TS EAMCET 2011 Mathematics Question Paper with Answer and Solution

(A) We know that for any natural number $n$,the expression $x^k + y^k$ is divisible by $x+y$ if $k$ is an odd number.
Since $n$ is a natural number,$2n-1$ is always an odd number.
Therefore,$a^{2n-1} + b^{2n-1}$ is divisible by $a+b$.

(D) Given,$\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$
Multiplying both sides by $(x-1)(x-2)(x-3)$,we get:
$x^2+x+1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$
To find $A$,put $x=1$:
$1^2+1+1 = A(1-2)(1-3) \Rightarrow 3 = A(-1)(-2) \Rightarrow 3 = 2A \Rightarrow A = \frac{3}{2}$
To find $C$,put $x=3$:
$3^2+3+1 = C(3-1)(3-2) \Rightarrow 13 = C(2)(1) \Rightarrow 13 = 2C \Rightarrow C = \frac{13}{2}$
Therefore,$A+C = \frac{3}{2} + \frac{13}{2} = \frac{16}{2} = 8$

(D) Given that $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$.
By the relation between roots and coefficients,we have:
$\tan A + \tan B = p$
$\tan A \tan B = q$
Using the trigonometric identity for $\tan(A+B)$:
$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{p}{1-q}$
Now,we know that $\sin^2 \theta = \frac{\tan^2 \theta}{1 + \tan^2 \theta}$.
Substituting $\tan(A+B) = \frac{p}{1-q}$:
$\sin^2(A+B) = \frac{(\frac{p}{1-q})^2}{1 + (\frac{p}{1-q})^2} = \frac{\frac{p^2}{(1-q)^2}}{\frac{(1-q)^2 + p^2}{(1-q)^2}} = \frac{p^2}{p^2+(1-q)^2}$

(B) Given,${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7$
$\Rightarrow \frac{15!}{7!} = A + 8 \cdot \frac{14!}{7!}$
$\Rightarrow A = \frac{15!}{7!} - 8 \cdot \frac{14!}{7!}$
$\Rightarrow A = \frac{15 \cdot 14!}{7!} - \frac{8 \cdot 14!}{7!}$
$\Rightarrow A = \frac{14!}{7!} (15 - 8)$
$\Rightarrow A = \frac{14!}{7!} \cdot 7$
$\Rightarrow A = \frac{14!}{6!} = { }^{14} P_8$

(B) five-digit number is divisible by $5$ if its unit place is either $0$ or $5$.
Case $I$: When $0$ is in the unit place,the remaining $4$ positions can be filled by the remaining $5$ digits $(1, 2, 3, 4, 5)$ in $^5P_4 = 120$ ways.
Case $II$: When $5$ is in the unit place,the first position (ten-thousands place) cannot be $0$. Thus,the first position can be filled by any of the $4$ digits $(1, 2, 3, 4)$. The remaining $3$ positions can be filled by the remaining $4$ digits (including $0$) in $^4P_3$ ways.
Number of ways for Case $II = 4 \times ^4P_3 = 4 \times 24 = 96$.
Total number of ways $= 120 + 96 = 216$.

(B) The total number of ways to draw $n$ pairs of balls such that each pair contains one white and one black ball is given by the product of the number of ways to choose one white and one black ball for each successive draw.
For the first draw,there are $n$ white and $n$ black balls. The number of ways to pick one white and one black ball is $\binom{n}{1} \times \binom{n}{1} = n^2$.
For the second draw,there are $(n-1)$ white and $(n-1)$ black balls remaining. The number of ways is $\binom{n-1}{1} \times \binom{n-1}{1} = (n-1)^2$.
Continuing this until the last pair,the total number of ways is $(n \times (n-1) \times \dots \times 1)^2 = (n!)^2$.
Given that $(n!)^2 = 14400$,we have $n! = \sqrt{14400} = 120$.
Since $5! = 120$,we find $n = 5$.

(D) Given,${ }^{n-1} C_3+{ }^{n-1} C_4>{ }^n C_3$
Using the identity ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$,we have:
${ }^n C_4>{ }^n C_3$
Expanding the combinations:
$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$
$\frac{1}{4(n-4)!} > \frac{1}{(n-3)(n-4)!}$
$\frac{1}{4} > \frac{1}{n-3}$
$n-3 > 4$
$n > 7$
Since $n$ must be an integer greater than $7$,the minimum value of $n$ is $8$.

(C) Given,$f(x) = \sin^6 x + \cos^6 x$.
We can write this as $f(x) = (\sin^2 x)^3 + (\cos^2 x)^3$.
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,where $a = \sin^2 x$ and $b = \cos^2 x$:
$f(x) = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$.
Since $\sin^2 x + \cos^2 x = 1$,we have:
$f(x) = \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x$.
Adding and subtracting $2 \sin^2 x \cos^2 x$ inside the expression:
$f(x) = (\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x = 1 - 3 \sin^2 x \cos^2 x$.
Multiply and divide by $4$:
$f(x) = 1 - \frac{3}{4}(4 \sin^2 x \cos^2 x) = 1 - \frac{3}{4}(\sin 2x)^2$.
Since $0 \leq \sin^2 2x \leq 1$,we have:
$1 - \frac{3}{4}(1) \leq f(x) \leq 1 - \frac{3}{4}(0)$.
$\frac{1}{4} \leq f(x) \leq 1$.
Thus,$f(x) \in \left[\frac{1}{4}, 1\right]$.

(D) Given expression: $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) = 16 \left[ 2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) \right]$
Using the formula $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$:
$= 16 \left[ \cos\left(\frac{5A}{2} - \frac{A}{2}\right) - \cos\left(\frac{5A}{2} + \frac{A}{2}\right) \right]$
$= 16 [ \cos(2A) - \cos(3A) ]$
Using $\cos 2A = 2 \cos^2 A - 1$ and $\cos 3A = 4 \cos^3 A - 3 \cos A$:
$= 16 [ (2 \cos^2 A - 1) - (4 \cos^3 A - 3 \cos A) ]$
Substitute $\cos A = \frac{3}{4}$:
$= 16 \left[ 2\left(\frac{3}{4}\right)^2 - 1 - 4\left(\frac{3}{4}\right)^3 + 3\left(\frac{3}{4}\right) \right]$
$= 16 \left[ 2\left(\frac{9}{16}\right) - 1 - 4\left(\frac{27}{64}\right) + \frac{9}{4} \right]$
$= 16 \left[ \frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4} \right]$
$= 16 \left[ \frac{18 - 16 - 27 + 36}{16} \right]$
$= 18 - 16 - 27 + 36 = 11$

(B) Given trigonometric equations are $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$.
Since $\tan \theta$ is negative and $\cos \theta$ is positive,$\theta$ must lie in the fourth quadrant.
The general solution for $\tan \theta = -1$ is $\theta = n \pi + \frac{3 \pi}{4}$.
The general solution for $\cos \theta = \frac{1}{\sqrt{2}}$ is $\theta = 2 n \pi \pm \frac{\pi}{4}$.
For $\theta$ to satisfy both,we look for the angle in the fourth quadrant where $\tan \theta = -1$ and $\cos \theta = \frac{1}{\sqrt{2}}$,which is $\theta = \frac{7 \pi}{4}$.
Thus,the general solution is $\theta = 2 n \pi + \frac{7 \pi}{4}$.

(C) The quadrilateral is bounded by the lines $x = 0$ ($y$-axis),$y = 0$ ($x$-axis),$2x + y = 2$,and $x + y = 5$.
For a point $(x, y)$ to be inside the quadrilateral with natural number coordinates $(x, y \in \{1, 2, 3, \dots\})$,it must satisfy the inequalities:
$1) \ x > 0$
$2) \ y > 0$
$3) \ 2x + y > 2$
$4) \ x + y < 5$
Testing integer values for $x$:
If $x = 1$: $2(1) + y > 2 \implies 2 + y > 2 \implies y > 0$. Also $1 + y < 5 \implies y < 4$. So $y \in \{1, 2, 3\}$. Points: $(1, 1), (1, 2), (1, 3)$.
If $x = 2$: $2(2) + y > 2 \implies 4 + y > 2 \implies y > -2$. Also $2 + y < 5 \implies y < 3$. So $y \in \{1, 2\}$. Points: $(2, 1), (2, 2)$.
If $x = 3$: $2(3) + y > 2 \implies 6 + y > 2 \implies y > -4$. Also $3 + y < 5 \implies y < 2$. So $y = 1$. Point: $(3, 1)$.
If $x \ge 4$: $x + y < 5$ cannot be satisfied for natural numbers $y \ge 1$.
The total number of such points is $3 + 2 + 1 = 6$.

