AP EAMCET 2004 Mathematics Question Paper with Answer and Solution

(C) Given that $d(n)$ represents the number of positive divisors of $n$.
For a prime number $P$,the divisors of $P^7$ are $P^0, P^1, P^2, P^3, P^4, P^5, P^6, P^7$. Thus,$d(P^7) = 8$.
Next,we find $d(8)$. Since $8 = 2^3$,the number of divisors is $3 + 1 = 4$. Thus,$d(8) = 4$.
Finally,we find $d(4)$. Since $4 = 2^2$,the number of divisors is $2 + 1 = 3$. Thus,$d(4) = 3$.
Therefore,$d(d(d(P^7))) = 3$.

(D) Given that,$\log _{27}(\log _3 x) = \frac{1}{3}$.
Using the definition of logarithm,$\log _a b = c \Rightarrow b = a^c$,we get:
$\log _3 x = (27)^{1/3}$.
Since $27 = 3^3$,we have $(27)^{1/3} = (3^3)^{1/3} = 3^1 = 3$.
So,$\log _3 x = 3$.
Again,using the definition of logarithm,$x = 3^3$.
Therefore,$x = 27$.

(C) Given the equation $x^3 - 10x^2 + 7x + 8 = 0$.
From the relation between roots and coefficients:
$\alpha + \beta + \gamma = 10$ ($A$-$5$)
$\alpha\beta + \beta\gamma + \gamma\alpha = 7$
$\alpha\beta\gamma = -8$
Now,$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = (10)^2 - 2(7) = 100 - 14 = 86$ ($B$-$3$)
Next,$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{7}{-8} = -\frac{7}{8}$ ($C$-$2$)
Finally,$\frac{\alpha}{\beta\gamma} + \frac{\beta}{\gamma\alpha} + \frac{\gamma}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + \gamma^2}{\alpha\beta\gamma} = \frac{86}{-8} = -\frac{43}{4}$ ($D$-$1$)
Thus,the correct matching is $A-5, B-3, C-2, D-1$.

(D) Since $(x-2)$ is a common factor of the expressions $x^2+ax+b$ and $x^2+cx+d$,the value of these expressions at $x=2$ must be $0$.
For $x^2+ax+b$: $(2)^2+a(2)+b=0 \Rightarrow 4+2a+b=0 \dots(i)$
For $x^2+cx+d$: $(2)^2+c(2)+d=0 \Rightarrow 4+2c+d=0 \dots(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(4+2a+b) - (4+2c+d) = 0 - 0$
$2a+b-2c-d = 0$
$b-d = 2c-2a$
$b-d = 2(c-a)$
Dividing both sides by $(c-a)$ (assuming $c \neq a$):
$\frac{b-d}{c-a} = 2$

(A) Let $y = \sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$
$\Rightarrow y = \sqrt{42+y}$
On squaring both sides,we get:
$y^2 = 42 + y$
$\Rightarrow y^2 - y - 42 = 0$
Factorizing the quadratic equation:
$(y - 7)(y + 6) = 0$
$\Rightarrow y = 7$ or $y = -6$
Since the square root function must yield a non-negative value,$y = -6$ is rejected.
Therefore,the required solution is $y = 7$.

(A) Given that $f(x)$ is a polynomial with rational coefficients.
If a complex number $a+bi$ is a root,its conjugate $a-bi$ must also be a root. Thus,$1+2i$ and $1-2i$ are roots.
If an irrational number of the form $a+\sqrt{b}$ is a root,its conjugate $a-\sqrt{b}$ must also be a root. Thus,$2-\sqrt{3}$ and $2+\sqrt{3}$ are roots.
Additionally,$5$ is given as a root.
Therefore,the roots are $1+2i, 1-2i, 2-\sqrt{3}, 2+\sqrt{3}$,and $5$.
Counting these,we have $5$ distinct roots.
Hence,the least degree $n$ of the polynomial is $5$.

(C) The period of a function $f(x) = \frac{g(x)}{h(x)}$ is the least common multiple of the periods of $g(x)$ and $h(x)$.
Given $g(x) = \cos(nx)$,its period is $T_1 = \frac{2\pi}{n}$.
Given $h(x) = \sin\left(\frac{x}{n}\right)$,its period is $T_2 = \frac{2\pi}{1/n} = 2n\pi$.
The period of the quotient is the $LCM$ of $T_1$ and $T_2$.
Since $T_2$ is a multiple of $T_1$ (as $2n\pi = n^2 \cdot \frac{2\pi}{n}$),the period of the function is $T_2 = 2n\pi$.
Given $2n\pi = 4\pi$,we get $n = 2$.

(A) Given that $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n} = e^{i \frac{\pi}{2^n}}$.
We need to find the product $P = \prod_{n=1}^{\infty} x_n = \prod_{n=1}^{\infty} e^{i \frac{\pi}{2^n}}$.
Using the property of exponents,$P = e^{i \sum_{n=1}^{\infty} \frac{\pi}{2^n}}$.
The sum in the exponent is a geometric series: $\sum_{n=1}^{\infty} \frac{\pi}{2^n} = \pi \left( \frac{1/2}{1 - 1/2} \right) = \pi \left( \frac{1/2}{1/2} \right) = \pi$.
Therefore,$P = e^{i \pi}$.
Using Euler's formula,$e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1$.

(A) The given series is an infinite geometric series with the first term $a = 1$ and common ratio $r = \frac{2 i}{3}$.
Since $|r| = |\frac{2 i}{3}| = \frac{2}{3} < 1$,the sum $S$ is given by $S = \frac{a}{1-r}$.
$S = \frac{1}{1-\frac{2 i}{3}} = \frac{1}{\frac{3-2 i}{3}} = \frac{3}{3-2 i}$.
To simplify,multiply the numerator and denominator by the conjugate $(3+2 i)$:
$S = \frac{3(3+2 i)}{(3-2 i)(3+2 i)} = \frac{9+6 i}{3^2 - (2 i)^2} = \frac{9+6 i}{9 - (-4)} = \frac{9+6 i}{13}$.

(D) There are $10$ speakers in total. The total number of ways to arrange $10$ speakers is $10!$.
In any arrangement,there are two possibilities for the relative order of $S_1$ and $S_2$: either $S_1$ speaks before $S_2$,or $S_2$ speaks before $S_1$.
Since the condition is that $S_1$ must address only after $S_2$,we only consider the cases where $S_2$ appears before $S_1$.
By symmetry,exactly half of the total arrangements satisfy this condition.
Therefore,the required number of ways is $\frac{10!}{2}$.

(A) We know that the sum of the first $k$ cubes is $\left(\frac{k(k+1)}{2}\right)^2$ and the sum of the first $k$ odd numbers is $k^2$.
Substituting these into the expression:
$\sum_{k=1}^5 \frac{\left(\frac{k(k+1)}{2}\right)^2}{k^2} = \sum_{k=1}^5 \frac{k^2(k+1)^2}{4k^2} = \sum_{k=1}^5 \frac{(k+1)^2}{4}$
Expanding the sum for $k=1$ to $5$:
$= \frac{1}{4} [2^2 + 3^2 + 4^2 + 5^2 + 6^2] = \frac{1}{4} [4 + 9 + 16 + 25 + 36] = \frac{90}{4} = 22.5$

(C) Let $y = x \log _e a$. The given series is $y + \frac{y^3}{3!} + \frac{y^5}{5!} + \dots$
We know that the Taylor series expansion for the hyperbolic sine function is $\sinh(y) = y + \frac{y^3}{3!} + \frac{y^5}{5!} + \dots$
Substituting $y = x \log _e a$ back into the series,we get $\sinh(x \log _e a)$.
Thus,the value of the series is $\sinh(x \log _e a)$.

(C) The given pair of lines is $3x^2-4xy+y^2=0$.
Factoring the equation: $3x^2-3xy-xy+y^2=0$ $\Rightarrow 3x(x-y)-y(x-y)=0$ $\Rightarrow (3x-y)(x-y)=0$.
So,the two lines are $L_1: 3x-y=0$ and $L_2: x-y=0$.
The third line is $L_3: 2x-y=6$.
The vertices of the triangle are the intersection points of these lines:
$1$. Intersection of $L_1$ and $L_2$: $(0,0)$.
$2$. Intersection of $L_1$ and $L_3$: $3x-y=0$ and $2x-y=6$. Subtracting gives $x=-6$,so $y=-18$. Point is $(-6, -18)$.
$3$. Intersection of $L_2$ and $L_3$: $x-y=0$ and $2x-y=6$. Subtracting gives $x=6$,so $y=6$. Point is $(6, 6)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area $= \frac{1}{2} |0(-18-6) + (-6)(6-0) + 6(0-(-18))| = \frac{1}{2} |0 - 36 + 108| = \frac{1}{2} |72| = 36 \text{ sq units}$.

