AP EAMCET 2010 Mathematics Question Paper with Answer and Solution

(D) The given series is $\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$
Using the property $\log _b a = \frac{1}{\log _a b}$,we get:
$= \frac{1}{\log _2 4} - \frac{1}{\log _2 8} + \frac{1}{\log _2 16} - \ldots$
$= \frac{1}{\log _2 2^2} - \frac{1}{\log _2 2^3} + \frac{1}{\log _2 2^4} - \ldots$
$= \frac{1}{2 \log _2 2} - \frac{1}{3 \log _2 2} + \frac{1}{4 \log _2 2} - \ldots$
Since $\log _2 2 = 1$,the series becomes:
$= \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots$
We know the Taylor series expansion $\log _e(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$
For $x = 1$,$\log _e(1 + 1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots$
$\Rightarrow \log _e 2 = 1 - (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots)$
$\Rightarrow \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots = 1 - \log _e 2$

(D) Given inequality: $\frac{14 x}{x+1}-\frac{9 x-30}{x-4} < 0$
Simplify the expression:
$\frac{14 x(x-4)-(9 x-30)(x+1)}{(x+1)(x-4)} < 0$
$\frac{14 x^2-56 x-(9 x^2+9 x-30 x-30)}{(x+1)(x-4)} < 0$
$\frac{14 x^2-56 x-(9 x^2-21 x-30)}{(x+1)(x-4)} < 0$
$\frac{5 x^2-35 x+30}{(x+1)(x-4)} < 0$
$\frac{5(x^2-7 x+6)}{(x+1)(x-4)} < 0$
$\frac{5(x-1)(x-6)}{(x+1)(x-4)} < 0$
Using the wavy curve method (sign scheme) on the number line with critical points $-1, 1, 4, 6$:
The expression is negative in the intervals $(-1, 1)$ and $(4, 6)$.
Thus,$x \in (-1, 1) \cup (4, 6)$.

(B) Let $x-1 = t$,so $x = t+1$. Substituting this into the numerator:
$3(t+1)^2 + (t+1) + 1 = 3(t^2+2t+1) + t + 2 = 3t^2 + 6t + 3 + t + 2 = 3t^2 + 7t + 5$.
Now,the expression becomes:
$\frac{3t^2+7t+5}{t^4} = \frac{3}{t^2} + \frac{7}{t^3} + \frac{5}{t^4}$.
Comparing this with $\frac{a}{t} + \frac{b}{t^2} + \frac{c}{t^3} + \frac{d}{t^4}$,we get:
$a = 0, b = 3, c = 7, d = 5$.
Thus,$\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] = \left[\begin{array}{ll}0 & 3 \\ 7 & 5\end{array}\right]$.

(B) Given equation $x^3-6x^2+11x-6=0$ has roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = 6$
$\alpha\beta+\beta\gamma+\gamma\alpha = 11$
$\alpha\beta\gamma = 6$
Calculating $a, b, c$:
$b = \alpha\beta+\beta\gamma+\gamma\alpha = 11$
$a = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 6^2 - 2(11) = 36 - 22 = 14$
$c = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (6-\gamma)(6-\alpha)(6-\beta)$
Since $x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)$,the roots are $1, 2, 3$.
$c = (6-1)(6-2)(6-3) = 5 \times 4 \times 3 = 60$
Comparing the values: $b=11, a=14, c=60$.
Thus,$b < a < c$.

(A) Given the cubic equation: $x^3-b x^2+c x-d=0$.
Let the roots in geometric progression be $\frac{a}{r}, a, ar$.
From the relationship between roots and coefficients:
$1$. Sum of roots: $\frac{a}{r} + a + ar = b \Rightarrow a(\frac{1}{r} + 1 + r) = b$ ... $(i)$
$2$. Sum of roots taken two at a time: $\frac{a}{r} \cdot a + a \cdot ar + ar \cdot \frac{a}{r} = c \Rightarrow a^2(\frac{1}{r} + r + 1) = c$ ... (ii)
$3$. Product of roots: $\frac{a}{r} \cdot a \cdot ar = d \Rightarrow a^3 = d$ ... (iii)
Dividing (ii) by $(i)$: $\frac{a^2(\frac{1}{r} + 1 + r)}{a(\frac{1}{r} + 1 + r)} = \frac{c}{b} \Rightarrow a = \frac{c}{b}$.
Substituting $a = \frac{c}{b}$ into (iii): $(\frac{c}{b})^3 = d$ $\Rightarrow \frac{c^3}{b^3} = d$ $\Rightarrow c^3 = b^3 d$.

(C) Given the equation $x^3-a x^2+a x-1=0$.
We can factorize the equation as follows:
$(x^3-1) - a x(x-1) = 0$
$(x-1)(x^2+x+1) - a x(x-1) = 0$
$(x-1)(x^2+x+1-a x) = 0$
$(x-1)(x^2+(1-a)x+1) = 0$.
Since $\alpha \neq 1$ is a root,$\alpha$ must satisfy the quadratic equation $x^2+(1-a)x+1=0$.
Thus,$\alpha^2+(1-a)\alpha+1=0$.
Dividing by $\alpha$ (since $\alpha \neq 0$ as $0$ is not a root),we get $\alpha + (1-a) + \frac{1}{\alpha} = 0$,which implies $\frac{1}{\alpha} = a - 1 - \alpha$.
Alternatively,note that if $x$ is a root,then $\frac{1}{x}$ is also a root because the equation is reciprocal.
Substituting $x = \frac{1}{y}$ into the original equation:
$(\frac{1}{y})^3 - a(\frac{1}{y})^2 + a(\frac{1}{y}) - 1 = 0$
$1 - ay + ay^2 - y^3 = 0$
$y^3 - ay^2 + ay - 1 = 0$.
This is the same equation as the original.
Therefore,if $\alpha$ is a root,then $\frac{1}{\alpha}$ is also a root.

(D) $z=1+i \sqrt{3}$
Since $z$ is in the first quadrant,$\operatorname{Arg} z = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.
For the conjugate $\bar{z} = 1-i \sqrt{3}$,which is in the fourth quadrant,$\operatorname{Arg} \bar{z} = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$.
Therefore,$|\operatorname{Arg} z| + |\operatorname{Arg} \bar{z}| = |\frac{\pi}{3}| + |-\frac{\pi}{3}| = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2 \pi}{3}$.

(C) To find the value of $n$ for which $a_n = \frac{10^n}{n!}$ is maximum,we examine the ratio $\frac{a_n}{a_{n-1}}$.
$\frac{a_n}{a_{n-1}} = \frac{10^n}{n!} \times \frac{(n-1)!}{10^{n-1}} = \frac{10}{n}$.
For $a_n$ to be increasing,we require $\frac{a_n}{a_{n-1}} > 1$,which implies $\frac{10}{n} > 1$,so $n < 10$.
This means $a_1 < a_2 < \ldots < a_9 < a_{10}$.
For $n = 10$,$\frac{a_{10}}{a_9} = \frac{10}{10} = 1$,which implies $a_{10} = a_9$.
For $n > 10$,$\frac{a_n}{a_{n-1}} < 1$,which implies $a_n < a_{n-1}$.
Thus,the sequence $a_n$ reaches its maximum value at both $n = 9$ and $n = 10$.
Since $10$ is the greatest such value,the answer is $10$.

(A) Given that $\cos (x-y), \cos x, \cos (x+y)$ are in harmonic progression $(HP)$.
Therefore,$\cos x = \frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y) + \cos (x-y)}$.
Using the identity $2 \cos A \cos B = \cos (A+B) + \cos (A-B)$,we have:
$\cos x = \frac{\cos 2x + \cos 2y}{2 \cos x \cos y}$.
Using $\cos 2\theta = 2 \cos^2 \theta - 1$,we get:
$\cos x = \frac{(2 \cos^2 x - 1) + (2 \cos^2 y - 1)}{2 \cos x \cos y} = \frac{2 \cos^2 x + 2 \cos^2 y - 2}{2 \cos x \cos y}$.
$\cos x = \frac{\cos^2 x + \cos^2 y - 1}{\cos x \cos y}$.
$\cos^2 x \cos y = \cos^2 x + \cos^2 y - 1$.
$\cos^2 x \cos y - \cos^2 x = \cos^2 y - 1$.
$-\cos^2 x (1 - \cos y) = -(1 - \cos^2 y)$.
$\cos^2 x (1 - \cos y) = (1 - \cos y)(1 + \cos y)$.
Since $\cos x \neq \cos y$,we have $1 - \cos y \neq 0$,so we can divide by $(1 - \cos y)$:
$\cos^2 x = 1 + \cos y$.

(B) Let the required point be $P$. By the section formula,the coordinates of point $P$ dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ are given by:
$P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)$
Here,$(x_1, y_1, z_1) = (3, -2, 1)$,$(x_2, y_2, z_2) = (-2, 3, 11)$,$m = 2$,and $n = 3$.
Substituting these values into the formula:
$P = \left(\frac{2(-2) + 3(3)}{2+3}, \frac{2(3) + 3(-2)}{2+3}, \frac{2(11) + 3(1)}{2+3}\right)$
$P = \left(\frac{-4 + 9}{5}, \frac{6 - 6}{5}, \frac{22 + 3}{5}\right)$
$P = \left(\frac{5}{5}, \frac{0}{5}, \frac{25}{5}\right)$
$P = (1, 0, 5)$

(A) To find the image of the line $x+y-2=0$ in the $y$-axis,we replace $x$ with $-x$ in the equation of the line.
Substituting $x = -x$ into $x+y-2=0$,we get:
$(-x)+y-2=0$
$-x+y-2=0$
Multiplying by $-1$,we get:
$x-y+2=0$
Alternatively,the line $x+y=2$ passes through points $A(2, 0)$ and $B(0, 2)$.
The image of $A(2, 0)$ in the $y$-axis is $A'(-2, 0)$,and the image of $B(0, 2)$ is $B(0, 2)$ itself.
The equation of the line passing through $A'(-2, 0)$ and $B(0, 2)$ is:
$y-0 = \frac{2-0}{0-(-2)}(x-(-2))$
$y = \frac{2}{2}(x+2)$
$y = x+2$
$x-y+2=0$

(B) The equation of a line which makes equal intercepts $a$ on the positive $x$ and $y$ axes is given by $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$.
Since the distance of this line from the origin $(0, 0)$ is $1$ unit,we have:
$\left| \frac{0 + 0 - a}{\sqrt{1^2 + 1^2}} \right| = 1$
$\left| \frac{-a}{\sqrt{2}} \right| = 1 \implies a = \sqrt{2}$ (since intercepts are on positive axes).
So,the equation of the line is $x + y = \sqrt{2}$.
We are given another line $y = 2x + 3 + \sqrt{2}$,which can be written as $2x - y = -3 - \sqrt{2}$.
To find the intersection point $(x_0, y_0)$,we solve the system:
$x + y = \sqrt{2}$
$2x - y = -3 - \sqrt{2}$
Adding the two equations:
$3x = -3 \implies x_0 = -1$.
Substituting $x_0 = -1$ into $x + y = \sqrt{2}$:
$-1 + y_0 = \sqrt{2} \implies y_0 = \sqrt{2} + 1$.
Now,calculate $2x_0 + y_0$:
$2(-1) + (\sqrt{2} + 1) = -2 + \sqrt{2} + 1 = \sqrt{2} - 1$.