(A) Let the coordinates of $A$ be $(x_1, y_1) = (2, 0)$ and $B$ be $(x_2, y_2) = (3, 1)$.
The length of $AB$ is $r = \sqrt{(3-2)^2 + (1-0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The angle $\theta$ that $AB$ makes with the positive $x$-axis is given by $\tan \theta = \frac{1-0}{3-2} = 1$,so $\theta = 45^{\circ}$.
When the line is rotated by $45^{\circ}$ in the anti-clockwise direction about $A$,the new angle $\theta'$ becomes $\theta + 45^{\circ} = 45^{\circ} + 45^{\circ} = 90^{\circ}$.
Let the new coordinates of $B$ be $(x', y')$.
Using the rotation formula,$x' = x_1 + r \cos \theta'$ and $y' = y_1 + r \sin \theta'$.
$x' = 2 + \sqrt{2} \cos 90^{\circ} = 2 + \sqrt{2}(0) = 2$.
$y' = 0 + \sqrt{2} \sin 90^{\circ} = 0 + \sqrt{2}(1) = \sqrt{2}$.
Thus,the new coordinates are $(2, \sqrt{2})$.

(C) The equation of the line perpendicular to $x+3y=7$ is of the form $3x-y+\lambda=0$.
Since this line passes through the point $(3,8)$,we have:
$3(3) - 8 + \lambda = 0$ $\Rightarrow 9 - 8 + \lambda = 0$ $\Rightarrow \lambda = -1$.
Thus,the equation of the perpendicular line is $3x-y-1=0$.
The foot of the perpendicular is the intersection of $x+3y=7$ and $3x-y-1=0$.
Solving these,we get $x=1$ and $y=2$.
Let the image of the point $(3,8)$ be $(x_1, y_1)$. Since $(1,2)$ is the midpoint of the segment joining $(3,8)$ and $(x_1, y_1)$,we have:
$\frac{3+x_1}{2} = 1 \Rightarrow x_1 = -1$
$\frac{8+y_1}{2} = 2 \Rightarrow y_1 = -4$
Therefore,the image is $(-1, -4)$.

(C) Let $P(x, y)$ be any point on the locus. According to the problem,the sum of distances from $P$ to $F_1(0, 2)$ and $F_2(0, -2)$ is $6$.
$\sqrt{x^2 + (y - 2)^2} + \sqrt{x^2 + (y + 2)^2} = 6$
$\sqrt{x^2 + (y - 2)^2} = 6 - \sqrt{x^2 + (y + 2)^2}$
Squaring both sides:
$x^2 + y^2 - 4y + 4 = 36 + x^2 + y^2 + 4y + 4 - 12\sqrt{x^2 + (y + 2)^2}$
$-8y - 36 = -12\sqrt{x^2 + (y + 2)^2}$
$2y + 9 = 3\sqrt{x^2 + (y + 2)^2}$
Squaring again:
$4y^2 + 36y + 81 = 9(x^2 + y^2 + 4y + 4)$
$4y^2 + 36y + 81 = 9x^2 + 9y^2 + 36y + 36$
$9x^2 + 5y^2 = 45$

(D) The $r$-th term in the expansion of $(3+7x)^{29}$ is $T_r = {}^{29}C_{r-1} (3)^{29-(r-1)} (7x)^{r-1}$. The coefficient is ${}^{29}C_{r-1} (3)^{30-r} (7)^{r-1}$.
The $(r+1)$-th term is $T_{r+1} = {}^{29}C_r (3)^{29-r} (7x)^r$. The coefficient is ${}^{29}C_r (3)^{29-r} (7)^r$.
Given that the coefficients are equal:
${}^{29}C_{r-1} (3)^{30-r} (7)^{r-1} = {}^{29}C_r (3)^{29-r} (7)^r$
Divide both sides by ${}^{29}C_{r-1} (3)^{29-r} (7)^{r-1}$:
$3 = {}^{29}C_r / {}^{29}C_{r-1} \times 7$
Using the formula ${}^nC_r / {}^nC_{r-1} = (n-r+1)/r$:
$3 = \frac{29-r+1}{r} \times 7$
$3r = 7(30-r)$
$3r = 210 - 7r$
$10r = 210$
$r = 21$

(A) We have the series $\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.
Rewrite the numerator as $(2n+1) - 1$:
$\sum_{n=1}^{\infty} \frac{2n+1-1}{(2n+1)!} = \sum_{n=1}^{\infty} \left( \frac{2n+1}{(2n+1)!} - \frac{1}{(2n+1)!} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right)$.
Expanding the summation:
$= \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{4!} - \frac{1}{5!} \right) + \dots$
Recall the Taylor series expansion for $e^x$ at $x = -1$:
$e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} = \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$
$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$
Thus,the sum is equal to $e^{-1} = 1/e$.

(A) Given,$P(A)=0.5, P(B)=0.4$ and $P(A \cup B)=0.6$.
$(i) \ P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.6 = 0.3$. Thus,$(i)-(3)$.
$(ii) \ P(A \cap \bar{B}) = P(A) - P(A \cap B) = 0.5 - 0.3 = 0.2$. Thus,$(ii)-(2)$.
$(iii) \ P(\bar{A} \cap B) = P(B) - P(A \cap B) = 0.4 - 0.3 = 0.1$. Thus,$(iii)-(4)$.
$(iv) \ P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$. Thus,$(iv)-(1)$.
Therefore,the correct match is $(i)-(3), (ii)-(2), (iii)-(4), (iv)-(1)$.

(A) Let the common root be $\alpha$. Then $\alpha^3 + a\alpha + 1 = 0$ and $\alpha^4 + a\alpha^2 + 1 = 0$.
From the first equation,$a\alpha = -\alpha^3 - 1$.
Substituting this into the second equation: $\alpha^4 + \alpha(a\alpha^2) + 1 = 0$ is not direct,so multiply the first by $\alpha$: $\alpha^4 + a\alpha^2 + \alpha = 0$.
Subtracting this from the second equation: $(\alpha^4 + a\alpha^2 + 1) - (\alpha^4 + a\alpha^2 + \alpha) = 0$.
This gives $1 - \alpha = 0$,so $\alpha = 1$.
Substituting $\alpha = 1$ into the first equation: $1^3 + a(1) + 1 = 0$.
$1 + a + 1 = 0$,which implies $a = -2$.

(D) Given the quadratic expression $y = ax^2 + bx + c$.
Since $a > 0$,the parabola opens upwards.
The discriminant $D = b^2 - 4ac = 0$ implies that the quadratic equation $ax^2 + bx + c = 0$ has two equal real roots.
This means the curve touches the $x$-axis at exactly one point.
Since $a > 0$,the vertex of the parabola is at its minimum value,which is $0$ on the $x$-axis,and the rest of the curve lies above the $x$-axis.
Therefore,the curve touches the $x$-axis and lies above it.

(D) We know that $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$.
Given expression: $E = \frac{(1+i)^{2011}}{(1-i)^{2009}} = \frac{(1+i)^{2009} \cdot (1+i)^2}{(1-i)^{2009}}$.
$E = \left(\frac{1+i}{1-i}\right)^{2009} \cdot (1+i)^2$.
Simplify $\frac{1+i}{1-i}$: $\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{1+i^2+2i}{1-i^2} = \frac{2i}{2} = i$.
Substitute back: $E = (i)^{2009} \cdot (2i) = 2 \cdot i^{2010}$.
Since $i^4 = 1$,$i^{2010} = (i^4)^{502} \cdot i^2 = 1^{502} \cdot (-1) = -1$.
Therefore,$E = 2 \cdot (-1) = -2$.

(B) Given,$z = a - \frac{i}{2}$.
We need to calculate $|i + z|^2 - |i - z|^2$.
First,substitute $z$ into the expressions:
$i + z = i + a - \frac{i}{2} = a + \frac{i}{2}$.
$i - z = i - (a - \frac{i}{2}) = -a + \frac{3i}{2}$.
Now,calculate the squared moduli:
$|i + z|^2 = |a + \frac{i}{2}|^2 = a^2 + (\frac{1}{2})^2 = a^2 + \frac{1}{4}$.
$|i - z|^2 = |-a + \frac{3i}{2}|^2 = (-a)^2 + (\frac{3}{2})^2 = a^2 + \frac{9}{4}$.
Finally,subtract the two values:
$|i + z|^2 - |i - z|^2 = (a^2 + \frac{1}{4}) - (a^2 + \frac{9}{4}) = \frac{1}{4} - \frac{9}{4} = -\frac{8}{4} = -2$.

