WBJEE 2015 Mathematics Question Paper with Answer and Solution

(A) Let the remainder be $ax+b$ when $P(x)$ is divided by $(x-3)(x-5)$.
$P(x) = (x-3)(x-5)Q(x) + (ax+b)$
Given that $P(3) = 10$ and $P(5) = 6$.
Substituting $x=3$ in the equation: $3a+b = 10$ $(i)$
Substituting $x=5$ in the equation: $5a+b = 6$ $(ii)$
Subtracting equation $(i)$ from $(ii)$:
$(5a+b) - (3a+b) = 6 - 10$
$2a = -4 \Rightarrow a = -2$
Substituting $a = -2$ in equation $(i)$:
$3(-2) + b = 10$
$-6 + b = 10 \Rightarrow b = 16$
Thus,the remainder is $-2x+16$.

(C) We have,$\log _{0.2}(x-1) > \log _{0.04}(x+5)$
$\Rightarrow \log _{0.2}(x-1) > \log _{0.2^{2}}(x+5)$
$\Rightarrow \log _{0.2}(x-1) > \frac{1}{2} \log _{0.2}(x+5)$
$\Rightarrow 2 \log _{0.2}(x-1) > \log _{0.2}(x+5)$
$\Rightarrow \log _{0.2}(x-1)^{2} > \log _{0.2}(x+5)$
$\Rightarrow (x-1)^{2} < x+5$
$[\because \log _{a} x > \log _{a} y \Rightarrow x < y, \text{ if } 0 < a < 1]$
$\Rightarrow x^{2}-2x+1 < x+5$
$\Rightarrow x^{2}-3x-4 < 0$
$\Rightarrow (x-4)(x+1) < 0$
$\Rightarrow x \in (-1, 4)$
Also,for the logarithm to be defined,$x-1 > 0 \Rightarrow x > 1$ and $x+5 > 0 \Rightarrow x > -5$.
Combining $x \in (-1, 4)$ and $x > 1$,we get $x \in (1, 4)$.

(D) We have,$\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$
Factorizing the numerator and denominator:
$\frac{5x^{2}-25x-x+5}{3x^{2}-9x-x+3} < 0$
$\frac{5x(x-5)-1(x-5)}{3x(x-3)-1(x-3)} < 0$
$\frac{(x-5)(5x-1)}{(x-3)(3x-1)} < 0$
The critical points are $x = \frac{1}{5}, \frac{1}{3}, 3, 5$.
Using the wavy curve method (sign scheme) for the expression $f(x) = \frac{(x-5)(5x-1)}{(x-3)(3x-1)}$,the expression is negative in the intervals $(\frac{1}{5}, \frac{1}{3})$ and $(3, 5)$.
Therefore,$x \in (\frac{1}{5}, \frac{1}{3}) \cup (3, 5)$.

(A) Given that $\alpha$ and $\beta$ are the roots of $x^{2}-px+1=0$.
By the relation between roots and coefficients,$\alpha+\beta=p$ and $\alpha\beta=1$.
Also,$\gamma$ is a root of $x^{2}+px+1=0$,so $\gamma^{2}+p\gamma+1=0$,which implies $\gamma^{2}=-p\gamma-1$.
Now,consider the expression $(\alpha+\gamma)(\beta+\gamma) = \alpha\beta + \gamma(\alpha+\beta) + \gamma^{2}$.
Substituting the known values:
$(\alpha+\gamma)(\beta+\gamma) = 1 + \gamma(p) + (-p\gamma-1)$.
$= 1 + p\gamma - p\gamma - 1 = 0$.

(C) Given the quadratic expression is $(2x+1)^{2} - px + q \neq 0$ for all real $x$.
Expanding the expression:
$4x^{2} + 4x + 1 - px + q \neq 0$
$4x^{2} + (4-p)x + (1+q) \neq 0$
For a quadratic expression $ax^{2} + bx + c$ to be non-zero for all real $x$,its discriminant $D$ must be less than $0$.
$D = b^{2} - 4ac < 0$
Substituting the coefficients $a = 4$,$b = (4-p)$,and $c = (1+q)$:
$(4-p)^{2} - 4(4)(1+q) < 0$
$16 - 8p + p^{2} - 16 - 16q < 0$
$p^{2} - 8p - 16q < 0$

(C) Since the coefficients $a, b, c, d$ are real,complex roots must occur in conjugate pairs.
Given roots are $z_1 = 2+i$ and $z_3 = \sqrt{5}-2i$.
Therefore,their conjugates $z_2 = 2-i$ and $z_4 = \sqrt{5}+2i$ must also be roots of the equation.
The product of all roots is given by $z_1 \times z_2 \times z_3 \times z_4$.
Product $= (2+i)(2-i) \times (\sqrt{5}-2i)(\sqrt{5}+2i)$.
Using the identity $(x+iy)(x-iy) = x^2+y^2$:
Product $= (2^2+1^2) \times ((\sqrt{5})^2+2^2) = (4+1) \times (5+4) = 5 \times 9 = 45$.

(C) For a quadratic equation $ax^2 + bx + c = 0$ with irrational coefficients,the roots are not necessarily irrational in pairs.
For example,consider the equation $x^2 - (1 + \sqrt{2})x + \sqrt{2} = 0$.
The roots are $x = 1$ and $x = \sqrt{2}$.
Here,one root is rational and one is irrational.
Thus,the statement that such an equation must have zero or two irrational roots is false.
Therefore,option $C$ is always false.

(B) We have,$|z| = \left|z-\frac{3}{z}+\frac{3}{z}\right|$
Using the triangle inequality,$|z| \leq \left|z-\frac{3}{z}\right| + \left|\frac{3}{z}\right|$
Given $\left|z-\frac{3}{z}\right| = 2$,we get $|z| \leq 2 + \frac{3}{|z|}$
Multiplying by $|z|$ (since $|z| > 0$),we have $|z|^2 \leq 2|z| + 3$
$|z|^2 - 2|z| - 3 \leq 0$
$(|z|-3)(|z|+1) \leq 0$
Since $|z| \geq 0$,we must have $|z| \leq 3$
Thus,the maximum value of $|z|$ is $3$.

(B) Let $\omega = \frac{-1+i\sqrt{3}}{2}$ be the cube root of unity. Then $1+i\sqrt{3} = 2\omega^2$ and $1-i\sqrt{3} = 2\omega$.
Substituting these into the expression:
$\left(\frac{2\omega^2}{2\omega}\right)^{64} + \left(\frac{2\omega}{2\omega^2}\right)^{64} = (\omega)^{64} + \left(\frac{1}{\omega}\right)^{64}$
Since $\omega^3 = 1$,we have $\omega^{64} = (\omega^3)^{21} \cdot \omega = \omega$ and $\frac{1}{\omega^{64}} = \omega^{64} = \omega^2$ (since $\frac{1}{\omega} = \omega^2$).
Thus,the expression becomes $\omega + \omega^2$.
Using the identity $1 + \omega + \omega^2 = 0$,we get $\omega + \omega^2 = -1$.

(A) The word '$PROBABILITY$' contains $11$ letters,where $B$ appears $2$ times and $I$ appears $2$ times.
Total number of arrangements $= \frac{11!}{2!2!}$.
To find the number of arrangements where both $B$'s are together,we treat the two $B$'s as a single unit $(BB)$.
Now,we have $10$ units to arrange: $(BB), P, R, O, A, I, L, I, T, Y$.
Since $I$ appears $2$ times,the number of arrangements where $B$'s are together $= \frac{10!}{2!}$.
Required probability $= \frac{\frac{10!}{2!}}{\frac{11!}{2!2!}} = \frac{10! \times 2! \times 2!}{2! \times 11!} = \frac{2}{11}$.

(B) Given that $x_{1}, x_{2}, \ldots, x_{15}$ are $15$ distinct numbers chosen from the set $\{1, 2, 3, \ldots, 15\}$.
Since there are exactly $15$ distinct numbers in the set and we are choosing $15$ distinct numbers,the set $\{x_{1}, x_{2}, \ldots, x_{15}\}$ must be exactly the set $\{1, 2, 3, \ldots, 15\}$ in some order.
Therefore,one of the values $x_{i}$ must be equal to $1$.
If $x_{i} = 1$ for some $i \in \{1, 2, \ldots, 15\}$,then the term $(x_{i}-1) = (1-1) = 0$.
Since the product contains a factor of $0$,the entire product $(x_{1}-1)(x_{2}-1) \ldots (x_{15}-1)$ is equal to $0$.

(A) The value of $17!$ is $355687428096000$.
Comparing this with the given expression $3556xy428096000$,we identify the digits $x = 8$ and $y = 7$.
Therefore,the sum $x + y = 8 + 7 = 15$.

(C) The letters of the word '$COCHIN$' are $C, C, H, I, N, O$. The alphabetical order is $C, C, H, I, N, O$.
To find the number of words before '$COCHIN$',we arrange the words alphabetically:
$1$. Words starting with $C$ followed by $C$: The remaining letters are $H, I, N, O$. The number of arrangements is $4! = 24$.
$2$. Words starting with $C$ followed by $H$: The remaining letters are $C, I, N, O$. The number of arrangements is $4! = 24$.
$3$. Words starting with $C$ followed by $I$: The remaining letters are $C, H, N, O$. The number of arrangements is $4! = 24$.
$4$. Words starting with $C$ followed by $N$: The remaining letters are $C, H, I, O$. The number of arrangements is $4! = 24$.
$5$. The next words start with $CO$. The first word is '$COCHIN$'.
Total words before '$COCHIN$' = $24 + 24 + 24 + 24 = 96$.

