TS EAMCET 2010 Chemistry Question Paper with Answer and Solution

(D) The number of diagonals in a polygon with $n$ sides is given by the formula $\frac{n(n-3)}{2}$.
Given that the number of diagonals is $54$,we have:
$\frac{n(n-3)}{2} = 54$
$n(n-3) = 108$
$n^2 - 3n - 108 = 0$
Factoring the quadratic equation:
$n^2 - 12n + 9n - 108 = 0$
$n(n - 12) + 9(n - 12) = 0$
$(n + 9)(n - 12) = 0$
Since the number of sides $n$ must be positive,we have $n = 12$.
Therefore,the number of sides in the polygon is $12$.

(A) Let $f(\theta) = \left(\tan \theta - \frac{1}{3} \tan^3 \theta\right) \left(\frac{1}{3} - \tan^2 \theta\right)^{-1}$.
Given $\tan^2 \theta \neq \frac{1}{3}$,we can rewrite the expression as:
$f(\theta) = \frac{\tan \theta - \frac{1}{3} \tan^3 \theta}{\frac{1}{3} - \tan^2 \theta} = \frac{\frac{3 \tan \theta - \tan^3 \theta}{3}}{\frac{1 - 3 \tan^2 \theta}{3}} = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$.
Using the identity $\tan(3\theta) = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we get $f(\theta) = \tan(3\theta)$.
The period of $\tan(x)$ is $\pi$. Therefore,the period of $\tan(3\theta)$ is $\frac{\pi}{3}$.

(B) Given: $a \sin^2 \theta + b \cos^2 \theta = c$
Divide both sides by $\cos^2 \theta$:
$a \tan^2 \theta + b = c \sec^2 \theta$
Since $\sec^2 \theta = 1 + \tan^2 \theta$,we have:
$a \tan^2 \theta + b = c(1 + \tan^2 \theta)$
$a \tan^2 \theta + b = c + c \tan^2 \theta$
Rearranging the terms to isolate $\tan^2 \theta$:
$a \tan^2 \theta - c \tan^2 \theta = c - b$
$(a - c) \tan^2 \theta = c - b$
$\tan^2 \theta = \frac{c - b}{a - c}$

(A) Given equation: $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$
Divide by $2\sqrt{2}$ on both sides:
$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin \theta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos \theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$
We know that $\cos(15^{\circ}) = \cos(\frac{\pi}{12}) = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin(15^{\circ}) = \sin(\frac{\pi}{12}) = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Substituting these values,we get:
$\sin(\frac{\pi}{12}) \sin \theta + \cos(\frac{\pi}{12}) \cos \theta = \frac{1}{\sqrt{2}}$
Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$\cos(\theta - \frac{\pi}{12}) = \cos(\frac{\pi}{4})$
The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$.
Therefore,$\theta - \frac{\pi}{12} = 2n\pi \pm \frac{\pi}{4}$
$\theta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$,where $n \in Z$.

(A) The given circle is $x^2+y^2-6x+12y+15=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$,$f=6$,and $c=15$.
The radius $r_1$ is $\sqrt{g^2+f^2-c} = \sqrt{(-3)^2+6^2-15} = \sqrt{9+36-15} = \sqrt{30}$.
The area of this circle is $A_1 = \pi r_1^2 = 30\pi$.
Let the concentric circle be $x^2+y^2-6x+12y+k=0$.
Its radius $r_2$ satisfies $r_2^2 = g^2+f^2-k = 45-k$.
The area of the new circle is $A_2 = \pi r_2^2 = \pi(45-k)$.
Given $A_2 = 2A_1$,we have $\pi(45-k) = 2(30\pi) = 60\pi$.
Thus,$45-k = 60$,which gives $k = -15$.
The equation of the circle is $x^2+y^2-6x+12y-15=0$.

(C) Given,$r_1 = 15$ units,$r_2 = 20$ units,and the distance between centers $C_1 C_2 = 25$ units.
Note that $r_1^2 + r_2^2 = 15^2 + 20^2 = 225 + 400 = 625 = 25^2 = (C_1 C_2)^2$.
Since the sum of the squares of the radii equals the square of the distance between the centers,the triangle $\triangle A C_1 C_2$ is a right-angled triangle with $\angle C_1 A C_2 = 90^\circ$.
The common chord $AB$ is perpendicular to the line joining the centers $C_1 C_2$ at point $D$.
In $\triangle A C_1 C_2$,the altitude $AD$ to the hypotenuse $C_1 C_2$ is given by $AD = \frac{r_1 \times r_2}{C_1 C_2}$.
$AD = \frac{15 \times 20}{25} = \frac{300}{25} = 12$ units.
The length of the common chord is $2 \times AD = 2 \times 12 = 24$ units.

(B) The equations of the circles are $S_1 \equiv x^2+y^2+2x+3y+1=0$ and $S_2 \equiv x^2+y^2+4x+3y+2=0$.
Since the circles intersect at $A$ and $B$,the equation of the common chord $AB$ is given by $S_1 - S_2 = 0$.
$(x^2+y^2+2x+3y+1) - (x^2+y^2+4x+3y+2) = 0$
$-2x - 1 = 0 \Rightarrow x = -\frac{1}{2}$.
The family of circles passing through the intersection of $S_1$ and $S_2$ is $S_1 + \lambda(S_1 - S_2) = 0$.
However,the circle with $AB$ as diameter is given by $S_1 + k(S_1 - S_2) = 0$ where the coefficient of $x^2$ and $y^2$ is $1$.
Alternatively,using the property that the circle with diameter $AB$ is $S_1 + k(S_1 - S_2) = 0$,we find the equation.
Substituting $x = -\frac{1}{2}$ into $S_1$: $\frac{1}{4} + y^2 - 1 + 3y + 1 = 0 \Rightarrow y^2 + 3y + \frac{1}{4} = 0$.
The circle with diameter $AB$ has the equation $S_1 + k(S_1 - S_2) = 0$.
$x^2+y^2+2x+3y+1 + k(2x+1) = 0$.
Since the center of this circle must lie on the line $x = -\frac{1}{2}$,we have $x + \frac{k}{1} = -\frac{1}{2}$ $\Rightarrow -1 + k = -\frac{1}{2}$ $\Rightarrow k = \frac{1}{2}$.
Substituting $k = \frac{1}{2}$: $x^2+y^2+2x+3y+1 + \frac{1}{2}(2x+1) = 0$.
$x^2+y^2+2x+3y+1 + x + \frac{1}{2} = 0 \Rightarrow x^2+y^2+3x+3y+\frac{3}{2} = 0$.
Multiplying by $2$: $2x^2+2y^2+2x+6y+1 = 0$.

