AIEEE 2002 Mathematics Question Paper with Answer and Solution

(D) Given inequality: $|z - 4| < |z - 2|$
Squaring both sides: $|z - 4|^2 < |z - 2|^2$
Let $z = x + iy$. Then $|(x - 4) + iy|^2 < |(x - 2) + iy|^2$
$(x - 4)^2 + y^2 < (x - 2)^2 + y^2$
$x^2 - 8x + 16 + y^2 < x^2 - 4x + 4 + y^2$
$-8x + 16 < -4x + 4$
$12 < 4x$
$x > 3$
Since $x = \text{Re}(z)$,the region is $\text{Re}(z) > 3$.
Thus,the correct option is $D$.

(D) Let $z = r(\cos \theta_1 + i \sin \theta_1)$ and $w = r(\cos \theta_2 + i \sin \theta_2)$,where $|z| = |w| = r$.
Given $arg(z) + arg(w) = \theta_1 + \theta_2 = \pi$,so $\theta_1 = \pi - \theta_2$.
Substituting this into $z$:
$z = r(\cos(\pi - \theta_2) + i \sin(\pi - \theta_2))$
$z = r(-\cos \theta_2 + i \sin \theta_2)$
Since $\overline{w} = r(\cos \theta_2 - i \sin \theta_2)$,we have $-\overline{w} = r(-\cos \theta_2 + i \sin \theta_2)$.
Thus,$z = -\overline{w}$.

(B) Let the $9$ terms of a $G.P.$ be $\frac{a}{r^4}, \frac{a}{r^3}, \frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2, ar^3, ar^4$.
The fifth term is $a = 2$.
The product of these $9$ terms is $P = \left(\frac{a}{r^4}\right) \times \left(\frac{a}{r^3}\right) \times \left(\frac{a}{r^2}\right) \times \left(\frac{a}{r}\right) \times a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)$.
Simplifying the product,we get $P = a^9$.
Since $a = 2$,the product is $P = 2^9 = 512$.

(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite $G.P.$ is given by $S = \frac{a}{1-r} = 20 \quad (i)$.
The sum of the squares of the terms is given by $\frac{a^2}{1-r^2} = 100 \quad (ii)$.
We can rewrite $(ii)$ as $\frac{a}{1-r} \times \frac{a}{1+r} = 100$.
Substituting $(i)$ into this equation: $20 \times \frac{a}{1+r} = 100$,which simplifies to $\frac{a}{1+r} = 5 \quad (iii)$.
From $(i)$,$a = 20(1-r)$. Substituting this into $(iii)$:
$\frac{20(1-r)}{1+r} = 5$
$4(1-r) = 1+r$
$4 - 4r = 1 + r$
$3 = 5r$
$r = 3/5$.

(B) Let the given expression be $P = 2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdot 16^{1/32} \cdots$
Expressing all terms with base $2$:
$P = 2^{1/4} \cdot (2^2)^{1/8} \cdot (2^3)^{1/16} \cdot (2^4)^{1/32} \cdots$
$P = 2^{1/4 + 2/8 + 3/16 + 4/32 + \cdots} = 2^S$,where $S = \sum_{n=1}^{\infty} \frac{n}{2^{n+1}}$.
$S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} + \cdots$ $(i)$
$\frac{1}{2}S = \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \cdots$ $(ii)$
Subtracting $(ii)$ from $(i)$:
$S - \frac{1}{2}S = \frac{1}{4} + (\frac{2}{8} - \frac{1}{8}) + (\frac{3}{16} - \frac{2}{16}) + \cdots$
$\frac{1}{2}S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$
This is an infinite geometric series with $a = 1/4$ and $r = 1/2$.
$\frac{1}{2}S = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.
Thus,$S = 1$.
Therefore,$P = 2^1 = 2$.

(A) For the quadratic equation $x^2 + px + q = 0$,the sum of the roots is $p + q = -p$ and the product of the roots is $pq = q$.
From $pq = q$,we have $q(p - 1) = 0$,which implies $q = 0$ or $p = 1$.
Case $1$: If $q = 0$,then $p + 0 = -p$,which gives $2p = 0$,so $p = 0$. This gives the roots $(0, 0)$.
Case $2$: If $p = 1$,then $1 + q = -1$,which gives $q = -2$. This gives the roots $(1, -2)$.
Comparing with the given options,the correct pair is $p = 1, q = -2$.

(D) Given $\alpha^2 - 5\alpha + 3 = 0$ and $\beta^2 - 5\beta + 3 = 0$.
This implies that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 5x + 3 = 0$.
From the properties of roots,we have:
Sum of roots: $\alpha + \beta = 5$
Product of roots: $\alpha \beta = 3$
We need to find the equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
Sum of new roots: $S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{5^2 - 2(3)}{3} = \frac{25 - 6}{3} = \frac{19}{3}$.
Product of new roots: $P = \frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1$.
The required equation is $x^2 - Sx + P = 0$.
Substituting the values: $x^2 - \frac{19}{3}x + 1 = 0$.
Multiplying by $3$,we get $3x^2 - 19x + 3 = 0$.

(C) Given the equation ${t^2}{x^2} + |x| + 9 = 0$.
For any real number $t$ and $x$,we know that ${t^2}{x^2} \ge 0$ and $|x| \ge 0$.
Therefore,${t^2}{x^2} + |x| + 9 \ge 9$.
Since the expression is always greater than or equal to $9$,it can never be equal to $0$.
Thus,the equation has no real roots.
Consequently,the product of the real roots does not exist.

(B) The numbers are $4$-digit numbers formed using the set $\{0, 1, 2, 3, 4\}$.
Since the number must be greater than $1000$ and less than or equal to $4000$,the first digit can be $1, 2, 3,$ or $4$.
Case $1$: First digit is $1, 2,$ or $3$.
For each of these $3$ choices,the remaining $3$ positions can each be filled in $5$ ways (digits $0, 1, 2, 3, 4$).
Total numbers for these cases $= 3 \times 5 \times 5 \times 5 = 375$.
However,we must exclude the case where the number is $1000$ (since it must be greater than $1000$).
So,$375 - 1 = 374$ numbers.
Case $2$: First digit is $4$.
The only number $\le 4000$ starting with $4$ is $4000$ itself.
Adding this to our count: $374 + 1 = 375$.
Thus,the total count is $375$.

(D) four-digit number is represented as $d_1 d_2 d_3 d_4$.
For the number to be odd,the last digit $d_4$ must be chosen from the set $\{1, 3, 5, 7\}$. There are $4$ ways to fill $d_4$.
For the first digit $d_1$,it cannot be $0$ (otherwise it would be a three-digit number). Thus,$d_1$ can be chosen from $\{1, 2, 3, 5, 7\}$,which gives $5$ ways.
For the second digit $d_2$ and third digit $d_3$,since repetition is allowed,each can be filled by any of the $6$ digits $\{0, 1, 2, 3, 5, 7\}$. Thus,there are $6$ ways for $d_2$ and $6$ ways for $d_3$.
Using the fundamental principle of counting,the total number of such odd numbers is $5 \times 6 \times 6 \times 4 = 720$.

