(a/\sqrt{2})}{a^{2}(b/\sqrt{2})} = -\frac{b}{a}$.
The equation of the tangent at $P$ is $y - \frac{b}{\sqrt{2}} = -\frac{b}{a}\left(x - \frac{a}{\sqrt{2}}\right)$,which simplifies to $bx + ay = \sqrt{2}ab$.
For the $x$-axis $(y=0)$,$x = \sqrt{2}a$. So,the first vertex is $A(\sqrt{2}a, 0)$.
The slope of the normal $m_{n} = \frac{a}{b}$.
The equation of the normal at $P$ is $y - \frac{b}{\sqrt{2}} = \frac{a}{b}\left(x - \frac{a}{\sqrt{2}}\right)$,which simplifies to $ax - by = \frac{a^{2}-b^{2}}{\sqrt{2}}$.
For the $x$-axis $(y=0)$,$x = \frac{a^{2}-b^{2}}{a\sqrt{2}}$. So,the second vertex is $B\left(\frac{a^{2}-b^{2}}{a\sqrt{2}}, 0\right)$.
The third vertex is $P\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)$.
The area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \left|\sqrt{2}a - \frac{a^{2}-b^{2}}{a\sqrt{2}}\right| \times \frac{b}{\sqrt{2}}$.
$= \frac{1}{2} \times \left|\frac{2a^{2}-a^{2}+b^{2}}{a\sqrt{2}}\right| \times \frac{b}{\sqrt{2}} = \frac{1}{2} \times \frac{a^{2}+b^{2}}{a\sqrt{2}} \times \frac{b}{\sqrt{2}} = \frac{b(a^{2}+b^{2})}{4a}$.

(D) The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 + b^2$.
Given the line $x \cos \alpha + y \sin \alpha = 4$,we can rewrite it as $y = -x \cot \alpha + 4 \operatorname{cosec} \alpha$.
Here,$m = -\cot \alpha$,$c = 4 \operatorname{cosec} \alpha$,$a^2 = 25$,and $b^2 = 9$.
Substituting these into the condition $c^2 = a^2m^2 + b^2$:
$16 \operatorname{cosec}^2 \alpha = 25 \cot^2 \alpha + 9$.
Using $\operatorname{cosec}^2 \alpha = 1 + \cot^2 \alpha$,we get $16(1 + \cot^2 \alpha) = 25 \cot^2 \alpha + 9$.
$16 + 16 \cot^2 \alpha = 25 \cot^2 \alpha + 9$.
$9 \cot^2 \alpha = 7 \implies \cot^2 \alpha = 7/9$.
Thus,$\tan^2 \alpha = 9/7$,which gives $\tan \alpha = 3/\sqrt{7}$.
Therefore,$\alpha = \tan^{-1}(3/\sqrt{7})$.

(C) The equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
The coordinates of the foci are $S(ae, 0)$ and $S^{\prime}(-ae, 0)$.
Let the point $P$ on the ellipse be $(a \cos \theta, b \sin \theta)$.
The area of $\triangle S P S^{\prime}$ is given by the determinant formula:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$\text{Area} = \frac{1}{2} |ae(b \sin \theta - 0) + a \cos \theta(0 - 0) + (-ae)(0 - b \sin \theta)|$
$\text{Area} = \frac{1}{2} |aeb \sin \theta + aeb \sin \theta| = |abe \sin \theta|$.
Since the maximum value of $\sin \theta$ is $1$,the maximum area of $\triangle S P S^{\prime}$ is $abe$.

(B) Let the point be $P(h, k)$.
$P_{1}$ is the length of the perpendicular from $P$ to the $x$-axis $(y=0)$,so $P_{1} = |k|$.
$P_{2}$ is the length of the perpendicular from $P$ to the line $x-y=0$,so $P_{2} = \frac{|h-k|}{\sqrt{1^{2}+(-1)^{2}}} = \frac{|h-k|}{\sqrt{2}}$.
Given that $P_{1} = 2P_{2}$,we have:
$|k| = 2 \cdot \frac{|h-k|}{\sqrt{2}}$
$|k| = \sqrt{2} |h-k|$
Squaring both sides:
$k^{2} = 2(h-k)^{2}$
$k^{2} = 2(h^{2} + k^{2} - 2hk)$
$k^{2} = 2h^{2} + 2k^{2} - 4hk$
$2h^{2} - 4hk + k^{2} = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is:
$2x^{2} - 4xy + y^{2} = 0$

(A) For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$,we have $a^2=16$ and $b^2=4$.
The eccentricity $e_1$ is given by $b^2 = a^2(1-e_1^2)$,so $4 = 16(1-e_1^2)$,which gives $1-e_1^2 = \frac{1}{4}$,so $e_1^2 = \frac{3}{4}$,$e_1 = \frac{\sqrt{3}}{2}$.
The foci are $(\pm a_1 e_1, 0) = (\pm 4 \cdot \frac{\sqrt{3}}{2}, 0) = (\pm 2\sqrt{3}, 0)$.
For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{9}=1$,we have $A^2=a^2$ and $B^2=9$.
The eccentricity $e_2$ is given by $B^2 = A^2(e_2^2-1)$,so $9 = a^2(e_2^2-1) = a^2 e_2^2 - a^2$.
The foci are $(\pm A e_2, 0) = (\pm a e_2, 0)$.
Since the foci coincide,$a e_2 = 2\sqrt{3}$,so $a^2 e_2^2 = 12$.
Substituting this into the hyperbola equation: $9 = 12 - a^2$.
Thus,$a^2 = 3$,which implies $a = \sqrt{3}$.