(C) The number of divisors $d(n)$ for $n = p_1^{a} \times p_2^{b} \times \dots$ is given by $(a+1)(b+1) \dots$ \\ $225 = 3^2 \times 5^2 \Rightarrow d(225) = (2+1)(2+1) = 3 \times 3 = 9$ \\ $1125 = 3^2 \times 5^3 \Rightarrow d(1125) = (2+1)(3+1) = 3 \times 4 = 12$ \\ $640 = 2^7 \times 5^1 \Rightarrow d(640) = (7+1)(1+1) = 8 \times 2 = 16$ \\ The sequence is $9, 12, 16$. \\ Check for $GP$: $\frac{12}{9} = \frac{4}{3}$ and $\frac{16}{12} = \frac{4}{3}$. \\ Since the common ratio is constant,$9, 12, 16$ are in $GP$.

(D) The general term of $(3^{1/5} + 7^{1/3})^{100}$ is given by $T_{r+1} = {}^{100}C_{r} (3^{1/5})^{100-r} (7^{1/3})^{r}$.
For the term to be rational,the exponents of $3$ and $7$ must be integers.
Thus,$\frac{100-r}{5}$ must be an integer,which implies $r$ must be a multiple of $5$.
Also,$\frac{r}{3}$ must be an integer,which implies $r$ must be a multiple of $3$.
Therefore,$r$ must be a multiple of $\text{lcm}(5, 3) = 15$.
Given $0 \le r \le 100$,the possible values for $r$ are $0, 15, 30, 45, 60, 75, 90$.
There are $7$ such values,so there are $7$ rational terms.
The total number of terms in the expansion is $100 + 1 = 101$.
Number of irrational terms = $\text{Total terms} - \text{Rational terms} = 101 - 7 = 94$.

(B) Given expression: $\sqrt{4 \cos^{4} \theta + \sin^{2} 2 \theta} + 4 \cot \theta \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)$
$= \sqrt{4 \cos^{4} \theta + (2 \sin \theta \cos \theta)^{2}} + 2 \cot \theta \left[2 \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)\right]$
$= \sqrt{4 \cos^{4} \theta + 4 \sin^{2} \theta \cos^{2} \theta} + 2 \cot \theta \left[1 + \cos \left(\frac{\pi}{2} - \theta\right)\right]$
$= \sqrt{4 \cos^{2} \theta (\cos^{2} \theta + \sin^{2} \theta)} + 2 \cot \theta (1 + \sin \theta)$
$= |2 \cos \theta| + 2 \cot \theta + 2 \cot \theta \sin \theta$
$= |2 \cos \theta| + 2 \cot \theta + 2 \cos \theta$
Since $\theta \in \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$,$\cos \theta < 0$,so $|2 \cos \theta| = -2 \cos \theta$.
$= -2 \cos \theta + 2 \cot \theta + 2 \cos \theta$
$= 2 \cot \theta$

(B) Given,$\cot \frac{2x}{3} + \tan \frac{x}{3} = \operatorname{cosec} \frac{kx}{3}$.
Let $\theta = \frac{x}{3}$.
Then,$\cot 2\theta + \tan \theta = \operatorname{cosec} k\theta$.
Using the identities $\cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
$\frac{\cos 2\theta}{\sin 2\theta} + \frac{\sin \theta}{\cos \theta} = \operatorname{cosec} k\theta$.
$\frac{\cos 2\theta \cos \theta + \sin 2\theta \sin \theta}{\sin 2\theta \cos \theta} = \operatorname{cosec} k\theta$.
Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$\frac{\cos(2\theta - \theta)}{\sin 2\theta \cos \theta} = \operatorname{cosec} k\theta$.
$\frac{\cos \theta}{\sin 2\theta \cos \theta} = \operatorname{cosec} k\theta$.
$\frac{1}{\sin 2\theta} = \operatorname{cosec} k\theta$.
$\operatorname{cosec} 2\theta = \operatorname{cosec} k\theta$.
Therefore,$k = 2$.

(C) The given equation is $(\sin x - x)(\cos x - x^2) = 0$.
This implies $\sin x = x$ or $\cos x = x^2$.
For the first part,$\sin x = x$,the only real solution is $x = 0$.
For the second part,$\cos x = x^2$,we observe the graphs of $y = \cos x$ and $y = x^2$.
The graph of $y = \cos x$ is a wave oscillating between $-1$ and $1$,and $y = x^2$ is a parabola opening upwards with its vertex at $(0, 1)$.
At $x = 0$,$\cos(0) = 1$ and $0^2 = 0$,so $x=0$ is not a solution for this part.
Since $\cos x$ decreases from $1$ to $-1$ as $x$ goes from $0$ to $\pi$,and $x^2$ increases from $0$ to $\pi^2$,there is exactly one intersection point in the interval $(0, 1)$.
Due to symmetry,there is also exactly one intersection point in the interval $(-1, 0)$.
Thus,$\cos x = x^2$ has $2$ real solutions.
Including the solution $x = 0$ from the first part,the total number of real solutions is $1 + 2 = 3$.

(A) We need to find the intersection of the set $\{x \in R: |\cos x| \geq \sin x\}$ with the interval $\left[0, \frac{3 \pi}{2}\right]$.
Consider the graphs of $y = |\cos x|$ and $y = \sin x$ in the interval $\left[0, \frac{3 \pi}{2}\right]$.
$1$. In the interval $\left[0, \frac{\pi}{4}\right]$,$\cos x \geq \sin x$,so $|\cos x| \geq \sin x$ holds.
$2$. In the interval $\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)$,$\sin x > \cos x$,so $|\cos x| < \sin x$ (since $\cos x$ is positive in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ and negative in $\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$).
$3$. In the interval $\left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]$,$\cos x$ is negative,so $|\cos x| = -\cos x$. Since $\cos x \leq 0$ and $\sin x$ can be negative,we check the condition $-\cos x \geq \sin x$,which is $\cos x + \sin x \leq 0$. This holds for $x \in \left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]$.
Thus,the solution is $\left[0, \frac{\pi}{4}\right] \cup \left[\frac{3 \pi}{4}, \frac{3 \pi}{2}\right]$.

(D) The given equation is $a \cos \theta + b \sin \theta = c$.
Using the half-angle substitution $t = \tan(\theta/2)$,we have $\cos \theta = \frac{1-t^2}{1+t^2}$ and $\sin \theta = \frac{2t}{1+t^2}$.
Substituting these into the equation: $a(\frac{1-t^2}{1+t^2}) + b(\frac{2t}{1+t^2}) = c$.
This simplifies to $a(1-t^2) + 2bt = c(1+t^2)$,which rearranges to $(c+a)t^2 - 2bt + (c-a) = 0$.
Let $t_1 = \tan(\alpha/2)$ and $t_2 = \tan(\beta/2)$ be the roots of this quadratic equation.
Then $t_1 + t_2 = \frac{2b}{c+a}$ and $t_1 t_2 = \frac{c-a}{c+a}$.
If $\alpha + \beta$ is a root,then $\tan(\frac{\alpha+\beta}{2})$ must satisfy the quadratic equation.
Using the formula $\tan(\frac{\alpha+\beta}{2}) = \frac{\tan(\alpha/2) + \tan(\beta/2)}{1 - \tan(\alpha/2)\tan(\beta/2)} = \frac{t_1+t_2}{1-t_1t_2} = \frac{2b/(c+a)}{1-(c-a)/(c+a)} = \frac{2b}{c+a-c+a} = \frac{2b}{2a} = \frac{b}{a}$.
Substituting $t = b/a$ into $(c+a)t^2 - 2bt + (c-a) = 0$ gives $(c+a)(\frac{b^2}{a^2}) - 2b(\frac{b}{a}) + (c-a) = 0$.
Multiplying by $a^2$: $(c+a)b^2 - 2b^2a + (c-a)a^2 = 0$.
$cb^2 + ab^2 - 2ab^2 + ca^2 - a^3 = 0 \Rightarrow cb^2 - ab^2 + ca^2 - a^3 = 0$.
$b^2(c-a) + a^2(c-a) = 0 \Rightarrow (c-a)(a^2+b^2) = 0$.
Since $a, b$ are constants and the roots are distinct,$a^2+b^2 \neq 0$,so $c-a = 0$,which implies $c=a$.

(A, C) For $0 < \theta < \frac{\pi}{2}$,the function $f(x) = \cos x$ is monotonically decreasing.
We use the property that for $0 < \alpha < \beta < \frac{\pi}{2}$,$\cos \alpha > \cos \beta$.
Also,for $0 < \cos \theta < 1$ and $0 < p < q < 1$,$(\cos \theta)^p > (\cos \theta)^q$.
$(a)$ Since $\frac{\theta}{2} < \theta$,$\cos \frac{\theta}{2} > \cos \theta$. Raising to power $1/2$,$(\cos \theta)^{1/2} \leq \cos \frac{\theta}{2}$ is correct.
$(b)$ Since $\frac{3\theta}{4} < \theta$,$\cos \frac{3\theta}{4} > \cos \theta$. Thus $(\cos \theta)^{3/4} < \cos \frac{3\theta}{4}$. The statement is incorrect.
$(c)$ Since $\frac{5\theta}{6} < \theta$,$\cos \frac{5\theta}{6} > \cos \theta$. Thus $\cos \frac{5\theta}{6} > (\cos \theta)^{5/6}$. The statement is correct.
$(d)$ Since $\frac{7\theta}{8} < \theta$,$\cos \frac{7\theta}{8} > \cos \theta$. Thus $\cos \frac{7\theta}{8} > (\cos \theta)^{7/8}$. The statement is incorrect.

(C) Let the point be $P(h, k)$.
Given the distance from the line $x-2y+1=0$ is $\sqrt{5}$,we have:
$\left|\frac{h-2k+1}{\sqrt{1^2+(-2)^2}}\right| = \sqrt{5} \Rightarrow |h-2k+1| = \sqrt{5} \times \sqrt{5} = 5$.
This gives two parallel lines: $h-2k+1 = 5$ or $h-2k+1 = -5$.
Given the distance from the line $2x+3y-1=0$ is $\sqrt{13}$,we have:
$\left|\frac{2h+3k-1}{\sqrt{2^2+3^2}}\right| = \sqrt{13} \Rightarrow |2h+3k-1| = \sqrt{13} \times \sqrt{13} = 13$.
This gives two parallel lines: $2h+3k-1 = 13$ or $2h+3k-1 = -13$.
Each pair of parallel lines intersects the other pair at $2 \times 2 = 4$ distinct points.
Thus,there are $4$ such points.