(A) number is divisible by $3$ if the sum of its digits is divisible by $3$. The sum of all given digits ${0, 1, 2, 3, 4, 5}$ is $15$. To form a $5$-digit number,we must exclude one digit such that the sum of the remaining $5$ digits is divisible by $3$.
Case $1$: Exclude $0$. The remaining digits are ${1, 2, 3, 4, 5}$. The sum is $15$,which is divisible by $3$. The number of $5$-digit numbers is $5! = 120$.
Case $2$: Exclude $3$. The remaining digits are ${0, 1, 2, 4, 5}$. The sum is $12$,which is divisible by $3$. The number of $5$-digit numbers is $5! - 4! = 120 - 24 = 96$ (subtracting cases where $0$ is in the first position).
Total ways = $120 + 96 = 216$.

(D) We know that $e = \lim_{n \to \infty} (1 + \frac{1}{n})^n$ and $2 < e < 3$.
For $n = 10000$,the expression $(1 + \frac{1}{10000})^{10000} = (1 + 0.0001)^{10000}$.
Since the sequence $(1 + \frac{1}{n})^n$ is strictly increasing and approaches $e$ from below,we have $(1 + 0.0001)^{10000} < e < 3$.
Using the binomial expansion,$(1 + 0.0001)^{10000} = 1 + 10000(0.0001) + \frac{10000 \times 9999}{2!} (0.0001)^2 + \dots = 1 + 1 + \frac{0.9999}{2} + \dots > 2$.
Thus,$2 < (1 + 0.0001)^{10000} < 3$.
The positive integer just greater than $(1 + 0.0001)^{10000}$ is $3$.
Therefore,the correct option is $D$.

(A) The general term in the expansion of $(1 + x)^n$ is given by $T_{r+1} = ^nC_r x^r$.
For the expansion of $(1 + x)^{p+q}$,the coefficient of $x^p$ is $^{p+q}C_p$.
Similarly,the coefficient of $x^q$ is $^{p+q}C_q$.
Using the property of binomial coefficients,$^nC_r = ^nC_{n-r}$,we have:
$^{p+q}C_p = ^{p+q}C_{(p+q)-p} = ^{p+q}C_q$.
Therefore,the coefficients of $x^p$ and $x^q$ are equal.

(A) We have the expression: ${\left( {1 + \frac{x}{a}} \right)^{-1/2}} + {\left( {1 - \frac{x}{a}} \right)^{-1/2}}$.
Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we expand both terms:
${\left( {1 + \frac{x}{a}} \right)^{-1/2}} = 1 - \frac{1}{2}\left( \frac{x}{a} \right) + \frac{(-1/2)(-3/2)}{2}\left( \frac{x}{a} \right)^2 + \dots = 1 - \frac{x}{2a} + \frac{3x^2}{8a^2} + \dots$
${\left( {1 - \frac{x}{a}} \right)^{-1/2}} = 1 - \frac{1}{2}\left( -\frac{x}{a} \right) + \frac{(-1/2)(-3/2)}{2}\left( -\frac{x}{a} \right)^2 + \dots = 1 + \frac{x}{2a} + \frac{3x^2}{8a^2} + \dots$
Adding these two expansions,the odd terms (involving $x/a$) cancel out:
$(1 + 1) + (-\frac{x}{2a} + \frac{x}{2a}) + (\frac{3x^2}{8a^2} + \frac{3x^2}{8a^2}) + \dots = 2 + \frac{6x^2}{8a^2} + \dots = 2 + \frac{3x^2}{4a^2} + \dots$

(B) The sum of the coefficients in the expansion of $(x + y)^n$ is given by putting $x = 1$ and $y = 1$,which is $(1 + 1)^n = 2^n$.
Given $2^n = 4096 = 2^{12}$,so $n = 12$.
The greatest coefficient in the expansion of $(x + y)^n$ when $n$ is even is given by the middle term coefficient,which is $^nC_{n/2}$.
For $n = 12$,the greatest coefficient is $^{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$.

(A) We know that $\sin^2 x = \frac{1 - \cos(2x)}{2}$.
The period of $\cos(kx)$ is given by $\frac{2\pi}{|k|}$.
Here,$k = 2$,so the period is $\frac{2\pi}{2} = \pi$.
Thus,the period of $\sin^2 x$ is $\pi$.

(C) Let the sides be $a = 3x + 4y$,$b = 4x + 3y$,and $c = 5x + 5y$.
Since $x, y > 0$,we compare the sides. Note that $c = 5x + 5y$ is the largest side because $5x + 5y > 3x + 4y$ and $5x + 5y > 4x + 3y$.
For a triangle to be obtuse,the square of the largest side must be greater than the sum of the squares of the other two sides $(c^2 > a^2 + b^2)$.
$a^2 + b^2 = (3x + 4y)^2 + (4x + 3y)^2 = (9x^2 + 24xy + 16y^2) + (16x^2 + 24xy + 9y^2) = 25x^2 + 48xy + 25y^2$.
$c^2 = (5x + 5y)^2 = 25x^2 + 50xy + 25y^2$.
Comparing $c^2$ and $a^2 + b^2$,we see that $25x^2 + 50xy + 25y^2 > 25x^2 + 48xy + 25y^2$ since $x, y > 0$.
Thus,$c^2 > a^2 + b^2$,which implies the triangle is obtuse angled.

(A) Let the vertices be $A(4, 0)$,$B(-1, -1)$,and $C(3, 5)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$AB = \sqrt{(-1 - 4)^2 + (-1 - 0)^2} = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26}$.
$AC = \sqrt{(3 - 4)^2 + (5 - 0)^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26}$.
$BC = \sqrt{(3 - (-1))^2 + (5 - (-1))^2} = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$.
Since $AB = AC = \sqrt{26}$,the triangle is isosceles.
Also,$AB^2 + AC^2 = 26 + 26 = 52$ and $BC^2 = 52$.
Since $AB^2 + AC^2 = BC^2$,the triangle is right-angled at $A$ by the converse of the Pythagorean theorem.
Thus,the triangle is isosceles and right-angled.