(D) The equation of the asymptotes of the hyperbola is given by $3x \pm 5y = 0$.
The equation of the hyperbola can be written as $(3x + 5y)(3x - 5y) = \lambda$,where $\lambda$ is a constant.
This simplifies to $9x^2 - 25y^2 = \lambda$.
Since the vertices of the hyperbola are $(\pm 5, 0)$,the point $(5, 0)$ must satisfy the equation of the hyperbola.
Substituting $(x, y) = (5, 0)$ into the equation $9x^2 - 25y^2 = \lambda$,we get:
$9(5)^2 - 25(0)^2 = \lambda$
$9(25) = \lambda$
$\lambda = 225$
Therefore,the equation of the hyperbola is $9x^2 - 25y^2 = 225$.

(D) To evaluate the limit $L = \lim _{x \rightarrow a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}$,we rationalize the numerator and the denominator:
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{a+2x} + \sqrt{3x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{3a+x} - 2\sqrt{x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{(a+2x) - 3x}{(3a+x) - 4x} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{a-x}{3a-3x} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{a-x}{3(a-x)} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \frac{1}{3} \cdot \frac{\sqrt{3a+a} + 2\sqrt{a}}{\sqrt{a+2a} + \sqrt{3a}} = \frac{1}{3} \cdot \frac{\sqrt{4a} + 2\sqrt{a}}{\sqrt{3a} + \sqrt{3a}}$
$L = \frac{1}{3} \cdot \frac{2\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \cdot \frac{4\sqrt{a}}{2\sqrt{3}\sqrt{a}} = \frac{2}{3\sqrt{3}}$

(D) To find the negation of $p \rightarrow (\sim p \vee q)$,we use the logical equivalence $\sim(A \rightarrow B) \equiv A \wedge \sim B$.
Here,$A = p$ and $B = (\sim p \vee q)$.
Therefore,$\sim[p \rightarrow (\sim p \vee q)] \equiv p \wedge \sim(\sim p \vee q)$.
Using De Morgan's Law,$\sim(\sim p \vee q) \equiv \sim(\sim p) \wedge \sim q \equiv p \wedge \sim q$.
Substituting this back,we get $p \wedge (p \wedge \sim q)$.
Since $p \wedge p \equiv p$,the expression simplifies to $p \wedge \sim q$.
Thus,the negation is $p \wedge \sim q$.

(A) Let the height of the tower be $h$ and the foot of the tower be $P$. Let $CP = x$,$BC = y$,and $AB = z$.
In $\triangle QCP$,$\tan 60^{\circ} = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$.
In $\triangle QBP$,$\tan 45^{\circ} = \frac{h}{x+y}$ $\Rightarrow x+y = h$ $\Rightarrow y = h - \frac{h}{\sqrt{3}} = h\left(1 - \frac{1}{\sqrt{3}}\right) = h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)$.
In $\triangle QAP$,$\tan 30^{\circ} = \frac{h}{x+y+z}$ $\Rightarrow x+y+z = h\sqrt{3}$ $\Rightarrow z = h\sqrt{3} - (x+y) = h\sqrt{3} - h = h(\sqrt{3}-1)$.
Now,the ratio $\frac{AB}{BC} = \frac{z}{y} = \frac{h(\sqrt{3}-1)}{h\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)} = \sqrt{3}$.
Thus,$AB: BC = \sqrt{3}: 1$.

(C) Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R = k$,so $\sin A = ak$ and $\sin B = bk$.
We know $\cos 2A = 1 - 2\sin^2 A$ and $\cos 2B = 1 - 2\sin^2 B$.
Substituting these into the expression:
$\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2}$
$= \frac{1 - 2(ak)^2}{a^2} - \frac{1 - 2(bk)^2}{b^2}$
$= \frac{1 - 2a^2k^2}{a^2} - \frac{1 - 2b^2k^2}{b^2}$
$= (\frac{1}{a^2} - 2k^2) - (\frac{1}{b^2} - 2k^2)$
$= \frac{1}{a^2} - \frac{1}{b^2}$

(D) The given equations of the circles are:
$2 x^{2}+2 y^{2}+4 x+5 y+1=0 \Rightarrow x^{2}+y^{2}+2 x+\frac{5}{2} y+\frac{1}{2}=0 \quad \dots (i)$
$3 x^{2}+3 y^{2}+6 x-7 y+3 k=0 \Rightarrow x^{2}+y^{2}+2 x-\frac{7}{3} y+k=0 \quad \dots (ii)$
Comparing these with the standard form $x^{2}+y^{2}+2 g x+2 f y+c=0$,we get:
For $(i): g_{1}=1, f_{1}=\frac{5}{4}, c_{1}=\frac{1}{2}$
For $(ii): g_{2}=1, f_{2}=-\frac{7}{6}, c_{2}=k$
The condition for two circles to be orthogonal is $2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2}$.
Substituting the values:
$2(1)(1)+2\left(\frac{5}{4}\right)\left(-\frac{7}{6}\right)=\frac{1}{2}+k$
$2-\frac{35}{12}=\frac{1}{2}+k$
$\frac{24-35}{12}=\frac{1}{2}+k$
$-\frac{11}{12}=\frac{1}{2}+k$
$k=-\frac{11}{12}-\frac{6}{12}=-\frac{17}{12}$

