AIEEE 2011 Mathematics Question Paper with Answer and Solution

(C) Given $A = \sin^2 x + \cos^4 x$.
Let $t = \sin^2 x$,then $0 \le t \le 1$. Since $\cos^2 x = 1 - \sin^2 x = 1 - t$,we have $\cos^4 x = (1 - t)^2$.
Substituting these into the expression for $A$:
$A = t + (1 - t)^2 = t + 1 - 2t + t^2 = t^2 - t + 1$.
This is a quadratic expression in $t$ where $t \in [0, 1]$.
The vertex of the parabola $f(t) = t^2 - t + 1$ is at $t = -(-1)/(2 \times 1) = 1/2$.
Since $1/2$ is within the interval $[0, 1]$,the minimum value is $f(1/2) = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4$.
The maximum value occurs at the boundaries $t=0$ or $t=1$:
$f(0) = 0^2 - 0 + 1 = 1$.
$f(1) = 1^2 - 1 + 1 = 1$.
Thus,the range of $A$ is $\frac{3}{4} \le A \le 1$.

(C) We know that for the cube roots of unity,$1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$.
Substituting this into the given expression:
$(1 + \omega)^7 = (-\omega^2)^7$
$= -\omega^{14}$
Since $\omega^3 = 1$,we have $\omega^{14} = \omega^{12} \cdot \omega^2 = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2$.
Thus,$(1 + \omega)^7 = -\omega^2$.
Using the identity $1 + \omega + \omega^2 = 0$,we get $-\omega^2 = 1 + \omega$.
Comparing $1 + \omega$ with $A + B\omega$,we get $A = 1$ and $B = 1$.

(D) For Statement-$1$: The number of ways to distribute $n$ identical items into $r$ distinct boxes such that no box is empty is given by the formula $^{n-1}C_{r-1}$.
Here,$n = 10$ and $r = 4$.
So,the number of ways is $^{10-1}C_{4-1} = ^9C_3$.
Thus,Statement-$1$ is true.
For Statement-$2$: The number of ways to choose $r$ items from $n$ distinct items is $^nC_r$.
Here,$n = 9$ and $r = 3$,so the number of ways is $^9C_3$.
Thus,Statement-$2$ is true.
Since the formula for distributing identical items into distinct boxes (stars and bars method) is derived using the concept of choosing positions,Statement-$2$ is the correct explanation for Statement-$1$.

(B) Given expression: $(1 - x - x^2 + x^3)^6 = ((1 - x)(1 - x^2))^6 = (1 - x)^6 (1 - x^2)^6 = (1 - x)^6 (1 - x)^6 (1 + x)^6 = (1 - x)^{12} (1 + x)^6$.
We need the coefficient of $x^7$ in $(1 - x)^{12} (1 + x)^6$.
Let $(1 - x)^{12} = \sum_{i=0}^{12} (-1)^i \binom{12}{i} x^i$ and $(1 + x)^6 = \sum_{j=0}^{6} \binom{6}{j} x^j$.
The coefficient of $x^7$ is $\sum_{i+j=7} (-1)^i \binom{12}{i} \binom{6}{j}$.
For $i=1, j=6: -\binom{12}{1} \binom{6}{6} = -12 \times 1 = -12$.
For $i=2, j=5: \binom{12}{2} \binom{6}{5} = 66 \times 6 = 396$.
For $i=3, j=4: -\binom{12}{3} \binom{6}{4} = -220 \times 15 = -3300$.
For $i=4, j=3: \binom{12}{4} \binom{6}{3} = 495 \times 20 = 9900$.
For $i=5, j=2: -\binom{12}{5} \binom{6}{2} = -792 \times 15 = -11880$.
For $i=6, j=1: \binom{12}{6} \binom{6}{1} = 924 \times 6 = 5544$.
For $i=7, j=0: -\binom{12}{7} \binom{6}{0} = -792 \times 1 = -792$.
Summing these: $-12 + 396 - 3300 + 9900 - 11880 + 5544 - 792 = -144$.

(D) For Statement $-2$: By Fermat's Little Theorem,for any prime $p$ and integer $n$,$n^p \equiv n \pmod{p}$. Here $p = 7$,so $n^7 \equiv n \pmod{7}$,which means $n^7 - n$ is divisible by $7$. Thus,Statement $-2$ is true.
For Statement $-1$: Expand $(n + 1)^7$ using the Binomial Theorem: $(n + 1)^7 = n^7 + 7n^6 + 21n^5 + 35n^4 + 35n^3 + 21n^2 + 7n + 1$.
Then $(n + 1)^7 - n^7 - 1 = 7n^6 + 21n^5 + 35n^4 + 35n^3 + 21n^2 + 7n = 7(n^6 + 3n^5 + 5n^4 + 5n^3 + 3n^2 + n)$.
This expression is clearly divisible by $7$. Thus,Statement $-1$ is true.
Note that Statement $-1$ can be written as $(n+1)^7 - (n+1) - (n^7 - n) = 7k$. Since both $(n+1)^7 - (n+1)$ and $n^7 - n$ are divisible by $7$ (by Statement $-2$),their difference is also divisible by $7$. Therefore,Statement $-2$ is a correct explanation for Statement $-1$.

