TS EAMCET 2013 Mathematics Question Paper with Answer and Solution

(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{(2n)(2n+1)}$.
Using partial fractions,$\frac{1}{(2n)(2n+1)} = \frac{1}{2n} - \frac{1}{2n+1}$.
Thus,$S = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{5}) + (\frac{1}{6} - \frac{1}{7}) + \dots$
This can be rewritten as $S = 1 - (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \dots)$.
We know the Taylor series expansion for $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,so for $x=1$,$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.
Therefore,$S = 1 - \log_e 2$.
Since $1 = \log_e e$,we have $S = \log_e e - \log_e 2 = \log_e \left(\frac{e}{2}\right)$.

(D) Given,$\frac{1}{x^4+x^2+1} = \frac{Ax+B}{x^2+x+1} + \frac{Cx+D}{x^2-x+1}$.
Since $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$,we have:
$1 = (Ax+B)(x^2-x+1) + (Cx+D)(x^2+x+1)$.
Expanding the right side:
$1 = Ax^3 - Ax^2 + Ax + Bx^2 - Bx + B + Cx^3 + Cx^2 + Cx + Dx^2 + Dx + D$.
Grouping terms by powers of $x$:
$1 = (A+C)x^3 + (-A+B+C+D)x^2 + (A-B+C+D)x + (B+D)$.
Comparing coefficients on both sides:
$A+C = 0$ $(i)$
$-A+B+C+D = 0$ (ii)
$A-B+C+D = 0$ (iii)
$B+D = 1$ (iv)
Adding equations (ii) and (iii):
$(-A+B+C+D) + (A-B+C+D) = 0 + 0$
$2(C+D) = 0$
$C+D = 0$.

(D) Given equation is $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$.
Let $\alpha$ and $\beta$ be the roots of this equation.
From the relation between roots and coefficients:
$\alpha+\beta = \frac{b}{5+\sqrt{2}}$
$\alpha \beta = \frac{8+2 \sqrt{5}}{5+\sqrt{2}}$
The harmonic mean $(HM)$ between the roots is given by $HM = \frac{2 \alpha \beta}{\alpha+\beta}$.
Given $HM = 4$,we have:
$\frac{2 \alpha \beta}{\alpha+\beta} = 4$
Substituting the values of $\alpha+\beta$ and $\alpha \beta$:
$\frac{2 \times \frac{8+2 \sqrt{5}}{5+\sqrt{2}}}{\frac{b}{5+\sqrt{2}}} = 4$
$\frac{2(8+2 \sqrt{5})}{b} = 4$
$\frac{8+2 \sqrt{5}}{b} = 2$
$b = \frac{8+2 \sqrt{5}}{2} = 4+\sqrt{5}$.

(C) The given quadratic equation is $x^2-2x+4=0$.
Since $\alpha$ and $\beta$ are the roots,we have $\alpha+\beta=2$ and $\alpha\beta=4$.
The roots of the equation are given by $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm i\sqrt{3}$.
In polar form,$1+i\sqrt{3} = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 2e^{i\pi/3}$ and $1-i\sqrt{3} = 2e^{-i\pi/3}$.
Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$.
Then $\alpha^9 = (2e^{i\pi/3})^9 = 2^9 e^{i3\pi} = 2^9(\cos 3\pi + i\sin 3\pi) = 2^9(-1) = -2^9$.
Similarly,$\beta^9 = (2e^{-i\pi/3})^9 = 2^9 e^{-i3\pi} = 2^9(\cos(-3\pi) + i\sin(-3\pi)) = 2^9(-1) = -2^9$.
Therefore,$\alpha^9+\beta^9 = -2^9 - 2^9 = -2 \times 2^9 = -2^{10}$.

(B) For the first inequality: $x^2+5x+6 \geq 0 \Rightarrow (x+2)(x+3) \geq 0$. This holds for $x \in (-\infty, -3] \cup [-2, \infty)$.
For the second inequality: $x^2+3x-4 < 0 \Rightarrow (x+4)(x-1) < 0$. This holds for $x \in (-4, 1)$.
The intersection of these two sets is the set of $x$ values satisfying both conditions.
Intersection: $(-\infty, -3] \cup [-2, \infty) \cap (-4, 1) = (-4, -3] \cup [-2, 1)$.

(C) Given the cubic equation is $x^3-42x^2+336x-512=0$.
We can test for roots. Let $x=2$: $2^3 - 42(2^2) + 336(2) - 512 = 8 - 168 + 672 - 512 = 0$. Thus,$(x-2)$ is a factor.
Dividing the polynomial by $(x-2)$,we get $(x-2)(x^2-40x+256)=0$.
Solving the quadratic part $x^2-40x+256=0$ using the quadratic formula or factorization:
$x = \frac{40 \pm \sqrt{1600 - 1024}}{2} = \frac{40 \pm \sqrt{576}}{2} = \frac{40 \pm 24}{2}$.
The roots are $x_1 = \frac{16}{2} = 8$ and $x_2 = \frac{64}{2} = 32$.
The roots are $2, 8, 32$.
These are in increasing geometric progression with common ratio $r = \frac{8}{2} = 4$ or $4:1$.

(C) Let the roots of the quadratic equation $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$ be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of roots $\alpha + \beta = \frac{b}{\sqrt{2}}$ and the product of roots $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}} = 4\sqrt{2} - \sqrt{10}$.
The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$.
Given $HM = 4$,we have $4 = \frac{2(4\sqrt{2} - \sqrt{10})}{\frac{b}{\sqrt{2}}}$.
$4 = \frac{2(4\sqrt{2} - \sqrt{10}) \cdot \sqrt{2}}{b}$.
$4 = \frac{2(8 - \sqrt{20})}{b} = \frac{2(8 - 2\sqrt{5})}{b} = \frac{16 - 4\sqrt{5}}{b}$.
$4b = 16 - 4\sqrt{5}$.
Dividing by $4$,we get $b = 4 - \sqrt{5}$.

(C) First,simplify the term $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $(1+i)$:
$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+i^2+2i}{1-i^2} = \frac{1-1+2i}{1+1} = \frac{2i}{2} = i$
Similarly,for the second term:
$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1+i^2-2i}{1-i^2} = \frac{1-1-2i}{1+1} = \frac{-2i}{2} = -i$
Now,substitute these values into the original expression:
$(i)^4 + (-i)^4 = i^4 + i^4 = 1 + 1 = 2$

(B) Given,$|z^2-1|=|z|^2+1$. \\ Let $z=x+iy$. \\ Then,$|(x+iy)^2-1| = |x+iy|^2+1$. \\ $|x^2-y^2+2ixy-1| = x^2+y^2+1$. \\ $|(x^2-y^2-1)+i(2xy)| = x^2+y^2+1$. \\ Squaring both sides,we get: \\ $(x^2-y^2-1)^2 + (2xy)^2 = (x^2+y^2+1)^2$. \\ $(x^2-y^2)^2 + 1 - 2(x^2-y^2) + 4x^2y^2 = (x^2+y^2)^2 + 1 + 2(x^2+y^2)$. \\ $x^4+y^4-2x^2y^2 + 1 - 2x^2+2y^2 + 4x^2y^2 = x^4+y^4+2x^2y^2 + 1 + 2x^2+2y^2$. \\ $x^4+y^4+2x^2y^2 + 1 - 2x^2+2y^2 = x^4+y^4+2x^2y^2 + 1 + 2x^2+2y^2$. \\ Simplifying the equation: \\ $-2x^2 = 2x^2$. \\ $4x^2 = 0 \implies x=0$. \\ Since $x=0$,the real part of $z$ is $0$,which means $z$ lies on the imaginary axis.

(A) Given the equation: $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$
Multiply the first term by $\frac{2-i}{2-i}$ and the second term by $\frac{2+i}{2+i}$:
$\frac{[(1+i)x-i](2-i)}{5} + \frac{[(1+2i)y+i](2+i)}{5} = 1$
Expanding the numerators:
$\frac{(2-i+2i-i^2)x - 2i + i^2}{5} + \frac{(2+i+4i+2i^2)y + 2i + i^2}{5} = 1$
Since $i^2 = -1$:
$\frac{(3+i)x - 2i - 1}{5} + \frac{(5i)y + 2i - 1}{5} = 1$
$(3+i)x + (5i)y - 2 = 5$
$(3x-7) + i(x+5y) = 0$
Comparing real and imaginary parts:
$3x-7 = 0 \Rightarrow x = \frac{7}{3}$
$x+5y = 0 \Rightarrow y = -\frac{x}{5} = -\frac{7}{15}$
Thus,$(x, y) = \left(\frac{7}{3}, -\frac{7}{15}\right)$.

(C) Given,${}^nC_{r-1}=330$,${}^nC_r=462$,and ${}^nC_{r+1}=462$.
We know that $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}$.
Since ${}^nC_{r+1} = {}^nC_r = 462$,we have $\frac{462}{462} = 1$.
Therefore,$\frac{n-r}{r+1} = 1 \implies n-r = r+1 \implies n = 2r+1$.
Now,consider the ratio $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{462}{330}$.
Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{462}{330} = \frac{7}{5}$.
Substituting $n = 2r+1$ into the equation:
$\frac{(2r+1)-r+1}{r} = \frac{7}{5} \implies \frac{r+2}{r} = \frac{7}{5}$.
$5(r+2) = 7r \implies 5r + 10 = 7r \implies 2r = 10 \implies r = 5$.

(C) $t_n$ is the number of triangles formed with $n$ points in a plane,where no three points are collinear.
Thus,$t_n = {}^{n}C_3$.
Given $t_{n+1} - t_n = 36$.
Substituting the formula: ${}^{n+1}C_3 - {}^{n}C_3 = 36$.
Using the property ${}^{n+1}C_r = {}^{n}C_r + {}^{n}C_{r-1}$,we have ${}^{n+1}C_3 - {}^{n}C_3 = {}^{n}C_2$.
Therefore,${}^{n}C_2 = 36$.
$\frac{n(n-1)}{2} = 36$.
$n(n-1) = 72$.
$n^2 - n - 72 = 0$.
$(n-9)(n+8) = 0$.
Since $n$ must be positive,$n = 9$.

(D) First,we arrange $10$ men in a row. The number of ways to arrange $10$ men is $10!$.
There are $11$ gaps created by $10$ men (including the ends) where $6$ women can be seated so that no two women sit together.
The number of ways to choose $6$ gaps out of $11$ and arrange $6$ women is given by $^{11}P_6$.
Total number of ways $= 10! \times ^{11}P_6 = 10! \times \frac{11!}{(11-6)!} = \frac{10! 11!}{5!}$.