(C) Given equations of the curves are:
$y^{2}=4x$ $(i)$
$x^{2}+y^{2}=12$ (ii)
First,find the intersection points by substituting $(i)$ into (ii):
$x^{2}+4x=12 \Rightarrow x^{2}+4x-12=0$
$(x+6)(x-2)=0$
Since $y^{2}=4x$,$x$ must be $\ge 0$,so $x=2$.
For $x=2$,$y^{2}=8 \Rightarrow y=\pm 2\sqrt{2}$.
Intersection points are $(2, 2\sqrt{2})$ and $(2, -2\sqrt{2})$.
Now,find the slopes $m_{1}$ and $m_{2}$ by differentiating:
For $(i)$: $2y \frac{dy}{dx} = 4 \Rightarrow m_{1} = \frac{2}{y}$.
For (ii): $2x + 2y \frac{dy}{dx} = 0 \Rightarrow m_{2} = -\frac{x}{y}$.
At $(2, 2\sqrt{2})$:
$m_{1} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$ and $m_{2} = -\frac{2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}$.
The angle $\theta$ is given by $\tan \theta = \left| \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right|$.
$\tan \theta = \left| \frac{\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})}{1 + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})} \right| = \left| \frac{\frac{2}{\sqrt{2}}}{1 - \frac{1}{2}} \right| = \left| \frac{\sqrt{2}}{\frac{1}{2}} \right| = 2\sqrt{2}$.
Thus,$\theta = \tan^{-1}(2\sqrt{2})$.

(B) Given,$\omega^{3}=1$ and $1+\omega+\omega^{2}=0$.
The determinant is $\Delta = \left|\begin{array}{ccc} 1 & \omega^{2} & 1-\omega \\ \omega & 1 & 1+\omega^{2} \\ 1 & \omega & \omega^{2} \end{array}\right|$.
Using $1+\omega^{2} = -\omega$,we get $\Delta = \left|\begin{array}{ccc} 1 & \omega^{2} & 1-\omega \\ \omega & 1 & -\omega \\ 1 & \omega & \omega^{2} \end{array}\right|$.
Expanding along $R_{1}$:
$\Delta = 1(\omega^{2} - (-\omega^{2})) - \omega^{2}(\omega^{3} - (-\omega)) + (1-\omega)(\omega^{2} - 1)$
$\Delta = 1(2\omega^{2}) - \omega^{2}(1+\omega) + (\omega^{2} - 1 - \omega^{3} + \omega)$
$\Delta = 2\omega^{2} - (\omega^{2} + \omega^{3}) + (\omega^{2} - 1 - 1 + \omega)$
$\Delta = 2\omega^{2} - \omega^{2} - 1 + \omega^{2} - 2 + \omega$
$\Delta = 2\omega^{2} + \omega - 3$.
Since $1+\omega+\omega^{2}=0$,we have $\omega = -1-\omega^{2}$.
$\Delta = 2\omega^{2} + (-1-\omega^{2}) - 3 = \omega^{2} - 4$.

(B) Given,$y = \sin x \cdot \sin 2x \cdot \sin 3x \cdot \ldots \cdot \sin nx$.
Taking the natural logarithm on both sides:
$\ln y = \ln(\sin x) + \ln(\sin 2x) + \ln(\sin 3x) + \ldots + \ln(\sin nx)$
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{1}{y} \cdot y^{\prime} = \frac{d}{dx}(\ln \sin x) + \frac{d}{dx}(\ln \sin 2x) + \ldots + \frac{d}{dx}(\ln \sin nx)$
$\frac{y^{\prime}}{y} = \frac{\cos x}{\sin x} + 2 \cdot \frac{\cos 2x}{\sin 2x} + 3 \cdot \frac{\cos 3x}{\sin 3x} + \ldots + n \cdot \frac{\cos nx}{\sin nx}$
$\frac{y^{\prime}}{y} = \sum_{k=1}^{n} k \cot kx$
Therefore,$y^{\prime} = y \cdot \sum_{k=1}^{n} k \cot kx$.

(C) Given binary operation is $a * b = \frac{3ab}{2}$.
First,we find the identity element $e$ such that $a * e = a$.
$\frac{3ae}{2} = a \implies e = \frac{2}{3}$.
Now,we find the inverse $3^{-1}$ such that $3 * 3^{-1} = e = \frac{2}{3}$.
$\frac{3 \cdot 3 \cdot 3^{-1}}{2} = \frac{2}{3} \implies \frac{9 \cdot 3^{-1}}{2} = \frac{2}{3} \implies 3^{-1} = \frac{4}{27}$.
Also,$2 * x = \frac{3 \cdot 2 \cdot x}{2} = 3x$.
Given equation: $(2 * x) * 3^{-1} = 4^{-1}$.
Note that $4^{-1}$ is the inverse of $4$,so $4 * 4^{-1} = \frac{2}{3} \implies \frac{3 \cdot 4 \cdot 4^{-1}}{2} = \frac{2}{3} \implies 6 \cdot 4^{-1} = \frac{2}{3} \implies 4^{-1} = \frac{1}{9}$.
Substituting these into the equation: $(3x) * \frac{4}{27} = \frac{1}{9}$.
$\frac{3 \cdot (3x) \cdot (4/27)}{2} = \frac{1}{9}$.
$\frac{36x}{54} = \frac{1}{9} \implies \frac{2x}{3} = \frac{1}{9}$.
$x = \frac{3}{18} = \frac{1}{6}$.