(C) The savings for the first few months are:
Month $1: 200$,Month $2: 200$,Month $3: 200$.
From month $4$ onwards,the savings form an Arithmetic Progression $(AP)$ with first term $a = 240$ and common difference $d = 40$.
Let the total number of months be $n$. The total savings is given by:
$600 + \sum_{k=1}^{n-3} [240 + (k-1)40] = 11040$
$600 + \frac{n-3}{2} [2(240) + (n-3-1)40] = 11040$
$600 + (n-3) [240 + (n-4)20] = 11040$
$(n-3) [240 + 20n - 80] = 10440$
$(n-3) [20n + 160] = 10440$
$(n-3)(n+8) = 522$
$n^2 + 5n - 24 = 522$
$n^2 + 5n - 546 = 0$
Solving the quadratic equation $n^2 + 5n - 546 = 0$:
$n = \frac{-5 \pm \sqrt{25 - 4(1)(-546)}}{2} = \frac{-5 \pm \sqrt{2209}}{2} = \frac{-5 \pm 47}{2}$
Taking the positive value,$n = \frac{42}{2} = 21$.
Thus,the total saving will be $11040$ after $21$ months.

(A) The given equations are $x^2 + y^2 - ax = 0$ and $x^2 + y^2 = c^2$.
For the first circle,the centre $C_1 = (\frac{a}{2}, 0)$ and radius $r_1 = |\frac{a}{2}|$.
For the second circle,the centre $C_2 = (0, 0)$ and radius $r_2 = |c|$.
The distance between the centres is $d = \sqrt{(\frac{a}{2} - 0)^2 + (0 - 0)^2} = |\frac{a}{2}|$.
Two circles touch each other if the distance between their centres is equal to the sum or difference of their radii,i.e.,$d = |r_1 \pm r_2|$.
$|\frac{a}{2}| = ||\frac{a}{2}| \pm |c||$.
Case $1$: $|\frac{a}{2}| = |\frac{a}{2}| + |c| \Rightarrow |c| = 0$ (not possible for a circle).
Case $2$: $|\frac{a}{2}| = | |\frac{a}{2}| - |c| |$.
This implies $|\frac{a}{2}| = |c| - |\frac{a}{2}|$ (assuming $|c| > |\frac{a}{2}|$) or $|\frac{a}{2}| = |\frac{a}{2}| - |c|$ (not possible).
Thus,$2|\frac{a}{2}| = |c|$,which simplifies to $|a| = |c|$ or $|a| = c$ (since $c$ is a radius,$c > 0$).

(D) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$.
Since it passes through $(-3, 1)$,we have $\frac{9}{a^{2}} + \frac{1}{b^{2}} = 1$.
Given $e^{2} = \frac{2}{5}$,we know $b^{2} = a^{2}(1 - e^{2}) = a^{2}(1 - \frac{2}{5}) = \frac{3}{5}a^{2}$.
Substituting $b^{2}$ into the equation: $\frac{9}{a^{2}} + \frac{1}{\frac{3}{5}a^{2}} = 1$.
$\frac{9}{a^{2}} + \frac{5}{3a^{2}} = 1$ $\Rightarrow \frac{27 + 5}{3a^{2}} = 1$ $\Rightarrow 3a^{2} = 32$ $\Rightarrow a^{2} = \frac{32}{3}$.
Then $b^{2} = \frac{3}{5} \times \frac{32}{3} = \frac{32}{5}$.
The equation is $\frac{x^{2}}{32/3} + \frac{y^{2}}{32/5} = 1$,which simplifies to $\frac{3x^{2}}{32} + \frac{5y^{2}}{32} = 1$.
Thus,$3x^{2} + 5y^{2} = 32$ or $3x^{2} + 5y^{2} - 32 = 0$.