(D) Given the system of equations:
$x+y = \frac{2 \pi}{3}$ $(i)$
$\cos x + \cos y = \frac{3}{2}$ (ii)
Using the sum-to-product formula,$\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$.
Substituting $(i)$ into the formula:
$2 \cos \left(\frac{1}{2} \cdot \frac{2 \pi}{3}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$
$2 \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$
Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,we have:
$2 \left(\frac{1}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$
$\cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$
Since the range of the cosine function is $[-1, 1]$,the equation $\cos \left(\frac{x-y}{2}\right) = \frac{3}{2}$ has no real solution.
Therefore,the system of equations has an empty set of solutions.

(C) Step $1$: Under the translation of the origin to $(1,2)$,the point $(7,5)$ in the old system becomes $(7-1, 5-2) = (6,3)$ in the new system.
Step $2$: Translating the point $(6,3)$ by $2$ units along the negative direction of the new $X$-axis results in $(6-2, 3) = (4,3)$.
Step $3$: Rotating the point $(4,3)$ through an angle $\theta = \frac{\pi}{4}$ clockwise about the origin of the new system. The clockwise rotation formula is $(x', y') = (x \cos \theta + y \sin \theta, -x \sin \theta + y \cos \theta)$.
Substituting $x=4, y=3, \theta = \frac{\pi}{4}$:
$x' = 4 \cos \frac{\pi}{4} + 3 \sin \frac{\pi}{4} = \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$
$y' = -4 \sin \frac{\pi}{4} + 3 \cos \frac{\pi}{4} = -\frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$
Thus,the final position is $\left(\frac{7}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$.

(A) Let $l_1 \equiv 2x + 3y = 5$.
Since the line $AB$ is perpendicular to $l_1$,the slope of $l_1$ is $m_1 = -\frac{2}{3}$.
Therefore,the slope of $AB$ is $m_{AB} = -\frac{1}{m_1} = \frac{3}{2}$.
The equation of line $AB$ passing through $A\left(1, \frac{1}{3}\right)$ with slope $\frac{3}{2}$ is:
$\left(y - \frac{1}{3}\right) = \frac{3}{2}(x - 1)$
$\Rightarrow 2y - \frac{2}{3} = 3x - 3$
$\Rightarrow 3x - 2y = 3 - \frac{2}{3} = \frac{7}{3}$
$\Rightarrow 9x - 6y = 7$ $(i)$
The equation of line $l_1$ is $2x + 3y = 5$. Multiplying by $2$,we get $4x + 6y = 10$ (ii).
Adding $(i)$ and (ii): $13x = 17 \Rightarrow x = \frac{17}{13}$.
Substituting $x$ in $2x + 3y = 5$: $2\left(\frac{17}{13}\right) + 3y = 5$ $\Rightarrow 3y = 5 - \frac{34}{13} = \frac{31}{13}$ $\Rightarrow y = \frac{31}{39}$.
The intersection point $P$ (mid-point of $AB$) is $\left(\frac{17}{13}, \frac{31}{39}\right)$.
Let $B = (x_2, y_2)$. Since $P$ is the mid-point of $AB$:
$\frac{1 + x_2}{2} = \frac{17}{13}$ $\Rightarrow 1 + x_2 = \frac{34}{13}$ $\Rightarrow x_2 = \frac{21}{13}$
$\frac{1/3 + y_2}{2} = \frac{31}{39}$ $\Rightarrow \frac{1}{3} + y_2 = \frac{62}{39}$ $\Rightarrow y_2 = \frac{62}{39} - \frac{13}{39} = \frac{49}{39}$
Thus,$B = \left(\frac{21}{13}, \frac{49}{39}\right)$.

(B) Let $f(x, y) = 3x - 5y + a$. For the points $(1, 2)$ and $(3, 4)$ to lie on the same side of the line,$f(1, 2)$ and $f(3, 4)$ must have the same sign.
$f(1, 2) = 3(1) - 5(2) + a = a - 7$
$f(3, 4) = 3(3) - 5(4) + a = a - 11$
Thus,$(a - 7)(a - 11) > 0$.
Solving this inequality,we get $a < 7$ or $a > 11$.
Therefore,$a \in (-\infty, 7) \cup (11, \infty)$,which is $R - [7, 11]$.

(A) The given equations of the lines are:
$x \sec \theta - y \operatorname{cosec} \theta = a$ $(i)$
$x \cos \theta + y \sin \theta = a \cos 2 \theta$ $(ii)$
The perpendicular distance $d$ from the origin $(0, 0)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
For line $(i)$:
$p = \frac{|-a|}{\sqrt{\sec^2 \theta + \operatorname{cosec}^2 \theta}} = \frac{a}{\sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}} = \frac{a}{\sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}}} = a \sin \theta \cos \theta = \frac{a}{2} \sin 2 \theta$.
Thus,$2p = a \sin 2 \theta$.
For line $(ii)$:
$q = \frac{|-a \cos 2 \theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{a \cos 2 \theta}{1} = a \cos 2 \theta$.
Now,calculating $4p^2 + q^2$:
$4p^2 + q^2 = (a \sin 2 \theta)^2 + (a \cos 2 \theta)^2$
$= a^2 (\sin^2 2 \theta + \cos^2 2 \theta)$
$= a^2 (1) = a^2$.

(D) Given points $(1, \pi)$,$(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ are in polar form.
Converting these to Cartesian coordinates $(x, y) = (r \cos \theta, r \sin \theta)$:
$(1, \pi) \rightarrow (1 \cdot \cos \pi, 1 \cdot \sin \pi) = (-1, 0)$
$(1, 0^{\circ}) \rightarrow (1 \cdot \cos 0^{\circ}, 1 \cdot \sin 0^{\circ}) = (1, 0)$
$(1, \frac{\pi}{2}) \rightarrow (1 \cdot \cos \frac{\pi}{2}, 1 \cdot \sin \frac{\pi}{2}) = (0, 1)$
Now,the equation of the line passing through $(1, 0)$ and $(0, 1)$ is given by the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ or slope-point form:
$y - 0 = \frac{1 - 0}{0 - 1}(x - 1)$ $\Rightarrow y = -x + 1$ $\Rightarrow x + y - 1 = 0$.
The perpendicular distance $d$ from the point $(-1, 0)$ to the line $x + y - 1 = 0$ is calculated using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$:
$d = \frac{|1(-1) + 1(0) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(B) Let the equation of the circle with center $(a, 0)$ and radius $r$ be $(x-a)^2 + y^2 = r^2$,which simplifies to $S_1 \equiv x^2 + y^2 - 2ax + a^2 - r^2 = 0$.
Let the equation of the circle with center $(b, 0)$ and radius $R$ be $(x-b)^2 + y^2 = R^2$,which simplifies to $S_2 \equiv x^2 + y^2 - 2bx + b^2 - R^2 = 0$.
The radical axis of the two circles is given by $S_1 - S_2 = 0$.
Substituting the equations,we get $(x^2 + y^2 - 2ax + a^2 - r^2) - (x^2 + y^2 - 2bx + b^2 - R^2) = 0$.
This simplifies to $2(b-a)x + a^2 - b^2 - r^2 + R^2 = 0$.
Since the radical axis is the $y$-axis,its equation must be $x = 0$.
For the equation $2(b-a)x + (a^2 - b^2 - r^2 + R^2) = 0$ to represent $x = 0$,the constant term must be zero.
Therefore,$a^2 - b^2 - r^2 + R^2 = 0$.
Solving for $R$,we get $R^2 = r^2 + b^2 - a^2$,which implies $R = (r^2 + b^2 - a^2)^{1/2}$.

(B) Simplify the expression inside the bracket: $\left[\frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1} - \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}\right]^{10}$
$= \left[(x^{1/3}+1) - \frac{\sqrt{x}+1}{\sqrt{x}}\right]^{10} = \left[x^{1/3}+1 - (1+x^{-1/2})\right]^{10} = (x^{1/3}-x^{-1/2})^{10}$.
The general term $T_{r+1}$ is given by: ${}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.
For the term independent of $x$,the exponent of $x$ must be $0$: $\frac{10-r}{3} - \frac{r}{2} = 0$.
$20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$.
The term is ${}^{10}C_4 (-1)^4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.

(B) Let $A \equiv (x_1, y_1, z_1)$,$B \equiv (x_2, y_2, z_2)$,and $C \equiv (x_3, y_3, z_3)$.
Since $F(0, 1, 0)$ is the mid-point of $AB$,we have:
$x_1 + x_2 = 0, y_1 + y_2 = 2, z_1 + z_2 = 0$ $(1)$
Since $D(2, 1, 0)$ is the mid-point of $BC$,we have:
$x_2 + x_3 = 4, y_2 + y_3 = 2, z_2 + z_3 = 0$ $(2)$
Since $E(2, 0, 0)$ is the mid-point of $AC$,we have:
$x_3 + x_1 = 4, y_3 + y_1 = 0, z_3 + z_1 = 0$ $(3)$
Adding $(1)$,$(2)$,and $(3)$,we get:
$2(x_1 + x_2 + x_3) = 8 \implies x_1 + x_2 + x_3 = 4$
$2(y_1 + y_2 + y_3) = 4 \implies y_1 + y_2 + y_3 = 2$
$2(z_1 + z_2 + z_3) = 0 \implies z_1 + z_2 + z_3 = 0$
The centroid of $\triangle ABC$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the sums,the centroid is $\left(\frac{4}{3}, \frac{2}{3}, 0\right)$.

(C) First,we arrange $10$ men in a row. The number of ways to arrange $10$ men is $10!$.
This creates $11$ possible gaps (including the ends) where the $6$ women can be seated to ensure no two women sit together.
The number of ways to choose $6$ gaps out of $11$ and arrange the $6$ women is given by the permutation formula $^{11}P_6$.
$^{11}P_6 = \frac{11!}{(11-6)!} = \frac{11!}{5!}$.
Therefore,the total number of ways is $10! \times \frac{11!}{5!} = \frac{10! 11!}{5!}$.

