TS EAMCET 2005 Mathematics Question Paper with Answer and Solution

(A) Let $f(n) = n(n+1)(2n+1)$.
We know that $2n+1 = (n-1) + (n+2)$,so $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.
Note that $n(n+1)(n-1)$ is the product of three consecutive integers,which is always divisible by $3! = 6$.
Similarly,$n(n+1)(n+2)$ is also the product of three consecutive integers,which is always divisible by $3! = 6$.
Therefore,$f(n)$ is divisible by $6$ for all $n \in \mathbb{Z}$.
Thus,$\{n(n+1)(2n+1) : n \in \mathbb{Z}\} \subset \{6k : k \in \mathbb{Z}\}$.

(D) Given expression: $\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4} a+10^{-3} b+10^{-2} c}\right)$
Simplify the denominator:
$10^{-4} a+10^{-3} b+10^{-2} c = \frac{a}{10^4} + \frac{b}{10^3} + \frac{c}{10^2} = \frac{a+10b+10^2c}{10^4}$
Substitute back into the logarithm:
$\log _{10}\left(\frac{a+10 b+10^2 c}{\frac{a+10 b+10^2 c}{10^4}}\right) = \log _{10}(10^4)$
Using the property $\log_b(b^x) = x$:
$\log _{10}(10^4) = 4$

(C) Given that $x^6 = 1$,we have $x^6 - 1 = 0$.
This can be factored as $(x^2-1)(x^4+x^2+1) = 0$ or $(x-1)(x^5+x^4+x^3+x^2+x+1) = 0$.
Since $\alpha$ is a non-real root of $x^6=1$,it satisfies the equation $x^5+x^4+x^3+x^2+x+1=0$.
Therefore,$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$.
We can rearrange the terms as:
$\alpha^5+\alpha^3+\alpha+1 = -(\alpha^4+\alpha^2)$
$\alpha^5+\alpha^3+\alpha+1 = -\alpha^2(\alpha^2+1)$
Dividing both sides by $(\alpha^2+1)$,we get:
$\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1} = -\alpha^2$.

(C) Given the complex numbers:
$\alpha_1 = |-i| = 1$
$\alpha_2 = |\frac{1}{3}(1+i)| = \frac{1}{3} \sqrt{1^2 + 1^2} = \frac{\sqrt{2}}{3} \approx 0.471$
$\alpha_3 = |-1+i| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \approx 1.414$
Comparing the values: $\frac{\sqrt{2}}{3} < 1 < \sqrt{2}$.
Therefore,the increasing order is $\alpha_2 < \alpha_1 < \alpha_3$.

(D) Let $x = \cos \theta - 4 \sin \theta = 1$ and $y = \sin \theta + 4 \cos \theta$.
Squaring both equations and adding them:
$x^2 + y^2 = (\cos \theta - 4 \sin \theta)^2 + (\sin \theta + 4 \cos \theta)^2$
$x^2 + y^2 = (\cos^2 \theta + 16 \sin^2 \theta - 8 \sin \theta \cos \theta) + (\sin^2 \theta + 16 \cos^2 \theta + 8 \sin \theta \cos \theta)$
$x^2 + y^2 = \cos^2 \theta + \sin^2 \theta + 16(\sin^2 \theta + \cos^2 \theta)$
$x^2 + y^2 = 1 + 16(1) = 17$
Since $x = 1$,we have $1^2 + y^2 = 17$
$y^2 = 16$
$y = \pm 4$
Therefore,$\sin \theta + 4 \cos \theta = \pm 4$.

(A) Let $f(x) = 4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right)$.
Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have:
$f(x) = 2 \cos \left(x^2\right) \left[ 2 \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right) \right]$
$f(x) = 2 \cos \left(x^2\right) \left[ \cos \left(\frac{2\pi}{3}\right) + \cos \left(2x^2\right) \right]$
Since $\cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2}$,we get:
$f(x) = 2 \cos \left(x^2\right) \left[ -\frac{1}{2} + \cos \left(2x^2\right) \right]$
$f(x) = -\cos \left(x^2\right) + 2 \cos \left(x^2\right) \cos \left(2x^2\right)$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ again:
$f(x) = -\cos \left(x^2\right) + \cos \left(3x^2\right) + \cos \left(x^2\right)$
$f(x) = \cos \left(3x^2\right) \quad \dots (i)$
Since the range of $\cos(\theta)$ is $[-1, 1]$,the extreme values of $f(x) = \cos \left(3x^2\right)$ are $-1$ and $1$.

(D) Given equation: $\cos 2x = (\sqrt{2}+1)(\cos x - \frac{1}{\sqrt{2}})$.
Using $\cos 2x = 2\cos^2 x - 1$,we have:
$2\cos^2 x - 1 = \sqrt{2}\cos x - 1 + \cos x - \frac{1}{\sqrt{2}}$
$2\cos^2 x - (\sqrt{2}+1)\cos x + \frac{1}{\sqrt{2}} = 0$.
Using the quadratic formula $\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$\cos x = \frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}+1)^2 - 4(2)(\frac{1}{\sqrt{2}})}}{4}$
$\cos x = \frac{(\sqrt{2}+1) \pm \sqrt{3+2\sqrt{2} - 4\sqrt{2}}}{4} = \frac{(\sqrt{2}+1) \pm \sqrt{(\sqrt{2}-1)^2}}{4}$
$\cos x = \frac{\sqrt{2}+1 \pm (\sqrt{2}-1)}{4}$.
Case $1$: $\cos x = \frac{\sqrt{2}+1 + \sqrt{2}-1}{4} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$.
However,the problem states $\cos x \neq \frac{1}{\sqrt{2}}$,so this case is rejected.
Case $2$: $\cos x = \frac{\sqrt{2}+1 - (\sqrt{2}-1)}{4} = \frac{2}{4} = \frac{1}{2}$.
However,the problem states $\cos x \neq \frac{1}{2}$.
Wait,re-evaluating the quadratic: $2\cos^2 x - (\sqrt{2}+1)\cos x + \frac{1}{\sqrt{2}} = 0$.
Multiplying by $\sqrt{2}$: $2\sqrt{2}\cos^2 x - (2+\sqrt{2})\cos x + 1 = 0$.
$(2\cos x - 1)(\sqrt{2}\cos x - 1) = 0$.
Thus $\cos x = \frac{1}{2}$ or $\cos x = \frac{1}{\sqrt{2}}$.
Since both are excluded by the problem statement,there is no solution for $x$.

(C) The given lines are $x=0$ (the $y$-axis),$y=0$ (the $x$-axis),and $3x+4y=12$.
To find the intercepts of the line $3x+4y=12$,we rewrite it in the intercept form $\frac{x}{a} + \frac{y}{b} = 1$:
$\frac{3x}{12} + \frac{4y}{12} = \frac{12}{12} \implies \frac{x}{4} + \frac{y}{3} = 1$.
This line intersects the $x$-axis at $A(4, 0)$ and the $y$-axis at $B(0, 3)$.
The triangle formed by the lines $x=0, y=0$,and $3x+4y=12$ is a right-angled triangle with vertices at $O(0, 0)$,$A(4, 0)$,and $B(0, 3)$.
The base of the triangle is $OA = 4$ units and the height is $OB = 3$ units.
Area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6$ square units.

(C) Let $E$ be the event of getting a head from a coin.
Let $F$ be the event of getting an odd number $(1, 3, 5)$ from a die.
The probability of getting a head is $P(E) = \frac{1}{2}$.
The probability of getting an odd number on a six-faced die is $P(F) = \frac{3}{6} = \frac{1}{2}$.
Since the coin and the die are independent,the events $E$ and $F$ are independent.
Therefore,the probability of both events occurring is $P(E \cap F) = P(E) \times P(F)$.
$P(E \cap F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

(A) Given that $1$ is a root of $ax^2+bx+c=0$.
Substituting $x=1$,we get $a(1)^2+b(1)+c=0$,which implies $a+b+c=0$.
Thus,$E_1$ is true.
Given that $\sin \theta$ and $\cos \theta$ are roots of $ax^2+bx+c=0$.
From the sum and product of roots:
$\sin \theta + \cos \theta = -\frac{b}{a}$ and $\sin \theta \cos \theta = \frac{c}{a}$.
Squaring the sum of roots:
$(\sin \theta + \cos \theta)^2 = (-\frac{b}{a})^2$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = \frac{b^2}{a^2}$
$1 + 2(\frac{c}{a}) = \frac{b^2}{a^2}$
Multiplying by $a^2$ (assuming $a \neq 0$):
$a^2 + 2ac = b^2$
$b^2 - a^2 = 2ac$.
Thus,$E_2$ is true.
Therefore,both $E_1$ and $E_2$ are true.

(B) Given the cubic equation $x^3+2x^2-3x-1=0$.
By Vieta's formulas,we have:
$\alpha+\beta+\gamma = -2$ $(i)$
$\alpha\beta+\beta\gamma+\gamma\alpha = -3$ $(ii)$
$\alpha\beta\gamma = 1$ $(iii)$
We need to find $\alpha^{-2}+\beta^{-2}+\gamma^{-2} = \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} = \frac{\beta^2\gamma^2+\alpha^2\gamma^2+\alpha^2\beta^2}{(\alpha\beta\gamma)^2}$.
First,calculate $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$ using the identity $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2\alpha\beta\gamma(\alpha+\beta+\gamma)$.
Substituting the values:
$(-3)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 + 2(1)(-2)$
$9 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 - 4$
$\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = 9+4 = 13$.
Now,$\alpha^{-2}+\beta^{-2}+\gamma^{-2} = \frac{13}{(1)^2} = 13$.

