IIT JEE 2005 Mathematics Question Paper with Answer and Solution

(C) The vertex of the rays $PQ$ and $PR$ is at $P(-1, 0)$,which corresponds to the complex number $z_0 = -1$.
The ray $PQ$ passes through $(-1 + \sqrt{2}, \sqrt{2}i)$,so its angle with the real axis is $\arg(z - (-1)) = \arg(z + 1) = \tan^{-1}(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{\pi}{4}$.
The ray $PR$ passes through $(-1 + \sqrt{2}, -\sqrt{2}i)$,so its angle is $\arg(z + 1) = -\frac{\pi}{4}$.
The shaded region lies between these rays,so $|\arg(z + 1)| < \frac{\pi}{4}$.
The arc $QAR$ is part of a circle centered at $P(-1, 0)$. The distance from $P(-1, 0)$ to $A(1, 0)$ is $|1 - (-1)| = 2$. Thus,the radius of the circle is $2$.
The shaded region is outside this circle,so $|z - (-1)| > 2$,which is $|z + 1| > 2$.
Therefore,the conditions are $|z + 1| > 2$ and $|\arg(z + 1)| < \frac{\pi}{4}$.

(C) Let $y = |a + b\omega + c\omega^2|$.
For $y$ to be minimum,$y^2$ must be minimum.
$y^2 = |a + b\omega + c\omega^2|^2 = (a + b\omega + c\omega^2)(a + b\bar{\omega} + c\bar{\omega}^2)$.
Using $\omega^2 = \bar{\omega}$ and $\omega = \bar{\omega}^2$,we get $y^2 = a^2 + b^2 + c^2 - ab - bc - ca$.
This can be written as $y^2 = \frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$.
Since $a, b, c$ are integers and not all equal,the minimum value occurs when two variables are equal and the third differs by $1$ (e.g.,$a=0, b=0, c=1$).
Substituting these values,$y^2 = \frac{1}{2}[(0-0)^2 + (0-1)^2 + (1-0)^2] = \frac{1}{2}[0 + 1 + 1] = 1$.
Thus,the minimum value of $y$ is $\sqrt{1} = 1$.

(D) rectangle is formed by choosing two horizontal lines and two vertical lines.
Let the horizontal lines be $y_0, y_1, \dots, y_{2n-1}$ and vertical lines be $x_0, x_1, \dots, x_{2m-1}$.
$A$ rectangle has odd side lengths if the difference between the indices of the chosen lines is odd.
For the horizontal side,we need to choose two lines $y_i, y_j$ such that $|i - j|$ is odd. This means one index must be even and the other must be odd.
In the set ${0, 1, \dots, 2n-1}$,there are $n$ even numbers ${0, 2, \dots, 2n-2}$ and $n$ odd numbers ${1, 3, \dots, 2n-1}$.
The number of ways to choose one even and one odd index is $n \times n = n^2$.
Similarly,for the vertical side,the number of ways to choose two lines such that the difference is odd is $m \times m = m^2$.
Therefore,the total number of rectangles with odd side lengths is $m^2 \times n^2 = m^2n^2$.

(B) We know that the given expression is the coefficient of $x^{20}$ in the product of two binomial expansions.
Consider $(1-x)^{30} = \sum_{r=0}^{30} (-1)^r \binom{30}{r} x^r = \binom{30}{0} - \binom{30}{1}x + \binom{30}{2}x^2 - ....... + \binom{30}{20}x^{20} - ....... + \binom{30}{30}x^{30}$.
Consider $(x+1)^{30} = \sum_{k=0}^{30} \binom{30}{k} x^{30-k} = \binom{30}{0}x^{30} + \binom{30}{1}x^{29} + ....... + \binom{30}{10}x^{20} + ....... + \binom{30}{30}$.
The given expression is the coefficient of $x^{20}$ in the product $(1-x)^{30}(x+1)^{30} = (1-x^2)^{30}$.
In the expansion of $(1-x^2)^{30} = \sum_{k=0}^{30} \binom{30}{k} (-x^2)^k = \sum_{k=0}^{30} \binom{30}{k} (-1)^k x^{2k}$.
To find the coefficient of $x^{20}$,we set $2k = 20$,which gives $k = 10$.
The coefficient is $\binom{30}{10} (-1)^{10} = \binom{30}{10}$.