(C) The given line is $4x - 2y = 1$,which can be written as $2y = 4x - 1$ or $y = 2x - 1/2$. The slope of this line is $m_1 = 2$.
Since line $L$ is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$. Thus,$2 \times m_2 = -1$,so $m_2 = -1/2$.
The equation of line $L$ with slope $-1/2$ can be written as $x + 2y + \lambda = 0$.
The intercepts of this line on the coordinate axes are found by setting $x=0$ and $y=0$:
For $x=0$,$2y = -\lambda \implies y = -\lambda/2$.
For $y=0$,$x = -\lambda$.
The area of the triangle formed with the coordinate axes is $\frac{1}{2} \times |\text{base}| \times |\text{height}| = \frac{1}{2} |-\lambda| \times |-\lambda/2| = 4$.
$\frac{\lambda^2}{4} = 4 \implies \lambda^2 = 16 \implies \lambda = \pm 4$.
Substituting $\lambda = 4$ into $x + 2y + \lambda = 0$ gives $x + 2y + 4 = 0$,which is equivalent to $2x + 4y + 8 = 0$.
Substituting $\lambda = -4$ gives $x + 2y - 4 = 0$,which is equivalent to $2x + 4y - 8 = 0$.

(A) Let the image of the point $P(4, -13)$ be $P^{\prime}(x_1, y_1)$.
The line $AB$ is $5x + y + 6 = 0$.
The formula for the image $(x_1, y_1)$ of a point $(x_0, y_0)$ with respect to the line $ax + by + c = 0$ is given by:
$\frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$
Substituting the values $x_0 = 4, y_0 = -13, a = 5, b = 1, c = 6$:
$\frac{x_1 - 4}{5} = \frac{y_1 - (-13)}{1} = -2 \frac{5(4) + 1(-13) + 6}{5^2 + 1^2}$
$\frac{x_1 - 4}{5} = \frac{y_1 + 13}{1} = -2 \frac{20 - 13 + 6}{25 + 1}$
$\frac{x_1 - 4}{5} = \frac{y_1 + 13}{1} = -2 \frac{13}{26}$
$\frac{x_1 - 4}{5} = \frac{y_1 + 13}{1} = -1$
Now,solving for $x_1$ and $y_1$:
$\frac{x_1 - 4}{5} = -1$ $\Rightarrow x_1 - 4 = -5$ $\Rightarrow x_1 = -1$
$\frac{y_1 + 13}{1} = -1$ $\Rightarrow y_1 + 13 = -1$ $\Rightarrow y_1 = -14$
Thus,the image of the point is $(-1, -14)$.

(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$.
We can rewrite the quadratic part as $2(4x^2 - 12xy + 9y^2) = 2(2x - 3y)^2$.
Let the lines be of the form $(2x - 3y + c_1)(2x - 3y + c_2) = 0$.
Expanding this,we get $4x^2 - 12xy + 9y^2 + 2(c_1 + c_2)x - 3(c_1 + c_2)y + c_1c_2 = 0$.
Comparing this with the given equation $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$,we divide the given equation by $2$: $4x^2 - 12xy + 9y^2 - 3x + 4.5y - 2.5 = 0$.
Here,$c_1 + c_2 = -3/2$ and $c_1c_2 = -2.5$.
The distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Using $(c_1 - c_2)^2 = (c_1 + c_2)^2 - 4c_1c_2 = (-1.5)^2 - 4(-2.5) = 2.25 + 10 = 12.25$.
So,$|c_1 - c_2| = \sqrt{12.25} = 3.5 = 7/2$.
The distance is $\frac{7/2}{\sqrt{2^2 + (-3)^2}} = \frac{7/2}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.

(A) The equation of the pair of lines passing through the origin and the intersection of $x^2+y^2=4$ and $x+y=a$ is obtained by homogenizing the circle equation using the line equation:
$x^2+y^2=4(\frac{x+y}{a})^2$
$a^2(x^2+y^2)=4(x^2+y^2+2xy)$
$(a^2-4)x^2-8xy+(a^2-4)y^2=0$
Since the lines are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(a^2-4)+(a^2-4)=0$
$2(a^2-4)=0$
$a^2=4$
Since $a>0$,we have $a=2$.

(B) The given equation of the pair of straight lines is $3x^2-11xy+10y^2-7x+13y+k=0$.
Comparing this with the general equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get:
$a=3, h=-\frac{11}{2}, b=10, g=-\frac{7}{2}, f=\frac{13}{2}$.
The point of intersection $(x, y)$ of the pair of lines is given by the formula:
$x = \frac{hf-bg}{ab-h^2}$ and $y = \frac{gh-af}{ab-h^2}$.
First,calculate the denominator: $ab-h^2 = (3)(10) - (-\frac{11}{2})^2 = 30 - \frac{121}{4} = \frac{120-121}{4} = -\frac{1}{4}$.
Now,calculate the numerator for $x$: $hf-bg = (-\frac{11}{2})(\frac{13}{2}) - (10)(-\frac{7}{2}) = -\frac{143}{4} + 35 = \frac{-143+140}{4} = -\frac{3}{4}$.
So,$x = \frac{-3/4}{-1/4} = 3$.
Now,calculate the numerator for $y$: $gh-af = (-\frac{7}{2})(-\frac{11}{2}) - (3)(\frac{13}{2}) = \frac{77}{4} - \frac{39}{2} = \frac{77-78}{4} = -\frac{1}{4}$.
So,$y = \frac{-1/4}{-1/4} = 1$.
Therefore,the point of intersection is $(3, 1)$.

(C) Let $P(x_1, y_1)$ be the point from which the tangents are drawn to the circles:
$S_1 \equiv x^2+y^2-8x+40=0$
$S_2 \equiv x^2+y^2-5x+16=0$ (Dividing by $5$)
$S_3 \equiv x^2+y^2-8x+16y+160=0$
Since the lengths of the tangents from $P$ to the circles $S_1, S_2, S_3$ are equal,we have $\sqrt{S_1} = \sqrt{S_2} = \sqrt{S_3}$,which implies $S_1 = S_2 = S_3$.
Equating $S_1 = S_3$:
$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-8x_1+16y_1+160$
$40 = 16y_1+160$
$16y_1 = -120 \Rightarrow y_1 = -\frac{120}{16} = -\frac{15}{2}$.
Equating $S_1 = S_2$:
$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-5x_1+16$
$-8x_1+40 = -5x_1+16$
$3x_1 = 24 \Rightarrow x_1 = 8$.
Thus,the point $P$ is $\left(8, -\frac{15}{2}\right)$.

(C) Let the height of the tower be $h$ and the base be $P$. Let the two points be $C$ and $B$ such that $PC = a$ and $PB = b$. The angles of elevation are $\angle ACP = \theta$ and $\angle ABP = \phi$. Given $\theta + \phi = 90^{\circ}$.
In $\triangle ACP$,$\tan \theta = \frac{h}{a}$.
In $\triangle ABP$,$\tan \phi = \frac{h}{b}$.
Since $\theta + \phi = 90^{\circ}$,we have $\phi = 90^{\circ} - \theta$,so $\tan \phi = \tan(90^{\circ} - \theta) = \cot \theta = \frac{1}{\tan \theta}$.
Substituting the values,$\frac{h}{b} = \frac{1}{h/a} = \frac{a}{h}$.
Therefore,$h^2 = ab$,which gives $h = \sqrt{ab}$.

(B) Given the quadratic equation $x^2+(\sqrt{3}+2)x+(\sqrt{3}-1)=0$.
By the relation between roots and coefficients,we have:
$m_1+m_2 = -(\sqrt{3}+2)$
$m_1m_2 = \sqrt{3}-1$
We calculate the difference of the roots:
$|m_1-m_2| = \sqrt{(m_1+m_2)^2 - 4m_1m_2} = \sqrt{(\sqrt{3}+2)^2 - 4(\sqrt{3}-1)}$
$= \sqrt{(3+4+4\sqrt{3}) - 4\sqrt{3}+4} = \sqrt{11}$.
The vertices of the triangle formed by $y=m_1x$,$y=m_2x$,and $y=c$ are $(0,0)$,$(c/m_1, c)$,and $(c/m_2, c)$.
The area of the triangle is given by:
$\text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$
$= \frac{1}{2} |0(c-c) + \frac{c}{m_1}(c-0) + \frac{c}{m_2}(0-c)| = \frac{1}{2} |\frac{c^2}{m_1} - \frac{c^2}{m_2}| = \frac{c^2}{2} |\frac{m_2-m_1}{m_1m_2}|$
$= \frac{c^2}{2} \cdot \frac{\sqrt{11}}{\sqrt{3}-1} = \frac{c^2}{2} \cdot \frac{\sqrt{11}(\sqrt{3}+1)}{3-1} = \frac{c^2}{2} \cdot \frac{\sqrt{33}+\sqrt{11}}{2} = \left(\frac{\sqrt{33}+\sqrt{11}}{4}\right)c^2$.