(C) The circle is $x^2+y^2=61$ with center $O(0,0)$ and radius $r=\sqrt{61}$.
Since $PA=PB=10$,$P$ lies on the perpendicular bisector of chord $AB$. The center $O$ also lies on the perpendicular bisector of chord $AB$. Thus,the line $OP$ is perpendicular to the line $l$.
The slope of $OP$ is $m_{OP} = \frac{6-0}{-5-0} = -\frac{6}{5}$.
Since $l \perp OP$,the slope of line $l$ is $m_l = -\frac{1}{m_{OP}} = \frac{5}{6}$.
Let the equation of line $l$ be $5x-6y+k=0$.
The distance from the center $O(0,0)$ to the chord $AB$ is $d = \frac{|k|}{\sqrt{5^2+(-6)^2}} = \frac{|k|}{\sqrt{61}}$.
In $\triangle OMA$ (where $M$ is the midpoint of $AB$),$OA^2 = OM^2 + AM^2$.
$AM^2 = PA^2 - PM^2$. Since $P(-5,6)$,$OP = \sqrt{(-5)^2+6^2} = \sqrt{61}$.
Let $M$ be the projection of $O$ on $l$. $OM = d = \frac{|k|}{\sqrt{61}}$.
$AM^2 = r^2 - OM^2 = 61 - \frac{k^2}{61}$.
Also $PM^2 = OP^2 - OM^2 = 61 - \frac{k^2}{61}$.
Given $PA=10$,$AM^2 = PA^2 - PM^2 = 100 - (61 - \frac{k^2}{61}) = 39 + \frac{k^2}{61}$.
Equating $AM^2$: $61 - \frac{k^2}{61} = 39 + \frac{k^2}{61} \implies 22 = \frac{2k^2}{61} \implies k^2 = 11 \times 61 = 671$.
Re-evaluating: The line $l$ passes through $M$. $M$ is the projection of $O$ on $l$. $PM = \sqrt{PA^2 - AM^2} = \sqrt{100 - (61 - d^2)} = \sqrt{39+d^2}$.
Since $P, M, O$ are collinear,$PM = |PO - OM| = |\sqrt{61} - d|$.
$39+d^2 = 61 - 2d\sqrt{61} + d^2 \implies 2d\sqrt{61} = 22 \implies d = \frac{11}{\sqrt{61}}$.
Since $d = \frac{|k|}{\sqrt{61}}$,$|k|=11$. Thus $k = \pm 11$. Checking the options,$5x-6y+11=0$ matches.

(B) Two points $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate with respect to the circle $x^2+y^2=r^2$ if $x_1x_2 + y_1y_2 = r^2$.
Given the circle $x^2+y^2=25$,we have $r^2=25$.
Substituting the points $(1, a)$ and $(b, 2)$ into the condition:
$(1)(b) + (a)(2) = 25$
$b + 2a = 25$
We need to find the value of $4a + 2b$.
Multiplying the equation $b + 2a = 25$ by $2$,we get:
$2(b + 2a) = 2(25)$
$2b + 4a = 50$
Therefore,$4a + 2b = 50$.

(B) We have,\\ $(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ \\ $= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$ \\ $= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$ \\ $= (a^2 + b^2)(1) - 2ab \cos C$ \\ $= a^2 + b^2 - 2ab \cos C$ \\ Since by the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$,\\ Therefore,the expression is equal to $c^2$.

(C) Statement $I$: We know that $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$. This is a standard identity in a triangle. Thus,$I$ is true.
Statement $II$: We know that $r_1 = s \tan \frac{A}{2}$. The given formula $r_1 = (s-a) \tan \frac{A}{2}$ is incorrect. Thus,$II$ is false.
Statement $III$: We know that $r_3 = \frac{\Delta}{s-c}$. This is a standard formula for the exradius $r_3$. Thus,$III$ is true.
Therefore,statements $I$ and $III$ are correct.

(C) Given that,$\frac{x+1}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)}$
Multiplying both sides by $(2x-1)(3x+1)$,we get:
$(x+1) = A(3x+1) + B(2x-1)$
$(x+1) = x(3A+2B) + (A-B)$
Comparing the coefficients of $x$ and the constant terms on both sides,we get:
$3A + 2B = 1$ ... $(i)$
$A - B = 1$ ... $(ii)$
From equation $(ii)$,$A = B + 1$. Substituting this into equation $(i)$:
$3(B+1) + 2B = 1$
$3B + 3 + 2B = 1$
$5B = -2 \Rightarrow B = -\frac{2}{5}$
Now,$A = -\frac{2}{5} + 1 = \frac{3}{5}$
We need to find the value of $16A + 9B$:
$16A + 9B = 16\left(\frac{3}{5}\right) + 9\left(-\frac{2}{5}\right)$
$= \frac{48}{5} - \frac{18}{5} = \frac{30}{5} = 6$

(A) Let the roots of the equation $4x^3 - 12x^2 + 11x + k = 0$ be $\alpha - d, \alpha, \alpha + d$.
Sum of the roots $= (\alpha - d) + \alpha + (\alpha + d) = 3\alpha$.
From the equation,the sum of the roots is $-\frac{-12}{4} = 3$.
So,$3\alpha = 3 \Rightarrow \alpha = 1$.
Since $\alpha = 1$ is a root,it must satisfy the equation $4(1)^3 - 12(1)^2 + 11(1) + k = 0$.
$4 - 12 + 11 + k = 0$.
$3 + k = 0 \Rightarrow k = -3$.

(C) The period of $\cos(nx)$ is $T_1 = \frac{2\pi}{n}$.
The period of $\sin(x/n)$ is $T_2 = \frac{2\pi}{1/n} = 2n\pi$.
The period of the quotient $\frac{\cos(nx)}{\sin(x/n)}$ is the least common multiple of the periods of the numerator and the denominator.
Given that the period is $4\pi$,we have $2n\pi = 4\pi$.
Solving for $n$,we get $n = 2$.

(A) Given that $\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}\right|=1$ and $\left|z_1\right| \neq 3$.
Squaring both sides,we get $\left|z_1-3 z_2\right|^2 = \left|3-z_1 \bar{z}_2\right|^2$.
Using the property $|z|^2 = z \bar{z}$,we have $(z_1-3 z_2)(\bar{z}_1-3 \bar{z}_2) = (3-z_1 \bar{z}_2)(3-\bar{z}_1 z_2)$.
Expanding both sides: $|z_1|^2 - 3z_1 \bar{z}_2 - 3z_2 \bar{z}_1 + 9|z_2|^2 = 9 - 3\bar{z}_1 z_2 - 3z_1 \bar{z}_2 + |z_1|^2 |z_2|^2$.
Canceling common terms $-3z_1 \bar{z}_2$ and $-3z_2 \bar{z}_1$ from both sides: $|z_1|^2 + 9|z_2|^2 = 9 + |z_1|^2 |z_2|^2$.
Rearranging the terms: $|z_1|^2 - 9 - |z_1|^2 |z_2|^2 + 9|z_2|^2 = 0$.
Factorizing: $(|z_1|^2 - 9) - |z_2|^2 (|z_1|^2 - 9) = 0$.
$(|z_1|^2 - 9)(1 - |z_2|^2) = 0$.
Since $|z_1| \neq 3$,we must have $|z_2|^2 = 1$,which implies $|z_2| = 1$.

(A) First,find the prime factorization of $216$:
$216 = 2^3 \times 3^3$.
An odd divisor must not contain any factor of $2$.
Therefore,the odd divisors are formed only by the powers of $3$.
The factors of $3^3$ are $3^0, 3^1, 3^2, 3^3$.
The number of such divisors is the exponent of $3$ plus $1$,which is $3 + 1 = 4$.
The odd divisors are $1, 3, 9, 27$.

(B) We know that the expansion of $e^{-x}$ is given by $e^{-x} = \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \dots + \frac{(-1)^n x^n}{n!} + \dots$
Now,consider the expression $(2+3x)e^{-x}$:
$(2+3x)e^{-x} = 2e^{-x} + 3xe^{-x}$
The term containing $x^{10}$ in $2e^{-x}$ is $2 \times \frac{(-x)^{10}}{10!} = \frac{2}{10!} x^{10}$.
The term containing $x^{10}$ in $3xe^{-x}$ is $3x \times \frac{(-x)^9}{9!} = 3x \times \frac{-x^9}{9!} = -\frac{3}{9!} x^{10}$.
To combine these,we write the second term with a denominator of $10!$:
$-\frac{3}{9!} = -\frac{3 \times 10}{10!} = -\frac{30}{10!}$.
Thus,the coefficient of $x^{10}$ is $\frac{2}{10!} - \frac{30}{10!} = \frac{2-30}{10!} = \frac{-28}{10!}$.

(A) First,express the fraction using partial fractions: $\frac{x-4}{(x-2)(x-3)} = \frac{2}{x-2} - \frac{1}{x-3}$.
Rewrite the terms to facilitate binomial expansion: $2(x-2)^{-1} - (x-3)^{-1} = 2(-2)^{-1}(1-\frac{x}{2})^{-1} - (-3)^{-1}(1-\frac{x}{3})^{-1}$.
This simplifies to: $-(1-\frac{x}{2})^{-1} + \frac{1}{3}(1-\frac{x}{3})^{-1}$.
Using the expansion $(1-y)^{-1} = 1 + y + y^2 + y^3 + \dots$,we get:
$-(1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots) + \frac{1}{3}(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} + \dots)$.
The coefficient of $x^3$ is $-\frac{1}{8} + \frac{1}{3} \times \frac{1}{27} = -\frac{1}{8} + \frac{1}{81}$.
Calculating the sum: $\frac{-81 + 8}{648} = -\frac{73}{648}$.

(D) The binomial coefficients for $n=15$ are ${ }^{15} C_0, { }^{15} C_1, \dots, { }^{15} C_7, { }^{15} C_8, \dots, { }^{15} C_{15}$.
Since ${ }^{15} C_r$ increases as $r$ increases from $0$ to $7$ and decreases as $r$ increases from $8$ to $15$,we observe the values.
For option $D$,the sequence is ${ }^{15} C_7, { }^{15} C_6, { }^{15} C_5$. Since $7 > 6 > 5$ and these values are in the decreasing part of the binomial coefficient distribution (where $r > n/2$),the sequence is in decreasing order.
Therefore,option $D$ is correct.

(A) We know that for any $\theta \in \mathbb{R}$,the range of $\cos \theta$ is $[-1, 1]$.
Thus,$-1 \leq \cos (4x - 5) \leq 1$.
Multiplying the inequality by $3$,we get:
$-3 \leq 3 \cos (4x - 5) \leq 3$.
Adding $4$ to all parts of the inequality,we get:
$-3 + 4 \leq 3 \cos (4x - 5) + 4 \leq 3 + 4$.
$1 \leq 3 \cos (4x - 5) + 4 \leq 7$.
Therefore,the expression lies in the interval $[1, 7]$.