(D) Given $\cos(\alpha - \beta) = 1$ and $-\pi < \alpha, \beta < \pi$.
Since $-\pi < \alpha < \pi$ and $-\pi < \beta < \pi$,we have $-2\pi < \alpha - \beta < 2\pi$.
The equation $\cos(\alpha - \beta) = 1$ implies $\alpha - \beta = 0$,so $\alpha = \beta$.
Substituting $\alpha = \beta$ into the second equation,we get $\cos(\alpha + \alpha) = \cos(2\alpha) = \frac{1}{e}$.
Since $-\pi < \alpha < \pi$,we have $-2\pi < 2\alpha < 2\pi$.
In the interval $(-2\pi, 2\pi)$,the equation $\cos(2\alpha) = \frac{1}{e}$ has solutions where $\cos(x) = \frac{1}{e}$ for $x \in (-2\pi, 2\pi)$.
Since $0 < \frac{1}{e} < 1$,there are two solutions for $2\alpha$ in $(0, 2\pi)$ and two solutions in $(-2\pi, 0)$.
Thus,there are $4$ distinct values for $\alpha$,and since $\alpha = \beta$,there are $4$ ordered pairs $(\alpha, \beta)$.

(C) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$a = k\sin A$,$b = k\sin B$,and $c = k\sin C$.
Consider the expression $\frac{b - c}{a} = \frac{\sin B - \sin C}{\sin A}$.
Using the sum-to-product formula,$\sin B - \sin C = 2\sin \frac{B - C}{2} \cos \frac{B + C}{2}$.
Since $A + B + C = \pi$,we have $\frac{B + C}{2} = \frac{\pi}{2} - \frac{A}{2}$,so $\cos \frac{B + C}{2} = \sin \frac{A}{2}$.
Also,$\sin A = 2\sin \frac{A}{2} \cos \frac{A}{2}$.
Substituting these into the expression:
$\frac{b - c}{a} = \frac{2\sin \frac{B - C}{2} \sin \frac{A}{2}}{2\sin \frac{A}{2} \cos \frac{A}{2}} = \frac{\sin \frac{B - C}{2}}{\cos \frac{A}{2}}$.
Rearranging gives $(b - c)\cos \frac{A}{2} = a\sin \frac{B - C}{2}$.

(A) Let the radius of each coin be $r = 1$. The centers of the three coins form an equilateral triangle with side length $2r = 2$.
Let the vertices of the large equilateral triangle be $A, B, C$ and the centers of the coins be $C_1, C_2, C_3$.
The distance from a vertex (e.g.,$B$) to the projection of the center of the nearest coin $(M)$ on the side $BC$ is given by $BM = r \cot(30^\circ) = 1 \times \sqrt{3} = \sqrt{3}$.
The distance between the centers of the two base coins is $MN = 2r = 2$.
By symmetry,the side length $s$ of the large equilateral triangle is $s = BM + MN + NC = \sqrt{3} + 2 + \sqrt{3} = 2 + 2\sqrt{3} = 2(1 + \sqrt{3})$.
The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.
Area $= \frac{\sqrt{3}}{4} [2(1 + \sqrt{3})]^2 = \frac{\sqrt{3}}{4} \times 4(1 + \sqrt{3})^2 = \sqrt{3}(1 + 3 + 2\sqrt{3}) = \sqrt{3}(4 + 2\sqrt{3}) = 4\sqrt{3} + 6 \; \text{sq. units}$.