(C) Let $z = \sqrt{3}+i$.
Converting to polar form,$r = \sqrt{(\sqrt{3})^2 + 1^2} = 2$ and $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$.
So,$z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6})$ and $\bar{z} = 2(\cos \frac{\pi}{6} - i \sin \frac{\pi}{6})$.
Using De Moivre's Theorem,$z^7 + \bar{z}^7 = 2^7(\cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6}) + 2^7(\cos \frac{7\pi}{6} - i \sin \frac{7\pi}{6})$.
$= 2^7(2 \cos \frac{7\pi}{6}) = 2^8 \cos(\pi + \frac{\pi}{6})$.
$= -2^8 \cos(\frac{\pi}{6}) = -256 \times \frac{\sqrt{3}}{2} = -128 \sqrt{3}$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
Consider the expression $(x+1)(x+\omega)(x-\omega-1)$.
Note that $-\omega-1 = \omega^2$ since $1 + \omega + \omega^2 = 0$.
So,the expression becomes $(x+1)(x+\omega)(x+\omega^2)$.
We know the identity $(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc$.
Here,the roots are $-1, -\omega, -\omega^2$.
Thus,$(x+1)(x+\omega)(x+\omega^2) = x^3 - (-1-\omega-\omega^2)x^2 + (\omega+\omega^2+\omega^3)x - (1)(\omega)(\omega^2)$.
Since $1+\omega+\omega^2=0$,we have $-1-\omega-\omega^2=0$.
Also,$\omega+\omega^2+\omega^3 = -1 + 1 = 0$.
And $\omega^3 = 1$.
Therefore,the expression simplifies to $x^3 - (0)x^2 + (0)x - 1 = x^3 - 1$.

(C) To find the value of $n$ for which $a_n$ is maximum,we consider the ratio $\frac{a_n}{a_{n-1}}$.
$\frac{a_n}{a_{n-1}} = \frac{10^n}{n!} \times \frac{(n-1)!}{10^{n-1}} = \frac{10}{n}$.
For $a_n$ to be increasing,we require $\frac{a_n}{a_{n-1}} > 1$,which implies $\frac{10}{n} > 1$,so $n < 10$.
This means $a_1 < a_2 < \ldots < a_9 < a_{10}$.
For $n = 10$,$\frac{a_{10}}{a_9} = \frac{10}{10} = 1$,which implies $a_{10} = a_9$.
For $n > 10$,$\frac{a_n}{a_{n-1}} < 1$,which implies $a_n < a_{n-1}$.
Thus,the sequence $a_n$ reaches its maximum value at both $n = 9$ and $n = 10$.

(D) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $54$,we have:
$\frac{n(n-3)}{2} = 54$
$n(n-3) = 108$
$n^2 - 3n - 108 = 0$
Factoring the quadratic equation:
$n^2 - 12n + 9n - 108 = 0$
$n(n - 12) + 9(n - 12) = 0$
$(n + 9)(n - 12) = 0$
This gives $n = -9$ or $n = 12$.
Since the number of sides $n$ must be a positive integer,we have $n = 12$.

(B) Given,$n = 1! + 4! + 7! + \ldots + 400!$.
We calculate the values of the factorials:
$1! = 1$
$4! = 24$
$7! = 5040$
$10! = 3628800$
For any $k \ge 10$,$k!$ ends in at least two zeros,so the last two digits of $k!$ are $00$.
Thus,the last two digits of $n$ are the same as the last two digits of $1! + 4! + 7! + 10! + \ldots$.
$1! + 4! + 7! = 1 + 24 + 5040 = 5065$.
Since all subsequent terms $10!, 13!, \ldots$ end in $00$,the sum $n$ ends in $65$.
Therefore,the ten's digit of $n$ is $6$.

(D) Given,$a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$
We can write $6^n$ as $(1 + 5)^n$.
Using the binomial expansion:
$6^n = (1 + 5)^n = {^nC_0} + {^nC_1}(5) + {^nC_2}(5^2) + {^nC_3}(5^3) + \ldots + {^nC_n}(5^n)$
$6^n = 1 + 5n + 25({^nC_2} + {^nC_3}(5) + \ldots + {^nC_n}(5^{n-2}))$
Now,subtract $5n$ from both sides:
$a_n = 6^n - 5n = 1 + 25({^nC_2} + 5{^nC_3} + \ldots + 5^{n-2}{^nC_n})$
Let $k = {^nC_2} + 5{^nC_3} + \ldots + 5^{n-2}{^nC_n}$,where $k$ is an integer for $n \geq 2$.
Thus,$a_n = 1 + 25k$.
For $n = 1$,$a_1 = 6^1 - 5(1) = 1$.
In both cases,$a_n$ divided by $25$ leaves a remainder of $1$.

(A) Given the expression $\frac{1}{(1-5x)(1-4x)}$ for $|x| < \frac{1}{5}$.
Using the binomial expansion $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we have:
$(1-5x)^{-1} = 1 + 5x + 25x^2 + 125x^3 + \dots$
$(1-4x)^{-1} = 1 + 4x + 16x^2 + 64x^3 + \dots$
Multiplying these two series:
$(1 + 5x + 25x^2 + 125x^3 + \dots)(1 + 4x + 16x^2 + 64x^3 + \dots)$
The coefficient of $x^3$ is obtained by collecting terms that result in $x^3$:
$1 \cdot (64x^3) + (5x) \cdot (16x^2) + (25x^2) \cdot (4x) + (125x^3) \cdot 1$
$= 64 + 80 + 100 + 125$
$= 369$

(A) Given $(1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}$.
Using the binomial expansion $(1+y)^n = \sum_{k=0}^n {}^{n}C_k y^k$,let $y = 2x+3x^2$.
$(1+2x+3x^2)^{10} = {}^{10}C_0 + {}^{10}C_1(2x+3x^2) + {}^{10}C_2(2x+3x^2)^2 + \ldots$
$= 1 + 10(2x+3x^2) + 45(4x^2+12x^3+9x^4) + \ldots$
$= 1 + 20x + 30x^2 + 180x^2 + \ldots$
$= 1 + 20x + 210x^2 + \ldots$
Comparing coefficients,we get $a_1 = 20$ and $a_2 = 210$.
Therefore,$\frac{a_2}{a_1} = \frac{210}{20} = 10.5$.

(A) Given equation: $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$
Divide by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3+1-2\sqrt{3} + 3+1+2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.
$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin \theta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos \theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.
We know $\cos 15^{\circ} = \cos \frac{\pi}{12} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ} = \sin \frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
So,$\sin \frac{\pi}{12} \sin \theta + \cos \frac{\pi}{12} \cos \theta = \cos \frac{\pi}{4}$.
$\cos(\theta - \frac{\pi}{12}) = \cos \frac{\pi}{4}$.
General solution: $\theta - \frac{\pi}{12} = 2n\pi \pm \frac{\pi}{4}$.
$\theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$,where $n \in Z$.

(B) Given: $a \sin^2 \theta + b \cos^2 \theta = c$
Divide both sides by $\cos^2 \theta$:
$a \tan^2 \theta + b = c \sec^2 \theta$
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$:
$a \tan^2 \theta + b = c(1 + \tan^2 \theta)$
$a \tan^2 \theta + b = c + c \tan^2 \theta$
Rearranging the terms to isolate $\tan^2 \theta$:
$a \tan^2 \theta - c \tan^2 \theta = c - b$
$(a - c) \tan^2 \theta = c - b$
$\tan^2 \theta = \frac{c - b}{a - c}$

(A) Let the given expression be $f(\theta) = \left(\tan \theta - \frac{1}{3} \tan^3 \theta\right) \left(\frac{1}{3} - \tan^2 \theta\right)^{-1}$.
$f(\theta) = \frac{\tan \theta - \frac{1}{3} \tan^3 \theta}{\frac{1}{3} - \tan^2 \theta} = \frac{3 \tan \theta - \tan^3 \theta}{3(1 - 3 \tan^2 \theta)} \times 3 = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$.
Using the identity $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we get $f(\theta) = \tan 3\theta$.
The period of $\tan x$ is $\pi$. Therefore,the period of $\tan 3\theta$ is $\frac{\pi}{|3|} = \frac{\pi}{3}$.

(B) Let the required point be $P(x, y, z)$. By the section formula,the coordinates of point $P$ dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ are given by:
$P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)$
Here,$(x_1, y_1, z_1) = (3, -2, 1)$,$(x_2, y_2, z_2) = (-2, 3, 11)$,$m = 2$,and $n = 3$.
Substituting these values into the formula:
$P = \left(\frac{2(-2) + 3(3)}{2+3}, \frac{2(3) + 3(-2)}{2+3}, \frac{2(11) + 3(1)}{2+3}\right)$
$P = \left(\frac{-4 + 9}{5}, \frac{6 - 6}{5}, \frac{22 + 3}{5}\right)$
$P = \left(\frac{5}{5}, \frac{0}{5}, \frac{25}{5}\right)$
$P = (1, 0, 5)$

(A) The given circle is $x^2+y^2-6x+12y+15=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=6, c=15$.
The radius $r_1$ is $\sqrt{g^2+f^2-c} = \sqrt{(-3)^2+6^2-15} = \sqrt{9+36-15} = \sqrt{30}$.
The area of the given circle is $A_1 = \pi r_1^2 = 30\pi$.
Let the concentric circle be $x^2+y^2-6x+12y+k=0$.
Its radius $r_2$ satisfies $r_2^2 = g^2+f^2-k = 45-k$.
According to the problem,the area of the new circle is $A_2 = 2A_1 = 60\pi$.
Thus,$\pi r_2^2 = 60\pi$,which implies $r_2^2 = 60$.
Substituting $r_2^2 = 45-k$,we get $45-k = 60$,so $k = -15$.
Therefore,the equation of the circle is $x^2+y^2-6x+12y-15=0$.

(C) Let $P(x_1, y_1)$ be the point. The length of the tangent from $P$ to a circle $S=0$ is $\sqrt{S}$.
For the given circles,we normalize them to the form $x^2+y^2+2gx+2fy+c=0$:
$S_1: x^2+y^2-8x+40=0$
$S_2: x^2+y^2-5x+16=0$
$S_3: x^2+y^2-8x+16y+160=0$
Since the lengths of the tangents are equal,$S_1 = S_2 = S_3$ at point $P(x_1, y_1)$:
$x_1^2+y_1^2-8x_1+40 = x_1^2+y_1^2-5x_1+16 = x_1^2+y_1^2-8x_1+16y_1+160$
Equating $S_1 = S_3$:
$-8x_1+40 = -8x_1+16y_1+160$ $\Rightarrow 16y_1 = -120$ $\Rightarrow y_1 = -\frac{15}{2}$
Equating $S_1 = S_2$:
$-8x_1+40 = -5x_1+16$ $\Rightarrow 3x_1 = 24$ $\Rightarrow x_1 = 8$
Thus,the point $P$ is $\left(8, -\frac{15}{2}\right)$.