(A) Let $z = x + iy$.
Given $\arg \left(\frac{z-2}{z+2}\right) = \frac{\pi}{3}$.
Using the property $\arg \left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$,we have:
$\arg(z-2) - \arg(z+2) = \frac{\pi}{3}$.
Substituting $z = x + iy$:
$\arg((x-2) + iy) - \arg((x+2) + iy) = \frac{\pi}{3}$.
Using $\arg(x+iy) = \tan^{-1}\left(\frac{y}{x}\right)$:
$\tan^{-1}\left(\frac{y}{x-2}\right) - \tan^{-1}\left(\frac{y}{x+2}\right) = \frac{\pi}{3}$.
Applying the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$:
$\tan^{-1}\left[\frac{\frac{y}{x-2} - \frac{y}{x+2}}{1 + \frac{y}{x-2} \cdot \frac{y}{x+2}}\right] = \frac{\pi}{3}$.
$\frac{\frac{y(x+2) - y(x-2)}{(x-2)(x+2)}}{\frac{(x^2-4) + y^2}{x^2-4}} = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.
$\frac{4y}{x^2 + y^2 - 4} = \sqrt{3}$.
$x^2 + y^2 - \frac{4}{\sqrt{3}}y - 4 = 0$.
This is the equation of a circle.

(D) The general term in the expansion of $(a+bx)^n$ is $T_{k+1} = {^nC_k} a^{n-k} (bx)^k$.
For the expansion of $(3+7x)^{29}$,the $r$ th term is $T_r = {^{29}C_{r-1}} (3)^{29-(r-1)} (7x)^{r-1}$.
The coefficient of the $r$ th term is ${^{29}C_{r-1}} (3)^{30-r} (7)^{r-1}$.
The $(r+1)$ th term is $T_{r+1} = {^{29}C_r} (3)^{29-r} (7x)^r$.
The coefficient of the $(r+1)$ th term is ${^{29}C_r} (3)^{29-r} (7)^r$.
Given that these coefficients are equal:
${^{29}C_{r-1}} (3)^{30-r} (7)^{r-1} = {^{29}C_r} (3)^{29-r} (7)^r$
Divide both sides by ${^{29}C_{r-1}} (3)^{29-r} (7)^{r-1}$:
$3 = {^{29}C_r} / {^{29}C_{r-1}} \times 7$
$3/7 = {^{29}C_r} / {^{29}C_{r-1}}$
Using the property ${^nC_r} / {^nC_{r-1}} = (n-r+1)/r$:
$3/7 = (29-r+1) / r$
$3/7 = (30-r) / r$
$3r = 7(30-r)$
$3r = 210 - 7r$
$10r = 210$
$r = 21$

(B) The given equation of the pair of lines is $3x^2 - 2xy - 15y^2 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we get $a = 3$,$2h = -2$ (so $h = -1$),and $b = -15$.
Let the slopes of the lines be $m_1$ and $m_2$.
The sum of the slopes is $s = m_1 + m_2 = -\frac{2h}{b} = -\frac{2(-1)}{-15} = -\frac{2}{15}$.
The product of the slopes is $p = m_1m_2 = \frac{a}{b} = \frac{3}{-15} = -\frac{3}{15}$.
Therefore,the ratio $s:p = \left(-\frac{2}{15}\right) : \left(-\frac{3}{15}\right) = 2:3$.

(B) The general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of parallel lines if $h^2=ab$ and $bg^2=af^2$.
From the condition $h^2=ab$,we have $h=\sqrt{ab}$.
From the condition $bg^2=af^2$,we can write $\frac{g^2}{f^2}=\frac{a}{b}$.
Taking the square root on both sides,we get $\frac{g}{f}=\sqrt{\frac{a}{b}}$.
For parallel lines,the distance between them is given by $2\sqrt{\frac{g^2-ac}{a(a+b)}} = 2\sqrt{\frac{f^2-bc}{b(a+b)}}$.
This implies $\frac{g^2-ac}{a} = \frac{f^2-bc}{b}$.
Rearranging the terms,we get $\frac{g^2-ac}{f^2-bc} = \frac{a}{b}$.
Therefore,$\sqrt{\frac{g^2-ac}{f^2-bc}} = \sqrt{\frac{a}{b}}$.

(C) The pair of straight lines is given by $ax^2+2hxy+by^2=0$.
If one of the lines bisects the angle between the coordinate axes,its equation is $y=x$ or $y=-x$.
This implies that the line satisfies the equation $y^2=x^2$,or $x^2-y^2=0$.
The condition for one of the lines of $ax^2+2hxy+by^2=0$ to be $y=mx$ is $am^2+2hm+b=0$.
For $y=x$,$m=1$,so $a+2h+b=0$. For $y=-x$,$m=-1$,so $a-2h+b=0$.
Combining these,we get $(a+b)^2 = (2h)^2 = 4h^2$.
Given $4x^2+6xy+ky^2=0$,we have $a=4, 2h=6 \Rightarrow h=3, b=k$.
Substituting into $(a+b)^2=4h^2$:
$(4+k)^2 = 4(3)^2 = 36$.
$4+k = \pm 6$.
If $4+k=6$,then $k=2$.
If $4+k=-6$,then $k=-10$.
Thus,$k \in \{-10, 2\}$.

(D) The given line is $y=2x+c$,which can be written as $2x-y+c=0$.
Comparing this with $y=mx+c_1$,we have $m=2$.
The equation of the circle is $x^2+y^2=5$,so the radius $r=\sqrt{5}$.
The condition for a line $y=mx+c_1$ to be a tangent to the circle $x^2+y^2=r^2$ is $c_1^2 = r^2(1+m^2)$.
Substituting the values,we get $c^2 = 5(1+2^2)$.
$c^2 = 5(1+4) = 5(5) = 25$.
Therefore,$c = \pm 5$.
Among the given options,the value of $c$ is $5$.

(C) The given lines are $3x + 4y - 14 = 0$ and $6x + 8y + 7 = 0$.
To make the coefficients of $x$ and $y$ the same,divide the second equation by $2$:
$3x + 4y + \frac{7}{2} = 0$.
Since these lines are parallel tangents to the circle,the diameter of the circle is equal to the distance between these two parallel lines.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 3$,$b = 4$,$c_1 = -14$,and $c_2 = \frac{7}{2}$.
$d = \frac{|-14 - \frac{7}{2}|}{\sqrt{3^2 + 4^2}} = \frac{|-\frac{28}{2} - \frac{7}{2}|}{\sqrt{9 + 16}} = \frac{|-\frac{35}{2}|}{5} = \frac{35}{2 \times 5} = \frac{7}{2}$.
The radius $r$ of the circle is half of the distance between the parallel tangents:
$r = \frac{d}{2} = \frac{7/2}{2} = \frac{7}{4}$.

(B) The given equations of the circles are:
$S_1 \equiv x^2+y^2+2x+2y+1=0$
$S_2 \equiv x^2+y^2-2x+2y+1=0$
For circle $S_1$,the centre is $C_1 = (-1, -1)$ and the radius is $r_1 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1+1-1} = 1$.
For circle $S_2$,the centre is $C_2 = (1, -1)$ and the radius is $r_2 = \sqrt{(1)^2 + (-1)^2 - 1} = \sqrt{1+1-1} = 1$.
The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-1))^2 + (-1 - (-1))^2} = \sqrt{2^2 + 0^2} = 2$.
Since $d = r_1 + r_2$ $(2 = 1 + 1)$,the circles touch each other externally.
The point of contact is the midpoint of the line segment joining the centres $C_1$ and $C_2$:
$\text{Point of contact} = \left(\frac{-1+1}{2}, \frac{-1-1}{2}\right) = (0, -1)$.

(B) Given the circle $x^2+y^2+8x-4y+c=0$,its center $C_1 = (-4, 2)$ and radius $r_1 = \sqrt{(-4)^2 + 2^2 - c} = \sqrt{20-c}$.
For the circle $x^2+y^2+2x+4y-11=0$,its center $C_2 = (-1, -2)$ and radius $r_2 = \sqrt{(-1)^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = 4$.
Since the circles touch externally,$C_1C_2 = r_1 + r_2$.
$C_1C_2 = \sqrt{(-4 - (-1))^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$.
So,$5 = \sqrt{20-c} + 4$,which implies $\sqrt{20-c} = 1$,so $20-c = 1$,hence $c = 19$.
Now,the circle $x^2+y^2+8x-4y+19=0$ cuts $x^2+y^2-6x+8y+k=0$ orthogonally.
The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Here,$g_1 = 4, f_1 = -2, c_1 = 19$ and $g_2 = -3, f_2 = 4, c_2 = k$.
$2(4)(-3) + 2(-2)(4) = 19 + k$.
$-24 - 16 = 19 + k$.
$-40 = 19 + k$.
$k = -59$.