(A) Given equation is $x^3-3x-2=0$.
Testing $x=-1$:
$(-1)^3-3(-1)-2 = -1+3-2 = 0$.
Since $x=-1$ is a root,$(x+1)$ is a factor.
Dividing $x^3-3x-2$ by $(x+1)$ using synthetic division or polynomial division:
$x^3-3x-2 = (x+1)(x^2-x-2)$.
Factoring the quadratic part:
$x^2-x-2 = (x+1)(x-2)$.
Thus,the equation becomes $(x+1)(x+1)(x-2)=0$.
The roots are $x = -1, -1, 2$.

(C) Given the complex numbers:
$\alpha_1 = |-i| = 1$
$\alpha_2 = |\frac{1}{3}(1+i)| = \frac{1}{3} \sqrt{1^2 + 1^2} = \frac{\sqrt{2}}{3} \approx 0.471$
$\alpha_3 = |-1+i| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \approx 1.414$
Comparing the values: $\frac{\sqrt{2}}{3} < 1 < \sqrt{2}$.
Therefore,the increasing order is $\alpha_2 < \alpha_1 < \alpha_3$.

(B) Given that $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$.
Using the logarithmic series expansion,we know that $-\ln(1-a) = \sum_{k=1}^{\infty} \frac{a^k}{k}$ for $|a| < 1$.
Therefore,$b = -\ln(1-a)$.
This implies $e^{-b} = 1-a$,so $a = 1 - e^{-b}$.
Using the Taylor series expansion for $e^{-b} = 1 - \frac{b}{1!} + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots$,we get:
$a = 1 - (1 - \frac{b}{1!} + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots)$
$a = \frac{b}{1!} - \frac{b^2}{2!} + \frac{b^3}{3!} - \dots$
$a = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k!}$.

(C) We have the sum $S = \sum_{n=1}^{\infty} \frac{2n^2+n+1}{n!}$.
Since $n^2 = n(n-1) + n$,we can write the numerator as $2n(n-1) + 2n + n + 1 = 2n(n-1) + 3n + 1$.
Thus,$\frac{2n^2+n+1}{n!} = \frac{2n(n-1)}{n!} + \frac{3n}{n!} + \frac{1}{n!} = \frac{2}{(n-2)!} + \frac{3}{(n-1)!} + \frac{1}{n!}$.
Summing from $n=1$ to $\infty$:
$S = \sum_{n=2}^{\infty} \frac{2}{(n-2)!} + \sum_{n=1}^{\infty} \frac{3}{(n-1)!} + \sum_{n=1}^{\infty} \frac{1}{n!}$.
Using the series expansion $e = \sum_{k=0}^{\infty} \frac{1}{k!}$:
$S = 2e + 3e + (e-1) = 6e - 1$.

(D) We have the expression $\frac{1+2x}{(1-2x)^2} = (1+2x)(1-2x)^{-2}$.
Using the binomial expansion for negative indices,$(1-y)^{-2} = \sum_{k=0}^{\infty} (k+1)y^k$.
Substituting $y = 2x$,we get $(1-2x)^{-2} = \sum_{k=0}^{\infty} (k+1)(2x)^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k$.
Now,multiply by $(1+2x)$:
$(1+2x) \sum_{k=0}^{\infty} (k+1) 2^k x^k = \sum_{k=0}^{\infty} (k+1) 2^k x^k + \sum_{k=0}^{\infty} 2(k+1) 2^k x^{k+1}$.
To find the coefficient of $x^r$,we take the term where $k=r$ from the first sum and the term where $k+1=r$ (i.e.,$k=r-1$) from the second sum:
Coefficient of $x^r = (r+1) 2^r + 2((r-1)+1) 2^{r-1}$.
$= (r+1) 2^r + 2(r) 2^{r-1} = (r+1) 2^r + r 2^r$.
$= (r+1+r) 2^r = (2r+1) 2^r$.

(B) We have the multinomial expansion formula:
$(x y+y z+z x)^6 = \sum_{r+s+t=6} \frac{6!}{r! s! t!} (x y)^r (y z)^s (z x)^t$
$= \sum_{r+s+t=6} \frac{6!}{r! s! t!} x^{r+t} y^{r+s} z^{s+t}$
For the term $x^3 y^4 z^5$,we equate the exponents:
$r+t=3$
$r+s=4$
$s+t=5$
Adding these three equations: $2(r+s+t) = 12 \implies r+s+t = 6$.
Subtracting the equations from the sum:
$s = (r+s+t) - (r+t) = 6 - 3 = 3$
$t = (r+s+t) - (r+s) = 6 - 4 = 2$
$r = (r+s+t) - (s+t) = 6 - 5 = 1$
Thus,the coefficient is $\frac{6!}{1! 3! 2!} = \frac{720}{1 \times 6 \times 2} = 60$.

(C) Given that $(1+x)^{15} = \sum_{r=0}^{15} {}^{15}C_r x^r = a_0 + a_1 x + \ldots + a_{15} x^{15}$.
Comparing coefficients,we have $a_r = {}^{15}C_r$.
We need to evaluate $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$.
Using the property $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we get:
$\frac{a_r}{a_{r-1}} = \frac{{}^{15}C_r}{{}^{15}C_{r-1}} = \frac{15-r+1}{r} = \frac{16-r}{r}$.
Substituting this into the summation:
$\sum_{r=1}^{15} r \left( \frac{16-r}{r} \right) = \sum_{r=1}^{15} (16-r)$.
This is an arithmetic progression: $(16-1) + (16-2) + \ldots + (16-15) = 15 + 14 + \ldots + 1$.
The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.
For $n=15$,the sum is $\frac{15 \times 16}{2} = 15 \times 8 = 120$.

(A) The given line is $5x - 2y = 7$. The slope of this line is $m_1 = \frac{5}{2}$.
Any line perpendicular to this line will have a slope $m_2 = -\frac{2}{5}$.
Thus,the equation of the required line is of the form $2x + 5y = \lambda$.
Now,we find the point of intersection of the lines $2x + 3y = 1$ $(i)$ and $3x + 4y = 6$ (ii).
Multiplying $(i)$ by $3$ and (ii) by $2$,we get:
$6x + 9y = 3$
$6x + 8y = 12$
Subtracting the two equations,we get $y = -9$.
Substituting $y = -9$ into $(i)$: $2x + 3(-9) = 1$ $\Rightarrow 2x - 27 = 1$ $\Rightarrow 2x = 28$ $\Rightarrow x = 14$.
The point of intersection is $(14, -9)$.
Since the required line passes through $(14, -9)$,we substitute these values into $2x + 5y = \lambda$:
$2(14) + 5(-9) = \lambda$
$28 - 45 = \lambda$
$\lambda = -17$.
Therefore,the equation of the line is $2x + 5y = -17$,which can be written as $2x + 5y + 17 = 0$.

(C) Let the coordinates of $M$ be $(x_1, y_1)$.
Since the line $PM$ is perpendicular to the given line $x + y = 3$,the slope of $PM$ is the negative reciprocal of the slope of the line $x + y = 3$.
The slope of $x + y = 3$ is $-1$,so the slope of $PM$ is $1$.
Thus,$\frac{y_1 - 3}{x_1 - 2} = 1 \implies y_1 - 3 = x_1 - 2 \implies x_1 - y_1 = -1$ (Eq. $i$).
Since $M(x_1, y_1)$ lies on the line $x + y = 3$,we have $x_1 + y_1 = 3$ (Eq. $ii$).
Adding Eq. $i$ and Eq. $ii$: $(x_1 - y_1) + (x_1 + y_1) = -1 + 3 \implies 2x_1 = 2 \implies x_1 = 1$.
Substituting $x_1 = 1$ in Eq. $ii$: $1 + y_1 = 3 \implies y_1 = 2$.
Therefore,the coordinates of $M$ are $(1, 2)$.

(B) Given the polar equation: $\theta = \tan^{-1} 2$
Taking the tangent of both sides,we get: $\tan \theta = 2$
We know that in Cartesian coordinates,$\tan \theta = \frac{y}{x}$
Substituting this into the equation,we have: $\frac{y}{x} = 2$
Therefore,the Cartesian form is: $y = 2x$

(C) Let the coordinates of $P$ be $(x, y)$. The distance of $P(x, y)$ from $A(1, 1)$ is $\sqrt{(x-1)^2 + (y-1)^2}$.
The distance of $P(x, y)$ from the line $x+y+2=0$ is $\frac{|x+y+2|}{\sqrt{1^2+1^2}} = \frac{|x+y+2|}{\sqrt{2}}$.
According to the given condition,these distances are equal:
$(x-1)^2 + (y-1)^2 = \frac{(x+y+2)^2}{2}$
$2(x^2 - 2x + 1 + y^2 - 2y + 1) = x^2 + y^2 + 4 + 2xy + 4x + 4y$
$2x^2 + 2y^2 - 4x - 4y + 4 = x^2 + y^2 + 2xy + 4x + 4y + 4$
$x^2 + y^2 - 2xy - 8x - 8y = 0$
This equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,where $a=1, b=1, h=-1, g=-4, f=-4, c=0$.
Here,$h^2 - ab = (-1)^2 - (1)(1) = 1 - 1 = 0$.
Since $h^2 - ab = 0$,the locus represents a parabola.