(C) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$.
First,normalize the equations so the coefficients of $x^2$ and $y^2$ are $1$.
For $S_1: x^2+y^2-x+2y+\frac{18}{7}=0$
For $S_2: x^2+y^2-\frac{7}{4}x+2y+5=0$
Subtracting $S_2$ from $S_1$:
$(x^2+y^2-x+2y+\frac{18}{7}) - (x^2+y^2-\frac{7}{4}x+2y+5) = 0$
$-x + \frac{7}{4}x + \frac{18}{7} - 5 = 0$
$\frac{3}{4}x + \frac{18-35}{7} = 0$
$\frac{3}{4}x - \frac{17}{7} = 0$
Multiplying by $28$:
$21x - 68 = 0$

(C) The general equation of a parabola with its axis parallel to the $y$-axis is given by $y = Ax^2 + Bx + C$ ...$(i)$
Since the parabola passes through the points $(0, 4), (1, 9),$ and $(4, 5)$,these points must satisfy the equation.
Substituting $(0, 4)$ in $(i)$: $4 = A(0)^2 + B(0) + C \Rightarrow C = 4$ ...(ii)
Substituting $(1, 9)$ in $(i)$: $9 = A(1)^2 + B(1) + 4 \Rightarrow A + B = 5$ ...(iii)
Substituting $(4, 5)$ in $(i)$: $5 = A(4)^2 + B(4) + 4 \Rightarrow 16A + 4B = 1$ ...(iv)
Dividing (iv) by $4$,we get $4A + B = 0.25$ or $4A + B = \frac{1}{4}$ ...$(v)$
Subtracting (iii) from $(v)$: $(4A + B) - (A + B) = \frac{1}{4} - 5$ $\Rightarrow 3A = -\frac{19}{4}$ $\Rightarrow A = -\frac{19}{12}$
Substituting $A$ in (iii): $-\frac{19}{12} + B = 5 \Rightarrow B = 5 + \frac{19}{12} = \frac{79}{12}$
Thus,the equation is $y = -\frac{19}{12}x^2 + \frac{79}{12}x + 4$.

(C) The given parabola is $y^2 = 8(x-3)$. Comparing this with $(y-k)^2 = 4a(x-h)$,we get $h=3, k=0$,and $4a=8$,so $a=2$.
The focus $S$ is $(h+a, k) = (3+2, 0) = (5, 0)$.
The directrix is $x = h-a = 3-2 = 1$. Wait,the directrix is $x = h-a = 3-2 = 1$. Let's re-evaluate.
Actually,for $y^2 = 8(x-3)$,the vertex is $(3,0)$ and $a=2$. The directrix is $x = 3-2 = 1$.
Let $P$ be $(x, y)$. Since $P$ is on the parabola,$PS = PM$,where $PM$ is the distance to the directrix $x=1$.
$PS = PM = |x-1|$.
In an equilateral triangle $\triangle SPM$,all sides are equal. Thus $SP = PM = SM = |x-1|$.
Since $P$ is on the parabola,$SP = PM$ is always true by definition.
For $\triangle SPM$ to be equilateral,we must have $SP = SM$.
Let $P = (3+2t^2, 4t)$. Then $S = (5, 0)$.
$SP = \sqrt{(3+2t^2-5)^2 + (4t-0)^2} = \sqrt{(2t^2-2)^2 + 16t^2} = \sqrt{4t^4 - 8t^2 + 4 + 16t^2} = \sqrt{4t^4 + 8t^2 + 4} = \sqrt{4(t^2+1)^2} = 2(t^2+1)$.
The directrix is $x=1$. The foot of the perpendicular $M$ from $P(3+2t^2, 4t)$ to $x=1$ is $M(1, 4t)$.
$SM = \sqrt{(5-1)^2 + (0-4t)^2} = \sqrt{4^2 + 16t^2} = \sqrt{16 + 16t^2} = 4\sqrt{1+t^2}$.
Equating $SP = SM$:
$2(t^2+1) = 4\sqrt{1+t^2}$.
Squaring both sides: $4(t^2+1)^2 = 16(1+t^2)$.
$(t^2+1)^2 = 4(t^2+1)$.
Since $t^2+1 \neq 0$,we have $t^2+1 = 4$,so $t^2 = 3$,$t = \sqrt{3}$ (taking positive $y$ coordinate).
Then $P = (3+2(3), 4\sqrt{3}) = (9, 4\sqrt{3})$.

(C) Given,the conjugate lines are $2x + 3y + 12 = 0$ $(i)$ and $x - y + k = 0$ $(ii)$.
Two lines are conjugate with respect to a parabola if the pole of one line lies on the other line.
Let the pole of the line $2x + 3y + 12 = 0$ be $(x_1, y_1)$ with respect to the parabola $y^2 = 8x$.
The equation of the polar of the point $(x_1, y_1)$ with respect to $y^2 = 8x$ is $yy_1 = 4(x + x_1)$,which simplifies to $4x - yy_1 + 4x_1 = 0$.
Comparing this with the given line $2x + 3y + 12 = 0$,we have:
$\frac{4}{2} = \frac{-y_1}{3} = \frac{4x_1}{12}$
$2 = \frac{-y_1}{3} \Rightarrow y_1 = -6$
$2 = \frac{x_1}{3} \Rightarrow x_1 = 6$
Thus,the pole is $(6, -6)$.
Since the lines are conjugate,the pole $(6, -6)$ must lie on the second line $x - y + k = 0$.
Substituting the coordinates,we get $6 - (-6) + k = 0$.
$12 + k = 0 \Rightarrow k = -12$.