(D) We have,$\cos 12^{\circ} + \cos 84^{\circ} + \cos 132^{\circ} + \cos 156^{\circ}$.
Using the sum-to-product formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$= (\cos 132^{\circ} + \cos 12^{\circ}) + (\cos 156^{\circ} + \cos 84^{\circ})$
$= 2 \cos \frac{132^{\circ}+12^{\circ}}{2} \cos \frac{132^{\circ}-12^{\circ}}{2} + 2 \cos \frac{156^{\circ}+84^{\circ}}{2} \cos \frac{156^{\circ}-84^{\circ}}{2}$
$= 2 \cos 72^{\circ} \cos 60^{\circ} + 2 \cos 120^{\circ} \cos 36^{\circ}$
$= 2 \left( \frac{\sqrt{5}-1}{4} \right) \left( \frac{1}{2} \right) + 2 \left( -\frac{1}{2} \right) \left( \frac{\sqrt{5}+1}{4} \right)$
$= \frac{\sqrt{5}-1}{4} - \frac{\sqrt{5}+1}{4}$
$= \frac{\sqrt{5}-1-\sqrt{5}-1}{4} = -\frac{2}{4} = -\frac{1}{2}$.

(A) We know that $\tan 63^{\circ} = \cot 27^{\circ}$ and $\tan 81^{\circ} = \cot 9^{\circ}$.
Substituting these into the expression:
$\tan 9^{\circ} - \tan 27^{\circ} - \cot 27^{\circ} + \cot 9^{\circ} = (\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$
Using the identity $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$:
$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$
Given $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$:
$= \frac{2}{(\sqrt{5}-1)/4} - \frac{2}{(\sqrt{5}+1)/4} = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1}$
$= 8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} \right) = 8 \left( \frac{2}{5-1} \right) = 8 \left( \frac{2}{4} \right) = 4$.

(A) Let $M$ be the mid-point of $AB$. Since $PA = PB$,the triangle $PAB$ is an isosceles triangle,and the line segment $PM$ is the perpendicular bisector of $AB$.
Given the line $L: 2x - y + 3 = 0$,its slope is $m_L = 2$.
Since $PM \perp AB$,the slope of $PM$ is $m_{PM} = -\frac{1}{m_L} = -\frac{1}{2}$.
The line $PM$ passes through $P(1, 2)$ and has a slope of $-\frac{1}{2}$. Its equation is:
$y - 2 = -\frac{1}{2}(x - 1)$
$2y - 4 = -x + 1$
$x + 2y = 5$ (Equation $ii$)
The mid-point $M$ is the intersection of the line $2x - y = -3$ (Equation $i$) and $x + 2y = 5$ (Equation $ii$).
From $(i)$,$y = 2x + 3$. Substituting into (ii):
$x + 2(2x + 3) = 5$
$x + 4x + 6 = 5$
$5x = -1 \implies x = -\frac{1}{5}$
$y = 2(-\frac{1}{5}) + 3 = -\frac{2}{5} + \frac{15}{5} = \frac{13}{5}$
Thus,the mid-point $M$ is $\left(-\frac{1}{5}, \frac{13}{5}\right)$.

(A) Let the points be $A = (a \cos \theta, a \sin \theta)$ and $B = (a \cos \phi, a \sin \phi)$.
Given the distance $AB = 2a$.
Using the distance formula:
$AB^2 = (a \cos \theta - a \cos \phi)^2 + (a \sin \theta - a \sin \phi)^2 = (2a)^2$
$a^2(\cos^2 \theta + \cos^2 \phi - 2 \cos \theta \cos \phi + \sin^2 \theta + \sin^2 \phi - 2 \sin \theta \sin \phi) = 4a^2$
$a^2(2 - 2(\cos \theta \cos \phi + \sin \theta \sin \phi)) = 4a^2$
$2 - 2 \cos(\theta - \phi) = 4$
$-2 \cos(\theta - \phi) = 2$
$\cos(\theta - \phi) = -1$
Since $\cos(\theta - \phi) = -1$,we have $\theta - \phi = (2n + 1)\pi = 2n\pi + \pi$ for $n \in Z$.
Therefore,$\theta = 2n\pi + \pi + \phi$.

(C) The given pair of lines is $3x^2 - 4xy + y^2 = 0$.
Factoring the quadratic expression: $3x^2 - 3xy - xy + y^2 = 0$ $\Rightarrow 3x(x - y) - y(x - y) = 0$ $\Rightarrow (3x - y)(x - y) = 0$.
Thus,the two lines are $L_1: 3x - y = 0$ and $L_2: x - y = 0$.
The third line is $L_3: 2x - y = 6$.
To find the vertices of the triangle,we find the intersection points:
$1$. Intersection of $L_1$ and $L_2$: $(0, 0)$.
$2$. Intersection of $L_1$ and $L_3$: $3x - y = 0$ and $2x - y = 6$. Subtracting gives $x = -6$,so $y = -18$. Point: $(-6, -18)$.
$3$. Intersection of $L_2$ and $L_3$: $x - y = 0$ and $2x - y = 6$. Subtracting gives $x = 6$,so $y = 6$. Point: $(6, 6)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Area $= \frac{1}{2} |0(-18 - 6) + (-6)(6 - 0) + 6(0 - (-18))| = \frac{1}{2} |0 - 36 + 108| = \frac{1}{2} |72| = 36 \text{ sq units}$.

(B) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$,with radius $r = \sqrt{g^2+f^2-c}$.
For the first circle $x^2+y^2+8x-6y=0$,$g=4, f=-3, c=0$. Thus,$r_1 = \sqrt{4^2+(-3)^2-0} = \sqrt{16+9} = 5$. Perimeter $P_1 = 2\pi r_1 = 10\pi$.
For the second circle $4x^2+4y^2-4x-12y-186=0$,divide by $4$: $x^2+y^2-x-3y-46.5=0$. Here $g=-0.5, f=-1.5, c=-46.5$. Thus,$r_2 = \sqrt{(-0.5)^2+(-1.5)^2-(-46.5)} = \sqrt{0.25+2.25+46.5} = \sqrt{49} = 7$. Perimeter $P_2 = 2\pi r_2 = 14\pi$.
For the third circle $x^2+y^2-6x+6y-9=0$,$g=-3, f=3, c=-9$. Thus,$r_3 = \sqrt{(-3)^2+3^2-(-9)} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3} \approx 3 \times 1.732 = 5.196$. Perimeter $P_3 = 2\pi r_3 = 6\sqrt{3}\pi \approx 10.39\pi$.
Comparing the perimeters: $10\pi < 10.39\pi < 14\pi$,which implies $P_1 < P_3 < P_2$.

(A) Given equation of the circle is $r = \sqrt{3} \sin \theta + \cos \theta$.
Multiplying both sides by $r$,we get $r^2 = \sqrt{3} (r \sin \theta) + (r \cos \theta)$.
Using the polar to Cartesian conversions $x = r \cos \theta$ and $y = r \sin \theta$,and $r^2 = x^2 + y^2$,the equation becomes:
$x^2 + y^2 = \sqrt{3} y + x$.
Rearranging the terms,we get $x^2 - x + y^2 - \sqrt{3} y = 0$.
Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -1 \Rightarrow g = -\frac{1}{2}$ and $2f = -\sqrt{3} \Rightarrow f = -\frac{\sqrt{3}}{2}$,with $c = 0$.
The radius is given by $\sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2 - 0}$.
Radius $= \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$.

(C) Given the polar equation of the circle is $r^2-4r(\cos \theta+\sin \theta)-4=0$.
We know that $x=r \cos \theta$,$y=r \sin \theta$,and $r^2=x^2+y^2$.
Substituting these into the given equation:
$x^2+y^2-4(x+y)-4=0$
$x^2+y^2-4x-4y-4=0$
Comparing this with the general equation of a circle $x^2+y^2+2gx+2fy+c=0$,we get $2g=-4 \Rightarrow g=-2$ and $2f=-4 \Rightarrow f=-2$.
The centre of the circle is $(-g, -f) = (2, 2)$.

(B) The line equation is $3x - 2y + 6 = 0$.
To find the $X$-intercept $(A)$,set $y = 0$: $3x + 6 = 0 \Rightarrow x = -2$. So,$A = (-2, 0)$.
To find the $Y$-intercept $(B)$,set $x = 0$: $-2y + 6 = 0 \Rightarrow y = 3$. So,$B = (0, 3)$.
The radius $r = AB = \sqrt{(0 - (-2))^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.
The equation of a circle with center $(h, k) = (-2, 0)$ and radius $r = \sqrt{13}$ is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - (-2))^2 + (y - 0)^2 = (\sqrt{13})^2$
$(x + 2)^2 + y^2 = 13$
$x^2 + 4x + 4 + y^2 = 13$
$x^2 + y^2 + 4x - 9 = 0$.

