TS EAMCET 2016 Mathematics Question Paper with Answer and Solution

(C) The expression $|z-1|+|z-5|$ represents the sum of the distances of a complex number $z$ from the points $z_1 = 1$ and $z_2 = 5$ in the complex plane.
By the triangle inequality,for any points $z, z_1, z_2$,we have $|z-z_1| + |z-z_2| \ge |z_1 - z_2|$.
Here,$|z_1 - z_2| = |1 - 5| = |-4| = 4$.
The minimum value occurs when $z$ lies on the line segment connecting $1$ and $5$.
Thus,the minimum value is $4$.

(D) First,we find the prime factorization of $7!$.
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$7! = 7 \times (2 \times 3) \times 5 \times 2^2 \times 3 \times 2 \times 1$
$7! = 2^{1+2+1} \times 3^{1+1} \times 5^1 \times 7^1$
$7! = 2^4 \times 3^2 \times 5^1 \times 7^1$
The number of divisors of a number $N = p_1^{a} \times p_2^{b} \times p_3^{c} \times p_4^{d}$ is given by $(a+1)(b+1)(c+1)(d+1)$.
Here,$a=4, b=2, c=1, d=1$.
Number of divisors $= (4+1)(2+1)(1+1)(1+1) = 5 \times 3 \times 2 \times 2 = 60$.

(D) The condition for two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to be orthogonal is $2g_1g_2+2f_1f_2=c_1+c_2$.
First,rewrite the equations in standard form $x^2+y^2+2gx+2fy+c=0$.
For the first circle: $x^2+y^2-2\lambda x-2y-7=0$,we have $g_1=-\lambda, f_1=-1, c_1=-7$.
For the second circle: $3(x^2+y^2)-8x+29y=0$,divide by $3$ to get $x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0$. Thus,$g_2=-\frac{4}{3}, f_2=\frac{29}{6}, c_2=0$.
Applying the condition $2g_1g_2+2f_1f_2=c_1+c_2$:
$2(-\lambda)(-\frac{4}{3}) + 2(-1)(\frac{29}{6}) = -7 + 0$
$\frac{8\lambda}{3} - \frac{29}{3} = -7$
Multiply by $3$: $8\lambda - 29 = -21$
$8\lambda = 8$
$\lambda = 1$.

(C) Given the inequality $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.
Multiplying by $x$ (since $x > 0$),we get $9mx^2 - x + 1 \geq 0$.
For a quadratic expression $ax^2 + bx + c \geq 0$ to hold for all $x > 0$,we analyze the discriminant $D = b^2 - 4ac$.
Here,$a = 9m$,$b = -1$,and $c = 1$.
The discriminant $D = (-1)^2 - 4(9m)(1) = 1 - 36m$.
For the quadratic to be non-negative for all $x$,we require $D \leq 0$.
$1 - 36m \leq 0 \implies 36m \geq 1 \implies m \geq \frac{1}{36}$.
Thus,the smallest value of $m$ is $\frac{1}{36}$.

(B) Given the equation $x^2-x+1=0$.
The roots of this equation are $\omega$ and $\omega^2$,where $\omega$ is the complex cube root of unity.
Let $\alpha = \omega$ and $\beta = \omega^2$.
We need to find the equation with roots $\alpha^{2015}$ and $\beta^{2015}$.
$\alpha^{2015} = \omega^{2015} = \omega^{3 \times 671 + 2} = \omega^2$.
$\beta^{2015} = (\omega^2)^{2015} = \omega^{4030} = \omega^{3 \times 1343 + 1} = \omega$.
The sum of the new roots is $\alpha^{2015} + \beta^{2015} = \omega^2 + \omega = -1$.
The product of the new roots is $\alpha^{2015} \cdot \beta^{2015} = \omega^2 \cdot \omega = \omega^3 = 1$.
The required quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

(C) Given equation: $|x|^2-5|x|+6=0$
Let $|x|=y$. Since $|x| \ge 0$,we have $y \ge 0$.
The equation becomes $y^2-5y+6=0$.
Factoring the quadratic: $(y-2)(y-3)=0$.
This gives $y=2$ or $y=3$.
Substituting back $|x|=y$:
Case $1$: $|x|=2 \Rightarrow x = \pm 2$.
Case $2$: $|x|=3 \Rightarrow x = \pm 3$.
Thus,the real roots are $x \in \{-3, -2, 2, 3\}$.
Therefore,the number of real roots is $4$.

(C) Given,the roots of $x^3+x^2+2x+3=0$ are $\alpha, \beta$ and $\gamma$.
From Vieta's formulas:
$\alpha+\beta+\gamma = -1$
$\alpha\beta+\beta\gamma+\gamma\alpha = 2$
$\alpha\beta\gamma = -3$
Let the roots of $f(x)=0$ be $S_1 = \alpha+\beta$,$S_2 = \beta+\gamma$,$S_3 = \gamma+\alpha$.
Note that $S_1 = -1-\gamma$,$S_2 = -1-\alpha$,$S_3 = -1-\beta$.
The sum of roots of $f(x)$ is $S_1+S_2+S_3 = 2(\alpha+\beta+\gamma) = 2(-1) = -2$.
The sum of roots taken two at a time is $S_1S_2+S_2S_3+S_3S_1 = (-1-\gamma)(-1-\alpha) + (-1-\alpha)(-1-\beta) + (-1-\beta)(-1-\gamma)$
$= (1+\alpha+\gamma+\alpha\gamma) + (1+\alpha+\beta+\alpha\beta) + (1+\beta+\gamma+\beta\gamma)$
$= 3 + 2(\alpha+\beta+\gamma) + (\alpha\beta+\beta\gamma+\gamma\alpha) = 3 + 2(-1) + 2 = 3$.
The product of roots is $S_1S_2S_3 = (-1-\gamma)(-1-\alpha)(-1-\beta) = -((1+\gamma)(1+\alpha)(1+\beta))$
$= -(1 + (\alpha+\beta+\gamma) + (\alpha\beta+\beta\gamma+\gamma\alpha) + \alpha\beta\gamma) = -(1 - 1 + 2 - 3) = -(-1) = 1$.
Thus,$f(x) = x^3 - (\text{sum of roots})x^2 + (\text{sum of roots taken two at a time})x - (\text{product of roots})$
$f(x) = x^3 - (-2)x^2 + 3x - 1 = x^3+2x^2+3x-1$.

(D) Given the equation $x^3-5x+4=0$ with roots $\alpha, \beta, \gamma$.
Comparing with $x^3+px^2+qx+r=0$,we have the sum of roots $\alpha+\beta+\gamma = 0$.
Since $\alpha+\beta+\gamma = 0$,we use the identity: if $a+b+c=0$,then $a^3+b^3+c^3 = 3abc$.
Here,$\alpha^3+\beta^3+\gamma^3 = 3\alpha\beta\gamma$.
From the given equation,the product of roots $\alpha\beta\gamma = -(\text{constant term}) = -4$.
Therefore,$\alpha^3+\beta^3+\gamma^3 = 3(-4) = -12$.
Finally,$(\alpha^3+\beta^3+\gamma^3)^2 = (-12)^2 = 144$.

(B) Given,$\bar{z}^{\frac{1}{3}} = a+ib$.
Taking the cube on both sides,we get $\bar{z} = (a+ib)^3$.
Since $\bar{z} = x-iy$,we have $x-iy = (a+ib)^3$.
Expanding the right side: $x-iy = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3$.
$x-iy = a^3 + 3a^2bi - 3ab^2 - ib^3$.
Grouping real and imaginary parts: $x-iy = (a^3-3ab^2) + i(3a^2b-b^3)$.
Comparing real and imaginary parts,we get $x = a^3-3ab^2$ and $-y = 3a^2b-b^3$,which implies $y = b^3-3a^2b$.
Now,$\frac{x}{a} = a^2-3b^2$ and $\frac{y}{b} = b^2-3a^2$.
Adding these,$\frac{x}{a} + \frac{y}{b} = (a^2-3b^2) + (b^2-3a^2) = -2a^2-2b^2 = -2(a^2+b^2)$.
Therefore,$\frac{1}{a^2+b^2}\left(\frac{x}{a}+\frac{y}{b}\right) = \frac{-2(a^2+b^2)}{a^2+b^2} = -2$.

(C) The equation of a circle with center at the origin and radius $r$ is given by $|z| = r$.
Given $z = (1+i)(1+2i)(1+3i) \ldots (1+10i)$.
Taking the modulus on both sides:
$|z| = |1+i| \cdot |1+2i| \cdot |1+3i| \ldots |1+10i| = r$.
Using the property $|a+bi| = \sqrt{a^2+b^2}$,we get:
$|z| = \sqrt{1^2+1^2} \cdot \sqrt{1^2+2^2} \cdot \sqrt{1^2+3^2} \ldots \sqrt{1^2+10^2} = r$.
$|z| = \sqrt{2} \cdot \sqrt{5} \cdot \sqrt{10} \ldots \sqrt{101} = r$.
Squaring both sides,we obtain:
$r^2 = 2 \times 5 \times 10 \times \ldots \times 101$.

(C) Let $z = x + iy$.
The given equation is $|z| + |z - 1| = 3$.
This represents the sum of distances of a point $z$ from the points $0$ and $1$ being constant $(3)$.
Since $3 > |0 - 1| = 1$,the locus is an ellipse with foci at $0$ and $1$.
Algebraically:
$\sqrt{x^2 + y^2} + \sqrt{(x - 1)^2 + y^2} = 3$
$\sqrt{(x - 1)^2 + y^2} = 3 - \sqrt{x^2 + y^2}$
Squaring both sides:
$(x - 1)^2 + y^2 = 9 + x^2 + y^2 - 6\sqrt{x^2 + y^2}$
$x^2 - 2x + 1 + y^2 = 9 + x^2 + y^2 - 6\sqrt{x^2 + y^2}$
$-2x + 1 = 9 - 6\sqrt{x^2 + y^2}$
$6\sqrt{x^2 + y^2} = 2x + 8$
$3\sqrt{x^2 + y^2} = x + 4$
Squaring again:
$9(x^2 + y^2) = (x + 4)^2$
$9x^2 + 9y^2 = x^2 + 8x + 16$
$8x^2 - 8x + 9y^2 = 16$
$8(x^2 - x + \frac{1}{4}) + 9y^2 = 16 + 2$
$8(x - \frac{1}{2})^2 + 9y^2 = 18$
$\frac{(x - 1/2)^2}{9/4} + \frac{y^2}{2} = 1$
This is the standard equation of an ellipse.