(D) Given,$a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$
We can write $6^n$ as $(1 + 5)^n$.
Using the binomial expansion:
$6^n = (1 + 5)^n = {^nC_0} + {^nC_1}(5) + {^nC_2}(5^2) + {^nC_3}(5^3) + \ldots$
$6^n = 1 + 5n + 25({^nC_2} + {^nC_3}(5) + \ldots)$
Now,substitute this into the expression for $a_n$:
$a_n = (1 + 5n + 25k) - 5n$,where $k = {^nC_2} + {^nC_3}(5) + \ldots$
$a_n = 1 + 25k$
This shows that when $a_n$ is divided by $25$,the remainder is $1$.

(B) We are given $n = 1! + 4! + 7! + \ldots + 400!$.
First,calculate the values of the factorials:
$1! = 1$
$4! = 24$
$7! = 5040$
$10! = 3628800$
For any $k \ge 10$,$k!$ ends with at least two zeros,meaning the last two digits of $k!$ are $00$.
Thus,the sum $n$ modulo $100$ is determined by the sum of the first few terms:
$n \equiv 1! + 4! + 7! + 10! + \ldots + 400! \pmod{100}$
$n \equiv 1 + 24 + 40 + 0 + \ldots + 0 \pmod{100}$
$n \equiv 65 \pmod{100}$
The last two digits of $n$ are $65$.
Therefore,the ten's digit of $n$ is $6$.

(A) Given $(1+2x+3x^2)^{10} = a_0+a_1x+a_2x^2+\ldots+a_{20}x^{20}$.
Using the binomial expansion $(1+y)^n = \sum_{k=0}^{n} {}^{n}C_k y^k$,where $y = 2x+3x^2$:
$(1+(2x+3x^2))^{10} = {}^{10}C_0 + {}^{10}C_1(2x+3x^2) + {}^{10}C_2(2x+3x^2)^2 + \ldots$
$= 1 + 10(2x+3x^2) + 45(4x^2+12x^3+9x^4) + \ldots$
$= 1 + 20x + 30x^2 + 180x^2 + 540x^3 + 405x^4 + \ldots$
$= 1 + 20x + 210x^2 + \ldots$
Comparing coefficients:
$a_1 = 20$
$a_2 = 210$
Therefore,$\frac{a_2}{a_1} = \frac{210}{20} = 10.5$.

(A) Given the expression $\frac{1}{(1-5 x)(1-4 x)}$ for $|x| < \frac{1}{5}$.
Using the binomial expansion formula $(1-z)^{-1} = 1 + z + z^2 + z^3 + \dots$,we have:
$(1-5x)^{-1} = 1 + 5x + 25x^2 + 125x^3 + \dots$
$(1-4x)^{-1} = 1 + 4x + 16x^2 + 64x^3 + \dots$
Multiplying these two series:
$(1 + 5x + 25x^2 + 125x^3 + \dots)(1 + 4x + 16x^2 + 64x^3 + \dots)$
The coefficient of $x^3$ is obtained by summing the products of terms whose powers add up to $3$:
$= (1 \cdot 64) + (5x \cdot 16x^2) + (25x^2 \cdot 4x) + (125x^3 \cdot 1)$
$= 64 + 80 + 100 + 125$
$= 369$

(C) Given the curve is $2 x^2+y^2=2 x$.
Rearranging terms: $2 x^2-2 x+y^2=0$.
Completing the square: $2(x^2-x)+y^2=0 \Rightarrow 2(x-\frac{1}{2})^2+y^2=\frac{1}{2}$.
Dividing by $\frac{1}{2}$: $\frac{(x-\frac{1}{2})^2}{1/4} + \frac{y^2}{1/2} = 1$.
This is an ellipse with center $(\frac{1}{2}, 0)$,$a^2 = \frac{1}{4} \Rightarrow a = \frac{1}{2}$,and $b^2 = \frac{1}{2} \Rightarrow b = \frac{1}{\sqrt{2}}$.
Let a point $P$ on the ellipse be $(\frac{1}{2} + \frac{1}{2} \cos \theta, \frac{1}{\sqrt{2}} \sin \theta)$.
The distance $PQ$ from $Q(a, 0)$ is $PQ^2 = (\frac{1}{2} + \frac{1}{2} \cos \theta - a)^2 + (\frac{1}{\sqrt{2}} \sin \theta)^2$.
$PQ^2 = (\frac{1}{2}-a)^2 + \frac{1}{4} \cos^2 \theta + (\frac{1}{2}-a) \cos \theta + \frac{1}{2} \sin^2 \theta$.
Substituting $\sin^2 \theta = 1 - \cos^2 \theta$:
$PQ^2 = (\frac{1}{2}-a)^2 + \frac{1}{4} \cos^2 \theta + (\frac{1}{2}-a) \cos \theta + \frac{1}{2} - \frac{1}{2} \cos^2 \theta$.
$PQ^2 = -\frac{1}{4} \cos^2 \theta + (\frac{1}{2}-a) \cos \theta + (\frac{1}{2}-a)^2 + \frac{1}{2}$.
This is a quadratic in $\cos \theta$. The maximum occurs at $\cos \theta = \frac{-b}{2A} = \frac{-(1/2-a)}{2(-1/4)} = 1-2a$.
Substituting $\cos \theta = 1-2a$ into the expression for $PQ^2$:
$PQ^2 = -\frac{1}{4}(1-2a)^2 + (\frac{1}{2}-a)(1-2a) + \frac{1}{4} - a + a^2 + \frac{1}{2}$.
$PQ^2 = -\frac{1}{4}(1-4a+4a^2) + (\frac{1}{2}-a-a+2a^2) + \frac{3}{4} - a + a^2$.
$PQ^2 = -\frac{1}{4} + a - a^2 + \frac{1}{2} - 2a + 2a^2 + \frac{3}{4} - a + a^2 = 2a^2 - 2a + 1$.
Thus,the longest distance is $\sqrt{2a^2 - 2a + 1}$.