(B) Let the coordinates of point $A$ be $(a \cos \theta, a \sin \theta)$.
Since $AM$ is parallel to the $X$-axis and has length $a$,the coordinates of point $M(x, y)$ are given by $(a \cos \theta + a, a \sin \theta)$ or $(a \cos \theta - a, a \sin \theta)$.
Case $1$: $x = a \cos \theta + a$ and $y = a \sin \theta$.
Then $x - a = a \cos \theta$ and $y = a \sin \theta$.
Squaring and adding,we get $(x - a)^2 + y^2 = a^2 \cos^2 \theta + a^2 \sin^2 \theta = a^2$.
$x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 = 2ax$.
Case $2$: $x = a \cos \theta - a$ and $y = a \sin \theta$.
Then $x + a = a \cos \theta$ and $y = a \sin \theta$.
Squaring and adding,we get $(x + a)^2 + y^2 = a^2 \cos^2 \theta + a^2 \sin^2 \theta = a^2$.
$x^2 + 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 = -2ax$.
Combining these,the locus is $x^2 + y^2 = \pm 2ax$. Among the given options,$x^2 + y^2 = 2ax$ is the correct one.

(D) The axis of the parabola $y^2=lx$ is the $x$-axis,which has the equation $y=0$.
Since the line $y=mx+c$ is parallel to the $x$-axis,its slope $m$ must be $0$.
Thus,the equation of the line becomes $y=c$.
Given that the line intersects the parabola at the point $\left(\frac{c^2}{8}, c\right)$,this point must satisfy the equation of the parabola $y^2=lx$.
Substituting $y=c$ and $x=\frac{c^2}{8}$ into the equation $y^2=lx$:
$c^2 = l \left(\frac{c^2}{8}\right)$
Assuming $c \neq 0$,we can divide both sides by $c^2$:
$1 = \frac{l}{8}$
$l = 8$
The length of the latus rectum of the parabola $y^2=lx$ is $l$.
Therefore,the length of the latus rectum is $8$.

(C) Let $P(t^2, 2t)$ be one end of a focal chord $PQ$ of the parabola $y^2=4x$. The coordinates of the other end $Q$ are $(\frac{1}{t^2}, \frac{-2}{t})$,since $tt' = -1$.
Given that the chord makes an angle $\theta$ with the positive direction of the $X$-axis,the slope of the chord is $\tan \theta$.
$\tan \theta = \frac{\frac{-2}{t} - 2t}{\frac{1}{t^2} - t^2} = \frac{-2(t + \frac{1}{t})}{\frac{1-t^4}{t^2}} = \frac{-2(t + \frac{1}{t})}{\frac{(1-t^2)(1+t^2)}{t^2}} = \frac{2t}{t^2-1}$.
Alternatively,using the slope formula for a focal chord: $\tan \theta = \frac{2}{t - \frac{1}{t}}$,so $t - \frac{1}{t} = 2 \cot \theta$.
The length of the focal chord $PQ$ is given by $a(t + \frac{1}{t})^2$,where $a=1$.
$PQ = (t + \frac{1}{t})^2 = (t - \frac{1}{t})^2 + 4$.
Substituting $t - \frac{1}{t} = 2 \cot \theta$:
$PQ = (2 \cot \theta)^2 + 4 = 4 \cot^2 \theta + 4 = 4(1 + \cot^2 \theta) = 4 \operatorname{cosec}^2 \theta$.

(A) Given equation of the ellipse: $x^2+4 y^2+2 x+16 y+13=0$
Completing the square for $x$ and $y$ terms:
$(x^2+2x+1) + 4(y^2+4y+4) + 13 - 1 - 16 = 0$
$(x+1)^2 + 4(y+2)^2 = 4$
Dividing by $4$:
$\frac{(x+1)^2}{4} + \frac{(y+2)^2}{1} = 1$
Comparing with the standard form $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$,we get $a^2 = 4$ and $b^2 = 1$.
Since $a^2 > b^2$,the eccentricity $e$ is given by $b^2 = a^2(1 - e^2)$.
$1 = 4(1 - e^2)$
$1 - e^2 = \frac{1}{4}$
$e^2 = 1 - \frac{1}{4} = \frac{3}{4}$
$e = \frac{\sqrt{3}}{2}$

(C) The given equation of the hyperbola is $x^2-3y^2=3$. Dividing by $3$,we get $\frac{x^2}{3}-\frac{y^2}{1}=1$.
Here,$a^2=3$ and $b^2=1$,so $a=\sqrt{3}$ and $b=1$.
The equations of the asymptotes are $y = \pm \frac{b}{a}x$,which gives $y = \frac{1}{\sqrt{3}}x$ and $y = -\frac{1}{\sqrt{3}}x$.
Let the slopes be $m_1 = \frac{1}{\sqrt{3}}$ and $m_2 = -\frac{1}{\sqrt{3}}$.
The angle $\theta$ between the asymptotes is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{\frac{1}{\sqrt{3}} - (-\frac{1}{\sqrt{3}})}{1 + (\frac{1}{\sqrt{3}})(-\frac{1}{\sqrt{3}})} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \right| = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = \frac{\pi}{3}$.

(C) To evaluate the limit $\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$,we rationalize the numerator:
$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8} \times \frac{\sqrt{1+\sqrt{1+x}}+2}{\sqrt{1+\sqrt{1+x}}+2}$
$= \lim _{x \rightarrow 8} \frac{1+\sqrt{1+x}-4}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$
$= \lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)}$
Now,rationalize the remaining radical in the numerator:
$= \lim _{x \rightarrow 8} \frac{\sqrt{1+x}-3}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)} \times \frac{\sqrt{1+x}+3}{\sqrt{1+x}+3}$
$= \lim _{x \rightarrow 8} \frac{1+x-9}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
$= \lim _{x \rightarrow 8} \frac{x-8}{(x-8)(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
$= \lim _{x \rightarrow 8} \frac{1}{(\sqrt{1+\sqrt{1+x}}+2)(\sqrt{1+x}+3)}$
Substitute $x = 8$:
$= \frac{1}{(\sqrt{1+\sqrt{9}}+2)(\sqrt{9}+3)}$
$= \frac{1}{(\sqrt{1+3}+2)(3+3)}$
$= \frac{1}{(2+2)(6)} = \frac{1}{4 \times 6} = \frac{1}{24}$

(A) We have the series $\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.
Rewrite the numerator as $(2n+1) - 1$:
$\sum_{n=1}^{\infty} \frac{2n+1-1}{(2n+1)!} = \sum_{n=1}^{\infty} \left( \frac{2n+1}{(2n+1)!} - \frac{1}{(2n+1)!} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right)$.
Expanding the summation:
$= \left( \frac{1}{2!} - \frac{1}{3!} \right) + \left( \frac{1}{4!} - \frac{1}{5!} \right) + \dots$
Recall the Taylor series for $e^x$ at $x=1$: $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ and $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$
Thus,$\frac{1}{e} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$
Therefore,the sum is equal to $\frac{1}{e}$.

(C) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
By the sine rule,we know that $a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k$ is a constant.
Substituting these values into the given equation:
$\frac{\cos A}{k \sin A} = \frac{\cos B}{k \sin B} = \frac{\cos C}{k \sin C}$
$\Rightarrow \cot A = \cot B = \cot C$
Since $A, B, C$ are angles of a triangle,$A = B = C$.
Since all angles are equal,the triangle is an equilateral triangle.

(A) Given: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$
Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$:
$a \cdot \frac{s(s-c)}{ab} + c \cdot \frac{s(s-a)}{bc} = \frac{3b}{2}$
$\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$
$\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$
Since $2s = a + b + c$,we have $2s - a - c = b$:
$\frac{s}{b} (b) = \frac{3b}{2}$
$s = \frac{3b}{2}$
$2s = 3b$
$a + b + c = 3b$
$a + c = 2b$
Therefore,$a, b, c$ are in arithmetic progression.

(A) We know that the formula for the inverse hyperbolic sine function is $\sinh ^{-1}(y) = \log \left(y + \sqrt{1 + y^2}\right)$.
Substituting $y = \cot x$ into the formula,we get:
$\sinh ^{-1}(\cot x) = \log \left(\cot x + \sqrt{1 + \cot ^2 x}\right)$
Since $1 + \cot ^2 x = \operatorname{cosec}^2 x$,we have:
$\sinh ^{-1}(\cot x) = \log \left(\cot x + \sqrt{\operatorname{cosec}^2 x}\right) = \log (\cot x + \operatorname{cosec} x)$
Using trigonometric identities $\cot x = \frac{\cos x}{\sin x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$,we get:
$\log \left(\frac{\cos x + 1}{\sin x}\right)$
Using half-angle formulas $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$\log \left(\frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) = \log \left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) = \log \left(\cot \frac{x}{2}\right)$.

(D) Let $y = \frac{x^2-3x+4}{x^2+3x+4}$.
$y(x^2+3x+4) = x^2-3x+4$
$yx^2 + 3yx + 4y = x^2 - 3x + 4$
$x^2(y-1) + x(3y+3) + (4y-4) = 0$.
Since $x$ is real,the discriminant $D \geq 0$.
$D = (3y+3)^2 - 4(y-1)(4y-4) \geq 0$
$9(y+1)^2 - 16(y-1)^2 \geq 0$
$9(y^2+2y+1) - 16(y^2-2y+1) \geq 0$
$9y^2 + 18y + 9 - 16y^2 + 32y - 16 \geq 0$
$-7y^2 + 50y - 7 \geq 0$
$7y^2 - 50y + 7 \leq 0$
$(7y-1)(y-7) \leq 0$.
Thus,the interval is $[\frac{1}{7}, 7]$.