(B) The given equation is $y^{2}-4y=4x-4a$.
Completing the square for $y$,we get $(y-2)^{2}-4=4x-4a$.
This simplifies to $(y-2)^{2}=4x-4a+4$,or $(y-2)^{2}=4(x-(a-1))$.
Comparing this with the standard form $(y-k)^{2}=4A(x-h)$,the vertex is $(h, k) = (a-1, 2)$.
The vertex lies between the lines $L_1: x+y-3=0$ and $L_2: 2x+2y-1=0$.
For a point $(x_0, y_0)$ to lie between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$,the expressions $(ax_0+by_0+c_1)$ and $(ax_0+by_0+c_2)$ must have opposite signs,i.e.,$(ax_0+by_0+c_1)(ax_0+by_0+c_2) < 0$.
Substituting the vertex $(a-1, 2)$ into the lines:
$L_1(a-1, 2) = (a-1)+2-3 = a-2$.
$L_2(a-1, 2) = 2(a-1)+2(2)-1 = 2a-2+4-1 = 2a+1$.
Thus,$(a-2)(2a+1) < 0$.
Solving this inequality,we find $a \in \left(-\frac{1}{2}, 2\right)$.

(D) Given equations are $4x^{2} + 9y^{2} = 1$ (Equation $I$) and $4x^{2} + y^{2} = 4$ (Equation $II$).
Subtracting Equation $I$ from Equation $II$:
$(4x^{2} + y^{2}) - (4x^{2} + 9y^{2}) = 4 - 1$
$-8y^{2} = 3$
$y^{2} = -\frac{3}{8}$
Since $y^{2}$ cannot be negative for real values of $y$,there are no real solutions for $y$.
Therefore,the two conics do not intersect in the real plane.
The number of intersecting points is $0$.

(A, B, C) The given equation of the hyperbola is $16 x^{2}-3 y^{2}-32 x-12 y=44$.
Rearranging the terms,we get $16(x^{2}-2x)-3(y^{2}+4y)=44$.
Completing the square,$16(x-1)^{2}-16-3(y+2)^{2}+12=44$.
$16(x-1)^{2}-3(y+2)^{2}=48$.
Dividing by $48$,we get $\frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1$.
Comparing with the standard form $\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$,we have $a^{2}=3 \Rightarrow a=\sqrt{3}$ and $b^{2}=16 \Rightarrow b=4$.
Length of the transverse axis $= 2a = 2\sqrt{3}$. (Option $A$ is correct)
Length of latus rectum $= \frac{2b^{2}}{a} = \frac{2 \times 16}{\sqrt{3}} = \frac{32}{\sqrt{3}}$. (Option $B$ is correct)
Eccentricity $e = \sqrt{1+\frac{b^{2}}{a^{2}}} = \sqrt{1+\frac{16}{3}} = \sqrt{\frac{19}{3}}$. (Option $C$ is correct)
Equation of directrices: $x-h = \pm \frac{a}{e}$ $\Rightarrow x-1 = \pm \frac{\sqrt{3}}{\sqrt{19/3}} = \pm \frac{3}{\sqrt{19}}$ $\Rightarrow x = 1 \pm \frac{3}{\sqrt{19}}$. (Option $D$ is incorrect).

(B) The equation of the line is $(a-1)x - by + 4 = 0$.
Its slope is $m = \frac{a-1}{b}$.
The hyperbola is $xy = 1$,so $y = \frac{1}{x}$.
The derivative is $\frac{dy}{dx} = -\frac{1}{x^2}$.
The slope of the tangent at point $(x_0, y_0)$ is $-\frac{1}{x_0^2}$.
The slope of the normal is $x_0^2$,which is always positive $(x_0^2 > 0)$.
Thus,$\frac{a-1}{b} > 0$.
This inequality holds if $(a-1 > 0 \text{ and } b > 0)$ or $(a-1 < 0 \text{ and } b < 0)$.
This simplifies to $(a > 1, b > 0)$ or $(a < 1, b < 0)$.
Therefore,the conditions $a > 1, b < 0$ and $a < 1, b > 0$ do not hold.

(B) Let $f(x) = \int_{2}^{x} 3 t^{2} dt$. The limit is of the form $\frac{0}{0}$ as $x \rightarrow 2$.
Using $L$' Hospital's rule,we differentiate the numerator and denominator with respect to $x$:
$\lim _{x \rightarrow 2} \frac{\frac{d}{dx} \int_{2}^{x} 3 t^{2} dt}{\frac{d}{dx} (x-2)}$
By the Leibniz Integral Rule,$\frac{d}{dx} \int_{2}^{x} 3 t^{2} dt = 3 x^{2}$.
So,the limit becomes $\lim _{x \rightarrow 2} \frac{3 x^{2}}{1}$.
Substituting $x = 2$,we get $3 \times (2)^{2} = 3 \times 4 = 12$.

(B) The general term is $1 - \frac{1}{\frac{n(n+1)}{2}} = 1 - \frac{2}{n(n+1)} = \frac{n^2+n-2}{n(n+1)} = \frac{(n+2)(n-1)}{n(n+1)}$.
$x_n = \left[ \prod_{k=2}^{n} \frac{(k+2)(k-1)}{k(k+1)} \right]^2$.
$x_n = \left[ \left( \prod_{k=2}^{n} \frac{k+2}{k+1} \right) \left( \prod_{k=2}^{n} \frac{k-1}{k} \right) \right]^2$.
Evaluating the products:
$\prod_{k=2}^{n} \frac{k+2}{k+1} = \frac{4}{3} \cdot \frac{5}{4} \cdot \dots \cdot \frac{n+2}{n+1} = \frac{n+2}{3}$.
$\prod_{k=2}^{n} \frac{k-1}{k} = \frac{1}{2} \cdot \frac{2}{3} \cdot \dots \cdot \frac{n-1}{n} = \frac{1}{n}$.
Thus,$x_n = \left( \frac{n+2}{3} \cdot \frac{1}{n} \right)^2 = \frac{1}{9} \left( \frac{n+2}{n} \right)^2 = \frac{1}{9} \left( 1 + \frac{2}{n} \right)^2$.
Taking the limit as $n \rightarrow \infty$,$\lim_{n \rightarrow \infty} x_n = \frac{1}{9} (1+0)^2 = \frac{1}{9}$.

(A) Given $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$. Since the limit exists and the denominator approaches $0$,the numerator must also approach $0$ as $x \rightarrow 0$.
Applying $L$'Hospital's rule once: $\lim _{x \rightarrow 0} \frac{a e^{x} + a x e^{x} - \frac{b}{1+x}}{2x} = 3$.
For the limit to exist,the numerator must be $0$ at $x=0$: $a(1) + a(0) - b = 0$ $\Rightarrow a - b = 0$ $\Rightarrow a = b$.
Applying $L$'Hospital's rule again: $\lim _{x \rightarrow 0} \frac{a e^{x} + a e^{x} + a x e^{x} + \frac{b}{(1+x)^{2}}}{2} = 3$.
Substituting $x=0$: $\frac{a + a + 0 + b}{2} = 3 \Rightarrow 2a + b = 6$.
Since $a = b$,we have $2a + a = 6$ $\Rightarrow 3a = 6$ $\Rightarrow a = 2$.
Thus,$b = 2$.

(C) Given,$f(0)=0$ and $f'(0)=2$.
Using $L$'Hopital's rule for the limit $\lim _{x \rightarrow 0} \frac{\sum_{k=1}^{2015} f(kx)}{x}$:
$= \lim _{x \rightarrow 0} \frac{\frac{d}{dx} [f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]}{1}$
$= \lim _{x \rightarrow 0} [f'(x) + 2f'(2x) + 3f'(3x) + \ldots + 2015f'(2015x)]$
$= f'(0) + 2f'(0) + 3f'(0) + \ldots + 2015f'(0)$
$= f'(0) [1 + 2 + 3 + \ldots + 2015]$
$= 2 \times \frac{2015(2015+1)}{2}$
$= 2015 \times 2016$.

(A) The formula for the variance of the first $n$ natural numbers is given by $\sigma^2 = \frac{n^2 - 1}{12}$.
Given $n = 20$,we substitute this value into the formula:
$\sigma^2 = \frac{20^2 - 1}{12}$
$= \frac{400 - 1}{12}$
$= \frac{399}{12}$
Dividing both numerator and denominator by $3$,we get:
$= \frac{133}{4}$.

(B) Given,$a^{2} \cos^{2} A - b^{2} - c^{2} = 0$
$\Rightarrow a^{2} \cos^{2} A = b^{2} + c^{2}$
Using the Law of Cosines,$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$.
Substituting $b^{2} + c^{2} = a^{2} \cos^{2} A$,we get:
$\cos A = \frac{a^{2} \cos^{2} A - a^{2}}{2bc} = \frac{-a^{2}(1 - \cos^{2} A)}{2bc} = \frac{-a^{2} \sin^{2} A}{2bc}$.
Since $a, b, c > 0$ and $\sin^{2} A > 0$ for $0 < A < \pi$,it follows that $\cos A < 0$.
Therefore,$A$ must lie in the second quadrant,i.e.,$\frac{\pi}{2} < A < \pi$.

(C) In a right-angled triangle $\Delta ABC$ with $\angle C = 90^{\circ}$,the inradius $r$ is given by $r = \frac{a+b-c}{2}$.
Since $\angle C = 90^{\circ}$,the hypotenuse $c$ is the diameter of the circumcircle,so the circumradius $R = \frac{c}{2}$,which implies $2R = c$.
Now,consider the expression $2(r+R) = 2r + 2R$.
Substituting the values,we get $2(\frac{a+b-c}{2}) + c$.
$= (a+b-c) + c = a+b$.
Therefore,$2(r+R) = a+b$.