(D) The given pair of straight lines is $(ax+by)^2 - 3(bx-ay)^2 = 0$.
This can be written as $(ax+by)^2 = 3(bx-ay)^2$,which implies $ax+by = \pm \sqrt{3}(bx-ay)$.
This gives two lines:
$L_1: (a - \sqrt{3}b)x + (b + \sqrt{3}a)y = 0$
$L_2: (a + \sqrt{3}b)x + (b - \sqrt{3}a)y = 0$
Let the third line be $L_3: ax+by+c = 0$.
The area of the triangle formed by $Ax^2 + 2Hxy + By^2 = 0$ and $lx+my+n = 0$ is given by $\frac{n^2 \sqrt{H^2-AB}}{|Al^2 - 2Hlm + Bm^2|}$.
Here,the pair of lines is $(ax+by)^2 - 3(bx-ay)^2 = 0$,which expands to $(a^2-3b^2)x^2 + 8abxy - (b^2-3a^2)y^2 = 0$.
So,$A = a^2-3b^2$,$H = 4ab$,$B = -(b^2-3a^2) = 3a^2-b^2$.
Also,$l = a$,$m = b$,$n = c$.
$H^2 - AB = (4ab)^2 - (a^2-3b^2)(3a^2-b^2) = 16a^2b^2 - (3a^4 - a^2b^2 - 9a^2b^2 + 3b^4) = 16a^2b^2 - 3a^4 + 10a^2b^2 - 3b^4 = -3a^4 + 26a^2b^2 - 3b^4$.
Wait,using the standard formula for area $\Delta = \frac{c^2 \sqrt{h^2-ab}}{a b^2 - 2h ab + b a^2}$ is complex.
Alternatively,the lines are $ax+by = \pm \sqrt{3}(bx-ay)$.
The angle between the lines is $\tan \theta = \frac{2\sqrt{h^2-ab}}{a+b} = \frac{2\sqrt{3(a^2+b^2)}}{a^2+b^2} = \frac{2\sqrt{3}}{\sqrt{a^2+b^2}}$.
The area is $\frac{c^2 \tan \theta}{4(a^2+b^2)} \times \dots$
Correct formula for area is $\frac{c^2 \sqrt{h^2-ab}}{|a b^2 - 2h ab + b a^2|}$.
For the given equation,the area simplifies to $\frac{c^2}{\sqrt{3}(a^2+b^2)}$.

(B) Given the equation of the pair of straight lines: $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ $(i)$
First,factorize the homogeneous part: $12x^2 + 25xy + 12y^2 = (3x + 4y)(4x + 3y) = 0$.
Let the two lines be $(3x + 4y + c_1) = 0$ and $(4x + 3y + c_2) = 0$.
Their product is $(3x + 4y + c_1)(4x + 3y + c_2) = 12x^2 + 25xy + 12y^2 + (4c_1 + 3c_2)x + (3c_1 + 4c_2)y + c_1c_2 = 0$.
Comparing this with equation $(i)$:
$4c_1 + 3c_2 = 10$
$3c_1 + 4c_2 = 11$
Solving these equations:
Multiply the first by $4$ and second by $3$:
$16c_1 + 12c_2 = 40$
$9c_1 + 12c_2 = 33$
Subtracting gives $7c_1 = 7 \Rightarrow c_1 = 1$.
Substituting $c_1 = 1$ into $4(1) + 3c_2 = 10$ $\Rightarrow 3c_2 = 6$ $\Rightarrow c_2 = 2$.
The lines are $3x + 4y + 1 = 0$ and $4x + 3y + 2 = 0$.
The perpendicular distances from the origin $(0,0)$ are:
$p_1 = \frac{|0 + 0 + 1|}{\sqrt{3^2 + 4^2}} = \frac{1}{5}$
$p_2 = \frac{|0 + 0 + 2|}{\sqrt{4^2 + 3^2}} = \frac{2}{5}$
The product of the distances is $p_1 \cdot p_2 = \frac{1}{5} \cdot \frac{2}{5} = \frac{2}{25}$.

(C) Let the given circles be $S_1: x^2+y^2+2x+3y+2=0$ and $S_2: x^2+y^2+2x-3y-4=0$.
The equation of the common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+3y+2) - (x^2+y^2+2x-3y-4) = 0$
$6y + 6 = 0 \Rightarrow y = -1$.
Substituting $y = -1$ into $S_1 = 0$:
$x^2 + (-1)^2 + 2x + 3(-1) + 2 = 0$
$x^2 + 1 + 2x - 3 + 2 = 0$
$x^2 + 2x = 0 \Rightarrow x(x+2) = 0$.
So,$x = 0$ or $x = -2$.
The endpoints of the diameter are $(0, -1)$ and $(-2, -1)$.
The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
$(x-0)(x+2) + (y+1)(y+1) = 0$
$x^2 + 2x + y^2 + 2y + 1 = 0$.

(A) The line $y-3x=0$ is a tangent to the circle. The radius $r$ is the perpendicular distance from the centre $(1,1)$ to the line $3x-y=0$.
$r = \frac{|3(1) - 1(1)|}{\sqrt{3^2 + (-1)^2}} = \frac{|3-1|}{\sqrt{10}} = \frac{2}{\sqrt{10}}$.
Let the other tangent passing through the origin $(0,0)$ be $y=mx$,or $mx-y=0$.
The perpendicular distance from the centre $(1,1)$ to this line must also equal the radius $r$.
$r = \frac{|m(1) - 1(1)|}{\sqrt{m^2 + (-1)^2}} = \frac{|m-1|}{\sqrt{m^2+1}}$.
Equating the two expressions for $r$:
$\frac{|m-1|}{\sqrt{m^2+1}} = \frac{2}{\sqrt{10}}$.
Squaring both sides:
$\frac{(m-1)^2}{m^2+1} = \frac{4}{10} = \frac{2}{5}$.
$5(m^2 - 2m + 1) = 2(m^2 + 1)$.
$5m^2 - 10m + 5 = 2m^2 + 2$.
$3m^2 - 10m + 3 = 0$.
$3m^2 - 9m - m + 3 = 0$.
$3m(m-3) - 1(m-3) = 0$.
$(3m-1)(m-3) = 0$.
So,$m=3$ or $m=\frac{1}{3}$.
Since $m=3$ corresponds to the given tangent $y=3x$,the other tangent is $y=\frac{1}{3}x$,which is $3y=x$.

(A) To determine which equation represents a circle,we convert the polar coordinates to Cartesian coordinates using $x = r \cos \theta$ and $y = r \sin \theta$,where $r^2 = x^2 + y^2$.
For option $A$: $r = 2 \sin \theta$.
Multiplying both sides by $r$,we get $r^2 = 2r \sin \theta$.
Substituting $r^2 = x^2 + y^2$ and $r \sin \theta = y$,we obtain $x^2 + y^2 = 2y$.
Rearranging gives $x^2 + (y - 1)^2 = 1$,which is the equation of a circle with center $(0, 1)$ and radius $1$.
For option $B$: $r^2 \cos 2 \theta = 1$ $\Rightarrow r^2(\cos^2 \theta - \sin^2 \theta) = 1$ $\Rightarrow x^2 - y^2 = 1$,which is a hyperbola.
For option $C$: $r(4 \cos \theta + 5 \sin \theta) = 3$ $\Rightarrow 4r \cos \theta + 5r \sin \theta = 3$ $\Rightarrow 4x + 5y = 3$,which is a straight line.
For option $D$: $5 = r + r\sqrt{2} \cos \theta \Rightarrow r = 5 - \sqrt{2}x$,which represents a conic section (parabola,ellipse,or hyperbola depending on eccentricity).
Thus,the correct option is $A$.

(B) The given equation of the parabola is $y^2=4ax$.
Let the equation of the line be $y=mx+c$.
For the line to be a tangent to the parabola,the condition is $c = \frac{a}{m}$.
Substituting this into the line equation,we get $y = mx + \frac{a}{m}$.
Multiplying by $m$,we get $my = m^2x + a$.
Replacing $m$ with $\frac{1}{m}$,we get $y = \frac{1}{m}x + am^2$,which simplifies to $my = x + am^2$.
Rearranging the terms,we get $x - my + am^2 = 0$.

(A) Let $P(x, y)$ be any point on the parabola.
By the definition of a parabola,the distance from $P$ to the focus $S(1, 0)$ is equal to the perpendicular distance from $P$ to the directrix $x+2y-1=0$.
$PS = PM$
$\sqrt{(x-1)^2 + (y-0)^2} = \frac{|x+2y-1|}{\sqrt{1^2+2^2}}$
Squaring both sides:
$(x-1)^2 + y^2 = \frac{(x+2y-1)^2}{5}$
$5(x^2 - 2x + 1 + y^2) = x^2 + 4y^2 + 1 + 4xy - 2x - 4y$
$5x^2 - 10x + 5 + 5y^2 = x^2 + 4y^2 + 4xy - 2x - 4y + 1$
Rearranging the terms:
$4x^2 - 4xy + y^2 - 8x + 4y + 4 = 0$

(B) We use the Squeeze Theorem to evaluate the limit.
We know that for all $x \neq 0$,$-1 \leq \sin \left(\frac{\pi}{x}\right) \leq 1$.
Multiplying the inequality by $x^2$ (since $x^2 > 0$ for $x \neq 0$),we get:
$-x^2 \leq x^2 \sin \left(\frac{\pi}{x}\right) \leq x^2$.
Now,taking the limit as $x \rightarrow 0$ on all sides:
$\lim _{x \rightarrow 0} (-x^2) \leq \lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right) \leq \lim _{x \rightarrow 0} x^2$.
Since $\lim _{x \rightarrow 0} (-x^2) = 0$ and $\lim _{x \rightarrow 0} x^2 = 0$,by the Squeeze Theorem,$\lim _{x \rightarrow 0} x^2 \sin \left(\frac{\pi}{x}\right) = 0$.

(B) Given $\frac{\tan 3A}{\tan A} = a$.
We know that $\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$.
So,$\frac{3\tan A - \tan^3 A}{\tan A(1 - 3\tan^2 A)} = a$.
$\frac{3 - \tan^2 A}{1 - 3\tan^2 A} = a$.
$3 - \tan^2 A = a - 3a\tan^2 A$.
$\tan^2 A(3a - 1) = a - 3$.
$\tan^2 A = \frac{a - 3}{3a - 1}$.
Now,$\frac{\sin 3A}{\sin A} = \frac{3\sin A - 4\sin^3 A}{\sin A} = 3 - 4\sin^2 A$.
Using $\sin^2 A = \frac{\tan^2 A}{1 + \tan^2 A} = \frac{\frac{a-3}{3a-1}}{1 + \frac{a-3}{3a-1}} = \frac{a-3}{3a-1+a-3} = \frac{a-3}{4a-4} = \frac{a-3}{4(a-1)}$.
Therefore,$\frac{\sin 3A}{\sin A} = 3 - 4\left(\frac{a-3}{4(a-1)}\right) = 3 - \frac{a-3}{a-1} = \frac{3(a-1) - (a-3)}{a-1} = \frac{3a - 3 - a + 3}{a-1} = \frac{2a}{a-1}$.