(C) The equation of a hyperbola with asymptotes $L_1=0$ and $L_2=0$ is given by $L_1 \cdot L_2 + k = 0$.
Here,the asymptotes are $(4x+3y-7)=0$ and $(x-2y-1)=0$.
So,the equation of the hyperbola is $(4x+3y-7)(x-2y-1)+k=0$ ...$(i)$
Since the hyperbola passes through the point $(2,3)$,we substitute $x=2$ and $y=3$ into Eq. $(i)$:
$(4(2)+3(3)-7)(2-2(3)-1)+k=0$
$(8+9-7)(2-6-1)+k=0$
$(10)(-5)+k=0$
$-50+k=0 \Rightarrow k=50$
Substituting $k=50$ into Eq. $(i)$:
$(4x+3y-7)(x-2y-1)+50=0$
$4x^2-8xy-4x+3xy-6y^2-3y-7x+14y+7+50=0$
$4x^2-5xy-6y^2-11x+11y+57=0$

(B) Let $(a \sec \theta, b \tan \theta)$ be any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
The equations of the asymptotes are $\frac{x}{a} + \frac{y}{b} = 0$ and $\frac{x}{a} - \frac{y}{b} = 0$.
The perpendicular distance $P_1$ from $(a \sec \theta, b \tan \theta)$ to the line $\frac{x}{a} + \frac{y}{b} = 0$ is given by:
$P_1 = \frac{|\sec \theta + \tan \theta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
The perpendicular distance $P_2$ from $(a \sec \theta, b \tan \theta)$ to the line $\frac{x}{a} - \frac{y}{b} = 0$ is given by:
$P_2 = \frac{|\sec \theta - \tan \theta|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
The product $P_1 P_2$ is:
$P_1 P_2 = \frac{|\sec^2 \theta - \tan^2 \theta|}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{a^2 + b^2}{a^2 b^2}} = \frac{a^2 b^2}{a^2 + b^2}$.

(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$.
Since this is a $\frac{0}{0}$ form,we use the Taylor series expansion:
$\tan x = x + \frac{x^3}{3} + O(x^5)$
$\sin x = x - \frac{x^3}{6} + O(x^5)$
Substituting these into the expression:
$\lim _{x \rightarrow 0} \frac{(x + \frac{x^3}{3}) - (x - \frac{x^3}{6})}{x^3} = \lim _{x \rightarrow 0} \frac{\frac{x^3}{3} + \frac{x^3}{6}}{x^3} = \lim _{x \rightarrow 0} \frac{\frac{x^3}{2}}{x^3} = \frac{1}{2}$.

(A) The Polydispersity Index $(PDI)$ is defined as the ratio of weight average molecular weight to number average molecular weight:
$PDI = \frac{\bar{M}_w}{\bar{M}_n}$
Given:
$\bar{M}_n = 40000$
$\bar{M}_w = 60000$
Substituting the values:
$PDI = \frac{60000}{40000} = 1.5$
Since $1.5 > 1$,the polydispersity index is greater than $1$.

(B) Given $\angle C = 90^{\circ}$. In a right-angled triangle,$c^2 = a^2 + b^2$.
Using the Sine Rule,$a = c \sin A$ and $b = c \sin B$.
Substituting these into the expression:
$\frac{a^2-b^2}{a^2+b^2} = \frac{(c \sin A)^2 - (c \sin B)^2}{c^2} = \sin^2 A - \sin^2 B$.
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$.
Since $A+B+C = 180^{\circ}$ and $C = 90^{\circ}$,we have $A+B = 90^{\circ}$.
Therefore,$\sin(A+B) = \sin 90^{\circ} = 1$.
Thus,$\sin^2 A - \sin^2 B = 1 \cdot \sin(A-B) = \sin(A-B)$.

(B) Given $\Delta = a^2 - (b - c)^2$.
Using the identity $x^2 - y^2 = (x - y)(x + y)$,we have $\Delta = (a - (b - c))(a + (b - c)) = (a - b + c)(a + b - c)$.
Since $2s = a + b + c$,we have $a + b - c = 2s - 2c$ and $a - b + c = 2s - 2b$.
Thus,$\Delta = (2s - 2b)(2s - 2c) = 4(s - b)(s - c)$.
We know that $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.
Equating the two expressions: $\sqrt{s(s - a)(s - b)(s - c)} = 4(s - b)(s - c)$.
Dividing both sides by $\sqrt{(s - b)(s - c)}$,we get $\sqrt{s(s - a)} = 4\sqrt{(s - b)(s - c)}$.
Therefore,$\sqrt{\frac{(s - b)(s - c)}{s(s - a)}} = \frac{1}{4}$.
Since $\tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}}$,we have $\tan \frac{A}{2} = \frac{1}{4}$.
Using the formula $\tan A = \frac{2 \tan \frac{A}{2}}{1 - \tan^2 \frac{A}{2}}$,we get $\tan A = \frac{2(1/4)}{1 - (1/4)^2} = \frac{1/2}{1 - 1/16} = \frac{1/2}{15/16} = \frac{8}{15}$.

(A) Given that $\operatorname{det}(I+A) \neq 0$,which implies that $(I+A)$ is an invertible matrix.
We are given $A^3 = O$.
We know the algebraic identity $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.
Substituting $x=A$ and $y=I$,we get:
$A^3 + I^3 = (A+I)(A^2 - AI + I^2)$.
Since $I^3 = I$ and $A^3 = O$,the equation becomes:
$O + I = (A+I)(A^2 - A + I)$.
$I = (A+I)(A^2 - A + I)$.
Since $(A+I)$ is invertible,we can multiply both sides by $(A+I)^{-1}$ from the left:
$(A+I)^{-1} I = (A+I)^{-1} (A+I)(A^2 - A + I)$.
$(A+I)^{-1} = I(A^2 - A + I)$.
$(A+I)^{-1} = I - A + A^2$.