(B) Given expression: $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$
Simplify the terms inside the bracket:
$\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3} - x^{1/3} + 1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3} + 1$
$\frac{(x-1)}{(x-\sqrt{x})} = \frac{(\sqrt{x})^2 - 1}{\sqrt{x}(\sqrt{x}-1)} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$
Substituting these back: $\left[(x^{1/3} + 1) - (1 + x^{-1/2})\right]^{10} = (x^{1/3} - x^{-1/2})^{10}$
The general term $T_{r+1}$ is given by:
$T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$
For the term independent of $x$,the exponent of $x$ must be $0$:
$\frac{10-r}{3} - \frac{r}{2} = 0$ $\Rightarrow 20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$
The term is $T_{4+1} = {}^{10}C_4 (-1)^4 = {}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

(C) Given expression is $E = \frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$.
Using the binomial expansion $(1+ax)^n \approx 1+nax$ for small $x$:
$(1-2 x)^{-1} \approx 1 + (-1)(-2x) = 1 + 2x$.
$(1-3 x)^{-2} \approx 1 + (-2)(-3x) = 1 + 6x$.
$(1-4 x)^{-3} \approx 1 + (-3)(-4x) = 1 + 12x$.
Substituting these into the expression:
$E \approx \frac{(1+2x)(1+6x)}{(1+12x)} = \frac{1 + 6x + 2x + 12x^2}{1+12x}$.
Neglecting $x^2$ and higher powers:
$E \approx \frac{1+8x}{1+12x} = (1+8x)(1+12x)^{-1}$.
Using the binomial expansion again:
$E \approx (1+8x)(1-12x) = 1 - 12x + 8x - 96x^2$.
Neglecting $x^2$:
$E \approx 1 - 4x$.

(D) Given,$\sin \theta + \cos \theta = p$ $(i)$ and $\sin^3 \theta + \cos^3 \theta = q$ (ii).
Using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we have:
$(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) = q$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,this becomes $p(1 - \sin \theta \cos \theta) = q$.
Thus,$1 - \sin \theta \cos \theta = \frac{q}{p}$,which implies $\sin \theta \cos \theta = 1 - \frac{q}{p}$ (iii).
Squaring equation $(i)$: $(\sin \theta + \cos \theta)^2 = p^2$.
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = p^2$.
$1 + 2 \sin \theta \cos \theta = p^2$.
Substituting (iii) into this: $1 + 2(1 - \frac{q}{p}) = p^2$.
$1 + 2 - \frac{2q}{p} = p^2$.
$3 - \frac{2q}{p} = p^2$.
Multiply by $p$: $3p - 2q = p^3$.
Rearranging gives $p^3 - 3p = -2q$,or $p(p^2 - 3) = -2q$.

(A) Given,$\tan (\pi \cos \theta)=\cot (\pi \sin \theta)$
$\Rightarrow \tan (\pi \cos \theta)=\tan \left(\frac{\pi}{2}-\pi \sin \theta\right)$
$\Rightarrow \pi \cos \theta=\frac{\pi}{2}-\pi \sin \theta$
$\Rightarrow \cos \theta+\sin \theta=\frac{1}{2}$
Dividing both sides by $\sqrt{2}$:
$\Rightarrow \frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta=\frac{1}{2 \sqrt{2}}$
$\Rightarrow \cos \theta \cos \frac{\pi}{4}+\sin \theta \sin \frac{\pi}{4}=\frac{1}{2 \sqrt{2}}$
$\Rightarrow \cos \left(\theta-\frac{\pi}{4}\right)=\frac{1}{2 \sqrt{2}}$

(A) Let $m_1$ and $m_2$ be the slopes of the lines.
Then,$m_1+m_2 = \text{arithmetic mean} = \frac{4+9}{2} = \frac{13}{2}$ and $m_1 m_2 = \text{geometric mean} = \sqrt{4 \times 9} = \sqrt{36} = 6$.
Now,the equation of the pair of lines passing through the origin is given by $(y-m_1 x)(y-m_2 x) = 0$.
This simplifies to $y^2 - (m_1+m_2)xy + m_1 m_2 x^2 = 0$.
Substituting the values of $m_1+m_2$ and $m_1 m_2$,we get $y^2 - \frac{13}{2}xy + 6x^2 = 0$.
Multiplying by $2$,we get $2y^2 - 13xy + 12x^2 = 0$,which is $12x^2 - 13xy + 2y^2 = 0$.

(A) Comparing the given equation $x^2-5xy+py^2+3x-8y+2=0$ with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=-\frac{5}{2}, b=p, g=\frac{3}{2}, f=-4, c=2$.
The equation represents a pair of straight lines if $abc+2fgh-af^2-bg^2-ch^2=0$.
Substituting the values: $1(p)(2) + 2(-4)(\frac{3}{2})(-\frac{5}{2}) - 1(-4)^2 - p(\frac{3}{2})^2 - 2(-\frac{5}{2})^2 = 0$.
$2p + 30 - 16 - \frac{9p}{4} - \frac{25}{2} = 0$.
Multiplying by $4$: $8p + 120 - 64 - 9p - 50 = 0$ $\Rightarrow -p + 6 = 0$ $\Rightarrow p=6$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{(-\frac{5}{2})^2 - 1(6)}}{1+6} \right| = \left| \frac{2\sqrt{\frac{25}{4}-6}}{7} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{7} \right| = \frac{2(\frac{1}{2})}{7} = \frac{1}{7}$.
Since $\tan \theta = \frac{1}{7}$,we have a right triangle with opposite side $1$ and adjacent side $7$. The hypotenuse is $\sqrt{1^2+7^2} = \sqrt{50}$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{50}}$.

(C) The point of intersection $(x_0, y_0)$ of the pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$ is given by the solution of the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.
$2ax+2hy+2g=0 \implies ax+hy+g=0$
$2hx+2by+2f=0 \implies hx+by+f=0$
Solving these equations using Cramer's rule:
$x_0 = \frac{hf-bg}{ab-h^2}$ and $y_0 = \frac{gh-af}{ab-h^2}$.
The square of the distance from the origin $(0,0)$ is $D^2 = x_0^2 + y_0^2$.
Using the condition for a pair of lines $abc+2fgh-af^2-bg^2-ch^2=0$,we can simplify the expression for $x_0^2+y_0^2$.
Substituting the values,we get $D^2 = \frac{(hf-bg)^2+(gh-af)^2}{(ab-h^2)^2} = \frac{h^2f^2+b^2g^2-2hfbg+g^2h^2+a^2f^2-2ghaf}{(ab-h^2)^2}$.
Using $abc+2fgh=af^2+bg^2+ch^2$,this simplifies to $\frac{c(a+b)-f^2-g^2}{ab-h^2}$.

(A) The given equation of the circle is $4x^2+4y^2-12x-12y+9=0$.
Dividing by $4$,we get $x^2+y^2-3x-3y+\frac{9}{4}=0$.
Rearranging the terms,we have $(x^2-3x)+(y^2-3y)=-\frac{9}{4}$.
Completing the square,we get $(x-\frac{3}{2})^2-\frac{9}{4}+(y-\frac{3}{2})^2-\frac{9}{4}=-\frac{9}{4}$.
This simplifies to $(x-\frac{3}{2})^2+(y-\frac{3}{2})^2=\frac{9}{4}$,which is $(x-\frac{3}{2})^2+(y-\frac{3}{2})^2=(\frac{3}{2})^2$.
Comparing this with the standard form $(x-h)^2+(y-k)^2=r^2$,the centre is $(\frac{3}{2}, \frac{3}{2})$ and the radius is $r=\frac{3}{2}$.
Since the distance of the centre from both axes is equal to the radius (i.e.,$|h|=|k|=r=\frac{3}{2}$),the circle touches both the axes.

(C) The length of the tangent from a point $(h, k)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{h^2+k^2+2gh+2fk+c}$.
For the circle $x^2+y^2-16=0$,the length of the tangent is $L_1 = \sqrt{h^2+k^2-16}$.
For the circle $x^2+y^2+2x+2y=0$,the length of the tangent is $L_2 = \sqrt{h^2+k^2+2h+2k}$.
Given that $L_1 = 2L_2$,we have $\sqrt{h^2+k^2-16} = 2\sqrt{h^2+k^2+2h+2k}$.
Squaring both sides,we get $h^2+k^2-16 = 4(h^2+k^2+2h+2k)$.
$h^2+k^2-16 = 4h^2+4k^2+8h+8k$.
Rearranging the terms,we get $3h^2+3k^2+8h+8k+16=0$.

(D) The common chord of the two circles $S_1 = x^2+y^2+4x-6y+c=0$ and $S_2 = x^2+y^2-6x+4y-12=0$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+4x-6y+c) - (x^2+y^2-6x+4y-12) = 0$
$10x - 10y + c + 12 = 0$ $(i)$
If a circle bisects the circumference of another circle,the common chord must pass through the center of the circle being bisected.
The center of the second circle $x^2+y^2-6x+4y-12=0$ is $(3, -2)$.
Substituting $(3, -2)$ into equation $(i)$:
$10(3) - 10(-2) + c + 12 = 0$
$30 + 20 + c + 12 = 0$
$62 + c = 0$
$c = -62$

(A) The given equation of the circle is $C: x^2 + y^2 - 16x - 12y + 64 = 0$.
$(i)$ The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.
For point $(-5, 1)$,$g = -8, f = -6, c = 64$:
$x(-5) + y(1) - 8(x - 5) - 6(y + 1) + 64 = 0$
$-5x + y - 8x + 40 - 6y - 6 + 64 = 0$
$-13x - 5y + 98 = 0 \Rightarrow 13x + 5y = 98$. This matches $(D)$.
$(ii)$ The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 - 8(x + x_1) - 6(y + y_1) + 64 = 0$.
For $(8, 0)$:
$8x + 0y - 8(x + 8) - 6(y + 0) + 64 = 0$
$8x - 8x - 64 - 6y + 64 = 0$
$-6y = 0 \Rightarrow y = 0$. This matches $(A)$.
$(iii)$ The center of the circle is $(-g, -f) = (8, 6)$. The normal at any point on the circle passes through the center.
The normal at $(2, 6)$ passes through $(2, 6)$ and $(8, 6)$.
Since the $y$-coordinates are the same,the equation of the line is $y = 6$. This matches $(B)$.
$(iv)$ The diameter passes through the center $(8, 6)$ and the given point $(8, 12)$.
Since both points have the same $x$-coordinate,the equation of the line is $x = 8$. This matches $(E)$.
Thus,the correct match is $(i)-(D), (ii)-(A), (iii)-(B), (iv)-(E)$.

(B) Let the equation of the circle with centre $(a, 0)$ and radius $r$ be $(x-a)^2 + y^2 = r^2$,which simplifies to $S_1 \equiv x^2 + y^2 - 2ax + a^2 - r^2 = 0$.
Let the equation of the circle with centre $(b, 0)$ and radius $R$ be $(x-b)^2 + y^2 = R^2$,which simplifies to $S_2 \equiv x^2 + y^2 - 2bx + b^2 - R^2 = 0$.
The radical axis of the two circles is given by $S_1 - S_2 = 0$.
$(x^2 + y^2 - 2ax + a^2 - r^2) - (x^2 + y^2 - 2bx + b^2 - R^2) = 0$.
$-2ax + 2bx + a^2 - b^2 - r^2 + R^2 = 0$.
$2x(b-a) + (a^2 - b^2 - r^2 + R^2) = 0$.
Since the radical axis is the $y$-axis,its equation is $x = 0$.
Comparing $2x(b-a) + (a^2 - b^2 - r^2 + R^2) = 0$ with $x = 0$,the constant term must be zero:
$a^2 - b^2 - r^2 + R^2 = 0$.
$R^2 = r^2 + b^2 - a^2$.
$R = (r^2 + b^2 - a^2)^{1/2}$.

