WBJEE 2010 Mathematics Question Paper with Answer and Solution

(D) Using the change of base formula $\log_{a} b = \frac{\log b}{\log a}$:
Numerator = $\log_{3} 5 \times \log_{25} 27 \times \log_{49} 7$
$= \frac{\log 5}{\log 3} \times \frac{\log 27}{\log 25} \times \frac{\log 7}{\log 49}$
$= \frac{\log 5}{\log 3} \times \frac{3 \log 3}{2 \log 5} \times \frac{\log 7}{2 \log 7}$
$= \frac{1}{1} \times \frac{3}{2} \times \frac{1}{2} = \frac{3}{4}$
Denominator = $\log_{81} 3 = \log_{3^4} 3 = \frac{1}{4} \log_{3} 3 = \frac{1}{4}$
Value = $\frac{3/4}{1/4} = 3$

(D) Given that $a, b, c$ are sides of a right-angled triangle with $c$ as the hypotenuse,we have $a^2 + b^2 = c^2$,which implies $a^2 = c^2 - b^2 = (c+b)(c-b)$.
Taking the logarithm on both sides,we get $\log(a^2) = \log((c+b)(c-b)) = \log(c+b) + \log(c-b)$.
Now,the given expression is $E = \frac{\log_{c+b} a + \log_{c-b} a}{2 \log_{c+b} a \cdot \log_{c-b} a}$.
Using the change of base formula $\log_x y = \frac{\log y}{\log x}$,we have:
$E = \frac{\frac{\log a}{\log(c+b)} + \frac{\log a}{\log(c-b)}}{2 \cdot \frac{\log a}{\log(c+b)} \cdot \frac{\log a}{\log(c-b)}}$.
Simplifying the numerator: $\frac{\log a (\log(c-b) + \log(c+b))}{\log(c+b) \log(c-b)}$.
Simplifying the denominator: $\frac{2 (\log a)^2}{\log(c+b) \log(c-b)}$.
Thus,$E = \frac{\log a (\log((c+b)(c-b)))}{2 (\log a)^2} = \frac{\log a \cdot \log(a^2)}{2 (\log a)^2} = \frac{\log a \cdot 2 \log a}{2 (\log a)^2} = \frac{2 (\log a)^2}{2 (\log a)^2} = 1$.

(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 need to find the equation whose roots are $\alpha^{19}$ and $\beta^7$.
Since $\omega^3 = 1$,we have $\alpha^{19} = \omega^{19} = (\omega^3)^6 \cdot \omega = 1^6 \cdot \omega = \omega$.
Similarly,$\beta^7 = (\omega^2)^7 = \omega^{14} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
The new roots are $\omega$ and $\omega^2$,which are the same as the original roots.
Therefore,the required equation is $x^2+x+1=0$.

(C) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.
Here,$a = 1$,$b = -2\sqrt{3}$,and $c = -22$.
$D = (-2\sqrt{3})^2 - 4(1)(-22) = 12 + 88 = 100$.
Since $D > 0$,the roots are real and unequal.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{2\sqrt{3} \pm \sqrt{100}}{2} = \frac{2\sqrt{3} \pm 10}{2} = \sqrt{3} \pm 5$.
Since $\sqrt{3}$ is an irrational number,the roots $\sqrt{3} + 5$ and $\sqrt{3} - 5$ are irrational.
Therefore,the roots are real,irrational,and unequal.

(A) In a triangle $PQR$,we have $P + Q + R = \pi$. Since $\angle R = \pi / 2$,it follows that $P + Q = \pi / 2$,which implies $\frac{P}{2} + \frac{Q}{2} = \frac{\pi}{4}$.
Given that $\tan(P/2)$ and $\tan(Q/2)$ are roots of $ax^2 + bx + c = 0$,by Vieta's formulas:
Sum of roots: $\tan(P/2) + \tan(Q/2) = -b/a$
Product of roots: $\tan(P/2) \tan(Q/2) = c/a$
Using the identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have:
$\tan(\frac{P}{2} + \frac{Q}{2}) = \tan(\frac{\pi}{4}) = 1$
$\frac{-b/a}{1 - c/a} = 1$
$\frac{-b}{a - c} = 1$
$-b = a - c$
$c = a + b$

(C) Given the equation $x^2+15|x|+14=0$.
Since $x^2 = |x|^2$,we can rewrite the equation as $|x|^2+15|x|+14=0$.
Let $|x| = t$,where $t \ge 0$.
The equation becomes $t^2+15t+14=0$.
Factoring the quadratic,we get $(t+1)(t+14)=0$.
This gives $t = -1$ or $t = -14$.
However,we defined $t = |x|$,which must be non-negative $(t \ge 0)$.
Since both $-1$ and $-14$ are less than $0$,there is no real value of $x$ that satisfies the equation.
Thus,the equation has no solution.

(D) Given $z = \frac{4}{1-i}$.
To find the conjugate $\bar{z}$,we take the conjugate of the expression:
$\bar{z} = \overline{\left(\frac{4}{1-i}\right)} = \frac{\bar{4}}{\overline{1-i}}$.
Since the conjugate of a real number $4$ is $4$ and the conjugate of $(1-i)$ is $(1+i)$,
$\bar{z} = \frac{4}{1+i}$.

(A) Let $\arg (z) = \theta$,where $-\pi < \theta < -\frac{\pi}{2}$.
Since $\bar{z}$ is the reflection of $z$ across the real axis,$\arg (\bar{z}) = -\arg (z) = -\theta$.
Since $-\pi < \theta < -\frac{\pi}{2}$,we have $\frac{\pi}{2} < -\theta < \pi$.
Now,$-\bar{z} = -1 \cdot \bar{z} = e^{i\pi} \cdot \bar{z}$.
Therefore,$\arg (-\bar{z}) = \arg (e^{i\pi}) + \arg (\bar{z}) = \pi + (-\theta) = \pi - \theta$.
We need to calculate $\arg (\bar{z}) - \arg (-\bar{z}) = -\theta - (\pi - \theta) = -\theta - \pi + \theta = -\pi$.
Since the argument is typically defined in the range $(-\pi, \pi]$,we can add $2\pi$ to get the principal value: $-\pi + 2\pi = \pi$.

(B) The general term of the series is $t_n = \frac{2n}{(2n+1)!}$.
We can rewrite this as $t_n = \frac{(2n+1) - 1}{(2n+1)!} = \frac{1}{(2n)!} - \frac{1}{(2n+1)!}$.
Summing from $n=1$ to $\infty$,we get $\sum_{n=1}^{\infty} \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right) = \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots \right)$.
Recall the expansion $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$. For $x = -1$,$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$.
Thus,the sum is $e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$.
Therefore,the value is $e^{-1}$.

(D) The word $COMBINE$ has $7$ letters: $C, O, M, B, I, N, E$.
The vowels are $O, I, E$ ($3$ vowels).
The consonants are $C, M, B, N$ ($4$ consonants).
There are $7$ positions in total. The odd positions are $1, 3, 5, 7$ ($4$ odd places).
We need to place $3$ vowels in $4$ odd places,which can be done in $^4P_3$ ways.
$^4P_3 = 4 \times 3 \times 2 = 24$ ways.
The remaining $4$ letters (consonants) can be arranged in the remaining $4$ positions in $4!$ ways.
$4! = 4 \times 3 \times 2 \times 1 = 24$ ways.
Total number of permutations $= ^4P_3 \times 4! = 24 \times 24 = 576$.

(D) The sum of an infinite geometric series is given by the formula $S = \frac{a}{1-r}$,where $a$ is the first term and $r$ is the common ratio.
Given $S = \frac{4}{5}$ and $a = \frac{3}{4}$.
Substituting the values into the formula: $\frac{4}{5} = \frac{\frac{3}{4}}{1-r}$.
Multiplying both sides by $(1-r)$,we get $\frac{4}{5}(1-r) = \frac{3}{4}$.
$1-r = \frac{3}{4} \times \frac{5}{4} = \frac{15}{16}$.
$r = 1 - \frac{15}{16} = \frac{1}{16}$.