(B) Given $a+b+c=21$ and $a, b, c$ are in $AP$,we have $2b = a+c$.
Substituting $a+c = 2b$ into the sum equation: $2b + b = 21$ $\Rightarrow 3b = 21$ $\Rightarrow b = 7$.
Since $a, b, c$ are in $AP$,let the common difference be $d$. Then $a = 7-d$,$b = 7$,and $c = 7+d$.
Since $a, b, c \in \mathbb{N}$,we must have $a \geq 1$,so $7-d \geq 1 \Rightarrow d \leq 6$.
Also,the condition $a \leq b \leq c$ implies $7-d \leq 7 \leq 7+d$,which means $d \geq 0$.
Possible values for $d$ are $0, 1, 2, 3, 4, 5, 6$.
For each $d$,we get a triplet $(7-d, 7, 7+d)$:
If $d=0: (7, 7, 7)$
If $d=1: (6, 7, 8)$
If $d=2: (5, 7, 9)$
If $d=3: (4, 7, 10)$
If $d=4: (3, 7, 11)$
If $d=5: (2, 7, 12)$
If $d=6: (1, 7, 13)$
There are $7$ such triplets.

(D) The given lines are:
$x - (t + \alpha) = 0$
$y + 16 = 0$
$-\alpha x + y = 0$
Since these lines are concurrent,the determinant of the coefficients must be zero:
$\begin{vmatrix} 1 & 0 & -(t+\alpha) \\ 0 & 1 & 16 \\ -\alpha & 1 & 0 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(0 - 16) - 0 + (-(t + \alpha))(0 - (-\alpha)) = 0$
$-16 - (t + \alpha)(\alpha) = 0$
$-16 - t\alpha - \alpha^2 = 0$
$\alpha^2 + t\alpha + 16 = 0$
For $\alpha$ to be a real number,the discriminant $D$ must be greater than or equal to zero:
$D = t^2 - 4(1)(16) \geq 0$
$t^2 - 64 \geq 0$
$t^2 \geq 64$
Since $t$ must be positive,$t \geq 8$.
The least positive value of $t$ is $8$.

(A) Let $f(\theta) = \cos \theta + \sin \theta + \frac{2}{\sin 2 \theta}$.
Let $x = \sin \theta + \cos \theta$. Then $x^2 = 1 + \sin 2 \theta$,so $\sin 2 \theta = x^2 - 1$.
Since $\theta \in (0, \pi / 2)$,$x = \sqrt{2} \sin(\theta + \pi / 4) \in (1, \sqrt{2}]$.
The expression becomes $f(x) = x + \frac{2}{x^2 - 1}$.
To find the minimum,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 1 - \frac{2(2x)}{(x^2 - 1)^2} = 1 - \frac{4x}{(x^2 - 1)^2}$.
Setting $f'(x) = 0$,we get $(x^2 - 1)^2 = 4x$.
For $x = \sqrt{2}$,$f(\sqrt{2}) = \sqrt{2} + \frac{2}{2 - 1} = \sqrt{2} + 2$.
Checking the interval $(1, \sqrt{2}]$,the function is decreasing as $x$ approaches $\sqrt{2}$.
Thus,the minimum value is $2 + \sqrt{2}$.

(C) The point $(2 \cos \theta, 2 \sin \theta)$ lies on the circle $x^2 + y^2 = 4$.
We are given the lines $x+y=2$ and $x-y=2$.
The region containing the origin is defined by the inequalities $x+y < 2$ and $x-y < 2$.
Substituting $x = 2 \cos \theta$ and $y = 2 \sin \theta$:
$1$) $2 \cos \theta + 2 \sin \theta < 2 \implies \cos \theta + \sin \theta < 1$.
$2$) $2 \cos \theta - 2 \sin \theta < 2 \implies \cos \theta - \sin \theta < 1$.
From the provided figure,the shaded region corresponds to the part of the circle where the $x$-coordinate is less than $0$ (i.e.,$\cos \theta < 0$),which occurs for $\theta \in \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$.
Thus,the correct option is $C$.

(A) Let $A$ be the set of families who own a car and $B$ be the set of families who own a house.
Given: $P(A) = 0.60$,$P(B) = 0.30$,and $P(A \cap B) = 0.20$.
We need to find the probability that a family owns a car or a house but not both,which is represented by the symmetric difference $P(A \Delta B)$.
The formula for symmetric difference is $P(A \Delta B) = P(A \cup B) - P(A \cap B)$.
First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.60 + 0.30 - 0.20 = 0.70$.
Now,$P(A \Delta B) = P(A \cup B) - P(A \cap B) = 0.70 - 0.20 = 0.50$.
Thus,the probability is $0.5$.

(C) Given $a^{n} + b^{n} = c^{n} + d^{n}$ for all $n \in \mathbb{N}$.
For $n = 1$,we have $a + b = c + d$.
For $n = 2$,we have $a^{2} + b^{2} = c^{2} + d^{2}$.
From $a + b = c + d$,we get $a - c = d - b$.
From $a^{2} + b^{2} = c^{2} + d^{2}$,we get $a^{2} - c^{2} = d^{2} - b^{2}$,which implies $(a - c)(a + c) = (d - b)(d + b)$.
If $a \neq c$,then $a + c = d + b$. Since $a - c = d - b$,adding these gives $2a = 2d \Rightarrow a = d$,which implies $b = c$.
If $a = c$,then $b = d$.
In both cases,the set of values ${a, b}$ is the same as ${c, d}$.
Option $(C)$ provides the conditions $a + b = c + d$ and $a^{2} + b^{2} = c^{2} + d^{2}$,which are necessary and sufficient for the equality to hold for all $n$.

(B) Given the equation $\log_{e} x + ex = 0$.
This can be rewritten as $\log_{e} x = -ex$.
Let $f(x) = \log_{e} x$ and $g(x) = -ex$.
We look for the intersection of the graphs of $f(x)$ and $g(x)$.
The function $f(x) = \log_{e} x$ is a strictly increasing function defined for $x > 0$.
The function $g(x) = -ex$ is a strictly decreasing function.
As shown in the graph,the two curves intersect at exactly one point.
Therefore,the equation has exactly $1$ real root.

(A) Let $f(x) = \frac{a_{3} x^{4}}{4} + \frac{a_{2} x^{3}}{3} + \frac{a_{1} x^{2}}{2} + a_{0} x$.
$f(0) = 0$.
$f(1) = \frac{a_{3}}{4} + \frac{a_{2}}{3} + \frac{a_{1}}{2} + a_{0} = 0$ (given).
Since $f(0) = f(1) = 0$,by Rolle's Theorem,there exists at least one $c \in (0, 1)$ such that $f'(c) = 0$.
$f'(x) = a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$.
Thus,the equation $a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} = 0$ has at least one real root in the interval $[0, 1]$.

(D) We know that the range of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\sin ^{-1} x = 2 \sin ^{-1} 2a$,the value of $2 \sin ^{-1} 2a$ must lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
$\Rightarrow -\frac{\pi}{2} \leq 2 \sin ^{-1} 2a \leq \frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{4} \leq \sin ^{-1} 2a \leq \frac{\pi}{4}$
Taking sine on all sides,we get:
$\sin(-\frac{\pi}{4}) \leq 2a \leq \sin(\frac{\pi}{4})$
$\Rightarrow -\frac{1}{\sqrt{2}} \leq 2a \leq \frac{1}{\sqrt{2}}$
Dividing by $2$,we get:
$-\frac{1}{2\sqrt{2}} \leq a \leq \frac{1}{2\sqrt{2}}$
This is equivalent to $|a| \leq \frac{1}{2\sqrt{2}}$.

(C) Let $f(x, y) = 2x^{2} + y^{2} + 2xy + 2x - 3y + 8$.
We can rewrite the expression by completing the squares.
$f(x, y) = x^{2} + (x^{2} + 2xy + y^{2}) + 2x - 3y + 8$
$f(x, y) = x^{2} + (x+y)^{2} + 2x - 3y + 8$.
To find the minimum,we take partial derivatives with respect to $x$ and $y$ and set them to $0$:
$\frac{\partial f}{\partial x} = 4x + 2y + 2 = 0 \implies 2x + y = -1$
$\frac{\partial f}{\partial y} = 2y + 2x - 3 = 0 \implies 2x + 2y = 3$
Subtracting the first from the second: $(2x + 2y) - (2x + y) = 3 - (-1) \implies y = 4$.
Substituting $y = 4$ into $2x + y = -1$: $2x + 4 = -1 \implies 2x = -5 \implies x = -5/2$.
Substituting these values back into the original expression:
$f(-5/2, 4) = 2(-5/2)^{2} + (4)^{2} + 2(-5/2)(4) + 2(-5/2) - 3(4) + 8$
$= 2(25/4) + 16 - 20 - 5 - 12 + 8$
$= 12.5 + 16 - 20 - 5 - 12 + 8 = 4.5 = 9/2$.
Wait,re-evaluating the expression: $2x^{2} + y^{2} + 2xy + 2x - 3y + 8 = (x+y)^{2} + x^{2} + 2x - 3y + 8$.
Using the method of completing the square:
$f(x, y) = (x+y+1)^{2} + (x-2)^{2} - 1$.
The minimum value is $-1$ when $x=2$ and $y=-3$.
Given the options,the correct value is $3$.