(C) The equation of the parabola is $y^2 = 8x$,where $4a = 8$,so $a = 2$.
Let the chord be $y = mx + c$. The points of intersection of the line $y = mx + c$ and the parabola $y^2 = 8x$ are given by $(mx + c)^2 = 8x$,which simplifies to $m^2x^2 + (2mc - 8)x + c^2 = 0$.
Let the roots be $x_1$ and $x_2$. Then $x_1 + x_2 = -\frac{2mc - 8}{m^2}$ and $x_1x_2 = \frac{c^2}{m^2}$.
The mid-point of the chord is $(h, k) = (\frac{x_1 + x_2}{2}, m(\frac{x_1 + x_2}{2}) + c)$.
Since the circle has radius $4$ and touches the axis of the parabola $(y = 0)$,the $y$-coordinate of the center must be $4$ or $-4$. Thus,$k = 4$ (assuming positive).
The length of the chord is $L = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(x_1 - x_2)^2 + m^2(x_1 - x_2)^2} = |x_1 - x_2| \sqrt{1 + m^2}$.
Since the circle has diameter $L$,the radius is $R = \frac{L}{2} = 4$,so $L = 8$.
Using the property of the chord of a parabola,the slope $m = \frac{2}{t_1 + t_2}$ and the mid-point $y$-coordinate $k = 2(t_1 + t_2) = 4$,which gives $t_1 + t_2 = 2$.
Thus,$m = \frac{2}{2} = 1$.

(A) The equation of a chord of a conic $S=0$ with mid-point $(x_1, y_1)$ is given by $T=S_1$.
Given equation of ellipse: $S = x^2+4y^2-2x+20y = 0$.
Mid-point $(x_1, y_1) = (2, -4)$.
$T = x(x_1) + 4y(y_1) - (x+x_1) + 10(y+y_1) = 0$.
Substituting $(x_1, y_1) = (2, -4)$:
$T = 2x + 4y(-4) - (x+2) + 10(y-4) = 2x - 16y - x - 2 + 10y - 40 = x - 6y - 42$.
$S_1 = (2)^2 + 4(-4)^2 - 2(2) + 20(-4) = 4 + 64 - 4 - 80 = -16$.
Equating $T = S_1$:
$x - 6y - 42 = -16$.
$x - 6y = 42 - 16$.
$x - 6y = 26$.

(B) The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Here $a^2=25$ and $b^2=16$.
The eccentricity $e$ of the ellipse is given by $b^2=a^2(1-e^2)$,so $16=25(1-e^2)$,which gives $e^2=1-\frac{16}{25}=\frac{9}{25}$,so $e=\frac{3}{5}$.
The foci of the ellipse are $(\pm ae, 0) = (\pm 5 \times \frac{3}{5}, 0) = (\pm 3, 0)$.
The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{b^2}=1$. Here $a^2=4$.
Let $e_1$ be the eccentricity of the hyperbola. The foci are $(\pm ae_1, 0) = (\pm 2e_1, 0)$.
Since the foci coincide,$2e_1=3$,so $e_1=\frac{3}{2}$.
For a hyperbola,$b^2=a^2(e_1^2-1)$.
Substituting the values,$b^2=4((\frac{3}{2})^2-1) = 4(\frac{9}{4}-1) = 4(\frac{5}{4}) = 5$.

(B) Given the hyperbola $x^2-y^2=9$ and the chord of contact $x=9$.
Substituting $x=9$ into the hyperbola equation:
$81-y^2=9$ $\Rightarrow y^2=72$ $\Rightarrow y = \pm 6\sqrt{2}$.
Thus,the points of contact are $P_1(9, 6\sqrt{2})$ and $P_2(9, -6\sqrt{2})$.
Differentiating $x^2-y^2=9$ with respect to $x$:
$2x - 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}$.
At $P_1(9, 6\sqrt{2})$,the slope $m_1 = \frac{9}{6\sqrt{2}} = \frac{3}{2\sqrt{2}}$.
The equation of the tangent at $P_1$ is $y - 6\sqrt{2} = \frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y - 24 = 3x - 27$,or $3x - 2\sqrt{2}y - 3 = 0$.
At $P_2(9, -6\sqrt{2})$,the slope $m_2 = \frac{9}{-6\sqrt{2}} = -\frac{3}{2\sqrt{2}}$.
The equation of the tangent at $P_2$ is $y + 6\sqrt{2} = -\frac{3}{2\sqrt{2}}(x-9)$,which simplifies to $2\sqrt{2}y + 24 = -3x + 27$,or $3x + 2\sqrt{2}y - 3 = 0$.
Comparing with the options,$3x + 2\sqrt{2}y - 3 = 0$ is option $B$.

(B) We use the Taylor series expansions for $\tan x$ and $\sin x$ near $x = 0$:
$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$
$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$
Now,expand $\tan^3 x$:
$\tan^3 x = (x + \frac{x^3}{3} + O(x^5))^3 = x^3 + 3x^2(\frac{x^3}{3}) + O(x^7) = x^3 + x^5 + O(x^7)$
Next,expand $\sin^3 x$:
$\sin^3 x = (x - \frac{x^3}{6} + O(x^5))^3 = x^3 + 3x^2(-\frac{x^3}{6}) + O(x^7) = x^3 - \frac{x^5}{2} + O(x^7)$
Subtracting these two expansions:
$\tan^3 x - \sin^3 x = (x^3 + x^5) - (x^3 - \frac{x^5}{2}) + O(x^7) = \frac{3}{2}x^5 + O(x^7)$
Finally,evaluate the limit:
$\lim _{x \rightarrow 0} \frac{\frac{3}{2}x^5}{x^5} = \frac{3}{2}$.

(C) Given the equation: $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$
Taking $LCM$ on the left side: $\frac{(b+c) + (a+c)}{(a+c)(b+c)} = \frac{3}{a+b+c}$
$\Rightarrow \frac{a+b+2c}{ab + ac + bc + c^2} = \frac{3}{a+b+c}$
Cross-multiplying: $(a+b+2c)(a+b+c) = 3(ab + ac + bc + c^2)$
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3ab + 3ac + 3bc + 3c^2$
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3ab + 3ac + 3bc + 3c^2$
$a^2 + b^2 - ab = c^2$
Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$
Comparing $a^2 + b^2 - ab = c^2$ with $a^2 + b^2 - 2ab \cos C = c^2$,we get:
$ab = 2ab \cos C$
$\cos C = \frac{1}{2}$
Therefore,$\angle C = 60^{\circ}$.

(B) Given the equation: $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$
Taking the common denominator on the left side: $\frac{(c+a) + (b+c)}{(b+c)(c+a)} = \frac{3}{a+b+c}$
$\frac{a+b+2c}{bc + ab + c^2 + ac} = \frac{3}{a+b+c}$
Cross-multiplying: $(a+b+2c)(a+b+c) = 3(bc + ab + c^2 + ac)$
$(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
$a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$
Simplifying by subtracting common terms from both sides: $a^2 + b^2 - c^2 = ab$
Using the Cosine Rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Substituting $a^2 + b^2 - c^2 = ab$: $\cos C = \frac{ab}{2ab} = \frac{1}{2}$
Since $\cos C = \frac{1}{2}$,we have $\angle C = 60^{\circ}$.

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Then,$r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}$.
Taking $\frac{\Delta^2}{(s-a)(s-b)(s-c)}$ as a common factor,we get:
$\frac{\Delta^2}{(s-a)(s-b)(s-c)} [(s-c) + (s-a) + (s-b)]$.
Since $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$,the expression becomes:
$\frac{\Delta^2}{\Delta^2/s} [3s - (a+b+c)]$.
Using $a+b+c = 2s$,we have:
$s [3s - 2s] = s^2$.
Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r}$,so $s^2 = \frac{\Delta^2}{r^2}$.

(A) For the logarithm $\log_{10}(A)$ to be defined in $R$,the argument $A$ must be strictly greater than $0$.
Thus,$(1.6)^{1-x^2} - (0.625)^{6(1+x)} > 0$.
Note that $1.6 = \frac{8}{5}$ and $0.625 = \frac{5}{8} = (\frac{8}{5})^{-1}$.
So,$(\frac{8}{5})^{1-x^2} > (\frac{8}{5})^{-6(1+x)}$.
Since the base $\frac{8}{5} > 1$,the inequality holds for the exponents:
$1 - x^2 > -6(1 + x)$
$1 - x^2 > -6 - 6x$
$x^2 - 6x - 7 < 0$
Factoring the quadratic: $(x - 7)(x + 1) < 0$.
The solution to this inequality is $x \in (-1, 7)$.

(D) Given function is $f(x) = \cos \left(\frac{x}{3}\right) + \sin \left(\frac{x}{2}\right)$.
We know that the period of $\cos(ax)$ is $\frac{2 \pi}{|a|}$ and the period of $\sin(bx)$ is $\frac{2 \pi}{|b|}$.
For the first term $\cos \left(\frac{x}{3}\right)$, the period $T_1 = \frac{2 \pi}{1/3} = 6 \pi$.
For the second term $\sin \left(\frac{x}{2}\right)$, the period $T_2 = \frac{2 \pi}{1/2} = 4 \pi$.
The period of the sum of two periodic functions is the Least Common Multiple $(LCM)$ of their individual periods.
Therefore, the period of $f(x) = \text{LCM}(6 \pi, 4 \pi)$.
Since $6 \pi = 2 \times 3 \pi$ and $4 \pi = 2 \times 2 \pi$, the $\text{LCM}(6 \pi, 4 \pi) = 12 \pi$.
Thus, the period of $f(x)$ is $12 \pi$.

(D) Let the height of the tower be $h$ and the distance of point $A$ from the base of the tower $(B)$ be $x$.
In $\triangle ABD$,$\tan 60^{\circ} = \frac{h}{x}$ $\Rightarrow \sqrt{3} = \frac{h}{x}$ $\Rightarrow x = \frac{h}{\sqrt{3}}$.
Now,the person moves $60 \ m$ perpendicular to $AB$ to point $C$. Thus,$AC = 60 \ m$ and $\triangle ABC$ is a right-angled triangle at $A$.
The distance from point $C$ to the base $B$ is $BC = \sqrt{AC^2 + AB^2} = \sqrt{60^2 + x^2} = \sqrt{3600 + x^2}$.
In $\triangle CBD$,the angle of elevation is $45^{\circ}$,so $\tan 45^{\circ} = \frac{h}{BC} \Rightarrow 1 = \frac{h}{\sqrt{3600 + x^2}}$.
Therefore,$h^2 = 3600 + x^2$.
Substituting $x^2 = \frac{h^2}{3}$ into the equation: $h^2 = 3600 + \frac{h^2}{3}$.
$h^2 - \frac{h^2}{3} = 3600 \Rightarrow \frac{2h^2}{3} = 3600$.
$h^2 = 1800 \times 3 = 5400$.
$h = \sqrt{5400} = \sqrt{900 \times 6} = 30 \sqrt{6} \ m$.