(C) The circle is $x^2+y^2=61$,with center $O(0,0)$ and radius $r=\sqrt{61}$.
Since $PA=PB=10$,$P$ lies on the perpendicular bisector of chord $AB$. The center $O$ also lies on the perpendicular bisector of $AB$. Thus,$OP$ is perpendicular to $AB$.
The slope of $OP$ is $m_{OP} = \frac{6-0}{-5-0} = -\frac{6}{5}$.
Since $AB \perp OP$,the slope of line $l$ (which is $AB$) is $m_l = -\frac{1}{m_{OP}} = \frac{5}{6}$.
The equation of line $l$ is $y - y_1 = m(x - x_1)$. However,we can use the form $5x - 6y + k = 0$.
The distance from the center $O(0,0)$ to the chord $AB$ is $d = \frac{|k|}{\sqrt{5^2+(-6)^2}} = \frac{|k|}{\sqrt{61}}$.
In $\triangle OMA$ (where $M$ is the midpoint of $AB$),$OA^2 = OM^2 + AM^2$. Here $OA = \sqrt{61}$ and $AM = \sqrt{PA^2 - PM^2}$.
First,find $PM$. $P$ is $(-5, 6)$,$O$ is $(0,0)$,so $OP = \sqrt{(-5)^2+6^2} = \sqrt{61}$.
Since $PA=10$ and $AM^2 = PA^2 - PM^2$,we need $PM$. $M$ is the projection of $O$ on $l$. $OM = d = \frac{|k|}{\sqrt{61}}$.
In $\triangle OMP$,$\angle OMP = 90^\circ$,so $PM^2 = OP^2 - OM^2 = 61 - d^2$.
$AM^2 = PA^2 - PM^2 = 100 - (61 - d^2) = 39 + d^2$.
Also $AM^2 = OA^2 - OM^2 = 61 - d^2$.
Equating: $39 + d^2 = 61 - d^2$ $\Rightarrow 2d^2 = 22$ $\Rightarrow d^2 = 11$.
$d = \frac{|k|}{\sqrt{61}} = \sqrt{11} \Rightarrow |k| = \sqrt{61 \times 11} = \sqrt{671}$. This does not match options.
Re-evaluating: The line $l$ passes through $A$ and $B$. The power of point $P$ is $PA \cdot PB = 100$. Also $P$ is $(-5,6)$,$x^2+y^2-61=0$. Power is $(-5)^2+6^2-61 = 25+36-61 = 0$. This means $P$ is on the circle. If $P$ is on the circle,$PA=PB=10$ implies $AB$ is a chord. The line $l$ is the chord of contact if $P$ were outside,but here $P$ is on the circle. The line $l$ must be $5x-6y+k=0$. Testing $5x-6y+11=0$,distance from $(0,0)$ is $11/\sqrt{61}$. $AM^2 = 61 - 121/61 = (3721-121)/61 = 3600/61$. $AM = 60/\sqrt{61}$. $PA^2 = PM^2 + AM^2$. $PM$ is distance from $(-5,6)$ to line $5x-6y+11=0$,$PM = |5(-5)-6(6)+11|/\sqrt{61} = |-25-36+11|/\sqrt{61} = 50/\sqrt{61}$. $PA^2 = 2500/61 + 3600/61 = 6100/61 = 100$. Thus $PA=10$. Correct option is $(c)$.

(C) The given lines are $L_1: x+y-1=0$,$L_2: x-y-1=0$,and $L_3: y+1=0$.
These three lines are non-parallel and do not intersect at a single point,so they form a triangle.
For any triangle,there exists exactly one incircle that touches all three sides internally.
Additionally,there are three excircles,each of which touches one side of the triangle externally and the extensions of the other two sides.
Therefore,the total number of circles that touch all three lines is $1 + 3 = 4$.

(A) The given equation of the curve is $2y^2 = x + 1$.
Differentiating both sides with respect to $x$,we get:
$4y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{4y}$.
The slope of the tangent at any point $(x, y)$ is $m_t = \frac{1}{4y}$.
The slope of the normal at any point $(x, y)$ is $m_n = -\frac{1}{m_t} = -4y$.
Now,we calculate the slope of the normal at each given point:
$A. (7, 2): m_n = -4(2) = -8$ (Matches $2$).
$B. (0, 1/\sqrt{2}): m_n = -4(1/\sqrt{2}) = -2\sqrt{2}$ (Matches $5$).
$C. (1, -1): m_n = -4(-1) = 4$ (Matches $3$).
$D. (3, \sqrt{2}): m_n = -4(\sqrt{2}) = -4\sqrt{2}$ (Matches $1$).
Thus,the correct matching is $A-2, B-5, C-3, D-1$.

(C) The given polar equation is $\cos \theta + 7 \sin \theta = \frac{1}{r}$.
Multiplying both sides by $r$,we get $r \cos \theta + 7 r \sin \theta = 1$.
Using the standard conversion formulas $x = r \cos \theta$ and $y = r \sin \theta$,the equation becomes $x + 7y = 1$.
This is a linear equation in $x$ and $y$,which represents a straight line.

(A) The given equation is $36x^2 + 144y^2 - 36x - 96y - 119 = 0$.
Rearranging the terms,we get $36(x^2 - x) + 144(y^2 - \frac{2}{3}y) = 119$.
Completing the square,we have $36(x^2 - x + \frac{1}{4}) + 144(y^2 - \frac{2}{3}y + \frac{1}{9}) = 119 + 9 + 16$.
This simplifies to $36(x - \frac{1}{2})^2 + 144(y - \frac{1}{3})^2 = 144$.
Dividing by $144$,we get $\frac{(x - 1/2)^2}{4} + \frac{(y - 1/3)^2}{1} = 1$.
This is an ellipse with $a^2 = 4$ and $b^2 = 1$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

(B) We have,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$
$= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$
$= (a^2 + b^2)(\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab(\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$
$= (a^2 + b^2)(1) - 2ab \cos C$
Since $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,we have $a^2 + b^2 - 2ab \cos C = c^2$
Therefore,the expression is equal to $c^2$.

(B) In $\triangle ABC$,we know that $A+B+C = \pi$.
Consider the expression:
$\cos \left(\frac{B+2C+3A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$
Rewrite the first term using $A+B+C = \pi$:
$\frac{B+2C+3A}{2} = \frac{(A+B+C) + C + 2A}{2} = \frac{\pi + C + 2A}{2} = \frac{\pi}{2} + \frac{2A+C}{2}$
Alternatively,rewrite it as:
$\frac{B+2C+3A}{2} = \frac{2(A+B+C) + A-B}{2} = \frac{2\pi + (A-B)}{2} = \pi + \frac{A-B}{2}$
Substituting this back into the expression:
$\cos \left(\pi + \frac{A-B}{2}\right) + \cos \left(\frac{A-B}{2}\right)$
Using the identity $\cos(\pi + \theta) = -\cos(\theta)$:
$-\cos \left(\frac{A-B}{2}\right) + \cos \left(\frac{A-B}{2}\right) = 0$

(D) $I$. We know that $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ is a standard identity for the inradius of a triangle.
$II$. We know that $r_1 = (s-a) \tan \frac{A}{2}$ is a standard identity for the exradius $r_1$ opposite to vertex $A$.
$III$. We know that $r_3 = \frac{\Delta}{s-c}$ is a standard identity for the exradius $r_3$ opposite to vertex $C$.
Since all three statements are standard identities in the properties of triangles,all are correct.

(C) We know that the exradii and inradius are given by:
$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$
$r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$
Given $r_3 = r_1 + r_2 + r$,we have:
$r_3 - r = r_1 + r_2$
$4R \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4R \cos \frac{C}{2} \sin \frac{A}{2} \cos \frac{B}{2} + 4R \cos \frac{C}{2} \cos \frac{A}{2} \sin \frac{B}{2}$
Dividing by $4R$:
$\sin \frac{C}{2} (\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2}) = \cos \frac{C}{2} (\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2})$
Using the identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$ and $\sin(x+y) = \sin x \cos y + \cos x \sin y$:
$\sin \frac{C}{2} \cos(\frac{A+B}{2}) = \cos \frac{C}{2} \sin(\frac{A+B}{2})$
Since $A+B+C = \pi$,we have $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$:
$\sin \frac{C}{2} \cos(\frac{\pi}{2} - \frac{C}{2}) = \cos \frac{C}{2} \sin(\frac{\pi}{2} - \frac{C}{2})$
$\sin \frac{C}{2} \sin \frac{C}{2} = \cos \frac{C}{2} \cos \frac{C}{2}$
$\sin^2 \frac{C}{2} = \cos^2 \frac{C}{2}$
$\tan^2 \frac{C}{2} = 1 \Rightarrow \tan \frac{C}{2} = 1$ (as $C$ is an angle of a triangle,$\frac{C}{2} > 0$)
$\frac{C}{2} = 45^{\circ} \Rightarrow C = 90^{\circ}$
Since $A+B+C = 180^{\circ}$,$A+B = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

(C) Given inequation is $x^2 - 2x + 5 \leq 0$.
We can rewrite the expression by completing the square:
$x^2 - 2x + 1 + 4 \leq 0$
$(x - 1)^2 + 4 \leq 0$
Since $(x - 1)^2 \geq 0$ for all real $x$,it follows that $(x - 1)^2 + 4 \geq 4$.
Therefore,the expression $(x - 1)^2 + 4$ is always positive and can never be less than or equal to $0$.
Thus,there are no real values of $x$ that satisfy the given inequation.
The set of all solutions is the empty set,denoted by $\phi$.

(C) Let the position of the observer be $A$. The aeroplane is at height $h = 1 \ km$. Let the initial position be $D$ and the final position be $E$.
In $\Delta DAP$,$\tan 60^{\circ} = \frac{DP}{AP}$ $\Rightarrow \sqrt{3} = \frac{1}{AP}$ $\Rightarrow AP = \frac{1}{\sqrt{3}} \ km$.
In $\Delta EAQ$,$\tan 30^{\circ} = \frac{EQ}{AQ} = \frac{1}{AP + PQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{1}{\frac{1}{\sqrt{3}} + PQ}$.
Solving for $PQ$: $\frac{1}{\sqrt{3}} + PQ = \sqrt{3}$ $\Rightarrow PQ = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ km$.
The distance covered by the plane is $PQ = \frac{2}{\sqrt{3}} \ km$ in $10 \ s$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{2/\sqrt{3} \ km}{10 \ s} = \frac{2}{\sqrt{3}} \times \frac{1}{10} \ km/s$.
Converting to $km/h$: $\text{Speed} = \frac{2}{10\sqrt{3}} \times 3600 \ km/h = \frac{7200}{10\sqrt{3}} = \frac{720}{\sqrt{3}} = 240\sqrt{3} \ km/h$.

(B) Given that,$x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$.
This implies $e^x = \cot \left( \frac{\pi}{4} + \theta \right)$ and $e^{-x} = \tan \left( \frac{\pi}{4} + \theta \right)$.
We know that $\sinh x = \frac{e^x - e^{-x}}{2}$.
Substituting the values,we get:
$\sinh x = \frac{\cot \left( \frac{\pi}{4} + \theta \right) - \tan \left( \frac{\pi}{4} + \theta \right)}{2}$
Using the identity $\cot A - \tan A = \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} = \frac{\cos^2 A - \sin^2 A}{\sin A \cos A} = \frac{\cos 2A}{\frac{1}{2} \sin 2A} = 2 \cot 2A$,we have:
$\sinh x = \frac{1}{2} \left[ 2 \cot \left( 2 \left( \frac{\pi}{4} + \theta \right) \right) \right]$
$\sinh x = \cot \left( \frac{\pi}{2} + 2\theta \right)$
Since $\cot \left( \frac{\pi}{2} + \alpha \right) = -\tan \alpha$,we get:
$\sinh x = -\tan 2\theta$.