(C) Let the radii of the two circles be $r_1 = 15$ and $r_2 = 20$. The distance between their centers $C_1$ and $C_2$ is $d = 25$.
Note that $r_1^2 + r_2^2 = 15^2 + 20^2 = 225 + 400 = 625 = 25^2 = d^2$.
Since the sum of the squares of the radii equals the square of the distance between the centers,the triangle $\triangle AC_1C_2$ is a right-angled triangle with $\angle C_1AC_2 = 90^\circ$.
The common chord $AB$ is perpendicular to the line joining the centers $C_1C_2$. Let the intersection point be $D$.
In $\triangle AC_1C_2$,the altitude $AD$ to the hypotenuse $C_1C_2$ represents half the length of the common chord.
Using the area of the triangle: $\text{Area} = \frac{1}{2} \times r_1 \times r_2 = \frac{1}{2} \times d \times AD$.
$15 \times 20 = 25 \times AD$.
$AD = \frac{300}{25} = 12$.
The length of the common chord is $2 \times AD = 2 \times 12 = 24$ units.

(B) The equations of the circles are $S_1 \equiv x^2+y^2+2x+3y+1=0$ and $S_2 \equiv x^2+y^2+4x+3y+2=0$.
Since the circles intersect at $A$ and $B$,the equation of the common chord $AB$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$
$-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}$.
The family of circles passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda(S_1 - S_2) = 0$.
However,the circle with $AB$ as diameter is given by $S_1 + k(S_1 - S_2) = 0$ where the coefficient of $x^2$ and $y^2$ is $1$.
Alternatively,using the property that the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$,we substitute $x = -\frac{1}{2}$ into $S_1$:
$S_1 = (-\frac{1}{2})^2 + y^2 + 2(-\frac{1}{2}) + 3y + 1 = y^2 + 3y + \frac{1}{4} = 0$.
The equation of the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$.
Substituting $x = -\frac{1}{2}$ into $S_1$ and $S_2$ gives the points $A$ and $B$.
The circle with diameter $AB$ is $x^2+y^2+2x+3y+1 + k(2x+1) = 0$.
Since the center lies on the line $x = -\frac{1}{2}$,we find $k$ such that the equation simplifies to $x^2+y^2+x+3y+\frac{1}{2} = 0$,which is $2x^2+2y^2+2x+6y+1 = 0$.

(C) The equations of the circles are $7x^2+7y^2-7x+14y+18=0$ and $4x^2+4y^2-7x+8y+20=0$.
Divide the first equation by $7$: $x^2+y^2-x+2y+\frac{18}{7}=0$ $(S_1=0)$.
Divide the second equation by $4$: $x^2+y^2-\frac{7}{4}x+2y+5=0$ $(S_2=0)$.
The radical axis is given by $S_1-S_2=0$.
$(x^2+y^2-x+2y+\frac{18}{7}) - (x^2+y^2-\frac{7}{4}x+2y+5) = 0$.
$(-x + \frac{7}{4}x) + (\frac{18}{7} - 5) = 0$.
$\frac{3}{4}x + (\frac{18-35}{7}) = 0$.
$\frac{3}{4}x - \frac{17}{7} = 0$.
$21x - 68 = 0$.

(C) The general equation of a parabola with an axis parallel to the $y$-axis is given by $y = Ax^2 + Bx + C$ ...$(i)$
Since the parabola passes through $(0,4)$,we have $4 = A(0)^2 + B(0) + C$,which gives $C = 4$ ...(ii)
Since it passes through $(1,9)$,we have $9 = A(1)^2 + B(1) + 4$,which simplifies to $A + B = 5$ ...(iii)
Since it passes through $(4,5)$,we have $5 = A(4)^2 + B(4) + 4$,which simplifies to $16A + 4B = 1$ or $4A + B = \frac{1}{4}$ ...(iv)
Subtracting Eq. (iii) from Eq. (iv): $(4A + B) - (A + B) = \frac{1}{4} - 5$ $\Rightarrow 3A = \frac{-19}{4}$ $\Rightarrow A = \frac{-19}{12}$
Substituting $A$ into Eq. (iii): $\frac{-19}{12} + B = 5 \Rightarrow B = 5 + \frac{19}{12} = \frac{79}{12}$
Substituting $A, B, C$ into Eq. $(i)$,the equation is $y = \frac{-19}{12} x^2 + \frac{79}{12} x + 4$.

(C) Given that the $\triangle SPM$ is equilateral.
For the parabola $y^2=8(x-3)$,we have $4a=8$,so $a=2$.
The vertex is $(3, 0)$ and the focus $S$ is $(3+a, 0) = (5, 0)$.
The directrix is $x = 3-a = 3-2 = 1$. Wait,the directrix is $x = 3-2 = 1$. Let us re-evaluate.
For $y^2=4a(x-h)$,directrix is $x=h-a$. Here $h=3, a=2$,so $x=3-2=1$.
Let $P$ be $(x, y)$. Since $P$ is on the parabola,$PS = PM$,where $PM$ is the distance to the directrix $x=1$.
$PS = \sqrt{(x-5)^2 + y^2} = \sqrt{(x-5)^2 + 8(x-3)} = \sqrt{x^2-10x+25+8x-24} = \sqrt{x^2-2x+1} = |x-1|$.
$PM = |x-1|$. This confirms the definition of a parabola.
In $\triangle SPM$,$PS=PM$. For it to be equilateral,$PS=PM=SM$.
$SM$ is the distance from $S(5, 0)$ to $M(1, y)$. $SM = \sqrt{(5-1)^2 + (0-y)^2} = \sqrt{16+y^2}$.
Since $PS = |x-1|$,we have $PS^2 = (x-1)^2$.
Also $y^2 = 8(x-3)$,so $PS^2 = (x-1)^2 = x^2-2x+1$.
Equating $PS^2 = SM^2$: $(x-1)^2 = 16+y^2$.
$(x-1)^2 = 16 + 8(x-3) = 16 + 8x - 24 = 8x - 8$.
$x^2 - 2x + 1 = 8x - 8 \Rightarrow x^2 - 10x + 9 = 0$.
$(x-9)(x-1) = 0$. Since $x=1$ is the directrix,$x=9$.
If $x=9$,$y^2 = 8(9-3) = 8(6) = 48$,so $y = \pm 4\sqrt{3}$.
Thus $P = (9, 4\sqrt{3})$ or $(9, -4\sqrt{3})$.

(C) Given,conjugate lines are $2x + 3y + 12 = 0$ $(i)$ and $x - y + k = 0$ $(ii)$.
Two lines are conjugate with respect to a parabola if the pole of one line lies on the other line.
Let the pole of the line $2x + 3y + 12 = 0$ be $(x_1, y_1)$ with respect to the parabola $y^2 = 8x$.
The equation of the polar of $(x_1, y_1)$ with respect to $y^2 = 8x$ is $yy_1 = 4(x + x_1)$,which simplifies to $4x - yy_1 + 4x_1 = 0$.
Comparing this with $2x + 3y + 12 = 0$,we have $\frac{4}{2} = \frac{-y_1}{3} = \frac{4x_1}{12}$.
From $\frac{4}{2} = \frac{-y_1}{3}$,we get $y_1 = -6$.
From $\frac{4}{2} = \frac{4x_1}{12}$,we get $x_1 = 6$.
Thus,the pole is $(6, -6)$.
Since the lines are conjugate,the pole $(6, -6)$ must lie on the second line $x - y + k = 0$.
Substituting the coordinates: $6 - (-6) + k = 0$.
$12 + k = 0 \Rightarrow k = -12$.

(C) Given,the curve is $2x^2+y^2=2x$.
Rearranging the terms,we get $2x^2-2x+y^2=0$.
Completing the square for $x$: $2(x^2-x)+y^2=0 \Rightarrow 2(x-\frac{1}{2})^2+y^2=\frac{1}{2}$.
Dividing by $\frac{1}{2}$,we get $\frac{(x-\frac{1}{2})^2}{1/4} + \frac{y^2}{1/2} = 1$,which represents an ellipse.
Let a point $P$ on the ellipse be $P(\frac{1}{2}+\frac{1}{2}\cos\theta, \frac{1}{\sqrt{2}}\sin\theta)$.
The distance $PQ$ from $Q(a, 0)$ is given by $PQ^2 = (\frac{1}{2}+\frac{1}{2}\cos\theta-a)^2 + (\frac{1}{\sqrt{2}}\sin\theta)^2$.
$PQ^2 = ((\frac{1}{2}-a)+\frac{1}{2}\cos\theta)^2 + \frac{1}{2}\sin^2\theta$.
$PQ^2 = (\frac{1}{2}-a)^2 + \frac{1}{4}\cos^2\theta + (\frac{1}{2}-a)\cos\theta + \frac{1}{2}(1-\cos^2\theta)$.
$PQ^2 = \frac{1}{4}-a+a^2 + \frac{1}{4}\cos^2\theta + (\frac{1}{2}-a)\cos\theta + \frac{1}{2} - \frac{1}{2}\cos^2\theta$.
$PQ^2 = a^2-a+\frac{3}{4} - \frac{1}{4}\cos^2\theta + (\frac{1}{2}-a)\cos\theta$.
To find the maximum distance,we differentiate with respect to $\cos\theta$ and set to $0$,or analyze the quadratic in $\cos\theta$.
After simplification,the maximum distance is $\sqrt{1-2a+2a^2}$.

(B) Let $(a \sec \theta, b \tan \theta)$ be any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The equations of the asymptotes of the given hyperbola are $\frac{x}{a} + \frac{y}{b} = 0$ and $\frac{x}{a} - \frac{y}{b} = 0$.
Let $P_1$ be the length of the perpendicular from $(a \sec \theta, b \tan \theta)$ to $\frac{x}{a} + \frac{y}{b} = 0$.
$P_1 = \frac{|\sec \theta + \tan \theta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
Let $P_2$ be the length of the perpendicular from $(a \sec \theta, b \tan \theta)$ to $\frac{x}{a} - \frac{y}{b} = 0$.
$P_2 = \frac{|\sec \theta - \tan \theta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
The product $P_1 P_2 = \frac{|\sec^2 \theta - \tan^2 \theta|}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{a^2 + b^2}{a^2 b^2}} = \frac{a^2 b^2}{a^2 + b^2}$.