(A) Let the two numbers be $a$ and $b$.
Given that the $G.M. = \sqrt{ab} = 10$,so $ab = 100$.
Given that the $H.M. = \frac{2ab}{a+b} = 8$.
Substituting $ab = 100$ into the $H.M.$ equation:
$\frac{2(100)}{a+b} = 8$ $\Rightarrow \frac{200}{a+b} = 8$ $\Rightarrow a+b = \frac{200}{8} = 25$.
Now we have $a+b = 25$ and $ab = 100$.
The quadratic equation with roots $a$ and $b$ is $x^2 - (a+b)x + ab = 0$.
$x^2 - 25x + 100 = 0$.
$(x-20)(x-5) = 0$.
Thus,the numbers are $5$ and $20$.

(A) Given that $\frac{x^{n+1}+y^{n+1}}{x^n+y^n} = \sqrt{xy}$.
This implies $x^{n+1} + y^{n+1} = (xy)^{1/2} (x^n + y^n)$.
$x^{n+1} + y^{n+1} = x^{n+1/2} y^{1/2} + x^{1/2} y^{n+1/2}$.
Rearranging the terms,we get $x^{n+1} - x^{n+1/2} y^{1/2} = x^{1/2} y^{n+1/2} - y^{n+1}$.
$x^{n+1/2} (x^{1/2} - y^{1/2}) = y^{n+1/2} (x^{1/2} - y^{1/2})$.
Assuming $x \neq y$,we can divide both sides by $(x^{1/2} - y^{1/2})$:
$x^{n+1/2} = y^{n+1/2}$.
$(x/y)^{n+1/2} = 1$.
Since $(x/y)^0 = 1$,we have $n + 1/2 = 0$.
Therefore,$n = -1/2$.

(A) The $n^{th}$ term of the series is $T_{n} = (2n-1)^{3}$.
Expanding this,we get $T_{n} = 8n^{3} - 12n^{2} + 6n - 1$.
The sum of $n$ terms is $S_{n} = \sum_{k=1}^{n} T_{k} = \sum_{k=1}^{n} (8k^{3} - 12k^{2} + 6k - 1)$.
Using standard summation formulas:
$S_{n} = 8 \left[ \frac{n(n+1)}{2} \right]^{2} - 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 6 \left[ \frac{n(n+1)}{2} \right] - n$.
$S_{n} = 2n^{2}(n+1)^{2} - 2n(n+1)(2n+1) + 3n(n+1) - n$.
$S_{n} = 2n^{2}(n^{2}+2n+1) - 2n(2n^{2}+3n+1) + 3n^{2} + 3n - n$.
$S_{n} = 2n^{4} + 4n^{3} + 2n^{2} - 4n^{3} - 6n^{2} - 2n + 3n^{2} + 2n$.
$S_{n} = 2n^{4} - n^{2} = n^{2}(2n^{2}-1)$.

(B) The general term in the expansion of $(a-2b)^{n}$ is given by $t_{r+1} = {}^{n}C_{r} (a)^{n-r} (-2b)^{r}$.
Given that the sum of the $5^{th}$ and $6^{th}$ term is zero,we have $t_5 + t_6 = 0$.
This implies ${}^{n}C_4 (a)^{n-4} (-2b)^4 + {}^{n}C_5 (a)^{n-5} (-2b)^5 = 0$.
Rearranging the terms,we get ${}^{n}C_4 (a)^{n-4} (16b^4) = -{}^{n}C_5 (a)^{n-5} (-32b^5)$.
Dividing both sides by ${}^{n}C_4 (a)^{n-5} (b^4)$,we get $a = -\frac{{}^{n}C_5}{{}^{n}C_4} (-2b)$.
Using the formula ${}^{n}C_r = \frac{n!}{r!(n-r)!}$,we have $\frac{{}^{n}C_5}{{}^{n}C_4} = \frac{n-4}{5}$.
Thus,$a = -\frac{n-4}{5} (-2b) = \frac{2(n-4)}{5} b$.
Therefore,$\frac{a}{b} = \frac{2(n-4)}{5}$.

(C) We have $2^{3n} = (2^3)^n = 8^n$.
Since $8 = (1 + 7)$,we can write $8^n = (1 + 7)^n$.
Using the Binomial Theorem,$(1 + 7)^n = {^nC_0} + {^nC_1}(7) + {^nC_2}(7^2) + \dots + {^nC_n}(7^n)$.
Since ${^nC_0} = 1$,we have $8^n = 1 + 7({^nC_1} + {^nC_2}(7) + \dots + {^nC_n}(7^{n-1}))$.
Subtracting $1$ from both sides,$8^n - 1 = 7({^nC_1} + {^nC_2}(7) + \dots + {^nC_n}(7^{n-1}))$.
This expression is clearly divisible by $7$ for all $n \in N$.

(D) Using the Pascal's identity,${}^{n-1}C_r + {}^{n-1}C_{r-1} = {}^{n}C_r$,we have:
${}^{n-1}C_3 + {}^{n-1}C_4 = {}^{n}C_4$.
Given the inequality:
${}^{n}C_4 > {}^{n}C_3$.
Expanding the combinations:
$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$.
Simplifying the factorials:
$\frac{1}{4(n-4)!} > \frac{1}{(n-3)(n-4)!}$.
$\frac{1}{4} > \frac{1}{n-3}$.
Since $n-3 > 0$,we get:
$n-3 > 4$,which implies $n > 7$.
Thus,$n$ is just greater than $7$.

(B) The expansion of $(1+x)^{59}$ has $59+1 = 60$ terms.
Let the coefficients be $C_0, C_1, C_2, \dots, C_{59}$.
The sum of all coefficients is $\sum_{r=0}^{59} C_r = (1+1)^{59} = 2^{59}$.
Since $C_r = C_{59-r}$,the sum of the first $30$ coefficients is equal to the sum of the last $30$ coefficients.
Let $S$ be the sum of the last $30$ coefficients.
Then $2S = \sum_{r=0}^{59} C_r = 2^{59}$.
Therefore,$S = \frac{2^{59}}{2} = 2^{58}$.

(D) Given the expansion: $(1-x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$.
Step $1$: Put $x = 1$ in the expansion:
$(1 - 1 + 1)^n = a_0 + a_1 + a_2 + \ldots + a_{2n}$
$1^n = a_0 + a_1 + a_2 + \ldots + a_{2n} \Rightarrow 1 = a_0 + a_1 + a_2 + \ldots + a_{2n} \quad (i)$
Step $2$: Put $x = -1$ in the expansion:
$(1 - (-1) + (-1)^2)^n = a_0 + a_1(-1) + a_2(-1)^2 + \ldots + a_{2n}(-1)^{2n}$
$(1 + 1 + 1)^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n}$
$3^n = a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n} \quad (ii)$
Step $3$: Add equations $(i)$ and $(ii)$:
$1 + 3^n = (a_0 + a_1 + a_2 + \ldots + a_{2n}) + (a_0 - a_1 + a_2 - \ldots + a_{2n})$
$1 + 3^n = 2(a_0 + a_2 + a_4 + \ldots + a_{2n})$
Therefore,$a_0 + a_2 + a_4 + \ldots + a_{2n} = \frac{3^n + 1}{2}$.

(C) Given $\frac{\cos A}{3} = \frac{1}{5} \implies \cos A = \frac{3}{5}$.
Since $-\frac{\pi}{2} < A < 0$,$A$ lies in the fourth quadrant,so $\sin A$ is negative.
$\sin A = -\sqrt{1 - \cos^2 A} = -\sqrt{1 - (\frac{3}{5})^2} = -\sqrt{1 - \frac{9}{25}} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$.
Given $\frac{\cos B}{4} = \frac{1}{5} \implies \cos B = \frac{4}{5}$.
Since $-\frac{\pi}{2} < B < 0$,$B$ lies in the fourth quadrant,so $\sin B$ is negative.
$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{4}{5})^2} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.
Now,calculate $2 \sin A + 4 \sin B = 2(-\frac{4}{5}) + 4(-\frac{3}{5}) = -\frac{8}{5} - \frac{12}{5} = -\frac{20}{5} = -4$.