(B) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $x$-axis,the radius $r = |k|$.
Since the circle touches the circle ${x^2} + {(y - 1)^2} = 1$ (centre $(0, 1)$,radius $1$) externally,the distance between their centres is equal to the sum of their radii:
$\sqrt {{{(h - 0)}^2} + {{(k - 1)}^2}} = 1 + |k|$
Squaring both sides:
${h^2} + {(k - 1)^2} = {(1 + |k|)^2}$
${h^2} + {k^2} - 2k + 1 = 1 + 2|k| + {k^2}$
${h^2} = 2k + 2|k|$
Case $1$: If $k > 0$,then $|k| = k$,so ${h^2} = 2k + 2k = 4k$. Thus,the locus is ${x^2} = 4y$ for $y > 0$.
Case $2$: If $k < 0$,then $|k| = -k$,so ${h^2} = 2k - 2k = 0$. Thus,the locus is $x = 0$ for $y < 0$.
Combining these,the locus is $\{ (x, y) : {x^2} = 4y, y > 0\} \cup \{ (0, y) : y < 0\}$.

(C) The equation of the tangent to the parabola $y = x^2 + 6$ at $(1, 7)$ is given by $\frac{1}{2}(y + 7) = x(1) + 6$.
Simplifying this,we get $y + 7 = 2x + 12$,which implies $y = 2x + 5$ $(i)$.
This tangent touches the circle $x^2 + y^2 + 16x + 12y + c = 0$ $(ii)$.
Substituting $(i)$ into $(ii)$:
$x^2 + (2x + 5)^2 + 16x + 12(2x + 5) + c = 0$
$x^2 + 4x^2 + 20x + 25 + 16x + 24x + 60 + c = 0$
$5x^2 + 60x + (85 + c) = 0$.
Since the line is tangent,the discriminant must be zero:
$D = (60)^2 - 4(5)(85 + c) = 0$
$3600 - 20(85 + c) = 0$
$180 - (85 + c) = 0 \implies c = 95$.
The quadratic equation becomes $5x^2 + 60x + 180 = 0$,or $x^2 + 12x + 36 = 0$.
$(x + 6)^2 = 0 \implies x = -6$.
Substituting $x = -6$ into $(i)$,$y = 2(-6) + 5 = -7$.
Thus,the point of contact is $(-6, -7)$.

(C) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos \theta, b \sin \theta)$ is given by $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$.
The $x$-intercept of this tangent is found by setting $y = 0$,which gives $P = (\frac{a}{\cos \theta}, 0)$.
The $y$-intercept of this tangent is found by setting $x = 0$,which gives $Q = (0, \frac{b}{\sin \theta})$.
The area of the triangle $OPQ$ formed by the tangent and the coordinate axes is $A = \frac{1}{2} \times |x_P| \times |y_Q| = \frac{1}{2} \left| \frac{a}{\cos \theta} \right| \left| \frac{b}{\sin \theta} \right| = \frac{ab}{|2 \sin \theta \cos \theta|} = \frac{ab}{|\sin 2\theta|}$.
Since the minimum value of $|\sin 2\theta|$ is $1$ (when $2\theta = 90^\circ$ or $270^\circ$),the minimum area is $ab$.

(D) Let $P(x) = bx^2 + ax + c$.
Given $P(0) = 0$,we have $c = 0$.
Given $P(1) = 1$,we have $a + b = 1$,which implies $b = 1 - a$.
Thus,$P(x) = (1 - a)x^2 + ax$.
The derivative is $P'(x) = 2(1 - a)x + a$.
We are given $P'(x) > 0$ for all $x \in (0, 1)$.
At $x = 0$,$P'(0) = a > 0$.
At $x = 1$,$P'(1) = 2(1 - a) + a = 2 - a > 0$,which implies $a < 2$.
Since $P'(x)$ is a linear function,if it is positive at the endpoints of the interval $(0, 1)$,it is positive for all $x \in (0, 1)$.
Thus,$0 < a < 2$.
Therefore,$S = \{ax + (1 - a)x^2 : a \in (0, 2)\}$.

(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & -2 & 4 \end{bmatrix}$.
First,we find the characteristic equation of $A$. The characteristic polynomial is $|A - \lambda I| = 0$.
$|A - \lambda I| = \begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & -2 & 4-\lambda \end{vmatrix} = (1-\lambda) [(1-\lambda)(4-\lambda) + 2] = (1-\lambda) [\lambda^2 - 5\lambda + 6] = 0$.
So,$\lambda^3 - 6\lambda^2 + 11\lambda - 6 = 0$.
By Cayley-Hamilton theorem,$A^3 - 6A^2 + 11A - 6I = 0$.
Multiplying by $A^{-1}$,we get $A^2 - 6A + 11I - 6A^{-1} = 0$.
$6A^{-1} = A^2 - 6A + 11I$.
$A^{-1} = \frac{1}{6}[A^2 - 6A + 11I]$.
Comparing this with $A^{-1} = \frac{1}{6}[A^2 + cA + dI]$,we get $c = -6$ and $d = 11$.
Thus,the pair $(c, d) = (-6, 11)$.