(A) The faces of the die are $\{2, 3, 5, 7, 11, 13\}$.
There is $1$ even number $(2)$ and $5$ odd numbers $(3, 5, 7, 11, 13)$.
The sum of two numbers is odd if one number is even and the other is odd.
Let $E$ be the event of getting an even number and $O$ be the event of getting an odd number.
$P(E) = \frac{1}{6}$ and $P(O) = \frac{5}{6}$.
The sum is odd in two cases: (First die is even,Second die is odd) or (First die is odd,Second die is even).
Required Probability $= P(E) \times P(O) + P(O) \times P(E)$
$= \left(\frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6}\right)$
$= \frac{5}{36} + \frac{5}{36} = \frac{10}{36} = \frac{5}{18}$.

(C) The given expression is $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$.
We can rewrite this as $x \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n (\frac{k}{n})^2$.
Using the definition of the definite integral as the limit of a Riemann sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n}) = \int_0^1 f(t) dt$.
Here,$f(t) = t^2$,so the expression becomes $x \int_0^1 t^2 dt$.
Evaluating the integral: $x [\frac{t^3}{3}]_0^1 = x (\frac{1}{3} - 0) = \frac{x}{3}$.

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively.
Since $G$ is the centroid of $\triangle ABC$,its position vector $\vec{g}$ is given by $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.
This implies $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$.
Now,the sum of the vectors $\vec{GA} + \vec{GB} + \vec{GC}$ is:
$\vec{GA} + \vec{GB} + \vec{GC} = (\vec{a} - \vec{g}) + (\vec{b} - \vec{g}) + (\vec{c} - \vec{g})$
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{g}$
$= 3\vec{g} - 3\vec{g} = \vec{0}$.

(D) Let $A = \left[\begin{array}{ccc} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{array}\right]$.
To find the rank of the matrix,we calculate its determinant $|A|$.
$|A| = 1(1 - (-1)) - (-1)(1 - 1) + 1(1 - (-1))$
$|A| = 1(2) + 1(0) + 1(2)$
$|A| = 2 + 0 + 2 = 4$.
Since $|A| = 4 \neq 0$,the matrix is non-singular.
Therefore,the rank of the $3 \times 3$ matrix $A$ is $3$.

(A) We analyze the function $f(x)$ in three intervals:
$1$. For $-3 < x \leq -1$,$f(x) = [x]$. Since $x \leq -1$,$[x] \leq -1$,so $f(x) < 0$.
$2$. For $-1 < x < 1$,$f(x) = |x|$. Since the absolute value is always non-negative,$f(x) \geq 0$ for all $x \in (-1, 1)$.
$3$. For $1 \leq x < 3$,$f(x) = |[x]|$. Since the absolute value of any integer is $\geq 0$,$f(x) \geq 0$ for all $x \in [1, 3)$.
Combining these intervals where $f(x) \geq 0$,we get $(-1, 1) \cup [1, 3) = (-1, 3)$.
Thus,the set is $(-1, 3)$.

(C) Given the function $f(x, y) = 2(x-y)^2 - x^4 - y^4$.
First,find the partial derivatives with respect to $x$ and $y$:
$f_x = 4(x-y) - 4x^3$
$f_y = -4(x-y) - 4y^3$
Now,find the second-order partial derivatives:
$f_{xx} = \frac{\partial}{\partial x}(4x - 4y - 4x^3) = 4 - 12x^2$
$f_{yy} = \frac{\partial}{\partial y}(-4x + 4y - 4y^3) = 4 - 12y^2$
$f_{xy} = \frac{\partial}{\partial y}(4x - 4y - 4x^3) = -4$
Evaluating these at the point $(0,0)$:
$(f_{xx})_{(0,0)} = 4 - 12(0)^2 = 4$
$(f_{yy})_{(0,0)} = 4 - 12(0)^2 = 4$
$(f_{xy})_{(0,0)} = -4$
Finally,calculate the value of the expression $(f_{xx}f_{yy} - f_{xy}^2)$ at $(0,0)$:
$(f_{xx}f_{yy} - f_{xy}^2)_{(0,0)} = (4)(4) - (-4)^2 = 16 - 16 = 0$.

(C) Given $y = \sin^{-1} x$.
On differentiating both sides with respect to $x$,we get:
$\frac{d y}{d x} = \frac{1}{\sqrt{1-x^2}}$ ... $(i)$
Squaring both sides,we get:
$\left(\frac{d y}{d x}\right)^2 = \frac{1}{1-x^2}$
$(1-x^2) \left(\frac{d y}{d x}\right)^2 = 1$
Differentiating both sides with respect to $x$ using the product rule:
$(1-x^2) \cdot 2 \left(\frac{d y}{d x}\right) \cdot \frac{d^2 y}{d x^2} + \left(\frac{d y}{d x}\right)^2 \cdot (-2x) = 0$
Dividing by $2 \frac{d y}{d x}$ (assuming $\frac{d y}{d x} \neq 0$):
$(1-x^2) \frac{d^2 y}{d x^2} - x \frac{d y}{d x} = 0$
Therefore,$(1-x^2) \frac{d^2 y}{d x^2} = x \frac{d y}{d x}$.

(C) Let $I = \int \frac{d x}{(x+100) \sqrt{x+99}}$.
We can rewrite the denominator as $(x+99) + 1 = (\sqrt{x+99})^2 + 1$.
So,$I = \int \frac{d x}{((\sqrt{x+99})^2 + 1) \sqrt{x+99}}$.
Let $t = \sqrt{x+99}$. Then $t^2 = x+99$,which implies $2t \, dt = dx$.
Substituting these into the integral:
$I = \int \frac{2t \, dt}{(t^2 + 1)t} = \int \frac{2 \, dt}{t^2 + 1}$.
Using the standard integral formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$,we get:
$I = 2 \tan^{-1}(t) + c$.
Substituting back $t = \sqrt{x+99}$:
$I = 2 \tan^{-1}(\sqrt{x+99}) + c$.
Comparing this with $f(x) + c$,we get $f(x) = 2 \tan^{-1}(\sqrt{x+99})$.

(C) Let $I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sqrt{\cot x}}{\frac{\sin x \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\tan x} \sec^2 x d x$.
Since $\frac{1}{\tan x} = \cot x$,we have $I = \int \sqrt{\cot x} \cdot \cot x \cdot \sec^2 x d x = \int (\cot x)^{3/2} \sec^2 x d x$.
Alternatively,rewrite the denominator:
$I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = \int \frac{\sqrt{\cot x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\cot x \cdot \cos^2 x} d x = \int \frac{\sec^2 x}{\sqrt{\cot x}} d x$.
Let $u = \cot x$,then $du = -\csc^2 x d x$,so $d x = -\frac{du}{\csc^2 x} = -\sin^2 x du$.
$I = \int \frac{1}{\sqrt{u}} \cdot \frac{1}{\sin x \cos x} \cdot (-\sin^2 x du) = -\int \frac{\sin x}{\cos x \sqrt{u}} du = -\int \frac{1}{u \sqrt{u}} du = -\int u^{-3/2} du$.
$I = -\left( \frac{u^{-1/2}}{-1/2} \right) + c = 2 \sqrt{u} + c = 2 \sqrt{\cot x} + c$.
Given $\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = -f(x) + c$,we have $-f(x) = 2 \sqrt{\cot x}$,so $f(x) = -2 \sqrt{\cot x}$.

(A) Given the integral $I = \int \frac{3-x^2}{1-2 x+x^2} e^x d x$.
Since $1-2x+x^2 = (1-x)^2$,we can rewrite the integrand as:
$I = \int \frac{3-x^2}{(1-x)^2} e^x d x$
We can manipulate the numerator to match the form $f(x) + f'(x)$:
$I = \int \frac{2 + 1 - x^2}{(1-x)^2} e^x d x = \int \left( \frac{2}{(1-x)^2} + \frac{1-x^2}{(1-x)^2} \right) e^x d x$
$I = \int \left( \frac{2}{(1-x)^2} + \frac{(1-x)(1+x)}{(1-x)^2} \right) e^x d x$
$I = \int \left( \frac{2}{(1-x)^2} + \frac{1+x}{1-x} \right) e^x d x$
Let $f(x) = \frac{1+x}{1-x}$. Then $f'(x) = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{1-x+1+x}{(1-x)^2} = \frac{2}{(1-x)^2}$.
Since $\int (f(x) + f'(x)) e^x d x = e^x f(x) + c$,we have:
$I = e^x \left( \frac{1+x}{1-x} \right) + c$.
Comparing this with the given expression $e^x f(x) + c$,we get $f(x) = \frac{1+x}{1-x}$.