(B) We know that $\cot \theta = \tan(90^{\circ} - \theta)$.
Using this identity:
$\cot 54^{\circ} = \tan(90^{\circ} - 54^{\circ}) = \tan 36^{\circ}$.
$\cot 70^{\circ} = \tan(90^{\circ} - 70^{\circ}) = \tan 20^{\circ}$.
Substituting these values into the expression:
$\frac{\tan 36^{\circ}}{\tan 36^{\circ}} + \frac{\tan 20^{\circ}}{\tan 20^{\circ}} = 1 + 1 = 2$.

(B) We know that $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x}$.
Using the identity $\cos 2x = \cos^2 x - \sin^2 x$ and $\sin 2x = 2 \sin x \cos x$,we get $\cot x - \tan x = \frac{\cos 2x}{\frac{1}{2} \sin 2x} = 2 \cot 2x$.
Therefore,$\frac{\cot x - \tan x}{\cot 2x} = \frac{2 \cot 2x}{\cot 2x} = 2$.

(D) We know that $\cos 55^{\circ} = \sin(90^{\circ} - 55^{\circ}) = \sin 35^{\circ}$.
Substituting this into the expression,we get $\frac{\sin 55^{\circ} - \sin 35^{\circ}}{\sin 10^{\circ}}$.
Using the formula $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$,we have:
$\sin 55^{\circ} - \sin 35^{\circ} = 2 \cos\left(\frac{55^{\circ} + 35^{\circ}}{2}\right) \sin\left(\frac{55^{\circ} - 35^{\circ}}{2}\right) = 2 \cos 45^{\circ} \sin 10^{\circ}$.
Now,the expression becomes $\frac{2 \cos 45^{\circ} \sin 10^{\circ}}{\sin 10^{\circ}} = 2 \cos 45^{\circ}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the value is $2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.

(D) Given equations are $2y = 1 \implies y = \frac{1}{2}$ and $y = \sin x$.
To find the points of intersection,we solve $\sin x = \frac{1}{2}$ for $x \in [-2\pi, 2\pi]$.
In the interval $[0, 2\pi]$,$\sin x = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$.
In the interval $[-2\pi, 0]$,$\sin x = \frac{1}{2}$ at $x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$ and $x = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6}$.
Thus,the solutions are $x \in \{\frac{\pi}{6}, \frac{5\pi}{6}, -\frac{7\pi}{6}, -\frac{11\pi}{6}\}$.
There are $4$ such points of intersection.

(A) Given equation: $\sin 6 \theta + \sin 4 \theta + \sin 2 \theta = 0$
Grouping terms: $(\sin 6 \theta + \sin 2 \theta) + \sin 4 \theta = 0$
Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:
$2 \sin 4 \theta \cos 2 \theta + \sin 4 \theta = 0$
Factoring out $\sin 4 \theta$:
$\sin 4 \theta (2 \cos 2 \theta + 1) = 0$
This gives two cases:
Case $1$: $\sin 4 \theta = 0$ $\Rightarrow 4 \theta = n \pi$ $\Rightarrow \theta = \frac{n \pi}{4}$
Case $2$: $2 \cos 2 \theta + 1 = 0 \Rightarrow \cos 2 \theta = -\frac{1}{2} = \cos \frac{2 \pi}{3}$
General solution for $\cos x = \cos \alpha$ is $x = 2 n \pi \pm \alpha$:
$2 \theta = 2 n \pi \pm \frac{2 \pi}{3} \Rightarrow \theta = n \pi \pm \frac{\pi}{3}$
Thus,$\theta = \frac{n \pi}{4}$ or $\theta = n \pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$.

(B) Let the line $3x + y - 9 = 0$ divide the line segment joining $A(1, 3)$ and $B(2, 7)$ in the ratio $k: 1$.
Using the section formula,the coordinates of the dividing point are $(\frac{2k+1}{k+1}, \frac{7k+3}{k+1})$.
Since this point lies on the line $3x + y = 9$,we have:
$3(\frac{2k+1}{k+1}) + (\frac{7k+3}{k+1}) = 9$
$6k + 3 + 7k + 3 = 9(k + 1)$
$13k + 6 = 9k + 9$
$4k = 3$
$k = \frac{3}{4}$.
Since $k > 0$,the division is internal in the ratio $3: 4$.

(A) The given equations are $y = \sqrt{3}x$,$y = -\sqrt{3}x$,and $y = 1$.
These can be written as $y = \tan(60^{\circ})x$,$y = \tan(120^{\circ})x$,and $y = 1$.
The lines $y = \sqrt{3}x$ and $y = -\sqrt{3}x$ pass through the origin $(0,0)$ and make angles of $60^{\circ}$ and $120^{\circ}$ with the positive $x$-axis,respectively.
The angle between these two lines is $120^{\circ} - 60^{\circ} = 60^{\circ}$.
The line $y = 1$ intersects $y = \sqrt{3}x$ at $(\frac{1}{\sqrt{3}}, 1)$ and $y = -\sqrt{3}x$ at $(-\frac{1}{\sqrt{3}}, 1)$.
The length of the base on the line $y = 1$ is $\frac{1}{\sqrt{3}} - (-\frac{1}{\sqrt{3}}) = \frac{2}{\sqrt{3}}$.
The lengths of the other two sides are $\sqrt{(\frac{1}{\sqrt{3}})^2 + 1^2} = \sqrt{\frac{1}{3} + 1} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
Since all three sides are equal to $\frac{2}{\sqrt{3}}$,the triangle is an equilateral triangle.

(D) The given line is $x+y=0$,which can be written as $y = -x$. The slope of this line is $m_1 = -1$.
Let the slope of the required lines be $m$. The angle between the lines is $45^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m m_1} \right|$,we have:
$\tan 45^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} \right|$
$1 = \left| \frac{m+1}{1-m} \right|$
This gives two cases:
Case $1$: $\frac{m+1}{1-m} = 1 \implies m+1 = 1-m \implies 2m = 0 \implies m = 0$.
The equation of the line passing through $(1,1)$ with slope $0$ is $y-1 = 0(x-1) \implies y-1 = 0$.
Case $2$: $\frac{m+1}{1-m} = -1 \implies m+1 = -1+m \implies 1 = -1$,which is impossible (this implies the line is vertical).
$A$ vertical line passing through $(1,1)$ is $x=1$,or $x-1=0$.
Thus,the equations of the lines are $x-1=0$ and $y-1=0$.

(C) Let the points be $A(3q, 0)$,$B(0, 3p)$,and $C(1, 1)$.
Since the points are collinear,the slope of $AC$ must be equal to the slope of $BC$.
Slope of $AC = \frac{1 - 0}{1 - 3q} = \frac{1}{1 - 3q}$.
Slope of $BC = \frac{3p - 1}{0 - 1} = \frac{3p - 1}{-1} = 1 - 3p$.
Equating the slopes: $\frac{1}{1 - 3q} = 1 - 3p$.
$1 = (1 - 3p)(1 - 3q)$.
$1 = 1 - 3q - 3p + 9pq$.
$3p + 3q = 9pq$.
Dividing both sides by $3pq$,we get $\frac{3p}{3pq} + \frac{3q}{3pq} = \frac{9pq}{3pq}$.
$\frac{1}{q} + \frac{1}{p} = 3$.

(C) Let the two mutually perpendicular lines be the coordinate axes,$x = 0$ and $y = 0$.
Let the point $P$ be $(x, y)$.
The distance of $P$ from the line $x = 0$ is $|x|$ and the distance of $P$ from the line $y = 0$ is $|y|$.
According to the problem,the sum of these distances is $1$,so $|x| + |y| = 1$.
This equation represents a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.
Since a square is a special type of rhombus and is not listed in the options,we must re-examine the problem statement. If the question implies the sum of the squares of the distances is constant,it would be a circle. However,for the sum of distances $|x| + |y| = 1$,the locus is a square. Given the standard nature of such problems,if this were a multiple-choice question where one must be selected,there is a discrepancy in the provided options as none represent a square.