(D) Let $f(x) = \frac{\sqrt{1 - \cos\{2(x - 2)\}}}{x - 2}$.
Using the identity $1 - \cos(2\theta) = 2\sin^2(\theta)$,we get $1 - \cos\{2(x - 2)\} = 2\sin^2(x - 2)$.
Thus,$f(x) = \frac{\sqrt{2\sin^2(x - 2)}}{x - 2} = \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$.
For the Left Hand Limit $(LHL)$: $\mathop {\lim }\limits_{x \to 2^-} \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$. As $x \to 2^-$,$(x - 2) < 0$,so $|\sin(x - 2)| = -\sin(x - 2)$.
$LHL = \mathop {\lim }\limits_{x \to 2^-} \frac{-\sqrt{2}\sin(x - 2)}{x - 2} = -\sqrt{2}(1) = -\sqrt{2}$.
For the Right Hand Limit $(RHL)$: $\mathop {\lim }\limits_{x \to 2^+} \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$. As $x \to 2^+$,$(x - 2) > 0$,so $|\sin(x - 2)| = \sin(x - 2)$.
$RHL = \mathop {\lim }\limits_{x \to 2^+} \frac{\sqrt{2}\sin(x - 2)}{x - 2} = \sqrt{2}(1) = \sqrt{2}$.
Since $LHL \neq RHL$,the limit does not exist.

(D) Given that $\lim_{x \to 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}} = 0$.
Let $L = \lim_{x \to 5} f(x)$.
Since the limit exists,we can write $\lim_{x \to 5} ((f(x))^2 - 9) = L^2 - 9$.
If $L^2 - 9 \neq 0$,then the limit $\lim_{x \to 5} \frac{(f(x))^2 - 9}{\sqrt{|x - 5|}}$ would be of the form $\frac{k}{0}$ (where $k \neq 0$),which would be $\infty$.
Since the given limit is $0$,the numerator must approach $0$ as $x \to 5$.
Therefore,$L^2 - 9 = 0$,which implies $L^2 = 9$.
Since the codomain of $f$ is $[0, \infty)$,$L$ must be non-negative.
Thus,$L = 3$.

(B) The given numbers are $a, 2a, 3a, \dots, 50a$. The total number of terms is $n = 50$.
Since $n$ is even,the median is the average of the $25^{th}$ and $26^{th}$ terms:
$\text{Median} = \frac{25a + 26a}{2} = 25.5a$.
Mean deviation about the median is given by $\frac{1}{n} \sum_{i=1}^{n} |x_i - \text{Median}| = 50$.
$\frac{1}{50} \sum_{i=1}^{50} |ia - 25.5a| = 50$.
$\sum_{i=1}^{50} |i - 25.5| |a| = 2500$.
$|a| (|25.5 - 1| + |25.5 - 2| + \dots + |25.5 - 50|) = 2500$.
$|a| (24.5 + 23.5 + \dots + 0.5 + 0.5 + \dots + 24.5) = 2500$.
$|a| \times 2 \times (0.5 + 1.5 + \dots + 24.5) = 2500$.
$|a| \times 2 \times \frac{25}{2} (0.5 + 24.5) = 2500$.
$|a| \times 25 \times 25 = 2500$.
$|a| \times 625 = 2500$.
$|a| = 4$.

(A) Let the quadratic equation be $ax^{2} + bx + c = 0$.
Sachin made a mistake in the constant term,so the sum of the roots is correct.
Sum of roots $= 4 + 3 = 7$.
Rahul made a mistake in the coefficient of $x$,so the product of the roots is correct.
Product of roots $= 3 \times 2 = 6$.
The correct quadratic equation is $x^{2} - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^{2} - 7x + 6 = 0$.
Factoring the equation: $x^{2} - 6x - x + 6 = 0 \implies x(x - 6) - 1(x - 6) = 0 \implies (x - 6)(x - 1) = 0$.
Thus,the correct roots are $6$ and $1$.

(D) Given $p(x) = f(x) - g(x) = (a - a_1)x^2 + (b - b_1)x + (c - c_1).$
Since $p(x) = 0$ has only one root at $x = -1,$ the quadratic $p(x)$ must be a perfect square of the form $p(x) = k(x + 1)^2$ for some constant $k = a - a_1 \ne 0.$
Given $p(-2) = 2,$
$k(-2 + 1)^2 = 2 \Rightarrow k(-1)^2 = 2 \Rightarrow k = 2.$
Thus,$p(x) = 2(x + 1)^2.$
Now,we need to find $p(2):$
$p(2) = 2(2 + 1)^2 = 2(3)^2 = 2 \times 9 = 18.$

(A) Given that $f(x)$ is an odd function,we have $f(-x) = -f(x)$.
For any odd function,$f(0) = -f(0)$,which implies $2f(0) = 0$,so $f(0) = 0$.
Given that $f(x)$ is a periodic function with period $T = 1$,we have $f(x + 1) = f(x)$ for all $x$.
By the property of periodicity,$f(x + nT) = f(x)$ for any integer $n$.
Therefore,$f(2) = f(0 + 2 \times 1) = f(0)$.
Since $f(0) = 0$,it follows that $f(2) = 0$.