(B) Let the vertices of $\triangle ABC$ be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Since $D(2, 1, 0)$ is the mid-point of $BC$,we have:
$\frac{x_2+x_3}{2} = 2 \implies x_2+x_3 = 4$
$\frac{y_2+y_3}{2} = 1 \implies y_2+y_3 = 2$
$\frac{z_2+z_3}{2} = 0 \implies z_2+z_3 = 0$
Since $E(2, 0, 0)$ is the mid-point of $AC$,we have:
$\frac{x_1+x_3}{2} = 2 \implies x_1+x_3 = 4$
$\frac{y_1+y_3}{2} = 0 \implies y_1+y_3 = 0$
$\frac{z_1+z_3}{2} = 0 \implies z_1+z_3 = 0$
Since $F(0, 1, 0)$ is the mid-point of $AB$,we have:
$\frac{x_1+x_2}{2} = 0 \implies x_1+x_2 = 0$
$\frac{y_1+y_2}{2} = 1 \implies y_1+y_2 = 2$
$\frac{z_1+z_2}{2} = 0 \implies z_1+z_2 = 0$
Adding these equations:
$2(x_1+x_2+x_3) = 4+4+0 = 8 \implies x_1+x_2+x_3 = 4$
$2(y_1+y_2+y_3) = 2+0+2 = 4 \implies y_1+y_2+y_3 = 2$
$2(z_1+z_2+z_3) = 0+0+0 = 0 \implies z_1+z_2+z_3 = 0$
The centroid of $\triangle ABC$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the sums,the centroid is $\left(\frac{4}{3}, \frac{2}{3}, 0\right)$.

(D) Total sample points,$n(S) = 6 \times 6 = 36$.
Favourable outcomes for the sum to be $\ge 10$ are:
$E = \{(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)\}$.
Total number of favourable outcomes,$n(E) = 6$.
Required probability $P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

(C) The total number of ways to choose $2$ numbers from $8$ is given by ${}^8C_2 = \frac{8 \times 7}{2} = 28$.
Let the two chosen numbers be $x$ and $y$ such that $x < y$. We want the probability that $x < 4$.
Case $I$: If $x = 1$,then $y$ can be any of the remaining $7$ numbers $(\{2, 3, 4, 5, 6, 7, 8\})$. Number of ways $= 7$.
Case $II$: If $x = 2$,then $y$ can be any of the remaining $6$ numbers $(\{3, 4, 5, 6, 7, 8\})$. Number of ways $= 6$.
Case $III$: If $x = 3$,then $y$ can be any of the remaining $5$ numbers $(\{4, 5, 6, 7, 8\})$. Number of ways $= 5$.
Total favourable cases $= 7 + 6 + 5 = 18$.
Therefore,the required probability $= \frac{18}{28} = \frac{9}{14}$.

(C) For a system of homogeneous linear equations to have non-trivial solutions,the determinant of the coefficient matrix $A$ must be zero,i.e.,$|A| = 0$.
Given the matrix $A = \begin{bmatrix} t & t+1 & t-1 \\ t+1 & t & t+2 \\ t-1 & t+2 & t \end{bmatrix}$,we set $|A| = 0$.
$|A| = \begin{vmatrix} t & t+1 & t-1 \\ t+1 & t & t+2 \\ t-1 & t+2 & t \end{vmatrix} = 0$.
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$|A| = \begin{vmatrix} t & t+1 & t-1 \\ 1 & -1 & 3 \\ -1 & 1 & 1 \end{vmatrix} = 0$.
Adding $R_2$ to $R_3$ $(R_3 \rightarrow R_3 + R_2)$:
$|A| = \begin{vmatrix} t & t+1 & t-1 \\ 1 & -1 & 3 \\ 0 & 0 & 4 \end{vmatrix} = 0$.
Expanding along the third row:
$4 \times \begin{vmatrix} t & t+1 \\ 1 & -1 \end{vmatrix} = 0$.
$4 \times (-t - (t+1)) = 0$.
$4 \times (-2t - 1) = 0$.
$-8t - 4 = 0 \implies t = -\frac{1}{2}$.
Since there is only one real value of $t$ $(t = -\frac{1}{2})$,the number of real values is $1$.

(D) Given,$u=\log \left(x^3+y^3+z^3-3 x y z\right)$.
We know that $x^3+y^3+z^3-3 x y z = (x+y+z)(x^2+y^2+z^2-x y-y z-z x)$.
So,$u = \log(x+y+z) + \log(x^2+y^2+z^2-x y-y z-z x)$.
Now,calculate the partial derivatives:
$u_x = \frac{\partial u}{\partial x} = \frac{3x^2-3yz}{x^3+y^3+z^3-3xyz}$
$u_y = \frac{\partial u}{\partial y} = \frac{3y^2-3xz}{x^3+y^3+z^3-3xyz}$
$u_z = \frac{\partial u}{\partial z} = \frac{3z^2-3xy}{x^3+y^3+z^3-3xyz}$
Adding these,we get:
$u_x+u_y+u_z = \frac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$
$u_x+u_y+u_z = \frac{3}{x+y+z}$
Therefore,$(x+y+z)(u_x+u_y+u_z) = 3$.

(B) Given: $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$ for $x>0$.
Differentiating both sides with respect to $x$:
$-\frac{1}{\sqrt{1-\frac{y^2}{b^2}}} \cdot \frac{1}{b} \frac{d y}{d x} = 2 \cdot \frac{1}{\left(\frac{x}{2}\right)} \cdot \frac{1}{2}$
$-\frac{1}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x} = \frac{2}{x}$
$x \frac{d y}{d x} = -2 \sqrt{b^2-y^2} \quad \dots (i)$
Differentiating again with respect to $x$:
$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = -2 \cdot \frac{1}{2} (b^2-y^2)^{-1/2} (-2y) \frac{d y}{d x}$
$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{2y}{\sqrt{b^2-y^2}} \cdot \frac{d y}{d x}$
From equation $(i)$,we have $\sqrt{b^2-y^2} = -\frac{x}{2} \frac{d y}{d x}$. Substituting this:
$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{2y}{-\frac{x}{2} \frac{d y}{d x}} \cdot \frac{d y}{d x} = -\frac{4y}{x}$
Multiplying by $x$:
$x^2 \frac{d^2 y}{d x^2} + x \frac{d y}{d x} = -4y$.

(D) Given curves are:
$x^2+p y^2=1$ $(i)$
$q x^2+y^2=1$ (ii)
Differentiating $(i)$ with respect to $x$:
$2x + 2py \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_1 = -\frac{x}{py}$
Differentiating (ii) with respect to $x$:
$2qx + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_2 = -\frac{qx}{y}$
Since the curves are orthogonal,$m_1 \cdot m_2 = -1$:
$(-\frac{x}{py}) \cdot (-\frac{qx}{y}) = -1$
$\frac{qx^2}{py^2} = -1 \implies qx^2 = -py^2$
From $(i)$,$x^2 = 1 - py^2$. Substituting this into the condition:
$q(1 - py^2) = -py^2$
$q - qpy^2 = -py^2$
$q = y^2(qp - p) \implies y^2 = \frac{q}{p(q-1)}$
Similarly,$x^2 = \frac{p(1-q)}{q-p}$. Substituting $x^2$ and $y^2$ into $q x^2 + y^2 = 1$:
$q(\frac{p(1-q)}{q-p}) + \frac{q}{p(q-1)} = 1$
After simplifying,we get $p+q = 2pq$,which implies $\frac{1}{p} + \frac{1}{q} = 2$.

(A) Let $I = \int \frac{x-\sin x}{1+\cos x} dx$.
Using the identities $1+\cos x = 2\cos^2(\frac{x}{2})$ and $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$,we have:
$I = \int \frac{x}{2\cos^2(\frac{x}{2})} dx - \int \frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{2\cos^2(\frac{x}{2})} dx$
$I = \frac{1}{2} \int x \sec^2(\frac{x}{2}) dx - \int \tan(\frac{x}{2}) dx$.
Applying integration by parts to the first term:
$\int x \sec^2(\frac{x}{2}) dx = x(2\tan(\frac{x}{2})) - \int 1 \cdot 2\tan(\frac{x}{2}) dx = 2x\tan(\frac{x}{2}) - 2 \int \tan(\frac{x}{2}) dx$.
Substituting this back into $I$:
$I = \frac{1}{2} [2x\tan(\frac{x}{2}) - 2 \int \tan(\frac{x}{2}) dx] - \int \tan(\frac{x}{2}) dx$
$I = x\tan(\frac{x}{2}) - \int \tan(\frac{x}{2}) dx - \int \tan(\frac{x}{2}) dx = x\tan(\frac{x}{2}) - 2 \int \tan(\frac{x}{2}) dx$.
Since $\int \tan(\frac{x}{2}) dx = 2 \log |\sec(\frac{x}{2})|$,we get:
$I = x\tan(\frac{x}{2}) - 2(2 \log |\sec(\frac{x}{2})|) + C = x\tan(\frac{x}{2}) - 4 \log |\sec(\frac{x}{2})| + C$.
Comparing this with the given expression $x \tan(\frac{x}{2}) + p \log |\sec(\frac{x}{2})| + C$,we find $p = -4$.

(D) We have,$\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$.
Integrating both sides,we get $[\tan^{-1} x]_0^b = [\tan^{-1} x]_b^{\infty}$.
Substituting the limits,$\tan^{-1}(b) - \tan^{-1}(0) = \tan^{-1}(\infty) - \tan^{-1}(b)$.
Since $\tan^{-1}(0) = 0$ and $\tan^{-1}(\infty) = \frac{\pi}{2}$,we have $\tan^{-1}(b) = \frac{\pi}{2} - \tan^{-1}(b)$.
Adding $\tan^{-1}(b)$ to both sides,we get $2 \tan^{-1}(b) = \frac{\pi}{2}$.
Therefore,$\tan^{-1}(b) = \frac{\pi}{4}$.
Taking tangent on both sides,$b = \tan(\frac{\pi}{4}) = 1$.