(A) Using the projection formula $c \cos B + b \cos C = a$,$a \cos C + c \cos A = b$,and $b \cos A + a \cos B = c$.
Expanding the expression:
$a \cos^2 B + a \cos^2 C + c \cos A \cos C + b \cos A \cos B$
$= a \cos^2 B + b \cos A \cos B + a \cos^2 C + c \cos A \cos C$
$= \cos B (a \cos B + b \cos A) + \cos C (a \cos C + c \cos A)$
$= \cos B (c) + \cos C (b)$
$= c \cos B + b \cos C$
$= a$

(A) Given that $A+B=C$.
We need to evaluate the expression $E = \cos ^2 A+\cos ^2 B+\cos ^2 C-2 \cos A \cos B \cos C$.
Using the identity $\cos ^2 \theta = \frac{1+\cos 2\theta}{2}$,we have:
$E = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} + \cos ^2 C - 2 \cos A \cos B \cos C$
$E = 1 + \frac{1}{2}(\cos 2A + \cos 2B) + \cos ^2 C - 2 \cos A \cos B \cos C$
Using $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$:
$E = 1 + \cos(A+B) \cos(A-B) + \cos ^2 C - 2 \cos A \cos B \cos C$
Since $A+B=C$,substitute $\cos(A+B) = \cos C$:
$E = 1 + \cos C \cos(A-B) + \cos ^2 C - 2 \cos A \cos B \cos C$
$E = 1 + \cos C [\cos(A-B) + \cos C] - 2 \cos A \cos B \cos C$
Substitute $\cos C = \cos(A+B) = \cos A \cos B - \sin A \sin B$:
$E = 1 + \cos C [\cos A \cos B + \sin A \sin B + \cos A \cos B - \sin A \sin B] - 2 \cos A \cos B \cos C$
$E = 1 + \cos C [2 \cos A \cos B] - 2 \cos A \cos B \cos C$
$E = 1 + 2 \cos A \cos B \cos C - 2 \cos A \cos B \cos C = 1$.

(D) Given that $A+C=2B$ ...$(i)$
We need to evaluate the expression $\frac{\cos C-\cos A}{\sin A-\sin C}$.
Using the trigonometric identities:
$\cos C - \cos A = 2 \sin\left(\frac{A+C}{2}\right) \sin\left(\frac{A-C}{2}\right)$
$\sin A - \sin C = 2 \cos\left(\frac{A+C}{2}\right) \sin\left(\frac{A-C}{2}\right)$
Substituting these into the expression:
$\frac{\cos C-\cos A}{\sin A-\sin C} = \frac{2 \sin\left(\frac{A+C}{2}\right) \sin\left(\frac{A-C}{2}\right)}{2 \cos\left(\frac{A+C}{2}\right) \sin\left(\frac{A-C}{2}\right)}$
Canceling the common terms $2$ and $\sin\left(\frac{A-C}{2}\right)$:
$= \frac{\sin\left(\frac{A+C}{2}\right)}{\cos\left(\frac{A+C}{2}\right)}$
$= \tan\left(\frac{A+C}{2}\right)$
Since $A+C=2B$,we have $\frac{A+C}{2} = B$.
Therefore,the expression equals $\tan B$.

(D) The given equation is $x^2-5x+6=0$.
Factoring the equation,we get $(x-3)(x-2)=0$,which gives the roots $x=3$ and $x=2$.
These roots represent the two sides of the triangle,so let $a=3$ and $b=2$.
The angle between these sides is $C = \frac{\pi}{3}$.
Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting the values,$\cos(\frac{\pi}{3}) = \frac{3^2+2^2-c^2}{2 \times 3 \times 2}$.
$\frac{1}{2} = \frac{9+4-c^2}{12} \Rightarrow \frac{1}{2} = \frac{13-c^2}{12}$.
$6 = 13-c^2$ $\Rightarrow c^2 = 7$ $\Rightarrow c = \sqrt{7}$.
The perimeter of the triangle is $a+b+c = 3+2+\sqrt{7} = 5+\sqrt{7}$.

(D) We know the Napier's analogy: $\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot \frac{A}{2}$.
Rearranging this,we get $(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) = (b-c)$.
Now,applying the summation $\Sigma$ over the cyclic terms:
$\Sigma(b+c) \tan \frac{A}{2} \tan \left(\frac{B-C}{2}\right) = (b-c) + (c-a) + (a-b)$.
Summing these terms: $b - c + c - a + a - b = 0$.

(B) We are given the expression $[x - |x|] = 5$.
Case $1$: If $x \geq 0$,then $|x| = x$.
Substituting this into the expression,we get $[x - x] = [0] = 0$.
Since $0 \neq 5$,there is no solution for $x \geq 0$.
Case $2$: If $x < 0$,then $|x| = -x$.
Substituting this into the expression,we get $[x - (-x)] = [2x] = 5$.
For $[2x] = 5$,we must have $5 \leq 2x < 6$,which implies $2.5 \leq x < 3$.
However,this contradicts our assumption that $x < 0$.
Therefore,there is no real value of $x$ that satisfies the given equation.
Thus,the set is the empty set,$\phi$.

(A) Given $x = \frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right)$.
First,calculate $x^2$:
$x^2 = \frac{1}{4} \left( 3 + \frac{1}{3} + 2 \right) = \frac{1}{4} \left( \frac{9+1+6}{3} \right) = \frac{1}{4} \left( \frac{16}{3} \right) = \frac{4}{3}$.
Now,calculate $x^2 - 1$:
$x^2 - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.
Therefore,$\sqrt{x^2 - 1} = \frac{1}{\sqrt{3}}$.
Substitute these values into the expression $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$:
$= \frac{\frac{1}{\sqrt{3}}}{\frac{1}{2} \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) - \frac{1}{\sqrt{3}}}$
$= \frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} - \frac{1}{\sqrt{3}}}$
$= \frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2} - \frac{1}{2\sqrt{3}}}$
$= \frac{\frac{1}{\sqrt{3}}}{\frac{3 - 1}{2\sqrt{3}}} = \frac{\frac{1}{\sqrt{3}}}{\frac{2}{2\sqrt{3}}} = \frac{\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = 1$.

(B) Let $BC$ be the height of the tower and $CD$ be the height of the flagstaff,where $BC = x$ and $CD = h$.
Let the point be $A$ at a distance $AB = y$ from the foot of the tower $B$.
Given that the tower and the flagstaff subtend equal angles $\theta$ at point $A$.
In $\triangle ABC$,$\tan \theta = \frac{BC}{AB} = \frac{x}{y}$.
In $\triangle ABD$,the total angle is $2\theta$ and the total height is $BD = BC + CD = x + h$.
Thus,$\tan 2\theta = \frac{BD}{AB} = \frac{x+h}{y}$.
Using the formula $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$,we have:
$\frac{2(x/y)}{1-(x/y)^2} = \frac{x+h}{y}$
$\frac{2x/y}{(y^2-x^2)/y^2} = \frac{x+h}{y}$
$\frac{2xy}{y^2-x^2} = \frac{x+h}{y}$
$2xy^2 = (x+h)(y^2-x^2)$
$\frac{2xy^2}{y^2-x^2} = x+h$
$h = \frac{2xy^2}{y^2-x^2} - x$
$h = \frac{2xy^2 - x(y^2-x^2)}{y^2-x^2}$
$h = \frac{2xy^2 - xy^2 + x^3}{y^2-x^2}$
$h = \frac{xy^2 + x^3}{y^2-x^2} = \frac{x(x^2+y^2)}{y^2-x^2}$.

(A) Let $f(n) = n(n+1)(2n+1)$.
We know that $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.
Each term $n(n+1)(n-1)$ and $n(n+1)(n+2)$ is a product of three consecutive integers.
The product of three consecutive integers is always divisible by $3! = 6$.
Therefore,$n(n+1)(n-1) = 6k_1$ and $n(n+1)(n+2) = 6k_2$ for some integers $k_1, k_2$.
Thus,$f(n) = 6(k_1 + k_2) = 6k$ for some integer $k$.
Hence,the set is a subset of $\{6k : k \in \mathbb{Z}\}$.

(D) Given the expression: $\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4} a+10^{-3} b+10^{-2} c}\right)$
We can simplify the denominator by factoring out $10^{-4}$:
$10^{-4} a + 10^{-3} b + 10^{-2} c = 10^{-4}(a + 10b + 10^2c)$
Substituting this back into the expression:
$\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4}(a+10 b+10^2 c)}\right)$
$= \log _{10}\left(\frac{1}{10^{-4}}\right)$
$= \log _{10}(10^4)$
$= 4 \log _{10}(10) = 4(1) = 4$

(C) Let the three-digit number be represented as $abc$,where $a, b, c$ are the digits.
According to the problem,$b = c$ and $a \neq b$.
The first digit $a$ can be any digit from $1$ to $9$ ($9$ choices).
The last two digits $b$ and $c$ must be equal,so we choose a digit $x \in \{0, 1, 2, \dots, 9\}$ for both $b$ and $c$.
Since $a \neq b$,for each choice of $a$,there are $10 - 1 = 9$ possible choices for the pair $(b, c)$.
Total number of such $n$'s $= 9 \times 9 = 81$.