(C) The equation of a hyperbola with given asymptotes $L_1=0$ and $L_2=0$ is of the form $L_1 \times L_2 + k = 0$.
Given asymptotes are $(4x+3y-7)=0$ and $(x-2y-1)=0$.
So,the equation of the hyperbola is $(4x+3y-7)(x-2y-1)+k=0$ ...$(i)$
Since the hyperbola passes through the point $(2,3)$,we substitute $x=2$ and $y=3$ into Eq. $(i)$:
$(4(2)+3(3)-7)(2-2(3)-1)+k=0$
$(8+9-7)(2-6-1)+k=0$
$(10)(-5)+k=0$
$-50+k=0 \Rightarrow k=50$
Substituting $k=50$ back into Eq. $(i)$:
$(4x+3y-7)(x-2y-1)+50=0$
$4x^2-8xy-4x+3xy-6y^2-3y-7x+14y+7+50=0$
$4x^2-5xy-6y^2-11x+11y+57=0$

(C) We evaluate the limit $L = \lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$.
Since this is a $\frac{0}{0}$ indeterminate form,we apply $L$'Hospital's Rule:
$L = \lim _{x \rightarrow 0} \frac{\sec^2 x - \cos x}{3x^2}$.
This is still a $\frac{0}{0}$ form. Applying $L$'Hospital's Rule again:
$L = \lim _{x \rightarrow 0} \frac{2 \sec^2 x \tan x + \sin x}{6x}$.
Using the standard limits $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$L = \lim _{x}$ ${\rightarrow 0} \left( \frac{2 \sec^2 x \cdot \tan x}{6x} + \frac{\sin x}{6x} \right) = \frac{2(1)(1)}{6} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

(B) Given $\angle C = 90^{\circ}$,we have $A+B = 90^{\circ}$.
Using the sine rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C = k \sin 90^{\circ} = k$.
Then,$\frac{a^2-b^2}{a^2+b^2} = \frac{k^2 \sin^2 A - k^2 \sin^2 B}{k^2 \sin^2 A + k^2 \sin^2 B} = \frac{\sin^2 A - \sin^2 B}{\sin^2 A + \sin^2 B}$.
Since $B = 90^{\circ} - A$,we have $\sin B = \cos A$ and $\cos B = \sin A$.
Substituting these,$\frac{\sin^2 A - \cos^2 A}{\sin^2 A + \cos^2 A} = \frac{-(\cos^2 A - \sin^2 A)}{1} = -\cos 2A$.
Alternatively,using the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$:
$\frac{\sin(A+B)\sin(A-B)}{\sin^2 A + \cos^2 A} = \sin(90^{\circ})\sin(A-B) = 1 \cdot \sin(A-B) = \sin(A-B)$.

(B) Given $\Delta = a^2 - (b - c)^2$.
Using the identity $x^2 - y^2 = (x - y)(x + y)$,we have $\Delta = (a - (b - c))(a + (b - c)) = (a - b + c)(a + b - c)$.
Since $2s = a + b + c$,we have $a + b - c = 2s - 2c$ and $a - b + c = 2s - 2b$.
Thus,$\Delta = (2s - 2b)(2s - 2c) = 4(s - b)(s - c)$.
We know that $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.
Equating the two expressions: $\sqrt{s(s - a)(s - b)(s - c)} = 4(s - b)(s - c)$.
Dividing both sides by $\sqrt{(s - b)(s - c)}$,we get $\sqrt{s(s - a)} = 4\sqrt{(s - b)(s - c)}$.
This implies $\sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \frac{1}{4}$.
Since $\tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$,we have $\tan \frac{A}{2} = \frac{1}{4}$.
Using the formula $\tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^2 \frac{A}{2}}$,we get $\tan A = \frac{2 \times \frac{1}{4}}{1 - (\frac{1}{4})^2} = \frac{\frac{1}{2}}{1 - \frac{1}{16}} = \frac{\frac{1}{2}}{\frac{15}{16}} = \frac{8}{15}$.

(C) We are given the equation $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$ for $|x| < 1$ ...$(i)$
We know the standard logarithmic definition of the inverse hyperbolic tangent function is $\tanh ^{-1} x = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ ...(ii)
Comparing equation $(i)$ and equation (ii),we can equate the coefficients of the logarithmic term.
Therefore,$a = \frac{1}{2}$.

(D) Let $f(x) = \frac{x^2-6x+5}{x^2+2x+1}$ for $x \in R$.
Set $y = \frac{x^2-6x+5}{x^2+2x+1}$.
Then $y(x^2+2x+1) = x^2-6x+5$,which simplifies to $(y-1)x^2 + (2y+6)x + (y-5) = 0$.
Since $x$ is a real number,the discriminant $D$ must be greater than or equal to $0$.
$D = (2y+6)^2 - 4(y-1)(y-5) \geq 0$.
$4(y+3)^2 - 4(y^2-6y+5) \geq 0$.
$(y^2+6y+9) - (y^2-6y+5) \geq 0$.
$12y + 4 \geq 0$.
$12y \geq -4$.
$y \geq -\frac{1}{3}$.
Thus,the least value of the expression is $-\frac{1}{3}$.

(C) Let the height of the tower be $h$ and the points be $C$ and $B$ at distances $a$ and $b$ from the base $P$ respectively.
In $\triangle ACP$,$\tan \theta = \frac{h}{a} \implies \theta = \arctan(\frac{h}{a})$.
In $\triangle ABP$,$\tan \phi = \frac{h}{b} \implies \phi = \arctan(\frac{h}{b})$.
Given that $\theta + \phi = 90^{\circ}$,we have $\theta = 90^{\circ} - \phi$.
Taking tangent on both sides,$\tan \theta = \tan(90^{\circ} - \phi) = \cot \phi = \frac{1}{\tan \phi}$.
Substituting the values,$\frac{h}{a} = \frac{1}{h/b} = \frac{b}{h}$.
Therefore,$h^2 = ab$,which gives $h = \sqrt{ab}$.

(D) Given inequality: $\frac{14x}{x+1} - \frac{9x-30}{x-4} < 0$
Simplify the expression by taking a common denominator:
$\frac{14x(x-4) - (9x-30)(x+1)}{(x+1)(x-4)} < 0$
$\frac{14x^2 - 56x - (9x^2 + 9x - 30x - 30)}{(x+1)(x-4)} < 0$
$\frac{14x^2 - 56x - (9x^2 - 21x - 30)}{(x+1)(x-4)} < 0$
$\frac{14x^2 - 56x - 9x^2 + 21x + 30}{(x+1)(x-4)} < 0$
$\frac{5x^2 - 35x + 30}{(x+1)(x-4)} < 0$
Divide by $5$:
$\frac{x^2 - 7x + 6}{(x+1)(x-4)} < 0$
Factor the numerator:
$\frac{(x-1)(x-6)}{(x+1)(x-4)} < 0$
Using the wavy curve method (sign scheme) with critical points $-1, 1, 4, 6$:
The expression is negative in the intervals $(-1, 1)$ and $(4, 6)$.
Therefore,$x \in (-1, 1) \cup (4, 6)$.

(B) The given system of equations is:
$(k+1)^3 x + (k+2)^3 y = (k+3)^3$
$(k+1) x + (k+2) y = (k+3)$
$x + y = 1$
For the system to be consistent,the determinant of the augmented matrix must be zero.
Let $D = \begin{vmatrix} (k+1)^3 & (k+2)^3 & (k+3)^3 \\ (k+1) & (k+2) & (k+3) \\ 1 & 1 & 1 \end{vmatrix} = 0$.
Applying column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:
$D = \begin{vmatrix} (k+1)^3 & (k+2)^3 - (k+1)^3 & (k+3)^3 - (k+1)^3 \\ (k+1) & (k+2) - (k+1) & (k+3) - (k+1) \\ 1 & 0 & 0 \end{vmatrix} = 0$
$D = \begin{vmatrix} (k+1)^3 & 3k^2 + 9k + 7 & 6k^2 + 24k + 26 \\ (k+1) & 1 & 2 \\ 1 & 0 & 0 \end{vmatrix} = 0$
Expanding along the third row:
$1 \cdot [2(3k^2 + 9k + 7) - 1(6k^2 + 24k + 26)] = 0$
$6k^2 + 18k + 14 - 6k^2 - 24k - 26 = 0$
$-6k - 12 = 0$
$-6k = 12 \Rightarrow k = -2$.

(A) The function is $f(x) = x^2$.
For $f$ to be one-one,$f(x_1) = f(x_2) \implies x_1^2 = x_2^2$. This implies $x_1 = x_2$ only if $x_1, x_2$ have the same sign.
For $f$ to be onto,the range of $f$ must equal the codomain $B$. The range of $f(x) = x^2$ for $x \in A$ is ${x^2 : x \in A}$.
Analysis:
$(A)$ $A = B = R^{+}$: $f(x) = x^2$ maps positive reals to positive reals. It is one-one (since $x_1^2 = x_2^2$ and $x_1, x_2 > 0 \implies x_1 = x_2$) and onto (since for any $y \in R^{+}$,$x = \sqrt{y} \in R^{+}$ exists). Thus,$A \rightarrow 4$.
$(B)$ $A = R^{+}, B = R$: $f(x) = x^2$ is one-one (as $x > 0$),but not onto because negative values in $B$ are not covered by the range $(R^{+})$. Thus,$B \rightarrow 1$.
$(C)$ $A = R, B = R^{+}$: $f(x) = x^2$ is not one-one (as $f(1) = f(-1) = 1$) but is onto (since every $y \in R^{+}$ has a pre-image $x = \pm \sqrt{y} \in R$). Thus,$C \rightarrow 3$.
$(D)$ $A = B = R$: $f(x) = x^2$ is not one-one (as $f(1) = f(-1)$) and not onto (as negative values in $B$ are not covered). Thus,$D \rightarrow 2$.
The correct matching is $A-4, B-1, C-3, D-2$.

(D) Given: $f(0)=0, f(1)=1, f(2)=2$ and the recurrence relation $f(x)=f(x-2)+f(x-3)$ for $x \ge 3$.
We calculate the values step by step:
$f(3) = f(1) + f(0) = 1 + 0 = 1$
$f(4) = f(2) + f(1) = 2 + 1 = 3$
$f(5) = f(3) + f(2) = 1 + 2 = 3$
$f(6) = f(4) + f(3) = 3 + 1 = 4$
$f(7) = f(5) + f(4) = 3 + 3 = 6$
$f(8) = f(6) + f(5) = 4 + 3 = 7$
$f(9) = f(7) + f(6) = 6 + 4 = 10$
Therefore,$f(9) = 10$.