(A) Let the given statements be:
$P :$ Suman is brilliant
$Q :$ Suman is rich
$R :$ Suman is honest
The statement "Suman is brilliant and dishonest" is represented as $(P \wedge \sim R)$.
The statement "Suman is brilliant and dishonest if and only if Suman is rich" is represented as $(P \wedge \sim R) \leftrightarrow Q$.
Since the biconditional operator is commutative,this is equivalent to $Q \leftrightarrow (P \wedge \sim R)$.
The negation of a statement $S$ is denoted by $\sim S$.
Therefore,the negation of the given statement is $\sim (Q \leftrightarrow (P \wedge \sim R))$.

(D) The line is given by $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda}$. The direction vector of the line is $\vec{b} = (1, 2, \lambda)$.
The normal vector to the plane $x + 2y + 3z = 4$ is $\vec{n} = (1, 2, 3)$.
Let $\theta$ be the angle between the line and the plane. The formula for the angle is $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.
Given $\theta = \cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,we have $\cos \theta = \sqrt{\frac{5}{14}}$.
Using $\sin^2 \theta + \cos^2 \theta = 1$,we get $\sin \theta = \sqrt{1 - \frac{5}{14}} = \sqrt{\frac{9}{14}} = \frac{3}{\sqrt{14}}$.
Now,$\frac{|(1)(1) + (2)(2) + (3)(\lambda)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}} = \frac{3}{\sqrt{14}}$.
$\frac{|1 + 4 + 3\lambda|}{\sqrt{5 + \lambda^2} \sqrt{14}} = \frac{3}{\sqrt{14}}$.
$|5 + 3\lambda| = 3\sqrt{5 + \lambda^2}$.
Squaring both sides: $(5 + 3\lambda)^2 = 9(5 + \lambda^2)$.
$25 + 9\lambda^2 + 30\lambda = 45 + 9\lambda^2$.
$30\lambda = 20$.
$\lambda = \frac{20}{30} = \frac{2}{3}$.

(D) Let the line be $L: \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = k$. Any point on the line is $P(k, 2k + 1, 3k + 2)$.
For $A(1, 0, 7)$ to be the mirror image of $B(1, 6, 3)$ in line $L$,the line $AB$ must be perpendicular to $L$ and the midpoint of $AB$ must lie on $L$.
$1$. Midpoint of $AB$ is $M = (\frac{1+1}{2}, \frac{0+6}{2}, \frac{7+3}{2}) = (1, 3, 5)$.
Checking if $M$ lies on $L$: $\frac{1}{1} = \frac{3-1}{2} = \frac{5-2}{3} \implies 1 = 1 = 1$. Thus,$M$ lies on $L$.
$2$. Direction ratios of $AB$ are $(1-1, 6-0, 3-7) = (0, 6, -4)$.
Direction ratios of $L$ are $(1, 2, 3)$.
Dot product: $(0)(1) + (6)(2) + (-4)(3) = 0 + 12 - 12 = 0$.
Since the dot product is $0$,$AB$ is perpendicular to $L$.
Both conditions are satisfied,so Statement $-1$ is true. Statement $-2$ is also true because the line $L$ passes through the midpoint of $AB$ and is perpendicular to $AB$,which is the definition of a mirror image in a line. Thus,Statement $-2$ is the correct explanation for Statement $-1$.

(D) Let $I = \int_{0}^{1} \frac{8 \log(1+x)}{1+x^{2}} dx$.
Substitute $x = \tan \theta$,so $dx = \sec^{2} \theta d\theta$.
When $x=0, \theta=0$ and when $x=1, \theta=\frac{\pi}{4}$.
$I = 8 \int_{0}^{\pi/4} \frac{\log(1+\tan \theta)}{1+\tan^{2} \theta} \sec^{2} \theta d\theta = 8 \int_{0}^{\pi/4} \log(1+\tan \theta) d\theta$ ... $(i)$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$:
$I = 8 \int_{0}^{\pi/4} \log(1+\tan(\frac{\pi}{4}-\theta)) d\theta$.
Since $\tan(\frac{\pi}{4}-\theta) = \frac{1-\tan \theta}{1+\tan \theta}$,we have:
$I = 8 \int_{0}^{\pi/4} \log(1 + \frac{1-\tan \theta}{1+\tan \theta}) d\theta = 8 \int_{0}^{\pi/4} \log(\frac{2}{1+\tan \theta}) d\theta$.
$I = 8 \int_{0}^{\pi/4} (\log 2 - \log(1+\tan \theta)) d\theta$.
$I = 8 \log 2 [\theta]_{0}^{\pi/4} - 8 \int_{0}^{\pi/4} \log(1+\tan \theta) d\theta$.
$I = 8 \log 2 (\frac{\pi}{4}) - I$.
$2I = 2\pi \log 2 \Rightarrow I = \pi \log 2$.

