AP EAMCET 2020 Mathematics Question Paper with Answer and Solution

(D) Given the equation of the parabola: $5x^2 = -12y$.
Dividing by $5$,we get $x^2 = -\frac{12}{5}y$.
Comparing this with the standard form $x^2 = 4ay$,we have $4a = -\frac{12}{5}$.
Solving for $a$,we get $a = -\frac{12}{5 \times 4} = -\frac{3}{5}$.
The focus of a parabola of the form $x^2 = 4ay$ is given by $(0, a)$.
Substituting the value of $a$,the focus is $\left(0, -\frac{3}{5}\right)$.
Therefore,option $D$ is correct.

(B) The given equation of the parabola is $(x+3)^2 = 2(y-5)$.
Comparing this with the standard form $(x-h)^2 = 4a(y-k)$,we get the vertex $(h, k) = (-3, 5)$.
Here,$4a = 2$,which implies $a = 1/2$.
The focus of a parabola of the form $(x-h)^2 = 4a(y-k)$ is given by $(h, k+a)$.
Substituting the values,the focus is $(-3, 5 + 1/2) = (-3, 11/2)$.
Thus,option $B$ is correct.

(A) For a parabola,the length of the semi-latus rectum is the harmonic mean of the segments of any focal chord $PSQ$.
Let $l$ be the semi-latus rectum. Then,$\frac{1}{PS} + \frac{1}{QS} = \frac{2}{l}$.
Given $PS = 3$ and $QS = 2$,we have $\frac{1}{3} + \frac{1}{2} = \frac{2}{l}$.
$\frac{2+3}{6} = \frac{2}{l} \implies \frac{5}{6} = \frac{2}{l}$.
$l = \frac{12}{5}$.
The length of the latus rectum is $2l = 2 \times \frac{12}{5} = \frac{24}{5}$.

(A) Given equation of the parabola is $2 y^2+25 x=0$.
Rearranging the equation,we get $2 y^2 = -25 x$.
Dividing by $2$,we get $y^2 = -\frac{25}{2} x$.
Comparing this with the standard form of the parabola $y^2 = -4 a x$,we have:
$4 a = \frac{25}{2}$
$a = \frac{25}{8}$
The equation of the directrix for the parabola $y^2 = -4 a x$ is $x = a$.
Substituting the value of $a$,we get $x = \frac{25}{8}$.
This can be written as $8 x = 25$,or $8 x - 25 = 0$.

(D) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{1} = 1$,where $a^2 = 9$ and $b^2 = 1$.
The equation of a tangent with slope $m$ is $y = mx \pm \sqrt{9m^2 - 1}$. Since it passes through $(3, 2)$,we have $2 = 3m \pm \sqrt{9m^2 - 1}$.
$(2 - 3m)^2 = 9m^2 - 1$ $\Rightarrow 4 - 12m + 9m^2 = 9m^2 - 1$ $\Rightarrow 12m = 5$ $\Rightarrow m = \frac{5}{12}$.
The other tangent is the vertical line $x = 3$ (since the point $(3, 2)$ lies on the line $x = 3$).
The equation of the first tangent is $y - 2 = \frac{5}{12}(x - 3)$ $\Rightarrow 12y - 24 = 5x - 15$ $\Rightarrow 5x - 12y + 9 = 0$.
The chord of contact for the point $(x_1, y_1) = (3, 2)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$ $\Rightarrow \frac{3x}{9} - \frac{2y}{1} = 1$ $\Rightarrow x - 6y = 3$.
The vertices of the triangle are the intersection of the two tangents $(3, 2)$,the intersection of the first tangent and the chord of contact $(3, 1)$,and the intersection of the second tangent $(x=3)$ and the chord of contact $(3, 0)$. Wait,let us re-evaluate the intersection points.
Intersection of $x=3$ and $x-6y=3$ is $(3, 0)$.
Intersection of $5x-12y+9=0$ and $x-6y=3$ (i.e.,$x=6y+3$): $5(6y+3) - 12y + 9 = 0$ $\Rightarrow 30y + 15 - 12y + 9 = 0$ $\Rightarrow 18y = -24$ $\Rightarrow y = -4/3$. Then $x = 6(-4/3) + 3 = -8 + 3 = -5$.
The vertices are $(3, 2)$,$(3, 0)$,and $(-5, -4/3)$.
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2} |3(0 - (-4/3)) + 3(-4/3 - 2) + (-5)(2 - 0)| = \frac{1}{2} |3(4/3) + 3(-10/3) - 10| = \frac{1}{2} |4 - 10 - 10| = \frac{1}{2} |-16| = 8$ sq units.

(D) The equation of the parabola is $y^2 = 12x$,which is of the form $y^2 = 4ax$,where $4a = 12$,so $a = 3$.
Let the point of contact be $(x_1, y_1)$.
The equation of the tangent to the parabola $y^2 = 4ax$ at $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.
Substituting $a = 3$,we get $yy_1 = 6(x + x_1)$,which simplifies to $6x - y_1y + 6x_1 = 0$.
We are given the tangent equation as $x - \sqrt{3}y + 9 = 0$.
Comparing the two equations: $\frac{6}{1} = \frac{-y_1}{-\sqrt{3}} = \frac{6x_1}{9}$.
From $\frac{6}{1} = \frac{y_1}{\sqrt{3}}$,we get $y_1 = 6\sqrt{3}$.
From $\frac{6}{1} = \frac{6x_1}{9}$,we get $x_1 = 9$.
Thus,the point of contact is $(9, 6\sqrt{3})$.

(C) The equation of the parabola is $y^2=16x$,so $4a=16$,which gives $a=4$.
The given line is $3x-4y+5=0$,which can be written as $y=\frac{3}{4}x+\frac{5}{4}$.
The slope of this line is $m_1=\frac{3}{4}$.
Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m \times m_1 = -1$.
Thus,$m = -\frac{4}{3}$.
The equation of a tangent to the parabola $y^2=4ax$ with slope $m$ is $y=mx+\frac{a}{m}$.
Substituting $a=4$ and $m=-\frac{4}{3}$,we get $y=-\frac{4}{3}x+\frac{4}{-4/3}$.
$y=-\frac{4}{3}x-3$.
Multiplying by $3$,we get $3y=-4x-9$,which simplifies to $4x+3y+9=0$.

(D) Given the curve $y^2 = 4x$ and the point $P(1, 2)$.
Differentiating both sides with respect to $x$:
$2y \cdot \frac{dy}{dx} = 4$
$\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}$
At the point $P(1, 2)$,the slope of the tangent $m$ is:
$m = \left. \frac{dy}{dx} \right|_{(1, 2)} = \frac{2}{2} = 1$
Since the slope $m = \tan \theta$,we have:
$\tan \theta = 1$
$\theta = 45^{\circ}$
Thus,the correct option is $D$.

(B) The equation of the tangent to the parabola $y^2=4ax$ at point $(x_1, y_1)$ is given by $yy_1 = 2a(x+x_1)$.
Here,$4a = 12$,so $a = 3$.
The point is $(x_1, y_1) = (3, -6)$.
Substituting these values into the formula:
$y(-6) = 2(3)(x+3)$
$-6y = 6(x+3)$
$-y = x+3$
$x+y+3 = 0$
Hence,option $B$ is correct.

(D) The equation of the tangent to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx+\frac{a}{m}$.
Here,the parabola is $y^2=8x$,so $4a=8$,which implies $a=2$.
The inclination is $30^{\circ}$,so the slope $m=\tan(30^{\circ})=\frac{1}{\sqrt{3}}$.
Substituting these values into the tangent equation:
$y=\frac{1}{\sqrt{3}}x+\frac{2}{1/\sqrt{3}}$
$y=\frac{x}{\sqrt{3}}+2\sqrt{3}$
Multiplying by $\sqrt{3}$ on both sides:
$\sqrt{3}y=x+6$
$x-\sqrt{3}y+6=0$.
Thus,the correct option is $D$.

(A) The given parabolas are $y^2=32x$ and $x^2=256y$.
The equation of a common tangent to the parabolas $y^2=4ax$ and $x^2=4by$ is given by $b^{1/3}y + a^{1/3}x + (a^2b^2)^{1/3} = 0$.
Comparing $y^2=32x$ with $y^2=4ax$,we get $4a=32 \Rightarrow a=8$.
Comparing $x^2=256y$ with $x^2=4by$,we get $4b=256 \Rightarrow b=64$.
Substituting these values into the formula:
$(64)^{1/3}y + (8)^{1/3}x + (8^2 \times 64^2)^{1/3} = 0$
$4y + 2x + (64 \times 4096)^{1/3} = 0$
$4y + 2x + (262144)^{1/3} = 0$
$4y + 2x + 64 = 0$
Dividing by $2$,we get $x + 2y + 32 = 0$. However,checking the options,the equation $2x+4y+64=0$ is equivalent to $x+2y+32=0$.

(D) The equation of a normal to the parabola $y^2=4ax$ with slope $m$ is given by $y=mx-2am-am^3$.
Comparing the given parabola $y^2=4x$ with $y^2=4ax$,we get $a=1$.
The given line is $y=2x+k$,so the slope $m=2$.
Substituting $a=1$ and $m=2$ into the normal equation:
$k = -2(1)(2) - (1)(2)^3$
$k = -4 - 8$
$k = -12$.
Thus,the correct option is $D$.

(B) The equation of the parabola is $y^2 = 4(x+4)$.
Let a point on the parabola be $P(x, y) = (t^2-4, 2t)$.
The distance squared from the origin $(0,0)$ to any point on the parabola is $d^2 = x^2 + y^2 = (t^2-4)^2 + (2t)^2$.
$d^2 = t^4 - 8t^2 + 16 + 4t^2 = t^4 - 4t^2 + 16$.
For the circle $x^2+y^2=k^2$ to lie inside the parabola,we must have $k^2 \leq d^2$ for all $t$.
Let $u = t^2$,where $u \geq 0$. Then $f(u) = u^2 - 4u + 16$.
The minimum value of $f(u)$ occurs at $u = -(-4)/(2 \times 1) = 2$.
Since $u=2 \geq 0$,the minimum value is $f(2) = 2^2 - 4(2) + 16 = 4 - 8 + 16 = 12$.
Thus,$k^2 \leq 12$,which implies $k \leq \sqrt{12} = 2\sqrt{3}$.
The largest value of $k$ is $2\sqrt{3}$.

(C) Let the expression be $E = \left(\frac{x}{2} + \frac{1}{x} + \sqrt{2}\right)^5$.
We can rewrite the expression as:
$E = \left(\frac{x^2 + 2 + 2\sqrt{2}x}{2x}\right)^5 = \frac{(x + \sqrt{2})^{10}}{32x^5}$.
The constant term in the expansion of $E$ is the coefficient of $x^5$ in the expansion of $(x + \sqrt{2})^{10}$ divided by $32$.
Using the Binomial Theorem,the general term in $(x + \sqrt{2})^{10}$ is $T_{r+1} = {}^{10}C_r \cdot x^{10-r} \cdot (\sqrt{2})^r$.
For the coefficient of $x^5$,we set $10-r = 5$,which gives $r = 5$.
The term is ${}^{10}C_5 \cdot (\sqrt{2})^5 = 252 \cdot 4\sqrt{2} = 1008\sqrt{2}$.
Thus,the constant term is $\frac{1008\sqrt{2}}{32} = \frac{63\sqrt{2}}{2}$.
Comparing this with $\frac{a\sqrt{2}}{2}$,we get $a = 63$.