(B) The given line is $x \cos \alpha + y \sin \alpha = p$.
To find the $x$-intercept,set $y = 0$: $x \cos \alpha = p \Rightarrow x = \frac{p}{\cos \alpha}$. So,point $A = (\frac{p}{\cos \alpha}, 0)$.
To find the $y$-intercept,set $x = 0$: $y \sin \alpha = p \Rightarrow y = \frac{p}{\sin \alpha}$. So,point $B = (0, \frac{p}{\sin \alpha})$.
Let $(h, k)$ be the mid-point of the segment $AB$.
Then $h = \frac{p}{2 \cos \alpha}$ and $k = \frac{p}{2 \sin \alpha}$.
This implies $\cos \alpha = \frac{p}{2h}$ and $\sin \alpha = \frac{p}{2k}$.
Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$,we get $(\frac{p}{2k})^2 + (\frac{p}{2h})^2 = 1$.
$\frac{p^2}{4k^2} + \frac{p^2}{4h^2} = 1 \Rightarrow \frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2}$.
Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

(A) The general equation of a pair of lines passing through the origin is given by $Ax^2 + 2Hxy + By^2 = 0$.
For these lines to be perpendicular,the condition is that the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$A + B = 0$.
In the given equation $3ax^2 + 5xy + (a^2 - 2)y^2 = 0$,we have $A = 3a$ and $B = a^2 - 2$.
Setting $A + B = 0$,we get $3a + a^2 - 2 = 0$,which simplifies to $a^2 + 3a - 2 = 0$.
This is a quadratic equation in $a$. The discriminant $D = b^2 - 4ac = (3)^2 - 4(1)(-2) = 9 + 8 = 17$.
Since $D > 0$,the quadratic equation $a^2 + 3a - 2 = 0$ has two distinct real roots for $a$.
Therefore,the lines are perpendicular to each other for two values of $a$.

(C) The angle subtended by a chord at the center is twice the angle subtended by it at the major segment.
Given the angle at the major segment is ${45^\circ}$,the angle at the center $C(0,0)$ is $2 \times {45^\circ} = {90^\circ}$.
Let $P$ be the foot of the perpendicular from the center $C(0,0)$ to the chord $y = mx + 1$.
In the right-angled triangle $CPR$,where $CR$ is the radius $r = 1$,the angle $\angle PCR = {45^\circ}$.
Thus,$CP = r \cos({45^\circ}) = 1 \times \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The perpendicular distance from $(0,0)$ to $mx - y + 1 = 0$ is given by $\frac{|m(0) - 0 + 1|}{\sqrt{m^2 + (-1)^2}} = \frac{1}{\sqrt{m^2 + 1}}$.
Equating the two expressions for $CP$:
$\frac{1}{\sqrt{m^2 + 1}} = \frac{1}{\sqrt{2}}$
$m^2 + 1 = 2$
$m^2 = 1$
$m = \pm 1$.
Since the options provided include $-1$ but not $1$,and the question asks for the value of $m$,the correct choice is $-1$.

(D) For an equilateral triangle,the centroid coincides with the circumcentre.
Given the origin $(0, 0)$ is the centre of the circle passing through the vertices,it is the circumcentre of the triangle.
The centroid divides the median in the ratio $2:1$.
Since the length of the median is $3a$,the distance from the centroid to the vertex (which is the radius $R$ of the circumcircle) is $R = \frac{2}{3} \times 3a = 2a$.
The equation of the circle with centre $(0, 0)$ and radius $R = 2a$ is $x^2 + y^2 = (2a)^2 = 4a^2$.
Since $4a^2$ is not among the given options,the correct answer is $(d)$.

(B) Let the required circle be $S_2$ with centre $(h, k)$ and radius $r$.
Since it passes through $(0, 0)$ and $(1, 0)$,the perpendicular bisector of the chord joining $(0, 0)$ and $(1, 0)$ must contain the centre. The midpoint is $(\frac{1}{2}, 0)$,so $h = \frac{1}{2}$.
The radius $r$ is the distance from $(\frac{1}{2}, k)$ to $(0, 0)$,so $r^2 = (\frac{1}{2})^2 + k^2 = \frac{1}{4} + k^2$.
The circle $S_2$ touches $x^2 + y^2 = 9$ (centre $(0, 0)$,radius $R = 3$).
For internal contact,the distance between centres $d = R - r$.
$d^2 = (\frac{1}{2})^2 + k^2 = r^2$.
So $r^2 = (3 - r)^2 = 9 - 6r + r^2$,which gives $6r = 9$,so $r = \frac{3}{2}$.
Substituting $r^2 = \frac{9}{4}$ into $r^2 = \frac{1}{4} + k^2$ gives $\frac{9}{4} = \frac{1}{4} + k^2$,so $k^2 = 2$,$k = \pm \sqrt{2}$.
Thus,the centre is $\left( \frac{1}{2}, \pm \sqrt{2} \right)$. Given the options,the correct choice is $\left( \frac{1}{2}, -\sqrt{2} \right)$.

(D) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this into the expression,we get $\sqrt{\frac{1}{2}(2 \sin^2 x)} = \sqrt{\sin^2 x} = |\sin x|$.
Thus,the limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{|\sin x|}{x}$.
For the right-hand limit $(x \to 0^+)$,$|\sin x| = \sin x$,so $\mathop {\lim }\limits_{x \to 0^+} \frac{\sin x}{x} = 1$.
For the left-hand limit $(x \to 0^-)$,$|\sin x| = -\sin x$,so $\mathop {\lim }\limits_{x \to 0^-} \frac{-\sin x}{x} = -1$.
Since the left-hand limit and right-hand limit are not equal,the limit does not exist.

(C) Let $L = \mathop {\lim }\limits_{x \to 2} \frac{{xf(2) - 2f(x)}}{{x - 2}}$.
Since the limit is of the form $\frac{0}{0}$ at $x = 2$,we can apply $L$'$H$ôpital's rule or manipulate the expression.
Adding and subtracting $2f(2)$ in the numerator:
$L = \mathop {\lim }\limits_{x \to 2} \frac{{xf(2) - 2f(2) + 2f(2) - 2f(x)}}{{x - 2}}$
$L = \mathop {\lim }\limits_{x \to 2} \left[ \frac{f(2)(x - 2)}{x - 2} - 2\frac{f(x) - f(2)}{x - 2} \right]$
$L = f(2) - 2 \mathop {\lim }\limits_{x \to 2} \frac{f(x) - f(2)}{x - 2}$
$L = f(2) - 2f'(2)$
Given $f(2) = 4$ and $f'(2) = 4$:
$L = 4 - 2(4) = 4 - 8 = -4$.

(A) We are given the limit $\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n} - [x]}}{{[x]}}.$
This can be rewritten as $\mathop {\lim }\limits_{x \to \infty } \left( \frac{{\log {x^n}}}{{[x]}} - \frac{{[x]}}{{[x]}} \right).$
Since $x - 1 < [x] \le x,$ as $x \to \infty,$ $[x] \approx x.$
Thus,$\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n}}}{{[x]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n \log x}}{x}.$
Using $L$'Hopital's rule or the standard limit $\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{x} = 0,$ we get $n \times 0 = 0.$
Therefore,the expression becomes $0 - 1 = -1.$

(A) Let $y = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$.
Multiplying the numerator and denominator by the conjugates $(\sqrt{f(x)} + 1)$ and $(\sqrt{x} + 1)$:
$y = \mathop {\lim }\limits_{x \to 1} \frac{(\sqrt{f(x)} - 1)(\sqrt{f(x)} + 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)} \times \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1}$
$y = \mathop {\lim }\limits_{x \to 1} \frac{f(x) - 1}{x - 1} \times \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1}$
Since $f(1) = 1$,we can write $f(x) - 1$ as $f(x) - f(1)$:
$y = \left( \mathop {\lim }\limits_{x \to 1} \frac{f(x) - f(1)}{x - 1} \right) \times \left( \mathop {\lim }\limits_{x \to 1} \frac{\sqrt{x} + 1}{\sqrt{f(x)} + 1} \right)$
$y = f'(1) \times \frac{\sqrt{1} + 1}{\sqrt{f(1)} + 1} = 2 \times \frac{2}{1 + 1} = 2 \times 1 = 2$.
Alternatively,using $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 1} \frac{\frac{1}{2\sqrt{f(x)}} f'(x)}{\frac{1}{2\sqrt{x}}} = \mathop {\lim }\limits_{x \to 1} \frac{f'(x) \sqrt{x}}{\sqrt{f(x)}} = \frac{f'(1) \sqrt{1}}{\sqrt{f(1)}} = \frac{2 \times 1}{1} = 2$.