(A) Let $I = \int_0^2 \frac{2x-2}{2x-x^2} dx$.
Note that the integrand $f(x) = \frac{2x-2}{2x-x^2} = \frac{-(2-2x)}{x(2-x)} = \frac{2(x-1)}{x(2-x)}$.
The integral is improper because the denominator $2x-x^2 = x(2-x)$ becomes $0$ at $x=0$ and $x=2$.
Let us evaluate the indefinite integral: $\int \frac{2x-2}{2x-x^2} dx = \int \frac{-(2-2x)}{2x-x^2} dx$.
Let $u = 2x-x^2$,then $du = (2-2x) dx$,so $-(2-2x) dx = -du$.
Thus,$\int \frac{-du}{u} = -\ln|u| + C = -\ln|2x-x^2| + C$.
Evaluating the definite integral as a limit: $\lim_{\epsilon \to 0^+} \int_{\epsilon}^{2-\epsilon} \frac{2x-2}{2x-x^2} dx = \lim_{\epsilon \to 0^+} [-\ln|2x-x^2|]_{\epsilon}^{2-\epsilon}$.
$= \lim_{\epsilon \to 0^+} [-\ln|2(2-\epsilon)-(2-\epsilon)^2| - (-\ln|2\epsilon-\epsilon^2|)]$.
$= \lim_{\epsilon \to 0^+} [-\ln|4-2\epsilon-(4-4\epsilon+\epsilon^2)| + \ln|2\epsilon-\epsilon^2|]$.
$= \lim_{\epsilon \to 0^+} [-\ln|2\epsilon-\epsilon^2| + \ln|2\epsilon-\epsilon^2|] = 0$.
However,since the integral diverges at both endpoints,the Cauchy Principal Value is $0$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$.
Define $f(\theta) = \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$.
Now,check for parity by calculating $f(-\theta)$:
$f(-\theta) = \log \left(\frac{2-\sin(-\theta)}{2+\sin(-\theta)}\right) = \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right)$.
Using the property $\log(x^{-1}) = -\log(x)$,we get:
$f(-\theta) = \log \left(\left(\frac{2-\sin \theta}{2+\sin \theta}\right)^{-1}\right) = -\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) = -f(\theta)$.
Since $f(-\theta) = -f(\theta)$,the function $f(\theta)$ is an odd function.
According to the property of definite integrals,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function.
Therefore,$I = 0$.

(A) Given,$y = A e^x + B e^{2x} + C e^{3x} \quad \dots(i)$
Differentiating with respect to $x$:
$y' = A e^x + 2B e^{2x} + 3C e^{3x} \quad \dots(ii)$
Differentiating again with respect to $x$:
$y'' = A e^x + 4B e^{2x} + 9C e^{3x} \quad \dots(iii)$
Differentiating again with respect to $x$:
$y''' = A e^x + 8B e^{2x} + 27C e^{3x} \quad \dots(iv)$
Alternatively,since the roots of the auxiliary equation are $m = 1, 2, 3$,the characteristic equation is $(m-1)(m-2)(m-3) = 0$.
$(m^2 - 3m + 2)(m-3) = 0$
$m^3 - 3m^2 - 3m^2 + 9m + 2m - 6 = 0$
$m^3 - 6m^2 + 11m - 6 = 0$
Replacing $m^k$ with $y^{(k)}$,we get the differential equation:
$y''' - 6y'' + 11y' - 6y = 0$.

(A) The given differential equation is $(x+2 y^3) \frac{d y}{d x}=y^2$.
Rearranging the equation to the form $\frac{d x}{d y} + P(y)x = Q(y)$,we get:
$y^2 \frac{d x}{d y} = x + 2y^3$
$\frac{d x}{d y} - \frac{1}{y^2} x = 2y$.
Here,$P(y) = -\frac{1}{y^2}$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy}$.
$IF = e^{\int -\frac{1}{y^2} dy} = e^{\int -y^{-2} dy} = e^{-(-y^{-1})} = e^{1/y}$.
Thus,the integrating factor is $e^{(1/y)}$.

(A) Let the sides of the parallelogram be represented by vectors $\vec{a} = \hat{i}+2 \hat{j}+3 \hat{k}$ and $\vec{b} = 3 \hat{i}+2 \hat{j}+\hat{k}$.
One diagonal of the parallelogram is given by the sum of the adjacent sides,$\vec{d_1} = \vec{a} + \vec{b}$.
$\vec{d_1} = (\hat{i}+2 \hat{j}+3 \hat{k}) + (3 \hat{i}+2 \hat{j}+\hat{k}) = 4\hat{i} + 4\hat{j} + 4\hat{k} = 4(\hat{i} + \hat{j} + \hat{k})$.
The unit vector parallel to this diagonal is $\hat{d_1} = \frac{\vec{d_1}}{|\vec{d_1}|}$.
$|\vec{d_1}| = \sqrt{4^2 + 4^2 + 4^2} = \sqrt{16+16+16} = \sqrt{48} = 4\sqrt{3}$.
Therefore,$\hat{d_1} = \frac{4(\hat{i} + \hat{j} + \hat{k})}{4\sqrt{3}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.
Alternatively,the other diagonal is $\vec{d_2} = \vec{a} - \vec{b} = (\hat{i}+2 \hat{j}+3 \hat{k}) - (3 \hat{i}+2 \hat{j}+\hat{k}) = -2\hat{i} + 2\hat{k} = 2(-\hat{i} + \hat{k})$.
The unit vector parallel to this diagonal is $\frac{-\hat{i} + \hat{k}}{\sqrt{2}}$.
Comparing with the given options,the correct unit vector is $\frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.

(C) Let $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$ and $\vec{b} = \lambda\hat{i} - 4\hat{j} + \hat{k}$.
Since the vectors $\vec{a}$ and $\vec{b}$ are orthogonal,their dot product must be zero,i.e.,$\vec{a} \cdot \vec{b} = 0$.
Substituting the components,we get:
$(\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (\lambda\hat{i} - 4\hat{j} + \hat{k}) = 0$
Performing the dot product:
$(1)(\lambda) + (3)(-4) + (4)(1) = 0$
$\lambda - 12 + 4 = 0$
$\lambda - 8 = 0$
$\lambda = 8$.

(A) Let $\vec{a} = 3 \hat{i} + 3 \hat{j} + \sqrt{3} \hat{k}$,$\vec{b} = \hat{i} + \hat{k}$,and $\vec{c} = \sqrt{3} \hat{i} + \sqrt{3} \hat{j} + \lambda \hat{k}$.
Since these vectors are coplanar,their scalar triple product must be zero,i.e.,$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
This is equivalent to the determinant of the components being zero:
$\begin{vmatrix} 3 & 3 & \sqrt{3} \\ 1 & 0 & 1 \\ \sqrt{3} & \sqrt{3} & \lambda \end{vmatrix} = 0$
Expanding the determinant along the first row:
$3(0 - \sqrt{3}) - 3(\lambda - \sqrt{3}) + \sqrt{3}(\sqrt{3} - 0) = 0$
$-3\sqrt{3} - 3\lambda + 3\sqrt{3} + 3 = 0$
$-3\lambda + 3 = 0$
$3\lambda = 3$
$\lambda = 1$

(A) Let the plane be $ax + by + cz + d = 0$.
Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, -1, 3)$,the normal vector $\vec{n}$ to the plane is the vector from the origin to the foot of the perpendicular,which is $\vec{n} = (2 - 0)\hat{i} + (-1 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} - \hat{j} + 3\hat{k}$.
Thus,the equation of the plane is $2x - y + 3z = D$.
Since the point $(2, -1, 3)$ lies on the plane,we substitute these coordinates into the equation:
$2(2) - (-1) + 3(3) = D$
$4 + 1 + 9 = D$
$D = 14$.
Therefore,the equation of the plane is $2x - y + 3z - 14 = 0$.

(D) The given equation of the plane is $3x - 2y - z = 18$.
Dividing both sides by $18$,we get the intercept form:
$\frac{3x}{18} - \frac{2y}{18} - \frac{z}{18} = 1$
$\frac{x}{6} + \frac{y}{-9} + \frac{z}{-18} = 1$.
Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,the intercepts on the coordinate axes are $a = 6, b = -9, c = -18$.
Thus,the coordinates of the points where the plane meets the axes are $A(6, 0, 0)$,$B(0, -9, 0)$,and $C(0, 0, -18)$.
The centroid $(G)$ of $\triangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
$G = \left(\frac{6+0+0}{3}, \frac{0-9+0}{3}, \frac{0+0-18}{3}\right)$
$G = (2, -3, -6)$.

(C) For a Poisson distribution with parameter $m$,the probability mass function is given by $P(X=x) = \frac{e^{-m} m^x}{x!}$.
Given $P(X=0) = 0.8$.
Substituting $x=0$ in the formula,we get $P(X=0) = \frac{e^{-m} m^0}{0!} = e^{-m}$.
So,$e^{-m} = 0.8$.
Taking the natural logarithm on both sides,$-m = \log_e(0.8) = \log_e(8/10) = \log_e(4/5)$.
Therefore,$m = -\log_e(4/5) = \log_e(5/4)$.
In a Poisson distribution,the variance is equal to the parameter $m$.
Thus,the variance is $\log_e(5/4)$.

(A) Let the position vectors of the vertices $A, B,$ and $C$ be $\vec{a}, \vec{b},$ and $\vec{c}$ respectively.
Since $G$ is the centroid of $\triangle ABC$,its position vector $\vec{g}$ is given by $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.
This implies $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$.
Now,the sum of the vectors $\vec{GA} + \vec{GB} + \vec{GC}$ can be written in terms of position vectors as:
$\vec{GA} + \vec{GB} + \vec{GC} = (\vec{a} - \vec{g}) + (\vec{b} - \vec{g}) + (\vec{c} - \vec{g})$
$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{g}$
Substituting $\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$,we get:
$= 3\vec{g} - 3\vec{g} = 0$.

(C) The area bounded by the curve $y = f(x)$,the $x$-axis,and the lines $x = a$ and $x = b$ is given by the definite integral $\int_a^b y \, dx$.
Here,$y = x^2 + 2$,$a = 1$,and $b = 2$.
$\text{Required Area} = \int_1^2 (x^2 + 2) \, dx$
$= \left[ \frac{x^3}{3} + 2x \right]_1^2$
$= \left( \frac{2^3}{3} + 2(2) \right) - \left( \frac{1^3}{3} + 2(1) \right)$
$= \left( \frac{8}{3} + 4 \right) - \left( \frac{1}{3} + 2 \right)$
$= \left( \frac{8 + 12}{3} \right) - \left( \frac{1 + 6}{3} \right)$
$= \frac{20}{3} - \frac{7}{3}$
$= \frac{13}{3} \text{ sq unit}$.

(D) Let $A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix}$.
To find the rank,we calculate the determinant of $A$:
$|A| = 1(1 - (-1)) - (-1)(1 - 1) + 1(1 - (-1))$
$|A| = 1(2) + 1(0) + 1(2)$
$|A| = 2 + 0 + 2 = 4$.
Since $|A| \neq 0$,the matrix $A$ is non-singular.
Therefore,the rank of the $3 \times 3$ matrix $A$ is $3$.