(B) Using the binomial expansion $(1+x)^n \approx 1+nx$ for small $|x|$:
$\sqrt{1+x} = (1+x)^{1/2} \approx 1+\frac{1}{2}x$
$(1-x)^{3/2} \approx 1-\frac{3}{2}x$
Substituting these into the expression:
$\frac{(1+\frac{1}{2}x) + (1-\frac{3}{2}x)}{(1+x) + (1+\frac{1}{2}x)} = \frac{2-x}{2+\frac{3}{2}x} = \frac{2-x}{\frac{4+3x}{2}} = \frac{2(2-x)}{4+3x}$
$= \frac{4-2x}{4+3x} = (4-2x)(4+3x)^{-1} = (4-2x) \cdot \frac{1}{4}(1+\frac{3}{4}x)^{-1}$
$\approx \frac{1}{4}(4-2x)(1-\frac{3}{4}x) = \frac{1}{4}(4 - 3x - 2x + \frac{6}{4}x^2)$
Neglecting $x^2$ terms:
$\approx \frac{1}{4}(4-5x) = 1-\frac{5x}{4}$

(C) Given the expansion: $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.
Differentiating both sides with respect to $x$:
$10(1-x+x^2)^9 \cdot (-1+2x) = a_1 + 2a_2 x + 3a_3 x^2 + \ldots + 20a_{20} x^{19}$.
Setting $x=1$:
$10(1-1+1)^9 \cdot (-1+2) = a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20}$.
$10(1)^9 \cdot (1) = a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20} = 10$.
To find $a_1$,differentiate the original expression and set $x=0$ or observe the coefficient of $x$ in $(1-x+x^2)^{10}$:
$(1-x+x^2)^{10} = 1 + 10(-x+x^2) + \ldots = 1 - 10x + \ldots$.
Thus,$a_1 = -10$.
Substituting $a_1 = -10$ into the equation $a_1 + 2a_2 + 3a_3 + \ldots + 20a_{20} = 10$:
$-10 + (2a_2 + 3a_3 + \ldots + 20a_{20}) = 10$.
Therefore,$2a_2 + 3a_3 + \ldots + 20a_{20} = 20$.

(C) The expansion of $\left(x^2-\frac{1}{2x}\right)^{20}$ contains $20+1 = 21$ terms.
Since the number of terms is odd,the middle term is the $\left(\frac{20}{2}+1\right)$-th term,which is the $11$-th term.
Thus,$m = 11$.
We need to find the coefficient of $T_{m+3} = T_{11+3} = T_{14} = T_{13+1}$.
The general term $T_{r+1}$ in the expansion of $(a+b)^n$ is given by ${}^nC_r a^{n-r} b^r$.
For $T_{13+1}$,we have $n=20$,$r=13$,$a=x^2$,and $b=-\frac{1}{2x}$.
$T_{14} = {}^{20}C_{13} (x^2)^{20-13} \left(-\frac{1}{2x}\right)^{13}$
$T_{14} = {}^{20}C_{13} (x^2)^7 \left(-\frac{1}{2^{13} x^{13}}\right)$
$T_{14} = -{}^{20}C_{13} \cdot \frac{x^{14}}{2^{13} x^{13}} = -{}^{20}C_{13} \cdot 2^{-13} x$.
Therefore,the coefficient of $T_{m+3}$ is $-{}^{20}C_{13} 2^{-13}$.

(D) The given expression is a geometric series with first term $a = (1+x)^{1000}$,common ratio $r = \frac{x}{1+x}$,and number of terms $n = 1001$.
Using the sum formula $S_n = \frac{a(1-r^n)}{1-r}$:
$f(x) = \frac{(1+x)^{1000} \left(1 - (\frac{x}{1+x})^{1001}\right)}{1 - \frac{x}{1+x}}$
$f(x) = \frac{(1+x)^{1000} - \frac{x^{1001}}{1+x}}{\frac{1+x-x}{1+x}} = \frac{(1+x)^{1001} - x^{1001}}{1} = (1+x)^{1001} - x^{1001}$.
The coefficient of $x^{50}$ in $(1+x)^{1001} - x^{1001}$ is the coefficient of $x^{50}$ in $(1+x)^{1001}$,which is ${}^{1001}C_{50}$.

(D) The general term of the expansion $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$ is given by $T_{r+1} = {}^{15}C_r (3^{1/5})^{15-r} (2^{1/3})^r = {}^{15}C_r 3^{(15-r)/5} 2^{r/3}$.
For the term to be rational,the exponents of $3$ and $2$ must be integers.
Thus,$(15-r)/5 = 3 - r/5$ must be an integer,implying $r$ is a multiple of $5$.
Also,$r/3$ must be an integer,implying $r$ is a multiple of $3$.
Therefore,$r$ must be a multiple of $\text{lcm}(5, 3) = 15$.
Given $0 \leq r \leq 15$,the possible values for $r$ are $0$ and $15$.
For $r=0$,$T_1 = {}^{15}C_0 3^3 2^0 = 1 \times 27 \times 1 = 27$.
For $r=15$,$T_{16} = {}^{15}C_{15} 3^0 2^5 = 1 \times 1 \times 32 = 32$.
Number of rational terms is $2$.
Sum of rational terms $= 27 + 32 = 59$.
Since the total sum of all terms is $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$,which is a very large number,the sum of irrational terms is significantly larger than the sum of rational terms $(59)$.

(C) The expression is $(a+1+\frac{1}{a})^n = \frac{(a^2+a+1)^n}{a^n}$.
For a trinomial expansion $(x+y+z)^n$,the number of terms is given by $\frac{(n+1)(n+2)}{2}$.
However,in this specific case,the expression is $(a^2+a+1)^n$ divided by $a^n$. The expansion of $(a^2+a+1)^n$ has $2n+1$ terms because the powers of $a$ range from $a^0$ to $a^{2n}$.
Given the number of terms is $2029$,we set $2n+1 = 2029$.
$2n = 2028$.
$n = 1014$.
Hence,option $C$ is correct.

(A) The general term in the expansion of $(3+\frac{x}{2})^n$ is given by $T_{r+1} = {}^nC_r (3)^{n-r} (\frac{x}{2})^r = {}^nC_r \frac{3^{n-r}}{2^r} x^r$.
The coefficient of $x^r$ is ${}^nC_r \frac{3^{n-r}}{2^r}$.
For $r=9$,the coefficient is ${}^nC_9 \frac{3^{n-9}}{2^9}$.
For $r=10$,the coefficient is ${}^nC_{10} \frac{3^{n-10}}{2^{10}}$.
Given that these coefficients are equal:
${}^nC_9 \frac{3^{n-9}}{2^9} = {}^nC_{10} \frac{3^{n-10}}{2^{10}}$.
Dividing both sides by ${}^nC_9 \frac{3^{n-10}}{2^{10}}$,we get:
$\frac{3^{n-9}}{3^{n-10}} \times \frac{2^{10}}{2^9} = \frac{{}^nC_{10}}{{}^nC_9}$.
$3^1 \times 2^1 = \frac{n-10+1}{10} = \frac{n-9}{10}$.
$6 = \frac{n-9}{10}$ $\Rightarrow n-9 = 60$ $\Rightarrow n = 69$.
Thus,the correct option is $A$.

(B) The general term in the expansion of $(x+y)^n$ is given by $T_{r+1} = {}^nC_r x^{n-r} y^r$.
For the expansion $\left(\frac{2p}{3} + \frac{3q}{2}\right)^9$,the $6^{th}$ term $(T_6)$ is $T_{5+1}$.
Here,$n=9$,$r=5$,$x=\frac{2p}{3}$,and $y=\frac{3q}{2}$.
$T_6 = {}^9C_5 \left(\frac{2p}{3}\right)^{9-5} \left(\frac{3q}{2}\right)^5$
$T_6 = {}^9C_5 \left(\frac{2p}{3}\right)^4 \left(\frac{3q}{2}\right)^5$
$T_6 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times \left(\frac{2^4 p^4}{3^4}\right) \times \left(\frac{3^5 q^5}{2^5}\right)$
$T_6 = 126 \times \frac{16 p^4}{81} \times \frac{243 q^5}{32}$
$T_6 = 126 \times \frac{1}{81} \times 243 \times \frac{16}{32} p^4 q^5$
$T_6 = 126 \times 3 \times \frac{1}{2} p^4 q^5 = 189 p^4 q^5$.
Comparing $189 p^4 q^5$ with $ap^bq^c$,we get $a=189, b=4, c=5$.

(A) Given expression: $(1+3x)^n \left(1+\frac{1}{3x}\right)^n$
$= (1+3x)^n \left(\frac{3x+1}{3x}\right)^n$
$= \frac{(1+3x)^n (1+3x)^n}{(3x)^n}$
$= \frac{(1+3x)^{2n}}{(3x)^n}$
The general term in the expansion of $(1+3x)^{2n}$ is $T_{r+1} = \binom{2n}{r} (3x)^r$.
To find the constant term,we need the term where the power of $x$ is $0$.
The expression is $\frac{1}{(3x)^n} \times \sum_{r=0}^{2n} \binom{2n}{r} (3x)^r = \sum_{r=0}^{2n} \binom{2n}{r} \frac{(3x)^r}{(3x)^n} = \sum_{r=0}^{2n} \binom{2n}{r} (3x)^{r-n}$.
The constant term occurs when $r-n = 0$,i.e.,$r = n$.
Substituting $r=n$,the constant term is $\binom{2n}{n} (3x)^{n-n} = \binom{2n}{n}$.
Thus,the constant term is $\binom{2n}{n}$.

(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$ is given by:
$T_{r+1} = \binom{10}{r} (x^{1/2})^{10-r} (-k x^{-2})^r$
$T_{r+1} = \binom{10}{r} (-k)^r x^{\frac{10-r}{2} - 2r}$
For the term to be independent of $x$,the exponent of $x$ must be $0$:
$\frac{10-r}{2} - 2r = 0$
$10 - r = 4r$
$5r = 10 \Rightarrow r = 2$
Now,substitute $r=2$ into the expression for the term:
$T_3 = \binom{10}{2} (-k)^2 = 405$
$45 k^2 = 405$
$k^2 = 9$
$k = \pm 3$

(D) The $r^{\text{th}}$ term from the end in the expansion of $(a + b)^n$ is the $(n - r + 2)^{\text{th}}$ term from the beginning.
Here,$n = 12$ and $r = 5$,so we need the $(12 - 5 + 2) = 9^{\text{th}}$ term from the beginning.
The general term $T_{k+1}$ is given by ${}^{12}C_k (\frac{x^3}{2})^{12-k} (-\frac{2}{x^2})^k$.
For the $9^{\text{th}}$ term,$k = 8$.
$T_9 = {}^{12}C_8 (\frac{x^3}{2})^4 (-\frac{2}{x^2})^8$
$T_9 = {}^{12}C_8 \cdot \frac{x^{12}}{2^4} \cdot \frac{2^8}{x^{16}}$
$T_9 = {}^{12}C_8 \cdot 2^4 \cdot x^{12-16}$
$T_9 = {}^{12}C_8 \cdot 16 \cdot x^{-4}$
Thus,the index of the power of $x$ is $-4$.

