WBJEE 2011 Mathematics Question Paper with Answer and Solution

(C) Given equations are:
$1$) $\log _3 x + \log _3 y = 2 + \log _3 2$
$2$) $\log _3(x+y) = 2$
From equation $(1)$,using the property $\log_a b + \log_a c = \log_a(bc)$:
$\log _3(xy) = \log _3(3^2) + \log _3 2$
$\log _3(xy) = \log _3 9 + \log _3 2$
$\log _3(xy) = \log _3(9 \times 2) = \log _3 18$
So,$xy = 18$.
From equation $(2)$,using the definition of logarithm:
$x + y = 3^2 = 9$.
We have a system of equations:
$x + y = 9$
$xy = 18$
These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$:
$t^2 - 9t + 18 = 0$
$(t-3)(t-6) = 0$
So,$t = 3$ or $t = 6$.
Thus,the solutions are $(x=3, y=6)$ or $(x=6, y=3)$.
Comparing with the given options,the correct choice is $x=3, y=6$.

(A) Given that $\log _{7} 2 = \lambda$.
We need to find the value of $\log _{49} (28)$.
Using the change of base formula $\log _{a^n} b = \frac{1}{n} \log _{a} b$,we have:
$\log _{49} (28) = \log _{7^2} (28) = \frac{1}{2} \log _{7} (28)$.
Since $28 = 4 \times 7 = 2^2 \times 7$,we can write:
$\frac{1}{2} \log _{7} (2^2 \times 7) = \frac{1}{2} [\log _{7} (2^2) + \log _{7} 7]$.
Using the property $\log _{a} (x^n) = n \log _{a} x$ and $\log _{a} a = 1$:
$\frac{1}{2} [2 \log _{7} 2 + 1] = \frac{1}{2} (2 \lambda + 1)$.

(B) Given that $\sin \theta$ and $\cos \theta$ are the roots of the quadratic equation $ax^2 - bx + c = 0$.
From the properties of roots,the sum of roots is $\sin \theta + \cos \theta = \frac{b}{a}$ and the product of roots is $\sin \theta \cdot \cos \theta = \frac{c}{a}$.
Squaring the sum of roots,we get $(\sin \theta + \cos \theta)^2 = (\frac{b}{a})^2$.
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = \frac{b^2}{a^2}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $1 + 2(\frac{c}{a}) = \frac{b^2}{a^2}$.
Multiplying by $a^2$,we get $a^2 + 2ac = b^2$.
Rearranging the terms,we obtain $a^2 - b^2 + 2ac = 0$.

(B) Let the roots of the equation $px^2+qx+r=0$ be $a\alpha$ and $b\alpha$.
From the relation between roots and coefficients:
Sum of roots: $a\alpha + b\alpha = (a+b)\alpha = -\frac{q}{p}$
Product of roots: $(a\alpha)(b\alpha) = ab\alpha^2 = \frac{r}{p}$
Now,consider the ratio:
$\frac{ab\alpha^2}{(a+b)^2\alpha^2} = \frac{ab}{(a+b)^2}$
Substituting the values:
$\frac{ab}{(a+b)^2} = \frac{r/p}{(-q/p)^2} = \frac{r}{p} \times \frac{p^2}{q^2} = \frac{pr}{q^2}$

(D) The roots of the equation $x^2+x+1=0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We know that $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$,which implies $\omega + \omega^2 = -1$.
Calculating the new roots:
$\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.
$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
The required quadratic equation with roots $\alpha^{19}$ and $\beta^7$ is $x^2 - (\alpha^{19} + \beta^7)x + (\alpha^{19} \cdot \beta^7) = 0$.
Substituting the values:
$x^2 - (\omega + \omega^2)x + (\omega \cdot \omega^2) = 0$.
Since $\omega + \omega^2 = -1$ and $\omega^3 = 1$,the equation becomes:
$x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

(A) Given $x+\frac{1}{x}=2 \cos \theta$.
Let $x = \cos \theta + i \sin \theta$.
Then by De Moivre's Theorem,$x^n = \cos n \theta + i \sin n \theta$.
Also,$\frac{1}{x} = x^{-1} = \cos(-\theta) + i \sin(-\theta) = \cos \theta - i \sin \theta$.
Thus,$\frac{1}{x^n} = \cos n \theta - i \sin n \theta$.
Adding these two expressions:
$x^n + \frac{1}{x^n} = (\cos n \theta + i \sin n \theta) + (\cos n \theta - i \sin n \theta) = 2 \cos n \theta$.

(C) The number of diagonals in a polygon with $n$ sides is given by the formula: $\frac{n(n-3)}{2} = 20$.
Multiplying by $2$,we get $n(n-3) = 40$.
$n^2 - 3n - 40 = 0$.
Factoring the quadratic equation: $(n-8)(n+5) = 0$.
Since the number of sides $n$ must be positive,we have $n = 8$.

(B) Let the terms of the sequence be $T_1, T_2, T_3, \ldots$ where $T_1 = \log a$,$T_2 = \log \frac{a^2}{b}$,and $T_3 = \log \frac{a^3}{b^2}$.
Using the logarithmic property $\log \frac{x}{y} = \log x - \log y$ and $\log x^n = n \log x$:
$T_1 = \log a$
$T_2 = \log a^2 - \log b = 2 \log a - \log b$
$T_3 = \log a^3 - \log b^2 = 3 \log a - 2 \log b$
Now,check the common difference $d = T_2 - T_1 = (2 \log a - \log b) - \log a = \log a - \log b$.
Check $T_3 - T_2 = (3 \log a - 2 \log b) - (2 \log a - \log b) = \log a - \log b$.
Since $T_2 - T_1 = T_3 - T_2$,the sequence is an $A$.$P$. with common difference $d = \log a - \log b$.

(B) Given that $\sin A, \sin B, \sin C$ are in $A.P.$
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,where $R$ is the circumradius.
Thus,$a, b, c$ are in $A.P.$
Let $p_1, p_2, p_3$ be the altitudes corresponding to sides $a, b, c$ respectively.
The area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
This implies $p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.
Since $a, b, c$ are in $A.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$
Therefore,$\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$
Hence,the altitudes $p_1, p_2, p_3$ are in $H.P.$

(A) Given the series $S = 1 + 2\omega + 3\omega^2 + \dots + 3n\omega^{3n-1}$.
Multiply by $\omega$: $S\omega = \omega + 2\omega^2 + 3\omega^3 + \dots + 3n\omega^{3n}$.
Subtracting the two equations: $S(1 - \omega) = 1 + \omega + \omega^2 + \dots + \omega^{3n-1} - 3n\omega^{3n}$.
Since $\omega^3 = 1$,the sum of the geometric series $1 + \omega + \dots + \omega^{3n-1}$ is $\frac{1 - (\omega^3)^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0$.
Thus,$S(1 - \omega) = 0 - 3n(1) = -3n$.
Therefore,$S = \frac{-3n}{1 - \omega} = \frac{3n}{\omega - 1}$.