(C) The word $EQUATION$ consists of $8$ distinct letters: $E, Q, U, A, T, I, O, N$.
Total number of $4$ letter words that can be formed using these $8$ letters (with repetition allowed) is $8^4 = 4096$.
Number of $4$ letter words that can be formed with distinct letters (no repetition) is given by the permutation formula $^8P_4 = 8 \times 7 \times 6 \times 5 = 1680$.
The number of $4$ letter words with at least one letter repeated is the total number of words minus the number of words with no letters repeated:
$= 8^4 - ^8P_4 = 4096 - 1680 = 2416$.

(B) The given sequence is an arithmetic progression with first term $a = 148$ and common difference $d = -2$.
The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values,$S_n = \frac{n}{2}[2(148) + (n-1)(-2)] = \frac{n}{2}[296 - 2n + 2] = \frac{n}{2}[298 - 2n] = n(149 - n)$.
The average of the first $n$ terms is $\frac{S_n}{n} = 149 - n$.
Given that the average is $125$,we have $149 - n = 125$.
Therefore,$n = 149 - 125 = 24$.

(B) We have $7^n = (1 + 6)^n$.
By the Binomial Theorem,$7^n = 1 + ^nC_1(6) + ^nC_2(6^2) + ^nC_3(6^3) + \dots + ^nC_n(6^n)$.
$7^n = 1 + 6n + 36(^nC_2 + ^nC_3(6) + \dots + ^nC_n(6^{n-2}))$.
Let $\lambda = ^nC_2 + ^nC_3(6) + \dots + ^nC_n(6^{n-2})$.
Then $7^n = 1 + 6n + 36\lambda$.
Rearranging,we get $7^n - 6n = 36\lambda + 1$.
Now,subtract $50$ from both sides:
$7^n - 6n - 50 = 36\lambda + 1 - 50 = 36\lambda - 49$.
We can write $-49$ as $-72 + 23$.
$7^n - 6n - 50 = 36\lambda - 72 + 23 = 36(\lambda - 2) + 23$.
Let $\mu = \lambda - 2$.
Then $7^n - 6n - 50 = 36\mu + 23$.
Therefore,when $7^n - 6n - 50$ is divided by $36$,the remainder is $23$.

(B) Consider the given expression: $S = \sum_{r=0}^n (3r-1) C_r^2$.
We know that $C_r^2 = C_r \cdot C_{n-r}$.
So,$S = 3 \sum_{r=0}^n r C_r^2 - \sum_{r=0}^n C_r^2$.
Using the identity $\sum_{r=0}^n C_r^2 = { }^{2n} C_n$ and $r C_r = n { }^{n-1} C_{r-1}$,we have:
$S = 3 \sum_{r=1}^n n { }^{n-1} C_{r-1} C_r - { }^{2n} C_n$.
$S = 3n \sum_{r=1}^n { }^{n-1} C_{r-1} { }^{n} C_{n-r} - { }^{2n} C_n$.
The summation $\sum_{r=1}^n { }^{n-1} C_{r-1} { }^{n} C_{n-r}$ represents the coefficient of $x^{n-1}$ in the expansion of $(1+x)^{n-1}(1+x)^n = (1+x)^{2n-1}$,which is ${ }^{2n-1} C_{n-1}$.
Thus,$S = 3n { }^{2n-1} C_{n-1} - { }^{2n} C_n$.
Since ${ }^{2n-1} C_{n-1} = \frac{n}{2n} { }^{2n} C_n = \frac{1}{2} { }^{2n} C_n$,we get:
$S = 3n \left( \frac{1}{2} { }^{2n} C_n \right) - { }^{2n} C_n = \left( \frac{3n}{2} - 1 \right) { }^{2n} C_n = \left( \frac{3n-2}{2} \right) { }^{2n} C_n$.

(A) The given series is of the form $1 + nx + \frac{n(n+1)}{2!}x^2 + \frac{n(n+1)(n+2)}{3!}x^3 + \dots = (1-x)^{-n}$.
Comparing the terms,we have $nx = \frac{2}{3} \times \frac{1}{8} = \frac{1}{12}$ and $x = \frac{1}{8}$.
Thus,$n \times \frac{1}{8} = \frac{1}{12} \implies n = \frac{8}{12} = \frac{2}{3}$.
The series is $(1 - \frac{1}{8})^{-\frac{2}{3}}$.
$= (\frac{7}{8})^{-\frac{2}{3}} = (\frac{8}{7})^{\frac{2}{3}} = \frac{8^{\frac{2}{3}}}{7^{\frac{2}{3}}} = \frac{(2^3)^{\frac{2}{3}}}{7^{\frac{2}{3}}} = \frac{2^2}{\sqrt[3]{49}} = \frac{4}{\sqrt[3]{49}}$.

(D) Given $f(x) = \cos^2 x + \cos^2 2x + \cos^2 3x = 1$.
Using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$,we have:
$f(x) = \frac{1+\cos 2x}{2} + \frac{1+\cos 4x}{2} + \cos^2 3x = 1$
$1 + \frac{1}{2}(\cos 2x + \cos 4x) + \cos^2 3x = 1$
$\frac{1}{2}(2\cos 3x \cos x) + \cos^2 3x = 0$
$\cos 3x (\cos x + \cos 3x) = 0$
$\cos 3x (2 \cos 2x \cos x) = 0$
$2 \cos 3x \cos 2x \cos x = 0$
This implies $\cos 3x = 0$ or $\cos 2x = 0$ or $\cos x = 0$.
For $x \in [0, 2\pi]$:
$1$. $\cos x = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}$
$2$. $\cos 2x = 0$ $\Rightarrow 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$ $\Rightarrow x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$
$3$. $\cos 3x = 0$ $\Rightarrow 3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$ $\Rightarrow x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$
Combining all unique values: $x \in \{\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, \frac{11\pi}{6}\}$.
Counting these,there are $10$ distinct values.

(D) Given,$\cos x+\cos y=-\cos \alpha$ $(i)$
Using the formula $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\cos \alpha$
Also,$\sin x+\sin y=-\sin \alpha$ (ii)
Using the formula $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get:
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\sin \alpha$
Dividing equation $(i)$ by equation (ii):
$\frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{-\cos \alpha}{-\sin \alpha}$
$\cot \left(\frac{x+y}{2}\right) = \cot \alpha$

(C) Given expression: $\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$
Divide the numerator and denominator of the first term by $\cos 13^{\circ}$:
$= \frac{1-\tan 13^{\circ}}{1+\tan 13^{\circ}} + \tan 148^{\circ}$
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B}$,where $A=45^{\circ}$ and $B=13^{\circ}$:
$= \tan(45^{\circ}-13^{\circ}) + \tan(180^{\circ}-32^{\circ})$
$= \tan 32^{\circ} - \tan 32^{\circ} = 0$

(B) Using the identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we have $\cos^3 x = \frac{\cos 3x + 3 \cos x}{4}$.
Applying this to each term:
$\frac{\cos 3 \theta + 3 \cos \theta}{4} + \frac{\cos(2 \pi + 3 \theta) + 3 \cos(\frac{2 \pi}{3} + \theta)}{4} + \frac{\cos(4 \pi + 3 \theta) + 3 \cos(\frac{4 \pi}{3} + \theta)}{4} = \alpha \cos 3 \theta$
Since $\cos(2 \pi + 3 \theta) = \cos 3 \theta$ and $\cos(4 \pi + 3 \theta) = \cos 3 \theta$,the equation becomes:
$\frac{3 \cos 3 \theta + 3 \cos \theta + 3 \cos(\frac{2 \pi}{3} + \theta) + 3 \cos(\frac{4 \pi}{3} + \theta)}{4} = \alpha \cos 3 \theta$
Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:
$\cos(\frac{2 \pi}{3} + \theta) + \cos(\frac{4 \pi}{3} + \theta) = 2 \cos(\pi + \theta) \cos(-\frac{\pi}{3}) = 2(-\cos \theta)(\frac{1}{2}) = -\cos \theta$
Substituting this back:
$\frac{3 \cos 3 \theta + 3 \cos \theta + 3(-\cos \theta)}{4} = \alpha \cos 3 \theta$
$\frac{3 \cos 3 \theta}{4} = \alpha \cos 3 \theta$
Thus,$\alpha = \frac{3}{4}$.

(B) Given points are $(3,3)$ and $(7,6)$.
The slope of the line $m = \frac{6-3}{7-3} = \frac{3}{4}$.
The equation of the line is $(y-3) = \frac{3}{4}(x-3)$,which simplifies to $4y - 12 = 3x - 9$,or $3x - 4y = -3$.
To find the $y$-intercept,set $x=0$: $3(0) - 4y = -3 \Rightarrow y = \frac{3}{4}$. The point is $(0, \frac{3}{4})$.
To find the $x$-intercept,set $y=0$: $3x - 4(0) = -3 \Rightarrow x = -1$. The point is $(-1, 0)$.
The length of the segment between $(0, \frac{3}{4})$ and $(-1, 0)$ is given by the distance formula:
$d = \sqrt{(0 - (-1))^2 + (\frac{3}{4} - 0)^2} = \sqrt{1^2 + (\frac{3}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(C) The sides of the triangle are given by the equations:
$(x-y)(x+y)(2x+3y-6)=0$
This implies the lines are $L_1: x-y=0$,$L_2: x+y=0$,and $L_3: 2x+3y-6=0$.
The vertices of the triangle are the intersection points of these lines:
Intersection of $L_1$ and $L_2$: $(0,0)$
Intersection of $L_1$ and $L_3$: $x-y=0$ and $2x+3y-6=0 \implies 2x+3x-6=0 \implies 5x=6 \implies x=6/5, y=6/5$
Intersection of $L_2$ and $L_3$: $x+y=0$ and $2x+3y-6=0 \implies 2(-y)+3y-6=0 \implies y=6, x=-6$
The triangle is formed by vertices $(0,0)$,$(6/5, 6/5)$,and $(-6, 6)$.
The point $(0, \alpha)$ lies on the $y$-axis $(x=0)$.
For the point $(0, \alpha)$ to be in the interior of the triangle,it must lie between the $x$-axis (which is not a side,but the region is bounded by the lines) and the line $2x+3y-6=0$ along the $y$-axis.
At $x=0$,the line $2x+3y-6=0$ gives $3y=6$,so $y=2$.
The triangle vertices are $(0,0)$,$(6/5, 6/5)$,and $(-6, 6)$. The interior points on the $y$-axis are between $y=0$ and $y=2$.
Thus,$0 < \alpha < 2$.