(C) We know that the sequence $a_n = \left(1 + \frac{1}{n}\right)^n$ is strictly increasing and converges to $e$ as $n \to \infty$.
Given $n = 100000$,we have $\left(1 + \frac{1}{100000}\right)^{100000} < e$.
Since $e \approx 2.71828$,the value of $\left(1 + \frac{1}{100000}\right)^{100000}$ is slightly less than $2.71828$.
Also,for $n=1$,$\left(1 + \frac{1}{1}\right)^1 = 2$.
Since the sequence is increasing,$\left(1 + \frac{1}{100000}\right)^{100000} > 2$.
Thus,$2 < \left(1 + \frac{1}{100000}\right)^{100000} < 2.71828$.
Therefore,the greatest integer not greater than this value is $[2.something] = 2$.

(B) The given expression is a geometric series: $S = 1 + (1+x) + (1+x)^2 + \cdots + (1+x)^n$.
Using the formula for the sum of a geometric series $S = \frac{a(r^n - 1)}{r - 1}$,where $a = 1$,$r = (1+x)$,and the number of terms is $n+1$:
$S = \frac{1((1+x)^{n+1} - 1)}{(1+x) - 1} = \frac{(1+x)^{n+1} - 1}{x}$.
We need the coefficient of $x^{100}$ in $S$,which is the coefficient of $x^{100}$ in $\frac{(1+x)^{n+1} - 1}{x}$.
This is equivalent to finding the coefficient of $x^{101}$ in $(1+x)^{n+1} - 1$.
The coefficient of $x^{101}$ in $(1+x)^{n+1}$ is given by ${ }^{n+1} C_{101}$.
Given that this coefficient is ${ }^{201} C_{101}$,we have ${ }^{n+1} C_{101} = { }^{201} C_{101}$.
Therefore,$n+1 = 201$,which implies $n = 200$.

(A) The general term in the expansion of $(a-b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} (-b)^r$.
For the $5^{\text{th}}$ term,$r=4$: $T_5 = \binom{n}{4} a^{n-4} (-b)^4 = \binom{n}{4} a^{n-4} b^4$.
For the $6^{\text{th}}$ term,$r=5$: $T_6 = \binom{n}{5} a^{n-5} (-b)^5 = -\binom{n}{5} a^{n-5} b^5$.
Given $T_5 + T_6 = 0$,we have $\binom{n}{4} a^{n-4} b^4 - \binom{n}{5} a^{n-5} b^5 = 0$.
This implies $\binom{n}{4} a^{n-4} b^4 = \binom{n}{5} a^{n-5} b^5$.
Dividing both sides by $\binom{n}{4} a^{n-5} b^4$,we get $\frac{a}{b} = \frac{\binom{n}{5}}{\binom{n}{4}}$.
Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have $\frac{\binom{n}{5}}{\binom{n}{4}} = \frac{n!}{5!(n-5)!} \times \frac{4!(n-4)!}{n!} = \frac{n-4}{5}$.
Thus,$\frac{a}{b} = \frac{n-4}{5}$.

(C) Using the remainder theorem,the remainder is $P(2)$.
Given $P(x) = 2x^3 - 5x^2 + 7$.
Substituting $x = 2$ into the polynomial:
$P(2) = 2(2)^3 - 5(2)^2 + 7$
$P(2) = 2(8) - 5(4) + 7$
$P(2) = 16 - 20 + 7$
$P(2) = 3$
Therefore,the remainder is $3$.

(D) For any odd positive integer $n$,the expression $a^n + b^n$ can be expanded using the binomial theorem or algebraic factorization.
Specifically,$a^n + b^n = (a + b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \dots + b^{n-1})$.
Therefore,$a^n + b^n$ is always divisible by $(a + b)$ when $n$ is an odd positive integer.

(B) Given that $x^4+x^2+1$ is divisible by both $x^2+mx+1$ and $x^2+nx+1$.
Since the polynomial is of degree $4$ and the divisors are of degree $2$,we can write:
$x^4+x^2+1 = (x^2+mx+1)(x^2+nx+1)$
Expanding the right side:
$x^4 + nx^3 + x^2 + mx^3 + mnx^2 + mx + x^2 + nx + 1$
$= x^4 + (m+n)x^3 + (mn+2)x^2 + (m+n)x + 1$
Comparing the coefficients of $x^3$ on both sides,we get:
$m+n = 0$

(A) Let $f(k) = 3^{3k} - 26k - 1 = 27^k - 26k - 1$.
Using the Binomial Theorem,we can write $27^k$ as $(1 + 26)^k$.
$27^k = (1 + 26)^k = \binom{k}{0} + \binom{k}{1}(26) + \binom{k}{2}(26)^2 + \dots + \binom{k}{k}(26)^k$.
$27^k = 1 + 26k + \binom{k}{2}(26)^2 + \dots + (26)^k$.
Substituting this into the expression for $f(k)$:
$f(k) = (1 + 26k + \binom{k}{2}(26)^2 + \dots + (26)^k) - 26k - 1$.
$f(k) = \binom{k}{2}(26)^2 + \binom{k}{3}(26)^3 + \dots + (26)^k$.
All terms in this expansion are divisible by $(26)^2 = 676$.
Therefore,$3^{3k} - 26k - 1$ is divisible by $676$.

(D) The general term in the expansion of $(1+x+x^2)^8$ is given by $\frac{8!}{n_1! n_2! n_3!} (1)^{n_1} (x)^{n_2} (x^2)^{n_3}$,where $n_1 + n_2 + n_3 = 8$ and $n_2 + 2n_3 = 5$.
We find the non-negative integer solutions for $(n_1, n_2, n_3)$:
$1$. If $n_3 = 0$,then $n_2 = 5$,so $n_1 = 8 - 5 - 0 = 3$. Coefficient: $\frac{8!}{3! 5! 0!} = 56$.
$2$. If $n_3 = 1$,then $n_2 = 3$,so $n_1 = 8 - 3 - 1 = 4$. Coefficient: $\frac{8!}{4! 3! 1!} = 280$.
$3$. If $n_3 = 2$,then $n_2 = 1$,so $n_1 = 8 - 1 - 2 = 5$. Coefficient: $\frac{8!}{5! 1! 2!} = 168$.
Summing these coefficients: $56 + 280 + 168 = 504$.
Thus,the coefficient of $x^5$ is $504$.

(C) The given expression is $(1+x)^{101} (1-x+x^2)^{100}$.
We can rewrite this as $(1+x) [(1+x)(1-x+x^2)]^{100}$.
Using the identity $(1+x)(1-x+x^2) = 1+x^3$,the expression becomes $(1+x)(1+x^3)^{100}$.
Expanding this,we get $(1+x^3)^{100} + x(1+x^3)^{100}$.
In the expansion of $(1+x^3)^{100}$,the powers of $x$ are of the form $3k$,where $k$ is an integer.
For the first part $(1+x^3)^{100}$,we need the coefficient of $x^{50}$. Since $50$ is not a multiple of $3$,the coefficient is $0$.
For the second part $x(1+x^3)^{100}$,we need the coefficient of $x^{49}$ in $(1+x^3)^{100}$. Since $49$ is not a multiple of $3$,the coefficient is $0$.
Therefore,the total coefficient of $x^{50}$ is $0 + 0 = 0$.
Hence,option $(C)$ is correct.

(A) Given the expression $(x+y)^{-5} = \frac{1}{x^5} \left(1 + \frac{y}{x}\right)^{-5}$ for $\left|\frac{y}{x}\right| < 1$.
Using the binomial expansion $(1+z)^{-n} = 1 - nz + \frac{n(n+1)}{2!}z^2 - \frac{n(n+1)(n+2)}{3!}z^3 + \dots$,where $z = \frac{y}{x}$ and $n = 5$.
The general term is given by $\binom{-5}{r} \left(\frac{y}{x}\right)^r$.
We want the coefficient of $\frac{y^3}{x^8} = \frac{1}{x^5} \cdot \left(\frac{y}{x}\right)^3$.
This corresponds to the term where $r = 3$ in the expansion of $\left(1 + \frac{y}{x}\right)^{-5}$.
The coefficient is $\binom{-5}{3} = \frac{(-5)(-6)(-7)}{3 \times 2 \times 1} = \frac{-210}{6} = -35$.
Thus,the coefficient is $-35$.
Hence,option $A$ is correct.

(D) Given,${}^n C_7 = {}^n C_6$.
Using the property: If ${}^n C_x = {}^n C_y$,then either $x = y$ or $x + y = n$.
Since $7 \neq 6$,we must have $n = 7 + 6 = 13$.
Therefore,${}^n C_2 = {}^{13} C_2 = \frac{13 \times 12}{2 \times 1} = 13 \times 6 = 78$.
Hence,option $D$ is correct.

(D) Given that,${ }^{12} C_{2 k-1}={ }^{12} C_{k+1}$.
We know that if ${ }^n C_x={ }^n C_y$,then either $x=y$ or $x+y=n$.
Case $1$: $2k-1 = k+1$
$k = 2$.
Case $2$: $(2k-1) + (k+1) = 12$
$3k = 12$
$k = 4$.
Since $k=4$ is one of the given options,the correct value is $4$.

(C) We know that $(1+x)^n = \sum_{r=0}^n C_r x^r = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$.
Differentiating both sides with respect to $x$:
$n(1+x)^{n-1} = C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1}$.
Multiplying both sides by $x$:
$nx(1+x)^{n-1} = C_1 x + 2C_2 x^2 + 3C_3 x^3 + \ldots + nC_n x^n$.
Differentiating again with respect to $x$:
$n[(1+x)^{n-1} + x(n-1)(1+x)^{n-2}] = C_1 + 2^2 C_2 x + 3^2 C_3 x^2 + \ldots + n^2 C_n x^{n-1}$.
Putting $x=1$:
$\sum_{r=1}^n r^2 C_r = n[2^{n-1} + (n-1)2^{n-2}] = n[2 \cdot 2^{n-2} + (n-1)2^{n-2}] = n(n+1)2^{n-2}$.
Thus,the missing term is $n(n+1)$.

(C) The sum of the coefficients of a polynomial $P(x)$ is obtained by setting all variables equal to $1$.
For the expression $(\alpha x^2 - 2x + 1)^{2019}$,the sum of coefficients is $(\alpha(1)^2 - 2(1) + 1)^{2019} = (\alpha - 1)^{2019}$.
For the expression $(x - \alpha y)^{2019}$,the sum of coefficients is $(1 - \alpha(1))^{2019} = (1 - \alpha)^{2019}$.
According to the problem,these sums are equal:
$(\alpha - 1)^{2019} = (1 - \alpha)^{2019}$.
Since the exponent $2019$ is an odd integer,we have:
$\alpha - 1 = 1 - \alpha$.
$2\alpha = 2$.
$\alpha = 1$.