(D) Given expression: $\frac{e^{7x}+e^x}{e^{3x}} = e^{4x} + e^{-2x}$.
Using the Taylor series expansion for $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$,we have:
$e^{4x} = \sum_{n=0}^{\infty} \frac{(4x)^n}{n!} = \sum_{n=0}^{\infty} \frac{4^n x^n}{n!}$.
$e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^n x^n}{n!}$.
Adding these two series,the coefficient of $x^n$ is $\frac{4^n + (-2)^n}{n!}$.

(C) Let the series be $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n(n+1)}$.
Using partial fractions,$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Thus,$S = (\frac{1}{1} - \frac{1}{2}) - (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) - (\frac{1}{4} - \frac{1}{5}) + \dots$
$S = 1 - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} - \frac{1}{4} - \frac{1}{4} + \dots$
$S = 1 - 2(\frac{1}{2}) + 2(\frac{1}{3}) - 2(\frac{1}{4}) + 2(\frac{1}{5}) - \dots$
$S = 1 - 2[\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots]$
We know the expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$
For $x=1$,$\log_{e} 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$
So,$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.
Substituting this back,$S = 1 - 2(1 - \log_{e} 2) = 1 - 2 + 2 \log_{e} 2 = 2 \log_{e} 2 - 1$.

(B) The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is given by $A = {}^{2n}C_{n}$.
The coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$ is given by $B = {}^{2n-1}C_{n}$.
Now,we calculate the ratio $A / B$:
$\frac{A}{B} = \frac{{}^{2n}C_{n}}{{}^{2n-1}C_{n}}$
Using the formula ${}^{n}C_{r} = \frac{n}{r} \times {}^{n-1}C_{r-1}$,we have ${}^{2n}C_{n} = \frac{2n}{n} \times {}^{2n-1}C_{n-1}$.
However,a simpler approach is:
$\frac{A}{B} = \frac{\frac{(2n)!}{n!n!}}{\frac{(2n-1)!}{n!(n-1)!}} = \frac{(2n)!}{n!n!} \times \frac{n!(n-1)!}{(2n-1)!} = \frac{2n \times (2n-1)!}{n \times (n-1)! \times n!} \times \frac{n!(n-1)!}{(2n-1)!} = \frac{2n}{n} = 2$.

(A) Using the Binomial Theorem,we can write $(101)^{100}-1$ as $(1+100)^{100}-1$.
Expanding this using the binomial expansion:
$(1+100)^{100}-1 = \left(1 + {}^{100}C_{1}(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}\right) - 1$.
Since ${}^{100}C_{1} = 100$,the expression becomes:
$100(100) + {}^{100}C_{2}(100)^{2} + {}^{100}C_{3}(100)^{3} + \dots + {}^{100}C_{100}(100)^{100}$.
$= 10^{4} + {}^{100}C_{2}(10^{4}) + {}^{100}C_{3}(10^{6}) + \dots + 10^{200}$.
$= 10^{4} \left(1 + {}^{100}C_{2} + {}^{100}C_{3}(10^{2}) + \dots + 10^{196}\right)$.
Thus,the expression is divisible by $10^{4}$.

(C) Using the Binomial Theorem,we have:
$(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + {^nC_3}x^3 + \dots + {^nC_n}x^n$
Since ${^nC_0} = 1$ and ${^nC_1} = n$,we can write:
$(1+x)^n = 1 + nx + {^nC_2}x^2 + {^nC_3}x^3 + \dots + x^n$
Now,subtract $nx$ and $1$ from both sides:
$(1+x)^n - nx - 1 = {^nC_2}x^2 + {^nC_3}x^3 + \dots + x^n$
Factoring out $x^2$ from the right side:
$(1+x)^n - nx - 1 = x^2 ({^nC_2} + {^nC_3}x + \dots + x^{n-2})$
Thus,the expression is divisible by $x^2$.

(A) Given that ${}^nC_4, {}^nC_5, {}^nC_6$ are in $A.P.$
Therefore,$2({}^nC_5) = {}^nC_4 + {}^nC_6$.
Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$2 \times \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$
Dividing by $n!$ and multiplying by $6!(n-4)!$:
$2 \times 6(n-4) = 30 + (n-4)(n-5)$
$12n - 48 = 30 + n^2 - 9n + 20$
$n^2 - 21n + 98 = 0$
$(n-7)(n-14) = 0$
Thus,$n = 7$ or $n = 14$.

(B) We know that the sum of binomial coefficients with odd indices is given by:
$^{n}C_1 + ^{n}C_3 + ^{n}C_5 + \ldots = 2^{n-1}$.
For $n = 15$,we have:
$^{15}C_1 + ^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = 2^{15-1} = 2^{14}$.
We need to find the sum $^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15}$.
This is equal to $2^{14} - ^{15}C_1$.
Since $^{15}C_1 = 15$,the sum is $2^{14} - 15$.

(C) Given $\sin \theta = \frac{2t}{1+t^2}$.
We know that $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{2t}{1+t^2}\right)^2$.
$\cos^2 \theta = 1 - \frac{4t^2}{(1+t^2)^2} = \frac{(1+t^2)^2 - 4t^2}{(1+t^2)^2} = \frac{1 + 2t^2 + t^4 - 4t^2}{(1+t^2)^2} = \frac{1 - 2t^2 + t^4}{(1+t^2)^2} = \frac{(1-t^2)^2}{(1+t^2)^2}$.
Taking the square root,$|\cos \theta| = \frac{|1-t^2|}{1+t^2}$.
Since $\theta$ lies in the second quadrant,$\cos \theta$ must be negative.
Therefore,$\cos \theta = -\frac{|1-t^2|}{1+t^2}$.

(D) Given the equation: $\sin \theta + \cos \theta = 0$
Dividing both sides by $\cos \theta$ (assuming $\cos \theta \neq 0$):
$\frac{\sin \theta}{\cos \theta} + 1 = 0$
$\tan \theta = -1$
Since $0 < \theta < \pi$,the value of $\theta$ where $\tan \theta = -1$ is in the second quadrant.
Therefore,$\theta = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.

(A) We use the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.
Substituting the given values:
$\tan(\alpha + \beta) = \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \left(\frac{a}{a+1}\right) \left(\frac{1}{2a+1}\right)}$
$= \frac{\frac{a(2a+1) + 1(a+1)}{(a+1)(2a+1)}}{\frac{(a+1)(2a+1) - a}{(a+1)(2a+1)}}$
$= \frac{2a^2 + a + a + 1}{2a^2 + a + 2a + 1 - a}$
$= \frac{2a^2 + 2a + 1}{2a^2 + 2a + 1} = 1$
Since $\tan(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

(B) Given that $\theta + \phi = \frac{\pi}{4}$.
Taking $\tan$ on both sides,we get $\tan(\theta + \phi) = \tan(\frac{\pi}{4}) = 1$.
Using the formula $\tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} = 1$.
This implies $\tan \theta + \tan \phi = 1 - \tan \theta \tan \phi$,or $\tan \theta + \tan \phi + \tan \theta \tan \phi = 1$.
Now,consider the expression $(1 + \tan \theta)(1 + \tan \phi) = 1 + \tan \phi + \tan \theta + \tan \theta \tan \phi$.
Substituting the value from the previous step: $1 + (\tan \theta + \tan \phi + \tan \theta \tan \phi) = 1 + 1 = 2$.