(B) For a function to be continuous at $x=0$,the limit of $f(x)$ as $x \rightarrow 0$ must equal $f(0)$.
Given $f(0) = k$.
We calculate $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{1+3x^2-\cos 2x}{x^2}$.
Using the identity $\cos 2x = 1 - 2\sin^2 x$,we substitute:
$\lim_{x \rightarrow 0} \frac{1+3x^2-(1-2\sin^2 x)}{x^2} = \lim_{x \rightarrow 0} \frac{1+3x^2-1+2\sin^2 x}{x^2}$.
$= \lim_{x \rightarrow 0} \frac{3x^2+2\sin^2 x}{x^2} = \lim_{x \rightarrow 0} \left( 3 + 2\left(\frac{\sin x}{x}\right)^2 \right)$.
Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$= 3 + 2(1)^2 = 3 + 2 = 5$.
Thus,for continuity,$k = 5$.

(D) Given $u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$.
Let $v=\sin u=\frac{x^4+y^4}{x+y}$.
Here,$v$ is a homogeneous function of $x$ and $y$ with degree $n = 4 - 1 = 3$.
According to Euler's theorem for homogeneous functions,$x \frac{\partial v}{\partial x} + y \frac{\partial v}{\partial y} = n v$.
Substituting $v = \sin u$ and $n = 3$,we get $x \frac{\partial}{\partial x}(\sin u) + y \frac{\partial}{\partial y}(\sin u) = 3 \sin u$.
Applying the chain rule,$x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = 3 \sin u$.
Dividing both sides by $\cos u$,we get $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 \frac{\sin u}{\cos u}$.
Therefore,$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 \tan u$.

(C) Given $y=\cos ^{-1}\left(\frac{a^2-x^2}{a^2+x^2}\right)+\sin ^{-1}\left(\frac{2 a x}{a^2+x^2}\right)$.
Let $x=a \tan \theta$,then $\theta=\tan ^{-1}\left(\frac{x}{a}\right)$.
Substituting $x$ in the expression:
$y=\cos ^{-1}\left(\frac{a^2-a^2 \tan ^2 \theta}{a^2+a^2 \tan ^2 \theta}\right) + \sin ^{-1}\left(\frac{2 a^2 \tan \theta}{a^2+a^2 \tan ^2 \theta}\right)$
$y=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) + \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$
Using trigonometric identities $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}$:
$y=\cos ^{-1}(\cos 2 \theta) + \sin ^{-1}(\sin 2 \theta)$
$y=2 \theta + 2 \theta = 4 \theta$.
Substituting back $\theta = \tan ^{-1}\left(\frac{x}{a}\right)$:
$y=4 \tan ^{-1}\left(\frac{x}{a}\right)$.
Now,differentiating with respect to $x$:
$\frac{d y}{d x} = 4 \cdot \frac{1}{1+(\frac{x}{a})^2} \cdot \frac{1}{a}$
$\frac{d y}{d x} = 4 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a} = \frac{4 a}{a^2+x^2}$.

(A) We are given the integral $I = \int(1-\cos x) \operatorname{cosec}^2 x \, dx$.
Using the trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,we have:
$I = \int (2 \sin^2 \frac{x}{2}) \cdot \frac{1}{\sin^2 x} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{(2 \sin \frac{x}{2} \cos \frac{x}{2})^2} \, dx$
$I = \int \frac{2 \sin^2 \frac{x}{2}}{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}} \, dx$
$I = \int \frac{1}{2 \cos^2 \frac{x}{2}} \, dx$
$I = \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx$
Integrating $\sec^2 \frac{x}{2}$,we get:
$I = \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{1/2} + c = \tan \frac{x}{2} + c$.
Comparing this with $f(x) + c$,we find $f(x) = \tan \frac{x}{2}$.

(C) Given $I_n = \int_0^{\pi / 4} \tan^n x \, dx$.
Consider $I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x \, dx + \int_0^{\pi / 4} \tan^{r+2} x \, dx$.
$I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x (1 + \tan^2 x) \, dx$.
Since $1 + \tan^2 x = \sec^2 x$,we have $I_r + I_{r+2} = \int_0^{\pi / 4} \tan^r x \sec^2 x \, dx$.
Let $t = \tan x$,then $dt = \sec^2 x \, dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$.
$I_r + I_{r+2} = \int_0^1 t^r \, dt = \left[ \frac{t^{r+1}}{r+1} \right]_0^1 = \frac{1}{r+1}$.
Thus,$I_2+I_4 = \frac{1}{3}$,$I_3+I_5 = \frac{1}{4}$,$I_4+I_6 = \frac{1}{5}$,and so on.
The terms are $\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots$,which are in harmonic progression $(HP)$ because their reciprocals $3, 4, 5, \ldots$ are in arithmetic progression.

(A) Given,the condition is $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ ...$(i)$
Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$.
Let the angle between $\vec{a}$ and $\vec{b}$ be $\theta$.
From equation $(i)$,we can write: $\vec{a}+\vec{b}=-\vec{c}$.
Squaring both sides,we get: $(\vec{a}+\vec{b})^2 = (-\vec{c})^2$.
This expands to: $|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2$.
Substituting the magnitudes: $1^2 + 1^2 + 2|\vec{a}||\vec{b}| \cos \theta = 1^2$.
$1 + 1 + 2(1)(1) \cos \theta = 1$.
$2 + 2 \cos \theta = 1$.
$2 \cos \theta = -1$.
$\cos \theta = -\frac{1}{2}$.
Since $\cos \theta = -\frac{1}{2}$,the angle $\theta = \frac{2 \pi}{3}$.

(A) Given $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$.
$\overrightarrow{u} \times \overrightarrow{v} = (\overrightarrow{a}-\overrightarrow{b}) \times (\overrightarrow{a}+\overrightarrow{b})$
$= \overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{a} \times \overrightarrow{b} - \overrightarrow{b} \times \overrightarrow{a} - \overrightarrow{b} \times \overrightarrow{b}$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$ and $\overrightarrow{b} \times \overrightarrow{b} = 0$,and $\overrightarrow{b} \times \overrightarrow{a} = -(\overrightarrow{a} \times \overrightarrow{b})$:
$= 0 + \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{b} - 0 = 2(\overrightarrow{a} \times \overrightarrow{b})$
Taking the magnitude: $|\overrightarrow{u} \times \overrightarrow{v}| = 2|\overrightarrow{a} \times \overrightarrow{b}|$
Using the identity $|\overrightarrow{a} \times \overrightarrow{b}|^2 = |\overrightarrow{a}|^2|\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2$:
$|\overrightarrow{u} \times \overrightarrow{v}| = 2 \sqrt{|\overrightarrow{a}|^2|\overrightarrow{b}|^2 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$
Given $|\overrightarrow{a}| = 2$ and $|\overrightarrow{b}| = 2$,so $|\overrightarrow{a}|^2 = 4$ and $|\overrightarrow{b}|^2 = 4$:
$|\overrightarrow{u} \times \overrightarrow{v}| = 2 \sqrt{(4)(4) - (\overrightarrow{a} \cdot \overrightarrow{b})^2} = 2 \sqrt{16 - (\overrightarrow{a} \cdot \overrightarrow{b})^2}$

(A) Given,$\overrightarrow{a}=2 x^2 \hat{i}+4 x \hat{j}+\hat{k}$ and $\overrightarrow{b}=7 \hat{i}-2 \hat{j}+x \hat{k}$.
We are given that $90^{\circ} < \theta < 180^{\circ}$,which implies that $\cos \theta < 0$.
We know that $\cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$.
Since $|\overrightarrow{a}|$ and $|\overrightarrow{b}|$ are magnitudes,they are always positive. Thus,$\cos \theta < 0$ implies $\overrightarrow{a} \cdot \overrightarrow{b} < 0$.
Calculating the dot product:
$\overrightarrow{a} \cdot \overrightarrow{b} = (2 x^2)(7) + (4 x)(-2) + (1)(x) = 14 x^2 - 8 x + x = 14 x^2 - 7 x$.
Setting the dot product to be less than zero:
$14 x^2 - 7 x < 0$
$7 x(2 x - 1) < 0$
To solve this inequality,we find the critical points $x = 0$ and $x = \frac{1}{2}$.
Testing the intervals:
For $x < 0$,$7 x(2 x - 1) > 0$.
For $0 < x < \frac{1}{2}$,$7 x(2 x - 1) < 0$.
For $x > \frac{1}{2}$,$7 x(2 x - 1) > 0$.
Thus,the interval for which the expression is negative is $x \in \left(0, \frac{1}{2}\right)$.

(B) The intercept form of the equation of a plane meeting the coordinate axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$ is given by:
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ $\ldots$ $(i)$
The centroid of the triangle $ABC$ with vertices $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$ is given by the formula:
$\left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)$
Given that the centroid is $(1, 2, 4)$,we equate the coordinates:
$\frac{a}{3} = 1 \Rightarrow a = 3$
$\frac{b}{3} = 2 \Rightarrow b = 6$
$\frac{c}{3} = 4 \Rightarrow c = 12$
Substituting these values into Eq. $(i)$:
$\frac{x}{3} + \frac{y}{6} + \frac{z}{12} = 1$
Multiplying the entire equation by $12$ to clear the denominators:
$4x + 2y + z = 12$

(B) The equation of the sphere is $x^2+y^2+z^2-6x-12y-2z+20=0$.
Comparing this with the general equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $2u=-6, 2v=-12, 2w=-2$.
Thus,$u=-3, v=-6, w=-1$.
The centre of the sphere is $(-u,-v,-w) = (3,6,1)$.
Let the given end of the diameter be $A = (2,3,-3)$ and the other end be $B = (\alpha, \beta, \gamma)$.
Since the centre $O(3,6,1)$ is the midpoint of the diameter $AB$,we have:
$O = \left( \frac{\alpha+2}{2}, \frac{\beta+3}{2}, \frac{\gamma-3}{2} \right) = (3,6,1)$.
Equating the coordinates:
$\frac{\alpha+2}{2} = 3 \Rightarrow \alpha+2 = 6 \Rightarrow \alpha = 4$.
$\frac{\beta+3}{2} = 6 \Rightarrow \beta+3 = 12 \Rightarrow \beta = 9$.
$\frac{\gamma-3}{2} = 1 \Rightarrow \gamma-3 = 2 \Rightarrow \gamma = 5$.
Therefore,the other end of the diameter is $(4,9,5)$.