(D) Using the Binomial Theorem,$(a+b)^n = \sum_{k=0}^{n} {^nC_k} a^{n-k} b^k$.
$(102)^4 = (100+2)^4$
$= {^4C_0}(100)^4(2)^0 + {^4C_1}(100)^3(2)^1 + {^4C_2}(100)^2(2)^2 + {^4C_3}(100)^1(2)^3 + {^4C_4}(100)^0(2)^4$
$= 1 \cdot 100000000 + 4 \cdot 1000000 \cdot 2 + 6 \cdot 10000 \cdot 4 + 4 \cdot 100 \cdot 8 + 1 \cdot 1 \cdot 16$
$= 100000000 + 8000000 + 240000 + 3200 + 16$
$= 108243216$
Thus,the correct option is $D$.

(D) Given expression: $\frac{(1-x)^{1/3}+(1-5x)^2}{(16-x)^{1/4}}$
$= \frac{1}{2} (1-x)^{1/3} (1-\frac{x}{16})^{-1/4} + \frac{1}{2} (1-5x)^2 (1-\frac{x}{16})^{-1/4}$
Using binomial expansion $(1+z)^n \approx 1+nz$ for small $z$:
$= \frac{1}{2} (1-\frac{1}{3}x)(1+\frac{x}{64}) + \frac{1}{2} (1-10x)(1+\frac{x}{64})$
$= \frac{1}{2} [ (1 - \frac{1}{3}x + \frac{x}{64}) + (1 - 10x + \frac{x}{64}) ]$
$= \frac{1}{2} [ 2 - x(\frac{1}{3} + 10 - \frac{2}{64}) ]$
$= 1 - \frac{x}{2} (\frac{1}{3} + 10 - \frac{1}{32})$
$= 1 - \frac{x}{2} (\frac{32 + 960 - 3}{96}) = 1 - \frac{989}{192}x$
Therefore,the coefficient of $x$ is $-\frac{989}{192}$.

(A) Given the equation of the ellipse: $2x^2 + 3y^2 - 4x - 12y + 13 = 0$.
Rearranging the terms to complete the square:
$2(x^2 - 2x) + 3(y^2 - 4y) = -13$
$2(x^2 - 2x + 1) + 3(y^2 - 4y + 4) = -13 + 2(1) + 3(4)$
$2(x - 1)^2 + 3(y - 2)^2 = 1$
Dividing by $1$,we get the standard form:
$\frac{(x - 1)^2}{1/2} + \frac{(y - 2)^2}{1/3} = 1$
Here,$a^2 = 1/2$ and $b^2 = 1/3$. Since $a^2 > b^2$,the major axis is horizontal.
The eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/3}{1/2}} = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.
The distance from the center to the foci is $ae = \sqrt{1/2} \times \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{6}}$.
The center is $(h, k) = (1, 2)$.
The foci are $(h \pm ae, k) = \left(1 \pm \frac{1}{\sqrt{6}}, 2\right)$.

(B) The standard form of the equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Given the vertices are $(5,0)$ and $(0,-4)$,these points represent the intercepts on the axes.
The $x$-intercepts are $\pm a = \pm 5$,so $a^2 = 25$.
The $y$-intercepts are $\pm b = \pm 4$,so $b^2 = 16$.
Substituting these values into the standard equation,we get $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

(D) Given,focus $S = (6, 2)$,centre $C = (1, 2) = (h, k)$,and the ellipse passes through $P = (4, 6)$.
The standard equation of the ellipse is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.
Substituting the centre $(1, 2)$,we get $\frac{(x-1)^2}{a^2} + \frac{(y-2)^2}{b^2} = 1$ ... $(i)$.
Since the ellipse passes through $P(4, 6)$,we have $\frac{(4-1)^2}{a^2} + \frac{(6-2)^2}{b^2} = 1$,which simplifies to $\frac{9}{a^2} + \frac{16}{b^2} = 1$ ... (ii).
The distance from the centre to the focus is $ae = 6 - 1 = 5$,so $a^2e^2 = 25$.
Using the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$,we get $b^2 = a^2 - 25$,or $a^2 = b^2 + 25$ ... (iii).
Substituting $a^2$ into (ii): $\frac{9}{b^2+25} + \frac{16}{b^2} = 1$.
$9b^2 + 16(b^2 + 25) = b^2(b^2 + 25) \implies 25b^2 + 400 = b^4 + 25b^2 \implies b^4 = 400 \implies b^2 = 20$.
From (iii),$a^2 = 20 + 25 = 45$.
Thus,the equation is $\frac{(x-1)^2}{45} + \frac{(y-2)^2}{20} = 1$.

(D) Given: Centre is $(0,0)$,eccentricity $e = 1/2$,and directrix $x = 4$.
For an ellipse with centre at origin and directrix $x = a/e$,we have $a/e = 4$.
Substituting $e = 1/2$,we get $a = 4 \times (1/2) = 2$.
Now,$b^2 = a^2(1 - e^2) = 4(1 - 1/4) = 4(3/4) = 3$.
The standard equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Substituting the values,we get $\frac{x^2}{4} + \frac{y^2}{3} = 1$.
Multiplying by $12$,we get $3x^2 + 4y^2 = 12$.
Thus,option $D$ is correct.

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
According to the problem,the length of the latus rectum is $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$ ... $(i)$.
The distance between the vertex $(a, 0)$ and the nearest focus $(ae, 0)$ is $a - ae = 3/2$,which implies $a(1 - e) = 3/2$ ... $(ii)$.
We know that for an ellipse,$b^2 = a^2(1 - e^2)$. Substituting this into equation $(i)$:
$\frac{2a^2(1 - e^2)}{a} = 4$ $\Rightarrow 2a(1 - e^2) = 4$ $\Rightarrow a(1 - e^2) = 2$.
Since $a(1 - e) = 3/2$,we can write $a(1 - e)(1 + e) = 2$.
Substituting $a(1 - e) = 3/2$ into this equation:
$\frac{3}{2}(1 + e) = 2$ $\Rightarrow 1 + e = \frac{4}{3}$ $\Rightarrow e = \frac{4}{3} - 1 = \frac{1}{3}$.
Thus,the eccentricity is $1/3$.

(A) Given equation of the ellipse is $4x^2 + 25y^2 = 100$.
Dividing both sides by $100$,we get $\frac{4x^2}{100} + \frac{25y^2}{100} = \frac{100}{100}$,which simplifies to $\frac{x^2}{25} + \frac{y^2}{4} = 1$.
Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 4$.
The eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$.
Substituting the values,$e = \sqrt{1 - \frac{4}{25}} = \sqrt{\frac{25 - 4}{25}} = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$.
Thus,the correct option is $A$.

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The foci are $P = (-ae, 0)$ and $Q = (ae, 0)$.
The end of the minor axis is $R = (0, b)$.
Since $\triangle PQR$ is an equilateral triangle,the distance $PQ = PR = QR$.
The distance $PQ = 2ae$.
The distance $PR = \sqrt{(ae - 0)^2 + (0 - b)^2} = \sqrt{a^2e^2 + b^2}$.
Equating $PQ^2 = PR^2$,we get $(2ae)^2 = a^2e^2 + b^2$.
$4a^2e^2 = a^2e^2 + b^2 \implies 3a^2e^2 = b^2$.
Using the relation $b^2 = a^2(1 - e^2)$,we have $3a^2e^2 = a^2(1 - e^2)$.
$3e^2 = 1 - e^2 \implies 4e^2 = 1 \implies e^2 = \frac{1}{4}$.
Thus,$e = \frac{1}{2}$.

(B) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$.
The foci are $S(ae, 0)$ and $S'(-ae, 0)$.
The upper vertex is $V(0, b)$.
The line segment joining the foci is $SS'$,which subtends an angle $2\alpha$ at $V(0, b)$.
This means $\angle SVS' = 2\alpha$,so $\angle SVO = \alpha$,where $O$ is the origin $(0, 0)$.
In the right-angled triangle $\triangle SVO$,we have $\tan \alpha = \frac{SO}{VO} = \frac{ae}{b}$.
Thus,$ae = b \tan \alpha$.
We know the relation $b^2 = a^2(1 - e^2)$,which implies $b^2 = a^2 - a^2e^2$.
Substituting $ae = b \tan \alpha$,we get $b^2 = a^2 - (b \tan \alpha)^2$,so $a^2 = b^2 + b^2 \tan^2 \alpha = b^2(1 + \tan^2 \alpha) = b^2 \sec^2 \alpha$.
Therefore,$a = b \sec \alpha$.
Now,the eccentricity $e = \frac{ae}{a} = \frac{b \tan \alpha}{b \sec \alpha} = \frac{\sin \alpha / \cos \alpha}{1 / \cos \alpha} = \sin \alpha$.

(A) The condition for the line $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2m^2+b^2$.
Given the ellipse equation $3x^2+5y^2=15$,we divide by $15$ to get the standard form: $\frac{x^2}{5}+\frac{y^2}{3}=1$.
Here,$a^2=5$,$b^2=3$,$m=2$,and $c=k$.
Substituting these values into the condition $c^2=a^2m^2+b^2$:
$k^2 = (5)(2^2) + 3$
$k^2 = 5 \times 4 + 3$
$k^2 = 20 + 3 = 23$
$k = \pm \sqrt{23}$.

(D) We know that the direction cosines of a line satisfy the relation $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,where $\alpha, \beta, \gamma$ are the angles made by the line with the $X, Y,$ and $Z$-axes respectively.
Given $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{4}$.
Substituting these values into the relation:
$\cos^2 \left(\frac{\pi}{3}\right) + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \gamma = 1$
$\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \gamma = 1$
$\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$
$\frac{3}{4} + \cos^2 \gamma = 1$
$\cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$
$\cos \gamma = \pm \frac{1}{2}$
Since $\gamma$ is the angle between $0$ and $\pi$,$\cos \gamma = \frac{1}{2}$ implies $\gamma = \frac{\pi}{3}$ (or $\cos \gamma = -\frac{1}{2}$ implies $\gamma = \frac{2\pi}{3}$).
Comparing with the options,the correct angle is $\frac{\pi}{3}$.

(B) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.
We know that the sum of the squares of the direction cosines is always $1$,i.e.,$\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma = 1$.
Using the trigonometric identity $\sin ^2 \theta = 1 - \cos ^2 \theta$,we can rewrite the expression as:
$\sin ^2 \alpha + \sin ^2 \beta + \sin ^2 \gamma = (1 - \cos ^2 \alpha) + (1 - \cos ^2 \beta) + (1 - \cos ^2 \gamma)$
$= 3 - (\cos ^2 \alpha + \cos ^2 \beta + \cos ^2 \gamma)$
$= 3 - 1 = 2$.
Thus,the value is $2$.