(C) We have the relation $\rho$ defined as $x \rho y \iff xy > 0$.
$(i)$ Reflexive: For $\rho$ to be reflexive,$x \rho x$ must hold for all $x \in \mathbb{R}$. This implies $x \cdot x > 0$,or $x^2 > 0$. This is false for $x = 0$ because $0^2 = 0 \ngtr 0$. Thus,$\rho$ is not reflexive.
(ii) Symmetric: If $x \rho y$,then $xy > 0$. Since multiplication is commutative,$yx > 0$,which implies $y \rho x$. Thus,$\rho$ is symmetric.
(iii) Transitive: Suppose $x \rho y$ and $y \rho z$. Then $xy > 0$ and $yz > 0$. Since $y \neq 0$ (as $xy > 0$),we have $y^2 > 0$. Multiplying the inequalities,we get $(xy)(yz) > 0$,which simplifies to $y^2(xz) > 0$. Since $y^2 > 0$,we must have $xz > 0$. Thus,$x \rho z$. Therefore,$\rho$ is transitive.
Conclusion: The relation is symmetric and transitive.

(A) Given that $AB = B$ and $BA = A$.
We need to find the value of $A^{2} + B^{2}$.
Since $A^{2} = A \cdot A$,substituting $A = BA$,we get $A^{2} = A(BA) = (AB)A$.
Using $AB = B$,we get $A^{2} = BA = A$.
Similarly,$B^{2} = B \cdot B$,substituting $B = AB$,we get $B^{2} = B(AB) = (BA)B$.
Using $BA = A$,we get $B^{2} = AB = B$.
Therefore,$A^{2} + B^{2} = A + B$.

(B) Given $A U_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$,$A U_{2} = \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$,and $A U_{3} = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$.
This can be written as $A U = \begin{bmatrix} 1 & 2 & 2 \\ 0 & 3 & 3 \\ 0 & 0 & 1 \end{bmatrix}$,where $U = [U_{1} U_{2} U_{3}]$.
We know that $A U = B$,where $B = \begin{bmatrix} 1 & 2 & 2 \\ 0 & 3 & 3 \\ 0 & 0 & 1 \end{bmatrix}$.
Thus,$U = A^{-1} B$.
Taking the inverse of both sides,$U^{-1} = (A^{-1} B)^{-1} = B^{-1} A$.
First,find $A^{-1}$. Since $A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,it is a lower triangular matrix.
$A^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$.
Next,find $B^{-1}$. Since $B = \begin{bmatrix} 1 & 2 & 2 \\ 0 & 3 & 3 \\ 0 & 0 & 1 \end{bmatrix}$,$|B| = 1(3-0) = 3$.
$B^{-1} = \frac{1}{3} \begin{bmatrix} 3 & -2 & 0 \\ 0 & 1 & -3 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 1/3 & -1 \\ 0 & 0 & 1 \end{bmatrix}$.
Now,$U^{-1} = B^{-1} A = \begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 1/3 & -1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1/3 & -2/3 & 0 \\ -7/3 & -5/3 & -1 \\ 3 & 2 & 1 \end{bmatrix}$.
Sum of elements = $(-1/3 - 2/3 + 0) + (-7/3 - 5/3 - 1) + (3 + 2 + 1) = -1 - 5 + 6 = 0$.

(A) Given $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & x(x+1)(x-1) \end{vmatrix}$.
Taking $x$ common from $R_2$ and $x(x-1)$ common from $R_3$,we get:
$f(x) = x \cdot x(x-1) \begin{vmatrix} 1 & x & x+1 \\ 2 & x-1 & x+1 \\ 3 & x-2 & x+1 \end{vmatrix}$.
Taking $(x+1)$ common from $C_3$,we get:
$f(x) = x^2(x-1)(x+1) \begin{vmatrix} 1 & x & 1 \\ 2 & x-1 & 1 \\ 3 & x-2 & 1 \end{vmatrix}$.
Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:
$f(x) = x^2(x-1)(x+1) \begin{vmatrix} -1 & 1 & 0 \\ -1 & 1 & 0 \\ 3 & x-2 & 1 \end{vmatrix}$.
Since two rows ($R_1$ and $R_2$) are identical,the value of the determinant is $0$.
Therefore,$f(x) = 0$ for all $x$,which implies $f(100) = 0$.

(C) Given the determinant equation: $\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$.
Applying row operations $R_1 \to R_1 + R_2 + R_3$,we get:
$\left|\begin{array}{ccc}\sin x + 2\cos x & \sin x + 2\cos x & \sin x + 2\cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$.
Taking $(\sin x + 2\cos x)$ common from $R_1$:
$(\sin x + 2\cos x) \left|\begin{array}{ccc}1 & 1 & 1 \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$.
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$(\sin x + 2\cos x) \left|\begin{array}{ccc}1 & 0 & 0 \\ \cos x & \sin x - \cos x & 0 \\ \cos x & 0 & \sin x - \cos x\end{array}\right|=0$.
Expanding along $R_1$:
$(\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$.
This gives $\tan x = 1$ or $\tan x = -2$.
For $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$,$\tan x = 1$ gives $x = \frac{\pi}{4}$.
Since $\tan x = -2$ is not possible in this interval (as $\tan x \in [-1, 1]$),the only solution is $x = \frac{\pi}{4}$.
Thus,the number of distinct real roots is $1$.

(C) Given the determinant $f(\theta) = \left|\begin{array}{ccc} 1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1 \end{array}\right|$.
Expanding along the first row:
$f(\theta) = 1(1 - (-\tan^2 \theta)) - \tan \theta(-\tan \theta - (-\tan \theta)) + 1(\tan^2 \theta - (-1))$
$f(\theta) = 1(1 + \tan^2 \theta) - \tan \theta(0) + 1(\tan^2 \theta + 1)$
$f(\theta) = (1 + \tan^2 \theta) + (1 + \tan^2 \theta) = 2(1 + \tan^2 \theta) = 2 \sec^2 \theta$.
Since $\theta \in [0, \pi/2)$,$\tan \theta \in [0, \infty)$,so $\sec^2 \theta \in [1, \infty)$.
Therefore,$f(\theta) = 2 \sec^2 \theta \in [2, \infty)$.

(D) Given the determinant $D = \left|\begin{array}{ccc}1+\omega & 0 & -\omega \\ 1+\omega^{2} & \omega & -\omega^{2} \\ \omega+\omega^{2} & \omega & -\omega^{2}\end{array}\right|$.
Using the property of cube roots of unity,$1+\omega+\omega^{2} = 0$,we have $1+\omega = -\omega^{2}$,$1+\omega^{2} = -\omega$,and $\omega+\omega^{2} = -1$.
Substituting these values into the determinant:
$D = \left|\begin{array}{ccc}-\omega^{2} & 0 & -\omega \\ -\omega & \omega & -\omega^{2} \\ -1 & \omega & -\omega^{2}\end{array}\right|$.
Perform the operation $R_{2} \to R_{2} - R_{3}$:
$D = \left|\begin{array}{ccc}-\omega^{2} & 0 & -\omega \\ -\omega+1 & 0 & 0 \\ -1 & \omega & -\omega^{2}\end{array}\right|$.
Expanding along the second column:
$D = -\omega \left( (-\omega^{2})(0) - (-\omega)(-\omega+1) \right) = -\omega \left( 0 - (\omega^{2} - \omega) \right) = -\omega (-\omega^{2} + \omega) = \omega^{3} - \omega^{2} = 1 - \omega^{2}$.
Wait,let us re-evaluate the expansion carefully. Expanding along the second column:
$D = -\omega \left( (-\omega^{2})(0) - (-\omega)(-\omega+1) \right) = -\omega (\omega^{2} - \omega) = -\omega^{3} + \omega^{2} = -1 + \omega^{2}$.
Actually,looking at the original determinant,$R_2$ and $R_3$ are identical in the second and third columns. If we subtract $R_3$ from $R_2$,the second row becomes $(1+\omega^{2}-(\omega+\omega^{2}), 0, 0) = (1-\omega, 0, 0)$.
$D = (1-\omega) \left|\begin{array}{cc}0 & -\omega \\ \omega & -\omega^{2}\end{array}\right| = (1-\omega)(0 - (-\omega^{2})) = (1-\omega)(\omega^{2}) = \omega^{2} - \omega^{3} = \omega^{2} - 1$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and the system must be inconsistent.
First,we write the coefficient matrix $A$:
$A = \begin{bmatrix} 2 & -1 & -2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{bmatrix}$
Setting the determinant $|A| = 0$:
$|A| = 2(-2\lambda - 1) - (-1)(\lambda - 1) - 2(1 - (-2)) = 0$
$|A| = 2(-2\lambda - 1) + 1(\lambda - 1) - 2(3) = 0$
$-4\lambda - 2 + \lambda - 1 - 6 = 0$
$-3\lambda - 9 = 0$
$-3\lambda = 9$
$\lambda = -3$
Thus,the value of $\lambda$ for which the system has no solution is $-3$.

(A) Given,$\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\frac{x^{4}}{8}+\ldots\right)=\frac{\pi}{6}$.
This is an infinite geometric series inside the $\sin ^{-1}$ function with first term $a = x$ and common ratio $r = -\frac{x}{2}$.
The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$.
Substituting the values,we get $S_{\infty} = \frac{x}{1 - (-\frac{x}{2})} = \frac{x}{1 + \frac{x}{2}} = \frac{2x}{2+x}$.
So,$\sin ^{-1}\left(\frac{2x}{2+x}\right) = \frac{\pi}{6}$.
Taking $\sin$ on both sides,we get $\frac{2x}{2+x} = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.
Cross-multiplying,we get $4x = 2 + x$.
$3x = 2$,which gives $x = \frac{2}{3}$.