(B) Given the determinant equation: $\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$.
We can split the determinant into two: $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|+\left|\begin{array}{ccc}x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3\end{array}\right|=0$.
Taking $xyz$ common from the second determinant: $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|+xyz\left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right|=0$.
By performing column swaps ($C_1 \leftrightarrow C_2$ then $C_2 \leftrightarrow C_3$),the second determinant becomes $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|$.
Thus,$(1+xyz)\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right|=0$.
Since $x \neq y \neq z$,the determinant $\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| \neq 0$.
Therefore,$1+xyz=0$,which implies $xyz = -1$.

(B) Given $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$.
Using the property of differentiation of a determinant,$f^{\prime}(x)$ is the sum of three determinants where each row is differentiated one at a time:
$f^{\prime}(x) = \left| \begin{array}{ccc} -2 \sin x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right| + \left| \begin{array}{ccc} 2 \cos x & 0 & 0 \\ x - \frac{\pi}{2} & -2 \sin x & 1 \\ 0 & 0 & 2 \cos x \end{array} \right| + \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 0 \\ 0 & 1 & -2 \sin x \end{array} \right|$.
Now,substitute $x = \pi$ into the expression. Since $\sin \pi = 0$ and $\cos \pi = -1$:
$f^{\prime}(\pi) = \left| \begin{array}{ccc} 0 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{array} \right| + \left| \begin{array}{ccc} -2 & 0 & 0 \\ \frac{\pi}{2} & 0 & 1 \\ 0 & 0 & -2 \end{array} \right| + \left| \begin{array}{ccc} -2 & 1 & 0 \\ \frac{\pi}{2} & -2 & 0 \\ 0 & 1 & 0 \end{array} \right|$.
Evaluating each determinant:
$1$. $\left| \begin{array}{ccc} 0 & 1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{array} \right| = 0 - 1(-2 - 0) + 0 = 2$.
$2$. $\left| \begin{array}{ccc} -2 & 0 & 0 \\ \frac{\pi}{2} & 0 & 1 \\ 0 & 0 & -2 \end{array} \right| = 0$ (since the second column is all zeros).
$3$. $\left| \begin{array}{ccc} -2 & 1 & 0 \\ \frac{\pi}{2} & -2 & 0 \\ 0 & 1 & 0 \end{array} \right| = 0$ (since the third column is all zeros).
Therefore,$f^{\prime}(\pi) = 2 + 0 + 0 = 2$.

(C) Given the equation: $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$,where $|x| < 1$ ...$(i)$
We know the standard logarithmic form of the inverse hyperbolic tangent function is: $\tanh ^{-1} x = \frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$ ...(ii)
Comparing equation $(i)$ and equation (ii),we can equate the coefficients of the logarithmic term.
Therefore,$a = \frac{1}{2}$.

(B) Given that $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$.
We know the formula $\tan ^{-1} x+\tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
So,$\tan ^{-1} \left( \frac{x+y}{1-xy} \right) + \tan ^{-1} z = \frac{\pi}{2}$.
Taking $\tan$ on both sides,we use $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Let $A = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$ and $B = \tan ^{-1} z$.
Then $\tan(A+B) = \tan \left( \frac{\pi}{2} \right) = \infty$.
For the expression to be undefined,the denominator must be zero:
$1 - \left( \frac{x+y}{1-xy} \right) z = 0$.
$1 - \frac{z(x+y)}{1-xy} = 0$.
$1 - xy - z(x+y) = 0$.
$1 - xy - zx - zy = 0$.
Therefore,$1 - xy - yz - zx = 0$.

(D) Let $f(x) = \frac{x^2-6x+5}{x^2+2x+1}$ for $x \in \mathbb{R}$.
Let $y = \frac{x^2-6x+5}{x^2+2x+1}$.
$y(x^2+2x+1) = x^2-6x+5$.
$yx^2 + 2yx + y = x^2 - 6x + 5$.
$(y-1)x^2 + (2y+6)x + (y-5) = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = (2y+6)^2 - 4(y-1)(y-5) \geq 0$.
$4(y+3)^2 - 4(y^2 - 6y + 5) \geq 0$.
$(y^2 + 6y + 9) - (y^2 - 6y + 5) \geq 0$.
$12y + 4 \geq 0$.
$12y \geq -4$.
$y \geq -\frac{4}{12} = -\frac{1}{3}$.
Thus,the least value is $-\frac{1}{3}$.

(B) Given $f(x) = \sin x + \cos x$.
First derivative: $f'(x) = \cos x - \sin x$.
Second derivative: $f''(x) = -\sin x - \cos x$.
Third derivative: $f'''(x) = -\cos x + \sin x$.
Fourth derivative: $f^{(iv)}(x) = \sin x + \cos x$.
Now,evaluate at $x = \frac{\pi}{4}$:
$f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Similarly,$f^{(iv)}\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \sqrt{2}$.
Therefore,$f\left(\frac{\pi}{4}\right) f^{(iv)}\left(\frac{\pi}{4}\right) = \sqrt{2} \times \sqrt{2} = 2$.