(A) Let the direction cosines of the lines be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.
Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$.
From the first equation,$l = -\frac{bm + cn}{a}$. Substituting this into the second equation:
$fmn + gn(-\frac{bm + cn}{a}) + hm(-\frac{bm + cn}{a}) = 0$
$afmn - bgnm - cgn^2 - bhm^2 - chmn = 0$
$bhm^2 + (ch + bg - af)mn + cgn^2 = 0$
Dividing by $n^2$,we get $bh(\frac{m}{n})^2 + (ch + bg - af)(\frac{m}{n}) + cg = 0$.
The product of the roots is $\frac{m_1 m_2}{n_1 n_2} = \frac{cg}{bh}$.
Similarly,by eliminating $m$,we get $ah(\frac{l}{n})^2 + (ch + af - bg)(\frac{l}{n}) + cf = 0$,so $\frac{l_1 l_2}{n_1 n_2} = \frac{cf}{ah}$.
Since the lines are perpendicular,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Dividing by $n_1 n_2$,we get $\frac{l_1 l_2}{n_1 n_2} + \frac{m_1 m_2}{n_1 n_2} + 1 = 0$.
Substituting the products: $\frac{cf}{ah} + \frac{cg}{bh} + 1 = 0$.
Dividing by $c$,we get $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

(C) The given Cartesian equation is $2x - 3 = 3y + 1 = 5 - 6z$.
To write this in standard form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$,we divide by the coefficients of $x, y, z$:
$2(x - \frac{3}{2}) = 3(y + \frac{1}{3}) = -6(z - \frac{5}{6})$.
Dividing by $6$,we get $\frac{x - 3/2}{3} = \frac{y + 1/3}{2} = \frac{z - 5/6}{-1}$.
The direction vector of this line is $\vec{v} = 3 \hat{i} + 2 \hat{j} - \hat{k}$.
The vector equation of a line passing through point $\vec{a} = 7 \hat{i} - 5 \hat{j} + 0 \hat{k}$ and parallel to $\vec{v}$ is given by $\vec{r} = \vec{a} + \lambda \vec{v}$.
Thus,$\vec{r} = (7 \hat{i} - 5 \hat{j}) + \lambda(3 \hat{i} + 2 \hat{j} - \hat{k})$.

(A) Given the sides of the parallelogram are $\vec{PQ} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{PS} = \hat{i} - 2\hat{k}$.
The diagonals of the parallelogram are given by $\vec{d_1} = \vec{PR} = \vec{PQ} + \vec{PS}$ and $\vec{d_2} = \vec{QS} = \vec{PS} - \vec{PQ}$.
Calculating $\vec{PR}$:
$\vec{PR} = (3\hat{i} - 2\hat{j} + 2\hat{k}) + (\hat{i} - 2\hat{k}) = 4\hat{i} - 2\hat{j} + 0\hat{k} = 4\hat{i} - 2\hat{j}$.
Calculating $\vec{QS}$:
$\vec{QS} = (\hat{i} - 2\hat{k}) - (3\hat{i} - 2\hat{j} + 2\hat{k}) = -2\hat{i} + 2\hat{j} - 4\hat{k}$.
Let $\theta$ be the angle between the diagonals $\vec{PR}$ and $\vec{QS}$.
$\cos \theta = \frac{\vec{PR} \cdot \vec{QS}}{|\vec{PR}| |\vec{QS}|}$.
$\vec{PR} \cdot \vec{QS} = (4\hat{i} - 2\hat{j} + 0\hat{k}) \cdot (-2\hat{i} + 2\hat{j} - 4\hat{k}) = (4)(-2) + (-2)(2) + (0)(-4) = -8 - 4 + 0 = -12$.
$|\vec{PR}| = \sqrt{4^2 + (-2)^2 + 0^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.
$|\vec{QS}| = \sqrt{(-2)^2 + 2^2 + (-4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
$\cos \theta = \frac{-12}{(2\sqrt{5})(2\sqrt{6})} = \frac{-12}{4\sqrt{30}} = \frac{-3}{\sqrt{30}} = -\sqrt{\frac{9}{30}} = -\sqrt{\frac{3}{10}}$.
Thus,$\cos \theta = -\sqrt{\frac{3}{10}}$.

(D) Let $A = (9, 13, 15)$,$B = (12, 21, 10)$,and $P = (5, 7, 3)$. Let $Q(x, y, z)$ be the foot of the perpendicular from $P$ to the line $AB$.
The direction ratios of line $AB$ are $(12 - 9, 21 - 13, 10 - 15) = (3, 8, -5)$.
The equation of line $AB$ is $\frac{x - 9}{3} = \frac{y - 13}{8} = \frac{z - 15}{-5} = \lambda$.
Any point $Q$ on the line $AB$ is given by $Q = (3\lambda + 9, 8\lambda + 13, -5\lambda + 15)$.
The direction ratios of $PQ$ are $(3\lambda + 9 - 5, 8\lambda + 13 - 7, -5\lambda + 15 - 3) = (3\lambda + 4, 8\lambda + 6, -5\lambda + 12)$.
Since $PQ \perp AB$,the dot product of their direction ratios is zero:
$3(3\lambda + 4) + 8(8\lambda + 6) - 5(-5\lambda + 12) = 0$
$9\lambda + 12 + 64\lambda + 48 + 25\lambda - 60 = 0$
$98\lambda = 0 \implies \lambda = 0$.
Substituting $\lambda = 0$ into the coordinates of $Q$,we get $Q = (9, 13, 15)$.
Thus,the foot of the perpendicular is $(9, 13, 15)$,which corresponds to option $D$.

(D) Let the direction ratios of the required line be $(a, b, c)$.
The line passes through the point $P(-1, 3, -2)$.
The direction ratios of the given lines are $\vec{v_1} = (1, 2, 3)$ and $\vec{v_2} = (-3, 2, 5)$.
Since the required line is perpendicular to both given lines,its direction vector $\vec{v} = (a, b, c)$ must be parallel to the cross product $\vec{v_1} \times \vec{v_2}$.
$\vec{v} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} = \hat{i}(10 - 6) - \hat{j}(5 - (-9)) + \hat{k}(2 - (-6)) = 4\hat{i} - 14\hat{j} + 8\hat{k}$.
Dividing by $2$,we get the direction ratios as $(2, -7, 4)$.
The equation of the line passing through $(-1, 3, -2)$ with direction ratios $(2, -7, 4)$ is $\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$,which simplifies to $\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$.

(B) Let the direction ratios of the two lines be $\vec{a} = (2, -2, 1)$ and $\vec{b} = (1, -2, 2)$.
The formula for the cosine of the angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the given values:
$\cos \theta = \frac{|(2)(1) + (-2)(-2) + (1)(2)|}{\sqrt{2^2 + (-2)^2 + 1^2} \sqrt{1^2 + (-2)^2 + 2^2}}$
Calculating the numerator:
$|2 + 4 + 2| = 8$
Calculating the denominators:
$\sqrt{4 + 4 + 1} = \sqrt{9} = 3$
$\sqrt{1 + 4 + 4} = \sqrt{9} = 3$
Thus,$\cos \theta = \frac{8}{3 \times 3} = \frac{8}{9}$.
Therefore,$\theta = \cos^{-1}\left(\frac{8}{9}\right)$.

(A) The distance of a plane $Ax + By + Cz + D = 0$ from a point $(x_1, y_1, z_1)$ is given by the formula:
$d = \left| \frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}} \right|$
Here,the plane is $2x - y - 2z - 9 = 0$ and the origin is $(0, 0, 0)$.
Substituting the values $A = 2, B = -1, C = -2, D = -9$ and $(x_1, y_1, z_1) = (0, 0, 0)$:
$d = \left| \frac{2(0) + (-1)(0) + (-2)(0) - 9}{\sqrt{2^2 + (-1)^2 + (-2)^2}} \right|$
$d = \left| \frac{-9}{\sqrt{4 + 1 + 4}} \right| = \left| \frac{-9}{\sqrt{9}} \right| = \left| \frac{-9}{3} \right| = 3 \text{ units.}$

(A) The equation of any plane passing through the intersection of two planes $P_1: x+2y+3z-4=0$ and $P_2: 4x+3y+2z+1=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+2y+3z-4) + \lambda(4x+3y+2z+1) = 0$ ...$(i)$
Since the plane passes through the origin $(0,0,0)$,we substitute $x=0, y=0, z=0$ into equation $(i)$:
$(0+0+0-4) + \lambda(0+0+0+1) = 0$
$-4 + \lambda = 0 \Rightarrow \lambda = 4$
Substituting $\lambda = 4$ back into equation $(i)$:
$(x+2y+3z-4) + 4(4x+3y+2z+1) = 0$
$x+2y+3z-4 + 16x+12y+8z+4 = 0$
$17x+14y+11z = 0$
Thus,the correct option is $A$.

(D) Let the direction ratios of the line of intersection of the planes $x-y=0$ and $2x+y+z=0$ be $a_1, b_1, c_1$. Since the line lies in both planes,it is perpendicular to the normals of both planes. The normals are $\vec{n_1} = (1, -1, 0)$ and $\vec{n_2} = (2, 1, 1)$.
The direction vector $\vec{v_1} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(-1-0) - \hat{j}(1-0) + \hat{k}(1+2) = -\hat{i} - \hat{j} + 3\hat{k}$.
Thus,the direction ratios are $(a_1, b_1, c_1) = (-1, -1, 3)$,which is equivalent to $(1, 1, -3)$.
Similarly,for the planes $2x-z=0$ and $x+y-3z=0$,the normals are $\vec{n_3} = (2, 0, -1)$ and $\vec{n_4} = (1, 1, -3)$.
The direction vector $\vec{v_2} = \vec{n_3} \times \vec{n_4} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -1 \\ 1 & 1 & -3 \end{vmatrix} = \hat{i}(0+1) - \hat{j}(-6+1) + \hat{k}(2-0) = \hat{i} + 5\hat{j} + 2\hat{k}$.
Wait,re-calculating $\vec{v_2}$: $\hat{i}(0 - (-1)) - \hat{j}(-6 - (-1)) + \hat{k}(2 - 0) = \hat{i} + 5\hat{j} + 2\hat{k}$.
Let us re-calculate the cross product for the second set: $\vec{n_3} = (2, 0, -1)$,$\vec{n_4} = (1, 1, -3)$. $\vec{v_2} = (0(-3) - (-1)(1), -((-1)(1) - 2(-3)), 2(1) - 0(1)) = (1, -5, 2)$.
Now,the angle $\theta$ between $\vec{v_1} = (1, 1, -3)$ and $\vec{v_2} = (1, -5, 2)$ is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
$\vec{v_1} \cdot \vec{v_2} = (1)(1) + (1)(-5) + (-3)(2) = 1 - 5 - 6 = -10$.
$\cos \theta = \frac{|-10|}{\sqrt{1^2+1^2+(-3)^2} \sqrt{1^2+(-5)^2+2^2}} = \frac{10}{\sqrt{11} \sqrt{30}}$.
Re-checking the original problem values: $x-y=0, 2x+y+z=0 \Rightarrow \vec{v_1} = (1, 1, -3)$. $2x-z=0, x+y-3z=0 \Rightarrow \vec{v_2} = (1, 5, 2)$.
$\vec{v_1} \cdot \vec{v_2} = 1 + 5 - 6 = 0$. Since the dot product is $0$,the angle is $90^{\circ}$.
Hence,option $(d)$ is correct.