(B) Given curves are $y = x$,$y = \frac{1}{x}$,$x = e$ and the positive $X$-axis $(y = 0)$.
The intersection point of $y = x$ and $y = \frac{1}{x}$ is found by setting $x = \frac{1}{x}$,which gives $x^2 = 1$. Since we are in the first quadrant,$x = 1$. Thus,the point of intersection is $(1, 1)$.
The region is bounded by $y = x$ from $x = 0$ to $x = 1$,and by $y = \frac{1}{x}$ from $x = 1$ to $x = e$.
Required Area = $\int_{0}^{1} x \, dx + \int_{1}^{e} \frac{1}{x} \, dx$
$= \left[ \frac{x^2}{2} \right]_{0}^{1} + [\ln |x|]_{1}^{e}$
$= \left( \frac{1}{2} - 0 \right) + (\ln e - \ln 1)$
$= \frac{1}{2} + (1 - 0) = \frac{1}{2} + 1 = \frac{3}{2} \text{ sq units}$.

(C) We need to evaluate the integral $I = \int_0^{1.5} x[x^2] dx$.
Since $[x^2]$ changes its value at $x^2 = 1$ and $x^2 = 2$,we split the integral at $x = 1$ and $x = \sqrt{2}$.
$I = \int_0^1 x[x^2] dx + \int_1^{\sqrt{2}} x[x^2] dx + \int_{\sqrt{2}}^{1.5} x[x^2] dx$.
For $0 \le x < 1$,$0 \le x^2 < 1$,so $[x^2] = 0$.
For $1 \le x < \sqrt{2}$,$1 \le x^2 < 2$,so $[x^2] = 1$.
For $\sqrt{2} \le x < 1.5$,$2 \le x^2 < 2.25$,so $[x^2] = 2$.
Thus,$I = \int_0^1 x(0) dx + \int_1^{\sqrt{2}} x(1) dx + \int_{\sqrt{2}}^{1.5} x(2) dx$.
$I = 0 + \left[ \frac{x^2}{2} \right]_1^{\sqrt{2}} + 2 \left[ \frac{x^2}{2} \right]_{\sqrt{2}}^{1.5}$.
$I = \left( \frac{2-1}{2} \right) + (1.5^2 - (\sqrt{2})^2) = \frac{1}{2} + (2.25 - 2) = 0.5 + 0.25 = 0.75 = \frac{3}{4}$.

(B) To find the intersection points,substitute $y = \frac{x^2}{4}$ into $y^2 = 4x$:
$(\frac{x^2}{4})^2 = 4x$
$\frac{x^4}{16} = 4x$
$x^4 = 64x$
$x(x^3 - 64) = 0$
So,$x = 0$ or $x = 4$.
When $x = 0, y = 0$. When $x = 4, y = 4$.
The intersection points are $(0,0)$ and $(4,4)$.
The area enclosed is given by:
$\text{Area} = \int_{0}^{4} (\sqrt{4x} - \frac{x^2}{4}) dx$
$= \int_{0}^{4} (2\sqrt{x} - \frac{x^2}{4}) dx$
$= [2 \cdot \frac{2}{3} x^{3/2} - \frac{1}{4} \cdot \frac{x^3}{3}]_{0}^{4}$
$= [\frac{4}{3} x^{3/2} - \frac{x^3}{12}]_{0}^{4}$
$= (\frac{4}{3} \cdot 8 - \frac{64}{12}) - 0$
$= \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \text{ sq. units}$.

(C) Given that $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
For Statement $-1$:
Consider $(A(BA))^{\prime} = (BA)^{\prime} A^{\prime} = (A^{\prime} B^{\prime}) A^{\prime} = (AB)A = A(BA)$.
Since $(A(BA))^{\prime} = A(BA)$,$A(BA)$ is symmetric.
Similarly,$((AB)A)^{\prime} = A^{\prime} (AB)^{\prime} = A(B^{\prime} A^{\prime}) = A(BA) = (AB)A$.
Thus,$(AB)A$ is also symmetric. So,Statement $-1$ is true.
For Statement $-2$:
Consider $(AB)^{\prime} = B^{\prime} A^{\prime} = BA$.
If $AB$ is symmetric,then $(AB)^{\prime} = AB$,which implies $BA = AB$.
Conversely,if $BA = AB$,then $(AB)^{\prime} = BA = AB$,so $AB$ is symmetric.
Thus,Statement $-2$ is true.
However,Statement $-2$ describes the condition for $AB$ to be symmetric,whereas Statement $-1$ deals with the symmetry of $A(BA)$ and $(AB)A$ regardless of commutativity. Therefore,Statement $-2$ is not a correct explanation for Statement $-1$.