(B) The given pair of lines is $3x^2-8xy+5y^2=0$.
The equation of the pair of lines passing through the origin and perpendicular to the given lines is obtained by interchanging the coefficients of $x^2$ and $y^2$ and changing the sign of the $xy$ term: $5x^2+8xy+3y^2=0$.
To find the equation of the pair of lines passing through $(1,1)$,we replace $x$ with $(x-1)$ and $y$ with $(y-1)$ in the equation $5x^2+8xy+3y^2=0$.
This gives $5(x-1)^2+8(x-1)(y-1)+3(y-1)^2=0$.
Expanding this,we get $5(x^2-2x+1)+8(xy-x-y+1)+3(y^2-2y+1)=0$.
Simplifying,$5x^2-10x+5+8xy-8x-8y+8+3y^2-6y+3=0$.
Combining like terms,$5x^2+8xy+3y^2-18x-14y+16=0$.

(A) Given lines are $y-3kx+4=0$ $(i)$ and $(2k-1)x-(8k-1)y-6=0$ (ii).
The slope of line $(i)$ is $m_1 = \frac{-(-3k)}{1} = 3k$.
The slope of line (ii) is $m_2 = \frac{-(2k-1)}{-(8k-1)} = \frac{2k-1}{8k-1}$.
Since the lines are perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 m_2 = -1$.
$3k \times \frac{2k-1}{8k-1} = -1$
$3k(2k-1) = -(8k-1)$
$6k^2 - 3k = -8k + 1$
$6k^2 + 5k - 1 = 0$
$6k^2 + 6k - k - 1 = 0$
$6k(k+1) - 1(k+1) = 0$
$(6k-1)(k+1) = 0$
Therefore,$k = 1/6$ or $k = -1$.

(A) Given equations are $y=kx+1$ $(i)$ and $3x+4y=9$ (ii).
Substituting $(i)$ into (ii):
$3x+4(kx+1)=9$
$3x+4kx+4=9$
$x(3+4k)=5$
$x=\frac{5}{3+4k}$
Since $x$ must be an integer,$(3+4k)$ must be a divisor of $5$. The divisors of $5$ are $\pm 1, \pm 5$.
Case $1$: $3+4k=1$ $\Rightarrow 4k=-2$ $\Rightarrow k=-0.5$ (Not an integer).
Case $2$: $3+4k=-1$ $\Rightarrow 4k=-4$ $\Rightarrow k=-1$.
Case $3$: $3+4k=5$ $\Rightarrow 4k=2$ $\Rightarrow k=0.5$ (Not an integer).
Case $4$: $3+4k=-5$ $\Rightarrow 4k=-8$ $\Rightarrow k=-2$.
Thus,the possible integer values for $k$ are $-1$ and $-2$.
The lines are $y=-x+1$ and $y=-2x+1$.
The combined equation is $(y+x-1)(y+2x-1)=0$.

(D) The given equation of the circle is $x^2+y^2-6x+8y-144=0$.
Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $2g = -6 \Rightarrow g = -3$ and $2f = 8 \Rightarrow f = 4$.
The centre of the circle is $(-g, -f) = (3, -4)$.
We know that the normal to a circle at any point always passes through its centre.
Let the point where the normal meets the circle again be $(x_1, y_1)$.
Since the centre $(3, -4)$ is the midpoint of the chord connecting $(8, 8)$ and $(x_1, y_1)$,we have:
$\frac{x_1+8}{2} = 3$ $\Rightarrow x_1+8 = 6$ $\Rightarrow x_1 = -2$
$\frac{y_1+8}{2} = -4$ $\Rightarrow y_1+8 = -8$ $\Rightarrow y_1 = -16$
Thus,the required point is $(-2, -16)$.

(C) Let $(a, b)$ be the point where the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.
The equation of the tangent at $(a, b)$ to the circle is given by $xa + yb - 3(x + a) + 2(y + b) - 12 = 0$.
Rearranging the terms,we get $(a - 3)x + (b + 2)y - (3a - 2b + 12) = 0$.
Comparing this with the given line $4x - 3y + 7 = 0$,we have the ratio of coefficients:
$\frac{a - 3}{4} = \frac{b + 2}{-3} = \frac{-(3a - 2b + 12)}{7} = k$.
From this,$a = 4k + 3$ and $b = -3k - 2$.
Substituting these into the third ratio: $-(3(4k + 3) - 2(-3k - 2) + 12) = 7k$.
$-(12k + 9 + 6k + 4 + 12) = 7k$ $\Rightarrow -(18k + 25) = 7k$ $\Rightarrow -25k = 25$ $\Rightarrow k = -1$.
Substituting $k = -1$ back into the expressions for $a$ and $b$:
$a = 4(-1) + 3 = -1$ and $b = -3(-1) - 2 = 1$.
Thus,the point of contact is $(-1, 1)$.

(D) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Given the circle $x^2+y^2-4x-6y+1=0$,we have $g=-2, f=-3, c=1$.
For the point $(2k, k-4)$,the polar equation is:
$x(2k) + y(k-4) - 2(x+2k) - 3(y+k-4) + 1 = 0$
$2kx + ky - 4y - 2x - 4k - 3y - 3k + 12 + 1 = 0$
$k(2x + y - 7) - 2x - 7y + 13 = 0$
For this line to pass through a fixed point regardless of $k$,the coefficients of $k$ and the constant term must both be zero:
$2x + y - 7 = 0$
$-2x - 7y + 13 = 0$
Adding these equations: $-6y + 6 = 0 \Rightarrow y = 1$.
Substituting $y=1$ into $2x + y - 7 = 0$: $2x + 1 - 7 = 0$ $\Rightarrow 2x = 6$ $\Rightarrow x = 3$.
Thus,the polar always passes through the point $(3, 1)$.

(D) Let the equations of the circles be:
$C_1: x^2+y^2-1=0$
$C_2: x^2+y^2-2x-3=0$
$C_3: x^2+y^2-2y-3=0$
The radical axis of $C_1$ and $C_2$ is given by $C_1 - C_2 = 0$:
$(x^2+y^2-1) - (x^2+y^2-2x-3) = 0$
$2x+2 = 0 \implies x = -1$
The radical axis of $C_1$ and $C_3$ is given by $C_1 - C_3 = 0$:
$(x^2+y^2-1) - (x^2+y^2-2y-3) = 0$
$2y+2 = 0 \implies y = -1$
The radical centre is the intersection of the radical axes,which is $(-1, -1)$.

(C) Given equations of parabolas:
$y^2=5x$ $(i)$
$x^2=5y$ (ii)
From equation (ii),we get $y = \frac{x^2}{5}$.
Substituting this into equation $(i)$:
$(\frac{x^2}{5})^2 = 5x$
$\frac{x^4}{25} = 5x$
$x^4 = 125x$
$x^4 - 125x = 0$
$x(x^3 - 125) = 0$
This gives $x = 0$ or $x^3 = 125$,which implies $x = 5$.
If $x = 0$,then $y = 0$. If $x = 5$,then $y = \frac{5^2}{5} = 5$.
The points of intersection are $(0,0)$ and $(5,5)$.
Checking the options,both points satisfy the line $x - y = 0$.

(C) The equation of the parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = \frac{1}{4}$.
Any normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.
Substituting $a = \frac{1}{4}$,the equation becomes $y = mx - \frac{m}{2} - \frac{m^3}{4}$.
If this normal passes through the point $(C, 0)$,then $0 = mC - \frac{m}{2} - \frac{m^3}{4}$.
This simplifies to $m(C - \frac{1}{2} - \frac{m^2}{4}) = 0$.
One root is $m = 0$,which corresponds to the normal along the $x$-axis.
For three distinct normals to be drawn,the quadratic equation $\frac{m^2}{4} = C - \frac{1}{2}$ must have two distinct non-zero real roots.
This requires $C - \frac{1}{2} > 0$,which implies $C > \frac{1}{2}$.

(A) The given equation of the parabola is $x^2=4ay$.
When the axes are rotated anticlockwise through an angle of $90^{\circ}$,the new coordinates $(x', y')$ are related to the old coordinates $(x, y)$ by the transformation:
$x = x' \cos(90^{\circ}) - y' \sin(90^{\circ}) = -y'$
$y = x' \sin(90^{\circ}) + y' \cos(90^{\circ}) = x'$
Substituting these into the original equation $x^2=4ay$:
$(-y')^2 = 4a(x')$
$y'^2 = 4ax'$
Thus,the new equation is $y^2=4ax$.

(A) Given the equation of the ellipse: $\frac{x^2}{25}+\frac{y^2}{16}=1$.
Here,$a^2=25 \implies a=5$ and $b^2=16 \implies b=4$.
The eccentricity $e$ is given by $e = \sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
By the definition of an ellipse,the sum of focal distances of any point $P$ on the ellipse is $PS+PS^{\prime} = 2a$.
Thus,$PS+PS^{\prime} = 2 \times 5 = 10$.
Given $SP=8$,we have $8+PS^{\prime} = 10 \implies PS^{\prime} = 2$.
The distance between the foci $SS^{\prime}$ is $2ae = 2 \times 5 \times \frac{3}{5} = 6$.
Checking the options:
$4+S^{\prime}P = 4+2 = 6$.
Therefore,$SS^{\prime} = 4+S^{\prime}P$.