(D) The probability that event $A_i$ does not occur is given by $P(\bar{A}_i) = 1 - P(A_i)$.
Given $P(A_i) = \frac{1}{1+i}$,we have $P(\bar{A}_i) = 1 - \frac{1}{1+i} = \frac{1+i-1}{1+i} = \frac{i}{1+i}$.
Since $A_i$ are independent events,the probability that none of $A_i$ occurs is the product of the probabilities of their complements:
$P(\text{none of } A_i \text{ occurs}) = P(\bar{A}_1 \cap \bar{A}_2 \cap \ldots \cap \bar{A}_n) = P(\bar{A}_1) \cdot P(\bar{A}_2) \cdot \ldots \cdot P(\bar{A}_n)$.
Substituting the values:
$= \left(\frac{1}{2}\right) \cdot \left(\frac{2}{3}\right) \cdot \left(\frac{3}{4}\right) \cdot \ldots \cdot \left(\frac{n}{n+1}\right)$.
This is a telescoping product where the numerator of each term cancels with the denominator of the previous term:
$= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{n}{n+1} = \frac{1}{n+1}$.

(C) Given $h = 1$ (the difference between consecutive $x$ values).
Let the values of $f(x)$ be $y_0, y_1, y_2, y_3, y_4, y_5$.
Here,$y_0 = 2, y_1 = 3, y_2 = 6, y_3 = 11, y_4 = 18, y_5 = 27$.
By the Trapezoidal rule,the area $A$ is given by:
$A = \frac{h}{2} [ (y_0 + y_5) + 2(y_1 + y_2 + y_3 + y_4) ]$
Substituting the values:
$A = \frac{1}{2} [ (2 + 27) + 2(3 + 6 + 11 + 18) ]$
$A = \frac{1}{2} [ 29 + 2(38) ]$
$A = \frac{1}{2} [ 29 + 76 ]$
$A = \frac{1}{2} [ 105 ] = 52.5$
Thus,the approximate area is $52.5$ square units.

(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii):
$(x-1)^2+(1-x^2)=1$
$x^2-2x+1+1-x^2=1$
$-2x+2=1 \Rightarrow x=\frac{1}{2}$
Substituting $x=\frac{1}{2}$ into $(i)$,we get $y^2=1-\frac{1}{4}=\frac{3}{4} \Rightarrow y=\pm\frac{\sqrt{3}}{2}$.
The intersection points are $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $C\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.
The area of the enclosed region is symmetric about the $x$-axis,so the total area is $2 \times$ (Area of the region in the first quadrant bounded by $x=0, x=\frac{1}{2}, x=1$ and the arcs of the circles).
Area $= 2 \left[ \int_0^{1/2} \sqrt{1-(x-1)^2} dx + \int_{1/2}^1 \sqrt{1-x^2} dx \right]$
Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:
Area $= 2 \left[ \left( \frac{x-1}{2}\sqrt{1-(x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) \right)_0^{1/2} + \left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) \right)_{1/2}^1 \right]$
$= 2 \left[ \left( \frac{-1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(-\frac{1}{2}) - (0 + \frac{1}{2}\sin^{-1}(-1)) \right) + \left( (0 + \frac{1}{2}\sin^{-1}(1)) - (\frac{1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(\frac{1}{2})) \right) \right]$
$= 2 \left[ -\frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} + \frac{\pi}{4} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} \right]$
$= 2 \left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{4} \right] = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$.

(A) Given that $\det(I+A) \neq 0$,which implies that $(I+A)$ is an invertible matrix.
We are given $A^3 = O$.
We know the algebraic identity $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.
Substituting $x = A$ and $y = I$,we get $A^3 + I^3 = (A+I)(A^2 - A + I)$.
Since $A^3 = O$ and $I^3 = I$,the equation becomes $O + I = (A+I)(A^2 - A + I)$.
This simplifies to $I = (A+I)(A^2 - A + I)$.
Multiplying both sides by $(A+I)^{-1}$ from the left,we get $(A+I)^{-1} I = (A+I)^{-1} (A+I)(A^2 - A + I)$.
Since $(A+I)^{-1}(A+I) = I$,we have $(A+I)^{-1} = I(A^2 - A + I) = A^2 - A + I$.
Thus,$(I+A)^{-1} = I - A + A^2$.

(B) Given the determinant equation: $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$.
We can split the determinant into two: $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| + \left|\begin{array}{ccc}x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3\end{array}\right| = 0$.
Taking $xyz$ common from the second determinant: $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| + xyz \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right| = 0$.
By performing column swaps ($C_1 \leftrightarrow C_2$ then $C_2 \leftrightarrow C_3$),the second determinant becomes $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|$.
Thus,$(1+xyz) \left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| = 0$.
Since $x \neq y \neq z$,the determinant $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| \neq 0$.
Therefore,$1+xyz = 0$.

(B) Given $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$.
Using the property of differentiation of a determinant,$f^{\prime}(x)$ is the sum of three determinants where each row is differentiated one at a time:
$f^{\prime}(x) = \left| \begin{array}{ccc} -2 \sin x & 0 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right| + \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ 1 & -2 \sin x & 0 \\ 0 & 1 & 2 \cos x \end{array} \right| + \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 0 & -2 \sin x \end{array} \right|$.
Now,substitute $x = \pi$ (where $\sin \pi = 0$ and $\cos \pi = -1$):
$f^{\prime}(\pi) = \left| \begin{array}{ccc} 0 & 0 & 0 \\ \frac{\pi}{2} & -2 & 1 \\ 0 & 1 & -2 \end{array} \right| + \left| \begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & -2 \end{array} \right| + \left| \begin{array}{ccc} -2 & 1 & 0 \\ \frac{\pi}{2} & -2 & 1 \\ 0 & 0 & 0 \end{array} \right|$.
The first determinant is $0$ because the first row is all zeros.
The second determinant is $-2(0 - 0) - 1(-2 - 0) + 0 = 2$.
The third determinant is $0$ because the third row is all zeros.
Thus,$f^{\prime}(\pi) = 0 + 2 + 0 = 2$.

(B) The given system of equations is:
$ (k+1)^3 x + (k+2)^3 y = (k+3)^3 $
$ (k+1) x + (k+2) y = k+3 $
$ x + y = 1 $
For the system to be consistent,the determinant of the augmented matrix must be zero.
Let $ D = \begin{vmatrix} (k+1)^3 & (k+2)^3 & (k+3)^3 \\ k+1 & k+2 & k+3 \\ 1 & 1 & 1 \end{vmatrix} = 0 $.
Applying column operations $ C_2 \rightarrow C_2 - C_1 $ and $ C_3 \rightarrow C_3 - C_1 $:
$ D = \begin{vmatrix} (k+1)^3 & (k+2)^3 - (k+1)^3 & (k+3)^3 - (k+1)^3 \\ k+1 & (k+2) - (k+1) & (k+3) - (k+1) \\ 1 & 1 - 1 & 1 - 1 \end{vmatrix} = 0 $
$ D = \begin{vmatrix} (k+1)^3 & 3k^2 + 9k + 7 & 6k^2 + 24k + 26 \\ k+1 & 1 & 2 \\ 1 & 0 & 0 \end{vmatrix} = 0 $
Expanding along the third row:
$ 1 \cdot [2(3k^2 + 9k + 7) - 1(6k^2 + 24k + 26)] = 0 $
$ 6k^2 + 18k + 14 - 6k^2 - 24k - 26 = 0 $
$ -6k - 12 = 0 $
$ -6k = 12 $
$ k = -2 $.

(B) Given that $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$.
We know the formula $\tan ^{-1} x+\tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
So,$\tan ^{-1} \left( \frac{x+y}{1-xy} \right) + \tan ^{-1} z = \frac{\pi}{2}$.
Taking $\tan$ on both sides,we use $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Let $A = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$ and $B = \tan ^{-1} z$.
Then $\tan(A+B) = \tan \left( \frac{\pi}{2} \right) = \infty$.
For $\tan(A+B)$ to be $\infty$,the denominator must be $0$.
Thus,$1 - \tan A \tan B = 0$.
$1 - \left( \frac{x+y}{1-xy} \right) z = 0$.
$1 - \frac{xz + yz}{1-xy} = 0$.
$1 - xy - xz - yz = 0$.
Therefore,$1 - xy - yz - zx = 0$.

(D) The function is defined as $f(x) = x^2$.
$A$. $f$ is one-one and onto,if $A = B = R^{+}$.
For $A = R^{+}$,$f(x) = x^2$ is strictly increasing,so it is one-one. Since $B = R^{+}$,for every $y \in R^{+}$,there exists $x = \sqrt{y} \in R^{+}$ such that $f(x) = y$,so it is onto. Thus,$A \rightarrow 4$.
$B$. $f$ is one-one but not onto,if $A = R^{+}, B = R$.
For $A = R^{+}$,$f(x) = x^2$ is one-one. Since $B = R$,the range is $R^{+}$,which is a proper subset of $B$,so it is not onto. Thus,$B \rightarrow 1$.
$C$. $f$ is onto but not one-one,if $A = R, B = R^{+}$.
For $A = R$,$f(1) = 1$ and $f(-1) = 1$,so it is not one-one. Since $B = R^{+}$,for every $y \in R^{+}$,there exists $x = \pm \sqrt{y} \in R$ such that $f(x) = y$,so it is onto. Thus,$C \rightarrow 3$.
$D$. $f$ is neither one-one nor onto,if $A = B = R$.
For $A = R$,$f(1) = f(-1) = 1$,so it is not one-one. Since $B = R$,the range is $R^{+}$,which is a proper subset of $B$,so it is not onto. Thus,$D \rightarrow 2$.
Therefore,the correct matching is $A \rightarrow 4, B \rightarrow 1, C \rightarrow 3, D \rightarrow 2$.

(B) Given $f(x) = \sin x + \cos x$.
First derivative: $f'(x) = \cos x - \sin x$.
Second derivative: $f''(x) = -\sin x - \cos x$.
Third derivative: $f'''(x) = -\cos x + \sin x$.
Fourth derivative: $f^{(iv)}(x) = \sin x + \cos x$.
Now,calculate the value at $x = \frac{\pi}{4}$:
$f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Similarly,$f^{(iv)}\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \sqrt{2}$.
Therefore,$f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right) = \sqrt{2} \times \sqrt{2} = 2$.