(B) Let $A = \begin{bmatrix} x & x \\ x & x \end{bmatrix}$. The identity element $E = \begin{bmatrix} e & e \\ e & e \end{bmatrix}$ must satisfy $AE = A$.
$\begin{bmatrix} x & x \\ x & x \end{bmatrix} \begin{bmatrix} e & e \\ e & e \end{bmatrix} = \begin{bmatrix} 2xe & 2xe \\ 2xe & 2xe \end{bmatrix} = \begin{bmatrix} x & x \\ x & x \end{bmatrix}$.
Thus,$2xe = x \implies e = 1/2$. The identity element is $E = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}$.
Let the inverse of $A = \begin{bmatrix} 1/3 & 1/3 \\ 1/3 & 1/3 \end{bmatrix}$ be $A^{-1} = \begin{bmatrix} y & y \\ y & y \end{bmatrix}$.
Then $AA^{-1} = E \implies \begin{bmatrix} 1/3 & 1/3 \\ 1/3 & 1/3 \end{bmatrix} \begin{bmatrix} y & y \\ y & y \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}$.
Multiplying the matrices: $\begin{bmatrix} 2y/3 & 2y/3 \\ 2y/3 & 2y/3 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}$.
Equating the elements: $2y/3 = 1/2 \implies y = 3/4$.
Therefore,$A^{-1} = \begin{bmatrix} 3/4 & 3/4 \\ 3/4 & 3/4 \end{bmatrix}$.

(A) Given set $A = \{1, 2, 3, 4\}$ and relation $x R y$ if $x$ divides $y$.
The relation $R$ is given by: $R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (1, 3), (1, 4), (2, 4)\}$.
$1$. Reflexive: For any $x \in A$,$x$ divides $x$ (i.e.,$x/x = 1$),so $(x, x) \in R$. Thus,$R$ is reflexive.
$2$. Symmetric: For $R$ to be symmetric,$(x, y) \in R$ must imply $(y, x) \in R$. Here,$(1, 2) \in R$ but $(2, 1) \notin R$. Thus,$R$ is not symmetric.
$3$. Transitive: If $(x, y) \in R$ and $(y, z) \in R$,then $x$ divides $y$ and $y$ divides $z$. This implies $x$ divides $z$,so $(x, z) \in R$. Thus,$R$ is transitive.
Therefore,$R$ is reflexive and transitive.

(A) Given,$A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
The transpose of matrix $A$ is $A^{\prime} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
Now,calculate the product $A \cdot A^{\prime}$:
$A \cdot A^{\prime} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$
Performing matrix multiplication:
$A \cdot A^{\prime} = \begin{bmatrix} (\cos \theta)(\cos \theta) + (\sin \theta)(\sin \theta) & (\cos \theta)(-\sin \theta) + (\sin \theta)(\cos \theta) \\ (-\sin \theta)(\cos \theta) + (\cos \theta)(\sin \theta) & (-\sin \theta)(-\sin \theta) + (\cos \theta)(\cos \theta) \end{bmatrix}$
Using the trigonometric identity $\cos^{2} \theta + \sin^{2} \theta = 1$:
$A \cdot A^{\prime} = \begin{bmatrix} \cos^{2} \theta + \sin^{2} \theta & -\sin \theta \cos \theta + \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \sin^{2} \theta + \cos^{2} \theta \end{bmatrix}$
$A \cdot A^{\prime} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
Thus,$A \cdot A^{\prime} = I$,which is the identity matrix.

(D) Given that $A$ and $B$ are symmetric matrices of the same order,we have $A^T = A$ and $B^T = B$.
$(i)$ $(A+B)^T = A^T + B^T = A+B$. Thus,$A+B$ is symmetric.
(ii) $(A-B)^T = A^T - B^T = A-B$. Thus,$A-B$ is symmetric.
(iii) $(AB+BA)^T = (AB)^T + (BA)^T = B^T A^T + A^T B^T = BA + AB = AB+BA$. Thus,$AB+BA$ is symmetric.
(iv) $(AB-BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T = BA - AB = -(AB-BA)$. Thus,$AB-BA$ is skew-symmetric,not symmetric.
Therefore,the statement '$AB-BA$ is symmetric' is not true.

(A) matrix $A$ is singular if its determinant $|A| = 0$.
Given $A = \begin{bmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{bmatrix}$.
Setting the determinant to zero:
$\begin{vmatrix} 1 & 2 & -1 \\ 1 & x-2 & 1 \\ x & 1 & 1 \end{vmatrix} = 0$
Expanding along the first row $(R_1)$:
$1((x-2)(1) - (1)(1)) - 2((1)(1) - (1)(x)) + (-1)((1)(1) - (x)(x-2)) = 0$
$1(x-2-1) - 2(1-x) - 1(1 - (x^2 - 2x)) = 0$
$(x-3) - 2 + 2x - 1 + x^2 - 2x = 0$
$x^2 + x - 6 = 0$
Factoring the quadratic equation:
$(x+3)(x-2) = 0$
Therefore,$x = -3$ or $x = 2$.