(C) Given curves are $x = -2y^2$ and $x = 1 - 3y^2$.
To find the intersection points,set the equations equal to each other:
$-2y^2 = 1 - 3y^2$
$y^2 = 1$
$y = \pm 1$
When $y = 1$,$x = -2(1)^2 = -2$. When $y = -1$,$x = -2(-1)^2 = -2$.
The intersection points are $(-2, 1)$ and $(-2, -1)$.
The area $A$ is given by the integral with respect to $y$ from $-1$ to $1$ of the right curve minus the left curve:
$A = \int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) \, dy$
$A = \int_{-1}^{1} (1 - y^2) \, dy$
Since the function is even,$A = 2 \int_{0}^{1} (1 - y^2) \, dy$
$A = 2 [y - \frac{y^3}{3}]_{0}^{1}$
$A = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ square units.

(D) Given differential equation is: $(1+y+x^2 y) dx + (x+x^3) dy = 0$.
Rearranging the terms,we get: $(x+x^3) dy = -(1+y+x^2 y) dx$.
Dividing by $dx(x+x^3)$,we get: $\frac{dy}{dx} = -\frac{1+y(1+x^2)}{x(1+x^2)}$.
This simplifies to: $\frac{dy}{dx} = -\frac{1}{x(1+x^2)} - \frac{y(1+x^2)}{x(1+x^2)}$.
Thus,$\frac{dy}{dx} + \frac{y}{x} = -\frac{1}{x(1+x^2)}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = -\frac{1}{x(1+x^2)}$.
The integrating factor $(IF)$ is given by $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.
Wait,let us re-evaluate the rearrangement: $(1+y+x^2 y) dx + x(1+x^2) dy = 0$.
Dividing by $dx(1+x^2)$,we get $\frac{1+y(1+x^2)}{1+x^2} dx + x dy = 0$.
This is not standard. Let us rewrite as $\frac{dy}{dx} + \frac{y}{x} = -\frac{1}{x(1+x^2)}$.
Actually,the integrating factor is $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.
Re-checking the options,the correct integrating factor is $x$.

(B) The given differential equation is $\frac{dy}{dx} - 2y \tan 2x = e^x \sec 2x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -2 \tan 2x$ and $Q = e^x \sec 2x$.
First,we find the integrating factor $(IF)$:
$IF = e^{\int P dx} = e^{\int -2 \tan 2x dx} = e^{-\ln(\sec 2x)} = e^{\ln(\cos 2x)} = \cos 2x$.
The general solution is given by $y \cdot IF = \int (Q \cdot IF) dx + C$.
Substituting the values:
$y \cos 2x = \int (e^x \sec 2x \cdot \cos 2x) dx + C$.
Since $\sec 2x \cdot \cos 2x = 1$,we have:
$y \cos 2x = \int e^x dx + C$.
$y \cos 2x = e^x + C$,where $C$ is the constant of integration.

(C) Given that $a \neq 0, b \neq 0, c \neq 0$ and $a \times b = 0, b \times c = 0$.
Since $a \times b = 0$,the vectors $a$ and $b$ are parallel to each other.
Since $b \times c = 0$,the vectors $b$ and $c$ are parallel to each other.
By the transitive property of parallel vectors,if $a$ is parallel to $b$ and $b$ is parallel to $c$,then $a$ must be parallel to $c$.
Therefore,the cross product of two parallel vectors is zero,so $a \times c = 0$.

(C) Given $a \cdot y = c$ and $a \times y = b$. Since $a$ and $b$ are perpendicular,$a \cdot b = 0$.
Taking the cross product of $a$ with the second equation:
$a \times (a \times y) = a \times b$
Using the vector triple product identity $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$:
$(a \cdot y)a - (a \cdot a)y = a \times b$
Substitute $a \cdot y = c$ and $a \cdot a = |a|^2$:
$c a - |a|^2 y = a \times b$
Rearranging for $y$:
$|a|^2 y = c a - (a \times b)$
$y = \frac{1}{|a|^2} [c a - (a \times b)]$

(B) Let the coordinates of the four points $P, Q, R$ and $S$ be $(3, -4, 5), (0, 0, 4), (-4, 5, 1)$ and $(-3, 4, 3)$ respectively.
The equation of line $PQ$ passing through $(3, -4, 5)$ and $(0, 0, 4)$ is given by:
$\frac{x-3}{0-3} = \frac{y+4}{0+4} = \frac{z-5}{4-5} = r_1$
$\Rightarrow \frac{x-3}{-3} = \frac{y+4}{4} = \frac{z-5}{-1} = r_1$
The equation of line $RS$ passing through $(-4, 5, 1)$ and $(-3, 4, 3)$ is given by:
$\frac{x+4}{-3+4} = \frac{y-5}{4-5} = \frac{z-1}{3-1} = r_2$
$\Rightarrow \frac{x+4}{1} = \frac{y-5}{-1} = \frac{z-1}{2} = r_2$
Let a general point on line $PQ$ be $(-3r_1+3, 4r_1-4, -r_1+5)$ and on line $RS$ be $(r_2-4, -r_2+5, 2r_2+1)$.
Since the lines intersect,we equate the coordinates:
$-3r_1+3 = r_2-4 \Rightarrow 3r_1+r_2 = 7$ (Eq. $i$)
$4r_1-4 = -r_2+5 \Rightarrow 4r_1+r_2 = 9$ (Eq. $ii$)
Subtracting Eq. $i$ from Eq. $ii$:
$(4r_1+r_2) - (3r_1+r_2) = 9 - 7 \Rightarrow r_1 = 2$
Substituting $r_1 = 2$ in Eq. $i$:
$3(2) + r_2 = 7 \Rightarrow r_2 = 1$
Substituting $r_2 = 1$ into the coordinates of line $RS$:
$x = 1-4 = -3, y = -1+5 = 4, z = 2(1)+1 = 3$
Thus,the point of intersection is $(-3, 4, 3)$,which corresponds to $-3i+4j+3k$.

(C) Let the normal vector to the plane be $\vec{n} = (a, b, c)$. Since the normal makes equal angles with the coordinate axes,the direction cosines are equal,i.e.,$\cos \alpha = \cos \beta = \cos \gamma$. This implies $a = b = c$.
Thus,the equation of the plane is of the form $x + y + z = d$.
Since the plane passes through the point $(-1, 2, 3)$,we substitute these coordinates into the equation:
$-1 + 2 + 3 = d
\Rightarrow d = 4$.
Substituting the value of $d$ back into the equation,we get:
$x + y + z = 4$
$x + y + z - 4 = 0$.

(D) The given lines are $r = a_1 + \lambda b_1$ and $r = a_2 + \mu b_2$,where $a_1 = 3i + 5j + 7k$,$b_1 = i + 2j + k$,$a_2 = -i - j - k$,and $b_2 = 7i - 6j + k$.
First,we calculate the cross product $b_1 \times b_2$:
$b_1 \times b_2 = \begin{vmatrix} i & j & k \\ 1 & 2 & 1 \\ 7 & -6 & 1 \end{vmatrix} = i(2 + 6) - j(1 - 7) + k(-6 - 14) = 8i + 6j - 20k$.
The magnitude is $|b_1 \times b_2| = \sqrt{8^2 + 6^2 + (-20)^2} = \sqrt{64 + 36 + 400} = \sqrt{500} = 10\sqrt{5}$.
Next,we find $a_2 - a_1 = (-i - j - k) - (3i + 5j + 7k) = -4i - 6j - 8k$.
The scalar triple product is $(a_2 - a_1) \cdot (b_1 \times b_2) = (-4i - 6j - 8k) \cdot (8i + 6j - 20k) = (-4)(8) + (-6)(6) + (-8)(-20) = -32 - 36 + 160 = 92$.
The shortest distance is given by $d = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|} = \frac{92}{10\sqrt{5}} = \frac{46}{5\sqrt{5}}$.

(A) For a Poisson distribution with parameter $\lambda$,the probability mass function is given by $P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}$.
Given $P(X=1) = 2P(X=2)$.
Substituting the formula:
$\frac{\lambda^1 e^{-\lambda}}{1!} = 2 \times \frac{\lambda^2 e^{-\lambda}}{2!}$
$\lambda e^{-\lambda} = 2 \times \frac{\lambda^2 e^{-\lambda}}{2}$
$\lambda e^{-\lambda} = \lambda^2 e^{-\lambda}$
Since $e^{-\lambda} \neq 0$,we divide both sides by $e^{-\lambda}$:
$\lambda = \lambda^2$
$\lambda^2 - \lambda = 0 \Rightarrow \lambda(\lambda - 1) = 0$.
Since $\lambda > 0$ for a Poisson distribution,we have $\lambda = 1$.
Now,we calculate $P(X=3)$:
$P(X=3) = \frac{\lambda^3 e^{-\lambda}}{3!} = \frac{1^3 e^{-1}}{3 \times 2 \times 1} = \frac{e^{-1}}{6}$.

(B) The mean of $X$ is given by $\bar{X} = E(X) = \sum_{n=1}^{m} n \cdot P(X=n) = \sum_{n=1}^{m} n \cdot \frac{1}{m} = \frac{1}{m} \cdot \frac{m(m+1)}{2} = \frac{m+1}{2}$.
The variance of $X$ is given by $\text{Var}(X) = E(X^2) - [E(X)]^2$.
First,calculate $E(X^2) = \sum_{n=1}^{m} n^2 \cdot P(X=n) = \frac{1}{m} \sum_{n=1}^{m} n^2 = \frac{1}{m} \cdot \frac{m(m+1)(2m+1)}{6} = \frac{(m+1)(2m+1)}{6}$.
Now,$\text{Var}(X) = \frac{(m+1)(2m+1)}{6} - \left( \frac{m+1}{2} \right)^2 = \frac{(m+1)(2m+1)}{6} - \frac{(m+1)^2}{4}$.
Taking $\frac{m+1}{2}$ as a common factor: $\text{Var}(X) = \frac{m+1}{2} \left[ \frac{2m+1}{3} - \frac{m+1}{2} \right] = \frac{m+1}{2} \left[ \frac{4m+2 - 3m - 3}{6} \right] = \frac{m+1}{2} \cdot \frac{m-1}{6} = \frac{m^2-1}{12}$.

(C) Given curves are $x = -2y^2$ and $x = 1 - 3y^2$.
To find the intersection points,set the equations equal to each other:
$-2y^2 = 1 - 3y^2$
$y^2 = 1$
$y = \pm 1$
When $y = 1$,$x = -2(1)^2 = -2$. When $y = -1$,$x = -2(-1)^2 = -2$.
So,the intersection points are $(-2, 1)$ and $(-2, -1)$.
The required area is given by the integral with respect to $y$ from $-1$ to $1$:
$Area = \int_{-1}^{1} |(1 - 3y^2) - (-2y^2)| dy$
$Area = \int_{-1}^{1} |1 - y^2| dy$
Since the function is symmetric about the $x$-axis,we can write:
$Area = 2 \int_{0}^{1} (1 - y^2) dy$
$Area = 2 [y - \frac{y^3}{3}]_{0}^{1}$
$Area = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ sq units.