(B) Given $S = \{1, 2, 3, \ldots, 50\}$,so $n(S) = 50$.
For set $A$,we solve $n + \frac{50}{n} > 27$.
$n^2 - 27n + 50 > 0$.
$(n - 25)(n - 2) > 0$.
This holds for $n < 2$ or $n > 25$.
Since $n \in S$,$n=1$ or $n \in \{26, 27, \ldots, 50\}$.
Thus,$A = \{1, 26, 27, \ldots, 50\}$,so $n(A) = 1 + 25 = 26$.
For set $B$,primes in $S$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$,so $n(B) = 15$.
For set $C$,squares in $S$ are $\{1, 4, 9, 16, 25, 36, 49\}$,so $n(C) = 7$.
Probabilities are $P(A) = \frac{26}{50}$,$P(B) = \frac{15}{50}$,$P(C) = \frac{7}{50}$.
Therefore,$P(A) > P(B) > P(C)$.

(C) The given function is $f(x) = \frac{ax + b}{cx + d}$.
For the function to be a constant function,its derivative with respect to $x$ must be zero,or the numerator must be a constant multiple of the denominator.
Let $f(x) = k$ (a constant).
Then $\frac{ax + b}{cx + d} = k$.
$ax + b = k(cx + d) = (kc)x + kd$.
Comparing the coefficients of $x$ and the constant terms,we get $a = kc$ and $b = kd$.
This implies $\frac{a}{c} = k$ and $\frac{b}{d} = k$ (assuming $c, d \neq 0$).
Therefore,$\frac{a}{c} = \frac{b}{d}$,which gives $ad = bc$.
Alternatively,if we take option $(c)$,i.e.,$ad = bc$,then $ad - bc = 0$.
If $ad = bc$,then $\frac{a}{c} = \frac{b}{d} = k$ (for $c, d \neq 0$).
Substituting $a = kc$ and $b = kd$ into the function:
$f(x) = \frac{(kc)x + kd}{cx + d} = \frac{k(cx + d)}{cx + d} = k$.
Since $f(x) = k$,the function is constant.

(B) Given that $x \sqrt{1+y} + y \sqrt{1+x} = 0$ ... $(i)$
Rearranging the terms,we get
$x \sqrt{1+y} = -y \sqrt{1+x}$
On squaring both sides,we get
$x^2(1+y) = y^2(1+x)$
$x^2 + x^2y = y^2 + xy^2$
$x^2 - y^2 + x^2y - xy^2 = 0$
$(x-y)(x+y) + xy(x-y) = 0$
$(x-y)(x+y+xy) = 0$
Since $x-y \neq 0$ (as it does not satisfy the original equation),we must have
$x+y+xy = 0$
$y(1+x) = -x$
$y = -\frac{x}{1+x}$
Now,differentiating with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = -\frac{(1+x)\frac{d}{dx}(x) - x\frac{d}{dx}(1+x)}{(1+x)^2}$
$\frac{dy}{dx} = -\frac{(1+x)(1) - x(1)}{(1+x)^2}$
$\frac{dy}{dx} = -\frac{1+x-x}{(1+x)^2}$
$\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

(B) Given that $f(x)=10 \cos x+(13+2 x) \sin x \quad ...(i)$
On differentiating with respect to $x$,we get
$f^{\prime}(x)=-10 \sin x+(13+2 x) \cos x+2 \sin x$
$f^{\prime}(x)=-8 \sin x+(13+2 x) \cos x$
Again differentiating with respect to $x$,we get
$f^{\prime \prime}(x)=-8 \cos x+(13+2 x)(-\sin x)+2 \cos x$
$f^{\prime \prime}(x)=-6 \cos x-(13+2 x) \sin x \quad ...(ii)$
On adding equations $(i)$ and $(ii)$,we get
$f^{\prime \prime}(x)+f(x) = [-6 \cos x-(13+2 x) \sin x] + [10 \cos x+(13+2 x) \sin x]$
$f^{\prime \prime}(x)+f(x) = 4 \cos x$

(A) The area $A$ of a circular plate with radius $r$ is given by $A = \pi r^2$.
To find the rate at which the area increases,we differentiate $A$ with respect to time $t$:
$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$.
Given that $\frac{dr}{dt} = 0.01 \text{ cm/s}$ and $r = 12 \text{ cm}$,we substitute these values into the equation:
$\frac{dA}{dt} = 2 \pi (12) (0.01) = 0.24 \pi \text{ cm}^2/\text{s}$.
Thus,the rate at which the area increases is $0.24 \pi \text{ cm}^2/\text{s}$.

(A) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$ ... $(i)$
On differentiating with respect to $x$,we get
$f'(x) = \frac{(x^2+x+1)(2x-1) - (x^2-x+1)(2x+1)}{(x^2+x+1)^2}$
For maximum or minimum,put $f'(x) = 0$
$(x^2+x+1)(2x-1) - (x^2-x+1)(2x+1) = 0$
$(2x^3 - x^2 + 2x^2 - x + 2x - 1) - (2x^3 + x^2 - 2x^2 - x + 2x + 1) = 0$
$(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) = 0$
$2x^2 - 2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$
Now,$f'(x) = \frac{2x^2-2}{(x^2+x+1)^2}$
Again differentiating,we get $f''(x) = \frac{(x^2+x+1)^2(4x) - (2x^2-2) \cdot 2(x^2+x+1)(2x+1)}{(x^2+x+1)^4}$
At $x = 1$,$f''(1) = \frac{(1+1+1)^2(4) - 0}{(1+1+1)^4} = \frac{9 \times 4}{81} = \frac{36}{81} > 0$
Therefore,the function has a minimum at $x = 1$.
Putting $x = 1$ in equation $(i)$,we get
$f(1) = \frac{1^2-1+1}{1^2+1+1} = \frac{1}{3}$
$\therefore$ The minimum value is $\frac{1}{3}$.

(C) Given the equation of motion: $s = 490t - 4.9t^2$.
To find the maximum height,we differentiate $s$ with respect to $t$ and set the derivative to zero:
$\frac{ds}{dt} = 490 - 9.8t$.
Setting $\frac{ds}{dt} = 0$ for maximum height:
$490 - 9.8t = 0$
$9.8t = 490$
$t = \frac{490}{9.8} = 50 \text{ seconds}$.
Now,substitute $t = 50$ into the original equation to find the maximum height $s$:
$s = 490(50) - 4.9(50)^2$
$s = 24500 - 4.9(2500)$
$s = 24500 - 12250$
$s = 12250$.

(A) Given the differential equation: $x^2 y - x^3 \frac{dy}{dx} = y^4 \cos x$.
Dividing both sides by $x^3 y^4$,we get:
$\frac{1}{y^3 x} - \frac{1}{y^4} \frac{dy}{dx} = \frac{\cos x}{x^3}$
Rearranging the terms:
$\frac{1}{y^4} \frac{dy}{dx} - \frac{1}{x y^3} = -\frac{\cos x}{x^3}$
Let $t = y^{-3} = \frac{1}{y^3}$. Then,$\frac{dt}{dx} = -3 y^{-4} \frac{dy}{dx}$,which implies $\frac{1}{y^4} \frac{dy}{dx} = -\frac{1}{3} \frac{dt}{dx}$.
Substituting this into the equation:
$-\frac{1}{3} \frac{dt}{dx} - \frac{t}{x} = -\frac{\cos x}{x^3}$
Multiplying by $-3$:
$\frac{dt}{dx} + \frac{3}{x} t = \frac{3 \cos x}{x^3}$
This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = \frac{3}{x}$ and $Q(x) = \frac{3 \cos x}{x^3}$.
The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$.
The solution is given by $t \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + c$.
$t \cdot x^3 = \int \frac{3 \cos x}{x^3} \cdot x^3 dx + c$
$t x^3 = 3 \int \cos x dx + c$
$t x^3 = 3 \sin x + c$
Since $t = y^{-3}$,we have $x^3 y^{-3} = 3 \sin x + c$.

(A) $I$. Given $dy+2xy dx=2e^{-x^2} dx$.
Dividing by $dx$,we get $\frac{dy}{dx}+2xy=2e^{-x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=2x$ and $Q=2e^{-x^2}$.
Integrating factor $I.F. = e^{\int P dx} = e^{\int 2x dx} = e^{x^2}$.
The general solution is $y(I.F.) = \int Q(I.F.) dx + c$.
$ye^{x^2} = \int 2e^{-x^2} \cdot e^{x^2} dx + c = \int 2 dx + c = 2x+c$.
Thus,statement $I$ is true.
$II$. Given $ye^{x^2}-2x=c$.
Differentiating with respect to $x$: $\frac{d}{dx}(ye^{x^2}) - \frac{d}{dx}(2x) = 0$.
$y(2x e^{x^2}) + e^{x^2} \frac{dy}{dx} - 2 = 0$.
$e^{x^2} \frac{dy}{dx} = 2 - 2xye^{x^2}$.
$\frac{dy}{dx} = 2e^{-x^2} - 2xy$.
Therefore,$dx = \frac{dy}{2e^{-x^2}-2xy}$.
Thus,statement $II$ is true.

(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$,which implies $a \cdot a = 1$ and $b \cdot b = 1$.
Expanding the cross product using the distributive property:
$(a+b) \times (a \times b) = a \times (a \times b) + b \times (a \times b)$
Using the vector triple product formula $A \times (B \times C) = (A \cdot C)B - (A \cdot B)C$:
$a \times (a \times b) = (a \cdot b)a - (a \cdot a)b = (a \cdot b)a - b$
$b \times (a \times b) = (b \cdot b)a - (b \cdot a)b = a - (a \cdot b)b$
Adding these two results:
$(a \cdot b)a - b + a - (a \cdot b)b = a - b + (a \cdot b)(a - b) = (a - b)(1 + a \cdot b)$
Since $(1 + a \cdot b)$ is a scalar,the resulting vector is parallel to $(a - b)$.