(C) Let $A = \left[\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 4 & 2 \\ 3 & -4 & 6 \end{array}\right]$.
The cofactor $C_{ij}$ of an element $a_{ij}$ is given by $(-1)^{i+j} M_{ij}$,where $M_{ij}$ is the minor.
$1$. For element $-1$ at position $(1, 2)$: $C_{12} = (-1)^{1+2} \left|\begin{array}{cc} 0 & 2 \\ 3 & 6 \end{array}\right| = -1(0 - 6) = 6$. So,$A-4$.
$2$. For element $1$ at position $(1, 1)$: $C_{11} = (-1)^{1+1} \left|\begin{array}{cc} 4 & 2 \\ -4 & 6 \end{array}\right| = 1(24 - (-8)) = 32$. So,$B-2$.
$3$. For element $3$ at position $(3, 1)$: $C_{31} = (-1)^{3+1} \left|\begin{array}{cc} -1 & 0 \\ 4 & 2 \end{array}\right| = 1(-2 - 0) = -2$. So,$C-1$.
$4$. For element $6$ at position $(3, 3)$: $C_{33} = (-1)^{3+3} \left|\begin{array}{cc} 1 & -1 \\ 0 & 4 \end{array}\right| = 1(4 - 0) = 4$. So,$D-3$.
Thus,the matching is $A-4, B-2, C-1, D-3$.
Therefore,option $C$ is correct.

(D) Let $A = \left|\begin{array}{lll}1990 & 1991 & 1992 \\ 1991 & 1992 & 1993 \\ 1992 & 1993 & 1994\end{array}\right|$.
Applying the column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$,we get:
$A = \left|\begin{array}{ccc}1990 & 1991-1990 & 1992-1991 \\ 1991 & 1992-1991 & 1993-1992 \\ 1992 & 1993-1992 & 1994-1993\end{array}\right|$
$A = \left|\begin{array}{ccc}1990 & 1 & 1 \\ 1991 & 1 & 1 \\ 1992 & 1 & 1\end{array}\right|$
Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.

(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$
Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we have:
$\sin ^{-1}(1-x) = \frac{\pi}{2} - \sin ^{-1} x - \sin ^{-1} x = \frac{\pi}{2} - 2 \sin ^{-1} x$
Taking $\sin$ on both sides:
$1-x = \sin(\frac{\pi}{2} - 2 \sin ^{-1} x) = \cos(2 \sin ^{-1} x)$
Using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$,where $\theta = \sin ^{-1} x$:
$1-x = 1 - 2(\sin(\sin ^{-1} x))^2$
$1-x = 1 - 2x^2$
$2x^2 - x = 0$
$x(2x - 1) = 0$
Thus,$x = 0$ or $x = \frac{1}{2}$.
Both values satisfy the domain of the inverse trigonometric functions.
Therefore,$x \in \{0, \frac{1}{2}\}$.

(B) Given that $f(x)$ is an even function,so $f(-x) = f(x)$.
By differentiating both sides with respect to $x$ using the chain rule:
$\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$
$-f^{\prime}(-x) = f^{\prime}(x)$,which implies $f^{\prime}(-x) = -f^{\prime}(x)$.
Thus,the first derivative $f^{\prime}(x)$ is an odd function.
Now,differentiating again:
$\frac{d}{dx}[f^{\prime}(-x)] = \frac{d}{dx}[-f^{\prime}(x)]$
$-f^{\prime \prime}(-x) = -f^{\prime \prime}(x)$,which implies $f^{\prime \prime}(-x) = f^{\prime \prime}(x)$.
Thus,the second derivative $f^{\prime \prime}(x)$ is an even function.
Similarly,the third derivative $f^{\prime \prime \prime}(x)$ will be an odd function.
Therefore,$f^{\prime}(x)$ is an odd function. Comparing with the options,option $B$ is correct.

(B) Given the function $f: N \rightarrow Z$ defined as:
$f(n) = \begin{cases} 2 & \text{if } n=3k, k \in Z \\ 10 & \text{if } n=3k+1, k \in Z \\ 0 & \text{if } n=3k+2, k \in Z \end{cases}$
We want to find the set $\{n \in N: f(n) > 2\}$.
Looking at the definition,$f(n) > 2$ only when $f(n) = 10$.
This occurs when $n = 3k + 1$ for $k \in Z$.
Since $n \in N$ (natural numbers),we consider $k \geq 0$ (assuming $N = \{1, 2, 3, \dots\}$):
For $k=0, n = 3(0) + 1 = 1$.
For $k=1, n = 3(1) + 1 = 4$.
For $k=2, n = 3(2) + 1 = 7$.
Thus,the set is $\{1, 4, 7, \dots\}$.

(C) Given the function $f: R \rightarrow R$ defined by $f(x)=3^{-x}$.
$I$. For one-one check: Let $f(x_1) = f(x_2)$.
$3^{-x_1} = 3^{-x_2} \Rightarrow -x_1 = -x_2 \Rightarrow x_1 = x_2$.
Since $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$,the function is one-one.
$II$. For onto check: The range of $f(x) = 3^{-x}$ is $(0, \infty)$,which is a subset of the codomain $R$. Since the range $\neq$ codomain,the function is not onto.
$III$. For decreasing check: Differentiating $f(x)$ with respect to $x$,we get $f'(x) = -3^{-x} \ln 3$. Since $3^{-x} > 0$ and $\ln 3 > 0$,$f'(x) < 0$ for all $x \in R$. Thus,the function is strictly decreasing.
Therefore,statements $I$ and $III$ are true.

(B) Given the equation $x^y = e^{x-y}$.
Taking the natural logarithm on both sides,we get:
$\ln(x^y) = \ln(e^{x-y})$
$y \ln x = x - y$
Rearranging the terms to isolate $y$:
$y \ln x + y = x$
$y(1 + \ln x) = x$
$y = \frac{x}{1 + \ln x}$
Now,differentiate $y$ with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:
Let $u = x$ and $v = 1 + \ln x$. Then $u' = 1$ and $v' = \frac{1}{x}$.
$\frac{dy}{dx} = \frac{(1 + \ln x)(1) - x(\frac{1}{x})}{(1 + \ln x)^2}$
$\frac{dy}{dx} = \frac{1 + \ln x - 1}{(1 + \ln x)^2}$
$\frac{dy}{dx} = \frac{\ln x}{(1 + \ln x)^2}$
Thus,the correct option is $B$.

(D) Given the equation of the curve is $y = 4 - 2x^2$.
On differentiating both sides with respect to time $t$,we get:
$\frac{dy}{dt} = -4x \frac{dx}{dt}$.
We are given that the $x$-coordinate is decreasing at the rate of $5 \text{ units/s}$,so $\frac{dx}{dt} = -5 \text{ units/s}$.
At the point $(1, 2)$,we have $x = 1$.
Substituting these values into the derivative equation:
$\frac{dy}{dt} = -4(1)(-5) = 20 \text{ units/s}$.
Thus,the $y$-coordinate is changing at the rate of $20 \text{ units/s}$.

(C) Given the equation of the curve is $y = x^2 + 2x$.
Since the $x$ and $y$ coordinates of the particle change at the same rate,we have $\frac{dy}{dt} = \frac{dx}{dt}$.
Differentiating the equation of the curve with respect to $t$,we get:
$\frac{dy}{dt} = (2x + 2) \frac{dx}{dt}$.
Substituting $\frac{dy}{dt} = \frac{dx}{dt}$ into the equation,we get:
$\frac{dx}{dt} = (2x + 2) \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} \neq 0$,we can divide both sides by $\frac{dx}{dt}$:
$1 = 2x + 2$
$2x = -1$
$x = -\frac{1}{2}$.
Now,substitute $x = -\frac{1}{2}$ into the curve equation to find $y$:
$y = \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right)$
$y = \frac{1}{4} - 1 = -\frac{3}{4}$.
Therefore,the required point is $\left(-\frac{1}{2}, -\frac{3}{4}\right)$.

(D) Let $I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x$.
Divide the numerator and denominator by $\cos^2 x$:
$I = \int \frac{\sqrt{\cot x}}{\frac{\sin x \cos x}{\cos^2 x}} \cdot \frac{1}{\cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\tan x} \sec^2 x d x$.
Since $\frac{1}{\tan x} = \cot x$,we have $I = \int \sqrt{\cot x} \cdot \cot x \cdot \sec^2 x d x = \int (\cot x)^{3/2} \sec^2 x d x$.
Alternatively,rewrite the integral as:
$I = \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x = \int \frac{\sqrt{\cot x}}{\frac{\sin x}{\cos x} \cdot \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\tan x \cos^2 x} d x = \int \frac{\sqrt{\cot x}}{\cot x} \cdot \frac{1}{\sin^2 x} d x$ is not correct.
Let $t = \cot x$,then $dt = -\csc^2 x d x$,so $dx = -\sin^2 x dt$.
$I = \int \frac{\sqrt{t}}{\sin x \cos x} (- \sin^2 x) dt = \int \frac{\sqrt{t}}{\cos x / \sin x} (-dt) = \int \frac{\sqrt{t}}{t} (-dt) = -\int t^{-1/2} dt$.
$I = -[2 t^{1/2}] + c = -2 \sqrt{\cot x} + c$.
Comparing with $-f(x) + c$,we get $-f(x) = -2 \sqrt{\cot x}$,so $f(x) = 2 \sqrt{\cot x}$.

(C) Let $I = \int \frac{d x}{(x+100) \sqrt{x+99}}$.
We can rewrite the denominator as $(x+99)+1$,so $I = \int \frac{d x}{((\sqrt{x+99})^2+1) \sqrt{x+99}}$.
Substitute $t = \sqrt{x+99}$,then $t^2 = x+99$,which implies $2t \, dt = dx$.
Substituting these into the integral:
$I = \int \frac{2t \, dt}{(t^2+1)t} = \int \frac{2 \, dt}{t^2+1}$.
Integrating,we get $I = 2 \tan^{-1}(t) + c$.
Substituting back $t = \sqrt{x+99}$,we get $I = 2 \tan^{-1}(\sqrt{x+99}) + c$.
Comparing this with $f(x) + c$,we find $f(x) = 2 \tan^{-1}(\sqrt{x+99})$.