(A) Given equations are:
$L_1: \sqrt{3} x - y = 4 \sqrt{3} \alpha$ $(1)$
$L_2: \alpha(\sqrt{3} x + y) = 4 \sqrt{3} \Rightarrow \sqrt{3} x + y = \frac{4 \sqrt{3}}{\alpha}$ $(2)$
Multiplying equation $(1)$ and $(2)$,we get:
$(\sqrt{3} x - y)(\sqrt{3} x + y) = (4 \sqrt{3} \alpha) \times \left(\frac{4 \sqrt{3}}{\alpha}\right)$
$3x^2 - y^2 = 16 \times 3 = 48$
Dividing by $48$,we get:
$\frac{x^2}{16} - \frac{y^2}{48} = 1$
This is the equation of a hyperbola where $a^2 = 16$ and $b^2 = 48$.
The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2$.
Thus,it is a hyperbola with eccentricity $2$.

(A) The equation of any circle passing through the intersection of the circle $S: x^2+y^2-6x-8y=0$ and the line $L: x+y-1=0$ is given by $S + \lambda L = 0$.
$x^2+y^2-6x-8y + \lambda(x+y-1) = 0$
$x^2+y^2 + (\lambda-6)x + (\lambda-8)y - \lambda = 0$.
Since $AB$ is a diameter,the center of this circle must lie on the line $x+y-1=0$.
The center of the circle is $(-\frac{\lambda-6}{2}, -\frac{\lambda-8}{2})$.
Substituting the center into the line equation:
$-\frac{\lambda-6}{2} - \frac{\lambda-8}{2} - 1 = 0$
$-(\lambda-6) - (\lambda-8) - 2 = 0$
$-\lambda + 6 - \lambda + 8 - 2 = 0$
$-2\lambda + 12 = 0 \implies \lambda = 6$.
Substituting $\lambda = 6$ into the circle equation:
$x^2+y^2 + (6-6)x + (6-8)y - 6 = 0$
$x^2+y^2-2y-6=0$.

(C) For a parabola $y^2 = 4ax$,the coordinates of any point on it can be represented as $(at^2, 2at)$.
Let the two end points of a focal chord be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.
The slope of the chord $PQ$ is $m = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$.
Since the chord passes through the focus $(a, 0)$,the slope is also $m = \frac{2at_1 - 0}{at_1^2 - a} = \frac{2at_1}{a(t_1^2 - 1)} = \frac{2t_1}{t_1^2 - 1}$.
Equating the slopes: $\frac{2}{t_1 + t_2} = \frac{2t_1}{t_1^2 - 1}$.
$t_1^2 - 1 = t_1(t_1 + t_2) = t_1^2 + t_1 t_2$.
$-1 = t_1 t_2$.
Therefore,$t_1 t_2 = -1$.

(C) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Foci are $S(ae, 0)$ and $T(-ae, 0)$.
$B(0, b)$ is the end point of the minor axis.
Since $\triangle STB$ is an equilateral triangle,the distance $SB = ST = TB$.
$ST = ae - (-ae) = 2ae$.
$SB = \sqrt{(ae-0)^{2} + (0-b)^{2}} = \sqrt{a^{2}e^{2} + b^{2}}$.
Since $SB = ST$,we have $SB^{2} = ST^{2}$.
$a^{2}e^{2} + b^{2} = (2ae)^{2} = 4a^{2}e^{2}$.
$b^{2} = 3a^{2}e^{2}$.
Using the relation $b^{2} = a^{2}(1-e^{2})$,we get:
$a^{2}(1-e^{2}) = 3a^{2}e^{2}$.
$1 - e^{2} = 3e^{2}$.
$4e^{2} = 1$.
$e^{2} = \frac{1}{4}$.
$e = \frac{1}{2}$ (since eccentricity $e > 0$).

(D) To find the limit $\lim _{x \rightarrow 0} \frac{\sin |x|}{x}$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.
For $x < 0$,$|x| = -x$,so $\lim _{x \rightarrow 0^-} \frac{\sin(-x)}{x} = \lim _{x \rightarrow 0^-} \frac{-\sin x}{x} = -1$.
For $x > 0$,$|x| = x$,so $\lim _{x \rightarrow 0^+} \frac{\sin x}{x} = 1$.
Since the left-hand limit $(-1)$ is not equal to the right-hand limit $(1)$,the limit does not exist.

(D) Let $L = \lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$.
Since the limit is in the $\frac{0}{0}$ form,we apply $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$:
$L = \lim_{x \rightarrow 5} \frac{\frac{d}{dx}(x f(5)-5 f(x))}{\frac{d}{dx}(x-5)}$
$L = \lim_{x \rightarrow 5} \frac{f(5)-5 f'(x)}{1}$
Substituting $x=5$:
$L = f(5)-5 f'(5)$
Given $f(5)=7$ and $f'(5)=7$:
$L = 7 - 5(7) = 7 - 35 = -28$.

(B) We need to evaluate the limit: $\operatorname{Lt}_{x \rightarrow 0} \frac{\sin^2 x + \cos x - 1}{x^2}$.
Using the identity $\sin^2 x = 1 - \cos^2 x$,the expression becomes:
$\operatorname{Lt}_{x \rightarrow 0} \frac{1 - \cos^2 x + \cos x - 1}{x^2} = \operatorname{Lt}_{x \rightarrow 0} \frac{\cos x - \cos^2 x}{x^2}$.
Factor out $\cos x$:
$\operatorname{Lt}_{x \rightarrow 0} \frac{\cos x(1 - \cos x)}{x^2} = \operatorname{Lt}_{x \rightarrow 0} \cos x \cdot \operatorname{Lt}_{x \rightarrow 0} \frac{1 - \cos x}{x^2}$.
We know that $\operatorname{Lt}_{x \rightarrow 0} \cos x = 1$ and $\operatorname{Lt}_{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.
Therefore,the limit is $1 \times \frac{1}{2} = \frac{1}{2}$.

(A) We use the standard limit formula $\operatorname{Lt}_{x \rightarrow a} [f(x)]^{g(x)} = e^{\operatorname{Lt}_{x \rightarrow a} g(x)[f(x)-1]}$ for the indeterminate form $1^{\infty}$.
Here,$f(x) = \frac{1+5x^2}{1+3x^2}$ and $g(x) = \frac{1}{x^2}$.
As $x \rightarrow 0$,$f(x) \rightarrow 1$ and $g(x) \rightarrow \infty$.
Thus,the limit is $e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2}{1+3x^2} - 1 \right)}$.
$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2 - (1+3x^2)}{1+3x^2} \right)}$.
$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{2x^2}{1+3x^2} \right)}$.
$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{2}{1+3x^2}}$.
$= e^{\frac{2}{1+0}} = e^2$.

(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A+C$. Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have $a = k \sin A, b = k \sin B, c = k \sin C$.
Therefore,$\frac{a+c}{b} = \frac{\sin A + \sin C}{\sin B}$.
Using the sum-to-product formula,$\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$.
Since $A+C = 2B$,we have $\frac{A+C}{2} = B$.
Thus,$\frac{a+c}{b} = \frac{2 \sin B \cos \left(\frac{A-C}{2}\right)}{\sin B} = 2 \cos \left(\frac{A-C}{2}\right)$.

(B) We know that in a $\triangle ABC$,$A+B+C = \pi$,so $A+C = \pi - B$.
Substituting this into the expression,we get $\frac{A+C-B}{2} = \frac{\pi-B-B}{2} = \frac{\pi}{2} - B$.
Thus,$2ac \sin \left(\frac{A-B+C}{2}\right) = 2ac \sin \left(\frac{\pi}{2}-B\right)$.
Using the identity $\sin \left(\frac{\pi}{2}-\theta\right) = \cos \theta$,we get $2ac \cos B$.
From the Law of Cosines,$\cos B = \frac{a^2+c^2-b^2}{2ac}$.
Substituting this,we get $2ac \left(\frac{a^2+c^2-b^2}{2ac}\right) = a^2+c^2-b^2$.