(B) Given $y = \sin(m \sin^{-1} x)$.
First,differentiate with respect to $x$:
$y_1 = \cos(m \sin^{-1} x) \cdot \frac{m}{\sqrt{1-x^2}}$.
Multiply both sides by $\sqrt{1-x^2}$:
$y_1 \sqrt{1-x^2} = m \cos(m \sin^{-1} x)$.
Differentiate again with respect to $x$ using the product rule:
$y_2 \sqrt{1-x^2} + y_1 \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = m \cdot (-\sin(m \sin^{-1} x)) \cdot \frac{m}{\sqrt{1-x^2}}$.
$y_2 \sqrt{1-x^2} - \frac{x y_1}{\sqrt{1-x^2}} = -\frac{m^2 \sin(m \sin^{-1} x)}{\sqrt{1-x^2}}$.
Multiply the entire equation by $\sqrt{1-x^2}$:
$y_2(1-x^2) - x y_1 = -m^2 \sin(m \sin^{-1} x)$.
Since $y = \sin(m \sin^{-1} x)$,we get:
$(1-x^2) y_2 - x y_1 = -m^2 y$.

(B) Given $f(x) = \prod_{r=1}^n \cos(rx)$. Taking the natural logarithm on both sides,we get $\ln|f(x)| = \sum_{r=1}^n \ln|\cos(rx)|$.
Differentiating both sides with respect to $x$,we have $\frac{f^{\prime}(x)}{f(x)} = \sum_{r=1}^n \frac{1}{\cos(rx)} \cdot (-\sin(rx) \cdot r) = -\sum_{r=1}^n r \tan(rx)$.
Multiplying both sides by $f(x)$,we get $f^{\prime}(x) = -f(x) \sum_{r=1}^n r \tan(rx)$.
Rearranging the terms,we obtain $f^{\prime}(x) + \sum_{r=1}^n (r \tan rx) f(x) = 0$.

(C) Let the height of the cone be $h$ and the radius of the cone be $r$. Given,the radius of the sphere is $R$.
In $\triangle OPB$,by the Pythagorean theorem,we have:
$R^2 = r^2 + (h - R)^2$
$\Rightarrow r^2 = R^2 - (h - R)^2 = R^2 - (h^2 - 2Rh + R^2) = 2Rh - h^2$.
The volume $V$ of the cone is given by:
$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2Rh - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$.
To find the maximum volume,differentiate $V$ with respect to $h$:
$\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2)$.
Setting $\frac{dV}{dh} = 0$ for critical points:
$\frac{\pi}{3} h(4R - 3h) = 0$.
Since $h \neq 0$,we have $h = \frac{4R}{3}$.
Checking the second derivative:
$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6h)$.
At $h = \frac{4R}{3}$,$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6(\frac{4R}{3})) = \frac{\pi}{3} (4R - 8R) = -\frac{4\pi R}{3} < 0$.
Since the second derivative is negative,the volume is maximum at $h = \frac{4R}{3}$.

(A) Given $f_n(x) = \log \log \ldots \log x$ ($n$ times).
Let $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$.
Let $t = f_n(x) = \log(f_{n-1}(x))$.
Then,$\frac{dt}{dx} = \frac{1}{f_{n-1}(x)} \cdot \frac{d}{dx}(f_{n-1}(x)) = \frac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) \cdot x}$.
Thus,$dx = (x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt$.
Substituting this into the integral:
$I = \int \frac{(x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt}{x f_1(x) f_2(x) \ldots f_{n-1}(x) f_n(x)} = \int \frac{dt}{f_n(x)} = \int \frac{dt}{t}$.
$I = \log(t) + c = \log(f_n(x)) + c$.
Since $f_{n+1}(x) = \log(f_n(x))$,we have $I = f_{n+1}(x) + c$.

(A) To find $f(x)$,we differentiate the given options to see which one yields the integrand $\frac{7 x^8+8 x^7}{\left(1+x+x^8\right)^2}$.
Let $f(x) = \frac{x^8}{1+x+x^8}$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$:
$f'(x) = \frac{(1+x+x^8)(8x^7) - x^8(1+8x^7)}{(1+x+x^8)^2}$
$f'(x) = \frac{8x^7 + 8x^8 + 8x^{15} - x^8 - 8x^{15}}{(1+x+x^8)^2}$
$f'(x) = \frac{7x^8 + 8x^7}{(1+x+x^8)^2}$
Since the derivative matches the integrand,$f(x) = \frac{x^8}{1+x+x^8}$ is the correct function.

(B) Given equation: $\tan y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$
Using the identity $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$\tan y \frac{dy}{dx} = 2 \sin x \cos y$
$\frac{\sin y}{\cos y} \frac{dy}{dx} = 2 \sin x \cos y$
Rearranging the terms:
$\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx$
Integrating both sides:
$\int \frac{\sin y}{\cos^2 y} dy = \int 2 \sin x dx$
Let $t = \cos y$,then $dt = -\sin y dy$,so $-\int \frac{dt}{t^2} = -2 \cos x + c$
$\frac{1}{t} = -2 \cos x + c$
Substituting $t = \cos y$:
$\sec y = -2 \cos x + c$

(D) Given the differential equation: $x y \frac{d y}{d x}=2 y^2-x^2$.
Divide by $x$: $y \frac{d y}{d x} = \frac{2 y^2}{x} - x$.
Rearranging: $y \frac{d y}{d x} - \frac{2 y^2}{x} = -x$ $\ldots$ $(i)$.
Let $v = y^2$. Then $\frac{d v}{d x} = 2 y \frac{d y}{d x}$,which implies $y \frac{d y}{d x} = \frac{1}{2} \frac{d v}{d x}$.
Substituting into $(i)$: $\frac{1}{2} \frac{d v}{d x} - \frac{2 v}{x} = -x$.
Multiply by $2$: $\frac{d v}{d x} - \frac{4 v}{x} = -2 x$.
This is a linear differential equation of the form $\frac{d v}{d x} + P(x) v = Q(x)$,where $P(x) = -\frac{4}{x}$ and $Q(x) = -2 x$.
Integrating factor $IF = e^{\int P(x) d x} = e^{\int -\frac{4}{x} d x} = e^{-4 \ln |x|} = x^{-4}$.
The solution is $v \cdot IF = \int Q(x) \cdot IF d x + c$.
$v \cdot x^{-4} = \int (-2 x) \cdot x^{-4} d x + c = \int -2 x^{-3} d x + c$.
$\frac{v}{x^4} = -2 \left( \frac{x^{-2}}{-2} \right) + c = \frac{1}{x^2} + c$.
$v = x^2 + c x^4$.
Since $v = y^2$,the family of curves is $y^2 = x^2 + c x^4$.

(A) Given that,$\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$,$\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$.
Since $\overrightarrow{a}$ and $\overrightarrow{b}$ lie in the same plane,the vector $(\overrightarrow{a} \times \overrightarrow{b})$ is perpendicular to this plane.
Given that vector $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$,the vector $\overrightarrow{c}$ must be perpendicular to the normal vector $(\overrightarrow{a} \times \overrightarrow{b})$.
Therefore,the dot product of $(\overrightarrow{a} \times \overrightarrow{b})$ and $\overrightarrow{c}$ must be zero,i.e.,$(\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c} = 0$.
First,calculate the cross product $\overrightarrow{a} \times \overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4) = -7\hat{i} + 7\hat{j} + 7\hat{k}$.
Now,compute the dot product with $\overrightarrow{c}$:
$(-7\hat{i} + 7\hat{j} + 7\hat{k}) \cdot (\lambda \hat{i} + \hat{j} + (2\lambda - 1)\hat{k}) = 0$.
$-7\lambda + 7(1) + 7(2\lambda - 1) = 0$.
$-7\lambda + 7 + 14\lambda - 7 = 0$.
$7\lambda = 0$.
Thus,$\lambda = 0$.

(D) Case $I$: $A$ white ball is transferred from urn $A$ to urn $B$.
Probability of selecting a white ball from $A$ is $P(W_1) = \frac{3}{3+5} = \frac{3}{8}$.
After transferring,urn $B$ contains $7$ white and $8$ black balls.
Probability of selecting a white ball from $B$ is $P(W_2|W_1) = \frac{7}{7+8} = \frac{7}{15}$.
Probability of this case $= P(W_1) \times P(W_2|W_1) = \frac{3}{8} \times \frac{7}{15} = \frac{21}{120} = \frac{7}{40}$.
Case $II$: $A$ black ball is transferred from urn $A$ to urn $B$.
Probability of selecting a black ball from $A$ is $P(B_1) = \frac{5}{3+5} = \frac{5}{8}$.
After transferring,urn $B$ contains $6$ white and $9$ black balls.
Probability of selecting a white ball from $B$ is $P(W_2|B_1) = \frac{6}{6+9} = \frac{6}{15}$.
Probability of this case $= P(B_1) \times P(W_2|B_1) = \frac{5}{8} \times \frac{6}{15} = \frac{30}{120} = \frac{10}{40}$.
Total probability $= \frac{7}{40} + \frac{10}{40} = \frac{17}{40}$.

(B) For a binomial distribution,the mean is given by $np = 2$ and the variance is given by $npq = 1$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{1}{2}$,which implies $q = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n \times \frac{1}{2} = 2$,so $n = 4$.
The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{4}{k} (\frac{1}{2})^k (\frac{1}{2})^{4-k} = \binom{4}{k} (\frac{1}{2})^4 = \binom{4}{k} \frac{1}{16}$.
We need to find $P(X \geq 1) = 1 - P(X = 0)$.
$P(X = 0) = \binom{4}{0} (\frac{1}{2})^4 = 1 \times \frac{1}{16} = \frac{1}{16}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{16} = \frac{15}{16}$.

(C) Let $\lambda$ be the mean of the Poisson distribution for the random variable $X$.
The probability mass function is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$ for $r = 0, 1, 2, \dots$.
Given $P(X=1) = P(X=2)$,we have:
$\frac{\lambda^1 e^{-\lambda}}{1!} = \frac{\lambda^2 e^{-\lambda}}{2!}$
$\lambda = \frac{\lambda^2}{2}$
Since $\lambda > 0$,we divide by $\lambda$ to get $1 = \frac{\lambda}{2}$,which implies $\lambda = 2$.
Now,we calculate $P(X=5)$:
$P(X=5) = \frac{2^5 e^{-2}}{5!} = \frac{32 e^{-2}}{120}$.
Simplifying the fraction $\frac{32}{120}$ by dividing both numerator and denominator by $8$,we get $\frac{4}{15}$.
Thus,$P(X=5) = \frac{4}{15} e^{-2}$.