(A) The given equation of the ellipse is $\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1$.
Comparing this with the standard form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$,we have $h=3, k=2, a^2=25, b^2=16$.
Thus,$a=5$ and $b=4$.
The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
$(i)$ The equation of the major axis is $y = k$,so $y = 2$. This matches $(q)$.
$(ii)$ The equations of the directrices are $x = h \pm \frac{a}{e}$.
$x = 3 \pm \frac{5}{3/5} = 3 \pm \frac{25}{3}$.
$x = 3 + \frac{25}{3} = \frac{34}{3} \Rightarrow 3x = 34$.
$x = 3 - \frac{25}{3} = -\frac{16}{3} \Rightarrow 3x = -16$.
Thus,$3x = 34$ matches $(p)$.
$(iii)$ The equations of the latera recta are $x = h \pm ae$.
$x = 3 \pm (5 \times \frac{3}{5}) = 3 \pm 3$.
$x = 3 + 3 = 6$ or $x = 3 - 3 = 0$.
Thus,$x = 6$ matches $(s)$.
Therefore,the correct matches are $(i) \rightarrow (q)$,$(ii) \rightarrow (p)$,$(iii) \rightarrow (s)$.

(A) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec \theta, b \tan \theta)$ is $ax \cos \theta + by \cot \theta = a^2 + b^2$.
For the given hyperbola $\frac{x^2}{4} - \frac{y^2}{9} = 1$,we have $a^2 = 4$ and $b^2 = 9$.
The normal at $A(2 \sec \theta, 3 \tan \theta)$ is $2x \cos \theta + 3y \cot \theta = 4 + 9 = 13$ $(i)$.
The normal at $B(2 \sec \phi, 3 \tan \phi)$ is $2x \cos \phi + 3y \cot \phi = 4 + 9 = 13$ (ii).
Since $\theta + \phi = \frac{\pi}{2}$,we have $\phi = \frac{\pi}{2} - \theta$. Thus,$\cos \phi = \sin \theta$ and $\cot \phi = \tan \theta$.
Substituting these into (ii): $2x \sin \theta + 3y \tan \theta = 13$ (iii).
To find the intersection $(\alpha, \beta)$,we solve the system of equations:
$2x \cos \theta + 3y \cot \theta = 13$
$2x \sin \theta + 3y \tan \theta = 13$
Subtracting the equations or using Cramer's rule,we find the $y$-coordinate $\beta$.
From $2x \cos \theta + 3y \frac{\cos \theta}{\sin \theta} = 13$,we get $2x \cos \theta \sin \theta + 3y \cos \theta = 13 \sin \theta$.
From $2x \sin \theta + 3y \frac{\sin \theta}{\cos \theta} = 13$,we get $2x \sin \theta \cos \theta + 3y \sin \theta = 13 \cos \theta$.
Subtracting the two resulting equations: $3y(\cos \theta - \sin \theta) = 13(\sin \theta - \cos \theta)$.
Since $\sin \theta \neq \cos \theta$ (for distinct points),we divide by $(\cos \theta - \sin \theta)$ to get $3y = -13$.
Therefore,$\beta = -\frac{13}{3}$.

(D) Given,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$.
This is of the form $1^\infty$.
Using the formula $\lim _{x \rightarrow \infty}[1+f(x)]^{g(x)} = e^{\lim _{x \rightarrow \infty} f(x) \cdot g(x)}$,we have:
$L = \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3}{x^2-x+2}-1\right]^x$
$= \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]^x$
$= \lim _{x \rightarrow \infty}\left[1+\frac{2x+1}{x^2-x+2}\right]^x$
$= e^{\lim _{x \rightarrow \infty} \frac{2x+1}{x^2-x+2} \cdot x}$
$= e^{\lim _{x \rightarrow \infty} \frac{2x^2+x}{x^2-x+2}}$
Dividing numerator and denominator by $x^2$:
$= e^{\lim _{x \rightarrow \infty} \frac{2+1/x}{1-1/x+2/x^2}} = e^{\frac{2+0}{1-0+0}} = e^2$.

(C) The given series is an arithmetic progression: $a, a+d, a+2d, \ldots, a+2nd$.
The number of terms $m$ is given by $a+(m-1)d = a+2nd$,which implies $m-1 = 2n$,so $m = 2n+1$.
The standard deviation $\sigma$ of an arithmetic progression with $m$ terms and common difference $d$ is given by the formula $\sigma = \sqrt{\frac{m^2-1}{12}} |d|$.
Substituting $m = 2n+1$:
$\sigma = \sqrt{\frac{(2n+1)^2-1}{12}} |d|$
$\sigma = \sqrt{\frac{4n^2+4n+1-1}{12}} |d|$
$\sigma = \sqrt{\frac{4n(n+1)}{12}} |d|$
$\sigma = \sqrt{\frac{n(n+1)}{3}} |d|$
Thus,the correct option is $C$.

(A) Given,$8R^2 = a^2 + b^2 + c^2$.
Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Substituting these in the given equation:
$8R^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2$
$8R^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C)$
$2 = \sin^2 A + \sin^2 B + \sin^2 C$
Using $\sin^2 \theta = 1 - \cos^2 \theta$:
$2 = (1 - \cos^2 A) + (1 - \cos^2 B) + \sin^2 C$
$2 = 2 - \cos^2 A - \cos^2 B + \sin^2 C$
$\cos^2 A + \cos^2 B = \sin^2 C$
$\cos^2 A + \cos^2 B = 1 - \cos^2 C$
$\cos^2 A + \cos^2 B + \cos^2 C = 1$
This is a known identity for a right-angled triangle where one of the angles is $90^\circ$.
Alternatively,$\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C = 1$ implies $\cos A \cos B \cos C = 0$.
Thus,$A = 90^\circ$ or $B = 90^\circ$ or $C = 90^\circ$.
Therefore,the triangle is a right-angled triangle.

(C) Given,$\angle A=90^{\circ}$ and $\angle B \neq \angle C$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have $b = k \sin B$ and $c = k \sin C$.
Substituting these into the expression:
$\frac{b^2+c^2}{b^2-c^2} \sin (B-C) = \frac{k^2 \sin^2 B + k^2 \sin^2 C}{k^2 \sin^2 B - k^2 \sin^2 C} \sin (B-C)$
$= \frac{\sin^2 B + \sin^2 C}{\sin^2 B - \sin^2 C} \sin (B-C)$
Since $\angle A = 90^{\circ}$,then $B+C = 90^{\circ}$,so $C = 90^{\circ}-B$.
Thus,$\sin C = \cos B$ and $\sin^2 C = \cos^2 B$.
Also,$\sin^2 B - \sin^2 C = \sin(B+C) \sin(B-C) = \sin(90^{\circ}) \sin(B-C) = 1 \cdot \sin(B-C)$.
Substituting these:
$= \frac{\sin^2 B + \cos^2 B}{\sin(B+C)} \cdot \sin(B-C) = \frac{1}{\sin(90^{\circ})} = 1$.

(D) We are given the relation $2R + r = r_2$.
Using the standard formulas for circumradius $R$,inradius $r$,and exradius $r_2$:
$r_2 - r = 4R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$r_2 - r = 4R \sin \frac{B}{2} [\cos \frac{A}{2} \cos \frac{C}{2} - \sin \frac{A}{2} \sin \frac{C}{2}]$
$r_2 - r = 4R \sin \frac{B}{2} \cos(\frac{A+C}{2})$
Since $A+B+C = \pi$,we have $\frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\cos(\frac{A+C}{2}) = \sin \frac{B}{2}$.
Thus,$r_2 - r = 4R \sin^2 \frac{B}{2}$.
Given $2R = r_2 - r$,we have $2R = 4R \sin^2 \frac{B}{2}$.
Dividing by $2R$,we get $1 = 2 \sin^2 \frac{B}{2}$,which implies $\sin^2 \frac{B}{2} = \frac{1}{2}$.
Therefore,$\sin \frac{B}{2} = \frac{1}{\sqrt{2}} = \sin \frac{\pi}{4}$.
This gives $\frac{B}{2} = \frac{\pi}{4}$,so $B = \frac{\pi}{2}$.

(D) Given curve is $x^2+y^2=16a^2$.
On differentiating the equation with respect to $x$,we get $2x + 2yy' = 0$,which implies $y' = -\frac{x}{y}$.
At the point $(2\sqrt{2}a, 2\sqrt{2}a)$,the slope of the tangent is $m_1 = -\frac{2\sqrt{2}a}{2\sqrt{2}a} = -1$.
The slope of the normal is $m_2 = -\frac{1}{m_1} = 1$.
The equation of the tangent is $y - 2\sqrt{2}a = -1(x - 2\sqrt{2}a)$,which simplifies to $x + y = 4\sqrt{2}a$.
The $X$-intercept of the tangent is found by setting $y=0$,giving $x = 4\sqrt{2}a$. Let this point be $B(4\sqrt{2}a, 0)$.
The equation of the normal is $y - 2\sqrt{2}a = 1(x - 2\sqrt{2}a)$,which simplifies to $y = x$.
The normal passes through the origin $O(0,0)$ and the point $A(2\sqrt{2}a, 2\sqrt{2}a)$.
The triangle is formed by the points $O(0,0)$,$A(2\sqrt{2}a, 2\sqrt{2}a)$,and $B(4\sqrt{2}a, 0)$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
Area $= \frac{1}{2} |0(2\sqrt{2}a-0) + 2\sqrt{2}a(0-0) + 4\sqrt{2}a(0-2\sqrt{2}a)| = \frac{1}{2} |4\sqrt{2}a(-2\sqrt{2}a)| = \frac{1}{2} |-16a^2| = 8a^2$.

(A) Given the partial fraction decomposition:
$\frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2}$
Multiplying both sides by $x^4(x+2)$,we get:
$x+1 = Ax^3(x+2) + Bx^2(x+2) + Cx(x+2) + D(x+2) + Ex^4$
To find the value of $B+D+E$,we substitute $x = -1$ into the equation:
$-1+1 = A(-1)^3(-1+2) + B(-1)^2(-1+2) + C(-1)(-1+2) + D(-1+2) + E(-1)^4$
$0 = A(-1)(1) + B(1)(1) + C(-1)(1) + D(1) + E(1)$
$0 = -A + B - C + D + E$
Rearranging the terms to isolate $B+D+E$:
$B+D+E = A+C$

(C) Let the point $B$ divide the line segment $AC$ in the ratio $k: 1$. Using the section formula,the coordinates of $B$ are given by:
$B = \left( \frac{k(9) + 1(3)}{k+1}, \frac{k(8) + 1(2)}{k+1}, \frac{k(10) + 1(4)}{k+1} \right)$
Given that $B = \left( \frac{33}{5}, \frac{28}{5}, \frac{38}{5} \right)$,we equate the $x$-coordinates:
$\frac{9k + 3}{k + 1} = \frac{33}{5}$
$5(9k + 3) = 33(k + 1)$
$45k + 15 = 33k + 33$
$45k - 33k = 33 - 15$
$12k = 18$
$k = \frac{18}{12} = \frac{3}{2}$
Thus,the ratio $k: 1$ is $3: 2$.