(A) Let $P(A), P(B),$ and $P(C)$ be the probabilities of students $A, B,$ and $C$ solving the problem respectively.
$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{4}$.
The probability that the problem is not solved by $A$ is $P(A') = 1 - \frac{1}{2} = \frac{1}{2}$.
The probability that the problem is not solved by $B$ is $P(B') = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability that the problem is not solved by $C$ is $P(C') = 1 - \frac{1}{4} = \frac{3}{4}$.
The probability that the problem is not solved by any of them is $P(\text{none}) = P(A') \times P(B') \times P(C') = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{6}{24} = \frac{1}{4}$.
The probability that the problem is solved is $P(\text{solved}) = 1 - P(\text{none}) = 1 - \frac{1}{4} = \frac{3}{4}$.

(A) Given: $P(A \cup B) = 3/4,$ $P(A \cap B) = 1/4,$ and $P(\bar{A}) = 2/3.$
First,find $P(A):$
$P(A) = 1 - P(\bar{A}) = 1 - 2/3 = 1/3.$
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B).$
$3/4 = 1/3 + P(B) - 1/4.$
$P(B) = 3/4 + 1/4 - 1/3 = 1 - 1/3 = 2/3.$
Now,calculate $P(\bar{A} \cap B):$
$P(\bar{A} \cap B) = P(B) - P(A \cap B).$
$P(\bar{A} \cap B) = 2/3 - 1/4 = (8 - 3)/12 = 5/12.$

(B) Let the average marks of the girls be $x$.
Total number of students = $100$.
Number of boys = $70$,so number of girls = $100 - 70 = 30$.
Total marks of the class = $100 \times 72 = 7200$.
Total marks of boys = $70 \times 75 = 5250$.
Total marks of girls = $7200 - 5250 = 1950$.
Average marks of girls = $\frac{1950}{30} = 65$.

(C) In the expansion of $(1 + x)^{2n}$,the coefficient of $x^k$ is given by $^{2n}C_k$,where $0 \le k \le 2n$.
Given that the coefficients of $x^{3r}$ and $x^{r+2}$ are equal,we have $^{2n}C_{3r} = ^{2n}C_{r+2}$.
Using the property $^{n}C_a = ^{n}C_b$,which implies $a = b$ or $a + b = n$,we get two cases:
Case $1$: $3r = r + 2$ $\Rightarrow 2r = 2$ $\Rightarrow r = 1$. However,it is given that $r > 1$,so this case is rejected.
Case $2$: $3r + (r + 2) = 2n$ $\Rightarrow 4r + 2 = 2n$ $\Rightarrow 2n = 4r + 2$ $\Rightarrow n = 2r + 1$.
Thus,the correct relation is $n = 2r + 1$.

(A) The given equation of the pair of lines is $f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
To find the point of intersection,we take partial derivatives with respect to $x$ and $y$:
$\frac{\partial f}{\partial x} = 2ax + 2hy + 2g = 0 \implies ax + hy + g = 0$
$\frac{\partial f}{\partial y} = 2hx + 2by + 2f = 0 \implies hx + by + f = 0$
Since the lines intersect on the $y$-axis,the $x$-coordinate of the point of intersection is $x = 0$.
Substituting $x = 0$ into the first partial derivative equation: $h(y) + g = 0 \implies y = -g/h$.
Now,substitute $(0, -g/h)$ into the original equation $f(x, y) = 0$:
$a(0)^2 + 2h(0)(-g/h) + b(-g/h)^2 + 2g(0) + 2f(-g/h) + c = 0$
$b(g^2/h^2) - 2fg/h + c = 0$
Multiplying by $h^2$,we get $bg^2 - 2fgh + ch^2 = 0$,which simplifies to $bg^2 + ch^2 = 2fgh$.

(A) Let $(h, k)$ be the center of a circle with radius $r = 3$. The equation of such a circle is $(x - h)^2 + (y - k)^2 = 3^2 = 9$.
Since the center $(h, k)$ lies on the circle $x^2 + y^2 = 25$,the distance of the center from the origin is $\sqrt{h^2 + k^2} = 5$.
Any point $(x, y)$ on a circle with center $(h, k)$ and radius $3$ satisfies the condition that its distance from $(h, k)$ is $3$.
By the triangle inequality,the distance $d$ of any point $(x, y)$ from the origin satisfies $|\sqrt{h^2 + k^2} - 3| \le \sqrt{x^2 + y^2} \le \sqrt{h^2 + k^2} + 3$.
Substituting $\sqrt{h^2 + k^2} = 5$,we get $|5 - 3| \le \sqrt{x^2 + y^2} \le 5 + 3$.
This simplifies to $2 \le \sqrt{x^2 + y^2} \le 8$.
Squaring the inequality,we get $4 \le x^2 + y^2 \le 64$.

(B) The equation of any tangent to the parabola ${y^2} = 8ax$ is given by $y = mx + \frac{2a}{m}$.
This line is also a tangent to the circle ${x^2} + {y^2} = 2{a^2}$.
The perpendicular distance from the center $(0, 0)$ of the circle to the line $mx - y + \frac{2a}{m} = 0$ must be equal to the radius $r = \sqrt{2}a$.
Thus,$\frac{|\frac{2a}{m}|}{\sqrt{m^2 + 1}} = \sqrt{2}a$.
Squaring both sides,we get $\frac{4a^2}{m^2(m^2 + 1)} = 2a^2$.
This simplifies to $m^2(m^2 + 1) = 2$,or $m^4 + m^2 - 2 = 0$.
Factoring the quadratic in $m^2$,we have $(m^2 - 1)(m^2 + 2) = 0$.
Since $m^2$ must be real and positive,we have $m^2 = 1$,which gives $m = \pm 1$.
Substituting $m = \pm 1$ into the tangent equation $y = mx + \frac{2a}{m}$,we get $y = \pm (x + 2a)$.