(A) The given equation of the circle is $x^2+y^2-2x+4y-5=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=2$.
The center of the circle $C$ is $(-g, -f) = (1, -2)$.
The normal to a circle at any point on its circumference always passes through the center of the circle.
Therefore,the normal is the line passing through the center $C(1, -2)$ and the given point $A(2, 1)$.
The slope $m$ of the line passing through $(1, -2)$ and $(2, 1)$ is given by $m = \frac{1 - (-2)}{2 - 1} = \frac{3}{1} = 3$.
The equation of the line passing through $(2, 1)$ with slope $m=3$ is $(y - 1) = 3(x - 2)$.
Simplifying this,we get $y - 1 = 3x - 6$,which results in $y = 3x - 5$.

(D) Let $A$ be the event of getting an even number on the first die. The possible outcomes for $A$ are $(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$. Thus,$|A| = 18$.
Let $B$ be the event of getting a sum of $8$. The possible outcomes for $B$ are $(2,6), (3,5), (4,4), (5,3), (6,2)$. Thus,$|B| = 5$.
The intersection $A \cap B$ contains outcomes where the first die is even and the sum is $8$,which are $(2,6), (4,4), (6,2)$. Thus,$|A \cap B| = 3$.
Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get:
$P(A \cup B) = \frac{18}{36} + \frac{5}{36} - \frac{3}{36} = \frac{20}{36} = \frac{5}{9}$.

(C) Given that $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.3$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
So,$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.6 + 0.3 = 0.9$.
We need to find $P(A') + P(B')$.
Since $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$,
$P(A') + P(B') = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the value,$P(A') + P(B') = 2 - 0.9 = 1.1$.

(D) For the function $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
Using the identity $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$,we have:
$\lim_{x \to 0} \frac{1 - \cos(1 - \cos x)}{x^4} = \lim_{x \to 0} \frac{2 \sin^2(\frac{1 - \cos x}{2})}{x^4}$
Since $1 - \cos x = 2 \sin^2(\frac{x}{2})$,this becomes:
$\lim_{x \to 0} \frac{2 \sin^2(\sin^2(\frac{x}{2}))}{x^4}$
Using the limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we approximate $\sin \theta \approx \theta$ for small $\theta$:
$\lim_{x \to 0} \frac{2 (\sin^2(\frac{x}{2}))^2}{x^4} = \lim_{x \to 0} \frac{2 (\frac{x}{2})^4}{x^4} = \lim_{x \to 0} \frac{2 \cdot \frac{x^4}{16}}{x^4} = \frac{2}{16} = \frac{1}{8}$.

(A) To find the product $AB$,we multiply the rows of matrix $A$ by the columns of matrix $B$:
$AB = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix}$
$AB = \begin{bmatrix} (2 \times 1 + 1 \times 0 + 3 \times 5) & (2 \times -1 + 1 \times 2 + 3 \times 0) \\ (4 \times 1 + 1 \times 0 + 0 \times 5) & (4 \times -1 + 1 \times 2 + 0 \times 0) \end{bmatrix}$
$AB = \begin{bmatrix} (2 + 0 + 15) & (-2 + 2 + 0) \\ (4 + 0 + 0) & (-4 + 2 + 0) \end{bmatrix}$
$AB = \begin{bmatrix} 17 & 0 \\ 4 & -2 \end{bmatrix}$

(A) Given matrix $A = \begin{bmatrix} 1 & 2 \\ -4 & -1 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = (1)(-1) - (2)(-4) = -1 + 8 = 7$.
Since $|A| \neq 0$,$A^{-1}$ exists.
The adjoint of a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Therefore,$\text{adj}(A) = \begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix}$.
The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
$A^{-1} = \frac{1}{7} \begin{bmatrix} -1 & -2 \\ 4 & 1 \end{bmatrix}$.

(B) Given the determinant equation: $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$.
Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:
Since $1 + \omega + \omega^2 = 0$,the first column becomes:
$C_1 = \begin{bmatrix} x + \omega^2 + \omega + 1 \\ \omega + \omega^2 + 1 + x \\ 1 + x + \omega + \omega^2 \end{bmatrix} = \begin{bmatrix} x \\ x \\ x \end{bmatrix}$.
Thus,the determinant is $x \left|\begin{array}{ccc} 1 & \omega & 1 \\ 1 & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2 \end{array}\right| = 0$.
This implies $x = 0$ is one of the solutions.

(B) Given expression is $\tan ^{-1}\left(\frac{\sin 2-1}{\cos 2}\right)$.
Using the identities $\sin 2 = 2\sin 1 \cos 1$,$1 = \sin^2 1 + \cos^2 1$,and $\cos 2 = \cos^2 1 - \sin^2 1$:
$\frac{\sin 2 - 1}{\cos 2} = \frac{2\sin 1 \cos 1 - (\sin^2 1 + \cos^2 1)}{\cos^2 1 - \sin^2 1} = \frac{-(\cos 1 - \sin 1)^2}{(\cos 1 - \sin 1)(\cos 1 + \sin 1)}$.
Since $\cos 1 > \sin 1$ (as $1 \text{ radian} \approx 57.3^\circ$),we can simplify this to $\frac{-(\cos 1 - \sin 1)}{\cos 1 + \sin 1} = \frac{\sin 1 - \cos 1}{\cos 1 + \sin 1}$.
Dividing numerator and denominator by $\cos 1$,we get $\frac{\tan 1 - 1}{1 + \tan 1} = \tan(1 - \frac{\pi}{4})$.
Thus,$\tan ^{-1}(\tan(1 - \frac{\pi}{4})) = 1 - \frac{\pi}{4}$.

(B) For the function $f(x) = \sqrt{\cos^{-1}\left(\frac{1-|x|}{2}\right)}$ to be defined,the expression inside the square root must be non-negative,and the argument of $\cos^{-1}$ must be in the interval $[-1, 1]$.
First,we require $\cos^{-1}\left(\frac{1-|x|}{2}\right) \geq 0$. Since the range of $\cos^{-1}(u)$ is $[0, \pi]$,this is always true for any $u \in [-1, 1]$.
Next,we solve for the domain of $\cos^{-1}\left(\frac{1-|x|}{2}\right)$ by setting $-1 \leq \frac{1-|x|}{2} \leq 1$.
Multiplying by $2$: $-2 \leq 1 - |x| \leq 2$.
Subtracting $1$: $-3 \leq -|x| \leq 1$.
Multiplying by $-1$ (and reversing the inequalities): $-1 \leq |x| \leq 3$.
Since $|x|$ is always non-negative,$|x| \leq 3$ implies $-3 \leq x \leq 3$.
Thus,the domain is $x \in [-3, 3]$.

(B) To determine if the function $f(x)$ is even or odd,we check $f(-x)$.
Let $g(x) = \log \left( x + \sqrt{1 + x^2} \right)$.
Then $g(-x) = \log \left( -x + \sqrt{1 + (-x)^2} \right) = \log \left( \sqrt{1 + x^2} - x \right)$.
Multiplying and dividing by $\sqrt{1 + x^2} + x$,we get $g(-x) = \log \left( \frac{(\sqrt{1 + x^2} - x)(\sqrt{1 + x^2} + x)}{\sqrt{1 + x^2} + x} \right) = \log \left( \frac{1 + x^2 - x^2}{\sqrt{1 + x^2} + x} \right) = \log \left( \frac{1}{\sqrt{1 + x^2} + x} \right) = -\log \left( x + \sqrt{1 + x^2} \right) = -g(x)$.
Thus,$g(x)$ is an odd function.
Now,$f(x) = \sec(g(x))$.
Since $\sec(- \theta) = \sec(\theta)$,we have $f(-x) = \sec(g(-x)) = \sec(-g(x)) = \sec(g(x)) = f(x)$.
Therefore,$f(x)$ is an even function.