(A) Given $Q = PAP^T$. Note that $P$ is an orthogonal matrix,so $PP^T = I$ and $P^T = P^{-1}$.
We want to compute $X = P^T Q^{2005} P$.
Since $Q = PAP^T$,we have $Q^n = (PAP^T)(PAP^T)...(PAP^T) = PA^n P^T$.
Substituting this into the expression:
$X = P^T (PA^{2005}P^T) P$
$X = (P^T P) A^{2005} (P^T P)$
Since $P^T P = I$,we get $X = I A^{2005} I = A^{2005}$.
Given $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$,we use the property of this specific matrix: $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.
Thus,$A^{2005} = \begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}$.
Therefore,the correct option is $A$.

(B) The given vectors are constructed using the Gram-Schmidt orthogonalization process.
For the set $\{a, b_1, c_2\}$:
$1$. $a \cdot b_1 = a \cdot (b - \frac{b \cdot a}{|a|^2} a) = a \cdot b - \frac{b \cdot a}{|a|^2} (a \cdot a) = a \cdot b - b \cdot a = 0$. Thus,$a \perp b_1$.
$2$. $a \cdot c_2 = a \cdot (c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1) = a \cdot c - \frac{c \cdot a}{|a|^2} (a \cdot a) - \frac{c \cdot b_1}{|b_1|^2} (a \cdot b_1) = a \cdot c - c \cdot a - 0 = 0$. Thus,$a \perp c_2$.
$3$. $b_1 \cdot c_2 = b_1 \cdot (c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1) = b_1 \cdot c - \frac{c \cdot a}{|a|^2} (b_1 \cdot a) - \frac{c \cdot b_1}{|b_1|^2} (b_1 \cdot b_1) = b_1 \cdot c - 0 - c \cdot b_1 = 0$. Thus,$b_1 \perp c_2$.
Since all pairs are orthogonal,the set $\{a, b_1, c_2\}$ is a set of mutually orthogonal vectors.
Therefore,option $(b)$ is the correct answer.

(C) For any subset $A \subseteq X$,we have $A \subseteq f^{-1}(f(A))$. Equality $f^{-1}(f(A)) = A$ holds if and only if $f$ is injective.
For any subset $B \subseteq Y$,we have $f(f^{-1}(B)) = B \cap f(X)$.
Therefore,$f(f^{-1}(B)) = B$ holds if and only if $B \subseteq f(X)$.
Thus,option $(c)$ is correct.

(A) Let $h(x) = (f - g)(x)$.
For $x \in \mathbb{Q}$,$h(x) = f(x) - g(x) = x - 0 = x$.
For $x \notin \mathbb{Q}$,$h(x) = f(x) - g(x) = 0 - x = -x$.
Thus,$h(x) = \begin{cases} x, & x \in \mathbb{Q} \\ -x, & x \notin \mathbb{Q} \end{cases}$.
To check if $h(x)$ is one-one: Let $h(x_1) = h(x_2)$. If $x_1, x_2 \in \mathbb{Q}$,then $x_1 = x_2$. If $x_1, x_2 \notin \mathbb{Q}$,then $-x_1 = -x_2 \implies x_1 = x_2$. If $x_1 \in \mathbb{Q}$ and $x_2 \notin \mathbb{Q}$,then $x_1 = -x_2$,which is possible (e.g.,$x_1 = 1, x_2 = -1$,but $-1$ is rational,so this case is restricted). Since $h(x)$ maps every real number to a unique value,it is one-one.
To check if $h(x)$ is onto: For any $y \in \mathbb{R}$,if $y \in \mathbb{Q}$,$h(y) = y$. If $y \notin \mathbb{Q}$,$h(-y) = -(-y) = y$. Thus,for every $y \in \mathbb{R}$,there exists an $x$ such that $h(x) = y$. Hence,it is onto.