(D) The function is given by $f(x) = 7 + \cos(5x + 3)$.
We know that the fundamental period of the function $\cos(ax + b)$ is given by $\frac{2\pi}{|a|}$.
Here,$a = 5$.
Therefore,the period of $\cos(5x + 3)$ is $\frac{2\pi}{|5|} = \frac{2\pi}{5}$.
Since adding a constant $7$ to the function does not change its period,the period of $f(x)$ is $\frac{2\pi}{5}$.

(C) Let $h = 2500 \ m$ be the height of the observation point above the lake. Let $H$ be the height of the cloud above the lake surface. The distance of the cloud from the observation point is $H-h$. The distance of the reflection of the cloud from the observation point is $H+h$.
Let $x$ be the horizontal distance from the observation point to the cloud.
In the triangle formed by the observation point,the cloud,and the horizontal line,we have $\cot 15^{\circ} = \frac{x}{H-h} \Rightarrow x = (H-h)(2+\sqrt{3}) \quad (i)$
In the triangle formed by the observation point,the reflection of the cloud,and the horizontal line,we have $\cot 45^{\circ} = \frac{x}{H+h} \Rightarrow x = H+h \quad (ii)$
Equating $(i)$ and $(ii)$:
$(H-h)(2+\sqrt{3}) = H+h$
$H(2+\sqrt{3}) - h(2+\sqrt{3}) = H+h$
$H(2+\sqrt{3}-1) = h(2+\sqrt{3}+1)$
$H(1+\sqrt{3}) = h(3+\sqrt{3})$
$H = h \frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = h\sqrt{3}$
Given $h = 2500 \ m$,so $H = 2500\sqrt{3} \ m$.

(A) We know that for any odd natural number $k$,the expression $a^k + b^k$ is divisible by $(a+b)$.
Given the expression $a^{2n-1} + b^{2n-1}$,where $n$ is a natural number,the exponent $(2n-1)$ is always an odd number.
Therefore,$a^{2n-1} + b^{2n-1}$ is divisible by $(a+b)$.

(A) The given polar equation of the line is $\sin \theta - \cos \theta = \frac{1}{r}$.
Multiplying by $r$,we get $r \sin \theta - r \cos \theta = 1$.
Using the conversion $x = r \cos \theta$ and $y = r \sin \theta$,the Cartesian equation is $y - x = 1$,or $x - y + 1 = 0$.
The slope of this line is $m_1 = 1$.
The line perpendicular to this line will have a slope $m_2 = -1$.
The point given is $\left(2, \frac{\pi}{6}\right)$. Converting to Cartesian coordinates:
$x = 2 \cos \left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$
$y = 2 \sin \left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1$
So the point is $(\sqrt{3}, 1)$.
The equation of the line with slope $-1$ passing through $(\sqrt{3}, 1)$ is:
$y - 1 = -1(x - \sqrt{3})$
$y - 1 = -x + \sqrt{3}$
$x + y = \sqrt{3} + 1$
Converting back to polar form using $x = r \cos \theta$ and $y = r \sin \theta$:
$r \cos \theta + r \sin \theta = \sqrt{3} + 1$
$r(\sin \theta + \cos \theta) = \sqrt{3} + 1$
$\sin \theta + \cos \theta = \frac{\sqrt{3} + 1}{r}$

(A) Given,$P(A)=0.5, P(B)=0.4$ and $P(A \cup B)=0.6$.
$(i) \ P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.5 + 0.4 - 0.6 = 0.3$. Thus,$(i) \rightarrow (3)$.
$(ii) \ P(A \cap \bar{B}) = P(A) - P(A \cap B) = 0.5 - 0.3 = 0.2$. Thus,$(ii) \rightarrow (2)$.
$(iii) \ P(\bar{A} \cap B) = P(B) - P(A \cap B) = 0.4 - 0.3 = 0.1$. Thus,$(iii) \rightarrow (4)$.
$(iv) \ P(\bar{A} \cap \bar{B}) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$. Thus,$(iv) \rightarrow (1)$.
Therefore,the correct match is $(i)-3, (ii)-2, (iii)-4, (iv)-1$.

(A) Total number of students = $15 + 5 = 20$.
Number of ways to select $3$ students from $20$ is $^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
Number of ways to select $2$ boys from $15$ and $1$ girl from $5$ is $^{15}C_2 \times ^5C_1$.
$^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.
$^{5}C_1 = 5$.
Number of favorable outcomes = $105 \times 5 = 525$.
Required probability = $\frac{525}{1140} = \frac{35}{76}$.

(C) We know that $1^{\circ} = \alpha$ radians,so $1^{\prime} = \frac{\alpha}{60}$ radians.
Using the differential approximation,$\cos(x + \Delta x) \approx \cos(x) - \sin(x) \Delta x$.
Here,$x = 60^{\circ} = \frac{\pi}{3}$ and $\Delta x = 1^{\prime} = \frac{\alpha}{60}$.
Thus,$\cos(60^{\circ} 1^{\prime}) \approx \cos(60^{\circ}) - \sin(60^{\circ}) \times \frac{\alpha}{60}$.
Substituting the values $\cos(60^{\circ}) = \frac{1}{2}$ and $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$,we get:
$\cos(60^{\circ} 1^{\prime}) \approx \frac{1}{2} - \frac{\sqrt{3}}{2} \times \frac{\alpha}{60} = \frac{1}{2} - \frac{\alpha \sqrt{3}}{120}$.

(D) Let $A = \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right]$.
Given the equation $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] = \left[\begin{array}{l} 1 \\ 0 \end{array}\right]$.
Multiplying the matrices on the left side:
$\left[\begin{array}{c} 2x_1 + x_2 \\ 3x_1 + 2x_2 \end{array}\right] = \left[\begin{array}{l} 1 \\ 0 \end{array}\right]$.
Equating the corresponding elements,we get the system of linear equations:
$2x_1 + x_2 = 1$ (Equation $1$)
$3x_1 + 2x_2 = 0$ (Equation $2$)
From Equation $1$,$x_2 = 1 - 2x_1$.
Substituting this into Equation $2$:
$3x_1 + 2(1 - 2x_1) = 0$
$3x_1 + 2 - 4x_1 = 0$
$-x_1 + 2 = 0 \Rightarrow x_1 = 2$.
Now,finding $x_2$:
$x_2 = 1 - 2(2) = 1 - 4 = -3$.
Therefore,$A = \left[\begin{array}{r} 2 \\ -3 \end{array}\right]$.

(B) Given,$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right]$.
First,calculate $A^2$:
$A^2 = \left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right] \left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right] = \left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1\end{array}\right]$.
Now,calculate $A^2 - 2A$:
$A^2 - 2A = \left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1\end{array}\right] - 2\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right] = \left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1\end{array}\right] - \left[\begin{array}{lll}2 & 0 & 2 \\ 0 & 2 & 2 \\ 0 & 2 & 0\end{array}\right] = \left[\begin{array}{rrr}-1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1\end{array}\right] \dots (i)$.
Next,find $A^{-1}$. The determinant $|A| = 1(0-1) - 0 + 1(0) = -1$.
The cofactor matrix $C$ is:
$C_{11} = -1, C_{12} = 0, C_{13} = 0$
$C_{21} = 1, C_{22} = 0, C_{23} = -1$
$C_{31} = -1, C_{32} = -1, C_{33} = 1$
$Adj(A) = C^T = \left[\begin{array}{rrr}-1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1\end{array}\right]$.
Since $A^{-1} = \frac{1}{|A|} Adj(A) = \frac{1}{-1} \left[\begin{array}{rrr}-1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1\end{array}\right] = -\left[\begin{array}{rrr}-1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1\end{array}\right]$.
Comparing this with $(i)$,we see that $A^2 - 2A = -A^{-1}$.

(C) Given $f(x) = |x| + |sin x|$.
For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,the left hand derivative $(LHD)$ at $x=0$ is defined as:
$LHD = \lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h}$
Since $f(0) = |0| + |sin 0| = 0$,we have:
$LHD = \lim_{h \to 0^+} \frac{|-h| + |sin(-h)| - 0}{-h}$
For small $h > 0$,$|-h| = h$ and $|sin(-h)| = |-sin h| = sin h$ (since $sin h > 0$ for $h \in (0, \pi/2)$).
$LHD = \lim_{h \to 0^+} \frac{h + sin h}{-h}$
$LHD = \lim_{h \to 0^+} -\left( \frac{h}{h} + \frac{sin h}{h} \right)$
$LHD = -(1 + 1) = -2$.