(A) Let the point be $P(1, 3, 4)$ and the plane be $2x - y + z + 3 = 0$.
The formula for the image $(x', y', z')$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by $\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = \frac{z' - z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$.
Substituting the values: $\frac{x' - 1}{2} = \frac{y' - 3}{-1} = \frac{z' - 4}{1} = -2 \frac{2(1) - 1(3) + 1(4) + 3}{2^2 + (-1)^2 + 1^2}$.
$\frac{x' - 1}{2} = \frac{y' - 3}{-1} = \frac{z' - 4}{1} = -2 \frac{2 - 3 + 4 + 3}{4 + 1 + 1} = -2 \frac{6}{6} = -2$.
Solving for coordinates:
$x' - 1 = 2(-2) \Rightarrow x' = -3$.
$y' - 3 = -1(-2) \Rightarrow y' = 5$.
$z' - 4 = 1(-2) \Rightarrow z' = 2$.
Thus,the image is $(-3, 5, 2)$.
Therefore,option $A$ is correct.

(A) The intercept form of the equation of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.
Given that the intercepts are $a = 1, b = 2, c = 4$.
Substituting these values into the formula,we get $\frac{x}{1} + \frac{y}{2} + \frac{z}{4} = 1$.
To simplify,multiply the entire equation by the least common multiple of the denominators,which is $4$.
$4 \times (\frac{x}{1}) + 4 \times (\frac{y}{2}) + 4 \times (\frac{z}{4}) = 4 \times 1$.
This simplifies to $4x + 2y + z = 4$.
Thus,the correct option is $A$.

(A) The distance of a point $P(x, y, z)$ from the $X$-axis is given by $\sqrt{y^2+z^2}$.
The distance of the point $P(x, y, z)$ from the plane $x+z-1=0$ is given by $\frac{|x+z-1|}{\sqrt{1^2+0^2+1^2}} = \frac{|x+z-1|}{\sqrt{2}}$.
According to the problem,these distances are equal:
$\sqrt{y^2+z^2} = \frac{|x+z-1|}{\sqrt{2}}$
Squaring both sides,we get:
$y^2+z^2 = \frac{(x+z-1)^2}{2}$
$2(y^2+z^2) = x^2+z^2+1+2xz-2x-2z$
$2y^2+2z^2 = x^2+z^2+1+2xz-2x-2z$
Rearranging the terms,we get:
$x^2-2y^2-z^2+2xz-2x-2z+1=0$
Thus,the correct option is $A$.

(B) The equations of the given planes are $x+2y+2z-5=0$ and $3x+3y+2z-8=0$.
The normal vectors to these planes are $\vec{n_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 3\hat{i} + 3\hat{j} + 2\hat{k}$.
The angle $\theta$ between two planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (1)(3) + (2)(3) + (2)(2) = 3 + 6 + 4 = 13$.
Calculating the magnitudes: $|\vec{n_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$ and $|\vec{n_2}| = \sqrt{3^2 + 3^2 + 2^2} = \sqrt{9+9+4} = \sqrt{22}$.
Thus,$\cos \theta = \frac{13}{3\sqrt{22}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{13}{3\sqrt{22}}\right)$.

(A) The given planes are $P_1: 2x - 3y + 6z + 21 = 0$ and $P_2: 2x - 3y + 6z - 14 = 0$.
Since the planes are parallel,the mid-parallel plane will have the same normal vector $(2, -3, 6)$.
Let the equation of the mid-parallel plane be $2x - 3y + 6z + d = 0$.
The constant term $d$ of the mid-parallel plane is the average of the constant terms of the two given planes,provided the coefficients of $x, y, z$ are identical.
First,ensure the coefficients are the same. Here they are already the same.
The constant $d$ is given by $d = \frac{d_1 + d_2}{2} = \frac{21 + (-14)}{2} = \frac{7}{2} = 3.5$.
Thus,the equation is $2x - 3y + 6z + 3.5 = 0$.
Multiplying by $2$ to clear the fraction,we get $4x - 6y + 12z + 7 = 0$.

(C) Let the general point $P$ on the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}=r$ be $P(2r+1, r-1, 1-r)$.
Since $P$ lies on the plane $x+2y+3z=4$,we substitute the coordinates of $P$ into the plane equation:
$(2r+1) + 2(r-1) + 3(1-r) = 4$
$2r + 1 + 2r - 2 + 3 - 3r = 4$
$r + 2 = 4 \Rightarrow r = 2$.
Substituting $r=2$ into the coordinates of $P$,we get the point of intersection as $P(5, 1, -1)$.
The direction vector of the required line is given by the cross product:
$\vec{v} = (2\hat{i}-3\hat{j}) \times (\hat{i}+2\hat{j}-\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 0 \\ 1 & 2 & -1 \end{vmatrix}$
$= \hat{i}(3-0) - \hat{j}(-2-0) + \hat{k}(4+3) = 3\hat{i} + 2\hat{j} + 7\hat{k}$.
The equation of the line passing through $(5, 1, -1)$ with direction vector $(3, 2, 7)$ is $\frac{x-5}{3} = \frac{y-1}{2} = \frac{z+1}{7}$.
Multiplying the denominators by $-1$,we get $\frac{x-5}{-3} = \frac{y-1}{-2} = \frac{z+1}{-7}$,which matches option $C$.

(B) Let the required vector be $\vec{v} = a(2 \hat{i}+\hat{j}+\hat{k}) + b(\hat{i}-\hat{j}+\hat{k})$.
This simplifies to $\vec{v} = (2a+b)\hat{i} + (a-b)\hat{j} + (a+b)\hat{k}$.
Since $\vec{v}$ is orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$,their dot product is zero:
$3(2a+b) + 2(a-b) + 6(a+b) = 0$.
$6a + 3b + 2a - 2b + 6a + 6b = 0$.
$14a + 7b = 0 \implies b = -2a$.
Substituting $b = -2a$ into the expression for $\vec{v}$:
$\vec{v} = (2a - 2a)\hat{i} + (a - (-2a))\hat{j} + (a + (-2a))\hat{k} = 0\hat{i} + 3a\hat{j} - a\hat{k} = a(3\hat{j} - \hat{k})$.
The unit vector is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{a(3\hat{j} - \hat{k})}{|a|\sqrt{3^2 + (-1)^2}} = \pm \frac{3\hat{j} - \hat{k}}{\sqrt{10}}$.

(A) For any vector $w$ in a plane to be represented as a linear combination $w = au + bv$ of two vectors $u$ and $v$ in that same plane,the vectors $u$ and $v$ must be linearly independent.
Two vectors are linearly independent if and only if they are not parallel to each other.
This means that neither $u$ can be written as $k \cdot v$ nor $v$ can be written as $k \cdot u$ for any scalar $k$.
Therefore,the condition is that none of $u$ and $v$ is a scalar multiple of the other.
Thus,option $A$ is the correct answer.

(A) Let the $yz$-plane intersect the line segment joining points $A(2, 3, 4)$ and $B(-3, 5, -4)$ in the ratio $\lambda : 1$ at point $M$.
Using the section formula,the coordinates of point $M$ are given by:
$M = \left( \frac{-3\lambda + 2}{\lambda + 1}, \frac{5\lambda + 3}{\lambda + 1}, \frac{-4\lambda + 4}{\lambda + 1} \right)$.
Since the point $M$ lies on the $yz$-plane,its $x$-coordinate must be $0$.
Therefore,$\frac{-3\lambda + 2}{\lambda + 1} = 0$,which implies $-3\lambda + 2 = 0$,so $\lambda = \frac{2}{3}$.
Substituting $\lambda = \frac{2}{3}$ into the coordinates of $M$:
$y = \frac{5(2/3) + 3}{(2/3) + 1} = \frac{10/3 + 9/3}{5/3} = \frac{19}{5}$.
$z = \frac{-4(2/3) + 4}{(2/3) + 1} = \frac{-8/3 + 12/3}{5/3} = \frac{4}{5}$.
Thus,the point of intersection is $\left(0, \frac{19}{5}, \frac{4}{5}\right)$.
Hence,option $A$ is correct.

(C) Let the equations of the given lines be:
$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}=r_1$
and
$\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}=r_2$.
For the intersection point,we equate the coordinates:
$x = r_1 + 1 = 3r_2 - 5$
$y = 2r_1 + 2 = -r_2 + 4$
$z = -3r_1 + 5 = 4r_2 - 3$
Solving the first two equations:
$r_1 - 3r_2 = -6$
$2r_1 + r_2 = 2$
Multiplying the second by $3$: $6r_1 + 3r_2 = 6$.
Adding to the first: $7r_1 = 0 \implies r_1 = 0$.
Then $r_2 = 2$.
Checking in the third equation: $-3(0) + 5 = 5$ and $4(2) - 3 = 5$.
Since $5 = 5$,the lines intersect at point $A(1, 2, 5)$.
$A$ plane parallel to the $xy$-plane has the equation $z = k$.
Since it passes through $(1, 2, 5)$,we have $z = 5$.
Thus,the correct option is $C$.

(C) The direction vector of the line passing through $P(1, 1, -1)$ and $Q(3, -1, 0)$ is $\vec{v} = (3-1)\hat{i} + (-1-1)\hat{j} + (0-(-1))\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k}$.
The normal vector to the plane $\sqrt{\lambda} x + 3y + 6z = 17$ is $\vec{n} = \sqrt{\lambda}\hat{i} + 3\hat{j} + 6\hat{k}$.
The angle $\theta$ between a line with direction $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin \theta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.
Given $\theta = \operatorname{Tan}^{-1}\left(\frac{1}{\sqrt{8}}\right)$,we have $\tan \theta = \frac{1}{\sqrt{8}}$,which implies $\sin \theta = \frac{1}{\sqrt{1^2 + (\sqrt{8})^2}} = \frac{1}{3}$.
Calculating the dot product: $\vec{v} \cdot \vec{n} = (2)(\sqrt{\lambda}) + (-2)(3) + (1)(6) = 2\sqrt{\lambda} - 6 + 6 = 2\sqrt{\lambda}$.
Calculating magnitudes: $|\vec{v}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$ and $|\vec{n}| = \sqrt{(\sqrt{\lambda})^2 + 3^2 + 6^2} = \sqrt{\lambda + 9 + 36} = \sqrt{\lambda + 45}$.
Substituting into the formula: $\frac{1}{3} = \frac{|2\sqrt{\lambda}|}{3 \sqrt{\lambda + 45}}$.
$1 = \frac{2\sqrt{\lambda}}{\sqrt{\lambda + 45}} \implies \sqrt{\lambda + 45} = 2\sqrt{\lambda}$.
Squaring both sides: $\lambda + 45 = 4\lambda \implies 3\lambda = 45 \implies \lambda = 15$.

(C) When a die is rolled three times,the total number of outcomes is $6 \times 6 \times 6 = 216$.
We are looking for the sum $S$ such that $S$ is a prime number of the form $4n+1$.
The possible sums range from $3$ to $18$.
The prime numbers in this range are $3, 5, 7, 11, 13, 17$.
Among these,the primes of the form $4n+1$ are $5, 13, 17$.
Number of ways to get sum $5$: $(1,1,3)$ in $3$ permutations,$(1,2,2)$ in $3$ permutations. Total = $6$ ways.
Number of ways to get sum $13$: $(1,6,6)$ in $3$ permutations,$(2,5,6)$ in $6$ permutations,$(3,4,6)$ in $6$ permutations,$(3,5,5)$ in $3$ permutations,$(4,4,5)$ in $3$ permutations. Total = $21$ ways.
Number of ways to get sum $17$: $(5,6,6)$ in $3$ permutations. Total = $3$ ways.
Total favorable outcomes = $6 + 21 + 3 = 30$.
Required probability = $\frac{30}{216} = \frac{5}{36}$.