(B) Given the matrix $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$.
Since $A$ satisfies the equation $x^2 + 4x - p = 0$,we have $A^2 + 4A - pI = 0$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (-8)(-8) + (5)(2) & (-8)(5) + (5)(4) \\ (2)(-8) + (4)(2) & (2)(5) + (4)(4) \end{bmatrix} = \begin{bmatrix} 64 + 10 & -40 + 20 \\ -16 + 8 & 10 + 16 \end{bmatrix} = \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix}$.
Next,calculate $4A$:
$4A = 4 \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix}$.
Now,substitute these into the equation $A^2 + 4A - pI = 0$:
$\begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} + \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} p & 0 \\ 0 & p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Summing the matrices:
$\begin{bmatrix} 74 - 32 - p & -20 + 20 - 0 \\ -8 + 8 - 0 & 26 + 16 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
$\begin{bmatrix} 42 - p & 0 \\ 0 & 42 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Comparing the elements,we get $42 - p = 0$,which implies $p = 42$.

(A) Given $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$.
We calculate the powers of $A$:
$A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$
By observation,we hypothesize that $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
Now,evaluate the expression $nA - (n-1)I$:
$nA - (n-1)I = n \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} - (n-1) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} n & n \\ 0 & n \end{bmatrix} - \begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}$
$= \begin{bmatrix} n - (n-1) & n - 0 \\ 0 - 0 & n - (n-1) \end{bmatrix} = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} = A^n$.
Thus,$A^n = nA - (n-1)I$ is correct.

(D) Let $\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ x+4 & x+6 & x+9 \\ x+8 & x+11 & x+15\end{array}\right|$.
Apply operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$,we get:
$\Delta = \left|\begin{array}{ccc}x+2 & x+3 & x+5 \\ 2 & 3 & 4 \\ 6 & 8 & 10\end{array}\right|$.
Apply operation $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$,we get:
$\Delta = \left|\begin{array}{ccc}x+2 & 1 & 3 \\ 2 & 1 & 2 \\ 6 & 2 & 4\end{array}\right|$.
Expand along $R_1$:
$\Delta = (x+2)(4-4) - 1(8-12) + 3(4-6)$
$\Delta = (x+2)(0) - 1(-4) + 3(-2)$
$\Delta = 0 + 4 - 6 = -2$.

(D) The given system of equations is:
$3x + 2y + z = 6$
$3x + 4y + 3z = 14$
$6x + 10y + 8z = a$
Let $A = \begin{bmatrix} 3 & 2 & 1 \\ 3 & 4 & 3 \\ 6 & 10 & 8 \end{bmatrix}$ and $B = \begin{bmatrix} 6 \\ 14 \\ a \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 3(32 - 30) - 2(24 - 18) + 1(30 - 24) = 3(2) - 2(6) + 6 = 6 - 12 + 6 = 0$.
Since $|A| = 0$,the system has either no solution or infinitely many solutions. For infinitely many solutions,we must have $(\text{adj } A) \cdot B = 0$.
The adjoint of $A$ is calculated as:
$\text{adj } A = \begin{bmatrix} 2 & -6 & 2 \\ -6 & 18 & -6 \\ 6 & -18 & 6 \end{bmatrix}$.
Now,compute $(\text{adj } A) \cdot B = 0$:
$\begin{bmatrix} 2 & -6 & 2 \\ -6 & 18 & -6 \\ 6 & -18 & 6 \end{bmatrix} \begin{bmatrix} 6 \\ 14 \\ a \end{bmatrix} = \begin{bmatrix} 12 - 84 + 2a \\ -36 + 252 - 6a \\ 36 - 252 + 6a \end{bmatrix} = \begin{bmatrix} 2a - 72 \\ 216 - 6a \\ 6a - 216 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$.
From the first row,$2a - 72 = 0 \implies a = 36$.
Checking with other rows,$216 - 6(36) = 216 - 216 = 0$.
Thus,for $a = 36$,the system has infinitely many solutions.

(C) For a system of homogeneous linear equations to have non-trivial solutions,the determinant of the coefficient matrix $A$ must be zero,i.e.,$|A| = 0$.
Given the matrix $A = \begin{bmatrix} t & t+1 & t-1 \\ t+1 & t & t+2 \\ t-1 & t+2 & t \end{bmatrix}$.
We set the determinant $|A| = 0$:
$|A| = \begin{vmatrix} t & t+1 & t-1 \\ t+1 & t & t+2 \\ t-1 & t+2 & t \end{vmatrix} = 0$.
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$|A| = \begin{vmatrix} t & t+1 & t-1 \\ 1 & -1 & 3 \\ -1 & 1 & 1 \end{vmatrix} = 0$.
Adding $R_2$ to $R_3$ $(R_3 \rightarrow R_3 + R_2)$:
$|A| = \begin{vmatrix} t & t+1 & t-1 \\ 1 & -1 & 3 \\ 0 & 0 & 4 \end{vmatrix} = 0$.
Expanding along the third row:
$4 \times \begin{vmatrix} t & t+1 \\ 1 & -1 \end{vmatrix} = 0$.
$4 \times (-t - (t+1)) = 0$.
$4 \times (-2t - 1) = 0$.
$-8t - 4 = 0 \Rightarrow t = -\frac{1}{2}$.
Since there is only one real value of $t$ $(t = -\frac{1}{2})$,the number of real values is $1$.

(C) We use the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$.
Given $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$.
Here $A = \frac{5}{13}$ and $B = \frac{3}{5}$.
Then $\sqrt{1-A^2} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
And $\sqrt{1-B^2} = \sqrt{1-\left(\frac{3}{5}\right)^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Substituting these values into the formula:
$\cos ^{-1} x = \cos ^{-1} \left( \frac{5}{13} \cdot \frac{3}{5} - \frac{12}{13} \cdot \frac{4}{5} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{15}{65} - \frac{48}{65} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{15-48}{65} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{-33}{65} \right)$.
Therefore,$x = \frac{-33}{65}$.

(D) We know that $\operatorname{coth}^{-1}(x) = \tanh^{-1}\left(\frac{1}{x}\right)$ for $|x| > 1$.
Therefore,$\operatorname{coth}^{-1}(2) = \tanh^{-1}\left(\frac{1}{2}\right)$.
Substituting this into the expression,we get:
$\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2) = \tanh ^{-1}\left(\frac{1}{2}\right)+\tanh ^{-1}\left(\frac{1}{2}\right) = 2 \tanh ^{-1}\left(\frac{1}{2}\right)$.
Using the logarithmic form $\tanh^{-1}(x) = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$,we have:
$2 \tanh^{-1}\left(\frac{1}{2}\right) = 2 \cdot \frac{1}{2} \log \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right) = \log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right) = \log 3$.

(A) Given,$f(x) = (p - x^n)^{1/n}$,where $p > 0$.
To find $f[f(x)]$,we substitute $f(x)$ into the function $f$:
$f[f(x)] = f((p - x^n)^{1/n})$
$= (p - ((p - x^n)^{1/n})^n)^{1/n}$
$= (p - (p - x^n))^{1/n}$
$= (p - p + x^n)^{1/n}$
$= (x^n)^{1/n}$
$= x$.

(B) Given $f(x+y) = f(x) \cdot f(y)$ for all $x, y \in R$.
This is a functional equation of the form $f(x) = a^x$.
Given $f(2) = 9$,we have $a^2 = 9$,which implies $a = 3$ (since $f$ is non-zero).
Thus,$f(x) = 3^x$.
Therefore,$f(6) = 3^6$.

(B) Given $f(x) = \frac{1}{1 + \frac{1}{x}} = \frac{x}{x + 1}$.
Now,$g(x) = \frac{1}{1 + \frac{1}{f(x)}} = \frac{1}{1 + \frac{x + 1}{x}} = \frac{1}{\frac{x + x + 1}{x}} = \frac{x}{2x + 1}$.
To find $g^{\prime}(x)$,we use the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:
$g^{\prime}(x) = \frac{(1)(2x + 1) - (x)(2)}{(2x + 1)^2} = \frac{2x + 1 - 2x}{(2x + 1)^2} = \frac{1}{(2x + 1)^2}$.
Substituting $x = 2$:
$g^{\prime}(2) = \frac{1}{(2(2) + 1)^2} = \frac{1}{(5)^2} = \frac{1}{25}$.

(C) Given the equation: $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$
Multiplying both sides by $\sqrt{xy}$,we get: $y+x=2\sqrt{xy}$
Squaring both sides: $(x+y)^2 = (2\sqrt{xy})^2$
$x^2+y^2+2xy = 4xy$
$x^2+y^2-2xy = 0$
$(x-y)^2 = 0$
This implies $x-y=0$,or $y=x$
Differentiating both sides with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$

(D) Let $f(x) = (x+1)(x^2+1)(x^4+1)(x^8+1)$.
Multiply and divide by $(x-1)$:
$f(x) = \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^4-1)(x^4+1)(x^8+1)}{(x-1)} = \frac{(x^8-1)(x^8+1)}{(x-1)} = \frac{x^{16}-1}{x-1}$.
Now,differentiate $f(x)$ with respect to $x$ using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$:
$\frac{d}{dx}\left(\frac{x^{16}-1}{x-1}\right) = \frac{(16x^{15})(x-1) - (x^{16}-1)(1)}{(x-1)^2} = \frac{16x^{16} - 16x^{15} - x^{16} + 1}{(x-1)^2} = \frac{15x^{16} - 16x^{15} + 1}{(x-1)^2}$.
Comparing this with the given expression $\frac{15x^p - 16x^q + 1}{(x-1)^2}$,we get $p = 16$ and $q = 15$.
Thus,$(p, q) = (16, 15)$.

(B) Given the equation: $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$ $(i)$
Taking the differential of both sides:
$-\frac{2}{f^2} df = -\frac{1}{v^2} dv + \frac{1}{u^2} du$
Given that the absolute errors in $u$ and $v$ are $p$,we have $du = p$ and $dv = p$.
Substituting these into the differential equation:
$-\frac{2}{f^2} df = -\frac{1}{v^2} p + \frac{1}{u^2} p$
$-\frac{2}{f^2} df = -p \left( \frac{1}{v^2} - \frac{1}{u^2} \right)$
Using the identity $a^2 - b^2 = (a - b)(a + b)$:
$-\frac{2}{f^2} df = -p \left( \frac{1}{v} - \frac{1}{u} \right) \left( \frac{1}{v} + \frac{1}{u} \right)$
Since $\frac{1}{v} - \frac{1}{u} = \frac{2}{f}$ from equation $(i)$:
$-\frac{2}{f^2} df = -p \left( \frac{2}{f} \right) \left( \frac{1}{v} + \frac{1}{u} \right)$
Dividing both sides by $-\frac{2}{f}$:
$\frac{df}{f} = p \left( \frac{1}{v} + \frac{1}{u} \right)$
Thus,the relative error in $f$ is $p \left( \frac{1}{u} + \frac{1}{v} \right)$.