(D) Let $B$ divide the line segment $\overline{AC}$ in the ratio $k: 1$. Using the section formula,the coordinates of $B$ are given by:
$B = \left( \frac{k(9) + 1(3)}{k+1}, \frac{k(8) + 1(2)}{k+1}, \frac{k(10) + 1(4)}{k+1} \right)$
Equating the $x$-coordinate:
$\frac{9k + 3}{k+1} = \frac{33}{5}$
$5(9k + 3) = 33(k + 1)$
$45k + 15 = 33k + 33$
$12k = 18$
$k = \frac{18}{12} = \frac{3}{2}$
Thus,the ratio $k: 1$ is $\frac{3}{2}: 1$,which is $3: 2$.

(A) Given,$A(5, -4)$ and $B(7, 6)$ are points in a plane. Let $P(x, y)$ be a point such that $AP:PB = 2:3$.
Then,$\frac{AP}{PB} = \frac{2}{3} \Rightarrow 3AP = 2PB$.
Squaring both sides,we get $9AP^2 = 4PB^2$.
Using the distance formula,$AP^2 = (x-5)^2 + (y+4)^2$ and $PB^2 = (x-7)^2 + (y-6)^2$.
Substituting these into the equation:
$9[(x-5)^2 + (y+4)^2] = 4[(x-7)^2 + (y-6)^2]$.
$9[x^2 - 10x + 25 + y^2 + 8y + 16] = 4[x^2 - 14x + 49 + y^2 - 12y + 36]$.
$9[x^2 + y^2 - 10x + 8y + 41] = 4[x^2 + y^2 - 14x - 12y + 85]$.
$9x^2 + 9y^2 - 90x + 72y + 369 = 4x^2 + 4y^2 - 56x - 48y + 340$.
$5x^2 + 5y^2 - 34x + 120y + 29 = 0$.
This is an equation of the form $x^2 + y^2 + 2gx + 2fy + c = 0$,which represents a circle.

(B) The total number of outcomes when two dice are thrown is $6 \times 6 = 36$.
The favorable outcomes where the sum is greater than $7$ are:
Sum $= 8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$ ($5$ outcomes)
Sum $= 9$: $(3,6), (4,5), (5,4), (6,3)$ ($4$ outcomes)
Sum $= 10$: $(4,6), (5,5), (6,4)$ ($3$ outcomes)
Sum $= 11$: $(5,6), (6,5)$ ($2$ outcomes)
Sum $= 12$: $(6,6)$ ($1$ outcome)
Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$.
The required probability is $\frac{15}{36} = \frac{5}{12}$.

(A) number is divisible by $4$ if the number formed by its last two digits is divisible by $4$.
Using the digits ${1, 2, 3, 4, 5}$,the possible two-digit combinations divisible by $4$ are $12, 24, 32, 52$.
There are $4$ such favorable combinations for the last two digits.
The remaining $3$ positions can be filled by the remaining $3$ digits in $3!$ ways.
Total number of favorable outcomes $= 4 \times 3!$.
Total number of possible five-digit numbers $= 5!$.
Probability $= \frac{4 \times 3!}{5!} = \frac{4 \times 3!}{5 \times 4 \times 3!} = \frac{1}{5}$.

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
Given equation is $\vec{a} + \vec{b} = -\sqrt{3} \vec{c}$.
Squaring both sides,we get $|\vec{a} + \vec{b}|^2 = |-\sqrt{3} \vec{c}|^2$.
$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = 3|\vec{c}|^2$.
Substituting the values,$1^2 + 1^2 + 2(\vec{a} \cdot \vec{b}) = 3(1)^2$.
$2 + 2(\vec{a} \cdot \vec{b}) = 3$.
$2(\vec{a} \cdot \vec{b}) = 1$.
$\vec{a} \cdot \vec{b} = \frac{1}{2}$.
Since $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
$\frac{1}{2} = (1)(1) \cos \theta$.
$\cos \theta = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(D) In a regular hexagon $ABCDEF$ with centre $O$,we have the following vector relations:
$\vec{AB} = \vec{ED}$ and $\vec{AF} = \vec{CD}$.
Now,consider the sum $\vec{S} = \vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF}$.
Substituting the relations,we get:
$\vec{S} = \vec{ED} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{CD}$
Rearranging the terms:
$\vec{S} = (\vec{AE} + \vec{ED}) + (\vec{AC} + \vec{CD}) + \vec{AD}$
Using the triangle law of vector addition,$\vec{AE} + \vec{ED} = \vec{AD}$ and $\vec{AC} + \vec{CD} = \vec{AD}$.
Therefore,$\vec{S} = \vec{AD} + \vec{AD} + \vec{AD} = 3 \vec{AD}$.
Since $O$ is the centre of the regular hexagon,$\vec{AD} = 2 \vec{AO}$.
Thus,$\vec{S} = 3(2 \vec{AO}) = 6 \vec{AO}$.

(B) Given curves are $y=ax^2$ $(i)$ and $x=ay^2$ $(ii)$.
Substituting $y$ from $(ii)$ into $(i)$,we get $x = a(ax^2)^2 = a^3x^4$,which implies $x(a^3x^3 - 1) = 0$.
Thus,$x=0$ or $x=\frac{1}{a}$.
For $x=0$,$y=0$,and for $x=\frac{1}{a}$,$y=\frac{1}{a}$.
The area bounded by the curves is given by $\int_0^{1/a} (\sqrt{x/a} - ax^2) dx = 3$.
$\int_0^{1/a} \frac{1}{\sqrt{a}} x^{1/2} dx - \int_0^{1/a} ax^2 dx = 3$.
$\frac{1}{\sqrt{a}} [\frac{2}{3} x^{3/2}]_0^{1/a} - a [\frac{x^3}{3}]_0^{1/a} = 3$.
$\frac{1}{\sqrt{a}} \cdot \frac{2}{3} \cdot (\frac{1}{a})^{3/2} - \frac{a}{3} \cdot (\frac{1}{a})^3 = 3$.
$\frac{2}{3a^2} - \frac{1}{3a^2} = 3$.
$\frac{1}{3a^2} = 3 \Rightarrow 9a^2 = 1 \Rightarrow a^2 = \frac{1}{9}$.
Since $a>0$,we have $a = \frac{1}{3}$.

(B) Since $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
Let $P = AB - BA$.
Taking the transpose of $P$:
$P^{\prime} = (AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$
Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$,we get:
$P^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$
Substituting $A^{\prime} = A$ and $B^{\prime} = B$:
$P^{\prime} = BA - AB = -(AB - BA) = -P$.
Since $P^{\prime} = -P$,the matrix $AB - BA$ is a skew-symmetric matrix.

(B) Given $A(x) = \begin{vmatrix} x+1 & 2x+1 & 3x+1 \\ 2x+1 & 3x+1 & x+1 \\ 3x+1 & x+1 & 2x+1 \end{vmatrix}$.
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$A(x) = \begin{vmatrix} x+1 & x & x \\ 2x+1 & x & -2x \\ 3x+1 & -2x & x \end{vmatrix}$.
Taking $x$ common from $C_2$ and $C_3$:
$A(x) = x^2 \begin{vmatrix} x+1 & 1 & 1 \\ 2x+1 & 1 & -2 \\ 3x+1 & -2 & 1 \end{vmatrix}$.
Applying $C_2 \rightarrow C_2 - C_3$:
$A(x) = x^2 \begin{vmatrix} x+1 & 0 & 1 \\ 2x+1 & 3 & -2 \\ 3x+1 & -3 & 1 \end{vmatrix}$.
Applying $R_2 \rightarrow R_2 + R_3$:
$A(x) = x^2 \begin{vmatrix} x+1 & 0 & 1 \\ 5x+2 & 0 & -1 \\ 3x+1 & -3 & 1 \end{vmatrix}$.
Expanding along $C_2$:
$A(x) = x^2 \cdot (-(-3)) \cdot \begin{vmatrix} x+1 & 1 \\ 5x+2 & -1 \end{vmatrix} = 3x^2 \{(-x-1) - (5x+2)\} = 3x^2(-6x-3) = -18x^3 - 9x^2$.
Now,$\int_0^1 A(x) dx = \int_0^1 (-18x^3 - 9x^2) dx = \left[ -\frac{18x^4}{4} - \frac{9x^3}{3} \right]_0^1 = \left[ -\frac{9}{2}x^4 - 3x^3 \right]_0^1 = -\frac{9}{2} - 3 = -\frac{15}{2}$.

(B) The system of equations is given by:
$ax + by + cz = 2$
$bx + cy + az = 2$
$cx + ay + bz = 2$
The determinant of the coefficient matrix is:
$\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$
Applying the row operation $R_1 \rightarrow R_1 + R_2 + R_3$:
$\Delta = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{vmatrix}$
Since $a+b+c = 0$,we have:
$\Delta = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \times \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} = 0$
Since $\Delta = 0$,the system either has no solution or infinitely many solutions.
Let us check the consistency by calculating the determinants $\Delta_x, \Delta_y, \Delta_z$.
$\Delta_x = \begin{vmatrix} 2 & b & c \\ 2 & c & a \\ 2 & a & b \end{vmatrix} = 2 \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 2(c^2 + ab + ab - c^2 - a^2 - b^2) = 2(2ab - a^2 - b^2 - c^2)$.
Since $a+b+c=0$,$c = -(a+b)$,so $c^2 = a^2 + b^2 + 2ab$.
Thus,$\Delta_x = 2(2ab - a^2 - b^2 - (a^2 + b^2 + 2ab)) = 2(-2a^2 - 2b^2) = -4(a^2 + b^2)$.
If $a, b, c$ are not all zero,$\Delta_x \neq 0$.
Since $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z \neq 0$,the system is inconsistent.