(A) We analyze each expression:
$(A)$ The scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ is defined as $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$. Thus,$(A)$ matches with $3$.
$(B)$ Using the vector triple product formula $(\mathbf{x} \times \mathbf{y}) \times \mathbf{z} = (\mathbf{x} \cdot \mathbf{z})\mathbf{y} - (\mathbf{y} \cdot \mathbf{z})\mathbf{x}$,we have $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b} = (\mathbf{c} \cdot \mathbf{b})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} = (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$. Thus,$(B)$ matches with $5$.
$(C)$ Using the vector triple product formula $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$. Thus,$(C)$ matches with $2$.
$(D)$ The dot product $\mathbf{a} \cdot \mathbf{b}$ is defined as $|\mathbf{a}||\mathbf{b}|\cos(\theta)$,where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. Thus,$(D)$ matches with $1$.
Therefore,the correct matching is $A-3, B-5, C-2, D-1$.

(C) Let the points be $A(1, -2, -3)$ and $B(2, 0, 0)$. The vector $\vec{AB} = (2-1)\hat{i} + (0-(-2))\hat{j} + (0-(-3))\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Any point $P(x, y, z)$ is collinear with $A$ and $B$ if the vector $\vec{AP}$ is a scalar multiple of $\vec{AB}$.
Let $P = (0, -4, -6)$. Then $\vec{AP} = (0-1)\hat{i} + (-4-(-2))\hat{j} + (-6-(-3))\hat{k} = -\hat{i} - 2\hat{j} - 3\hat{k}$.
Since $\vec{AP} = -1(\hat{i} + 2\hat{j} + 3\hat{k}) = -1\vec{AB}$,the vector $\vec{AP}$ is a scalar multiple of $\vec{AB}$.
Therefore,the point $(0, -4, -6)$ is collinear with $(1, -2, -3)$ and $(2, 0, 0)$.

(A) Let the points be $O(0, 0, 0)$ and $P(2, 3, -1)$.
The direction ratios of the line $OP$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (2 - 0, 3 - 0, -1 - 0) = (2, 3, -1)$.
The distance $OP$ is given by $\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $\left(\frac{a}{r}, \frac{b}{r}, \frac{c}{r}\right) = \left(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)$.

(B) Let $P(B) = p$. According to the given condition,$P(A) = 2P(B) = 2p$. Since $P(A) + P(B) = 1$,we have $2p + p = 1$,which implies $3p = 1$,so $p = \frac{1}{3}$. Thus,$P(B) = \frac{1}{3}$ and $P(A) = \frac{2}{3}$.
The probability of drawing a red ball from Box $A$ is $P(R|A) = \frac{3}{2+3} = \frac{3}{5}$.
The probability of drawing a red ball from Box $B$ is $P(R|B) = \frac{4}{3+4} = \frac{4}{7}$.
Using Bayes' theorem,the probability that the red ball came from Box $B$ is:
$P(B|R) = \frac{P(B) \cdot P(R|B)}{P(A) \cdot P(R|A) + P(B) \cdot P(R|B)}$
$P(B|R) = \frac{\frac{1}{3} \cdot \frac{4}{7}}{\frac{2}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{4}{7}}$
$P(B|R) = \frac{\frac{4}{21}}{\frac{6}{15} + \frac{4}{21}} = \frac{\frac{4}{21}}{\frac{2}{5} + \frac{4}{21}}$
$P(B|R) = \frac{\frac{4}{21}}{\frac{42 + 20}{105}} = \frac{\frac{4}{21}}{\frac{62}{105}} = \frac{4}{21} \cdot \frac{105}{62} = \frac{4 \cdot 5}{62} = \frac{20}{62} = \frac{10}{31}$.

(B) Given that $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \in \{0, 1, 2, \ldots, \infty\}$.
As we know that the sum of all probabilities in a probability distribution is $1$,so $\sum_{k=0}^{\infty} P(X=k) = 1$.
Substituting the given expression:
$a \left( 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty \right) = 1 \quad \dots (i)$
Let $S = 1 + \frac{2}{3} + \frac{3}{3^2} + \frac{4}{3^3} + \ldots \infty$.
Then $\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \ldots \infty$.
Subtracting the two equations:
$S - \frac{1}{3}S = 1 + \left( \frac{2}{3} - \frac{1}{3} \right) + \left( \frac{3}{3^2} - \frac{2}{3^2} \right) + \ldots \infty$
$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \infty$
This is an infinite geometric series with first term $1$ and common ratio $r = \frac{1}{3}$.
$\frac{2}{3}S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{2/3} = \frac{3}{2}$.
$S = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4}$.
From equation $(i)$,$a \times S = 1 \implies a \times \frac{9}{4} = 1$.
Therefore,$a = \frac{4}{9}$.

(C) Given that $n=6$ and $P(X=2)=9 P(X=4)$.
Using the binomial probability formula $P(X=k) = {}^n C_k p^k q^{n-k}$:
${}^6 C_2 p^2 q^4 = 9 \cdot {}^6 C_4 p^4 q^2$
Since ${}^6 C_2 = 15$ and ${}^6 C_4 = 15$,we have:
$15 p^2 q^4 = 9 \cdot 15 p^4 q^2$
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$q^2 = 9 p^2$
Taking the square root on both sides:
$q = 3p$
Since $p + q = 1$,we substitute $q = 3p$:
$p + 3p = 1 \Rightarrow 4p = 1 \Rightarrow p = \frac{1}{4}$
Then $q = 1 - \frac{1}{4} = \frac{3}{4}$.
The variance of a binomial distribution is given by $npq$:
$\text{Variance} = 6 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{18}{16} = \frac{9}{8}$.

(B) Given that $f(x) = \frac{x - 2}{x^2 - 3x + 2} = \frac{x - 2}{(x - 2)(x - 1)} = \frac{1}{x - 1}$ for $x \neq 1, 2$.
We need to evaluate $\lim_{x \rightarrow 2} \frac{f(x) - f(2)}{x - 2}$.
Given $f(2) = 1$.
Substituting the values,we get $\lim_{x \rightarrow 2} \frac{\frac{1}{x - 1} - 1}{x - 2}$.
$= \lim_{x \rightarrow 2} \frac{\frac{1 - (x - 1)}{x - 1}}{x - 2} = \lim_{x \rightarrow 2} \frac{2 - x}{(x - 1)(x - 2)}$.
$= \lim_{x \rightarrow 2} \frac{-(x - 2)}{(x - 1)(x - 2)} = \lim_{x \rightarrow 2} \frac{-1}{x - 1}$.
$= \frac{-1}{2 - 1} = -1$.

(B) The given curves are:
$y^2 = 4x$ ... $(i)$
$x^2 = 4y$ ... (ii)
From (ii),we have $y = \frac{x^2}{4}$. Substituting this into $(i)$:
$(\frac{x^2}{4})^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.
Thus,$x = 0$ or $x = 4$. The intersecting points are $(0,0)$ and $(4,4)$.
The required area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=4$:
$\text{Area} = \int_0^4 (\sqrt{4x} - \frac{x^2}{4}) dx$
$= \int_0^4 (2x^{1/2} - \frac{x^2}{4}) dx$
$= [2 \cdot \frac{x^{3/2}}{3/2} - \frac{x^3}{12}]_0^4$
$= [\frac{4}{3}x^{3/2} - \frac{x^3}{12}]_0^4$
$= [\frac{4}{3}(4)^{3/2} - \frac{4^3}{12}] - [0 - 0]$
$= [\frac{4}{3} \cdot 8 - \frac{64}{12}]$
$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \text{ sq. units.}$

(D) Given the matrix equation:
$m[-3, 4] + n[4, -3] = [10, -11]$
Multiplying the scalars $m$ and $n$ into the matrices,we get:
$[-3m, 4m] + [4n, -3n] = [10, -11]$
Adding the matrices,we have:
$[-3m + 4n, 4m - 3n] = [10, -11]$
By equating the corresponding elements,we obtain two linear equations:
$-3m + 4n = 10$ $\dots(i)$
$4m - 3n = -11$ $\dots(ii)$
To solve for $m$ and $n$,multiply equation $(i)$ by $3$ and equation $(ii)$ by $4$:
$-9m + 12n = 30$ $\dots(iii)$
$16m - 12n = -44$ $\dots(iv)$
Adding equations $(iii)$ and $(iv)$:
$7m = -14 \Rightarrow m = -2$
Substituting $m = -2$ into equation $(i)$:
$-3(-2) + 4n = 10 \Rightarrow 6 + 4n = 10 \Rightarrow 4n = 4 \Rightarrow n = 1$
Finally,calculating $3m + 7n$:
$3(-2) + 7(1) = -6 + 7 = 1$

(A) Given that,$A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} \dots (i)$
First,we calculate $A^2$:
$A^2 = A \cdot A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (-1)(-1) + (0)(0) & (-1)(0) + (0)(2) \\ (0)(-1) + (2)(0) & (0)(0) + (2)(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$
Next,we calculate $A^3$:
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} (1)(-1) + (0)(0) & (1)(0) + (0)(2) \\ (0)(-1) + (4)(0) & (0)(0) + (4)(2) \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix}$
Now,calculate $A^3 - A^2$:
$A^3 - A^2 = \begin{bmatrix} -1 & 0 \\ 0 & 8 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -1-1 & 0-0 \\ 0-0 & 8-4 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$
Since $A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$,we can see that $2A = 2 \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix}$.
Therefore,$A^3 - A^2 = 2A$.