(A) Given functions are $f(x) = 5 - x^2$ and $g(x) = 3x - 4$.
To find $(f \circ g)(-1)$,we use the definition of composition of functions: $(f \circ g)(x) = f(g(x))$.
First,calculate $g(-1)$:
$g(-1) = 3(-1) - 4 = -3 - 4 = -7$.
Now,substitute this value into $f(x)$:
$(f \circ g)(-1) = f(g(-1)) = f(-7)$.
Using the definition of $f(x)$:
$f(-7) = 5 - (-7)^2 = 5 - 49 = -44$.
Thus,the value of $(f \circ g)(-1)$ is $-44$.

(C) Given $f(x) = x + 2$ where $x \in \{1, 2, 3, 4\}$.
Calculating the values of the function:
$f(1) = 1 + 2 = 3$
$f(2) = 2 + 2 = 4$
$f(3) = 3 + 2 = 5$
$f(4) = 4 + 2 = 6$
Since each element in set $A$ has a distinct image in set $B$,the function is one-one.
Since the range $\{3, 4, 5, 6\}$ is not equal to the codomain $\{1, 2, 3, 4, 5, 6\}$,the function is not onto.
Therefore,the function is one-one.

(C) Given,$y=(1+x)(1+x^{2})(1+x^{4}) \ldots (1+x^{2^{n}})$.
Taking the natural logarithm on both sides:
$\log y = \log(1+x) + \log(1+x^{2}) + \log(1+x^{4}) + \ldots + \log(1+x^{2^{n}})$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{d y}{d x} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}}$.
Thus,$\frac{d y}{d x} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \ldots + \frac{2^{n}x^{2^{n}-1}}{1+x^{2^{n}}} \right]$.
At $x=0$,$y = (1+0)(1+0) \ldots (1+0) = 1$.
Substituting $x=0$ in the derivative expression:
$\left(\frac{d y}{d x}\right)_{x=0} = 1 \left[ \frac{1}{1+0} + 0 + 0 + \ldots + 0 \right] = 1 \times 1 = 1$.

(B) Given the curve $y = x^2 - 3x + 2$.
Find the derivative to get the slope of the tangent: $\frac{dy}{dx} = 2x - 3$.
The given line is $y = x$,which has a slope $m_1 = 1$.
Since the tangent is perpendicular to the line $y = x$,the slope of the tangent $m_2$ must satisfy $m_1 \times m_2 = -1$.
Therefore,$1 \times (2x - 3) = -1$,which implies $2x - 3 = -1$.
Solving for $x$: $2x = 2$,so $x = 1$.
Substitute $x = 1$ into the curve equation to find the $y$-coordinate: $y = (1)^2 - 3(1) + 2 = 1 - 3 + 2 = 0$.
Thus,the point is $(1, 0)$.

(C) The equation of the curve is $xy = 4$,which can be written as $y = \frac{4}{x}$.
Taking the derivative with respect to $x$,we get $\frac{dy}{dx} = -\frac{4}{x^2}$.
The equation of the line is $ax + by + c = 0$,which can be rewritten as $y = -\frac{a}{b}x - \frac{c}{b}$.
The slope of this line is $m = -\frac{a}{b}$.
Since the line is a tangent to the curve,the slope of the tangent at any point $(x, y)$ on the curve must equal the slope of the line:
$-\frac{4}{x^2} = -\frac{a}{b} \implies \frac{a}{b} = \frac{4}{x^2}$.
Since $x^2 > 0$ for all $x \neq 0$,it follows that $\frac{a}{b} > 0$.
This implies that $a$ and $b$ must have the same sign.
Looking at the options,the condition $a < 0$ and $b < 0$ satisfies $\frac{a}{b} > 0$.

(B) We know that velocity $v = \frac{dx}{dt}$,so $\frac{dt}{dx} = \frac{1}{v}$.
Now,differentiate with respect to $x$:
$\frac{d^2 t}{dx^2} = \frac{d}{dx} \left( \frac{1}{v} \right) = -\frac{1}{v^2} \frac{dv}{dx}$.
Using the chain rule,$\frac{dv}{dx} = \frac{dv}{dt} \times \frac{dt}{dx} = f \times \frac{1}{v} = \frac{f}{v}$.
Substituting this back into the equation:
$\frac{d^2 t}{dx^2} = -\frac{1}{v^2} \times \frac{f}{v} = -\frac{f}{v^3}$.
Rearranging for $f$,we get $f = -v^3 \frac{d^2 t}{dx^2}$.

(B) Given the displacement equation: $x = At^2 + Bt + C$.
Velocity $v$ is the rate of change of displacement with respect to time $t$: $v = \frac{dx}{dt} = 2At + B$.
Now,calculate $v^2$: $v^2 = (2At + B)^2 = 4A^2t^2 + 4ABt + B^2$.
Next,calculate $4Ax$: $4Ax = 4A(At^2 + Bt + C) = 4A^2t^2 + 4ABt + 4AC$.
Now,subtract $v^2$ from $4Ax$: $4Ax - v^2 = (4A^2t^2 + 4ABt + 4AC) - (4A^2t^2 + 4ABt + B^2)$.
Simplifying the expression: $4Ax - v^2 = 4AC - B^2$.

(B) To find the intervals where the function $f(x) = x^4 - 4x^3 + 4x^2 + 40$ is monotonically decreasing,we first find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^4 - 4x^3 + 4x^2 + 40) = 4x^3 - 12x^2 + 8x$.
Next,we factor the derivative:
$f'(x) = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2)$.
$A$ function is monotonically decreasing when $f'(x) < 0$.
We analyze the sign of $f'(x) = 4x(x - 1)(x - 2)$ using the wavy curve method (sign scheme):
- For $x < 0$,$f'(x) < 0$.
- For $0 < x < 1$,$f'(x) > 0$.
- For $1 < x < 2$,$f'(x) < 0$.
- For $x > 2$,$f'(x) > 0$.
Thus,the function is monotonically decreasing for $x \in (-\infty, 0) \cup (1, 2)$.
Comparing this with the given options,the correct interval is $1 < x < 2$.

(C) Given function is $f(x) = e^{(x^4 - x^3 + x^2)}$.
To find the minimum value,we find the derivative $f'(x)$:
$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot \frac{d}{dx}(x^4 - x^3 + x^2)$
$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot (4x^3 - 3x^2 + 2x)$
$f'(x) = e^{(x^4 - x^3 + x^2)} \cdot x(4x^2 - 3x + 2)$.
For the quadratic factor $4x^2 - 3x + 2$,the discriminant $D = (-3)^2 - 4(4)(2) = 9 - 32 = -23 < 0$.
Since the leading coefficient $4 > 0$ and $D < 0$,the expression $4x^2 - 3x + 2$ is always positive for all real $x$.
Thus,the sign of $f'(x)$ depends only on $x$.
For $x < 0$,$f'(x) < 0$,so the function is decreasing.
For $x > 0$,$f'(x) > 0$,so the function is increasing.
Therefore,the function attains its minimum value at $x = 0$.
The minimum value is $f(0) = e^{(0^4 - 0^3 + 0^2)} = e^0 = 1$.

(A) The displacement is given by $x = t^4 - k t^3$.
The velocity $v$ is the derivative of displacement with respect to time $t$:
$v = \frac{dx}{dt} = 4t^3 - 3kt^2$.
To find the condition for minimum velocity,we find the acceleration $a = \frac{dv}{dt}$:
$a = \frac{dv}{dt} = 12t^2 - 6kt$.
For the velocity to be minimum at $t = 2$,the derivative of velocity with respect to time must be zero at $t = 2$:
$\frac{dv}{dt} \big|_{t=2} = 12(2)^2 - 6k(2) = 0$.
$12(4) - 12k = 0$.
$48 - 12k = 0$.
$12k = 48$.
$k = 4$.
Thus,the value of $k$ is $4$.