(A) We know that $\mathop {\lim }\limits_{x \to \infty } {(1 + f(x))^{g(x)}} = e^{\mathop {\lim }\limits_{x \to \infty } f(x)g(x)}$ if the form is $1^\infty$.
Here,$\mathop {\lim }\limits_{x \to \infty } \frac{x^2 + 5x + 3}{x^2 + x + 3} = 1$.
So,the expression is of the form $1^\infty$.
Let $f(x) = \frac{x^2 + 5x + 3}{x^2 + x + 3} - 1 = \frac{x^2 + 5x + 3 - x^2 - x - 3}{x^2 + x + 3} = \frac{4x}{x^2 + x + 3}$.
Then,$\mathop {\lim }\limits_{x \to \infty } f(x)g(x) = \mathop {\lim }\limits_{x \to \infty } \left( \frac{4x}{x^2 + x + 3} \cdot x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{4x^2}{x^2 + x + 3}$.
Dividing numerator and denominator by $x^2$,we get $\mathop {\lim }\limits_{x \to \infty } \frac{4}{1 + \frac{1}{x} + \frac{3}{x^2}} = \frac{4}{1 + 0 + 0} = 4$.
Therefore,the limit is $e^4$.

(D) Let $\alpha, \beta$ be the roots of $x^2+ax+b=0$ and $\gamma, \delta$ be the roots of $x^2+bx+a=0$.
From the relations between roots and coefficients:
$\alpha+\beta = -a, \alpha\beta = b$
$\gamma+\delta = -b, \gamma\delta = a$
Given that the difference between the roots is the same:
$|\alpha-\beta| = |\gamma-\delta|$
Squaring both sides:
$(\alpha-\beta)^2 = (\gamma-\delta)^2$
$(\alpha+\beta)^2 - 4\alpha\beta = (\gamma+\delta)^2 - 4\gamma\delta$
$(-a)^2 - 4b = (-b)^2 - 4a$
$a^2 - 4b = b^2 - 4a$
$a^2 - b^2 + 4a - 4b = 0$
$(a-b)(a+b) + 4(a-b) = 0$
$(a-b)(a+b+4) = 0$
Since $a \neq b$,we have $a-b \neq 0$.
Therefore,$a+b+4 = 0$.

(B) Given that $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$
Since $2b = a + c$ for an $A.P.$,we have:
$2 \log _9(3^{1-x}+2) = 1 + \log _3(4 \cdot 3^x-1)$
Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we get $\log_9(y) = \frac{1}{2} \log_3(y)$:
$2 \cdot \frac{1}{2} \log _3(3^{1-x}+2) = \log _3 3 + \log _3(4 \cdot 3^x-1)$
$\log _3(3^{1-x}+2) = \log _3(3(4 \cdot 3^x-1))$
$3^{1-x}+2 = 12 \cdot 3^x - 3$
Let $3^x = t$. Then $\frac{3}{t} + 2 = 12t - 3$
$3 + 2t = 12t^2 - 3t$
$12t^2 - 5t - 3 = 0$
$(4t - 3)(3t + 1) = 0$
Since $t = 3^x > 0$,we have $t = \frac{3}{4}$.
$3^x = \frac{3}{4} \Rightarrow x = \log_3 \left(\frac{3}{4}\right) = \log_3 3 - \log_3 4 = 1 - \log_3 4$.

(A) Let $f(x) = ax^2 + bx + c$.
Define $F(x) = \int_{0}^{x} f(t) dt = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx$.
Clearly,$F(0) = 0$.
Also,$F(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6}$.
Given $2a + 3b + 6c = 0$,we have $F(1) = 0$.
Since $F(0) = F(1) = 0$ and $F(x)$ is a polynomial (and thus continuous and differentiable),by Rolle's Theorem,there exists at least one $x \in (0, 1)$ such that $F'(x) = 0$.
Since $F'(x) = f(x) = ax^2 + bx + c$,there exists at least one root of $ax^2 + bx + c = 0$ in the interval $(0, 1)$.

(C) Let $\Delta = \left| \begin{array}{ccc} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0 \end{array} \right|$.
Applying the row operation $R_3 \to R_3 - xR_1 - R_2$,we get:
$\Delta = \left| \begin{array}{ccc} a & b & ax + b \\ b & c & bx + c \\ 0 & 0 & -(ax^2 + 2bx + c) \end{array} \right|$.
Expanding along the third row:
$\Delta = -(ax^2 + 2bx + c) \times (ac - b^2) = (b^2 - ac)(ax^2 + 2bx + c)$.
Given that the discriminant $D = (2b)^2 - 4ac = 4(b^2 - ac) < 0$,it follows that $b^2 - ac < 0$.
Since $a > 0$ and the discriminant is negative,the quadratic expression $ax^2 + 2bx + c$ is always positive for all $x \in \mathbb{R}$.
Therefore,$\Delta = (b^2 - ac)(ax^2 + 2bx + c)$ is the product of a negative value and a positive value,which results in a negative value.

(D) The system of equations is given by:
$x + y + z = 6$
$x + 2y + 3z = 10$
$x + 2y + \lambda z = 12$
$A$ system of linear equations $AX = B$ is inconsistent if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.
First,calculate the determinant $D$:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix}$
Expanding along the first row:
$D = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2)$
$D = 2\lambda - 6 - \lambda + 3 + 0$
$D = \lambda - 3$
For the system to be inconsistent,we set $D = 0$:
$\lambda - 3 = 0 \Rightarrow \lambda = 3$
Now,check for consistency at $\lambda = 3$ by calculating $D_z$:
$D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 2 & 12 \end{vmatrix}$
$D_z = 1(24 - 20) - 1(12 - 10) + 6(2 - 2)$
$D_z = 1(4) - 1(2) + 6(0) = 4 - 2 = 2$
Since $D = 0$ and $D_z \neq 0$,the system is inconsistent when $\lambda = 3$.

(D) Let $A$ be the first term and $R$ be the common ratio of the $G$.$P$. Then,
$l = A R^{p-1} \Rightarrow \log l = \log A + (p-1) \log R$ ... $(i)$
$m = A R^{q-1} \Rightarrow \log m = \log A + (q-1) \log R$ ... $(ii)$
$n = A R^{r-1} \Rightarrow \log n = \log A + (r-1) \log R$ ... $(iii)$
Let the determinant be $\Delta = \left| \begin{array}{ccc} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1 \end{array} \right|$.
Applying the column operations $C_1 \to C_1 - (\log A) C_3$ and then $C_1 \to C_1 / (\log R)$,we observe that the first column becomes proportional to the second column shifted by a constant. More simply,applying $C_1 \to C_1 - (\log A) C_3$,we get $\log l - \log A = (p-1) \log R$,etc. Since the columns become linearly dependent (specifically,$C_1$ is a linear combination of $C_2$ and $C_3$),the value of the determinant is $0$.