(C) Given the distance function: $s = t^2 - 2t + 5$.
Velocity $v$ is the first derivative of distance with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(t^2 - 2t + 5) = 2t - 2$.
Acceleration $a$ is the derivative of velocity with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(2t - 2) = 2$.
Therefore,the acceleration of the particle is $2$ units.

(A) Given $u = \sin(y + ax) - (y + ax)^2$ ... $(i)$
Differentiating partially with respect to $x$:
$u_x = \cos(y + ax) \cdot a - 2(y + ax) \cdot a$
$u_x = a[\cos(y + ax) - 2(y + ax)]$
Differentiating again with respect to $x$:
$u_{xx} = a[-\sin(y + ax) \cdot a - 2 \cdot a]$
$u_{xx} = -a^2[\sin(y + ax) + 2]$ ... (ii)
Differentiating $(i)$ partially with respect to $y$:
$u_y = \cos(y + ax) - 2(y + ax)$
Differentiating again with respect to $y$:
$u_{yy} = -\sin(y + ax) - 2$
$u_{yy} = -[\sin(y + ax) + 2]$ ... (iii)
From (ii) and (iii),we can see that:
$u_{xx} = a^2 \cdot [-\sin(y + ax) - 2]$
$u_{xx} = a^2 \cdot u_{yy}$

(A) Let $I = \int \frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx$.
We know that $\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$.
Also,$1 - 2 \sin^2 x \cos^2 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = \sin^4 x + \cos^4 x$.
Substituting these into the integral,we get:
$I = \int \frac{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}{\sin^4 x + \cos^4 x} dx$.
$I = \int (\sin^4 x - \cos^4 x) dx$.
Using the identity $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = -\cos 2x(1) = -\cos 2x$.
$I = \int -\cos 2x dx = -\frac{\sin 2x}{2} + B$.
Comparing this with $I = A \sin 2x + B$,we find $A = -\frac{1}{2}$.

(C) Given,$I_n = \int_0^{\pi/4} \tan^n \theta \, d\theta$.
We need to find $I_{n-1} + I_{n+1}$.
$I_{n-1} + I_{n+1} = \int_0^{\pi/4} \tan^{n-1} \theta \, d\theta + \int_0^{\pi/4} \tan^{n+1} \theta \, d\theta$.
Factor out $\tan^{n-1} \theta$:
$I_{n-1} + I_{n+1} = \int_0^{\pi/4} \tan^{n-1} \theta (1 + \tan^2 \theta) \, d\theta$.
Since $1 + \tan^2 \theta = \sec^2 \theta$,we have:
$I_{n-1} + I_{n+1} = \int_0^{\pi/4} \tan^{n-1} \theta \sec^2 \theta \, d\theta$.
Let $u = \tan \theta$,then $du = \sec^2 \theta \, d\theta$.
When $\theta = 0$,$u = 0$. When $\theta = \pi/4$,$u = 1$.
$I_{n-1} + I_{n+1} = \int_0^1 u^{n-1} \, du = \left[ \frac{u^n}{n} \right]_0^1 = \frac{1}{n} - 0 = \frac{1}{n}$.

(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \frac{\phi(y/x)}{\phi'(y/x)}$.
Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v + \frac{\phi(v)}{\phi'(v)}$.
Subtracting $v$ from both sides:
$x \frac{dv}{dx} = \frac{\phi(v)}{\phi'(v)}$.
Separating the variables:
$\frac{\phi'(v)}{\phi(v)} dv = \frac{dx}{x}$.
Integrating both sides:
$\int \frac{\phi'(v)}{\phi(v)} dv = \int \frac{dx}{x}$.
$\ln|\phi(v)| = \ln|x| + C$,where $C = \ln|k|$.
$\ln|\phi(v)| = \ln|kx|$.
Taking the exponential of both sides:
$\phi(v) = kx$.
Substituting $v = \frac{y}{x}$ back:
$\phi\left(\frac{y}{x}\right) = kx$.

(A) Two vectors are orthogonal if their dot product is equal to $0$.
Given vectors are $\vec{a} = \hat{i}-2x\hat{j}-3y\hat{k}$ and $\vec{b} = \hat{i}+3x\hat{j}+2y\hat{k}$.
Taking the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (-2x)(3x) + (-3y)(2y) = 0$.
$1 - 6x^2 - 6y^2 = 0$.
$6x^2 + 6y^2 = 1$.
$x^2 + y^2 = \frac{1}{6}$.
This equation represents a circle with center at the origin $(0, 0)$ and radius $\frac{1}{\sqrt{6}}$.

(C) Let the required ratio be $\lambda : 1$.
Using the section formula,the coordinates of the point $P$ that divides the line segment joining $(2, -4, 3)$ and $(-4, 5, -6)$ in the ratio $\lambda : 1$ are given by:
$P = \left( \frac{-4\lambda + 2}{\lambda + 1}, \frac{5\lambda - 4}{\lambda + 1}, \frac{-6\lambda + 3}{\lambda + 1} \right)$
Since this point $P$ lies on the plane $3x + 2y + z - 4 = 0$,it must satisfy the equation:
$3 \left( \frac{-4\lambda + 2}{\lambda + 1} \right) + 2 \left( \frac{5\lambda - 4}{\lambda + 1} \right) + \left( \frac{-6\lambda + 3}{\lambda + 1} \right) - 4 = 0$
Multiplying by $(\lambda + 1)$,we get:
$3(-4\lambda + 2) + 2(5\lambda - 4) + (-6\lambda + 3) - 4(\lambda + 1) = 0$
$-12\lambda + 6 + 10\lambda - 8 - 6\lambda + 3 - 4\lambda - 4 = 0$
$-12\lambda - 3 = 0$
$-12\lambda = 3$
$\lambda = -\frac{3}{12} = -\frac{1}{4}$
Thus,the ratio is $-\frac{1}{4} : 1$,which is $-1 : 4$.
The negative sign indicates that the division is external.

(D) The equation of a plane passing through the point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
Given the point $(2, 3, -1)$,the equation is $a(x-2) + b(y-3) + c(z+1) = 0 \dots (i)$.
Since the plane is perpendicular to the line with direction ratios $(3, -4, 7)$,the normal vector of the plane is parallel to this line.
Thus,we can take $a=3, b=-4, c=7$.
Substituting these into equation $(i)$:
$3(x-2) - 4(y-3) + 7(z+1) = 0$
$3x - 6 - 4y + 12 + 7z + 7 = 0$
$3x - 4y + 7z + 13 = 0$.
The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.
Here,$A=3, B=-4, C=7, D=13$.
$d = \frac{|13|}{\sqrt{3^2 + (-4)^2 + 7^2}} = \frac{13}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}}$.

(B) The equation of the sphere is $x^2+y^2+z^2+2x-2y-4z-19=0$.
Comparing this with the general equation $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get the centre $C = (-u, -v, -w) = (-1, 1, 2)$.
The radius of the sphere $R$ is given by $R = \sqrt{u^2+v^2+w^2-d} = \sqrt{(-1)^2 + 1^2 + 2^2 - (-19)} = \sqrt{1+1+4+19} = \sqrt{25} = 5$.
The perpendicular distance $p$ from the centre $C(-1, 1, 2)$ to the plane $x+2y+2z+7=0$ is:
$p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2 + 2^2 + 2^2}} = \frac{|-1 + 2 + 4 + 7|}{\sqrt{1+4+4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.
Let $r$ be the radius of the circle. In the right-angled triangle formed by the centre of the sphere,the centre of the circle,and a point on the circle,we have $R^2 = p^2 + r^2$.
$r^2 = R^2 - p^2 = 5^2 - 4^2 = 25 - 16 = 9$.
Therefore,$r = \sqrt{9} = 3$.

(A) The mean $E(X) = \sum x_i P(x_i) = 0 \times \frac{1}{10} + 1 \times \frac{2}{10} + 2 \times \frac{3}{10} + 3 \times \frac{4}{10} = 0 + \frac{2}{10} + \frac{6}{10} + \frac{12}{10} = \frac{20}{10} = 2$.
The variance $Var(X) = E(X^2) - [E(X)]^2$.
$E(X^2) = \sum x_i^2 P(x_i) = 0^2 \times \frac{1}{10} + 1^2 \times \frac{2}{10} + 2^2 \times \frac{3}{10} + 3^2 \times \frac{4}{10} = 0 + \frac{2}{10} + \frac{12}{10} + \frac{36}{10} = \frac{50}{10} = 5$.
Therefore,$Var(X) = 5 - (2)^2 = 5 - 4 = 1$.

(D) Given: $p = 0.001$,$n = 2000$.
Since $n$ is large and $p$ is very small,we use the Poisson distribution with parameter $\lambda = np$.
$\lambda = 2000 \times 0.001 = 2$.
The probability mass function for Poisson distribution is $P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}$.
We need to find the probability for exactly $x = 3$ individuals:
$P(X = 3) = \frac{2^3 e^{-2}}{3!} = \frac{8 \times e^{-2}}{6} = \frac{4}{3 e^2}$.
Thus,the correct option is $D$.