(B) This is a problem of Bayes' theorem.
Let $A$ be the event that all $3$ balls are of different colours.
Let $E_1, E_2, E_3$ be the events that Box $1$,Box $2$,and Box $3$ are chosen respectively.
Since one box is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.
The probability of drawing $3$ balls of different colours from each box is:
$P(A|E_1) = \frac{{}^1C_1 \times {}^2C_1 \times {}^3C_1}{{}^6C_3} = \frac{1 \times 2 \times 3}{20} = \frac{6}{20} = \frac{3}{10}$
$P(A|E_2) = \frac{{}^1C_1 \times {}^1C_1 \times {}^2C_1}{{}^4C_3} = \frac{1 \times 1 \times 2}{4} = \frac{2}{4} = \frac{1}{2}$
$P(A|E_3) = \frac{{}^5C_1 \times {}^4C_1 \times {}^1C_1}{{}^{10}C_3} = \frac{5 \times 4 \times 1}{120} = \frac{20}{120} = \frac{1}{6}$
By Bayes' theorem,the probability that the balls came from Box $2$ is:
$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2) + P(E_3) \cdot P(A|E_3)}$
$P(E_2|A) = \frac{\frac{1}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{6}}$
$P(E_2|A) = \frac{\frac{1}{6}}{\frac{1}{10} + \frac{1}{6} + \frac{1}{18}} = \frac{\frac{1}{6}}{\frac{9+15+5}{90}} = \frac{\frac{1}{6}}{\frac{29}{90}} = \frac{1}{6} \times \frac{90}{29} = \frac{15}{29}$

(D) Total number of outcomes when a coin is tossed $6$ times is $2^6 = 64$.
Let $X$ be the number of heads. We want the probability $P(X > 3)$,which means $X$ can be $4, 5,$ or $6$.
The number of ways to get $r$ heads in $n$ tosses is given by $\binom{n}{r}$.
Number of ways to get $4$ heads $= \binom{6}{4} = \frac{6!}{4!2!} = 15$.
Number of ways to get $5$ heads $= \binom{6}{5} = \frac{6!}{5!1!} = 6$.
Number of ways to get $6$ heads $= \binom{6}{6} = \frac{6!}{6!0!} = 1$.
Total favorable outcomes $= 15 + 6 + 1 = 22$.
Required Probability $= \frac{22}{64} = \frac{11}{32}$.
Hence,option $D$ is correct.

(D) Let $X$ be the number of heads obtained by $P$ and $Y$ be the number of heads obtained by $Q$. Both $X$ and $Y$ follow a binomial distribution $B(n=3, p=0.5)$.
The probability of getting $r$ heads in $3$ tosses is given by $P(X=r) = ^{3}C_{r} (0.5)^3$.
We want to find $P(X=Y) = \sum_{r=0}^{3} P(X=r) \times P(Y=r)$.
Since $P(X=r) = P(Y=r)$,this becomes $\sum_{r=0}^{3} [P(X=r)]^2$.
$P(X=0) = ^{3}C_{0} (0.5)^3 = \frac{1}{8}$,$P(X=1) = ^{3}C_{1} (0.5)^3 = \frac{3}{8}$,$P(X=2) = ^{3}C_{2} (0.5)^3 = \frac{3}{8}$,$P(X=3) = ^{3}C_{3} (0.5)^3 = \frac{1}{8}$.
$P(X=Y) = (\frac{1}{8})^2 + (\frac{3}{8})^2 + (\frac{3}{8})^2 + (\frac{1}{8})^2 = \frac{1+9+9+1}{64} = \frac{20}{64} = \frac{5}{16}$.
Thus,option $D$ is correct.

(B) The total number of ways to distribute $5$ different books among $4$ students is $4^5 = 1024$.
To ensure each student gets at least one book,we must distribute the books such that one student gets $2$ books and the other three students get $1$ book each.
The number of ways to choose which student gets $2$ books is $\binom{4}{1} = 4$.
The number of ways to choose $2$ books out of $5$ is $\binom{5}{2} = 10$.
The number of ways to distribute the remaining $3$ books to the remaining $3$ students is $3! = 6$.
Total favorable ways = $4 \times 10 \times 6 = 240$.
The required probability is $\frac{240}{1024} = \frac{15}{64}$.
Thus,option $B$ is correct.

(B) $2 \times 2$ determinant has $4$ positions,each can be filled by $0$ or $1$. Thus,the total number of possible determinants is $2^4 = 16$.
Let the determinant be $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
The determinant is non-zero if $ad - bc \neq 0$,which means $ad \neq bc$.
Since $a, b, c, d \in \{0, 1\}$,the possible values for $ad$ and $bc$ are $0$ or $1$.
The condition $ad \neq bc$ holds in the following cases:
$1$. $ad = 1$ and $bc = 0$: This implies $a=1, d=1$ and $(b=0$ or $c=0)$. The pairs $(b, c)$ can be $(0, 0), (0, 1), (1, 0)$. There are $3$ such matrices.
$2$. $ad = 0$ and $bc = 1$: This implies $b=1, c=1$ and $(a=0$ or $d=0)$. The pairs $(a, d)$ can be $(0, 0), (0, 1), (1, 0)$. There are $3$ such matrices.
Total number of matrices with non-zero determinant $= 3 + 3 = 6$.
Therefore,the required probability $= \frac{6}{16} = \frac{3}{8}$.

(D) Let $X_1, X_2, X_3$ be the numbers on the three books drawn with replacement. Each $X_i \in \{1, 2, \dots, 20\}$.
Total number of outcomes $= 20^3$.
We want the largest number to be $7$. This means all three numbers must be $\le 7$,and at least one number must be $7$.
Let $E$ be the event that the largest number is $\le 7$. Then $P(E) = \left(\frac{7}{20}\right)^3$.
Let $F$ be the event that the largest number is $\le 6$. Then $P(F) = \left(\frac{6}{20}\right)^3$.
The probability that the largest number is exactly $7$ is $P(E) - P(F) = \left(\frac{7}{20}\right)^3 - \left(\frac{6}{20}\right)^3$.
Therefore,option $D$ is correct.

(C) The total number of possible outcomes when tossing $6$ coins is $2^6 = 64$.
We need to find the probability of getting at least $4$ heads,which is $P(X \ge 4) = P(X=4) + P(X=5) + P(X=6)$.
Using the binomial distribution formula $P(X=r) = \binom{n}{r} p^r q^{n-r}$,where $n=6$ and $p=q=1/2$:
$P(X=4) = \binom{6}{4} (1/2)^6 = \frac{15}{64}$
$P(X=5) = \binom{6}{5} (1/2)^6 = \frac{6}{64}$
$P(X=6) = \binom{6}{6} (1/2)^6 = \frac{1}{64}$
Summing these probabilities: $P(X \ge 4) = \frac{15+6+1}{64} = \frac{22}{64} = \frac{11}{32}$.
Thus,the correct option is $C$.

(B) Let $p$ be the probability of a climber returning safely,so $p = \frac{9}{10}$.
Let $q$ be the probability of a climber not returning safely,so $q = 1 - p = \frac{1}{10}$.
We use the binomial distribution formula $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$,where $n = 5$.
We need to find the probability that at least $4$ climbers return safely,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.
$P(X = 4) = {}^5C_4 \cdot (\frac{9}{10})^4 \cdot (\frac{1}{10})^1 = 5 \cdot \frac{9^4}{10^5}$.
$P(X = 5) = {}^5C_5 \cdot (\frac{9}{10})^5 = 1 \cdot \frac{9^5}{10^5}$.
$P(X \ge 4) = \frac{5 \cdot 9^4}{10^5} + \frac{9^5}{10^5} = \frac{9^4(5 + 9)}{10^5} = \frac{9^4 \cdot 14}{100000} = \frac{9^4 \cdot 7}{50000}$.

(B) Let $S$ be the set of outcomes where the sum of three dice is $8$. The possible combinations (unordered) are:
$(1, 1, 6) \rightarrow \frac{3!}{2!} = 3$ permutations
$(1, 2, 5) \rightarrow 3! = 6$ permutations
$(1, 3, 4) \rightarrow 3! = 6$ permutations
$(2, 2, 4) \rightarrow \frac{3!}{2!} = 3$ permutations
$(2, 3, 3) \rightarrow \frac{3!}{2!} = 3$ permutations
Total number of outcomes $n(S) = 3 + 6 + 6 + 3 + 3 = 21$.
Let $E$ be the event that at least one die shows a $4$. The favorable outcomes are from $(1, 3, 4)$ and $(2, 2, 4)$.
Number of favorable outcomes $n(E) = 6 + 3 = 9$.
The probability $P(E) = \frac{n(E)}{n(S)} = \frac{9}{21} = \frac{3}{7}$.

(C) Given,$P(A) = 0.7$.
Probability that exactly one of them is selected is $P(A \cap \overline{B}) + P(\overline{A} \cap B) = 0.6$.
Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) \cdot P(B)$.
The probability that exactly one is selected is given by $P(A) + P(B) - 2P(A \cap B) = 0.6$.
Substituting the values: $0.7 + P(B) - 2(0.7)P(B) = 0.6$.
$0.7 + P(B) - 1.4P(B) = 0.6$.
$-0.4P(B) = 0.6 - 0.7$.
$-0.4P(B) = -0.1$.
$P(B) = \frac{0.1}{0.4} = 0.25$.
Thus,the probability that $B$ is selected is $0.25$.

(B) Let $B$ be the event that Bill hits the target and $G$ be the event that George hits the target.
Given $P(B) = 0.7$ and $P(G) = 0.4$.
The probability that Bill misses is $P(B') = 1 - 0.7 = 0.3$.
The probability that George misses is $P(G') = 1 - 0.4 = 0.6$.
Since the events are independent,the probability that exactly one shot hits the target is the sum of the probabilities of two mutually exclusive cases: (Bill hits and George misses) or (Bill misses and George hits).
$P(\text{Exactly one hit}) = P(B \cap G') + P(B' \cap G) = P(B)P(G') + P(B')P(G)$
$= (0.7 \times 0.6) + (0.3 \times 0.4) = 0.42 + 0.12 = 0.54$.
We want to find the conditional probability that it was George's shot,given that exactly one hit occurred.
$P(G \text{ hit} | \text{Exactly one hit}) = \frac{P(B' \cap G)}{P(\text{Exactly one hit})} = \frac{0.12}{0.54} = \frac{12}{54} = \frac{2}{9}$.
Thus,the correct option is $B$.

(C) Given that in $3$ trials of a binomial distribution,the probability of $2$ successes is $9$ times the probability of $3$ successes.
Let $p$ be the probability of success and $q$ be the probability of failure,where $p + q = 1$.
The probability of $r$ successes in $n$ trials is given by $P(X = r) = {}^nC_r p^r q^{n-r}$.
For $n = 3$:
$P(X = 2) = 9 \times P(X = 3)$
${}^3C_2 p^2 q^1 = 9 \times {}^3C_3 p^3 q^0$
$3 p^2 q = 9 p^3$
Since $p \neq 0$,we can divide by $3p^2$:
$q = 3p$
Substitute $q = 1 - p$:
$1 - p = 3p$
$1 = 4p$
$p = \frac{1}{4}$
Thus,the probability of success in each trial is $\frac{1}{4}$.
Hence,option $C$ is correct.