(A) Given: $\cot^{-1}[(\cos \alpha)^{1/2}] - \tan^{-1}[(\cos \alpha)^{1/2}] = x$
Using the identity $\cot^{-1}(y) = \tan^{-1}(\frac{1}{y})$,we get:
$\tan^{-1}[\frac{1}{\sqrt{\cos \alpha}}] - \tan^{-1}[\sqrt{\cos \alpha}] = x$
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}(\frac{A-B}{1+AB})$:
$\tan^{-1}[\frac{\frac{1}{\sqrt{\cos \alpha}} - \sqrt{\cos \alpha}}{1 + (\frac{1}{\sqrt{\cos \alpha}})(\sqrt{\cos \alpha})}] = x$
$\tan^{-1}[\frac{\frac{1-\cos \alpha}{\sqrt{\cos \alpha}}}{1+1}] = x$
$\tan x = \frac{1-\cos \alpha}{2\sqrt{\cos \alpha}}$
Now,we need to find $\sin x$. Using the identity $\sin x = \frac{\tan x}{\sqrt{1+\tan^2 x}}$ or by constructing a right triangle where the opposite side is $1-\cos \alpha$ and the adjacent side is $2\sqrt{\cos \alpha}$:
Hypotenuse = $\sqrt{(1-\cos \alpha)^2 + (2\sqrt{\cos \alpha})^2} = \sqrt{1 - 2\cos \alpha + \cos^2 \alpha + 4\cos \alpha} = \sqrt{1 + 2\cos \alpha + \cos^2 \alpha} = \sqrt{(1+\cos \alpha)^2} = 1+\cos \alpha$
Therefore,$\sin x = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1-\cos \alpha}{1+\cos \alpha} = \frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)} = \tan^2(\frac{\alpha}{2})$.

(A) Let the two forces be $P$ and $Q$,where $Q > P$.
Given that the sum of the forces is $P + Q = 18 \ N$.
Let the resultant $R = 12 \ N$ be perpendicular to the smaller force $P$.
The angle between the resultant $R$ and the force $P$ is $90^o$.
The formula for the direction of the resultant is given by $\tan \alpha = \frac{Q \sin \theta}{P + Q \cos \theta}$,where $\theta$ is the angle between the forces.
Since the resultant is perpendicular to $P$,$\tan 90^o = \infty$,which implies $P + Q \cos \theta = 0$,or $\cos \theta = -\frac{P}{Q}$.
Using the magnitude formula for the resultant: $R^2 = P^2 + Q^2 + 2PQ \cos \theta$.
Substituting $\cos \theta = -\frac{P}{Q}$:
$12^2 = P^2 + Q^2 + 2PQ(-\frac{P}{Q}) = P^2 + Q^2 - 2P^2 = Q^2 - P^2$.
$144 = (Q - P)(Q + P)$.
Since $Q + P = 18$,we have $144 = (Q - P) \times 18$,so $Q - P = 8$.
Solving the system $Q + P = 18$ and $Q - P = 8$:
Adding the equations: $2Q = 26 \Rightarrow Q = 13 \ N$.
Subtracting the equations: $2P = 10 \Rightarrow P = 5 \ N$.
Thus,the magnitudes of the two forces are $13 \ N$ and $5 \ N$.

(D) Given that $|a| = 3$,$|b| = 4$,and $|c| = 5$.
Also,$a + b + c = 0$.
Squaring both sides of the equation $a + b + c = 0$,we get:
$|a + b + c|^2 = 0^2$
$|a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
Substituting the given magnitudes:
$3^2 + 4^2 + 5^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
$9 + 16 + 25 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
$50 + 2(a \cdot b + b \cdot c + c \cdot a) = 0$
$2(a \cdot b + b \cdot c + c \cdot a) = -50$
$a \cdot b + b \cdot c + c \cdot a = -25$.

(C) Given,$a + b + c = 0 \Rightarrow a + b = -c$.
Squaring both sides,we get $|a + b|^2 = |-c|^2$.
Using the property $|u|^2 = u \cdot u$,we have $|a|^2 + |b|^2 + 2(a \cdot b) = |c|^2$.
Since $a \cdot b = |a||b| \cos \theta$,where $\theta$ is the angle between $a$ and $b$,we have $|a|^2 + |b|^2 + 2|a||b| \cos \theta = |c|^2$.
Substituting the given values $|a| = 3, |b| = 5, |c| = 7$:
$3^2 + 5^2 + 2(3)(5) \cos \theta = 7^2$.
$9 + 25 + 30 \cos \theta = 49$.
$34 + 30 \cos \theta = 49$.
$30 \cos \theta = 49 - 34 = 15$.
$\cos \theta = \frac{15}{30} = \frac{1}{2}$.
Therefore,$\theta = 60^\circ$.

(C) Given that $a + b + c = 0.$
Taking the cross product with $a$ on both sides:
$a \times (a + b + c) = a \times 0$
$a \times a + a \times b + a \times c = 0$
Since $a \times a = 0,$ we have $a \times b + a \times c = 0,$ which implies $a \times b = - (a \times c) = c \times a$ .....$(i)$
Similarly,taking the cross product with $b$ on both sides:
$b \times (a + b + c) = b \times 0$
$b \times a + b \times b + b \times c = 0$
Since $b \times b = 0,$ we have $b \times a + b \times c = 0,$ which implies $-(a \times b) = b \times c,$ or $a \times b = b \times c$ .....$(ii)$
From $(i)$ and $(ii),$ we get $a \times b = b \times c = c \times a.$

(B) We know that the magnitude of the cross product of two vectors $a$ and $b$ is given by $|a \times b| = |a||b| \sin \theta$,where $\theta$ is the angle between the vectors.
Given $|a| = 4$,$|b| = 2$,and $\theta = \frac{\pi}{6}$.
Substituting these values,we get $|a \times b| = 4 \times 2 \times \sin\left(\frac{\pi}{6}\right)$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $|a \times b| = 8 \times \frac{1}{2} = 4$.
Therefore,$|a \times b|^2 = 4^2 = 16$.

(A) We know that the scalar triple product of three vectors $x, y, z$ is given by $[x, y, z] = x \cdot (y \times z)$.
Let $x = a \times b$,$y = b \times c$,and $z = c \times a$.
Then $[a \times b, b \times c, c \times a] = (a \times b) \cdot ((b \times c) \times (c \times a))$.
Using the vector triple product identity $(p \times q) \times r = (p \cdot r)q - (q \cdot r)p$,we have:
$(b \times c) \times (c \times a) = ([b, c, a]c - [b, c, c]a)$.
Since $[b, c, c] = 0$ and $[b, c, a] = [a, b, c]$,we get:
$(b \times c) \times (c \times a) = [a, b, c]c$.
Substituting this back into the expression:
$[a \times b, b \times c, c \times a] = (a \times b) \cdot ([a, b, c]c) = [a, b, c] (a \times b) \cdot c = [a, b, c] [a, b, c]$.
Given $[a, b, c] = 4$,we have:
$[a \times b, b \times c, c \times a] = 4 \times 4 = 16$.