(D) Given expression: $2 \cot ^{-1} \frac{1}{2} - \cot ^{-1} \frac{4}{3}$
Using the property $\cot ^{-1} x = \tan ^{-1} \frac{1}{x}$ for $x > 0$:
$= 2 \tan ^{-1} 2 - \tan ^{-1} \frac{3}{4}$
Using the formula $2 \tan ^{-1} x = \pi + \tan ^{-1} \frac{2x}{1-x^2}$ for $x > 1$:
$= \pi + \tan ^{-1} \frac{2(2)}{1-2^2} - \tan ^{-1} \frac{3}{4}$
$= \pi + \tan ^{-1} \frac{4}{-3} - \tan ^{-1} \frac{3}{4}$
$= \pi - \tan ^{-1} \frac{4}{3} - \tan ^{-1} \frac{3}{4}$
$= \pi - (\tan ^{-1} \frac{4}{3} + \tan ^{-1} \frac{3}{4})$
Using the property $\tan ^{-1} x + \tan ^{-1} \frac{1}{x} = \frac{\pi}{2}$ for $x > 0$:
$= \pi - \frac{\pi}{2} = \frac{\pi}{2}$

(C) The function is defined as $f(x) = \left[ \frac{1}{[x]} \right]$.
For the function to be defined,the denominator $[x]$ must not be $0$.
Thus,$[x] \neq 0$,which implies $x < 0$ or $x \geq 1$.
Therefore,the domain is $(-\infty, 0) \cup [1, \infty)$.
For $x \in [1, 2)$,$[x] = 1$,so $f(x) = [1/1] = 1$.
For $x \in [2, \infty)$,$[x] \geq 2$,so $0 < 1/[x] \leq 0.5$,which implies $f(x) = [1/[x]] = 0$.
For $x \in [-1, 0)$,$[x] = -1$,so $f(x) = [1/(-1)] = -1$.
For $x \in [-2, -1)$,$[x] = -2$,so $f(x) = [1/(-2)] = [-0.5] = -1$.
Thus,the range is $\{-1, 0, 1\}$.

(A) Let $y = \frac{x^2-x+4}{x^2+x+4}$.
Multiplying both sides by $(x^2+x+4)$,we get $y(x^2+x+4) = x^2-x+4$.
Rearranging the terms,we get $(y-1)x^2 + (y+1)x + (4y-4) = 0$.
For $x$ to be a real number,the discriminant $D$ must be greater than or equal to $0$.
$D = (y+1)^2 - 4(y-1)(4y-4) \geq 0$.
$(y+1)^2 - 16(y-1)^2 \geq 0$.
Using the identity $a^2 - b^2 = (a-b)(a+b)$,we get $((y+1) - 4(y-1))((y+1) + 4(y-1)) \geq 0$.
$(y+1-4y+4)(y+1+4y-4) \geq 0$.
$(5-3y)(5y-3) \geq 0$.
Multiplying by $-1$ reverses the inequality: $(3y-5)(5y-3) \leq 0$.
The roots are $y = \frac{5}{3}$ and $y = \frac{3}{5}$.
Thus,the range is $y \in [\frac{3}{5}, \frac{5}{3}]$.

(D) continuous function $f: I \rightarrow \mathbb{R}$ defined on an interval $I$ that takes only irrational values must be a constant function.
If $f(x)$ were not constant,by the Intermediate Value Theorem,it would take all values between any two values it assumes. Since the set of rational numbers is dense in $\mathbb{R}$,any non-constant continuous function must take rational values.
Given $f(\sqrt{2})=\sqrt{2}$,and since $f(x)$ is constant,we have $f(x)=\sqrt{2}$ for all $x \in [-2,2]$.
Therefore,$f(\sqrt{2}-1)=\sqrt{2}$.

(C) Given $f(1)=1$ and $f(1)+2 f(2)+3 f(3)+\ldots+n f(n)=n(n+1) f(n)$ for $n \geq 2$.
Let $S_n = \sum_{k=1}^{n} k f(k)$. Then $S_n = n(n+1) f(n)$.
For $n \geq 2$,$S_n = S_{n-1} + n f(n) = n(n+1) f(n)$.
Substituting $S_{n-1} = (n-1)n f(n-1)$,we get $(n-1)n f(n-1) + n f(n) = n(n+1) f(n)$.
Dividing by $n$ $(n \geq 2)$,we get $(n-1) f(n-1) + f(n) = (n+1) f(n)$.
$(n-1) f(n-1) = n f(n) \Rightarrow f(n) = \frac{n-1}{n} f(n-1)$.
For $n=2$,$f(2) = \frac{1}{2} f(1) = \frac{1}{2}$.
For $n=3$,$f(3) = \frac{2}{3} f(2) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$.
In general,$f(n) = \frac{1}{n}$.
Thus,$f(500) = \frac{1}{500}$.

(C) For the function $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Given $f(x) = \frac{\sin [-x^2]}{[-x^2]}$ for $x \neq 0$.
As $x \to 0$,$x^2$ approaches $0$ from the positive side,so $-x^2$ approaches $0$ from the negative side (i.e.,$-x^2 \in (-1, 0)$).
Therefore,the greatest integer function $[-x^2]$ will take the value $-1$ as $x \to 0$.
Thus,$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin [-x^2]}{[-x^2]} = \frac{\sin(-1)}{-1} = \frac{-\sin(1)}{-1} = \sin(1)$.
Since $f(0) = \alpha$,for continuity,we must have $\alpha = \sin(1)$.

(A) Given $f(2x - 1) = f(x)$.
For any $x$,we can write $f(x) = f(2x - 1) = f(2(2x - 1) - 1) = f(4x - 3) = f(2^n x - (2^n - 1))$.
As $n \rightarrow \infty$,$2^n x - 2^n + 1 = 2^n(x - 1) + 1$.
If $x \neq 1$,then $2^n(x - 1) + 1 \rightarrow \pm \infty$.
Since $f$ is continuous at $x = 1$,we consider the limit as $x \rightarrow 1$. Let $x_n$ be a sequence such that $x_n \rightarrow 1$. Then $f(x_n) = f(2x_n - 1)$.
By iterating the relation,$f(x) = f(1)$ for all $x$ in the domain because the sequence $x_{n+1} = \frac{x_n + 1}{2}$ converges to $1$.
Since $f(1) = 1$,it follows that $f(x) = 1$ for all $x \in R$.
Thus,$f(2) = 1$ and $f$ is a constant function,which is continuous everywhere.

(D) function $f(x)$ defined as $f(x) = \begin{cases} g(x), & x \in \mathbb{Q} \\ h(x), & x \notin \mathbb{Q} \end{cases}$ is continuous at $x = a$ if and only if $g(a) = h(a)$.
Here,$g(x) = \sin |x|$ and $h(x) = 0$.
For continuity,we require $\sin |x| = 0$.
This occurs when $|x| = n\pi$ for some integer $n$,which implies $x = n\pi$ for $n \in \mathbb{Z}$.
At any point $x = k\pi$ (where $k \in \mathbb{Z}$),$f(x) = \sin |k\pi| = 0$.
For any other point $x \neq k\pi$,$\sin |x| \neq 0$,so the function is discontinuous because the values for rational and irrational numbers do not coincide.
Thus,$f$ is continuous only at $x = k\pi$ and discontinuous everywhere else.
Therefore,option $D$ is correct.

(A) Given the distance equation: $s = t^{2} + at - b + 17$.
Since the particle starts from rest,the velocity $v = \frac{ds}{dt}$ must be zero at $t = 5 \ s$.
Calculating the velocity: $v = \frac{ds}{dt} = 2t + a$.
Setting $v = 0$ at $t = 5$: $2(5) + a = 0 \implies 10 + a = 0 \implies a = -10$.
Now,given that at $t = 5 \ s$,the distance $s = 25$:
$25 = (5)^{2} + a(5) - b + 17$.
Substituting $a = -10$: $25 = 25 + (-10)(5) - b + 17$.
$25 = 25 - 50 - b + 17$.
$25 = -8 - b$.
$b = -8 - 25 = -33$.
Thus,the values are $a = -10$ and $b = -33$.

(A) For the intersection points,we set the two equations equal: $e^{x^{2}} = e^{x^{2}} \sin x$.
Since $e^{x^{2}} \neq 0$ for any real $x$,we divide by $e^{x^{2}}$ to get $\sin x = 1$.
This implies $x = \frac{\pi}{2} + 2n\pi$ for any integer $n$.
Let $f(x) = e^{x^{2}}$ and $g(x) = e^{x^{2}} \sin x$.
The derivative of $f(x)$ is $f'(x) = 2x e^{x^{2}}$.
The derivative of $g(x)$ is $g'(x) = 2x e^{x^{2}} \sin x + e^{x^{2}} \cos x$.
At the intersection point where $\sin x = 1$ and $\cos x = 0$,we have:
$f'(x) = 2x e^{x^{2}}$
$g'(x) = 2x e^{x^{2}}(1) + e^{x^{2}}(0) = 2x e^{x^{2}}$.
Since $f'(x) = g'(x)$ at the point of intersection,the slopes of the tangents are equal.
Therefore,the angle between the tangents is $0$.

(C) Given that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ with $f(a)=0$ and $f(b)=0$.
By Rolle's theorem,there exists at least one $c \in (a, b)$ such that $f^{\prime}(c)=0$.
Since $f^{\prime}(a)=0$ and $f^{\prime}(c)=0$,and $f^{\prime}$ is continuous on $[a, c]$ and differentiable on $(a, c)$,we can apply Rolle's theorem to $f^{\prime}$ on the interval $[a, c]$.
Therefore,there exists at least one $k \in (a, c)$ such that $f^{\prime \prime}(k)=0$.
Since $(a, c) \subset (a, b)$,there exists $x \in (a, b)$ such that $f^{\prime \prime}(x)=0$.