(B) We know that $\cos 15^{\circ} = \cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
Also,$\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.
Subtracting the two values:
$\cos 15^{\circ} - \sin 15^{\circ} = \frac{\sqrt{3} + 1}{2\sqrt{2}} - \frac{\sqrt{3} - 1}{2\sqrt{2}} = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2\sqrt{2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

(D) The given equation is of the form $a \sin x + b \cos x = c$,where $a = 2$,$b = 1$,and $c = 3$.
For the equation $a \sin x + b \cos x = c$ to have a solution,the condition $|c| \leq \sqrt{a^2 + b^2}$ must be satisfied.
Here,$\sqrt{a^2 + b^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$.
Since $\sqrt{5} \approx 2.236$,we have $c = 3 > \sqrt{5}$.
Because the maximum value of $2 \sin x + \cos x$ is $\sqrt{5}$,it can never reach the value $3$.
Therefore,the equation has no solution.

(B) Let the point on the line $x+y=4$ be $P(h, 4-h)$.
The perpendicular distance from $P(h, 4-h)$ to the line $4x+3y-10=0$ is given by $d = \frac{|4h + 3(4-h) - 10|}{\sqrt{4^2 + 3^2}}$.
Given $d=1$,we have $\frac{|4h + 12 - 3h - 10|}{5} = 1$.
$|h + 2| = 5$.
This gives two cases:
Case $1$: $h+2 = 5 \implies h = 3$. The point is $(3, 4-3) = (3, 1)$.
Case $2$: $h+2 = -5 \implies h = -7$. The point is $(-7, 4-(-7)) = (-7, 11)$.
Thus,the coordinates are $(3, 1)$ and $(-7, 11)$.

(A) For three points $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$ to be collinear,the area of the triangle formed by them must be zero,or the slope of $AB$ must equal the slope of $BC$.
Using the determinant method:
$\begin{vmatrix} 1 & 6 & 1 \\ 3 & -4 & 1 \\ x & y & 1 \end{vmatrix} = 0$
Expanding along the first row:
$1(-4 - y) - 6(3 - x) + 1(3y - (-4x)) = 0$
$-4 - y - 18 + 6x + 3y + 4x = 0$
$10x + 2y - 22 = 0$
Dividing by $2$:
$5x + y - 11 = 0$

(D) Given equations are:
$x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ $(1)$
$x \sin \theta - (1 + \cos \theta) y = -a \sin \theta$ $(2)$
Subtracting $(2)$ from $(1)$:
$(1 - \cos \theta + 1 + \cos \theta) y = a \sin \theta + a \sin \theta$
$2y = 2a \sin \theta \implies y = a \sin \theta$
Adding $(1)$ and $(2)$:
$2x \sin \theta + (1 - \cos \theta - 1 - \cos \theta) y = 0$
$2x \sin \theta - 2y \cos \theta = 0$
$x \sin \theta = y \cos \theta$
Substitute $y = a \sin \theta$:
$x \sin \theta = (a \sin \theta) \cos \theta$
$x = a \cos \theta$
Now,$x^2 + y^2 = (a \cos \theta)^2 + (a \sin \theta)^2 = a^2(\cos^2 \theta + \sin^2 \theta) = a^2$
Thus,the locus is $x^2 + y^2 = a^2$.

(A) The general equation of a second-degree curve is given by $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$.
Comparing the given equation $8 x^2+12 y^2-4 x+4 y-1=0$ with the general form,we have $a=8$,$b=12$,and $h=0$.
The nature of the conic section is determined by the discriminant $h^2-a b$.
Here,$h^2-a b = (0)^2 - (8)(12) = -96$.
Since $h^2-a b < 0$ and $a \neq b$,the equation represents an ellipse.

(B) To find the points of intersection $A$ and $B$,substitute $y=x$ into the circle equation $x^2+y^2-2x=0$.
$x^2+x^2-2x=0$
$2x^2-2x=0$
$2x(x-1)=0$
Thus,$x=0$ or $x=1$. Since $y=x$,the points are $A(0,0)$ and $B(1,1)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.
Substituting $(0,0)$ and $(1,1)$:
$(x-0)(x-1)+(y-0)(y-1)=0$
$x(x-1)+y(y-1)=0$
$x^2-x+y^2-y=0$
$x^2+y^2-x-y=0$.

(C) The given equation of the circle is $x^2+y^2+4x-8y+5=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $2g=4 \Rightarrow g=2$ and $2f=-8 \Rightarrow f=-4$.
The center of the circle is $(-g, -f) = (-2, 4)$.
Let the coordinates of the other end of the diameter be $(h, k)$.
Since the center of the circle is the midpoint of the diameter,we have:
$\frac{h+2}{2} = -2$ $\Rightarrow h+2 = -4$ $\Rightarrow h = -6$
$\frac{k+1}{2} = 4$ $\Rightarrow k+1 = 8$ $\Rightarrow k = 7$
Thus,the coordinates of the other end are $(-6, 7)$.

(B) The equation of the circle is $x^2 + y^2 - 20y + 90 = 0$.
Substituting $y = mx$ into the circle equation:
$x^2 + (mx)^2 - 20(mx) + 90 = 0$
$x^2(1 + m^2) - 20mx + 90 = 0$.
For the line to lie outside the circle,the discriminant $D$ must be less than $0$:
$D = (-20m)^2 - 4(1 + m^2)(90) < 0$
$400m^2 - 360(1 + m^2) < 0$
$400m^2 - 360 - 360m^2 < 0$
$40m^2 - 360 < 0$
$40m^2 < 360$
$m^2 < 9$
$|m| < 3$.

(D) Let the centre of the circle be $(h, k)$.
Since the circle passes through the points $(a, 0)$ and $(-a, 0)$,the distance from the centre $(h, k)$ to both points must be equal to the radius $r$.
Therefore,$\sqrt{(h-a)^2 + (k-0)^2} = \sqrt{(h-(-a))^2 + (k-0)^2}$.
Squaring both sides,we get $(h-a)^2 + k^2 = (h+a)^2 + k^2$.
Simplifying this,$h^2 - 2ah + a^2 + k^2 = h^2 + 2ah + a^2 + k^2$.
This leads to $-2ah = 2ah$,which implies $4ah = 0$.
Since $a \neq 0$,we must have $h = 0$.
Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $x = 0$,which is the $y$-axis.

(B) Given the equation of the parabola: $y^2 + 6x - 2y + 13 = 0$.
Rearrange the terms to group $y$ variables: $y^2 - 2y = -6x - 13$.
Complete the square for the $y$ terms: $(y^2 - 2y + 1) = -6x - 13 + 1$.
This simplifies to: $(y - 1)^2 = -6x - 12$.
Factor out $-6$ on the right side: $(y - 1)^2 = -6(x + 2)$.
Comparing this with the standard form $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex:
Here,$h = -2$ and $k = 1$.
Therefore,the vertex is $(-2, 1)$.