(A) The equation of a plane passing through the point $(3, 2, 0)$ is given by $A(x - 3) + B(y - 2) + C(z - 0) = 0 \dots (i)$.
Since the plane contains the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$,it must pass through the point $(3, 6, 4)$ on the line.
Substituting $(3, 6, 4)$ into $(i)$,we get $A(3 - 3) + B(6 - 2) + C(4 - 0) = 0$,which simplifies to $4B + 4C = 0$,or $B + C = 0 \dots (ii)$.
Also,the normal vector of the plane $(A, B, C)$ must be perpendicular to the direction vector of the line $(1, 5, 4)$. Thus,$1A + 5B + 4C = 0 \dots (iii)$.
From $(ii)$,$C = -B$. Substituting into $(iii)$,we get $A + 5B - 4B = 0$,so $A = -B$.
Let $B = -1$,then $A = 1$ and $C = 1$.
Substituting these into $(i)$,we get $1(x - 3) - 1(y - 2) + 1(z - 0) = 0$,which simplifies to $x - y + z = 1$.

(B) The equation of a plane passing through $(1, 0, 0)$ and $(0, 1, 0)$ can be written in intercept form as $\frac{x}{1} + \frac{y}{1} + \frac{z}{c} = 1$,where $c$ is the $z$-intercept.
This simplifies to $x + y + \frac{z}{c} = 1$.
The direction ratios of the normal to this plane are $(1, 1, \frac{1}{c})$.
The given plane is $x + y = 3$,and its normal vector has direction ratios $(1, 1, 0)$.
The angle $\theta$ between two planes with normals $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Given $\theta = \frac{\pi}{4}$,we have $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.
Substituting the values: $\frac{1}{\sqrt{2}} = \frac{|1(1) + 1(1) + \frac{1}{c}(0)|}{\sqrt{1^2 + 1^2 + (\frac{1}{c})^2} \sqrt{1^2 + 1^2 + 0^2}}$.
$\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2 + \frac{1}{c^2}} \cdot \sqrt{2}}$.
$\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2} \sqrt{2 + \frac{1}{c^2}}}$.
$1 = \frac{2}{\sqrt{2 + \frac{1}{c^2}}}$.
$\sqrt{2 + \frac{1}{c^2}} = 2$.
Squaring both sides: $2 + \frac{1}{c^2} = 4$.
$\frac{1}{c^2} = 2 \Rightarrow c^2 = \frac{1}{2} \Rightarrow c = \frac{1}{\sqrt{2}}$.
The direction ratios are $(1, 1, \frac{1}{c}) = (1, 1, \sqrt{2})$.

(A) The given function is $f(x) = \sin^{-1} \left[ \log_3 \left( \frac{x}{3} \right) \right]$.
For the function $\sin^{-1}(u)$ to be defined,the argument $u$ must satisfy $-1 \le u \le 1$.
Therefore,we must have $-1 \le \log_3 \left( \frac{x}{3} \right) \le 1$.
Applying the definition of the logarithm,we exponentiate with base $3$:
$3^{-1} \le \frac{x}{3} \le 3^1$.
This simplifies to $\frac{1}{3} \le \frac{x}{3} \le 3$.
Multiplying the entire inequality by $3$,we get $1 \le x \le 9$.
Thus,the domain is $x \in [1, 9]$.

(A) Given $y = (x + \sqrt{1 + x^2})^n$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot (1 + \frac{x}{\sqrt{1 + x^2}})$
$\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \cdot \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}}$
$\frac{dy}{dx} = \frac{n(x + \sqrt{1 + x^2})^n}{\sqrt{1 + x^2}}$
$\sqrt{1 + x^2} \frac{dy}{dx} = ny$.
Differentiating both sides with respect to $x$ again:
$\frac{d}{dx}(\sqrt{1 + x^2} \frac{dy}{dx}) = \frac{d}{dx}(ny)$
$\sqrt{1 + x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x}{\sqrt{1 + x^2}} = n \frac{dy}{dx}$
Multiply both sides by $\sqrt{1 + x^2}$:
$(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n \sqrt{1 + x^2} \frac{dy}{dx}$
Since $\sqrt{1 + x^2} \frac{dy}{dx} = ny$,we substitute this into the equation:
$(1 + x^2) \frac{d^2y}{dx^2} + x \frac{dy}{dx} = n(ny) = n^2y$.

(D) Given $f''(x) - g''(x) = 0$.
Integrating with respect to $x$,we get $f'(x) - g'(x) = c$,where $c$ is a constant.
At $x = 1$,$f'(1) - g'(1) = c \implies 2 - 4 = c \implies c = -2$.
Thus,$f'(x) - g'(x) = -2$.
Integrating again with respect to $x$,we get $f(x) - g(x) = -2x + c_1$,where $c_1$ is a constant.
At $x = 2$,$f(2) - g(2) = -2(2) + c_1 \implies 3 - 9 = -4 + c_1 \implies -6 = -4 + c_1 \implies c_1 = -2$.
Therefore,$f(x) - g(x) = -2x - 2$.
At $x = 3/2$,$f(3/2) - g(3/2) = -2(3/2) - 2 = -3 - 2 = -5$.

(C) Given the functional equation $f(x + y) = f(x)f(y)$.
Setting $y = 0$,we get $f(x + 0) = f(x)f(0)$,which implies $f(x) = f(x)f(0)$. Since $f(5) = 2$,$f(x)$ is not identically zero,so $f(0) = 1$.
By the definition of the derivative,$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$.
Substituting $f(x + h) = f(x)f(h)$,we get $f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$.
Since $f(0) = 1$,this is $f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - f(0)}{h} = f(x)f'(0)$.
Given $f(5) = 2$ and $f'(0) = 3$,we have $f'(5) = f(5)f'(0) = 2 \times 3 = 6$.

(C) Let $I = \int_{0}^{\sqrt{2}} [x^2] \, dx$.
Since the function $[x^2]$ changes its value at points where $x^2$ is an integer,we split the interval $[0, \sqrt{2}]$.
For $0 \le x < 1$,$0 \le x^2 < 1$,so $[x^2] = 0$.
For $1 \le x \le \sqrt{2}$,$1 \le x^2 \le 2$,so $[x^2] = 1$.
Thus,$I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx$.
$I = 0 + [x]_{1}^{\sqrt{2}}$.
$I = \sqrt{2} - 1$.

(A) We are given the limit: $\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}}$
This can be written in summation notation as: $\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} \sum\limits_{r = 1}^n {\left( {\frac{r}{n}} \right)^p}$
Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n} \sum\limits_{r = 1}^n {f\left( {\frac{r}{n}} \right)} = \int_0^1 {f(x)dx}$,where $f(x) = x^p$.
Thus,the integral becomes: $\int_0^1 {x^p dx}$
Evaluating the integral: $\left[ {\frac{{{x^{p + 1}}}}{{p + 1}}} \right]_0^1 = \frac{1^{p+1}}{p+1} - \frac{0^{p+1}}{p+1} = \frac{1}{{p + 1}}$.