(A) Let $E_1$ be the event that die $A$ is chosen (coin shows head) and $E_2$ be the event that die $B$ is chosen (coin shows tail).
Let $R$ be the event that a red face appears in all $n$ throws.
Given $P(E_1) = \frac{1}{2}$ and $P(E_2) = \frac{1}{2}$.
For die $A$,the probability of getting a red face in one throw is $\frac{4}{6} = \frac{2}{3}$. Thus,$P(R \mid E_1) = \left(\frac{2}{3}\right)^n$.
For die $B$,the probability of getting a red face in one throw is $\frac{2}{6} = \frac{1}{3}$. Thus,$P(R \mid E_2) = \left(\frac{1}{3}\right)^n$.
By Bayes' theorem,the probability that die $A$ was used given that red appeared $n$ times is:
$P(E_1 \mid R) = \frac{P(E_1)P(R \mid E_1)}{P(E_1)P(R \mid E_1) + P(E_2)P(R \mid E_2)}$
Substituting the values:
$\frac{32}{33} = \frac{\frac{1}{2} \times (\frac{2}{3})^n}{\frac{1}{2} \times (\frac{2}{3})^n + \frac{1}{2} \times (\frac{1}{3})^n} = \frac{2^n}{2^n + 1^n} = \frac{2^n}{2^n + 1}$.
Solving for $n$:
$32(2^n + 1) = 33(2^n) \implies 32 \cdot 2^n + 32 = 33 \cdot 2^n \implies 2^n = 32$.
Since $32 = 2^5$,we get $n = 5$.

(D) Given,mean of a binomial distribution $= 20$.
$\Rightarrow np = 20$ $(i)$.
Standard deviation of a binomial distribution $= 4$.
$\Rightarrow \sqrt{np(1-p)} = 4$.
Squaring both sides,we get $np(1-p) = 16$ $(ii)$.
Substitute $np = 20$ from equation $(i)$ into equation $(ii)$:
$20(1-p) = 16$.
$1-p = \frac{16}{20} = \frac{4}{5}$.
$p = 1 - \frac{4}{5} = \frac{1}{5}$.
Substitute the value of $p$ into equation $(i)$:
$n \times \frac{1}{5} = 20$.
$n = 20 \times 5 = 100$.
Therefore,the number of trials is $100$.

(A) The sample space $S$ for tossing three fair coins is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
The total number of outcomes is $n(S) = 8$.
$X$ is a random variable representing the number of heads.
We need to find $P(X = 2)$,which is the probability of getting exactly two heads.
The outcomes with exactly two heads are $\{HHT, HTH, THH\}$.
Thus,the number of favorable outcomes is $n(X = 2) = 3$.
Therefore,$P(X = 2) = \frac{n(X = 2)}{n(S)} = \frac{3}{8}$.

(B) The expected value $E[X^2]$ is calculated using the formula $E[X^2] = \sum x_i^2 P(X=x_i)$.
Given the probability distribution:

$X$	$0$	$1$	$2$
$P(X)$	$0.2$	$0.5$	$0.3$

$E[X^2] = (0^2 \times 0.2) + (1^2 \times 0.5) + (2^2 \times 0.3)$
$E[X^2] = (0 \times 0.2) + (1 \times 0.5) + (4 \times 0.3)$
$E[X^2] = 0 + 0.5 + 1.2$
$E[X^2] = 1.7$

(C) Let $W$ denote winning (getting a number $> 4$,probability $p = \frac{2}{6} = \frac{1}{3}$) and $L$ denote losing (getting a number $\le 4$,probability $q = \frac{4}{6} = \frac{2}{3}$). The player quits when $W$ occurs or after $3$ throws. The possible outcomes are:
$(i)$ $W$: Win $5$ rupees. Probability $P_1 = \frac{1}{3}$.
(ii) $LW$: Lose $1$ rupee,then win $5$ rupees. Net gain $= 4$ rupees. Probability $P_2 = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$.
(iii) $LLW$: Lose $1$ rupee,lose $1$ rupee,then win $5$ rupees. Net gain $= 3$ rupees. Probability $P_3 = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{4}{27}$.
(iv) $LLL$: Lose $1$ rupee,lose $1$ rupee,lose $1$ rupee. Net gain $= -3$ rupees. Probability $P_4 = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27}$.
The expected value $E = \sum x_i P_i = 5(\frac{1}{3}) + 4(\frac{2}{9}) + 3(\frac{4}{27}) + (-3)(\frac{8}{27})$.
$E = \frac{5}{3} + \frac{8}{9} + \frac{12}{27} - \frac{24}{27} = \frac{45 + 24 + 12 - 24}{27} = \frac{57}{27} = \frac{19}{9}$.

(B) We know that the sum of all probabilities in a probability distribution is $1$.
Thus,$\sum_{i=1}^{100} P(X=x_i) = 1$.
Substituting the given probability $P(X=x_i) = K i(i+1)$,we get:
$K \sum_{i=1}^{100} (i^2 + i) = 1$.
Using the summation formulas $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ for $n=100$:
$K \left[ \frac{100(101)(201)}{6} + \frac{100(101)}{2} \right] = 1$.
$K \left[ \frac{100 \times 101}{2} \left( \frac{201}{3} + 1 \right) \right] = 1$.
$K \left[ 5050 \times (67 + 1) \right] = 1$.
$K \times 5050 \times 68 = 1$.
$K = \frac{1}{5050 \times 68} = \frac{1}{343400}$.
Therefore,$200 K = \frac{200}{343400} = \frac{2}{3434} = \frac{1}{1717}$.
Hence,option $(B)$ is correct.

(A) Given the probability distribution:
$\sum P(X) = 1$
$K + 2K + K^2 + 2K^2 + 5K^2 + K + K + 2K = 1$
$8K^2 + 7K - 1 = 0$
$(8K - 1)(K + 1) = 0$
Since $K > 0$,we have $K = \frac{1}{8}$.
The events are $E = \{2, 3, 5, 7\}$ and $F = \{1, 2, 3\}$.
$E \cup F = \{1, 2, 3, 5, 7\}$.
$P(E \cup F) = P(1) + P(2) + P(3) + P(5) + P(7)$
$P(E \cup F) = K + 2K + K^2 + 5K^2 + K = 6K^2 + 4K$
Substituting $K = \frac{1}{8}$:
$P(E \cup F) = 6(\frac{1}{64}) + 4(\frac{1}{8}) = \frac{6}{64} + \frac{32}{64} = \frac{38}{64}$.

(A) The number of accidents follows a Poisson distribution with parameter $\lambda = 3$.
The probability mass function of a Poisson random variable $X$ is given by $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$,where $x = 0, 1, 2, \dots$.
We want to find the probability that no accidents occur today,which corresponds to $P(X = 0)$.
Substituting $\lambda = 3$ and $x = 0$ into the formula:
$P(X = 0) = \frac{3^0 e^{-3}}{0!}$.
Since $3^0 = 1$ and $0! = 1$,we get:
$P(X = 0) = \frac{1 \times e^{-3}}{1} = e^{-3} = \frac{1}{e^3}$.
Thus,the probability that no accidents occur today is $\frac{1}{e^3}$.

(C) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$0.1 + k + 0.2 + 2k + 0.3 + k = 1$.
Combining the terms,we get $0.6 + 4k = 1$.
$4k = 0.4$,which implies $k = 0.1$.
The mean of $X$,denoted by $E(X)$,is given by $\sum x_i P(x_i)$.
$E(X) = (-2 \times 0.1) + (-1 \times 0.1) + (0 \times 0.2) + (1 \times 0.2) + (2 \times 0.3) + (3 \times 0.1)$.
$E(X) = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3$.
$E(X) = 0.8$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
Therefore,$\sum P(X=x) = 1$.
$0.1 + k + 0.2 + 2k + 0.3 + k = 1$
Combine the terms involving $k$ and the constant terms:
$(0.1 + 0.2 + 0.3) + (k + 2k + k) = 1$
$0.6 + 4k = 1$
$4k = 1 - 0.6$
$4k = 0.4$
$k = \frac{0.4}{4}$
$k = 0.1$

(D) square matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$|A| = 0$.
Given the matrix $A = \begin{bmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{bmatrix}$.
We calculate the determinant:
$|A| = 1(x - (-1)) - (-1)(1 - x) + x(-1 - x^2) = 0$
$|A| = 1(x + 1) + 1(1 - x) + x(-1 - x^2) = 0$
$|A| = x + 1 + 1 - x - x - x^3 = 0$
$|A| = -x^3 - x + 2 = 0$
$x^3 + x - 2 = 0$
By inspection,if $x = 1$,then $1^3 + 1 - 2 = 0$,which satisfies the equation.
Alternatively,if $x = 1$,the matrix becomes $\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{bmatrix}$.
Since the first and third rows (or columns) are identical,the determinant is $0$.
Thus,the real value of $x$ is $1$.

(C) To check for injectivity,let $f(x_1) = f(x_2)$ for $x_1, x_2 \in R$.
$\frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}}$
Squaring both sides:
$\frac{x_1^2}{1+x_1^2} = \frac{x_2^2}{1+x_2^2}$
$x_1^2(1+x_2^2) = x_2^2(1+x_1^2)$
$x_1^2 + x_1^2x_2^2 = x_2^2 + x_1^2x_2^2$
$x_1^2 = x_2^2$
Since $f'(x) = \frac{1}{(1+x^2)^{3/2}} > 0$ for all $x \in R$,the function is strictly increasing.
Therefore,$f(x_1) = f(x_2) \implies x_1 = x_2$. Thus,$f$ is injective.
To check for surjectivity,let $y = \frac{x}{\sqrt{1+x^2}}$.
Since $x^2 < 1+x^2$,we have $\frac{|x|}{\sqrt{1+x^2}} < 1$.
Thus,the range of $f$ is $(-1, 1)$,which is not equal to the codomain $R$.
Therefore,$f$ is not surjective.
Hence,the function is injective but not surjective.

(C) Given,$a+b+\sqrt{3} c=0$.
Rearranging the terms,we get $a+b = -\sqrt{3} c$.
Taking the dot product of both sides with themselves:
$(a+b) \cdot (a+b) = (-\sqrt{3} c) \cdot (-\sqrt{3} c)$.
Expanding the left side and simplifying the right side:
$|a|^2 + 2(a \cdot b) + |b|^2 = 3|c|^2$.
Since $a, b, c$ are unit vectors,$|a| = |b| = |c| = 1$.
Substituting these values:
$1^2 + 2(1)(1) \cos \theta + 1^2 = 3(1)^2$,where $\theta$ is the angle between $a$ and $b$.
$1 + 2 \cos \theta + 1 = 3$.
$2 + 2 \cos \theta = 3$.
$2 \cos \theta = 1$.
$\cos \theta = \frac{1}{2}$.
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,the angle $\theta = \frac{\pi}{3}$.