(C) Given the coordinates of the point $P$ are $x = 2t^2 + 4$ and $y = 4t + 6$.
From the second equation,we have $4t = y - 6$,which implies $t = \frac{y - 6}{4}$.
Substituting this value of $t$ into the equation for $x$:
$x = 2\left(\frac{y - 6}{4}\right)^2 + 4$
$x - 4 = 2 \cdot \frac{(y - 6)^2}{16}$
$x - 4 = \frac{(y - 6)^2}{8}$
$(y - 6)^2 = 8(x - 4)$
This equation is of the form $(y - k)^2 = 4a(x - h)$,which represents a parabola.

(B) Given the complex number $z = (1 - t^2) + i \sqrt{1 + t^2}$.
Let $z = x + iy$,where $x$ and $y$ are real numbers.
Equating the real and imaginary parts,we get:
$x = 1 - t^2$
$y = \sqrt{1 + t^2}$
Squaring the imaginary part,we have $y^2 = 1 + t^2$.
From the first equation,$t^2 = 1 - x$.
Substituting this into the equation for $y^2$:
$y^2 = 1 + (1 - x)$
$y^2 = 2 - x$
$y^2 = -(x - 2)$
This is the standard equation of a parabola of the form $Y^2 = -4aX$ with vertex at $(2, 0)$.

(C) Let the midpoint of a chord passing through the vertex $(0, 0)$ be $(h, k)$.
Since the chord passes through the vertex $(0, 0)$ and the midpoint $(h, k)$,the equation of the chord is $y - 0 = \frac{k - 0}{h - 0}(x - 0)$,which simplifies to $y = \frac{k}{h}x$.
Substituting this into the parabola equation $y^2 = 4ax$,we get $(\frac{k}{h}x)^2 = 4ax$.
$\frac{k^2}{h^2}x^2 = 4ax$.
Since $x=0$ is the vertex,for the chord,we consider $x \neq 0$,so $\frac{k^2}{h^2}x = 4a$,which gives $x = \frac{4ah^2}{k^2}$.
However,the midpoint $(h, k)$ must satisfy the chord equation $k = \frac{k}{h}h$,which is trivial.
The condition for $(h, k)$ to be the midpoint of a chord of $y^2 = 4ax$ is $k^2 = 2a(h - x_1)$ where $x_1$ is the other end. For a chord through the vertex,the equation of the chord with midpoint $(h, k)$ is $T = S_1$,i.e.,$yk - 2a(x + h) = k^2 - 4ah$.
Since it passes through $(0, 0)$,we have $0 - 2ah = k^2 - 4ah$,which simplifies to $k^2 = 2ah$.
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = 2ax$,which is a parabola.

(B) Given the equation of the ellipse: $16x^2 + 25y^2 = 400$.
Divide both sides by $400$ to get the standard form: $\frac{16x^2}{400} + \frac{25y^2}{400} = 1$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{16} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 16$.
Thus,$a = 5$ and $b = 4$.
The length of the latus rectum of an ellipse is given by the formula $\frac{2b^2}{a}$.
Substituting the values: $\text{Length} = \frac{2 \times 16}{5} = \frac{32}{5}$ units.

(C) Given the equation of the hyperbola: $4x^2 - 9y^2 = 36$.
Dividing both sides by $36$,we get: $\frac{x^2}{9} - \frac{y^2}{4} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 4$.
Thus,$a = 3$ and $b = 2$.
The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{\frac{a^2 + b^2}{a^2}}$.
Substituting the values: $e = \sqrt{\frac{9 + 4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.

(C) We are given the limit $L = \lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$.
Since the form is $\frac{0}{0}$ at $x=1$,we can rewrite the numerator as $(x-1) + (x^2-1) + \ldots + (x^n-1)$.
Thus,$L = \lim _{x \rightarrow 1} \left( \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \ldots + \frac{x^n-1}{x-1} \right)$.
Using the standard limit formula $\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1}$,each term becomes $\lim _{x \rightarrow 1} \frac{x^k-1}{x-1} = k(1)^{k-1} = k$.
Therefore,$L = 1 + 2 + 3 + \ldots + n$.
The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

(D) We use the limit formula $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$.
Given expression: $\lim_{x \rightarrow 0} \frac{\sin(\pi \sin^2 x)}{x^2}$.
Multiply and divide by $\pi \sin^2 x$:
$= \lim_{x \rightarrow 0} \left[ \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \times \frac{\pi \sin^2 x}{x^2} \right]$.
Since $\pi \sin^2 x \rightarrow 0$ as $x \rightarrow 0$,the first part approaches $1$.
$= 1 \times \pi \times \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2$.
$= \pi \times (1)^2 = \pi$.

(C) function $f(x)$ is even if $f(-x) = f(x)$.
Let us check option $C$: $f(x) = x \cdot \frac{a^x - 1}{a^x + 1}$.
$f(-x) = (-x) \cdot \frac{a^{-x} - 1}{a^{-x} + 1}$
$f(-x) = (-x) \cdot \frac{\frac{1}{a^x} - 1}{\frac{1}{a^x} + 1}$
$f(-x) = (-x) \cdot \frac{\frac{1 - a^x}{a^x}}{\frac{1 + a^x}{a^x}}$
$f(-x) = (-x) \cdot \frac{-(a^x - 1)}{a^x + 1}$
$f(-x) = x \cdot \frac{a^x - 1}{a^x + 1} = f(x)$.
Since $f(-x) = f(x)$,the function in option $C$ is an even function.

(A) The period of $\cos(ax)$ is $\frac{2\pi}{|a|}$. Thus,the period of $\cos 4x$ is $T_1 = \frac{2\pi}{4} = \frac{\pi}{2}$.
The period of $\tan(bx)$ is $\frac{\pi}{|b|}$. Thus,the period of $\tan 3x$ is $T_2 = \frac{\pi}{3}$.
The period of the sum of two periodic functions is the least common multiple $(LCM)$ of their individual periods.
We need to find $\text{LCM}\left(\frac{\pi}{2}, \frac{\pi}{3}\right)$.
To find the $LCM$ of fractions $\frac{a}{b}$ and $\frac{c}{d}$,we use the formula $\frac{\text{LCM}(a, c)}{\text{HCF}(b, d)}$.
Here,$\text{LCM}(\pi, \pi) = \pi$ and $\text{HCF}(2, 3) = 1$.
Therefore,the period is $\frac{\pi}{1} = \pi$.

(C) Given the equation $ax^2 + bx + c = 0$ and the condition $a + 2b + 4c = 0$.
Divide the condition $a + 2b + 4c = 0$ by $4$:
$\frac{a}{4} + \frac{2b}{4} + \frac{4c}{4} = 0$
$\frac{a}{4} + \frac{b}{2} + c = 0$
This can be rewritten as:
$a(\frac{1}{2})^2 + b(\frac{1}{2}) + c = 0$
Comparing this with the quadratic equation $f(x) = ax^2 + bx + c = 0$,we see that $f(\frac{1}{2}) = 0$.
Therefore,$x = \frac{1}{2}$ is a root of the equation.