(A) Given equation is $\sin \left(\cot ^{-1} x\right)=\cos \left(\tan ^{-1}(1+x)\right)$.
We know that $\sin \left(\cot ^{-1} x\right) = \sin \left(\sin ^{-1} \frac{1}{\sqrt{1+x^2}}\right) = \frac{1}{\sqrt{1+x^2}}$.
Also,$\cos \left(\tan ^{-1}(1+x)\right) = \cos \left(\cos ^{-1} \frac{1}{\sqrt{1+(1+x)^2}}\right) = \frac{1}{\sqrt{1+(1+x)^2}}$.
Equating both sides,we get $\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+(1+x)^2}}$.
Squaring both sides,$1+x^2 = 1+(1+x)^2$.
$1+x^2 = 1+1+x^2+2x$.
$1+x^2 = 2+x^2+2x$.
Subtracting $x^2$ from both sides,$1 = 2+2x$.
$2x = -1$.
$x = -\frac{1}{2}$.

(C) We know that the inverse hyperbolic secant function is defined as $\operatorname{sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)$.
Substituting $x = \cos \theta$,we get:
$\operatorname{sech}^{-1}(\cos \theta) = \log \left( \frac{1 + \sqrt{1 - \cos^2 \theta}}{\cos \theta} \right)$
Since $\theta \in \left(0, \frac{\pi}{2}\right)$,$\sin \theta > 0$,so $\sqrt{1 - \cos^2 \theta} = \sin \theta$.
Thus,the expression becomes $\log \left( \frac{1 + \sin \theta}{\cos \theta} \right)$.
This can be rewritten as $\log (\sec \theta + \tan \theta)$.
Using the identity $\sec \theta + \tan \theta = \tan \left( \frac{\pi}{4} + \frac{\theta}{2} \right)$,we obtain:
$\operatorname{sech}^{-1}(\cos \theta) = \log \left| \tan \left( \frac{\pi}{4} + \frac{\theta}{2} \right) \right|$.

(C) Given,$f(x)=x^2-2x+4$.
We need to solve $f(x-1)=f(x+1)$.
Substituting $(x-1)$ and $(x+1)$ into the function:
$(x-1)^2-2(x-1)+4 = (x+1)^2-2(x+1)+4$.
Expanding both sides:
$(x^2-2x+1) - 2x + 2 + 4 = (x^2+2x+1) - 2x - 2 + 4$.
Simplifying:
$x^2-4x+7 = x^2-1+4$.
$x^2-4x+7 = x^2+3$.
Subtracting $x^2$ from both sides:
$-4x+7 = 3$.
$-4x = 3-7$.
$-4x = -4$.
$x = 1$.
Thus,the set of values is $\{1\}$.

(D) Let $f(x) = ax + b$.
Given the equation $f(f(x)) = x + f(x)$.
Substituting $f(x)$ into the equation:
$a(ax + b) + b = x + (ax + b)$
$a^2x + ab + b = x + ax + b$
$a^2x + ab = x + ax$
$x(a^2 - a - 1) + ab = 0$.
For this to hold for all $x$,the coefficients must be zero:
$a^2 - a - 1 = 0$ and $ab = 0$.
Since $a^2 - a - 1 = 0$,$a$ cannot be $0$,therefore $b = 0$.
Solving $a^2 - a - 1 = 0$ using the quadratic formula:
$a = \frac{1 \pm \sqrt{5}}{2}$.
Thus,the functions are $f(x) = \left(\frac{1 + \sqrt{5}}{2}\right)x$ and $f(x) = \left(\frac{1 - \sqrt{5}}{2}\right)x$.
There are $2$ such real linear functions.

(D) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
First,calculate the Left Hand Limit $(LHL)$:
$\lim_{x \to 0^-} (1+|\sin x|)^{\frac{p}{|\sin x|}} = \lim_{x \to 0^-} (1-\sin x)^{\frac{p}{-\sin x}}$ (since $|\sin x| = -\sin x$ for $x < 0$).
Let $h = -x$,as $x \to 0^-$,$h \to 0^+$. Then $\sin x = -\sin h$.
$\lim_{h \to 0^+} (1+\sin h)^{\frac{p}{\sin h}} = e^{\lim_{h \to 0^+} \sin h \cdot \frac{p}{\sin h}} = e^p$.
Next,calculate the Right Hand Limit $(RHL)$:
$\lim_{x \to 0^+} e^{\frac{\sin 2x}{\sin 3x}} = e^{\lim_{x \to 0^+} \frac{\sin 2x}{\sin 3x}} = e^{\lim_{x \to 0^+} \frac{\sin 2x}{2x} \cdot \frac{3x}{\sin 3x} \cdot \frac{2}{3}} = e^{2/3}$.
Since $f(0) = q$,we equate the limits:
$e^p = q = e^{2/3}$.
Thus,$p = 2/3$ and $q = e^{2/3}$.
Wait,checking the $LHL$ again: $(1+|\sin x|)^{\frac{p}{|\sin x|}} = (1-\sin x)^{\frac{p}{-\sin x}} = ((1-\sin x)^{\frac{1}{-\sin x}})^p = e^p$.
If the original question implies $f(x) = (1+|\sin x|)^{\frac{p}{\sin x}}$,then $LHL$ $= e^{-p}$.
Equating $e^{-p} = e^{2/3} = q$,we get $p = -2/3$ and $q = e^{2/3}$.

(C) Let $y = \tan ^{-1}\left[\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\right]$.
We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$.
Similarly,$1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.
Substituting these into the expression:
$y = \tan ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}\right]$
$y = \tan ^{-1}\left[\frac{2 \sin \frac{x}{2}}{2 \cos \frac{x}{2}}\right] = \tan ^{-1}(\tan \frac{x}{2}) = \frac{x}{2}$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.

(B) Given,$y = \tan^{-1} \left[ \frac{5 \cos x - 12 \sin x}{12 \cos x + 5 \sin x} \right]$.
Divide the numerator and denominator by $12 \cos x$:
$y = \tan^{-1} \left[ \frac{\frac{5}{12} - \tan x}{1 + \frac{5}{12} \tan x} \right]$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A - B}{1 + AB} \right)$,we get:
$y = \tan^{-1} \left( \frac{5}{12} \right) - \tan^{-1}(\tan x)$.
$y = \tan^{-1} \left( \frac{5}{12} \right) - x$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \tan^{-1} \left( \frac{5}{12} \right) \right) - \frac{d}{dx}(x)$.
Since $\tan^{-1} \left( \frac{5}{12} \right)$ is a constant,its derivative is $0$.
$\frac{dy}{dx} = 0 - 1 = -1$.

(C) Given,$y=a \cos (\sin 2 x)+b \sin (\sin 2 x)$.
On differentiating both sides with respect to $x$,we get:
$y^{\prime} = -a \sin(\sin 2x) \cdot \cos 2x \cdot 2 + b \cos(\sin 2x) \cdot \cos 2x \cdot 2$
$y^{\prime} = 2 \cos 2x \{-a \sin(\sin 2x) + b \cos(\sin 2x)\}$.
Again,differentiating with respect to $x$ using the product rule:
$y^{\prime \prime} = -4 \sin 2x \{-a \sin(\sin 2x) + b \cos(\sin 2x)\} + 2 \cos 2x \{-a \cos(\sin 2x) \cdot \cos 2x \cdot 2 - b \sin(\sin 2x) \cdot \cos 2x \cdot 2\}$.
Substituting $y^{\prime} = 2 \cos 2x \{-a \sin(\sin 2x) + b \cos(\sin 2x)\}$ into the first part:
$y^{\prime \prime} = -4 \sin 2x \cdot \frac{y^{\prime}}{2 \cos 2x} - 4 \cos^2 2x \{a \cos(\sin 2x) + b \sin(\sin 2x)\}$.
Since $y = a \cos(\sin 2x) + b \sin(\sin 2x)$,we have:
$y^{\prime \prime} = -2 \tan 2x \cdot y^{\prime} - 4 \cos^2 2x \cdot y$.
Therefore,$y^{\prime \prime} + (2 \tan 2x) y^{\prime} = -4 \cos^2 2x \cdot y$.

(B) Given,the parametric equations of the curve are:
$x = a \cos^3 t$
$y = a \sin^3 t$
Differentiating with respect to $t$:
$\frac{dx}{dt} = -3a \cos^2 t \sin t$
$\frac{dy}{dt} = 3a \sin^2 t \cos t$
The slope of the tangent is:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\sin t}{\cos t}$
The equation of the tangent at $(a \cos^3 t, a \sin^3 t)$ is:
$y - a \sin^3 t = -\frac{\sin t}{\cos t} (x - a \cos^3 t)$
$y \cos t - a \sin^3 t \cos t = -x \sin t + a \cos^3 t \sin t$
$x \sin t + y \cos t = a \sin t \cos t (\sin^2 t + \cos^2 t)$
$x \sin t + y \cos t = a \sin t \cos t$
To find the intercepts,set $y=0$ to get $x = a \cos t$,and set $x=0$ to get $y = a \sin t$.
The points are $A(a \cos t, 0)$ and $B(0, a \sin t)$.
The length of the segment $AB$ is:
$L = \sqrt{(a \cos t - 0)^2 + (0 - a \sin t)^2} = \sqrt{a^2 \cos^2 t + a^2 \sin^2 t} = \sqrt{a^2(1)} = a$.