(A) Given the integral $I = \int \frac{3-x^2}{1-2x+x^2} e^x dx$.
We can rewrite the denominator as $(1-x)^2$.
So,$I = \int \frac{3-x^2}{(1-x)^2} e^x dx$.
We manipulate the numerator to match the form $f(x) + f'(x)$:
$I = \int \frac{-(x^2-1) + 2}{(1-x)^2} e^x dx = \int \frac{-(x-1)(x+1) + 2}{(x-1)^2} e^x dx$
$I = \int \left( \frac{-(x+1)}{x-1} + \frac{2}{(x-1)^2} \right) e^x dx = \int \left( \frac{x+1}{1-x} + \frac{2}{(1-x)^2} \right) e^x dx$.
Let $f(x) = \frac{1+x}{1-x}$. Then $f'(x) = \frac{(1)(1-x) - (1+x)(-1)}{(1-x)^2} = \frac{1-x+1+x}{(1-x)^2} = \frac{2}{(1-x)^2}$.
Since $\int (f(x) + f'(x)) e^x dx = e^x f(x) + c$,we have $I = e^x \left( \frac{1+x}{1-x} \right) + c$.
Comparing this with $e^x f(x) + c$,we get $f(x) = \frac{1+x}{1-x}$.

(C) Given that,$f(x) = \frac{1}{x^2} \int_3^x (2t - 3f'(t)) dt$.
Multiplying both sides by $x^2$,we get $x^2 f(x) = \int_3^x (2t - 3f'(t)) dt$.
Differentiating both sides with respect to $x$ using the product rule on the left and the Fundamental Theorem of Calculus on the right:
$x^2 f'(x) + 2x f(x) = 2x - 3f'(x)$.
Rearranging the terms: $f'(x)(x^2 + 3) = 2x - 2x f(x)$.
At $x = 3$,note that $f(3) = \frac{1}{3^2} \int_3^3 (2t - 3f'(t)) dt = 0$.
Substituting $x = 3$ into the differentiated equation:
$f'(3)(3^2 + 3) = 2(3) - 2(3) f(3)$.
$f'(3)(9 + 3) = 6 - 6(0)$.
$12 f'(3) = 6$.
$f'(3) = \frac{6}{12} = \frac{1}{2}$.

(C) For statement $A$:
The given differential equation is $\frac{dy}{dx} + y = x^2$.
Comparing this with the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$,we get $P(x) = 1$.
The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int 1 dx} = e^x$.
Thus,statement $A$ is true.
For statement $R$:
The standard form of a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.
The integrating factor is defined as $e^{\int P(x) dx}$.
Thus,statement $R$ is true.
Since statement $A$ is derived directly from the formula given in statement $R$,$R \Rightarrow A$ is true.
Therefore,both statements are true and $R$ is the correct explanation for $A$.

(A) The given differential equation is $\left(x+2 y^3\right) \frac{d y}{d x}=y^2$.
Rearranging the equation to the form $\frac{dx}{dy} + P(y)x = Q(y)$,we get:
$\frac{dx}{dy} = \frac{x+2y^3}{y^2} = \frac{x}{y^2} + 2y$
$\frac{dx}{dy} - \frac{1}{y^2}x = 2y$
Here,$P(y) = -\frac{1}{y^2}$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy}$.
$IF = e^{\int -\frac{1}{y^2} dy} = e^{\int -y^{-2} dy} = e^{-(-y^{-1})} = e^{\frac{1}{y}}$.

(A) We are given the expression $c \cdot ((b+c) \times (a+b+c))$.
Using the distributive property of the cross product,we have:
$(b+c) \times (a+b+c) = b \times a + b \times b + b \times c + c \times a + c \times b + c \times c$
Since the cross product of any vector with itself is zero ($b \times b = 0$ and $c \times c = 0$),the expression simplifies to:
$b \times a + b \times c + c \times a + c \times b$
Now,taking the dot product with $c$:
$c \cdot (b \times a + b \times c + c \times a + c \times b)$
$= c \cdot (b \times a) + c \cdot (b \times c) + c \cdot (c \times a) + c \cdot (c \times b)$
Using the property of the scalar triple product $[x \ y \ z] = x \cdot (y \times z)$,where the product is zero if any two vectors are identical:
$= [c \ b \ a] + [c \ b \ c] + [c \ c \ a] + [c \ c \ b]$
$= [c \ b \ a] + 0 + 0 + 0$
$= c \cdot (b \times a)$
Thus,the correct option is $A$.

(C) Given equations are:
$l+m+n=0 \quad \dots(i)$
$mn-2ln+lm=0 \quad \dots(ii)$
From $(i)$,$l = -(m+n)$. Substituting this into $(ii)$:
$mn - 2n(-(m+n)) + m(-(m+n)) = 0$
$mn + 2mn + 2n^2 - m^2 - mn = 0$
$2n^2 + 2mn - m^2 = 0$
Dividing by $m^2$,we get $2(\frac{n}{m})^2 + 2(\frac{n}{m}) - 1 = 0$. Let the roots be $\frac{n_1}{m_1}$ and $\frac{n_2}{m_2}$.
Then $\frac{n_1 n_2}{m_1 m_2} = -\frac{1}{2} \implies n_1 n_2 = -\frac{1}{2} m_1 m_2 \quad \dots(iii)$
Similarly,from $(i)$,$m = -(l+n)$. Substituting into $(ii)$:
$n(-(l+n)) - 2ln + l(-(l+n)) = 0$
$-ln - n^2 - 2ln - l^2 - ln = 0$
$l^2 + 4ln + n^2 = 0$
Dividing by $n^2$,we get $(\frac{l}{n})^2 + 4(\frac{l}{n}) + 1 = 0$. Let the roots be $\frac{l_1}{n_1}$ and $\frac{l_2}{n_2}$.
Then $\frac{l_1 l_2}{n_1 n_2} = 1 \implies l_1 l_2 = n_1 n_2 \quad \dots(iv)$
From $(iii)$ and $(iv)$,$l_1 l_2 = -\frac{1}{2} m_1 m_2 \implies 2l_1 l_2 + m_1 m_2 + 0n_1 n_2 = 0$.
Since the condition for perpendicularity is $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$,and here $l_1 l_2 + m_1 m_2 + n_1 n_2 = 2l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ is not directly satisfied,we check the dot product $l_1 l_2 + m_1 m_2 + n_1 n_2$.
Actually,for these lines,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ holds true. Thus,the angle is $\frac{\pi}{2}$.

(C) Let $H$ denote a head and $T$ denote a tail. For each $H$,the gain is $+2$ points,and for each $T$,the loss is $-1$ point.
When three coins are tossed,the possible outcomes for the number of heads $(n_H)$ and tails $(n_T)$ are as follows:
$1$. Three tails $(0H, 3T)$: Score $X = 0(2) + 3(-1) = -3$.
$2$. Two tails and one head $(1H, 2T)$: Score $X = 1(2) + 2(-1) = 2 - 2 = 0$.
$3$. One tail and two heads $(2H, 1T)$: Score $X = 2(2) + 1(-1) = 4 - 1 = 3$.
$4$. Three heads $(3H, 0T)$: Score $X = 3(2) + 0(-1) = 6$.
Thus,the possible values for the total score $X$ are $\{-3, 0, 3, 6\}$.
Therefore,the range of $X$ is $\{-3, 0, 3, 6\}$.

(C) Given that,$P(E) = 1/5$ and $P(F|E) = 1/10$.
We know that the probability of both events occurring is given by $P(E \cap F) = P(E) \cdot P(F|E)$.
Substituting the values,we get $P(E \cap F) = (1/5) \cdot (1/10) = 1/50$.
The probability of non-occurrence of at least one of the events $E$ and $F$ is equivalent to the complement of the event that both $E$ and $F$ occur.
Thus,the required probability is $1 - P(E \cap F)$.
Calculating this,we get $1 - 1/50 = 49/50$.

(A) Let $H$ denote a head and $T$ denote a tail. The points awarded are $2$ for $H$ and $1$ for $T$.
When three coins are tossed,let $n_H$ be the number of heads and $n_T$ be the number of tails.
The total points $S = 2n_H + 1n_T$.
Since $n_H + n_T = 3$,we have $n_T = 3 - n_H$.
Substituting this,$S = 2n_H + (3 - n_H) = n_H + 3$.
For $S$ to be an odd number,$n_H + 3$ must be odd,which means $n_H$ must be even.
Possible values for $n_H$ are $0$ and $2$.
Case $1$: $n_H = 0$ (all tails). The probability is $\binom{3}{0} (\frac{1}{2})^0 (\frac{1}{2})^3 = \frac{1}{8}$.
Case $2$: $n_H = 2$ (two heads,one tail). The probability is $\binom{3}{2} (\frac{1}{2})^2 (\frac{1}{2})^1 = 3 \times \frac{1}{8} = \frac{3}{8}$.
The total probability is $\frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$.

(C) For a Poisson distribution,the probability mass function is given by $P(X=x) = \frac{e^{-m} m^x}{x!}$,where $m$ is the parameter (mean and variance).
Given $P(X=0) = 0.8$.
Substituting $x=0$ in the formula:
$P(X=0) = \frac{e^{-m} m^0}{0!} = e^{-m} = 0.8$.
Taking the natural logarithm on both sides:
$-m = \ln(0.8) = \ln(\frac{8}{10}) = \ln(\frac{4}{5})$.
Therefore,$m = -\ln(\frac{4}{5}) = \ln((\frac{4}{5})^{-1}) = \ln(\frac{5}{4}) = \ln(1.25)$.
Since the variance of a Poisson distribution is equal to its parameter $m$,the variance is $\ln(1.25)$ or $\log _e 1.25$.