(C) Total number of people = $4 + 2 = 6$.
The total number of ways to arrange $6$ people in a row is $n(S) = 6! = 720$.
To find the number of ways where the $2$ girls sit side by side,we treat the $2$ girls as a single unit.
Now,we have $4$ boys and $1$ unit (the $2$ girls),which makes $5$ units in total.
These $5$ units can be arranged in $5!$ ways.
The $2$ girls within their unit can be arranged in $2!$ ways.
So,the number of favorable outcomes is $n(E) = 5! \times 2! = 120 \times 2 = 240$.
The probability $P$ is given by $\frac{n(E)}{n(S)} = \frac{5! \times 2!}{6!} = \frac{120 \times 2}{720} = \frac{240}{720} = \frac{1}{3}$.

(B) The tossing of a coin is an independent event.
Each toss of a fair coin has two equally likely outcomes: Head $(H)$ and Tail $(T)$.
The probability of getting a head in any single toss is $P(H) = \frac{1}{2}$.
The outcome of the fourth toss does not depend on the outcomes of the previous three tosses.
Therefore,the probability of getting a head on the fourth toss remains $\frac{1}{2}$.

(C) We express the sum as a definite integral using the limit of a sum formula: $\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4} = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n^3(r/n)^3}{n^4((r/n)^4+1)} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{(r/n)^3}{(r/n)^4+1}$.
Let $x = r/n$,then as $n \rightarrow \infty$,the sum becomes the integral $\int_0^1 \frac{x^3}{x^4+1} dx$.
To solve this,let $u = x^4+1$,then $du = 4x^3 dx$,or $x^3 dx = \frac{1}{4} du$.
The integral becomes $\frac{1}{4} \int_1^2 \frac{1}{u} du = \frac{1}{4} [\log_e u]_1^2$.
$= \frac{1}{4} (\log_e 2 - \log_e 1) = \frac{1}{4} \log_e 2$.

(C) For the sum $A+B$ to be defined,the matrices $A$ and $B$ must have the same order,say $m \times n$.
For the product $AB$ to be defined,the number of columns in $A$ must be equal to the number of rows in $B$.
Since $A$ is of order $m \times n$,it has $n$ columns. Since $B$ is of order $m \times n$,it has $m$ rows.
Therefore,for $AB$ to be defined,we must have $n = m$.
Since $m = n$,both matrices have the same order $n \times n$,which means they are square matrices of the same order.

(C) matrix $A$ is symmetric if $A = A^T$,where $A^T$ is the transpose of matrix $A$.
Given $A = \begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}$.
The transpose $A^T$ is obtained by interchanging rows and columns: $A^T = \begin{bmatrix} 3 & 2x+3 \\ x-1 & x+2 \end{bmatrix}$.
Since $A = A^T$,we equate the corresponding elements:
$x-1 = 2x+3$.
Subtracting $x$ from both sides: $-1 = x+3$.
Subtracting $3$ from both sides: $x = -4$.
Therefore,the value of $x$ is $-4$.

(A) matrix $A$ is Hermitian if $A = \bar{A}^T$. Let us check the given matrix $z$.
The transpose of $z$ is $z^T = \begin{bmatrix} 1 & 1-2i & 5i \\ 1+2i & -3 & 5-3i \\ -5i & 5+3i & 7 \end{bmatrix}$.
The conjugate of $z$ is $\bar{z} = \begin{bmatrix} 1 & 1-2i & 5i \\ 1+2i & -3 & 5-3i \\ -5i & 5+3i & 7 \end{bmatrix}$.
Since $z^T = \bar{z}$,the matrix $z$ is a Hermitian matrix.
For any Hermitian matrix,the determinant is always a real number.
Let $D = \det(z)$. Since $D$ is real,$D = \bar{D}$.
Thus,$z$ (as a determinant value) is purely real.

(D) Let the determinant be $\Delta = \left|\begin{array}{ccc}1 & 1+\omega^2 & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega & -1\end{array}\right|$.
Since $1+\omega+\omega^2 = 0$,we have $1+\omega^2 = -\omega$.
Substituting this,$\Delta = \left|\begin{array}{ccc}1 & -\omega & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega & -1\end{array}\right|$.
Applying $C_2 \rightarrow C_2 + C_1$:
$\Delta = \left|\begin{array}{ccc}1 & 1-\omega & \omega^2 \\ 1-i & -i & \omega^2-1 \\ -i & -1 & -1\end{array}\right|$.
By evaluating the determinant directly or simplifying further,we find that the rows/columns are linearly dependent or the expansion results in $0$.
Thus,the value of the determinant is $0$.

(A) Let $\Delta = \left| \begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array} \right|$.
Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:
$\Delta = \left| \begin{array}{lll} (a-b)+(b-c)+(c-a) & b-c & c-a \\ (b-c)+(c-a)+(a-b) & c-a & a-b \\ (c-a)+(a-b)+(b-c) & a-b & b-c \end{array} \right|$.
Simplifying the elements in the first column:
$\Delta = \left| \begin{array}{lll} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{array} \right|$.
Since all elements in the first column are $0$,the value of the determinant is $0$.

(D) We are given the inequation $\cos ^{-1} x < \sin ^{-1} x$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the inequation,we get:
$\frac{\pi}{2} - \sin ^{-1} x < \sin ^{-1} x$
$\frac{\pi}{2} < 2 \sin ^{-1} x$
$\sin ^{-1} x > \frac{\pi}{4}$
Taking the sine of both sides (since $\sin x$ is an increasing function in its domain):
$x > \sin\left(\frac{\pi}{4}\right)$
$x > \frac{1}{\sqrt{2}}$
Since the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $[-1, 1]$,we must have $x \le 1$.
Therefore,the solution set is $x \in \left(\frac{1}{\sqrt{2}}, 1\right]$.

(C) Let $u = x+2y$ and $v = x-2y$.
Adding the two equations: $u+v = (x+2y) + (x-2y) = 2x$,which implies $x = \frac{u+v}{2}$.
Subtracting the two equations: $u-v = (x+2y) - (x-2y) = 4y$,which implies $y = \frac{u-v}{4}$.
Given $f(x+2y, x-2y) = xy$,we substitute $u$ and $v$ to get $f(u, v) = \left(\frac{u+v}{2}\right) \left(\frac{u-v}{4}\right)$.
Simplifying the expression: $f(u, v) = \frac{u^2-v^2}{8}$.
Replacing $u$ with $x$ and $v$ with $y$,we get $f(x, y) = \frac{x^2-y^2}{8}$.