(B) To find the point where the slope is maximum,we need to maximize $f'(x)$. Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.
For $g(x)$ to have a maximum,we set $g'(x) = f''(x) = 0$.
$f'(x) = e^x(\sin x + \cos x)$
$f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.
Setting $f''(x) = 0$ gives $2e^x \cos x = 0$. Since $e^x \neq 0$,we have $\cos x = 0$.
In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.
We check the second derivative of $g(x)$,which is $g'(x) = f''(x) = 2e^x \cos x$.
$g''(x) = f'''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.
At $x = \frac{\pi}{2}$,$g''(\frac{\pi}{2}) = 2e^{\pi/2}(0 - 1) = -2e^{\pi/2} < 0$ (Local maximum).
At $x = \frac{3\pi}{2}$,$g''(\frac{3\pi}{2}) = 2e^{3\pi/2}(0 - (-1)) = 2e^{3\pi/2} > 0$ (Local minimum).
Thus,the slope is maximum at $x = \frac{\pi}{2}$.

(D) For Rolle's theorem to be applicable to a function $f(x)$ on $[a, b]$,the following conditions must be met:
$1$. $f(x)$ is continuous on $[a, b]$.
$2$. $f(x)$ is differentiable on $(a, b)$.
$3$. $f(a) = f(b)$.
Checking the options:
$(A)$ $f(x)=|x|$ is not differentiable at $x=0$,which lies in $(-2, 2)$. Thus,Rolle's theorem is not applicable.
$(B)$ $f(x)=\tan x$ is not continuous at $x=\frac{\pi}{2}$,which lies in $[0, \pi]$. Thus,Rolle's theorem is not applicable.
$(C)$ $f(x)=1+(x-2)^{\frac{2}{3}}$ is not differentiable at $x=2$,which lies in $(1, 3)$ because $f'(x) = \frac{2}{3}(x-2)^{-\frac{1}{3}}$. Thus,Rolle's theorem is not applicable.
$(D)$ $f(x)=x(x-2)^2$ is a polynomial function,so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$. Also,$f(0) = 0(0-2)^2 = 0$ and $f(2) = 2(2-2)^2 = 0$. Since $f(0) = f(2)$,all conditions are satisfied. Therefore,Rolle's theorem is applicable.

(B) We know that $1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)$.
Substituting this into the integral,we get:
$\int \sqrt{2 \cos^2 \left(\frac{x}{2}\right)} \, dx = \int \sqrt{2} \left| \cos \left(\frac{x}{2}\right) \right| \, dx$.
Assuming $\cos \left(\frac{x}{2}\right) > 0$,we have:
$\sqrt{2} \int \cos \left(\frac{x}{2}\right) \, dx$.
Integrating with respect to $x$:
$\sqrt{2} \cdot \frac{\sin \left(\frac{x}{2}\right)}{1/2} + C = 2 \sqrt{2} \sin \left(\frac{x}{2}\right) + C$.

(A) Let $I = \int \frac{\log \sqrt{x}}{3 x} d x$.
Substitute $z = \log \sqrt{x} = \frac{1}{2} \log x$.
Then,$dz = \frac{1}{2} \cdot \frac{1}{x} dx$,which implies $\frac{dx}{x} = 2 dz$.
Substituting these into the integral:
$I = \int \frac{z}{3} (2 dz) = \frac{2}{3} \int z dz$.
Integrating $z$ with respect to $z$:
$I = \frac{2}{3} \cdot \frac{z^2}{2} + C = \frac{1}{3} z^2 + C$.
Substituting back $z = \log \sqrt{x}$:
$I = \frac{1}{3} (\log \sqrt{x})^2 + C$.

(C) Let $I = \int \frac{dx}{(e^x + e^{-x})^2}$.
Multiply the numerator and denominator by $e^{2x}$:
$I = \int \frac{e^{2x} dx}{(e^x \cdot e^x + e^{-x} \cdot e^x)^2} = \int \frac{e^{2x} dx}{(e^{2x} + 1)^2}$.
Let $u = e^{2x} + 1$. Then $du = 2e^{2x} dx$,which implies $e^{2x} dx = \frac{du}{2}$.
Substituting these into the integral:
$I = \int \frac{du/2}{u^2} = \frac{1}{2} \int u^{-2} du$.
Integrating with respect to $u$:
$I = \frac{1}{2} \left( \frac{u^{-1}}{-1} \right) + C = -\frac{1}{2u} + C$.
Substituting $u = e^{2x} + 1$ back:
$I = -\frac{1}{2(e^{2x} + 1)} + C = -\frac{1}{2}(e^{2x} + 1)^{-1} + C$.

(C) We know the standard integral formula: $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Given the integral: $I = \int e^{x} \left(\frac{2}{x} - \frac{2}{x^2}\right) dx$.
Let $f(x) = \frac{2}{x}$.
Then,$f'(x) = \frac{d}{dx} (2x^{-1}) = -2x^{-2} = -\frac{2}{x^2}$.
Substituting these into the formula: $I = \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Therefore,$I = e^{x} \left(\frac{2}{x}\right) + C = \frac{2 e^{x}}{x} + C$.

(A) Given that $\frac{d}{dx}\{f(x)\} = g(x)$.
Let $I = \int_a^b f(x) g(x) dx$.
Substitute $g(x) dx = df(x)$ into the integral:
$I = \int_{f(a)}^{f(b)} f(x) df(x)$.
Using the power rule for integration $\int u du = \frac{u^2}{2} + C$:
$I = \left[ \frac{\{f(x)\}^2}{2} \right]_a^b$.
$I = \frac{1}{2} \left[ f^2(b) - f^2(a) \right]$.

(C) To evaluate the integral $I = \int_0^{\pi / 2} \sin^5 x \, dx$,we use the Wallis formula for $\int_0^{\pi / 2} \sin^n x \, dx$,which is $\frac{(n-1)!!}{n!!} \times \frac{\pi}{2}$ if $n$ is even,and $\frac{(n-1)!!}{n!!}$ if $n$ is odd.
For $n = 5$ (which is odd),the formula gives:
$I = \frac{(5-1) \times (5-3)}{5 \times 3 \times 1} = \frac{4 \times 2}{5 \times 3 \times 1} = \frac{8}{15}$.
Alternatively,using substitution:
$I = \int_0^{\pi / 2} \sin^4 x \sin x \, dx = \int_0^{\pi / 2} (1 - \cos^2 x)^2 \sin x \, dx$.
Let $u = \cos x$,then $du = -\sin x \, dx$. When $x = 0, u = 1$; when $x = \pi / 2, u = 0$.
$I = -\int_1^0 (1 - u^2)^2 \, du = \int_0^1 (1 - 2u^2 + u^4) \, du$.
$I = [u - \frac{2u^3}{3} + \frac{u^5}{5}]_0^1 = 1 - \frac{2}{3} + \frac{1}{5} = \frac{15 - 10 + 3}{15} = \frac{8}{15}$.

(B) Given the integral $I = \int_{-\pi / 2}^{\pi / 2} |\sin x| \, dx$.
Since $|\sin x|$ is an even function,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$.
Thus,$I = 2 \int_{0}^{\pi / 2} |\sin x| \, dx$.
In the interval $[0, \pi / 2]$,$\sin x \geq 0$,so $|\sin x| = \sin x$.
Therefore,$I = 2 \int_{0}^{\pi / 2} \sin x \, dx$.
Evaluating the integral: $I = 2 [-\cos x]_{0}^{\pi / 2}$.
$I = 2 [-\cos(\pi / 2) - (-\cos(0))]$.
$I = 2 [0 - (-1)] = 2(1) = 2$.

(C) Let $g(x) = f(\cos^2 x)$.
Since $\cos^2(x + \pi) = (-\cos x)^2 = \cos^2 x$,the function $g(x)$ is periodic with period $\pi$.
Using the property of definite integrals for periodic functions,$\int_0^{n T} g(x) dx = n \int_0^T g(x) dx$,where $T$ is the period.
Here,$T = \pi$ and $n = 3$.
Therefore,$I_1 = \int_0^{3 \pi} f(\cos^2 x) dx = 3 \int_0^\pi f(\cos^2 x) dx$.
Since $I_2 = \int_0^\pi f(\cos^2 x) dx$,we have $I_1 = 3 I_2$.