(B) Given ${I_n} = \int_{0}^{\pi /4} {\tan^n x} \,dx$.
We know that $\tan^2 x = \sec^2 x - 1$.
So,${I_n} = \int_{0}^{\pi /4} {\tan^{n-2} x (\sec^2 x - 1)} \,dx$.
${I_n} = \int_{0}^{\pi /4} {\tan^{n-2} x \sec^2 x} \,dx - \int_{0}^{\pi /4} {\tan^{n-2} x} \,dx$.
${I_n} = \left[ \frac{\tan^{n-1} x}{n-1} \right]_{0}^{\pi /4} - {I_{n-2}}$.
Since $\tan(\pi/4) = 1$ and $\tan(0) = 0$,we get ${I_n} = \frac{1}{n-1} - {I_{n-2}}$.
Thus,${I_n} + {I_{n-2}} = \frac{1}{n-1}$.
Now,we need to find $\lim_{n \to \infty} n[{I_n} + {I_{n-2}}]$.
$\lim_{n \to \infty} n \left( \frac{1}{n-1} \right) = \lim_{n \to \infty} \frac{n}{n-1} = \lim_{n \to \infty} \frac{1}{1 - 1/n} = 1$.

(A) We know that $\ln x$ is defined for $x > 0$ and $\ln |x|$ is defined for all $x \in \mathbb{R} - \{0\}$.
Also,$|\ln x| \ge 0$ and $|\ln |x|| \ge 0$.
The region bounded by these curves is symmetric about both the $x$-axis and the $y$-axis.
The area is given by $4 \times \int_{0}^{1} |\ln x| \, dx$.
Since for $x \in (0, 1)$,$\ln x < 0$,we have $|\ln x| = -\ln x$.
Area $= -4 \int_{0}^{1} \ln x \, dx$.
Using integration by parts,$\int \ln x \, dx = x \ln x - x$.
Area $= -4 [x \ln x - x]_{0}^{1} = -4 [(1 \ln 1 - 1) - (\lim_{x \to 0^+} x \ln x - 0)]$.
Since $\lim_{x \to 0^+} x \ln x = 0$,the area $= -4 [0 - 1 - 0] = -4(-1) = 4 \, sq. \, units$.

(D) We know that the function $f(x) = |\sin x|$ is periodic with period $\pi$.
Using the property $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$,where $T$ is the period,we have:
$\int_{\,\pi }^{\,10\pi } {|\sin x|dx} = \int_{\,0}^{\,9\pi } {|\sin x|dx}$ (by shifting the limits by $-\pi$).
Since the period is $\pi$,we can write this as:
$9 \int_{\,0}^{\,\pi } {|\sin x|dx} = 9 \int_{\,0}^{\,\pi } {\sin x dx}$ (since $\sin x \ge 0$ in $[0, \pi]$).
Evaluating the integral:
$9 [-\cos x]_{0}^{\pi} = 9 (-(\cos \pi) + \cos 0) = 9 (-(-1) + 1) = 9(1 + 1) = 18$.

(B) Let $I = \int_{ - \pi }^\pi {\frac{{2x(1 + \sin x)}}{{1 + {{\cos }^2}x}}dx} $.
We can split the integral into two parts:
$I = \int_{ - \pi }^\pi {\frac{{2x}}{{1 + {{\cos }^2}x}}dx} + \int_{ - \pi }^\pi {\frac{{2x\sin x}}{{1 + {{\cos }^2}x}}dx} $.
For the first part,let $f(x) = \frac{2x}{1 + \cos^2 x}$. Since $f(-x) = \frac{-2x}{1 + \cos^2(-x)} = -f(x)$,the function is odd. Thus,$\int_{-\pi}^{\pi} f(x) dx = 0$.
For the second part,let $g(x) = \frac{2x\sin x}{1 + \cos^2 x}$. Since $g(-x) = \frac{2(-x)\sin(-x)}{1 + \cos^2(-x)} = \frac{2x\sin x}{1 + \cos^2 x} = g(x)$,the function is even. Thus,$\int_{-\pi}^{\pi} g(x) dx = 2 \int_0^{\pi} g(x) dx$.
So,$I = 2 \int_0^{\pi} \frac{2x\sin x}{1 + \cos^2 x} dx = 4 \int_0^{\pi} \frac{x\sin x}{1 + \cos^2 x} dx$ ... $(i)$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:
$I = 4 \int_0^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} dx = 4 \int_0^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x} dx$ ... $(ii)$.
Adding $(i)$ and $(ii)$:
$2I = 4 \int_0^{\pi} \frac{x\sin x + (\pi - x)\sin x}{1 + \cos^2 x} dx = 4\pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
$I = 2\pi \int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$.
Let $t = \cos x$,then $dt = -\sin x dx$. When $x=0, t=1$; when $x=\pi, t=-1$.
$I = 2\pi \int_1^{-1} \frac{-dt}{1 + t^2} = 2\pi \int_{-1}^1 \frac{dt}{1 + t^2} = 2\pi [\tan^{-1} t]_{-1}^1$.
$I = 2\pi (\tan^{-1}(1) - \tan^{-1}(-1)) = 2\pi (\frac{\pi}{4} - (-\frac{\pi}{4})) = 2\pi (\frac{\pi}{2}) = \pi^2$.

(B) Given the differential equation: $\frac{d^2y}{dx^2} = e^{-2x}$.
Integrating both sides with respect to $x$ once:
$\int \frac{d^2y}{dx^2} dx = \int e^{-2x} dx$
$\frac{dy}{dx} = \frac{e^{-2x}}{-2} + c = -\frac{1}{2}e^{-2x} + c$.
Integrating both sides with respect to $x$ again:
$\int \frac{dy}{dx} dx = \int (-\frac{1}{2}e^{-2x} + c) dx$
$y = -\frac{1}{2} \cdot \frac{e^{-2x}}{-2} + cx + d$
$y = \frac{1}{4}e^{-2x} + cx + d$.
Thus,the correct option is $B$.

(C) To find the order and degree of the differential equation,we must first eliminate the fractional exponent by raising both sides to the power of $3$.
Given equation: ${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^3}y}}{{d{x^3}}}$
Cubing both sides:
${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^2} = {\left( {4\frac{{{d^3}y}}{{d{x^3}}}} \right)^3}$
Now,the equation is a polynomial in terms of derivatives.
The order of a differential equation is the highest derivative present,which is $\frac{{{d^3}y}}{{d{x^3}}}$,so the order is $3$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{{{d^3}y}}{{d{x^3}}}$ is $3$.
Thus,the order is $3$ and the degree is $3$.

(D) This is a binomial distribution problem where $n = 5$ trials are performed.
The probability of getting an odd number (success) in a single toss of a die is $p = \frac{3}{6} = \frac{1}{2}$.
The probability of failure is $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
The variance of a binomial distribution is given by the formula $\text{Variance} = npq$.
Substituting the values: $\text{Variance} = 5 \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{4}$.

(B) Given vectors are $a = 3i - 5j$ and $b = 6i + 3j$.
First,we calculate the cross product $c = a \times b$:
$c = \begin{vmatrix} i & j & k \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{vmatrix} = i(0 - 0) - j(0 - 0) + k(9 - (-30)) = 39k$.
Now,we find the magnitudes of the vectors:
$|a| = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$.
$|b| = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45}$.
$|c| = \sqrt{0^2 + 0^2 + 39^2} = 39$.
Therefore,the ratio $|a|:|b|:|c|$ is $\sqrt{34} : \sqrt{45} : 39$.