(A) For the function $f(x)$ to be continuous at $x=2$,the limit of $f(x)$ as $x \to 2$ must exist and be equal to $f(2)$.
Given $f(2) = 2$.
We evaluate the limit: $\lim_{x \to 2} \frac{x^2-(A+2)x+A}{x-2}$.
For the limit to exist,the numerator must be zero at $x=2$ because the denominator is zero.
Substituting $x=2$ in the numerator: $2^2 - (A+2)(2) + A = 4 - 2A - 4 + A = -A$.
Setting $-A = 0$,we get $A = 0$.
Now,check the limit with $A=0$: $\lim_{x \to 2} \frac{x^2-2x}{x-2} = \lim_{x \to 2} \frac{x(x-2)}{x-2} = \lim_{x \to 2} x = 2$.
Since the limit $2$ equals $f(2) = 2$,the function is continuous at $x=2$ when $A=0$.

(A) For $f(x)$ to be continuous at $x = 2$,the limit of $f(x)$ as $x \to 2$ must equal $f(2)$.
Given $f(x) = [x] + [-x]$ for $x \neq 2$.
We know that for any integer $n$,$[n] + [-n] = 0$,but for any non-integer $x$,$[x] + [-x] = -1$.
As $x \to 2$,$x$ takes values close to $2$ but not equal to $2$. Since $x$ is not an integer in the neighborhood of $2$,we have $[x] + [-x] = -1$.
Therefore,$\lim_{x \to 2} f(x) = -1$.
Since the function is continuous at $x = 2$,we must have $f(2) = \lim_{x \to 2} f(x)$.
Thus,$\lambda = -1$.

(B) Given the equation $x^2+y^2=4$.
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4)$
$2x + 2y \frac{dy}{dx} = 0$
Dividing the entire equation by $2$,we get:
$x + y \frac{dy}{dx} = 0$
Therefore,$y \frac{dy}{dx} + x = 0$.

(A) Given the equation: $y = A x^{-1} + B x^2$.
First,differentiate $y$ with respect to $x$:
$\frac{d y}{d x} = -A x^{-2} + 2 B x$.
Next,differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2 y}{d x^2} = 2 A x^{-3} + 2 B$.
Now,multiply by $x^2$:
$x^2 \frac{d^2 y}{d x^2} = x^2 (2 A x^{-3} + 2 B) = 2 A x^{-1} + 2 B x^2$.
Factor out $2$:
$x^2 \frac{d^2 y}{d x^2} = 2 (A x^{-1} + B x^2)$.
Since $y = A x^{-1} + B x^2$,we substitute $y$ back into the expression:
$x^2 \frac{d^2 y}{d x^2} = 2 y$.

(B) Given $f(x) = \tan^{-1} x$.
First derivative: $f'(x) = \frac{1}{1+x^2}$.
Second derivative: $f''(x) = \frac{d}{dx} (1+x^2)^{-1} = -1(1+x^2)^{-2} \cdot (2x) = \frac{-2x}{(1+x^2)^2}$.
We are given $f'(x) + f''(x) = 0$.
Substituting the derivatives: $\frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} = 0$.
Multiply by $(1+x^2)^2$: $(1+x^2) - 2x = 0$.
This simplifies to $x^2 - 2x + 1 = 0$,which is $(x-1)^2 = 0$.
Therefore,$x = 1$.

(A) Given: $y = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right)$.
Substitute $x = \tan \theta$,which implies $\theta = \tan^{-1} x$.
Then,$y = \tan^{-1} \left( \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta} \right) = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right)$.
Using trigonometric identities,$\frac{\sec \theta - 1}{\tan \theta} = \frac{1-\cos \theta}{\sin \theta} = \frac{2 \sin^2 (\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)} = \tan(\theta/2)$.
So,$y = \tan^{-1} (\tan(\theta/2)) = \theta/2 = \frac{1}{2} \tan^{-1} x$.
Differentiating with respect to $x$,we get $y' = \frac{1}{2(1+x^2)}$.
At $x = 1$,$y'(1) = \frac{1}{2(1+1^2)} = \frac{1}{2(2)} = \frac{1}{4}$.

(B) Given the function $y = 2x^3 - 2x^2 + 3x - 5$.
To find the approximate change $\Delta y$,we use the formula $\Delta y \approx \frac{dy}{dx} \cdot \Delta x$.
First,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(2x^3 - 2x^2 + 3x - 5) = 6x^2 - 4x + 3$.
Now,evaluate the derivative at $x = 2$:
$\left(\frac{dy}{dx}\right)_{x=2} = 6(2)^2 - 4(2) + 3 = 6(4) - 8 + 3 = 24 - 8 + 3 = 19$.
Given $\Delta x = 0.1$,we calculate $\Delta y$:
$\Delta y \approx 19 \times 0.1 = 1.9$.

(C) Let $f(x) = x^{1/5}$. We need to find the value of $f(33)$.
Let $x = 32$ and $\Delta x = 1$,so that $x + \Delta x = 33$.
We know that $f(x) = x^{1/5} \implies f'(x) = \frac{1}{5} x^{-4/5} = \frac{1}{5 x^{4/5}}$.
At $x = 32$,$f(32) = (32)^{1/5} = 2$.
$f'(32) = \frac{1}{5(32)^{4/5}} = \frac{1}{5(2^4)} = \frac{1}{5 \times 16} = \frac{1}{80} = 0.0125$.
Using the differential approximation formula: $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
$f(33) \approx f(32) + f'(32) \times 1$.
$f(33) \approx 2 + 0.0125 = 2.0125$.

(A) Given: Initial velocity $u = 0$,acceleration $a = 8 \text{ m/s}^2$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$,since $u = 0$,we have $S = \frac{1}{2}at^2$.
$1$. Time taken to cover the first metre $(S_1 = 1 \text{ m})$:
$1 = \frac{1}{2} \times 8 \times t_1^2 \implies 1 = 4t_1^2 \implies t_1^2 = \frac{1}{4} \implies t_1 = \frac{1}{2} \text{ s}$.
$2$. Time taken to cover the first two metres $(S_2 = 2 \text{ m})$:
$2 = \frac{1}{2} \times 8 \times t_2^2 \implies 2 = 4t_2^2 \implies t_2^2 = \frac{1}{2} \implies t_2 = \frac{1}{\sqrt{2}} \text{ s}$.
$3$. Time taken to move the second metre is the difference between the time taken to cover $2 \text{ m}$ and $1 \text{ m}$:
$\Delta t = t_2 - t_1 = \frac{1}{\sqrt{2}} - \frac{1}{2} = \frac{\sqrt{2}-1}{2} \text{ s}$.

(B) The given function is $f(x) = \begin{cases} 0, & x = 0 \\ x - 3, & x > 0 \end{cases}$.
First,let us check the continuity at $x = 0$.
$f(0) = 0$.
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x - 3) = -3$.
Since $f(0) \neq \lim_{x \to 0^+} f(x)$,the function is discontinuous at $x = 0$.
Now,consider the interval $x > 0$. For any $x_1, x_2$ such that $0 < x_1 < x_2$,we have $f(x_1) = x_1 - 3$ and $f(x_2) = x_2 - 3$. Since $x_1 < x_2$,it follows that $x_1 - 3 < x_2 - 3$,which means $f(x_1) < f(x_2)$. Therefore,the function is strictly increasing for $x > 0$.
However,for the function to be increasing on the interval $[0, \infty)$,it must be continuous on $[0, \infty)$. Since it is discontinuous at $x = 0$,it is not increasing on $[0, \infty)$.
Thus,the function is strictly increasing when $x > 0$.