(A) Given the determinant $\Delta = \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin (\alpha+\delta) \\ \sin \beta & \cos \beta & \sin (\beta+\delta) \\ \sin \gamma & \cos \gamma & \sin (\gamma+\delta)\end{array}\right|$.
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can expand the third column:
$\Delta = \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cos \delta + \cos \alpha \sin \delta \\ \sin \beta & \cos \beta & \sin \beta \cos \delta + \cos \beta \sin \delta \\ \sin \gamma & \cos \gamma & \sin \gamma \cos \delta + \cos \gamma \sin \delta\end{array}\right|$.
By the property of determinants,we can split this into the sum of two determinants:
$\Delta = \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \cos \delta \\ \sin \beta & \cos \beta & \sin \beta \cos \delta \\ \sin \gamma & \cos \gamma & \sin \gamma \cos \delta\end{array}\right| + \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos \alpha \sin \delta \\ \sin \beta & \cos \beta & \cos \beta \sin \delta \\ \sin \gamma & \cos \gamma & \cos \gamma \sin \delta\end{array}\right|$.
Factoring out $\cos \delta$ from the third column of the first determinant and $\sin \delta$ from the third column of the second determinant:
$\Delta = \cos \delta \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \sin \alpha \\ \sin \beta & \cos \beta & \sin \beta \\ \sin \gamma & \cos \gamma & \sin \gamma\end{array}\right| + \sin \delta \left|\begin{array}{ccc}\sin \alpha & \cos \alpha & \cos \alpha \\ \sin \beta & \cos \beta & \cos \beta \\ \sin \gamma & \cos \gamma & \cos \gamma\end{array}\right|$.
In the first determinant,column $1$ and column $3$ are identical,so its value is $0$.
In the second determinant,column $2$ and column $3$ are identical,so its value is $0$.
Therefore,$\Delta = \cos \delta (0) + \sin \delta (0) = 0$.

(C) The given function is $f(x) = \sin^{-1}\left[\log_{2}\left(\frac{x}{2}\right)\right]$.
For the function $\sin^{-1}(u)$ to be defined,the argument $u$ must satisfy $-1 \leq u \leq 1$.
Therefore,we must have $-1 \leq \log_{2}\left(\frac{x}{2}\right) \leq 1$.
Applying the definition of the logarithm,this inequality is equivalent to $2^{-1} \leq \frac{x}{2} \leq 2^{1}$.
This simplifies to $\frac{1}{2} \leq \frac{x}{2} \leq 2$.
Multiplying the entire inequality by $2$,we get $1 \leq x \leq 4$.
Thus,the domain of the function is $x \in [1, 4]$.

(B) Given equation: $\tan ^{-1} x = \frac{\pi}{4} - \tan ^{-1} \left( \frac{1}{3} \right)$
Since $\frac{\pi}{4} = \tan ^{-1}(1)$,we can write:
$\tan ^{-1} x = \tan ^{-1}(1) - \tan ^{-1} \left( \frac{1}{3} \right)$
Using the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left( \frac{A - B}{1 + AB} \right)$:
$\tan ^{-1} x = \tan ^{-1} \left( \frac{1 - \frac{1}{3}}{1 + 1 \times \frac{1}{3}} \right)$
$\tan ^{-1} x = \tan ^{-1} \left( \frac{\frac{2}{3}}{\frac{4}{3}} \right)$
$\tan ^{-1} x = \tan ^{-1} \left( \frac{2}{4} \right)$
$\tan ^{-1} x = \tan ^{-1} \left( \frac{1}{2} \right)$
Therefore,$x = \frac{1}{2}$.

(C) Given that the function $f(x)$ is continuous at $x=1$,the condition for continuity is $f(1) = \lim_{x \to 1} f(x)$.
From the definition of the function,$f(1) = k$.
Therefore,we need to evaluate the limit: $k = \lim_{x \to 1} \frac{\log x}{x-1}$.
Substituting $x=1$,we get the indeterminate form $\frac{0}{0}$.
Applying $L$'$H$ôpital's rule by differentiating the numerator and the denominator with respect to $x$:
$k = \lim_{x \to 1} \frac{\frac{d}{dx}(\log x)}{\frac{d}{dx}(x-1)} = \lim_{x \to 1} \frac{1/x}{1}$.
Evaluating the limit as $x \to 1$:
$k = \frac{1/1}{1} = 1$.
Thus,the value of $k$ is $1$.

(A) Let $y = \cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)$.
Put $x = 2 \cos \theta$,then $\cos \theta = \frac{x}{2}$.
Then,$\sqrt{\frac{2+x}{2-x}} = \sqrt{\frac{2+2\cos \theta}{2-2\cos \theta}} = \sqrt{\frac{2(1+\cos \theta)}{2(1-\cos \theta)}} = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}} = \cot(\theta/2)$.
Substituting this into the expression,we get $y = \cos^2(\cot^{-1}(\cot(\theta/2))) = \cos^2(\theta/2)$.
Using the identity $\cos^2(\theta/2) = \frac{1+\cos \theta}{2}$,we have $y = \frac{1}{2} + \frac{1}{2}\cos \theta$.
Since $\cos \theta = \frac{x}{2}$,we have $y = \frac{1}{2} + \frac{x}{4}$.
Now,differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2} + \frac{x}{4}) = 0 + \frac{1}{4} = \frac{1}{4}$.