(C) Given,$f(x) = \frac{1}{2}[|\sin x| + \sin x]$ for $0 < x \leq 2\pi$.
Case $I$: When $0 < x \leq \pi$,$\sin x \geq 0$,so $|\sin x| = \sin x$.
Thus,$f(x) = \frac{1}{2}[\sin x + \sin x] = \sin x$.
The derivative is $f'(x) = \cos x$.
For $0 < x < \frac{\pi}{2}$,$f'(x) = \cos x > 0$,so $f(x)$ is increasing.
For $\frac{\pi}{2} < x < \pi$,$f'(x) = \cos x < 0$,so $f(x)$ is decreasing.
Case $II$: When $\pi < x \leq 2\pi$,$\sin x \leq 0$,so $|\sin x| = -\sin x$.
Thus,$f(x) = \frac{1}{2}[-\sin x + \sin x] = 0$.
Therefore,$f(x)$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

(A) Let $I = \int \frac{x^2+1}{x^4+7 x^2+1} d x$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{1+\frac{1}{x^2}}{x^2+7+\frac{1}{x^2}} d x = \int \frac{1+\frac{1}{x^2}}{(x^2+\frac{1}{x^2}+2)+5} d x$.
$I = \int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+9} d x$.
Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + 3^2} = \frac{1}{3} \tan^{-1}(\frac{t}{3}) + C$.
Substituting $t = x - \frac{1}{x}$ back:
$I = \frac{1}{3} \tan^{-1}(\frac{x - \frac{1}{x}}{3}) + C = \frac{1}{3} \tan^{-1}(\frac{x^2-1}{3x}) + C$.

(C) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} dx$.
We can rewrite the integral as $I = \int x^2 \cdot \frac{x}{\sqrt{1+x^2}} dx$.
Using integration by parts,let $u = x^2$ and $dv = \frac{x}{\sqrt{1+x^2}} dx$.
Then $du = 2x dx$ and $v = \int (1+x^2)^{-1/2} x dx = \sqrt{1+x^2}$.
Applying the formula $\int u dv = uv - \int v du$:
$I = x^2 \sqrt{1+x^2} - \int 2x \sqrt{1+x^2} dx$.
For the remaining integral,let $t = 1+x^2$,so $dt = 2x dx$.
$I = x^2 \sqrt{1+x^2} - \int t^{1/2} dt = x^2 \sqrt{1+x^2} - \frac{t^{3/2}}{3/2} + C$.
$I = x^2 \sqrt{1+x^2} - \frac{2}{3}(1+x^2)^{3/2} + C$.

(A) Let $I = \int \frac{2x+2}{\sqrt{x^2-4x-5}} dx$.
We express the numerator as $2x+2 = A \frac{d}{dx}(x^2-4x-5) + B$.
$2x+2 = A(2x-4) + B$.
Comparing coefficients of $x$ and constants,we get $2A = 2 \Rightarrow A = 1$ and $-4A + B = 2 \Rightarrow -4(1) + B = 2 \Rightarrow B = 6$.
Thus,$I = \int \frac{2x-4}{\sqrt{x^2-4x-5}} dx + 6 \int \frac{dx}{\sqrt{(x-2)^2 - 3^2}}$.
For the first integral,let $t = x^2-4x-5$,then $dt = (2x-4)dx$.
$I = \int t^{-1/2} dt + 6 \int \frac{dx}{\sqrt{(x-2)^2 - 3^2}}$.
$I = 2\sqrt{t} + 6 \log |(x-2) + \sqrt{(x-2)^2 - 3^2}| + C$.
$I = 2\sqrt{x^2-4x-5} + 6 \log |(x-2) + \sqrt{x^2-4x-5}| + C$.

(C) Let $I = \int \frac{d x}{\cos (x+4) \cos (x+2)}$.
Multiply and divide by $\sin 2$:
$I = \frac{1}{\sin 2} \int \frac{\sin 2}{\cos (x+4) \cos (x+2)} d x$.
Since $2 = (x+4) - (x+2)$,we have:
$I = \frac{1}{\sin 2} \int \frac{\sin [(x+4)-(x+2)]}{\cos (x+4) \cos (x+2)} d x$.
Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \frac{1}{\sin 2} \int \frac{\sin (x+4) \cos (x+2) - \cos (x+4) \sin (x+2)}{\cos (x+4) \cos (x+2)} d x$.
$I = \frac{1}{\sin 2} \int [\tan (x+4) - \tan (x+2)] d x$.
Integrating $\tan x$,we get $\log |\sec x|$:
$I = \frac{1}{\sin 2} [\log |\sec (x+4)| - \log |\sec (x+2)|] + C$.
$I = \frac{1}{\sin 2} \log \left|\frac{\sec (x+4)}{\sec (x+2)}\right| + C$.

(A) Let $I = \int_0^{\pi / 4} (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$.
$I = \int_0^{\pi / 4} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right) \, dx$.
$I = \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$.
Multiply numerator and denominator by $\sqrt{2}$:
$I = \int_0^{\pi / 4} \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi / 4} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}} \, dx$.
Let $t = \sin x - \cos x$,then $dt = (\cos x + \sin x) \, dx$.
When $x = 0$,$t = -1$. When $x = \pi/4$,$t = 0$.
$I = \sqrt{2} \int_{-1}^0 \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^0$.
$I = \sqrt{2} [\sin^{-1}(0) - \sin^{-1}(-1)] = \sqrt{2} [0 - (-\pi/2)] = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.

(D) Let $I = \int_0^{\pi / 4} \frac{\sin x+\cos x}{7+9 \sin 2 x} d x$.
We know that $(\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$.
Therefore,$\sin 2x = 1 - (\sin x - \cos x)^2$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$.
Substituting these into the integral:
$I = \int_{-1}^0 \frac{dt}{7 + 9(1 - t^2)} = \int_{-1}^0 \frac{dt}{16 - 9t^2}$.
$I = \frac{1}{9} \int_{-1}^0 \frac{dt}{(4/3)^2 - t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$,we get:
$I = \frac{1}{9} \cdot \frac{1}{2(4/3)} \left[ \log \left| \frac{4/3 + t}{4/3 - t} \right| \right]_{-1}^0 = \frac{1}{24} \left[ \log \left| \frac{4+3t}{4-3t} \right| \right]_{-1}^0$.
$I = \frac{1}{24} [\log(1) - \log| (4-3)/(4+3) |] = \frac{1}{24} [0 - \log(1/7)] = \frac{1}{24} \log 7$.

(B) Given the family of curves: $y=(\alpha+\beta x) e^{p x}$ $(i)$
Differentiating with respect to $x$:
$y^{\prime} = \beta e^{p x} + p(\alpha+\beta x) e^{p x}$
$y^{\prime} = \beta e^{p x} + p y$ (ii)
From (ii),we have $\beta e^{p x} = y^{\prime} - p y$.
Differentiating (ii) again with respect to $x$:
$y^{\prime \prime} = \beta p e^{p x} + p y^{\prime}$
Substitute $\beta e^{p x} = y^{\prime} - p y$ into the equation:
$y^{\prime \prime} = p(y^{\prime} - p y) + p y^{\prime}$
$y^{\prime \prime} = p y^{\prime} - p^2 y + p y^{\prime}$
$y^{\prime \prime} = 2 p y^{\prime} - p^2 y$
Rearranging the terms,we get:
$y^{\prime \prime} - 2 p y^{\prime} + p^2 y = 0$

(A) Given differential equation: $3 x y' - 3 y + \sqrt{x^2 - y^2} = 0$.
Dividing by $x$: $3 y' - 3 \frac{y}{x} + \sqrt{1 - (\frac{y}{x})^2} = 0$.
Let $y = vx$,then $y' = v + x \frac{dv}{dx}$.
Substituting: $3(v + x \frac{dv}{dx}) - 3v + \sqrt{1 - v^2} = 0$.
$3v + 3x \frac{dv}{dx} - 3v + \sqrt{1 - v^2} = 0$.
$3x \frac{dv}{dx} = -\sqrt{1 - v^2}$.
Separating variables: $\frac{-dv}{\sqrt{1 - v^2}} = \frac{1}{3} \frac{dx}{x}$.
Integrating both sides: $\int \frac{-dv}{\sqrt{1 - v^2}} = \int \frac{1}{3} \frac{dx}{x}$.
$\cos^{-1}(v) = \frac{1}{3} \ln |x| + C$.
Since $y(1) = 1$,$v = \frac{y}{x} = 1$ when $x = 1$.
$\cos^{-1}(1) = \frac{1}{3} \ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Thus,$\cos^{-1}(\frac{y}{x}) = \frac{1}{3} \ln |x|$,which implies $3 \cos^{-1}(\frac{y}{x}) = \ln |x|$.

(A) Given differential equation is $\frac{dy}{dx} = \frac{1}{e^y - x}$.
Taking the reciprocal on both sides,we get:
$\frac{dx}{dy} = e^y - x$.
Rearranging the terms,we get the linear differential equation in $x$:
$\frac{dx}{dy} + x = e^y$.
Comparing this with the standard form $\frac{dx}{dy} + Px = Q$,where $P = 1$ and $Q = e^y$.
The integrating factor $(IF)$ is given by:
$IF = e^{\int P dy} = e^{\int 1 dy} = e^y$.
The solution is given by $x \cdot (IF) = \int (Q \cdot IF) dy + C$.
$x \cdot e^y = \int (e^y \cdot e^y) dy + C$.
$x \cdot e^y = \int e^{2y} dy + C$.
$x \cdot e^y = \frac{e^{2y}}{2} + C$.
Dividing by $e^y$,we get $x = \frac{e^y}{2} + Ce^{-y}$.
Wait,re-evaluating the original equation $y' = \frac{1}{e^y - x}$ as $\frac{dx}{dy} = e^y - x$.
If the equation was $y' = \frac{1}{e^{-y} - x}$,then $\frac{dx}{dy} = e^{-y} - x \Rightarrow \frac{dx}{dy} + x = e^{-y}$.
$IF$ $= e^{\int 1 dy} = e^y$.
$x \cdot e^y = \int e^{-y} \cdot e^y dy + C = \int 1 dy + C = y + C$.
$x = e^{-y}(y + C)$.
Thus,the correct option is $A$.

(B) Let the three vectors be $\vec{a} = 2 \hat{i}-3 \hat{j}+\hat{k}$,$\vec{b} = \hat{i}-2 \hat{j}+3 \hat{k}$,and $\vec{c} = 3 \hat{i}+\hat{j}-2 \hat{k}$.
To check if they are linearly independent,we calculate the scalar triple product $[\vec{a} \vec{b} \vec{c}]$,which is the determinant of the matrix formed by these vectors:
$\Delta = \begin{vmatrix} 2 & -3 & 1 \\ 1 & -2 & 3 \\ 3 & 1 & -2 \end{vmatrix}$
Expanding along the first row:
$\Delta = 2((-2)(-2) - (3)(1)) - (-3)((1)(-2) - (3)(3)) + 1((1)(1) - (-2)(3))$
$\Delta = 2(4 - 3) + 3(-2 - 9) + 1(1 + 6)$
$\Delta = 2(1) + 3(-11) + 1(7)$
$\Delta = 2 - 33 + 7 = -24$
Since the scalar triple product $\Delta \neq 0$,the vectors are linearly independent.