(A) For reflexive property:
$(A, A) \in R$ because $A = I^{-1}AI$,where $I$ is the identity matrix,which is invertible. Thus,$R$ is reflexive.
For symmetric property:
If $(A, B) \in R$,then $A = P^{-1}BP$ for some invertible matrix $P$.
Multiplying by $P$ on the left and $P^{-1}$ on the right: $PAP^{-1} = B$.
Let $Q = P^{-1}$. Since $P$ is invertible,$Q$ is also invertible.
Then $B = Q^{-1}AQ$,so $(B, A) \in R$. Thus,$R$ is symmetric.
For transitive property:
If $(A, B) \in R$ and $(B, C) \in R$,then $A = P^{-1}BP$ and $B = Q^{-1}CQ$ for some invertible matrices $P$ and $Q$.
Substituting $B$: $A = P^{-1}(Q^{-1}CQ)P = (QP)^{-1}C(QP)$.
Since $QP$ is invertible,$(A, C) \in R$. Thus,$R$ is transitive.
Therefore,$R$ is an equivalence relation. Statement-$1$ is true.
Statement-$2$ is a standard property of invertible matrices: $(MN)^{-1} = N^{-1}M^{-1}$. This is true.
However,Statement-$2$ is a general property of matrix inversion and is not the specific reason why the relation $R$ (similarity) is an equivalence relation. Thus,Statement-$2$ is not the correct explanation for Statement-$1$.

(A) For Statement-$1$: $A$ skew-symmetric matrix $A$ satisfies $A^T = -A$. Taking the determinant on both sides,we get $\det(A^T) = \det(-A)$. Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$ for a matrix of order $n$,we have $\det(A) = (-1)^n \det(A)$. For $n=3$,$\det(A) = -\det(A)$,which implies $2 \det(A) = 0$,so $\det(A) = 0$. Thus,Statement-$1$ is true.
For Statement-$2$: The property $\det(A^T) = \det(A)$ is always true. The property $\det(-A) = (-1)^n \det(A)$ is also true for a matrix of order $n$. Therefore,Statement-$2$ is true.
Since Statement-$2$ provides the mathematical properties used to prove Statement-$1$,it is the correct explanation for Statement-$1$.

(C) Given $H = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$.
We calculate the powers of $H$:
$H^2 = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = \begin{bmatrix} \omega^2 & 0 \\ 0 & \omega^2 \end{bmatrix}$.
By induction,$H^n = \begin{bmatrix} \omega^n & 0 \\ 0 & \omega^n \end{bmatrix}$.
For $n = 70$,$H^{70} = \begin{bmatrix} \omega^{70} & 0 \\ 0 & \omega^{70} \end{bmatrix}$.
Since $\omega^3 = 1$,we have $\omega^{70} = (\omega^3)^{23} \cdot \omega = (1)^{23} \cdot \omega = \omega$.
Therefore,$H^{70} = \begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix} = H$.

(A) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = q$.
First,calculate the Right Hand Limit ($R$.$H$.$L$.):
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x \cdot \sqrt{x}} = \lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x}$.
Multiplying by the conjugate $\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$,we get:
$\lim_{x \to 0^+} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0^+} \frac{x}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0^+} \frac{1}{\sqrt{1+x} + 1} = \frac{1}{1+1} = \frac{1}{2}$.
Thus,$q = 1/2$.
Next,calculate the Left Hand Limit ($L$.$H$.$L$.):
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(p+1)x + \sin x}{x} = \lim_{x \to 0^-} \left( \frac{\sin(p+1)x}{x} + \frac{\sin x}{x} \right)$.
Using $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we get:
$(p+1) + 1 = p+2$.
Since $L.H.L. = q$,we have $p+2 = 1/2$,which implies $p = 1/2 - 2 = -3/2$.
Therefore,the values are $(p, q) = (-3/2, 1/2)$.