(C) Given,$f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+\tan x}$.
Simplify the expression:
$f(x) = \frac{\sin^{2} x}{1+\frac{\cos x}{\sin x}} + \frac{\cos^{2} x}{1+\frac{\sin x}{\cos x}}$
$f(x) = \frac{\sin^{3} x}{\sin x + \cos x} + \frac{\cos^{3} x}{\cos x + \sin x}$
$f(x) = \frac{\sin^{3} x + \cos^{3} x}{\sin x + \cos x}$
Using the identity $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$:
$f(x) = \frac{(\sin x + \cos x)(\sin^{2} x - \sin x \cos x + \cos^{2} x)}{\sin x + \cos x}$
$f(x) = \sin^{2} x - \sin x \cos x + \cos^{2} x$
$f(x) = 1 - \sin x \cos x = 1 - \frac{1}{2} \sin(2x)$.
Now,differentiate with respect to $x$:
$f^{\prime}(x) = -\frac{1}{2} \cdot \cos(2x) \cdot 2 = -\cos(2x)$.
Substitute $x = \frac{\pi}{4}$:
$f^{\prime}\left(\frac{\pi}{4}\right) = -\cos\left(2 \cdot \frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{2}\right) = 0$.

(B) Let $u = \tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)$ and $v = \tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$.
For $u$:
$u = \tan ^{-1}\left(\frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)}\right) = \tan ^{-1}(\tan(x/2)) = x/2$.
Thus,$\frac{du}{dx} = \frac{1}{2}$.
For $v$:
$v = \tan ^{-1}\left(\frac{\cos^2(x/2) - \sin^2(x/2)}{(\cos(x/2) + \sin(x/2))^2}\right) = \tan ^{-1}\left(\frac{(\cos(x/2) - \sin(x/2))(\cos(x/2) + \sin(x/2))}{(\cos(x/2) + \sin(x/2))^2}\right)$.
$v = \tan ^{-1}\left(\frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)}\right) = \tan ^{-1}\left(\frac{1 - \tan(x/2)}{1 + \tan(x/2)}\right) = \tan ^{-1}\left(\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) = \frac{\pi}{4} - \frac{x}{2}$.
Thus,$\frac{dv}{dx} = -\frac{1}{2}$.
Now,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/2}{-1/2} = -1$.

(D) Given,$\sqrt{r} = a e^{\theta \cot \alpha}$
Squaring both sides,we get $r = a^{2} e^{2 \theta \cot \alpha}$.
Differentiating with respect to $\theta$:
$\frac{dr}{d\theta} = a^{2} \cdot e^{2 \theta \cot \alpha} \cdot (2 \cot \alpha) = 2 a^{2} \cot \alpha \cdot e^{2 \theta \cot \alpha}$.
Differentiating again with respect to $\theta$:
$\frac{d^{2}r}{d\theta^{2}} = 2 a^{2} \cot \alpha \cdot e^{2 \theta \cot \alpha} \cdot (2 \cot \alpha) = 4 a^{2} \cot^{2} \alpha \cdot e^{2 \theta \cot \alpha}$.
Since $r = a^{2} e^{2 \theta \cot \alpha}$,we can substitute this into the expression:
$\frac{d^{2}r}{d\theta^{2}} = 4 \cot^{2} \alpha \cdot (a^{2} e^{2 \theta \cot \alpha}) = 4 r \cot^{2} \alpha$.
Therefore,$\frac{d^{2}r}{d\theta^{2}} - 4 r \cot^{2} \alpha = 0$.

(C) The given differential equation is $\left(1+\left(\frac{dy}{dx}\right)^{2}\right)^{2/3} = \frac{d^{2}y}{dx^{2}}$.
To find the degree,we must eliminate the fractional exponent by cubing both sides:
$\left(1+\left(\frac{dy}{dx}\right)^{2}\right)^{2} = \left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.
The order $n$ is the highest derivative present,which is $\frac{d^{2}y}{dx^{2}}$,so $n = 2$.
The degree $m$ is the power of the highest derivative after rationalizing,which is $3$,so $m = 3$.
Now,calculate the required value: $\frac{m+n}{m-n} = \frac{3+2}{3-2} = \frac{5}{1} = 5$.

(A) Given the rate of increase in the volume of the sphere is $\frac{dV}{dt} = \pi \text{ cm}^3/\text{s}$.
We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting $\frac{dV}{dt} = \pi$,we have $\pi = 4 \pi r^2 \frac{dr}{dt}$,which implies $\frac{dr}{dt} = \frac{1}{4r^2}$.
The surface area of a sphere is $S = 4 \pi r^2$.
Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.
Substituting $\frac{dr}{dt} = \frac{1}{4r^2}$,we get $\frac{dS}{dt} = 8 \pi r \left( \frac{1}{4r^2} \right) = \frac{2 \pi}{r}$.
When $r = 1 \text{ cm}$,the rate of increase in surface area is $\frac{dS}{dt} = \frac{2 \pi}{1} = 2 \pi \text{ cm}^2/\text{s}$.