(B) Given $y = \sin(m \sin^{-1} x)$.
Differentiating with respect to $x$:
$y_1 = \cos(m \sin^{-1} x) \cdot \frac{m}{\sqrt{1 - x^2}}$
$y_1 \sqrt{1 - x^2} = m \cos(m \sin^{-1} x)$
Squaring both sides:
$y_1^2 (1 - x^2) = m^2 \cos^2(m \sin^{-1} x) = m^2 (1 - \sin^2(m \sin^{-1} x)) = m^2 (1 - y^2)$
$y_1^2 (1 - x^2) = m^2 - m^2 y^2$
Differentiating again with respect to $x$:
$2 y_1 y_2 (1 - x^2) + y_1^2 (-2x) = -2 m^2 y y_1$
Dividing by $2 y_1$ (assuming $y_1 \neq 0$):
$y_2 (1 - x^2) - x y_1 = -m^2 y$

(B) Given $f(x) = \prod_{r=1}^n \cos(rx)$.
Taking the natural logarithm on both sides:
$\ln|f(x)| = \sum_{r=1}^n \ln|\cos(rx)|$.
Differentiating both sides with respect to $x$:
$\frac{1}{f(x)} f^{\prime}(x) = \sum_{r=1}^n \frac{1}{\cos(rx)} \cdot (-\sin(rx) \cdot r)$.
$\frac{f^{\prime}(x)}{f(x)} = -\sum_{r=1}^n r \tan(rx)$.
Multiplying both sides by $f(x)$:
$f^{\prime}(x) = -f(x) \sum_{r=1}^n r \tan(rx)$.
Rearranging the terms:
$f^{\prime}(x) + \sum_{r=1}^n (r \tan(rx)) f(x) = 0$.

(C) Let the height of the cone be $h$ and the radius of the base of the cone be $r$.
Given,the radius of the sphere is $R$.
In the right-angled triangle $\triangle OPB$,where $O$ is the center of the sphere,$P$ is the center of the base of the cone,and $B$ is a point on the circumference of the base of the cone:
$R^2 = r^2 + (h - R)^2$
$r^2 = R^2 - (h - R)^2 = R^2 - (h^2 - 2Rh + R^2) = 2Rh - h^2$.
The volume of the cone $V$ is given by:
$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2Rh - h^2) h = \frac{\pi}{3} (2Rh^2 - h^3)$.
To find the maximum volume,differentiate $V$ with respect to $h$:
$\frac{dV}{dh} = \frac{\pi}{3} (4Rh - 3h^2)$.
Setting $\frac{dV}{dh} = 0$:
$\frac{\pi}{3} h(4R - 3h) = 0$.
Since $h \neq 0$,we have $h = \frac{4R}{3}$.
Checking the second derivative:
$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6h)$.
At $h = \frac{4R}{3}$,$\frac{d^2V}{dh^2} = \frac{\pi}{3} (4R - 6(\frac{4R}{3})) = \frac{\pi}{3} (4R - 8R) = -\frac{4\pi R}{3} < 0$.
Since the second derivative is negative,the volume is maximum at $h = \frac{4R}{3}$.

(A) To find $f(x)$,we differentiate the given options to see which one yields the integrand $\frac{7x^8+8x^7}{(1+x+x^8)^2}$.
Let $f(x) = \frac{x^8}{1+x+x^8}$.
Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}$,where $u = x^8$ and $v = 1+x+x^8$:
$f'(x) = \frac{(1+x+x^8)(8x^7) - (x^8)(1+8x^7)}{(1+x+x^8)^2}$
$f'(x) = \frac{8x^7 + 8x^8 + 8x^{15} - x^8 - 8x^{15}}{(1+x+x^8)^2}$
$f'(x) = \frac{7x^8 + 8x^7}{(1+x+x^8)^2}$
Since the derivative of option $A$ matches the integrand,$f(x) = \frac{x^8}{1+x+x^8}$.

(A) Given $f_n(x) = \log(\log(\ldots \log x))$ ($n$ times).
Let $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$.
Let $t = f_n(x) = \log(f_{n-1}(x))$.
Then,$\frac{dt}{dx} = \frac{1}{f_{n-1}(x)} \cdot \frac{d}{dx}(f_{n-1}(x)) = \frac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) \cdot x}$.
Thus,$dx = (x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt$.
Substituting this into the integral:
$I = \int \frac{(x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt}{x f_1(x) f_2(x) \ldots f_{n-1}(x) f_n(x)} = \int \frac{dt}{f_n(x)}$.
Wait,the substitution $t = f_n(x)$ implies $dt = \frac{dx}{x f_1(x) \ldots f_{n-1}(x)}$.
Therefore,$I = \int \frac{dt}{t} = \log|t| + c = \log|f_n(x)| + c$.
Since $f_{n+1}(x) = \log(f_n(x))$,we have $I = f_{n+1}(x) + c$.

(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii):
$(x-1)^2+(1-x^2)=1$
$x^2-2x+1+1-x^2=1$
$2-2x=1 \Rightarrow x=\frac{1}{2}$
Substituting $x=\frac{1}{2}$ into $(i)$,we get $y^2=1-\frac{1}{4}=\frac{3}{4}$,so $y=\pm\frac{\sqrt{3}}{2}$.
The intersection points are $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $C\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.
The total area is symmetric about the $x$-axis,so Area $= 2 \times \int_{0}^{1/2} \sqrt{1-(x-1)^2} dx + 2 \times \int_{1/2}^{1} \sqrt{1-x^2} dx$.
Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$,
Area $= 2 \left[ \frac{x-1}{2}\sqrt{1-(x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) \right]_0^{1/2} + 2 \left[ \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) \right]_{1/2}^1$
$= 2 \left[ (-\frac{1}{4}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(-\frac{1}{2})) - (0 + \frac{1}{2}\sin^{-1}(-1)) \right] + 2 \left[ (0 + \frac{1}{2}\sin^{-1}(1)) - (\frac{1}{4}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(\frac{1}{2})) \right]$
$= 2 \left[ -\frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} \right] + 2 \left[ \frac{\pi}{4} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} \right]$
$= 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right] + 2 \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right] = 4 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{8} \right) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$.