(B) Given that $a$ is perpendicular to both $b$ and $c$, $a$ is parallel to the vector $b \times c$. Thus, we can write $a = k(b \times c)$ for some scalar $k$.
The scalar triple product $c \cdot (a \times b)$ can be rewritten using the property of cyclic permutation: $c \cdot (a \times b) = a \cdot (b \times c)$.
Since $a$ is parallel to $b \times c$, the angle between $a$ and $b \times c$ is $0$ or $\pi$. Given $|a|=2$, we have $a = \pm 2 \frac{b \times c}{|b \times c|}$.
First, calculate the magnitude of the cross product $b \times c$:
$|b \times c| = |b| |c| \sin\left(\frac{2 \pi}{3}\right) = 3 \times 4 \times \frac{\sqrt{3}}{2} = 6 \sqrt{3}$.
Now, substitute this into the scalar triple product expression:
$c \cdot (a \times b) = a \cdot (b \times c) = |a| |b \times c| \cos(0) = 2 \times 6 \sqrt{3} = 12 \sqrt{3}$.

(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 $\theta = \frac{\pi}{3}$.

(C) Given that,$|a|=1, |b|=2, |c|=3$.
Since $b$ and $c$ are perpendicular,we have $b \cdot c = 0$.
The projection of $b$ on $a$ is $\frac{a \cdot b}{|a|}$ and the projection of $c$ on $a$ is $\frac{a \cdot c}{|a|}$.
Given that these projections are equal,we have $\frac{a \cdot b}{|a|} = \frac{a \cdot c}{|a|}$,which implies $a \cdot b = a \cdot c$.
Now,we calculate $|a-b+c|^2$:
$|a-b+c|^2 = |a|^2 + |b|^2 + |c|^2 - 2(a \cdot b) + 2(a \cdot c) - 2(b \cdot c)$.
Substituting the known values:
$|a-b+c|^2 = (1)^2 + (2)^2 + (3)^2 - 2(a \cdot b) + 2(a \cdot b) - 2(0)$.
$|a-b+c|^2 = 1 + 4 + 9 = 14$.
Therefore,$|a-b+c| = \sqrt{14}$.

(A) Let the position vectors of vertices $A, B, D$ be $\vec{0}, \vec{b}, \vec{d}$ respectively. Then the position vector of $C$ is $\vec{b}+\vec{d}$.
Since $P$ is the mid-point of $AD$,the position vector of $P$ is $\frac{\vec{d}}{2}$.
Let $Q$ divide $AC$ in the ratio $\lambda:1$ and $BP$ in the ratio $\mu:1$.
The position vector of $Q$ on $AC$ is $\frac{\lambda(\vec{b}+\vec{d}) + 1(\vec{0})}{\lambda+1} = \frac{\lambda}{\lambda+1}\vec{b} + \frac{\lambda}{\lambda+1}\vec{d}$.
The position vector of $Q$ on $BP$ is $\frac{\mu(\frac{\vec{d}}{2}) + 1(\vec{b})}{\mu+1} = \frac{1}{\mu+1}\vec{b} + \frac{\mu}{2(\mu+1)}\vec{d}$.
Comparing the coefficients of $\vec{b}$ and $\vec{d}$:
$\frac{\lambda}{\lambda+1} = \frac{1}{\mu+1}$ and $\frac{\lambda}{\lambda+1} = \frac{\mu}{2(\mu+1)}$.
From these,$\frac{1}{\mu+1} = \frac{\mu}{2(\mu+1)} \Rightarrow \mu = 2$.
Substituting $\mu=2$ into the first equation: $\frac{\lambda}{\lambda+1} = \frac{1}{2+1} = \frac{1}{3}$.
$3\lambda = \lambda+1 \Rightarrow 2\lambda = 1 \Rightarrow \lambda = \frac{1}{2}$.
Thus,the ratio $AQ:QC = \lambda:1 = \frac{1}{2}:1 = 1:2$.

(B) Given the direction cosines of two lines as $(l_1, m_1, n_1) = \left(-\frac{2}{\sqrt{21}}, \frac{C}{\sqrt{21}}, \frac{1}{\sqrt{21}}\right)$ and $(l_2, m_2, n_2) = \left(\frac{3}{\sqrt{54}}, \frac{3}{\sqrt{54}}, -\frac{6}{\sqrt{54}}\right)$.
Since the angle between the lines is $\frac{\pi}{2}$,the lines are perpendicular.
For perpendicular lines,the condition is $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Substituting the values:
$\left(-\frac{2}{\sqrt{21}}\right) \left(\frac{3}{\sqrt{54}}\right) + \left(\frac{C}{\sqrt{21}}\right) \left(\frac{3}{\sqrt{54}}\right) + \left(\frac{1}{\sqrt{21}}\right) \left(-\frac{6}{\sqrt{54}}\right) = 0$
$\frac{-6}{\sqrt{21}\sqrt{54}} + \frac{3C}{\sqrt{21}\sqrt{54}} - \frac{6}{\sqrt{21}\sqrt{54}} = 0$
Multiplying by $\sqrt{21}\sqrt{54}$ on both sides:
$-6 + 3C - 6 = 0$
$3C - 12 = 0$
$3C = 12$
$C = 4$.

(C) Let the given point be $P(x_1, y_1, z_1) = (5, 2, 6)$ and the plane be $ax + by + cz + d = 0$,where $x + y + z - 9 = 0$.
Here,$a = 1, b = 1, c = 1$ and $d = -9$.
The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ with respect to 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:
$\frac{x - 5}{1} = \frac{y - 2}{1} = \frac{z - 6}{1} = -2 \frac{(1)(5) + (1)(2) + (1)(6) - 9}{1^2 + 1^2 + 1^2}$
$\frac{x - 5}{1} = \frac{y - 2}{1} = \frac{z - 6}{1} = -2 \frac{5 + 2 + 6 - 9}{3} = -2 \frac{4}{3} = -\frac{8}{3}$
Now,solving for $x, y, z$:
$x - 5 = -\frac{8}{3} \Rightarrow x = 5 - \frac{8}{3} = \frac{15 - 8}{3} = \frac{7}{3}$
$y - 2 = -\frac{8}{3} \Rightarrow y = 2 - \frac{8}{3} = \frac{6 - 8}{3} = -\frac{2}{3}$
$z - 6 = -\frac{8}{3} \Rightarrow z = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3}$
Thus,the image of the point $(5, 2, 6)$ is $(\frac{7}{3}, -\frac{2}{3}, \frac{10}{3})$.

(D) Total number of outcomes $= 2^4 = 16$.
Let $X$ be the number of girls.
We need to find the probability of at least two girls,i.e.,$P(X \ge 2)$.
This can be calculated as $P(X \ge 2) = 1 - \{P(X=0) + P(X=1)\}$.
For $P(X=0)$ (no girls,all boys): The only outcome is $\{BBBB\}$,so $P(X=0) = \frac{1}{16}$.
For $P(X=1)$ (exactly one girl): The outcomes are $\{GBBB, BGBB, BBGB, BBBG\}$,so $P(X=1) = \frac{4}{16}$.
Therefore,$P(X \ge 2) = 1 - \left(\frac{1}{16} + \frac{4}{16}\right) = 1 - \frac{5}{16} = \frac{11}{16}$.

(A) Given: $P(A)=\frac{1}{4}$,$P(A|B)=\frac{1}{4}$,$P(B|A)=\frac{1}{2}$.
We know $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{4} \implies P(A \cap B) = \frac{1}{4}P(B)$.
Also $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{2} \implies P(A \cap B) = \frac{1}{2}P(A) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$.
Equating the two,$\frac{1}{4}P(B) = \frac{1}{8} \implies P(B) = \frac{1}{2}$.
$(I)$ $P(\bar{A}|\bar{B}) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})} = \frac{1-P(A \cup B)}{1-P(B)} = \frac{1-[P(A)+P(B)-P(A \cap B)]}{1-P(B)} = \frac{1-[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}]}{1-\frac{1}{2}} = \frac{1-\frac{5}{8}}{\frac{1}{2}} = \frac{3/8}{1/2} = \frac{3}{4}$. Statement $(I)$ is correct.
$(II)$ Since $P(A \cap B) = \frac{1}{8} \neq 0$,$A$ and $B$ are not mutually exclusive. Statement $(II)$ is incorrect.
$(III)$ $P(A|B)+P(A|\bar{B}) = \frac{P(A \cap B)}{P(B)} + \frac{P(A \cap \bar{B})}{P(\bar{B})} = \frac{1/8}{1/2} + \frac{P(A)-P(A \cap B)}{1-P(B)} = \frac{1}{4} + \frac{1/4-1/8}{1/2} = \frac{1}{4} + \frac{1/8}{1/2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \neq 1$. Statement $(III)$ is incorrect.
Thus,only $(I)$ is correct.

(D) Let $p$ denote the probability that a ship departed at $A$ reaches $B$ safely.
Then,$p = \frac{9}{10} = 0.9$.
Therefore,$q = 1 - p = 1 - 0.9 = 0.1$.
Let $X$ denote the number of ships that reach $B$ safely out of $n = 5$ ships.
This follows a binomial distribution $B(n, p) = B(5, 0.9)$.
The required probability is $P(X \geq 4) = P(X = 4) + P(X = 5)$.
Using the formula $P(X = k) = {}^nC_k p^k q^{n-k}$:
$P(X = 4) = {}^5C_4 (0.9)^4 (0.1)^1 = 5 \times (0.9)^4 \times 0.1 = 0.5 \times (0.9)^4$.
$P(X = 5) = {}^5C_5 (0.9)^5 (0.1)^0 = 1 \times (0.9)^5 \times 1 = 0.9 \times (0.9)^4$.
Adding these probabilities:
$P(X \geq 4) = 0.5(0.9)^4 + 0.9(0.9)^4 = (0.5 + 0.9)(0.9)^4 = 1.4(0.9)^4$.