(A) We are given the integral $I = \int_{0}^{1} \frac{dx}{1+x^{\pi / 2}}$.
Since $1 < \pi / 2 < 2$,for $x \in (0, 1)$,we have $x^2 < x^{\pi / 2} < x^1$.
Adding $1$ to all sides,we get $1+x^2 < 1+x^{\pi / 2} < 1+x$.
Taking the reciprocal reverses the inequality: $\frac{1}{1+x^2} > \frac{1}{1+x^{\pi / 2}} > \frac{1}{1+x}$.
Integrating from $0$ to $1$ with respect to $x$:
$\int_{0}^{1} \frac{dx}{1+x^2} > \int_{0}^{1} \frac{dx}{1+x^{\pi / 2}} > \int_{0}^{1} \frac{dx}{1+x}$.
Evaluating the integrals:
$[\tan^{-1}(x)]_{0}^{1} > I > [\log_{e}(1+x)]_{0}^{1}$.
$\frac{\pi}{4} > I > \log_{e}(2)$.
Thus,the correct relation is $\log_{e} 2 < I < \pi / 4$.

(B) The given curves are $y^2 = x$ (a parabola opening to the right) and $y = |x|$ (a $V$-shaped graph).
To find the points of intersection,we set $y^2 = x$ and $y = |x|$.
Since $y = |x|$,we have $y^2 = x^2$.
Substituting $y^2 = x$ into $x^2 = y^2$,we get $x^2 = x$,which implies $x(x - 1) = 0$.
Thus,the intersection points are at $x = 0$ and $x = 1$.
In the interval $[0, 1]$,the curve $y = \sqrt{x}$ lies above the line $y = x$.
The area $A$ is given by the integral:
$A = \int_{0}^{1} (\sqrt{x} - x) \, dx$
$A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{1}$
$A = \left( \frac{2}{3}(1)^{3/2} - \frac{1^2}{2} \right) - (0 - 0)$
$A = \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6} \text{ sq. unit}$.

(D) The area $A$ enclosed by the curve $y=3x-5$,the $x$-axis $(y=0)$,and the lines $x=3$ and $x=5$ is given by the definite integral:
$A = \int_{3}^{5} (3x-5) \, dx$
Evaluating the integral:
$A = \left[ \frac{3x^2}{2} - 5x \right]_{3}^{5}$
Substitute the upper limit $x=5$:
$\left( \frac{3(5)^2}{2} - 5(5) \right) = \left( \frac{75}{2} - 25 \right) = \frac{75-50}{2} = \frac{25}{2} = 12.5$
Substitute the lower limit $x=3$:
$\left( \frac{3(3)^2}{2} - 5(3) \right) = \left( \frac{27}{2} - 15 \right) = \frac{27-30}{2} = -\frac{3}{2} = -1.5$
Calculate the final area:
$A = 12.5 - (-1.5) = 12.5 + 1.5 = 14 \text{ sq. units}$

(D) Given parabolas are $y=4x^2 \implies x^2 = \frac{y}{4} \implies x = \pm \frac{\sqrt{y}}{2}$ and $y=\frac{x^2}{9} \implies x^2 = 9y \implies x = \pm 3\sqrt{y}$.
Since the region is symmetric about the $y$-axis,we calculate the area in the first quadrant and multiply by $2$.
In the first quadrant,the region is bounded by $x = 3\sqrt{y}$ and $x = \frac{\sqrt{y}}{2}$ from $y=0$ to $y=2$.
Area $A = 2 \int_{0}^{2} (3\sqrt{y} - \frac{\sqrt{y}}{2}) dy$.
$A = 2 \int_{0}^{2} (\frac{6\sqrt{y} - \sqrt{y}}{2}) dy = 2 \int_{0}^{2} \frac{5\sqrt{y}}{2} dy$.
$A = 5 \int_{0}^{2} y^{1/2} dy = 5 [\frac{y^{3/2}}{3/2}]_{0}^{2}$.
$A = 5 \times \frac{2}{3} [y^{3/2}]_{0}^{2} = \frac{10}{3} (2^{3/2} - 0)$.
$A = \frac{10}{3} (2\sqrt{2}) = \frac{20\sqrt{2}}{3}$ sq. units.

(C) The given differential equation is $x = 1 + \left(\frac{dy}{dx}\right) + \frac{1}{2!} \left(\frac{dy}{dx}\right)^2 + \frac{1}{3!} \left(\frac{dy}{dx}\right)^3 + \dots$
This series is the expansion of the exponential function $e^{\frac{dy}{dx}}$.
Thus,the equation can be written as $x = e^{\frac{dy}{dx}}$.
Taking the natural logarithm on both sides,we get $\ln(x) = \frac{dy}{dx}$.
The differential equation is $\frac{dy}{dx} = \ln(x)$.
In this equation,the highest order derivative is $\frac{dy}{dx}$,which is of order $1$.
The power of the highest order derivative is $1$.
Therefore,the degree of the differential equation is $1$.

(B) The slope of the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = y + 2x$.
This is a linear differential equation of the form $\frac{dy}{dx} - y = 2x$.
The integrating factor is $IF = e^{\int -1 dx} = e^{-x}$.
Multiplying both sides by $e^{-x}$,we get $e^{-x} \frac{dy}{dx} - y e^{-x} = 2x e^{-x}$.
This can be written as $\frac{d}{dx}(y e^{-x}) = 2x e^{-x}$.
Integrating both sides with respect to $x$:
$y e^{-x} = \int 2x e^{-x} dx$.
Using integration by parts: $\int 2x e^{-x} dx = 2x(-e^{-x}) - \int 2(-e^{-x}) dx = -2x e^{-x} - 2e^{-x} + C$.
So,$y e^{-x} = -2x e^{-x} - 2e^{-x} + C$.
Multiplying by $e^x$,we get $y = -2x - 2 + C e^x$.
For a curve passing through the origin $(0,0)$,$0 = 0 - 2 + C(1) \Rightarrow C = 2$.
Thus,$y = 2e^x - 2x - 2 = 2(e^x - x - 1)$.

(D) Given differential equation is $x \, dy - y \, dx = 0$.
Rearranging the terms,we get $x \, dy = y \, dx$.
Separating the variables,we have $\frac{dy}{y} = \frac{dx}{x}$.
Integrating both sides,we get $\int \frac{dy}{y} = \int \frac{dx}{x} + C$.
This results in $\ln|y| = \ln|x| + \ln|c|$,where $\ln|c|$ is the constant of integration.
Using logarithmic properties,$\ln|y| = \ln|cx|$,which implies $y = cx$.
The equation $y = cx$ represents a straight line passing through the origin.

(C) The given differential equation is $100 \frac{d^2 y}{dx^2}-20 \frac{dy}{dx}+y=0$.
To find the general solution,we write the auxiliary equation as $100 m^2 - 20 m + 1 = 0$.
This can be factored as $(10 m - 1)^2 = 0$.
Solving for $m$,we get $m = \frac{1}{10}$ as a repeated root.
For a second-order linear differential equation with a repeated root $m$,the general solution is given by $y = (c_1 + c_2 x) e^{mx}$.
Substituting $m = \frac{1}{10}$,we get $y = (c_1 + c_2 x) e^{\frac{x}{10}}$.

(D) The given differential equation is $y'' - 3y' + 2y = 0$.
The characteristic equation is $m^2 - 3m + 2 = 0$.
Factoring the quadratic,we get $(m - 1)(m - 2) = 0$,so the roots are $m = 1$ and $m = 2$.
The general solution is $y(x) = Ae^x + Be^{2x}$.
Differentiating with respect to $x$,we get $y'(x) = Ae^x + 2Be^{2x}$.
Using the initial conditions:
At $x = 0$,$y(0) = A + B = 1$.
At $x = 0$,$y'(0) = A + 2B = 0$.
Subtracting the first equation from the second,we get $B = -1$.
Substituting $B = -1$ into $A + B = 1$,we get $A = 2$.
Thus,the specific solution is $y(x) = 2e^x - e^{2x}$.
Now,evaluate $y$ at $x = \log_{e} 2$:
$y(\log_{e} 2) = 2e^{\log_{e} 2} - e^{2\log_{e} 2} = 2(2) - (e^{\log_{e} 2})^2 = 4 - (2)^2 = 4 - 4 = 0$.