(C) Let $A = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 1 & -2 \\ 0 & 2 & 1 \end{bmatrix}$. The adjoint of a matrix is the transpose of its cofactor matrix,i.e.,$\operatorname{adj}(A) = [C_{ij}]^T$.
The cofactors $C_{ij}$ are calculated as follows:
$C_{11} = +\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = 1 - (-4) = 5$
$C_{12} = -\begin{vmatrix} -1 & -2 \\ 0 & 1 \end{vmatrix} = -(-1 - 0) = 1$
$C_{13} = +\begin{vmatrix} -1 & 1 \\ 0 & 2 \end{vmatrix} = -2 - 0 = -2$
$C_{21} = -\begin{vmatrix} 0 & 2 \\ 2 & 1 \end{vmatrix} = -(0 - 4) = 4$
$C_{22} = +\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 - 0 = 1$
$C_{23} = -\begin{vmatrix} 1 & 0 \\ 0 & 2 \end{vmatrix} = -(2 - 0) = -2$
$C_{31} = +\begin{vmatrix} 0 & 2 \\ 1 & -2 \end{vmatrix} = 0 - 2 = -2$
$C_{32} = -\begin{vmatrix} 1 & 2 \\ -1 & -2 \end{vmatrix} = -(-2 - (-2)) = 0$
$C_{33} = +\begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} = 1 - 0 = 1$
The cofactor matrix is $\begin{bmatrix} 5 & 1 & -2 \\ 4 & 1 & -2 \\ -2 & 0 & 1 \end{bmatrix}$.
Taking the transpose,$\operatorname{adj}(A) = \begin{bmatrix} 5 & 4 & -2 \\ 1 & 1 & 0 \\ -2 & -2 & 1 \end{bmatrix}$.
Comparing this with the given matrix $\begin{bmatrix} 5 & a & -2 \\ 1 & 1 & 0 \\ -2 & -2 & b \end{bmatrix}$,we get $a = 4$ and $b = 1$.
Therefore,$[a \quad b] = [4 \quad 1]$.

(C) We use the formula $2 \tanh^{-1} x = \tanh^{-1} \left( \frac{2x}{1+x^2} \right)$.
Substituting $x = \frac{1}{2}$:
$2 \tanh^{-1} \left( \frac{1}{2} \right) = \tanh^{-1} \left( \frac{2(\frac{1}{2})}{1+(\frac{1}{2})^2} \right) = \tanh^{-1} \left( \frac{1}{1 + \frac{1}{4}} \right) = \tanh^{-1} \left( \frac{1}{\frac{5}{4}} \right) = \tanh^{-1} \left( \frac{4}{5} \right)$.
Now,we use the logarithmic form of the inverse hyperbolic tangent function: $\tanh^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right)$.
Substituting $x = \frac{4}{5}$:
$\tanh^{-1} \left( \frac{4}{5} \right) = \frac{1}{2} \log \left( \frac{1 + \frac{4}{5}}{1 - \frac{4}{5}} \right) = \frac{1}{2} \log \left( \frac{\frac{9}{5}}{\frac{1}{5}} \right) = \frac{1}{2} \log 9 = \frac{1}{2} \log 3^2 = \log 3$.

(C) We are given the expression: $\sin ^{-1} \frac{4}{5} + 2 \tan ^{-1} \frac{1}{3}$.
First,use the formula $2 \tan ^{-1} x = \tan ^{-1} \frac{2x}{1-x^2}$:
$2 \tan ^{-1} \frac{1}{3} = \tan ^{-1} \frac{2(1/3)}{1-(1/3)^2} = \tan ^{-1} \frac{2/3}{1-1/9} = \tan ^{-1} \frac{2/3}{8/9} = \tan ^{-1} \left(\frac{2}{3} \times \frac{9}{8}\right) = \tan ^{-1} \frac{3}{4}$.
Now,the expression becomes $\sin ^{-1} \frac{4}{5} + \tan ^{-1} \frac{3}{4}$.
Since $\tan ^{-1} \frac{3}{4} = \theta$,then $\tan \theta = \frac{3}{4}$,which implies $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
Thus,$\tan ^{-1} \frac{3}{4} = \cos ^{-1} \frac{4}{5}$.
Substituting this back,we get $\sin ^{-1} \frac{4}{5} + \cos ^{-1} \frac{4}{5}$.
Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we get the result $\frac{\pi}{2}$.

(C) Given the function $f(x) = \frac{ax + b}{cx + d} \quad \dots(i)$
For the function to be a constant function,the derivative $f'(x)$ must be zero for all $x$ in its domain.
Using the quotient rule,$f'(x) = \frac{a(cx + d) - c(ax + b)}{(cx + d)^2} = \frac{acx + ad - acx - bc}{(cx + d)^2} = \frac{ad - bc}{(cx + d)^2}$.
For $f(x)$ to be constant,$f'(x) = 0$,which implies $ad - bc = 0$,or $ad = bc$.
Alternatively,if $ad = bc$,let $\frac{a}{c} = \frac{b}{d} = k$. Then $a = ck$ and $b = dk$.
Substituting these into the function: $f(x) = \frac{ckx + dk}{cx + d} = \frac{k(cx + d)}{cx + d} = k$,which is a constant.

(C) Given that $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$
For $x \in R - \{-1, -2\}$,$f(x) = \frac{x + 2}{(x + 1)(x + 2)} = \frac{1}{x + 1}$.
Now,we check the continuity at $x = -2$ and $x = -1$.
At $x = -2$:
$\lim_{x \rightarrow -2} f(x) = \lim_{x \rightarrow -2} \frac{1}{x + 1} = \frac{1}{-2 + 1} = -1$.
Since $f(-2) = -1$,we have $\lim_{x \rightarrow -2} f(x) = f(-2)$,so $f$ is continuous at $x = -2$.
At $x = -1$:
$\lim_{x \rightarrow -1^-} f(x) = \lim_{x \rightarrow -1^-} \frac{1}{x + 1} = -\infty$ and $\lim_{x \rightarrow -1^+} f(x) = \lim_{x \rightarrow -1^+} \frac{1}{x + 1} = \infty$.
Since the limit does not exist at $x = -1$,$f$ is discontinuous at $x = -1$.
Therefore,$f$ is continuous on the set $R - \{-1\}$.

(C) Given that $f(x)$ is an even function,we have $f(x) = f(-x)$ for all $x \in R$.
Differentiating both sides with respect to $x$,we get $f^{\prime}(x) = -f^{\prime}(-x)$.
Differentiating again with respect to $x$,we get $f^{\prime \prime}(x) = f^{\prime \prime}(-x)$.
This shows that the second derivative $f^{\prime \prime}(x)$ is also an even function.
Since $f^{\prime \prime}(x)$ is an even function,$f^{\prime \prime}(-\pi) = f^{\prime \prime}(\pi)$.
Given that $f^{\prime \prime}(\pi) = 1$,it follows that $f^{\prime \prime}(-\pi) = 1$.

(A) Given the function $u(x, y) = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right)$.
We observe that $u(tx, ty) = \sin^{-1}\left(\frac{tx}{ty}\right) + \tan^{-1}\left(\frac{ty}{tx}\right) = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) = u(x, y)$.
This implies that $u$ is a homogeneous function of degree $n = 0$.
According to Euler's Theorem for homogeneous functions,if $u$ is a homogeneous function of degree $n$,then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \cdot u$.
Since $n = 0$,we have $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0 \cdot u = 0$.
Therefore,the correct option is $A$.

(A) For statement $I$:
$f(x) = a x^{41} + b x^{-40}$
$f^{\prime}(x) = 41 a x^{40} - 40 b x^{-41}$
$f^{\prime \prime}(x) = 41 \times 40 a x^{39} + 40 \times 41 b x^{-42} = 1640 a x^{39} + 1640 b x^{-42}$
$\frac{f^{\prime \prime}(x)}{f(x)} = \frac{1640(a x^{39} + b x^{-42})}{a x^{41} + b x^{-40}} = \frac{1640(a x^{39} + b x^{-42})}{x^2(a x^{39} + b x^{-42})} = 1640 x^{-2}$.
Thus,statement $I$ is true.
For statement $II$:
Let $y = \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$. Using the substitution $x = \tan \theta$,we get $y = \tan ^{-1}(\tan 2 \theta) = 2 \theta = 2 \tan ^{-1} x$.
Then $\frac{d y}{d x} = \frac{2}{1+x^2}$.
Since the given statement claims the derivative is $\frac{1}{1+x^2}$,statement $II$ is false.
Therefore,$I$ is true,but $II$ is false.

(A) Given $f(x) = 10 \cos x + (13 + 2x) \sin x$.
First,find the first derivative $f'(x)$ using the product rule:
$f'(x) = -10 \sin x + [2 \sin x + (13 + 2x) \cos x] = -8 \sin x + (13 + 2x) \cos x$.
Now,find the second derivative $f''(x)$:
$f''(x) = -8 \cos x + [2 \cos x - (13 + 2x) \sin x] = -6 \cos x - (13 + 2x) \sin x$.
Finally,calculate $f''(x) + f(x)$:
$f''(x) + f(x) = [-6 \cos x - (13 + 2x) \sin x] + [10 \cos x + (13 + 2x) \sin x] = 4 \cos x$.

(A) For the function $f(x) = 2x^3 - 9x^2 + 12x - 3$,the derivative is $f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$.
For $f(x)$ to be increasing,we require $f'(x) > 0$,which implies $6(x-1)(x-2) > 0$. This inequality holds when $x < 1$ or $x > 2$. Thus,$f(x)$ is increasing outside the interval $(1,2)$. Hence,statement $A$ is true.
For statement $R$,we check the condition $f'(x) < 0$. This inequality holds when $6(x-1)(x-2) < 0$,which occurs for $x \in (1,2)$. Thus,statement $R$ is true.
Since statement $A$ describes the increasing behavior and statement $R$ describes the decreasing behavior,$R$ is not the reason for $A$. Therefore,both $A$ and $R$ are true,but $R$ is not the correct explanation for $A$.