(C) The Trapezoidal rule for approximating the integral $\int_{x_0}^{x_n} f(x) dx$ is given by:
$\int_{x_0}^{x_n} f(x) dx \approx \frac{h}{2} [ (y_0 + y_n) + 2(y_1 + y_2 + \dots + y_{n-1}) ]$
Here,$h = 1$ (the difference between consecutive $x$ values).
The values are $y_0=2, y_1=3, y_2=6, y_3=11, y_4=18, y_5=27$.
Substituting these values into the formula:
Area $\approx \frac{1}{2} [ (2 + 27) + 2(3 + 6 + 11 + 18) ]$
Area $\approx \frac{1}{2} [ 29 + 2(38) ]$
Area $\approx \frac{1}{2} [ 29 + 76 ]$
Area $\approx \frac{1}{2} [ 105 ] = 52.5$ square units.

(C) For an optical fibre,the acceptance angle $\theta_a$ is given by the formula: $\sin \theta_a = \sqrt{\mu_1^2 - \mu_2^2}$,where $\mu_1$ is the refractive index of the core and $\mu_2$ is the refractive index of the cladding,assuming the surrounding medium is air $(\mu_0 = 1)$.
Given $\mu_1 = 1.5$ and $\mu_2 = 1.414$.
Substituting the values: $\sin \theta_a = \sqrt{(1.5)^2 - (1.414)^2}$.
Since $(1.414)^2 \approx 2$,we have $\sin \theta_a = \sqrt{2.25 - 2} = \sqrt{0.25} = 0.5$.
Therefore,$\theta_a = \sin^{-1}(0.5) = 30^{\circ}$.
This means the incident light must enter the fibre within an angle range of $0^{\circ}$ to $30^{\circ}$ with the axis to undergo total internal reflection.
Hence,option $(C)$ is correct.

(A) The resolving limit $(d\theta)$ of a telescope is given by the formula:
$d\theta = \frac{1.22 \lambda}{a}$
Where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens.
Given:
$\lambda = 4538 \ \mathring{A} = 4538 \times 10^{-10} \ \text{m}$
$a = 1 \ \text{m}$
Substituting the values:
$d\theta = \frac{1.22 \times 4538 \times 10^{-10}}{1}$
$d\theta = 5536.36 \times 10^{-10} \ \text{rad}$
$d\theta \approx 5.54 \times 10^{-7} \ \text{rad}$

(D) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and both are equal to $3/4$ of the angle of the prism:
$i = e = \frac{3}{4} \times A = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
The relationship between the angle of incidence,angle of emergence,angle of the prism,and the angle of deviation $\delta$ is given by:
$i + e = A + \delta$
Substituting the known values into the equation:
$45^{\circ} + 45^{\circ} = 60^{\circ} + \delta$
$90^{\circ} = 60^{\circ} + \delta$
$\delta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Thus,the angle of deviation is $30^{\circ}$.

(B) When chlorine reacts with hot and concentrated sodium hydroxide,it undergoes a disproportionation reaction to form sodium chloride,sodium chlorate,and water.
The balanced chemical equation is:
$3Cl_{2(g)} + 6NaOH_{(aq, \text{hot conc})} \longrightarrow 5NaCl_{(aq)} + NaClO_{3(aq)} + 3H_2O_{(l)}$
Thus,apart from sodium chloride,sodium chlorate $(NaClO_3)$ is formed.

(D) The current gain $\beta$ of a common-emitter transistor is defined as the ratio of the change in collector current $(\Delta i_C)$ to the change in base current $(\Delta i_B)$.
$\beta = \frac{\Delta i_C}{\Delta i_B}$
Given: $\beta = 80$ and $\Delta i_B = 250 \mu A = 250 \times 10^{-6} \text{ A}$.
Substituting the values into the formula:
$80 = \frac{\Delta i_C}{250 \times 10^{-6} \text{ A}}$
$\Delta i_C = 80 \times 250 \times 10^{-6} \text{ A}$
$\Delta i_C = 20,000 \times 10^{-6} \text{ A}$
$\Delta i_C = 20 \times 10^{-3} \text{ A} = 20 \text{ mA}$.
Therefore,the change in collector current is $20 \text{ mA}$.

(B) Given: $\tan y \frac{dy}{dx} = \sin(x+y) + \sin(x-y)$
Using the trigonometric identity $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
$\tan y \frac{dy}{dx} = 2 \sin x \cos y$
Rearranging the terms to separate variables:
$\frac{\sin y}{\cos^2 y} dy = 2 \sin x dx$
Integrating both sides:
$\int \frac{\sin y}{\cos^2 y} dy = \int 2 \sin x dx$
Let $t = \cos y$,then $dt = -\sin y dy$,so $\sin y dy = -dt$.
Substituting this into the integral:
$-\int \frac{dt}{t^2} = -2 \cos x + c$
$-(- \frac{1}{t}) = -2 \cos x + c$
$\frac{1}{\cos y} = -2 \cos x + c$
$\sec y = -2 \cos x + c$

(D) Given differential equation is $x y \frac{d y}{d x}=2 y^2-x^2$.
Dividing by $x$,we get $\frac{d y}{d x}=\frac{2 y}{x}-\frac{x}{y}$.
Rearranging the terms: $y \frac{d y}{d x}-\frac{2 y^2}{x}=-x$ $\ldots$ $(i)$.
Let $v=y^2$,then differentiating with respect to $x$,we get $\frac{d v}{d x}=2 y \frac{d y}{d x}$,which implies $y \frac{d y}{d x}=\frac{1}{2} \frac{d v}{d x}$.
Substituting this into $(i)$: $\frac{1}{2} \frac{d v}{d x}-\frac{2 v}{x}=-x$.
Multiplying by $2$: $\frac{d v}{d x}-\frac{4 v}{x}=-2 x$.
This is a linear differential equation of the form $\frac{d v}{d x}+P v=Q$,where $P=-\frac{4}{x}$ and $Q=-2 x$.
Integrating factor $IF=e^{\int P d x}=e^{\int-\frac{4}{x} d x}=e^{-4 \log x}=x^{-4}$.
The solution is $v \cdot IF = \int Q \cdot IF d x + c$.
$v \cdot x^{-4} = \int (-2 x) \cdot x^{-4} d x + c = -2 \int x^{-3} d x + c$.
$v x^{-4} = -2 \left( \frac{x^{-2}}{-2} \right) + c = x^{-2} + c$.
$v = x^2 + c x^4$.
Since $v=y^2$,the family of curves is $y^2 = x^2 + c x^4$.