(B) The function is defined as $f(x) = ||x| - 1|$.
We can break this down based on the sign of the inner expression:
$f(x) = \begin{cases} |x| - 1, & |x| - 1 \ge 0 \\ -(|x| - 1), & |x| - 1 < 0 \end{cases}$
$f(x) = \begin{cases} |x| - 1, & |x| \ge 1 \\ 1 - |x|, & |x| < 1 \end{cases}$
This can be further expanded for different intervals of $x$:
$f(x) = \begin{cases} -x - 1, & x \le -1 \\ x + 1, & -1 < x < 0 \\ 1 - x, & 0 \le x < 1 \\ x - 1, & x \ge 1 \end{cases}$
$A$ function is not differentiable at points where there are sharp corners (cusps) in its graph.
By observing the graph,the function has sharp corners at $x = -1$,$x = 0$,and $x = 1$.
Therefore,$f(x)$ is not differentiable at $x \in \{-1, 0, 1\}$.

(B) Let $g(x) = f(x) - x^2$.
Since $f(1) = 1, f(2) = 4, f(3) = 9$,we have $g(1) = 1 - 1^2 = 0$,$g(2) = 4 - 2^2 = 0$,and $g(3) = 9 - 3^2 = 0$.
Thus,$g(x)$ has at least $3$ real roots at $x = 1, 2, 3$.
By Rolle's Theorem,$g'(x)$ has at least one root in $(1, 2)$ and at least one root in $(2, 3)$,meaning $g'(x)$ has at least $2$ roots in $(1, 3)$.
Applying Rolle's Theorem again to $g'(x)$,$g''(x)$ must have at least $1$ root in $(1, 3)$.
Since $g(x) = f(x) - x^2$,we have $g'(x) = f'(x) - 2x$ and $g''(x) = f''(x) - 2$.
Since $g''(c) = 0$ for some $c \in (1, 3)$,it follows that $f''(c) - 2 = 0$,or $f''(c) = 2$ for at least one $c \in (1, 3)$.

(B) Given that $f\left( \frac{1}{n} \right) = 0$ for all $n \in \{1, 2, 3, \dots\}$.
Since $f(x)$ is continuous (as it is differentiable),we have $f(0) = \lim_{n \to \infty} f\left( \frac{1}{n} \right) = 0$.
Now,consider the sequence $x_n = \frac{1}{n}$. We have $f(x_n) = 0$ and $f(0) = 0$.
By the definition of the derivative at $x = 0$:
$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{n \to \infty} \frac{f(1/n) - f(0)}{1/n - 0} = \lim_{n \to \infty} \frac{0 - 0}{1/n} = 0$.
Thus,$f(0) = 0$ and $f'(0) = 0$.

(B) Let $I = \int_{-2}^{0} [x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)] \, dx$.
Substitute $t = x + 1$,so $dt = dx$.
When $x = -2$,$t = -1$. When $x = 0$,$t = 1$.
Also,$x^3 + 3x^2 + 3x + 3 = (x+1)^3 + 2 = t^3 + 2$.
Thus,$I = \int_{-1}^{1} [t^3 + 2 + t\cos(t)] \, dt$.
We can split this into $I = \int_{-1}^{1} t^3 \, dt + \int_{-1}^{1} 2 \, dt + \int_{-1}^{1} t\cos(t) \, dt$.
Since $t^3$ and $t\cos(t)$ are odd functions,their integrals over the symmetric interval $[-1, 1]$ are $0$.
Therefore,$I = 0 + \int_{-1}^{1} 2 \, dt + 0 = [2t]_{-1}^{1} = 2(1 - (-1)) = 2(2) = 4$.