(D) Given,for $x > 0$,we need to evaluate $I = \int \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) dx$.
Using the trigonometric substitution $x = \tan \theta$,we know that for $x > 0$,$\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) = 2 \tan^{-1} x$.
Substituting this into the integral,we get $I = \int 2 \tan^{-1} x dx = 2 \int \tan^{-1} x dx$.
Using integration by parts,let $u = \tan^{-1} x$ and $dv = dx$. Then $du = \frac{1}{1+x^{2}} dx$ and $v = x$.
$I = 2 \left( x \tan^{-1} x - \int \frac{x}{1+x^{2}} dx \right)$.
To solve $\int \frac{x}{1+x^{2}} dx$,let $t = 1+x^{2}$,then $dt = 2x dx$,so $x dx = \frac{1}{2} dt$.
Thus,$\int \frac{x}{1+x^{2}} dx = \frac{1}{2} \int \frac{1}{t} dt = \frac{1}{2} \log(1+x^{2})$.
Substituting this back,$I = 2x \tan^{-1} x - 2 \left( \frac{1}{2} \log(1+x^{2}) \right) + C = 2x \tan^{-1} x - \log(1+x^{2}) + C$.

(C) We are given the integral $I = \int e^{x} \left( \frac{\sin x + \cos x}{\cos^2 x} \right) dx$.
Using the identity $1 - \sin^2 x = \cos^2 x$,we rewrite the expression:
$I = \int e^{x} \left( \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\cos^2 x} \right) dx$
$I = \int e^{x} (\tan x \sec x + \sec x) dx$
Let $f(x) = \sec x$. Then $f'(x) = \sec x \tan x$.
We know the standard integral formula $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Applying this to our integral:
$I = \int e^{x} (\sec x + \sec x \tan x) dx = e^{x} \sec x + C$.

(B) To evaluate $\int_{0}^{4}|x-1| dx$,we first identify the point where the expression inside the modulus changes sign,which is $x = 1$.
We split the integral at $x = 1$:
$\int_{0}^{4}|x-1| dx = \int_{0}^{1}-(x-1) dx + \int_{1}^{4}(x-1) dx$
Evaluating the first part:
$\int_{0}^{1}(-x+1) dx = [-\frac{x^2}{2} + x]_{0}^{1} = (-\frac{1}{2} + 1) - (0) = \frac{1}{2}$
Evaluating the second part:
$\int_{1}^{4}(x-1) dx = [\frac{x^2}{2} - x]_{1}^{4} = (\frac{16}{2} - 4) - (\frac{1}{2} - 1) = (8 - 4) - (-\frac{1}{2}) = 4 + \frac{1}{2} = \frac{9}{2}$
Adding both parts:
$\frac{1}{2} + \frac{9}{2} = \frac{10}{2} = 5$

(A) Given integral is $I_{n} = \int_{0}^{\pi / 4} \tan ^{n} x d x$.
We can write $\tan^{n} x$ as $\tan^{n-2} x \cdot \tan^{2} x$.
Using the identity $\tan^{2} x = \sec^{2} x - 1$,we have:
$I_{n} = \int_{0}^{\pi / 4} \tan^{n-2} x (\sec^{2} x - 1) d x$.
$I_{n} = \int_{0}^{\pi / 4} \tan^{n-2} x \sec^{2} x d x - \int_{0}^{\pi / 4} \tan^{n-2} x d x$.
Let $t = \tan x$,then $dt = \sec^{2} x dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$.
$I_{n} = \int_{0}^{1} t^{n-2} dt - I_{n-2}$.
$I_{n} = \left[ \frac{t^{n-1}}{n-1} \right]_{0}^{1} - I_{n-2}$.
$I_{n} = \frac{1}{n-1} - I_{n-2}$.
Therefore,$I_{n} + I_{n-2} = \frac{1}{n-1}$.
For $n=10$,we get $I_{10} + I_{8} = \frac{1}{10-1} = \frac{1}{9}$.

(NONE) The given curves are $y=m x^{2}$ and $x=m y^{2}$ where $m>0$.
To find the intersection points,substitute $y=m x^{2}$ into $x=m y^{2}$:
$x=m(m x^{2})^{2} = m^{3} x^{4}$
$m^{3} x^{4}-x=0 \Rightarrow x(m^{3} x^{3}-1)=0$
The intersection points are $x=0$ and $x=1/m$.
The area $A$ between the curves is given by:
$A = \int_{0}^{1/m} (\sqrt{x/m} - m x^{2}) dx$
$A = [\frac{1}{\sqrt{m}} \cdot \frac{x^{3/2}}{3/2} - m \cdot \frac{x^{3}}{3}]_{0}^{1/m}$
$A = [\frac{2}{3 \sqrt{m}} \cdot (1/m)^{3/2} - \frac{m}{3} \cdot (1/m)^{3}]$
$A = \frac{2}{3 m^{2}} - \frac{1}{3 m^{2}} = \frac{1}{3 m^{2}}$
Given $A = 1/4$,so $\frac{1}{3 m^{2}} = \frac{1}{4}$
$3 m^{2} = 4 \Rightarrow m^{2} = 4/3$
Since $m>0$,$m = \frac{2}{\sqrt{3}}$.
Note: None of the provided options match the calculated result. The correct value is $m = 2/\sqrt{3}$.

(B) Given,$\cos ^{-1}\left(\frac{y}{b}\right)=n \log \left(\frac{x}{n}\right)$.
Differentiating both sides with respect to $x$:
$-\frac{1}{\sqrt{1-(y/b)^2}} \cdot \frac{1}{b} \cdot y_1 = n \cdot \frac{1}{(x/n)} \cdot \frac{1}{n}$
Simplifying the expression:
$-\frac{1}{\sqrt{(b^2-y^2)/b^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$
$-\frac{b}{\sqrt{b^2-y^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$
$-\frac{y_1}{\sqrt{b^2-y^2}} = \frac{n}{x}$
Cross-multiplying:
$-x y_1 = n \sqrt{b^2-y^2}$
$x y_1 + n \sqrt{b^2-y^2} = 0$.