(C) We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$.
Now,differentiate with respect to $y$:
$\frac{d^2x}{dy^2} = \frac{d}{dy} \left( \frac{dx}{dy} \right) = \frac{d}{dy} \left( \left( \frac{dy}{dx} \right)^{-1} \right)$.
Using the chain rule,$\frac{d}{dy} = \frac{dx}{dy} \cdot \frac{d}{dx} = \frac{1}{\frac{dy}{dx}} \cdot \frac{d}{dx}$:
$\frac{d^2x}{dy^2} = \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^{-1} \right)$.
Applying the power rule:
$\frac{d^2x}{dy^2} = \left( \frac{1}{\frac{dy}{dx}} \right) \cdot \left( -1 \left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} \right)$.
Simplifying the expression:
$\frac{d^2x}{dy^2} = - \left( \frac{dy}{dx} \right)^{-1} \cdot \left( \frac{dy}{dx} \right)^{-2} \cdot \frac{d^2y}{dx^2} = - \left( \frac{dy}{dx} \right)^{-3} \left( \frac{d^2y}{dx^2} \right)$.

(D) Given the differential equation $\frac{dy}{dx} = y + 3$.
Separating the variables,we get $\frac{dy}{y + 3} = dx$.
Integrating both sides,we obtain $\int \frac{dy}{y + 3} = \int dx$,which gives $\ln|y + 3| = x + C$.
Since $y + 3 > 0$,we have $y + 3 = e^{x + C} = e^C \cdot e^x$.
Let $e^C = A$,so $y + 3 = A e^x$.
Using the initial condition $y(0) = 2$,we substitute $x = 0$ and $y = 2$:
$2 + 3 = A e^0 \Rightarrow A = 5$.
Thus,the particular solution is $y + 3 = 5 e^x$,or $y = 5 e^x - 3$.
To find $y(\ln 2)$,substitute $x = \ln 2$:
$y(\ln 2) = 5 e^{\ln 2} - 3$.
Since $e^{\ln 2} = 2$,we have $y(\ln 2) = 5(2) - 3 = 10 - 3 = 7$.

(A) Given the differential equation: $\frac{dV}{dt} = -k(T - t)$.
Integrating both sides with respect to $t$:
$V(t) = \int -k(T - t) dt = k \int (T - t) d(T - t) = \frac{k(T - t)^2}{2} + C$.
At $t = 0$,the value of the equipment is the purchase value $I$,so $V(0) = I$:
$I = \frac{k(T - 0)^2}{2} + C \implies I = \frac{kT^2}{2} + C \implies C = I - \frac{kT^2}{2}$.
Substituting $C$ back into the equation for $V(t)$:
$V(t) = \frac{k(T - t)^2}{2} + I - \frac{kT^2}{2}$.
The scrap value $V(T)$ is the value at $t = T$:
$V(T) = \frac{k(T - T)^2}{2} + I - \frac{kT^2}{2} = 0 + I - \frac{kT^2}{2} = I - \frac{kT^2}{2}$.

(C) Given the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$.
Dividing by $y^2 dy$,we get $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{y^2}$ and $Q(y) = \frac{1}{y^3}$.
The Integrating Factor $(IF)$ is $e^{\int P(y) dy} = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}}$.
The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.
$x e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy + C$.
Let $t = -\frac{1}{y}$,then $dt = \frac{1}{y^2} dy$. Also,$\frac{1}{y} = -t$.
Substituting these,$x e^{-\frac{1}{y}} = \int (-t) e^t dt + C = - (t e^t - e^t) + C = (1 - t) e^t + C$.
$x e^{-\frac{1}{y}} = (1 + \frac{1}{y}) e^{-\frac{1}{y}} + C$.
Using $y(1) = 1$,we have $1 \cdot e^{-1} = (1 + 1) e^{-1} + C \Rightarrow e^{-1} = 2e^{-1} + C \Rightarrow C = -e^{-1} = -\frac{1}{e}$.
Thus,$x e^{-\frac{1}{y}} = (1 + \frac{1}{y}) e^{-\frac{1}{y}} - \frac{1}{e}$.
Dividing by $e^{-\frac{1}{y}}$,we get $x = 1 + \frac{1}{y} - \frac{e^{\frac{1}{y}}}{e}$.

(A) The conditional probability of an event $C$ given that event $D$ has occurred is defined as $P(C|D) = \frac{P(C \cap D)}{P(D)}$.
If $C \subset D$,then $C \cap D = C$,which implies $P(C \cap D) = P(C)$.
Substituting this into the formula,we get $P(C|D) = \frac{P(C)}{P(D)}$.
Since $D$ is an event,$P(D) \le 1$. Therefore,$\frac{1}{P(D)} \ge 1$.
Multiplying both sides by $P(C)$,we get $\frac{P(C)}{P(D)} \ge P(C)$.
Thus,$P(C|D) \ge P(C)$.