(A) Given the equation: $\int_{\sin x}^1 {t^2}f(t) dt = 1 - \sin x$.
Applying the Leibniz integral rule to differentiate both sides with respect to $x$:
$\frac{d}{dx} \left( \int_{\sin x}^1 {t^2}f(t) dt \right) = \frac{d}{dx} (1 - \sin x)$.
Using the Fundamental Theorem of Calculus,the derivative of $\int_{g(x)}^a h(t) dt$ is $-h(g(x)) \cdot g'(x)$.
So,$-(\sin x)^2 f(\sin x) \cdot \cos x = -\cos x$.
Since $x \in (0, \frac{\pi}{2})$,$\cos x \neq 0$,we can divide both sides by $-\cos x$:
$(\sin x)^2 f(\sin x) = 1$.
Thus,$f(\sin x) = \frac{1}{\sin^2 x}$.
Replacing $\sin x$ with $t$,we get $f(t) = \frac{1}{t^2}$.
Therefore,$f\left( \frac{1}{\sqrt{3}} \right) = \frac{1}{(1/\sqrt{3})^2} = \frac{1}{1/3} = 3$.

(D) The curves are $y = (x + 1)^2$ and $y = (x - 1)^2$. The line is $y = \frac{1}{4}$.
By symmetry,the area is twice the area bounded by $y = (x - 1)^2$,the $y$-axis $(x=0)$,and the line $y = \frac{1}{4}$ in the region $x \ge 0$.
For $y = (x - 1)^2$,we have $x - 1 = \pm \sqrt{y}$,so $x = 1 \pm \sqrt{y}$. Since we are considering the right branch,$x = 1 - \sqrt{y}$ for $x \in [0, 1]$.
The intersection of $y = (x - 1)^2$ and $y = \frac{1}{4}$ gives $(x - 1)^2 = \frac{1}{4}$,so $x - 1 = -1/2$ (as $x < 1$),which means $x = 1/2$.
The area $A = 2 \int_{1/4}^{1} (1 - \sqrt{y}) dy$ is incorrect based on the graph. The correct integral for the region bounded by $y = (x - 1)^2$,$x=0$,and $y=1/4$ is $\int_{0}^{1/2} (1/4 - (x-1)^2) dx + \int_{1/2}^{1} (1/4 - 0) dx$ is not right.
Let's use horizontal strips: The area is $2 \int_{1/4}^{1} (1 - \sqrt{y}) dy = 2 [y - \frac{2}{3} y^{3/2}]_{1/4}^{1} = 2 [(1 - 2/3) - (1/4 - \frac{2}{3} \cdot \frac{1}{8})] = 2 [1/3 - (1/4 - 1/12)] = 2 [1/3 - 2/12] = 2 [1/3 - 1/6] = 2 [1/6] = 1/3$ sq. units.

(B) Given the differential equation $xdy = y(dx + ydy)$.
Dividing both sides by $y^2$,we get $\frac{xdy - ydx}{y^2} = dy$.
This can be written as $-d(\frac{x}{y}) = dy$.
Integrating both sides,we get $-\frac{x}{y} = y + C$,which simplifies to $\frac{x}{y} + y = C$.
Given $y(1) = 1$,substituting $x=1$ and $y=1$ gives $\frac{1}{1} + 1 = C$,so $C = 2$.
The equation becomes $\frac{x}{y} + y = 2$,or $x + y^2 = 2y$,which is $y^2 - 2y + x = 0$.
For $x = -3$,the equation is $y^2 - 2y - 3 = 0$.
Factoring the quadratic,we get $(y - 3)(y + 1) = 0$.
Thus,$y = 3$ or $y = -1$.
Since the condition $y > 0$ is given,we have $y = 3$.