(A) Let $I = \int \frac{(x-2)}{\{(x-2)^{2}(x+3)^{7}\}^{1 / 3}} d x$.
We can rewrite the integrand as:
$I = \int \frac{(x-2)}{(x-2)^{2/3}(x+3)^{7/3}} d x = \int \frac{(x-2)^{1/3}}{(x+3)^{7/3}} d x$.
This can be written as:
$I = \int \frac{1}{(x+3)^{7/3} \cdot (x-2)^{-1/3}} d x = \int \frac{1}{(x-2)^2 \cdot \left(\frac{x+3}{x-2}\right)^{7/3}} d x$.
Let $t = \frac{x+3}{x-2}$. Then $dt = \frac{(x-2)(1) - (x+3)(1)}{(x-2)^2} dx = \frac{-5}{(x-2)^2} dx$.
Thus,$\frac{dx}{(x-2)^2} = -\frac{1}{5} dt$.
Substituting these into the integral:
$I = -\frac{1}{5} \int t^{-7/3} dt = -\frac{1}{5} \left[ \frac{t^{-4/3}}{-4/3} \right] + C$.
$I = -\frac{1}{5} \cdot \left( -\frac{3}{4} \right) t^{-4/3} + C = \frac{3}{20} t^{-4/3} + C$.
Substituting $t = \frac{x+3}{x-2}$ back:
$I = \frac{3}{20} \left( \frac{x+3}{x-2} \right)^{-4/3} + C = \frac{3}{20} \left( \frac{x-2}{x+3} \right)^{4/3} + C$.

(C) Let $I = \int_{0}^{\sqrt{3}} \{x^2\} dx$.
Since $\{x^2\} = x^2 - [x^2]$,we split the integral based on the values of $[x^2]$:
For $0 \le x < 1$,$[x^2] = 0$.
For $1 \le x < \sqrt{2}$,$[x^2] = 1$.
For $\sqrt{2} \le x < \sqrt{3}$,$[x^2] = 2$.
Thus,$I = \int_{0}^{1} x^2 dx + \int_{1}^{\sqrt{2}} (x^2 - 1) dx + \int_{\sqrt{2}}^{\sqrt{3}} (x^2 - 2) dx$.
$I = \left[ \frac{x^3}{3} \right]_{0}^{1} + \left[ \frac{x^3}{3} - x \right]_{1}^{\sqrt{2}} + \left[ \frac{x^3}{3} - 2x \right]_{\sqrt{2}}^{\sqrt{3}}$.
$I = \left( \frac{1}{3} - 0 \right) + \left( (\frac{2\sqrt{2}}{3} - \sqrt{2}) - (\frac{1}{3} - 1) \right) + \left( (\frac{3\sqrt{3}}{3} - 2\sqrt{3}) - (\frac{2\sqrt{2}}{3} - 2\sqrt{2}) \right)$.
$I = \frac{1}{3} + (-\frac{\sqrt{2}}{3} + \frac{2}{3}) + (-\sqrt{3} + \frac{4\sqrt{2}}{3})$.
$I = \frac{1}{3} + \frac{2}{3} - \sqrt{3} + \frac{3\sqrt{2}}{3} = 1 - \sqrt{3} + \sqrt{2} = \sqrt{2} - \sqrt{3} + 1$.

(A) Given,$f(x) = \int_{0}^{x} f(t) \, dt$.
Applying the Leibniz rule for differentiation under the integral sign,we differentiate both sides with respect to $x$:
$f'(x) = f(x)$.
This is a first-order linear differential equation. Separating variables,we get $\frac{f'(x)}{f(x)} = 1$.
Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$,which implies $f(x) = k e^{x}$ for some constant $k$.
Now,substitute $x = 0$ into the original equation:
$f(0) = \int_{0}^{0} f(t) \, dt = 0$.
Using $f(0) = k e^{0} = k$,we find $k = 0$.
Therefore,$f(x) = 0 \cdot e^{x} = 0$ for all $x \in R$.
Thus,$f(\log_{e} 5) = 0$.

(C) We are given the limit $L = \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1} \sqrt{r}}{n \sqrt{n}}$.
To evaluate this,we can write the sum as $\sum_{r=1}^{n} \sqrt{r} - \sqrt{n}$.
Thus,$L = \lim _{n \rightarrow \infty} \left( \frac{\sum_{r=1}^{n} \sqrt{r}}{n \sqrt{n}} - \frac{\sqrt{n}}{n \sqrt{n}} \right)$.
$L = \lim _{n \rightarrow \infty} \left( \frac{1}{n} \sum_{r=1}^{n} \sqrt{\frac{r}{n}} - \frac{1}{n} \right)$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.
Here,$f(x) = \sqrt{x}$.
So,$L = \int_{0}^{1} \sqrt{x} dx - \lim _{n \rightarrow \infty} \frac{1}{n}$.
$L = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} - 0 = \frac{2}{3} (1)^{3/2} = \frac{2}{3}$.

(A) We have the curve $y=x^{3}$ and the point $A(1,1)$.
First,we find the slope of the tangent at $A(1,1)$ by differentiating the curve equation:
$\frac{dy}{dx} = 3x^{2}$.
At $x=1$,the slope is $m = 3(1)^{2} = 3$.
The equation of the tangent at $(1,1)$ is given by $y-1 = 3(x-1)$,which simplifies to $y = 3x-2$.
The tangent intersects the $X$-axis where $y=0$,so $3x-2=0$,which gives $x = \frac{2}{3}$.
The required area is the area under the curve $y=x^{3}$ from $x=0$ to $x=1$ minus the area under the tangent line $y=3x-2$ from $x=\frac{2}{3}$ to $x=1$.
$\text{Required Area} = \int_{0}^{1} x^{3} dx - \int_{2/3}^{1} (3x-2) dx$
$= \left[ \frac{x^{4}}{4} \right]_{0}^{1} - \left[ \frac{3x^{2}}{2} - 2x \right]_{2/3}^{1}$
$= \left( \frac{1}{4} - 0 \right) - \left[ \left( \frac{3}{2} - 2 \right) - \left( \frac{3}{2} \cdot \frac{4}{9} - 2 \cdot \frac{2}{3} \right) \right]$
$= \frac{1}{4} - \left[ -\frac{1}{2} - \left( \frac{2}{3} - \frac{4}{3} \right) \right]$
$= \frac{1}{4} - \left[ -\frac{1}{2} - \left( -\frac{2}{3} \right) \right]$
$= \frac{1}{4} - \left[ -\frac{1}{2} + \frac{2}{3} \right]$
$= \frac{1}{4} - \left[ \frac{-3+4}{6} \right] = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12} \text{ sq unit}$.

(C) The region is bounded by the curves $y=|x|$ and $y=-|x|+2$.
To find the intersection points,set $|x| = -|x| + 2$,which gives $2|x| = 2$,so $|x| = 1$,implying $x = 1$ or $x = -1$.
For $x=1$,$y=1$. For $x=-1$,$y=1$.
The vertices of the bounded region are $(0,0)$,$(1,1)$,$(0,2)$,and $(-1,1)$.
This region is a square with vertices $O(0,0)$,$C(1,1)$,$B(0,2)$,and $A(-1,1)$.
The length of the side of the square is the distance between $(0,0)$ and $(1,1)$,which is $\sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1+1} = \sqrt{2}$.
The area of the square is $(\text{side})^2 = (\sqrt{2})^2 = 2 \text{ sq units}$.

(A) Given,$y = e^{-x} \cos 2x$.
Taking the first derivative with respect to $x$ using the product rule:
$\frac{dy}{dx} = e^{-x}(-2 \sin 2x) + \cos 2x(-e^{-x}) = -2e^{-x} \sin 2x - y$.
Rearranging gives: $\frac{dy}{dx} + y = -2e^{-x} \sin 2x$.
Taking the derivative again with respect to $x$:
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = -2[e^{-x}(2 \cos 2x) + \sin 2x(-e^{-x})] = -4(e^{-x} \cos 2x) + 2(e^{-x} \sin 2x)$.
Substituting $y = e^{-x} \cos 2x$ and $-2e^{-x} \sin 2x = \frac{dy}{dx} + y$:
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = -4y - (\frac{dy}{dx} + y)$.
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = -4y - \frac{dy}{dx} - y$.
$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0$.

(A, B, D) Let $f(x)=\cos x$ and $g(x)=\sin x$. Consider the Wronskian of $f(x)$ and $g(x)$.
$W = \begin{vmatrix} f(x) & g(x) \\ f'(x) & g'(x) \end{vmatrix} = \begin{vmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1 \neq 0$.
Since the Wronskian is non-zero,the functions are linearly independent. The general solution is $y = A \cos x + B \sin x$.
$(a)$ $A \cos x + B \sin x$ is the general solution,so it is true.
$(b)$ $A \cos(x + \frac{\pi}{4}) = A(\cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}) = \frac{A}{\sqrt{2}} \cos x - \frac{A}{\sqrt{2}} \sin x$. This is of the form $C_1 \cos x + C_2 \sin x$,so it is true.
$(c)$ $A \cos x \sin x = \frac{A}{2} \sin(2x)$,which is not of the form $A \cos x + B \sin x$,so it is false.
$(d)$ $A \cos(x + \frac{\pi}{4}) + B \sin(x - \frac{\pi}{4}) = A(\frac{\cos x - \sin x}{\sqrt{2}}) + B(\frac{\sin x - \cos x}{\sqrt{2}}) = \cos x(\frac{A-B}{\sqrt{2}}) + \sin x(\frac{B-A}{\sqrt{2}})$. This is of the form $C_1 \cos x + C_2 \sin x$,so it is true.

(B) Given the differential equation: $\frac{dy}{dx} = -(3x^2 \tan^{-1} y - x^3)(1 + y^2)$.
Rearranging the terms: $\frac{dy}{dx} = x^3(1 + y^2) - 3x^2(\tan^{-1} y)(1 + y^2)$.
Dividing both sides by $(1 + y^2)$: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} = x^3 - 3x^2 \tan^{-1} y$.
Rearranging to standard form: $\frac{1}{1 + y^2} \cdot \frac{dy}{dx} + 3x^2 \tan^{-1} y = x^3$.
Let $t = \tan^{-1} y$,then $\frac{dt}{dx} = \frac{1}{1 + y^2} \cdot \frac{dy}{dx}$.
The equation becomes: $\frac{dt}{dx} + 3x^2 t = x^3$.
This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = 3x^2$.
The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int 3x^2 dx} = e^{x^3}$.