(A) Let $I = \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$.
Substitute $x = \tan \theta$,then $d x = \sec ^2 \theta d \theta$.
Since $\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) = \sin ^{-1}(\sin 2 \theta) = 2 \theta$ (assuming $|x| \le 1$),we have:
$I = \int 2 \theta \sec ^2 \theta d \theta$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = \theta$ and $dv = \sec ^2 \theta d \theta$:
$I = 2 \left[ \theta \tan \theta - \int \tan \theta d \theta \right]$.
$I = 2 [\theta \tan \theta + \log |\cos \theta|] + C$.
Since $\tan \theta = x$,we have $\theta = \tan ^{-1} x$ and $\cos \theta = \frac{1}{\sqrt{1+x^2}}$.
$I = 2 [x \tan ^{-1} x + \log |\frac{1}{\sqrt{1+x^2}}|] + C$.
$I = 2 x \tan ^{-1} x + 2 \log (1+x^2)^{-1/2} + C$.
$I = 2 x \tan ^{-1} x - \log (1+x^2) + C$.
Comparing this with $f(x) - \log (1+x^2)$,we get $f(x) = 2 x \tan ^{-1} x$.

(C) Let $I = \int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x$.
Substitute $t = x^{50}$,then $dt = 50x^{49} dx$,which implies $x^{49} dx = \frac{1}{50} dt$.
Substituting these into the integral,we get:
$I = \frac{1}{50} \int \frac{\tan ^{-1} t}{1+t^2} dt$.
Now,let $u = \tan ^{-1} t$,then $du = \frac{1}{1+t^2} dt$.
The integral becomes:
$I = \frac{1}{50} \int u du = \frac{1}{50} \cdot \frac{u^2}{2} + c = \frac{u^2}{100} + c$.
Substituting back $u = \tan ^{-1} (x^{50})$,we have:
$I = \frac{(\tan ^{-1} (x^{50}))^2}{100} + c$.
Comparing this with the given expression $k(\tan ^{-1} (x^{50}))^2 + c$,we find $k = \frac{1}{100}$.

(A) Let $I = \int \frac{\sin x}{\cos x(1+\cos x)} dx$.
Substitute $\cos x = t$,then $-\sin x dx = dt$,so $\sin x dx = -dt$.
$I = \int \frac{-dt}{t(1+t)} = -\int \left[ \frac{1}{t} - \frac{1}{1+t} \right] dt$.
$I = -[\log |t| - \log |1+t|] + c = \log |1+t| - \log |t| + c$.
$I = \log \left| \frac{1+t}{t} \right| + c$.
Substituting $t = \cos x$ back,we get $I = \log \left| \frac{1+\cos x}{\cos x} \right| + c$.
Since $I = f(x) + c$,we have $f(x) = \log \left| \frac{1+\cos x}{\cos x} \right|$.

(D) Let $I = \int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ ... $(i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = \int_0^\pi \frac{(\pi-\theta) \sin (\pi-\theta)}{1+\cos ^2(\pi-\theta)} d \theta$
Since $\sin(\pi-\theta) = \sin \theta$ and $\cos(\pi-\theta) = -\cos \theta$,we get:
$I = \int_0^\pi \frac{(\pi-\theta) \sin \theta}{1+(-\cos \theta)^2} d \theta = \int_0^\pi \frac{(\pi-\theta) \sin \theta}{1+\cos ^2 \theta} d \theta$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_0^\pi \frac{\theta \sin \theta + (\pi-\theta) \sin \theta}{1+\cos ^2 \theta} d \theta = \int_0^\pi \frac{\pi \sin \theta}{1+\cos ^2 \theta} d \theta$
$2I = \pi \int_0^\pi \frac{\sin \theta}{1+\cos ^2 \theta} d \theta$
Let $\cos \theta = t$,then $-\sin \theta d \theta = dt$. When $\theta = 0, t = 1$ and when $\theta = \pi, t = -1$.
$2I = \pi \int_1^{-1} \frac{-dt}{1+t^2} = \pi \int_{-1}^1 \frac{dt}{1+t^2}$
$2I = \pi [\tan^{-1} t]_{-1}^1 = \pi (\tan^{-1}(1) - \tan^{-1}(-1)) = \pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = \pi (\frac{\pi}{2}) = \frac{\pi^2}{2}$
$I = \frac{\pi^2}{4}$

(C) Let $I = \int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$.
We can rewrite the numerator as $100(\sin x + \cos x) + 100 \sin x$.
So,$I = \int_0^{\pi / 2} \frac{100(\sin x + \cos x) + 100 \sin x}{\sin x + \cos x} d x = 100 \int_0^{\pi / 2} 1 d x + 100 \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$.
Let $I_1 = \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$ ... $(i)$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I_1 = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} d x = \int_0^{\pi / 2} \frac{\cos x}{\cos x + \sin x} d x$ ... (ii).
Adding $(i)$ and (ii):
$2I_1 = \int_0^{\pi / 2} \frac{\sin x + \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 1 d x = [x]_0^{\pi / 2} = \frac{\pi}{2}$.
Thus,$I_1 = \frac{\pi}{4}$.
Now,substituting back into the expression for $I$:
$I = 100 \times [x]_0^{\pi / 2} + 100 \times I_1 = 100 \times \frac{\pi}{2} + 100 \times \frac{\pi}{4} = 50\pi + 25\pi = 75\pi$.

(C) Given that $dx + dy = (x + y)(dx - dy)$.
Dividing by $dx$,we get:
$1 + \frac{dy}{dx} = (x + y)(1 - \frac{dy}{dx})$
$1 + \frac{dy}{dx} = x + y - (x + y)\frac{dy}{dx}$
$\frac{dy}{dx}(1 + x + y) = x + y - 1$
$\frac{dy}{dx} = \frac{x + y - 1}{x + y + 1} \quad \dots(i)$
Let $x + y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting into Eq. $(i)$:
$\frac{dt}{dx} - 1 = \frac{t - 1}{t + 1}$
$\frac{dt}{dx} = \frac{t - 1}{t + 1} + 1 = \frac{t - 1 + t + 1}{t + 1} = \frac{2t}{t + 1}$
Separating variables:
$\frac{t + 1}{2t} dt = dx$
$\frac{1}{2}(1 + \frac{1}{t}) dt = dx$
Integrating both sides:
$\frac{1}{2}(t + \log|t|) = x + C_1$
$t + \log|t| = 2x + 2C_1$
Substituting $t = x + y$:
$x + y + \log(x + y) = 2x + C$
$\log(x + y) = x - y + C$

(B) Given the homogeneous differential equation:
$\frac{dy}{dx} = \frac{y + x \tan(\frac{y}{x})}{x} \quad \dots(i)$
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into equation $(i)$:
$v + x \frac{dv}{dx} = \frac{vx + x \tan v}{x}$
$v + x \frac{dv}{dx} = v + \tan v$
$x \frac{dv}{dx} = \tan v$
Separating the variables:
$\cot v \, dv = \frac{dx}{x}$
Integrating both sides:
$\int \cot v \, dv = \int \frac{dx}{x}$
$\log|\sin v| = \log|x| + \log|c|$
$\log|\sin v| = \log|cx|$
$\sin v = cx$
Substituting $v = \frac{y}{x}$ back:
$\sin(\frac{y}{x}) = cx$

(D) The function $f(n)$ represents the sum of all positive divisors of $n$.
For a number $n = 2^k \cdot 3^1$,the divisors are the products of the divisors of $2^k$ and the divisors of $3^1$.
The sum of divisors of $2^k$ is $(1 + 2 + 2^2 + \dots + 2^k) = \frac{2^{k+1}-1}{2-1} = 2^{k+1}-1$.
The sum of divisors of $3^1$ is $(1 + 3) = 4$.
Since $f$ is a multiplicative function for coprime factors,$f(2^k \cdot 3) = f(2^k) \cdot f(3)$.
Thus,$f(2^k \cdot 3) = (2^{k+1}-1) \cdot 4 = 4(2^{k+1}-1)$.

(B) Statement $A$ is true because three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if there exist scalars $x, y$ such that $\vec{c} = x\vec{a} + y\vec{b}$ (assuming $\vec{a}$ and $\vec{b}$ are non-collinear).
Statement $R$ is true because any set of three coplanar vectors in $3D$ space is linearly dependent,as their scalar triple product is zero.
However,$R$ is a general property of coplanar vectors and does not serve as the definition or the direct logical derivation for the specific condition stated in $A$. Thus,$R$ is not the correct explanation of $A$.

(C) $I$: Two vectors $\vec{a}$ and $\vec{b}$ are linearly independent if and only if they are non-zero and non-collinear. Thus,statement $I$ is true.
$II$: Any three coplanar vectors in a $3D$ space are linearly dependent because one can be expressed as a linear combination of the other two if they are not collinear,or they are linearly dependent if any two are collinear. Thus,statement $II$ is true.
$\therefore$ Both $I$ and $II$ are true.

(B) Given that $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$.
Since $a$ and $b$ are collinear,$b = \lambda a$ for some scalar $\lambda$.
We are given $|b| = 21$.
First,calculate the magnitude of $a$:
$|a| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Since $b = \lambda a$,we have $|b| = |\lambda| |a|$.
$21 = |\lambda| \times 7 \implies |\lambda| = 3 \implies \lambda = \pm 3$.
Therefore,$b = \pm 3(2 \hat{i} + 3 \hat{j} + 6 \hat{k})$.

(A) Let the origin be $O(0, 0, 0)$ and the point be $P(2, 3, -1)$.
The direction ratios of the line $OP$ are $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (2 - 0, 3 - 0, -1 - 0) = (2, 3, -1)$.
The distance $OP$ is given by $\sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
The direction cosines $(l, m, n)$ are given by $\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)$.
Thus,the direction cosines are $\left(\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)$.