(A) function $f(x)$ is strictly increasing if its derivative $f'(x) > 0$ for all $x$ in its domain.
Given the function $f(x) = ax + b$.
Taking the derivative with respect to $x$,we get $f'(x) = \frac{d}{dx}(ax + b) = a$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
Substituting the derivative,we get $a > 0$.
Therefore,the function is strictly increasing if $a > 0$.

(A) Given the function $f(x) = x^3 e^{-3x}$ for $x > 0$.
To find the maximum value,we first find the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx}(x^3) \cdot e^{-3x} + x^3 \cdot \frac{d}{dx}(e^{-3x})$
$f'(x) = 3x^2 e^{-3x} + x^3 (-3 e^{-3x})$
$f'(x) = 3x^2 e^{-3x} (1 - x)$
Setting $f'(x) = 0$ for critical points:
$3x^2 e^{-3x} (1 - x) = 0$
Since $x > 0$ and $e^{-3x} \neq 0$,we have $1 - x = 0$,which gives $x = 1$.
To confirm it is a maximum,we check the sign of $f'(x)$ around $x = 1$. For $x < 1$,$f'(x) > 0$ and for $x > 1$,$f'(x) < 0$.
Thus,$f(x)$ has a local maximum at $x = 1$.
The maximum value is $f(1) = (1)^3 e^{-3(1)} = e^{-3}$.

(A) Rolle's theorem states that for a function $f(x)$ to be applicable on an interval $[a, b]$,it must satisfy three conditions:
$1$. $f(x)$ is continuous on $[a, b]$.
$2$. $f(x)$ is differentiable on $(a, b)$.
$3$. $f(a) = f(b)$.
For $f(x) = e^{\cos x}$,the function is continuous and differentiable everywhere.
We check the condition $f(a) = f(b)$ for the given options:
For option $A$: $f(\frac{\pi}{2}) = e^{\cos(\pi/2)} = e^0 = 1$ and $f(\frac{3\pi}{2}) = e^{\cos(3\pi/2)} = e^0 = 1$.
Since $f(\frac{\pi}{2}) = f(\frac{3\pi}{2})$,Rolle's theorem is applicable on the interval $[\frac{\pi}{2}, \frac{3\pi}{2}]$.

(C) We know that $\cos 2x = 2 \cos^2 x - 1$.
Substituting this into the integral,we get:
$\int \frac{2 \cos^2 x - 1}{\cos x} dx = \int (2 \cos x - \frac{1}{\cos x}) dx$.
This simplifies to $\int (2 \cos x - \sec x) dx$.
Integrating term by term,we get:
$2 \int \cos x dx - \int \sec x dx$.
The integral of $\cos x$ is $\sin x$ and the integral of $\sec x$ is $\log |\sec x + \tan x|$.
Thus,the final result is $2 \sin x - \log |\sec x + \tan x| + C$.

(A) Let $I = \int \frac{\sin ^8 x - \cos ^8 x}{1 - 2 \sin ^2 x \cos ^2 x} \, dx$.
Using the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$,we can write the numerator as:
$\sin ^8 x - \cos ^8 x = (\sin ^4 x - \cos ^4 x)(\sin ^4 x + \cos ^4 x) = (\sin ^2 x - \cos ^2 x)(\sin ^2 x + \cos ^2 x)(\sin ^4 x + \cos ^4 x)$.
Since $\sin ^2 x + \cos ^2 x = 1$,the numerator becomes $(\sin ^2 x - \cos ^2 x)(\sin ^4 x + \cos ^4 x)$.
Also,note that $\sin ^4 x + \cos ^4 x = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x = 1 - 2 \sin ^2 x \cos ^2 x$.
Substituting this into the integral:
$I = \int \frac{(\sin ^2 x - \cos ^2 x)(1 - 2 \sin ^2 x \cos ^2 x)}{1 - 2 \sin ^2 x \cos ^2 x} \, dx$.
$I = \int (\sin ^2 x - \cos ^2 x) \, dx = \int -(\cos ^2 x - \sin ^2 x) \, dx$.
Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$,we get:
$I = -\int \cos 2x \, dx = -\frac{\sin 2x}{2} + C$.

(B) To evaluate the integral $I = \int \frac{x^3 \, dx}{1+x^8}$,we can rewrite the denominator as $1 + (x^4)^2$.
Let $u = x^4$.
Then,differentiating both sides with respect to $x$,we get $du = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{du}{4}$.
Substituting these into the integral,we get:
$I = \int \frac{du/4}{1+u^2} = \frac{1}{4} \int \frac{du}{1+u^2}$.
Using the standard integral formula $\int \frac{du}{1+u^2} = \tan^{-1}(u) + c$,we obtain:
$I = \frac{1}{4} \tan^{-1}(u) + c$.
Substituting $u = x^4$ back,we get the final result:
$I = \frac{1}{4} \tan^{-1}(x^4) + c$.

(B) Let $I = \int 2^x (f^{\prime}(x) + f(x) \log 2) \, dx$.
We can rewrite the integrand as:
$I = \int (2^x f^{\prime}(x) + 2^x \log 2 \cdot f(x)) \, dx$.
Recall the product rule for differentiation: $\frac{d}{dx} [u(x)v(x)] = u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$.
Let $u(x) = 2^x$ and $v(x) = f(x)$.
Then $u^{\prime}(x) = \frac{d}{dx} (2^x) = 2^x \log 2$.
Thus,the integrand is of the form $u(x)v^{\prime}(x) + v(x)u^{\prime}(x)$,which is $\frac{d}{dx} [2^x f(x)]$.
Therefore,$I = \int \frac{d}{dx} [2^x f(x)] \, dx = 2^x f(x) + C$.

(D) Let $I = \int_{-2}^2(x \cos x+\sin x+1) d x$.
We can split the integral as $I = \int_{-2}^2 x \cos x d x + \int_{-2}^2 \sin x d x + \int_{-2}^2 1 d x$.
Recall the property of definite integrals: $\int_{-a}^a f(x) d x = 0$ if $f(x)$ is an odd function,and $2 \int_0^a f(x) d x$ if $f(x)$ is an even function.
Let $f_1(x) = x \cos x$. Since $f_1(-x) = (-x) \cos(-x) = -x \cos x = -f_1(x)$,$f_1(x)$ is an odd function. Thus,$\int_{-2}^2 x \cos x d x = 0$.
Let $f_2(x) = \sin x$. Since $f_2(-x) = \sin(-x) = -\sin x = -f_2(x)$,$f_2(x)$ is an odd function. Thus,$\int_{-2}^2 \sin x d x = 0$.
Therefore,$I = 0 + 0 + \int_{-2}^2 1 d x = [x]_{-2}^2 = 2 - (-2) = 4$.