(A) Given the differential equation $(x^2 + y^2)dy = xy dx$.
Rearranging the terms,we get $x^2 dy + y^2 dy = xy dx$.
$x^2 dy - xy dx = -y^2 dy$.
Dividing by $xy^2$,we get $\frac{x dy - y dx}{y^2} = -\frac{dy}{x}$.
This is equivalent to $d(\frac{x}{y}) = -\frac{dy}{x}$ (Note: The standard form is $x dy - y dx = x^2 d(\frac{y}{x})$ or $y dx - x dy = y^2 d(\frac{x}{y})$).
Let us rewrite: $x^2 dy = xy dx - y^2 dy = y(x dx - y dy)$ is not correct. Let's use substitution $y = vx$,$dy = v dx + x dv$.
$(x^2 + v^2 x^2)(v dx + x dv) = x(vx) dx$.
$x^2(1 + v^2)(v dx + x dv) = vx^2 dx$.
$(1 + v^2)(v dx + x dv) = v dx$.
$v dx + x dv + v^3 dx + v^2 x dv = v dx$.
$x dv(1 + v^2) = -v^3 dx$.
$\frac{1 + v^2}{v^3} dv = -\frac{dx}{x}$.
Integrating both sides: $\int (v^{-3} + v^{-1}) dv = -\int \frac{dx}{x}$.
$-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|\frac{y}{x}| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| - \ln|x| = -\ln|x| + C$.
$-\frac{x^2}{2y^2} + \ln|y| = C$.
Given $y(1) = 1$,so $-\frac{1^2}{2(1)^2} + \ln(1) = C \implies C = -\frac{1}{2}$.
So,$-\frac{x^2}{2y^2} + \ln|y| = -\frac{1}{2}$.
For $y(x_0) = e$,$-\frac{x_0^2}{2e^2} + \ln(e) = -\frac{1}{2}$.
$-\frac{x_0^2}{2e^2} + 1 = -\frac{1}{2}$.
$-\frac{x_0^2}{2e^2} = -\frac{3}{2}$.
$x_0^2 = 3e^2 \implies x_0 = \sqrt{3}e$.

(B) Given the equation: $y \cos x + x \cos y = \pi$.
At $x = 0$,we have $y \cos(0) + 0 \cos y = \pi$,which implies $y = \pi$.
Differentiating both sides with respect to $x$:
$-y \sin x + y' \cos x + \cos y - x \sin y \cdot y' = 0$.
At $x = 0$ and $y = \pi$:
$-(\pi) \sin(0) + y'(0) \cos(0) + \cos(\pi) - 0 \sin(\pi) \cdot y'(0) = 0$.
$0 + y'(0) \cdot 1 - 1 - 0 = 0 \implies y'(0) = 1$.
Now,differentiate the first derivative expression again:
$-y' \sin x - y \cos x + y'' \cos x - y' \sin x - \sin y \cdot y' - x(\cos y \cdot (y')^2 + \sin y \cdot y'') - \sin y \cdot y' = 0$.
Substitute $x = 0, y = \pi, y' = 1$:
$-1 \cdot 0 - \pi \cdot 1 + y''(0) \cdot 1 - 1 \cdot 0 - 0 - 0 - 0 = 0$.
$-\pi + y''(0) = 0 \implies y''(0) = \pi$.

(B) Let $p$ be the probability of getting a $1$ in a single throw,so $p = \frac{1}{6}$.
Let $q$ be the probability of not getting a $1$,so $q = 1 - \frac{1}{6} = \frac{5}{6}$.
We want to find the probability of getting a $1$ for the first time on an even throw (i.e.,$2^{nd}, 4^{th}, 6^{th}, \dots$ throw).
The probability is given by the sum of the infinite geometric series:
$P = qp + q^3p + q^5p + \dots$
This is a geometric series with first term $a = qp = \frac{5}{6} \times \frac{1}{6} = \frac{5}{36}$ and common ratio $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.
The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.
$P = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{36 - 25}{36}} = \frac{5}{36} \times \frac{36}{11} = \frac{5}{11}$.

(B) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
Since the plane is at a unit distance from the origin $(0, 0, 0)$,the perpendicular distance $d$ is given by $d = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 1$.
Squaring both sides,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = 1$ ... $(i)$.
The coordinates of the points $P, Q,$ and $R$ are $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively.
The centroid $(x, y, z)$ of $\Delta PQR$ is given by $x = \frac{a+0+0}{3} = \frac{a}{3}$,$y = \frac{0+b+0}{3} = \frac{b}{3}$,and $z = \frac{0+0+c}{3} = \frac{c}{3}$.
Thus,$a = 3x, b = 3y, c = 3z$.
Substituting these into equation $(i)$,we get $\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = 1$.
This simplifies to $\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = 1$,which implies $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = 9$.
Comparing this with the given locus $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = k$,we find $k = 9$.