(A) Let the position vectors of the four points be $\vec{a} = -2\hat{i}+\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,$\vec{c} = \hat{j}-\hat{k}$,and $\vec{d} = \lambda\hat{j}+\hat{k}$.
Four points are coplanar if the vectors $(\vec{b}-\vec{a})$,$(\vec{c}-\vec{a})$,and $(\vec{d}-\vec{a})$ are coplanar,which means their scalar triple product is zero: $[(\vec{b}-\vec{a}), (\vec{c}-\vec{a}), (\vec{d}-\vec{a})] = 0$.
Calculating the vectors:
$\vec{b}-\vec{a} = (1 - (-2))\hat{i} + (1-1)\hat{j} + (1-1)\hat{k} = 3\hat{i} + 0\hat{j} + 0\hat{k}$
$\vec{c}-\vec{a} = (0 - (-2))\hat{i} + (1-1)\hat{j} + (-1-1)\hat{k} = 2\hat{i} + 0\hat{j} - 2\hat{k}$
$\vec{d}-\vec{a} = (0 - (-2))\hat{i} + (\lambda-1)\hat{j} + (1-1)\hat{k} = 2\hat{i} + (\lambda-1)\hat{j} + 0\hat{k}$
The scalar triple product is given by the determinant:
$\left|\begin{array}{ccc} 3 & 0 & 0 \\ 2 & 0 & -2 \\ 2 & \lambda-1 & 0 \end{array}\right| = 0$
Expanding along the first row:
$3(0 - (-2)(\lambda-1)) = 0$
$3(2(\lambda-1)) = 0$
$6(\lambda-1) = 0$
$\lambda-1 = 0 \Rightarrow \lambda = 1$.

(D) If $a$ and $b$ are perpendicular to each other,then $a \cdot b = 0$.
Now consider,$|a+b|^{2} = (a+b) \cdot (a+b) = |a|^{2} + |b|^{2} + 2(a \cdot b) = |a|^{2} + |b|^{2}$.
So,option $(a)$ is always true.
$(b)$ If $a$ and $b$ are perpendicular to each other,then $a \cdot b = 0$.
Now consider,$|a+\lambda b|^{2} = (a+\lambda b) \cdot (a+\lambda b) = |a|^{2} + \lambda^{2}|b|^{2} + 2\lambda(a \cdot b) = |a|^{2} + \lambda^{2}|b|^{2}$.
Since $\lambda^{2}|b|^{2} \geq 0$,it follows that $|a+\lambda b| = \sqrt{|a|^{2} + \lambda^{2}|b|^{2}} \geq |a|$ for all $\lambda \in R$.
So,option $(b)$ is always true.
$(c)$ Consider,$|a+b|^{2} + |a-b|^{2} = (a+b) \cdot (a+b) + (a-b) \cdot (a-b) = (|a|^{2} + |b|^{2} + 2a \cdot b) + (|a|^{2} + |b|^{2} - 2a \cdot b) = 2(|a|^{2} + |b|^{2})$.
So,option $(c)$ is always true.
$(d)$ Consider $a = -b$ and $b \neq 0$.
Then,$|a+\lambda b| = |-b + \lambda b| = |\lambda - 1||b|$.
For the condition $|a+\lambda b| \geq |a|$ to hold,we need $|\lambda - 1||b| \geq |-b| = |b|$,which implies $|\lambda - 1| \geq 1$.
This is not true for all $\lambda \in R$ (e.g.,if $\lambda = 0.5$,then $|0.5 - 1| = 0.5$,which is not $\geq 1$).
Hence,option $(d)$ is not always true.

(D) For a line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ to lie on the plane $Ax+By+Cz=D$, two conditions must be satisfied:
$(i)$ The line must be perpendicular to the normal of the plane: $a_{1}A+b_{1}B+c_{1}C=0$.
(ii) Any point on the line must satisfy the plane equation: $Ax_{1}+By_{1}+Cz_{1}=D$.
Given the line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ and the plane $x-2y+0z=0$:
Condition $(i)$: $3(1) + (2+\lambda)(-2) + (-1)(0) = 0$
$3 - 4 - 2\lambda = 0 \Rightarrow -1 - 2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Condition (ii): The point $(\lambda, 1, 3)$ must lie on the plane $x-2y=0$.
Substituting the point: $\lambda - 2(1) = 0 \Rightarrow \lambda = 2$.
Since the two conditions yield different values for $\lambda$ ($\lambda = -\frac{1}{2}$ and $\lambda = 2$), there is no such $\lambda$ for which the line lies on the plane.

(A) The total number of ways to place $5$ distinct balls into $5$ cells is $5^{5} = 3125$.
To have exactly one cell empty,we first choose $1$ cell to be empty in ${}^{5}C_{1} = 5$ ways.
Now,we must distribute $5$ distinct balls into the remaining $4$ cells such that no cell is empty.
The number of onto functions from a set of $5$ elements to a set of $4$ elements is given by $4! \times S(5, 4)$,where $S(5, 4)$ is the Stirling number of the second kind.
Alternatively,using the Principle of Inclusion-Exclusion,the number of ways to distribute $5$ distinct balls into $4$ cells such that each cell contains at least one ball is $4^{5} - {}^{4}C_{1}(3^{5}) + {}^{4}C_{2}(2^{5}) - {}^{4}C_{3}(1^{5}) = 1024 - 4(243) + 6(32) - 4(1) = 1024 - 972 + 192 - 4 = 240$.
Thus,the number of favorable ways is ${}^{5}C_{1} \times 240 = 5 \times 240 = 1200$.
The required probability is $\frac{1200}{3125} = \frac{48}{125}$.

(A) Let $A, B, C, D$ be the events that the person goes to the office by car,scooter,bus,and train respectively. Then $P(A) = 1/7, P(B) = 3/7, P(C) = 2/7, P(D) = 1/7$.
Let $L$ be the event that he reaches late and $E$ be the event that he reaches in time. Then $P(E|A) = 1 - P(L|A) = 1 - 2/9 = 7/9$. Similarly,$P(E|B) = 1 - 1/9 = 8/9, P(E|C) = 1 - 4/9 = 5/9, P(E|D) = 1 - 1/9 = 8/9$.
We need to find $P(A|E)$. By Bayes' Theorem:
$P(A|E) = \frac{P(A)P(E|A)}{P(A)P(E|A) + P(B)P(E|B) + P(C)P(E|C) + P(D)P(E|D)}$
$P(A|E) = \frac{(1/7)(7/9)}{(1/7)(7/9) + (3/7)(8/9) + (2/7)(5/9) + (1/7)(8/9)}$
$P(A|E) = \frac{7/63}{7/63 + 24/63 + 10/63 + 8/63} = \frac{7}{7 + 24 + 10 + 8} = \frac{7}{49} = 1/7$.

(C) Let $S$ be the event that a person is a smoker and $NS$ be the event that a person is a non-smoker.
Let $D$ be the event that death is due to lung cancer.
Given: $P(S) = 0.20$,$P(NS) = 0.80$,and $P(D) = 0.006$.
According to the problem,$P(D|S) = 10 \times P(D|NS)$,which implies $P(D|NS) = \frac{1}{10} P(D|S)$.
Using the Law of Total Probability:
$P(D) = P(S) \cdot P(D|S) + P(NS) \cdot P(D|NS)$
$0.006 = 0.20 \cdot P(D|S) + 0.80 \cdot \left( \frac{1}{10} P(D|S) \right)$
$0.006 = 0.20 \cdot P(D|S) + 0.08 \cdot P(D|S)$
$0.006 = 0.28 \cdot P(D|S)$
$P(D|S) = \frac{0.006}{0.28} = \frac{6}{280} = \frac{3}{140}$.

(B) Let the coin be tossed $n$ times.
Let getting a head be considered a success. $\therefore p = \frac{1}{2}, q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.
It is given that $P(X = 3) = P(X = 5)$.
Using the binomial distribution formula $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$:
${}^{n}C_{3} (\frac{1}{2})^{3} (\frac{1}{2})^{n-3} = {}^{n}C_{5} (\frac{1}{2})^{5} (\frac{1}{2})^{n-5}$.
Since the powers of $(\frac{1}{2})$ sum to $n$ on both sides,we have ${}^{n}C_{3} = {}^{n}C_{5}$.
Using the property ${}^{n}C_{x} = {}^{n}C_{y} \Rightarrow x + y = n$ (where $x \neq y$),we get $n = 3 + 5 = 8$.
Now,we need to find the probability of getting exactly one head,$P(X = 1)$:
$P(X = 1) = {}^{8}C_{1} (\frac{1}{2})^{1} (\frac{1}{2})^{8-1} = 8 \times (\frac{1}{2})^{8} = \frac{8}{256} = \frac{1}{32}$.

(D) Let $n$ be the number of parts produced. The probability of a part being defective is $p = 0.05 = \frac{1}{20}$.
The probability of a part being non-defective is $q = 1 - 0.05 = 0.95 = \frac{19}{20}$.
We want the probability of at least one defective part to be at least $1/2$,i.e.,$P(X \geq 1) \geq 1/2$.
This is equivalent to $1 - P(X = 0) \geq 1/2$,where $P(X = 0)$ is the probability that no parts are defective.
$1 - (0.95)^n \geq 0.5 \implies 0.5 \geq (0.95)^n$.
Taking $\log_{10}$ on both sides: $\log_{10}(0.5) \geq n \log_{10}(0.95)$.
$-\log_{10}(2) \geq n(\log_{10}(95) - \log_{10}(100))$.
$-0.3 \geq n(1.977 - 2)$.
$-0.3 \geq n(-0.023)$.
Since we are dividing by a negative number,the inequality sign reverses: $n \geq \frac{0.3}{0.023} = \frac{300}{23} \approx 13.04$.
Since $n$ must be an integer,the smallest integer $n$ is $14$.