(B) Let $X$ be the number of successes in $5$ independent Bernoulli trials. The probability of success is $p$ and the probability of failure is $q = 1 - p$.
The probability of at least one failure is given by $1 - P(\text{no failure}) = 1 - P(X = 5)$.
Since the trials are independent,$P(X = 5) = p^5$.
According to the problem,the probability of at least one failure is at least $\frac{31}{32}$:
$1 - p^5 \geq \frac{31}{32}$
Rearranging the inequality:
$1 - \frac{31}{32} \geq p^5$
$\frac{1}{32} \geq p^5$
Taking the fifth root on both sides:
$p \leq (\frac{1}{32})^{1/5}$
$p \leq \frac{1}{2}$
Since $p$ is a probability,$p \geq 0$. Therefore,$p \in [0, \frac{1}{2}]$.

(A) We need to find $P(A' \cap B' | C)$. By the definition of conditional probability:
$P(A' \cap B' | C) = \frac{P(A' \cap B' \cap C)}{P(C)}$
Using the property $P(X \cap Y \cap Z) = P(Z) - P(X' \cap Z) - P(Y' \cap Z) + P(X' \cap Y' \cap Z)$ is not direct,so we use the inclusion-exclusion principle for sets:
$P(A' \cap B' \cap C) = P(C \setminus ((A \cup B) \cap C)) = P(C) - P((A \cup B) \cap C)$
$P(A' \cap B' \cap C) = P(C) - [P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)]$
Given $P(A \cap B \cap C) = 0$ and $A, B, C$ are pairwise independent,we have $P(A \cap C) = P(A)P(C)$ and $P(B \cap C) = P(B)P(C)$.
Substituting these values:
$P(A' \cap B' \cap C) = P(C) - P(A)P(C) - P(B)P(C) + 0$
$P(A' \cap B' \cap C) = P(C)(1 - P(A) - P(B))$
Therefore,$P(A' \cap B' | C) = \frac{P(C)(1 - P(A) - P(B))}{P(C)} = 1 - P(A) - P(B)$
Since $1 - P(A) = P(A')$,we get $P(A' \cap B' | C) = P(A') - P(B)$.

(A) The function is defined as $f(x) = \frac{1}{\sqrt{x-[x]}}$.
Domain of $f$:
We know that $0 \leq x-[x] < 1$ for all $x \in \mathbb{R}$.
Also,$x-[x] = 0$ when $x \in \mathbb{Z}$.
Since the denominator cannot be zero,$x-[x] > 0$,which implies $x \notin \mathbb{Z}$.
Thus,the domain is $\mathbb{R} - \mathbb{Z}$.
Range of $f$:
For $x \in \mathbb{R} - \mathbb{Z}$,we have $0 < x-[x] < 1$.
Taking the square root,$0 < \sqrt{x-[x]} < 1$.
Taking the reciprocal,$\frac{1}{\sqrt{x-[x]}} > 1$.
Therefore,$f(x) \in (1, \infty)$.
Hence,the range of $f$ is $(1, \infty)$.

(D) Given $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.
To find $f^{-1}(x)$,set $y = (x - 1)^2 + 1$.
Then $y - 1 = (x - 1)^2$. Since $x \ge 1$,we have $x - 1 = \sqrt{y - 1}$,so $x = 1 + \sqrt{y - 1}$.
Thus,$f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$. So,Statement-$2$ is true.
To find $S = \{x : f(x) = f^{-1}(x)\}$,we solve $f(x) = x$.
$(x - 1)^2 + 1 = x \implies x^2 - 2x + 1 + 1 = x \implies x^2 - 3x + 2 = 0$.
$(x - 1)(x - 2) = 0$,so $x = 1$ or $x = 2$.
Since both values satisfy $x \ge 1$,$S = \{1, 2\}$. So,Statement-$1$ is true.
Since $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ for increasing functions,Statement-$2$ explains Statement-$1$.

(C) Given $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$.
Taking the cross product with $\vec{a}$ on both sides: $\vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times (\vec{b} \times \vec{d})$.
Using the vector triple product formula $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$,we get:
$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (\vec{a} \cdot \vec{d}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{d}$.
Since $\vec{a} \cdot \vec{d} = 0$,the equation simplifies to:
$(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = 0 - (\vec{a} \cdot \vec{b}) \vec{d}$.
Rearranging for $\vec{d}$:
$(\vec{a} \cdot \vec{b}) \vec{d} = (\vec{a} \cdot \vec{b}) \vec{c} - (\vec{a} \cdot \vec{c}) \vec{b}$.
Dividing by $(\vec{a} \cdot \vec{b})$ (which is non-zero as $\vec{a}$ and $\vec{b}$ are not perpendicular):
$\vec{d} = \vec{c} - \left( \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \right) \vec{b}$.