(A) Given the differential equation: $\left(\frac{dy}{dx}\right)^{2} = 1 - x^{2} - y^{2} + x^{2}y^{2}$
Factor the right side: $\left(\frac{dy}{dx}\right)^{2} = (1 - x^{2}) - y^{2}(1 - x^{2}) = (1 - x^{2})(1 - y^{2})$
Taking the square root on both sides: $\frac{dy}{dx} = \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}$
Separate the variables: $\frac{dy}{\sqrt{1 - y^{2}}} = \sqrt{1 - x^{2}} dx$
Integrate both sides: $\int \frac{dy}{\sqrt{1 - y^{2}}} = \int \sqrt{1 - x^{2}} dx$
Using standard integrals $\int \frac{1}{\sqrt{1 - y^{2}}} dy = \sin^{-1} y$ and $\int \sqrt{1 - x^{2}} dx = \frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{-1} x + C_1$:
$\sin^{-1} y = \frac{x}{2} \sqrt{1 - x^{2}} + \frac{1}{2} \sin^{-1} x + C_1$
Multiply by $2$: $2 \sin^{-1} y = x \sqrt{1 - x^{2}} + \sin^{-1} x + 2C_1$
Let $C = 2C_1$,then: $2 \sin^{-1} y = x \sqrt{1 - x^{2}} + \sin^{-1} x + C$

(C) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$.
Also,$a + b + c = 0$,which implies $a + b = -c$.
Squaring both sides,we get $(a + b) \cdot (a + b) = (-c) \cdot (-c)$.
$|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.
Substituting the values,$1^2 + 1^2 + 2(a \cdot b) = 1^2$.
$2 + 2(a \cdot b) = 1$.
$2(a \cdot b) = -1$,so $a \cdot b = -\frac{1}{2}$.
Since $a \cdot b = |a||b| \cos \theta$,we have $1 \cdot 1 \cdot \cos \theta = -\frac{1}{2}$.
$\cos \theta = -\frac{1}{2} = \cos \left(\frac{2\pi}{3}\right)$.
Therefore,$\theta = \frac{2\pi}{3}$.

(C) Given that $a, b, c$ are non-coplanar,so the scalar triple product $[a b c] \neq 0$.
We know that $[a b c] = [b c a] = [c a b]$.
Let $V = [a b c]$.
The expression is $E = a \cdot \left\{ \frac{b \times c}{3 b \cdot (c \times a)} \right\} - b \cdot \left\{ \frac{c \times a}{2 c \cdot (a \times b)} \right\}$.
Using the definition of scalar triple product,$b \cdot (c \times a) = [b c a] = [a b c] = V$ and $c \cdot (a \times b) = [c a b] = [a b c] = V$.
Substituting these into the expression:
$E = \frac{a \cdot (b \times c)}{3 [a b c]} - \frac{b \cdot (c \times a)}{2 [a b c]}$
$E = \frac{[a b c]}{3 [a b c]} - \frac{[b c a]}{2 [a b c]}$
Since $[a b c] = [b c a] = V$,we have:
$E = \frac{V}{3V} - \frac{V}{2V} = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6}$.

(D) Let the vectors be $\vec{a} = 2i + 3j + 0k$,$\vec{b} = i + j + k$,and $\vec{c} = \lambda i + 4j + 2k$.
The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \vec{c}$ is given by the absolute value of the scalar triple product $|[\vec{a} \vec{b} \vec{c}]| = |(\vec{a} \times \vec{b}) \cdot \vec{c}|$.
First,calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ 2 & 3 & 0 \\ 1 & 1 & 1 \end{vmatrix} = i(3-0) - j(2-0) + k(2-3) = 3i - 2j - k$.
Now,calculate the scalar triple product:
$(\vec{a} \times \vec{b}) \cdot \vec{c} = (3i - 2j - k) \cdot (\lambda i + 4j + 2k) = 3\lambda - 8 - 2 = 3\lambda - 10$.
Given the volume is $2$,we have $|3\lambda - 10| = 2$.
Case $1$: $3\lambda - 10 = 2 \Rightarrow 3\lambda = 12 \Rightarrow \lambda = 4$.
Case $2$: $3\lambda - 10 = -2 \Rightarrow 3\lambda = 8 \Rightarrow \lambda = 8/3$.
Since $4$ is the provided option,the value is $4$.

(B) Let $\vec{u} = i+j+k$ and $\vec{v} = 2i+j+3k$.
$A$ vector perpendicular to both $\vec{u}$ and $\vec{v}$ is given by the cross product $\vec{w} = \vec{u} \times \vec{v}$.
$\vec{w} = \begin{vmatrix} i & j & k \\ 1 & 1 & 1 \\ 2 & 1 & 3 \end{vmatrix} = i(3-1) - j(3-2) + k(1-2) = 2i - j - k$.
The magnitude of $\vec{w}$ is $|\vec{w}| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4+1+1} = \sqrt{6}$.
The unit vector perpendicular to both is $\pm \frac{\vec{w}}{|\vec{w}|} = \pm \frac{2i-j-k}{\sqrt{6}}$.
Comparing with the given options,the correct answer is $\frac{2i-j-k}{\sqrt{6}}$.