(A) Given vectors are $\overrightarrow{a}=\hat{i}-2 \hat{j}+3 \hat{k}$,$\overrightarrow{b}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\overrightarrow{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1) \hat{k}$.
Since $\overrightarrow{a}$ and $\overrightarrow{b}$ lie in a plane,their cross product $\overrightarrow{n} = \overrightarrow{a} \times \overrightarrow{b}$ is a vector perpendicular to that plane.
If $\overrightarrow{c}$ is parallel to the plane containing $\overrightarrow{a}$ and $\overrightarrow{b}$,then $\overrightarrow{c}$ must be perpendicular to the normal vector $\overrightarrow{n}$.
Therefore,the scalar triple product $(\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c} = 0$.
First,calculate $\overrightarrow{a} \times \overrightarrow{b}$:
$\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = \hat{i}(2-9) - \hat{j}(-1-6) + \hat{k}(3+4) = -7\hat{i} + 7\hat{j} + 7\hat{k}$.
Now,compute the dot product with $\overrightarrow{c}$:
$(-7\hat{i} + 7\hat{j} + 7\hat{k}) \cdot (\lambda \hat{i} + \hat{j} + (2\lambda-1)\hat{k}) = 0$.
$-7\lambda + 7(1) + 7(2\lambda-1) = 0$.
$-7\lambda + 7 + 14\lambda - 7 = 0$.
$7\lambda = 0 \Rightarrow \lambda = 0$.
Thus,the value of $\lambda$ is $0$.

(A) Given that $P(A \cap B) = \frac{3}{25}$ and $P(B - A) = \frac{8}{25}$.
From the properties of sets and probability,the event $B$ can be expressed as the union of two disjoint sets: $(B - A)$ and $(A \cap B)$.
Therefore,$P(B) = P(B - A) + P(A \cap B)$.
Substituting the given values:
$P(B) = \frac{8}{25} + \frac{3}{25} = \frac{11}{25}$.

(D) Case $I$: $A$ white ball is transferred from urn $A$ to urn $B$.
The probability of selecting a white ball from urn $A$ is $P(W_A) = \frac{3}{3+5} = \frac{3}{8}$.
After transferring,urn $B$ contains $7$ white and $8$ black balls (total $15$).
The probability of selecting a white ball from urn $B$ is $P(W_B|W_A) = \frac{7}{15}$.
The probability of this case is $P(W_A) \times P(W_B|W_A) = \frac{3}{8} \times \frac{7}{15} = \frac{21}{120} = \frac{7}{40}$.
Case $II$: $A$ black ball is transferred from urn $A$ to urn $B$.
The probability of selecting a black ball from urn $A$ is $P(B_A) = \frac{5}{3+5} = \frac{5}{8}$.
After transferring,urn $B$ contains $6$ white and $9$ black balls (total $15$).
The probability of selecting a white ball from urn $B$ is $P(W_B|B_A) = \frac{6}{15}$.
The probability of this case is $P(B_A) \times P(W_B|B_A) = \frac{5}{8} \times \frac{6}{15} = \frac{30}{120} = \frac{10}{40}$.
The total probability of drawing a white ball from urn $B$ is $\frac{7}{40} + \frac{10}{40} = \frac{17}{40}$.

(C) Let $\lambda$ be the parameter (mean) of the Poisson distribution for the random variable $X$.
The probability mass function is given by $P(X=r) = \frac{\lambda^r e^{-\lambda}}{r!}$ for $r = 0, 1, 2, \dots$.
Given that $P(X=1) = P(X=2)$,we have:
$\frac{\lambda^1 e^{-\lambda}}{1!} = \frac{\lambda^2 e^{-\lambda}}{2!}$
$\lambda = \frac{\lambda^2}{2}$
Since $\lambda \neq 0$,we divide by $\lambda$ to get $1 = \frac{\lambda}{2}$,which implies $\lambda = 2$.
Now,we calculate $P(X=5)$:
$P(X=5) = \frac{\lambda^5 e^{-\lambda}}{5!} = \frac{2^5 e^{-2}}{120}$
$P(X=5) = \frac{32 e^{-2}}{120} = \frac{4}{15} e^{-2}$.

(B) For a binomial distribution,the mean is given by $np = 2$ and the variance is given by $npq = 1$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{1}{2}$.
Since $p = 1 - q$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n(\frac{1}{2}) = 2$,which implies $n = 4$.
The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
We need to find $P(X \geq 1) = 1 - P(X = 0)$.
$P(X = 0) = \binom{4}{0} (\frac{1}{2})^0 (\frac{1}{2})^4 = 1 \times 1 \times \frac{1}{16} = \frac{1}{16}$.
Therefore,$P(X \geq 1) = 1 - \frac{1}{16} = \frac{15}{16}$.

(C) White tin,an allotropic form of tin,is an example of a tetragonal crystal system.
For a tetragonal system,the unit cell parameters are $a = b \neq c$ and $\alpha = \beta = \gamma = 90^{\circ}$.
The reasoning provided states $a=b=c$ and $\alpha=\beta=\gamma \neq 90^{\circ}$,which is incorrect.
Therefore,Assertion $(A)$ is true,but Reason $(R)$ is false.

(D) Given,the radius ratio of anion to cation $= 10 : 9.3$.
$\therefore$ The radius ratio of cation to anion $= 9.3 : 10 = 0.93$.
When the radius ratio of cation to anion lies between $0.732$ and $1.00$,the coordination number is $8$.
Thus,the coordination number of the cation in the crystal is $8$.