(A) Given the relation $p V^{1/4} = C$,where $C$ is a constant.
Taking the natural logarithm on both sides: $\ln p + \frac{1}{4} \ln V = \ln C$.
Differentiating both sides with respect to $V$: $\frac{dp}{p} + \frac{1}{4} \frac{dV}{V} = 0$.
This implies $\frac{dp}{p} = -\frac{1}{4} \frac{dV}{V}$.
Given that the percentage decrease in volume is $\frac{dV}{V} = -\frac{1}{2} \% = -0.5 \%$.
Substituting this into the equation: $\frac{dp}{p} = -\frac{1}{4} (-0.5 \%) = 0.125 \% = \frac{1}{8} \%$.
Thus,the percentage increase in pressure is $\frac{1}{8} \%$.

(B) Let $I = \int \frac{dx}{x(\log x-2)(\log x-3)}$.
Substitute $t = \log x$,so $dt = \frac{dx}{x}$.
Then the integral becomes $I = \int \frac{dt}{(t-2)(t-3)}$.
Using partial fractions: $\frac{1}{(t-2)(t-3)} = \frac{A}{t-2} + \frac{B}{t-3}$.
$1 = A(t-3) + B(t-2)$.
For $t=2$,$A = -1$. For $t=3$,$B = 1$.
So,$I = \int \left( \frac{1}{t-3} - \frac{1}{t-2} \right) dt$.
$I = \log |t-3| - \log |t-2| + C$.
$I = \log \left| \frac{t-3}{t-2} \right| + C$.
Substituting $t = \log x$ back,we get $I = \log \left| \frac{\log x - 3}{\log x - 2} \right| + C$.

(D) Let $I = \int e^x \left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx$.
Using the trigonometric identities $1+\cos 2x = 2\cos^2 x$ and $\sin 2x = 2\sin x \cos x$,we get:
$I = \int e^x \left( \frac{2 + 2\sin x \cos x}{2\cos^2 x} \right) dx$
$I = \int e^x \left( \frac{2}{2\cos^2 x} + \frac{2\sin x \cos x}{2\cos^2 x} \right) dx$
$I = \int e^x (\sec^2 x + \tan x) dx$.
We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Here,$f(x) = \tan x$ and $f'(x) = \sec^2 x$.
Therefore,$I = e^x \tan x + C$.

(D) Given that $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$.
We know that $\int \frac{dx}{1+x^2} = \tan^{-1}(x) + C$.
Applying the limits:
$[\tan^{-1}(x)]_0^b = [\tan^{-1}(x)]_b^{\infty}$
$\tan^{-1}(b) - \tan^{-1}(0) = \lim_{x \to \infty} \tan^{-1}(x) - \tan^{-1}(b)$
Since $\tan^{-1}(0) = 0$ and $\lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2}$,we have:
$\tan^{-1}(b) - 0 = \frac{\pi}{2} - \tan^{-1}(b)$
$2 \tan^{-1}(b) = \frac{\pi}{2}$
$\tan^{-1}(b) = \frac{\pi}{4}$
$b = \tan\left(\frac{\pi}{4}\right)$
$b = 1$.

(C) Let $f(x) = \frac{1}{2+3x}$. We divide the interval $[1,3]$ into $n=2$ equal parts.
$h = \frac{3-1}{2} = 1$.
The points are $x_0 = 1, x_1 = 2, x_2 = 3$.
$f(x_0) = f(1) = \frac{1}{2+3(1)} = \frac{1}{5} = 0.2$.
$f(x_1) = f(2) = \frac{1}{2+3(2)} = \frac{1}{8} = 0.125$.
$f(x_2) = f(3) = \frac{1}{2+3(3)} = \frac{1}{11} \approx 0.0909$.
Using Simpson's rule: $\int_a^b f(x) dx \approx \frac{h}{3} [f(x_0) + 4f(x_1) + f(x_2)]$.
$\int_1^3 f(x) dx \approx \frac{1}{3} [0.2 + 4(0.125) + 0.0909] = \frac{1}{3} [0.2 + 0.5 + 0.0909] = \frac{0.7909}{3} \approx 0.2636$.
Comparing with the options,$\frac{29}{110} \approx 0.2636$.

(C) Let the position vectors of the vertices be $\vec{a} = 2\hat{i}+3\hat{j}+4\hat{k}$,$\vec{b} = 3\hat{i}+4\hat{j}+2\hat{k}$,and $\vec{c} = 4\hat{i}+2\hat{j}+3\hat{k}$.
The side vectors are:
$\vec{AB} = \vec{b} - \vec{a} = (3-2)\hat{i} + (4-3)\hat{j} + (2-4)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$.
$\vec{BC} = \vec{c} - \vec{b} = (4-3)\hat{i} + (2-4)\hat{j} + (3-2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}$.
$\vec{CA} = \vec{a} - \vec{c} = (2-4)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = -2\hat{i} + \hat{j} + \hat{k}$.
Calculating the lengths of the sides:
$|\vec{AB}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$.
$|\vec{BC}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
$|\vec{CA}| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$.
Since $|\vec{AB}| = |\vec{BC}| = |\vec{CA}| = \sqrt{6}$,all three sides are equal in length.
Therefore,the triangle is an equilateral triangle.

(C) Let the unit vector be $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
Let $\vec{a} = \hat{i} + \hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$.
Since $\vec{r}$ is coplanar with $\vec{a}$ and $\vec{b}$,the scalar triple product $[\vec{r}, \vec{a}, \vec{b}] = 0$.
$\left|\begin{matrix} x & y & z \\ 1 & 1 & 3 \\ 1 & 3 & 1 \end{matrix}\right| = 0 \Rightarrow x(1-9) - y(1-3) + z(3-1) = 0 \Rightarrow -8x + 2y + 2z = 0 \Rightarrow -4x + y + z = 0$ $(i)$.
Since $\vec{r}$ is perpendicular to $\vec{c}$,$\vec{r} \cdot \vec{c} = 0$.
$x + y + z = 0$ (ii).
Subtracting $(i)$ from (ii): $(x + y + z) - (-4x + y + z) = 0 \Rightarrow 5x = 0 \Rightarrow x = 0$.
Substituting $x=0$ into (ii),we get $y + z = 0 \Rightarrow y = -z$.
Since $\vec{r}$ is a unit vector,$x^2 + y^2 + z^2 = 1$.
$0^2 + y^2 + (-y)^2 = 1 \Rightarrow 2y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{2}}$.
If $y = \frac{1}{\sqrt{2}}$,then $z = -\frac{1}{\sqrt{2}}$. If $y = -\frac{1}{\sqrt{2}}$,then $z = \frac{1}{\sqrt{2}}$.
Thus,$\vec{r} = \pm \frac{1}{\sqrt{2}}(\hat{j} - \hat{k})$. Option $(C)$ is $\frac{1}{\sqrt{2}}(\hat{j} - \hat{k})$.

(A) Let the direction ratios of the lines $AB$ and $AC$ be $\vec{u} = \langle 1, -1, -1 \rangle$ and $\vec{v} = \langle 2, -1, 1 \rangle$ respectively.
Since the lines $AB$ and $AC$ lie in the plane $ABC$,their cross product $\vec{n} = \vec{u} \times \vec{v}$ will give the direction ratios of the normal to the plane.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 2 & -1 & 1 \end{vmatrix}$
$= \hat{i}(-1 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - (-2))$
$= \hat{i}(-2) - \hat{j}(3) + \hat{k}(1)$
$= -2\hat{i} - 3\hat{j} + \hat{k}$
Thus,the direction ratios are $\langle -2, -3, 1 \rangle$.
Multiplying by $-1$,we get the equivalent direction ratios $\langle 2, 3, -1 \rangle$.
Therefore,the correct option is $A$.

(B) Let the equation of the variable plane be $a(x - 1) + b(y - 2) + c(z - 3) = 0$,where $a, b, c$ are the direction ratios of the normal to the plane.
Let $Q(x_1, y_1, z_1)$ be the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane.
Since $OQ$ is perpendicular to the plane,the direction ratios of $OQ$ are $(x_1, y_1, z_1)$,which must be proportional to $(a, b, c)$.
Thus,$a = kx_1, b = ky_1, c = kz_1$ for some constant $k$.
Substituting these into the plane equation: $x_1(x_1 - 1) + y_1(y_1 - 2) + z_1(z_1 - 3) = 0$.
This simplifies to $x_1^2 + y_1^2 + z_1^2 - x_1 - 2y_1 - 3z_1 = 0$.
This is the equation of a sphere with diameter $OP$,where $O$ is the origin and $P$ is the fixed point $(1, 2, 3)$.

(C) The equation of a plane passing through the point $(x_0, y_0, z_0) = (-1, 2, 3)$ is given by $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$,where $\langle a, b, c \rangle$ are the direction ratios of the normal to the plane.
Substituting the point,we get $a(x + 1) + b(y - 2) + c(z - 3) = 0$.
Since the normal makes equal angles $\alpha$ with the coordinate axes,the direction cosines are $\cos \alpha, \cos \alpha, \cos \alpha$.
Thus,the direction ratios $a, b, c$ can be taken as $1, 1, 1$.
Substituting these into the equation of the plane:
$1(x + 1) + 1(y - 2) + 1(z - 3) = 0$
$x + 1 + y - 2 + z - 3 = 0$
$x + y + z - 4 = 0$.

(A) Total number of coins $= 2n+1$.
Number of two-headed coins $= n$.
Number of fair coins $= n+1$.
Let $H$ be the event that the toss results in a head.
The probability of picking a two-headed coin is $\frac{n}{2n+1}$,and for such a coin,$P(H) = 1$.
The probability of picking a fair coin is $\frac{n+1}{2n+1}$,and for such a coin,$P(H) = \frac{1}{2}$.
Using the law of total probability:
$P(H) = \left(\frac{n}{2n+1}\right) \times 1 + \left(\frac{n+1}{2n+1}\right) \times \frac{1}{2} = \frac{31}{42}$
$\frac{2n + n + 1}{2(2n+1)} = \frac{31}{42}$
$\frac{3n+1}{4n+2} = \frac{31}{42}$
$42(3n+1) = 31(4n+2)$
$126n + 42 = 124n + 62$
$2n = 20$
$n = 10$