(D) Given curves are $x=y^2$ and $x=3-2y^2$.
To find the points of intersection,set $y^2 = 3-2y^2$,which gives $3y^2 = 3$,so $y^2 = 1$,meaning $y = \pm 1$.
When $y = 1$,$x = 1$. When $y = -1$,$x = 1$.
Thus,the points of intersection are $(1, 1)$ and $(1, -1)$.
The area is symmetric about the $x$-axis.
$\text{Required Area} = 2 \int_0^1 (x_2 - x_1) dy$
$= 2 \int_0^1 ((3-2y^2) - y^2) dy$
$= 2 \int_0^1 (3-3y^2) dy$
$= 2 [3y - y^3]_0^1$
$= 2 [3(1) - (1)^3 - (0)]$
$= 2 [3 - 1] = 2 \times 2 = 4$ square units.

(D) Let $A = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.
Given the equation $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \begin{bmatrix} 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$.
First,compute the product $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 2x_1 + x_2 \\ 3x_1 + 2x_2 \end{bmatrix}$.
Now,multiply by $\begin{bmatrix} 1 & 1 \end{bmatrix}$:
$\begin{bmatrix} 2x_1 + x_2 \\ 3x_1 + 2x_2 \end{bmatrix} \begin{bmatrix} 1 & 1 \end{bmatrix} = \begin{bmatrix} 2x_1 + x_2 & 2x_1 + x_2 \\ 3x_1 + 2x_2 & 3x_1 + 2x_2 \end{bmatrix}$.
Equating this to the given matrix $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$,we get:
$2x_1 + x_2 = 1$ and $3x_1 + 2x_2 = 0$.
From the second equation,$x_2 = -\frac{3}{2}x_1$.
Substituting into the first: $2x_1 - \frac{3}{2}x_1 = 1 \Rightarrow \frac{1}{2}x_1 = 1 \Rightarrow x_1 = 2$.
Then $x_2 = -\frac{3}{2}(2) = -3$.
Thus,$A = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$.

(B) Given,$A(\alpha, \beta) = \begin{bmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^\beta \end{bmatrix}$.
First,we find the determinant $|A(\alpha, \beta)| = e^\beta(\cos^2 \alpha + \sin^2 \alpha) = e^\beta$.
Next,we find the cofactor matrix. The cofactors are:
$C_{11} = e^\beta \cos \alpha, C_{12} = e^\beta \sin \alpha, C_{13} = 0$
$C_{21} = -e^\beta \sin \alpha, C_{22} = e^\beta \cos \alpha, C_{23} = 0$
$C_{31} = 0, C_{32} = 0, C_{33} = \cos^2 \alpha + \sin^2 \alpha = 1$
Thus,$\text{adj}(A(\alpha, \beta)) = \begin{bmatrix} e^\beta \cos \alpha & -e^\beta \sin \alpha & 0 \\ e^\beta \sin \alpha & e^\beta \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Then,$[A(\alpha, \beta)]^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{e^\beta} \begin{bmatrix} e^\beta \cos \alpha & -e^\beta \sin \alpha & 0 \\ e^\beta \sin \alpha & e^\beta \cos \alpha & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & e^{-\beta} \end{bmatrix}$.
Since $\cos(-\alpha) = \cos \alpha$ and $\sin(-\alpha) = -\sin \alpha$,this matrix is equivalent to $A(-\alpha, -\beta)$.

(B) Given,$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix}$.
Now,$A^2 - 2A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 2 \\ 0 & 2 & 2 \\ 0 & 2 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$.
Next,find $|A| = 1(0-1) - 0 + 1(0-0) = -1$.
The cofactor matrix $C$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$.
$C_{11} = -1, C_{12} = 0, C_{13} = 0$.
$C_{21} = 1, C_{22} = 0, C_{23} = -1$.
$C_{31} = -1, C_{32} = -1, C_{33} = 1$.
$adj(A) = C^T = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$.
Since $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-1} \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{bmatrix}$.
Comparing $A^2 - 2A = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$ with $-A^{-1} = - \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$.
Thus,$A^2 - 2A = -A^{-1}$.

(C) Let $\Delta = \left|\begin{array}{lll}24 & 25 & 26 \\ 25 & 26 & 27 \\ 26 & 27 & 27\end{array}\right|$.
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_2$:
$\Delta = \left|\begin{array}{ccc} 24 & 25 & 26 \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{array}\right|$.
Expanding along the third row $(R_3)$:
$\Delta = 1(25 \times 1 - 26 \times 1) - 1(24 \times 1 - 26 \times 1) + 0(24 \times 1 - 25 \times 1)$.
$\Delta = 1(25 - 26) - 1(24 - 26) + 0$.
$\Delta = -1 - (-2) = -1 + 2 = 1$.

(A) Given equation is $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$.
Using the identity $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$,we can write $(\tan ^{-1} x)^2 + (\cot ^{-1} x)^2 = (\tan ^{-1} x + \cot ^{-1} x)^2 - 2 \tan ^{-1} x \cot ^{-1} x$.
Substituting the identity: $(\frac{\pi}{2})^2 - 2 \tan ^{-1} x (\frac{\pi}{2} - \tan ^{-1} x) = \frac{5 \pi^2}{8}$.
$\frac{\pi^2}{4} - \pi \tan ^{-1} x + 2(\tan ^{-1} x)^2 = \frac{5 \pi^2}{8}$.
$2(\tan ^{-1} x)^2 - \pi \tan ^{-1} x + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.
$2(\tan ^{-1} x)^2 - \pi \tan ^{-1} x - \frac{3 \pi^2}{8} = 0$.
Let $u = \tan ^{-1} x$. Then $2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.
Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$u = \frac{\pi \pm \sqrt{\pi^2 - 4(2)(-\frac{3 \pi^2}{8})}}{4} = \frac{\pi \pm \sqrt{\pi^2 + 3 \pi^2}}{4} = \frac{\pi \pm 2 \pi}{4}$.
So,$u = \frac{3 \pi}{4}$ or $u = -\frac{\pi}{4}$.
Since $x = \tan u$,we have $x = \tan(\frac{3 \pi}{4}) = -1$ or $x = \tan(-\frac{\pi}{4}) = -1$.
Therefore,$x = -1$.

(D) Given,$f(x) = [\frac{x}{5}]$ where $|x| < 71$.
This implies $-71 < x < 71$.
Dividing by $5$,we get $-\frac{71}{5} < \frac{x}{5} < \frac{71}{5}$.
$-14.2 < \frac{x}{5} < 14.2$.
Now,we find the range of the greatest integer function $[\frac{x}{5}]$.
The minimum value is $[\frac{x}{5}] = [-14.2] = -15$.
The maximum value is $[\frac{x}{5}] = [14.2] = 14$.
Since $x$ can take any real value in the interval $(-71, 71)$,the function $f(x)$ will take all integer values from $-15$ to $14$.
Thus,the set is $\{-15, -14, \ldots, 0, \ldots, 13, 14\}$.

(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0)$.
First,consider the $LHL$: $\lim_{x \rightarrow 0^-} \frac{a+2 \cos x}{x^2}$. For this limit to exist and be finite,the numerator must approach $0$ as $x \rightarrow 0$. Thus,$a + 2 \cos(0) = 0 \Rightarrow a + 2 = 0 \Rightarrow a = -2$.
Substituting $a = -2$ into the limit: $\lim_{x \rightarrow 0^-} \frac{-2 + 2 \cos x}{x^2} = \lim_{x \rightarrow 0^-} \frac{-2(1 - \cos x)}{x^2} = \lim_{x \rightarrow 0^-} \frac{-2(2 \sin^2(x/2))}{x^2} = -2 \lim_{x \rightarrow 0^-} \frac{\sin^2(x/2)}{(x/2)^2 \times 2} = -2 \times \frac{1}{2} = -1$.
Now,consider the $RHL$: $\lim_{x \rightarrow 0^+} b \tan \frac{\pi}{[x+4]}$. As $x \rightarrow 0^+$,$[x+4] = 4$.
So,$RHL = b \tan \frac{\pi}{4} = b(1) = b$.
Since $LHL = RHL$,we have $-1 = b$.
Thus,the ordered pair $(a, b)$ is $(-2, -1)$.