(A) Let $I = \int_0^\pi \sin^{50} x \cos^{49} x \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we have:
$I = \int_0^\pi \sin^{50}(\pi - x) \cos^{49}(\pi - x) \, dx$.
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we get:
$I = \int_0^\pi \sin^{50} x (-\cos x)^{49} \, dx$.
$I = -\int_0^\pi \sin^{50} x \cos^{49} x \, dx$.
$I = -I$.
Adding $I$ to both sides,we get $2I = 0$,which implies $I = 0$.

(C) The function $f(x) = |\sin x|$ is a periodic function with a period of $\pi$.
We know that $\int_{a}^{a+nT} f(x) dx = n \int_{0}^{T} f(x) dx$ for a periodic function with period $T$.
Here,the interval length is $16\pi - \pi = 15\pi$.
Thus,$\int_{\pi}^{16\pi} |\sin x| dx = 15 \int_{0}^{\pi} |\sin x| dx$.
Since $\sin x \ge 0$ for $x \in [0, \pi]$,we have $|\sin x| = \sin x$.
Therefore,$15 \int_{0}^{\pi} \sin x dx = 15 [-\cos x]_{0}^{\pi}$.
$= 15 [-\cos(\pi) - (-\cos(0))] = 15 [-(-1) - (-1)] = 15 [1 + 1] = 15 \times 2 = 30$.

(D) To find the area enclosed between the curves $y^2=x$ and $y=x$,we first find their points of intersection by solving the equations simultaneously:
$y^2 = y \implies y^2 - y = 0 \implies y(y-1) = 0$.
Thus,the points of intersection are at $y=0$ and $y=1$. Correspondingly,$x=0$ and $x=1$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:
$A = \int_0^1 (\sqrt{x} - x) \, dx$
$A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_0^1$
$A = \left[ \frac{2}{3}x^{3/2} - \frac{1}{2}x^2 \right]_0^1$
$A = \left( \frac{2}{3}(1)^{3/2} - \frac{1}{2}(1)^2 \right) - (0 - 0)$
$A = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6} \text{ sq. units}$.

(B) The given curves are $y^2 = 4x$ and $x^2 = 4y$.
To find the points of intersection,substitute $y = \frac{x^2}{4}$ into $y^2 = 4x$:
$(\frac{x^2}{4})^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.
Thus,$x = 0$ or $x = 4$. The points of intersection are $(0, 0)$ and $(4, 4)$.
The area $A$ bounded by the curves is given by the integral of the upper curve minus the lower curve from $x = 0$ to $x = 4$:
$A = \int_0^4 (\sqrt{4x} - \frac{x^2}{4}) dx$
$A = \int_0^4 (2\sqrt{x} - \frac{x^2}{4}) dx$
$A = [2 \cdot \frac{2}{3} x^{3/2} - \frac{1}{4} \cdot \frac{x^3}{3}]_0^4$
$A = [\frac{4}{3} x^{3/2} - \frac{x^3}{12}]_0^4$
$A = (\frac{4}{3} \cdot 4^{3/2} - \frac{4^3}{12}) - (0 - 0)$
$A = (\frac{4}{3} \cdot 8 - \frac{64}{12}) = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \text{ sq. unit}$.

(B) The given differential equation is $y = x(\frac{dy}{dx})^2 + \frac{dy}{dx}$.
To find the order and degree,we first express the equation in a form where the derivatives are free from radicals and fractions.
The highest order derivative present in the equation is $\frac{dy}{dx}$,which is of order $1$.
Since the highest order derivative is $\frac{dy}{dx}$,the order of the differential equation is $1$.
The degree of a differential equation is the power of the highest order derivative when the equation is written as a polynomial in derivatives.
Here,the highest power of the derivative $\frac{dy}{dx}$ is $2$.
Therefore,the degree is $2$ and the order is $1$.

(B) Given equation: $y = ae^{bx} \dots (i)$
Differentiating with respect to $x$:
$y_1 = \frac{dy}{dx} = abe^{bx}$
Since $y = ae^{bx}$,we can write:
$y_1 = by \dots (ii)$
Differentiating $y_1 = by$ with respect to $x$:
$y_2 = \frac{d^2y}{dx^2} = by_1 \dots (iii)$
From equation $(ii)$,we have $b = \frac{y_1}{y}$.
Substituting $b$ into equation $(iii)$:
$y_2 = \left(\frac{y_1}{y}\right)y_1$
$y_2 = \frac{y_1^2}{y}$
Therefore,$yy_2 = y_1^2$.

(A) Given the differential equation: $\log_{e}\left(\frac{dy}{dx}\right) = x + y$.
By the definition of logarithm,we can write this as: $\frac{dy}{dx} = e^{x+y}$.
Using the property of exponents,we have: $\frac{dy}{dx} = e^x \cdot e^y$.
Separating the variables,we get: $e^{-y} dy = e^x dx$.
Integrating both sides: $\int e^{-y} dy = \int e^x dx$.
This yields: $-e^{-y} = e^x + C_1$.
Rearranging the terms,we get: $e^x + e^{-y} = -C_1$.
Letting $-C_1 = C$,the general solution is: $e^x + e^{-y} = C$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \tan \frac{y}{x}$.
This is a homogeneous differential equation.
Let $v = \frac{y}{x}$,so $y = vx$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = v + \tan v$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = \tan v$.
Separating the variables: $\frac{dv}{\tan v} = \frac{dx}{x}$,which is $\cot v \, dv = \frac{dx}{x}$.
Integrating both sides: $\int \cot v \, dv = \int \frac{dx}{x}$.
This gives $\ln |\sin v| = \ln |x| + \ln |c|$.
Using logarithmic properties: $\ln |\sin v| = \ln |cx|$.
Taking the exponential of both sides: $\sin v = cx$.
Substituting back $v = \frac{y}{x}$: $\sin(\frac{y}{x}) = cx$,or $x = c \sin(\frac{y}{x})$.

(B) The given differential equation is $\frac{d^2 y}{d x^2}+8 \frac{d y}{d x}+16 y=0$.
The auxiliary equation is $m^2+8m+16=0$.
This can be written as $(m+4)^2=0$.
Thus,the roots are $m = -4, -4$.
Since the roots are real and equal,the general solution is given by $y = (A+Bx)e^{mx}$.
Substituting $m = -4$,we get $y = (A+Bx)e^{-4x}$.
Therefore,the correct option is $B$.

(D) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{3x^2}{1+x^3}$ and $Q = \frac{\sin^2(x)}{1+x}$.
The Integrating Factor $(I.F.)$ is given by the formula $I.F. = e^{\int P dx}$.
Substituting the value of $P$:
$I.F. = e^{\int -\frac{3x^2}{1+x^3} dx}$.
Let $u = 1+x^3$,then $du = 3x^2 dx$. Thus,$\int \frac{3x^2}{1+x^3} dx = \int \frac{1}{u} du = \log|1+x^3|$.
Therefore,$I.F. = e^{-\log(1+x^3)} = e^{\log(1+x^3)^{-1}} = (1+x^3)^{-1} = \frac{1}{1+x^3}$.