(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.
Multiply and divide by $(1-x)$:
$y = \frac{(1-x)(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x}$
Using the identity $(a-b)(a+b) = a^2-b^2$ repeatedly:
$y = \frac{(1-x^2)(1+x^2)(1+x^4) \dots (1+x^{2^n})}{1-x} = \frac{(1-x^4)(1+x^4) \dots (1+x^{2^n})}{1-x}$
Continuing this process,we get:
$y = \frac{1-x^{2^{n+1}}}{1-x}$
Now,differentiate using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}$:
$\frac{dy}{dx} = \frac{(1-x)(-2^{n+1}x^{2^{n+1}-1}) - (1-x^{2^{n+1}})(-1)}{(1-x)^2}$
Substitute $x=0$:
$\left(\frac{dy}{dx}\right)_{x=0} = \frac{(1-0)(0) - (1-0)(-1)}{(1-0)^2} = \frac{0 + 1}{1} = 1$.

(D) Given,$y=\frac{\log _e x}{x}$ and $z=\log _e x$.
Since $z=\log _e x$,we have $x=e^z$.
Substituting $x$ in $y$,we get $y=\frac{z}{e^z} = z e^{-z}$.
Now,differentiating $y$ with respect to $z$:
$\frac{d y}{d z} = \frac{d}{d z}(z e^{-z}) = e^{-z} + z(-e^{-z}) = e^{-z}(1-z)$.
Again,differentiating with respect to $z$:
$\frac{d^2 y}{d z^2} = \frac{d}{d z}(e^{-z}(1-z)) = -e^{-z}(1-z) + e^{-z}(-1) = e^{-z}(-1+z-1) = e^{-z}(z-2)$.
Now,calculating $\frac{d^2 y}{d z^2} + \frac{d y}{d z}$:
$\frac{d^2 y}{d z^2} + \frac{d y}{d z} = e^{-z}(z-2) + e^{-z}(1-z) = e^{-z}(z-2+1-z) = e^{-z}(-1) = -e^{-z}$.

(A) Given,$\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$
$\Rightarrow \frac{x^2-y^2}{x^2+y^2}=\cos k$
Let $\cos k = C$ (a constant).
Then,$x^2 - y^2 = C(x^2 + y^2)$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(C(x^2 + y^2))$
$2x - 2y \frac{dy}{dx} = C(2x + 2y \frac{dy}{dx})$
$x - y \frac{dy}{dx} = C(x + y \frac{dy}{dx})$
$x - Cx = Cy \frac{dy}{dx} + y \frac{dy}{dx}$
$x(1 - C) = y \frac{dy}{dx}(C + 1)$
$\frac{dy}{dx} = \frac{x(1 - C)}{y(1 + C)}$
Substituting $C = \frac{x^2-y^2}{x^2+y^2}$ back:
$\frac{dy}{dx} = \frac{x(1 - \frac{x^2-y^2}{x^2+y^2})}{y(1 + \frac{x^2-y^2}{x^2+y^2})} = \frac{x(\frac{x^2+y^2-x^2+y^2}{x^2+y^2})}{y(\frac{x^2+y^2+x^2-y^2}{x^2+y^2})} = \frac{x(2y^2)}{y(2x^2)} = \frac{xy^2}{yx^2} = \frac{y}{x}$

(A) Let $v = y + ax$. Then $u = \sin(v) - v^2$.
First,we find the partial derivatives with respect to $x$:
$u_x = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial x} = (\cos(v) - 2v) \cdot a$.
$u_{xx} = \frac{\partial}{\partial x} [a(\cos(v) - 2v)] = a(-\sin(v) - 2) \cdot \frac{\partial v}{\partial x} = a^2(-\sin(v) - 2)$.
Next,we find the partial derivatives with respect to $y$:
$u_y = \frac{\partial u}{\partial v} \cdot \frac{\partial v}{\partial y} = (\cos(v) - 2v) \cdot 1$.
$u_{yy} = \frac{\partial}{\partial y} [\cos(v) - 2v] = (-\sin(v) - 2) \cdot \frac{\partial v}{\partial y} = -\sin(v) - 2$.
Comparing $u_{xx}$ and $u_{yy}$,we see that $u_{xx} = a^2 u_{yy}$.

(C) Given the curve $y = 5^x$.
First,find the derivative with respect to $x$:
$\frac{dy}{dx} = 5^x \log_e 5$.
At the point $(x_1, y_1)$,the slope of the tangent is:
$\left(\frac{dy}{dx}\right)_{(x_1, y_1)} = 5^{x_1} \log_e 5$.
The formula for the length of the subtangent is given by:
$L = \left| \frac{y_1}{\frac{dy}{dx}} \right|$.
Substituting the values:
$L = \frac{y_1}{5^{x_1} \log_e 5}$.
Since $(x_1, y_1)$ lies on the curve,$y_1 = 5^{x_1}$.
Therefore,$L = \frac{5^{x_1}}{5^{x_1} \log_e 5} = \frac{1}{\log_e 5}$.

(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.
To find the range $B$ for the function to be a bijection,we need to determine the set of all possible values of $f(x)$.
First,we rewrite the function by completing the square:
$f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.
Since the domain is $x \in [2, \infty)$,we have $x - 2 \geq 0$.
Therefore,$(x - 2)^2 \geq 0$.
Adding $1$ to both sides,we get $(x - 2)^2 + 1 \geq 1$.
Thus,$f(x) \geq 1$.
The range of the function is $[1, \infty)$.
Since the function is strictly increasing on $[2, \infty)$,it is injective. For it to be a bijection,the codomain $B$ must be equal to the range.
Therefore,$B = [1, \infty)$.

(D) Let $I = \int \frac{1+\cos 4 x}{\cot x-\tan x} dx$.
Using the identity $1+\cos 4x = 2\cos^2 2x$ and $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x$.
Substituting these into the integral:
$I = \int \frac{2\cos^2 2x}{2\cot 2x} dx = \int \frac{\cos^2 2x}{\frac{\cos 2x}{\sin 2x}} dx = \int \sin 2x \cos 2x dx$.
Using the identity $\sin 2x \cos 2x = \frac{1}{2} \sin 4x$:
$I = \int \frac{1}{2} \sin 4x dx = \frac{1}{2} \left( -\frac{\cos 4x}{4} \right) + C = -\frac{1}{8} \cos 4x + C$.

(B) Let $I = \int \left( \sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} \right) dx$.
Simplify the integrand:
$\sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} = \frac{a+x + a-x}{\sqrt{(a-x)(a+x)}} = \frac{2a}{\sqrt{a^2-x^2}}$.
Now,the integral becomes:
$I = \int \frac{2a}{\sqrt{a^2-x^2}} dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C$:
$I = 2a \int \frac{1}{\sqrt{a^2-x^2}} dx = 2a \sin^{-1}\left(\frac{x}{a}\right) + C$.

(A) Given $a=0, b=2$ and $n=4$ intervals.
The step size $h = \frac{b-a}{n} = \frac{2-0}{4} = 0.5$.
The values are $y_0=f(0)=1, y_1=f(0.5)=\frac{5}{4}, y_2=f(1)=2, y_3=f(1.5)=\frac{13}{4}, y_4=f(2)=5$.
According to Simpson's $1/3$ rule:
$\int_a^b f(x) dx = \frac{h}{3} [ (y_0 + y_4) + 4(y_1 + y_3) + 2(y_2) ]$
Substituting the values:
$\int_0^2 f(x) dx = \frac{0.5}{3} [ (1 + 5) + 4(\frac{5}{4} + \frac{13}{4}) + 2(2) ]$
$= \frac{0.5}{3} [ 6 + 4(\frac{18}{4}) + 4 ]$
$= \frac{0.5}{3} [ 6 + 18 + 4 ]$
$= \frac{0.5}{3} [ 28 ] = \frac{14}{3}$.

(A) The given differential equation is $\left(\frac{2+\sin x}{y+1}\right) \frac{d y}{d x}+\cos x=0$.
Rearranging the terms to separate the variables,we get:
$\frac{d y}{1+y} + \frac{\cos x}{2+\sin x} dx = 0$.
Integrating both sides:
$\int \frac{d y}{1+y} + \int \frac{\cos x}{2+\sin x} dx = C_0$.
Let $t = 2+\sin x$,then $dt = \cos x dx$. The integral becomes:
$\ln|1+y| + \ln|2+\sin x| = C_1$.
This simplifies to $(1+y)(2+\sin x) = C$.
Given $y(0)=1$,we substitute $x=0$ and $y=1$:
$(1+1)(2+\sin 0) = C \Rightarrow 2(2) = C \Rightarrow C=4$.
So,the particular solution is $(1+y)(2+\sin x) = 4$.
Now,find $y\left(\frac{\pi}{2}\right)$ by substituting $x=\frac{\pi}{2}$:
$(1+y(\frac{\pi}{2}))(2+\sin(\frac{\pi}{2})) = 4$.
$(1+y(\frac{\pi}{2}))(2+1) = 4$.
$3(1+y(\frac{\pi}{2})) = 4$.
$1+y(\frac{\pi}{2}) = \frac{4}{3}$.
$y(\frac{\pi}{2}) = \frac{4}{3} - 1 = \frac{1}{3}$.

(B) Let $D$ be the midpoint of side $BC$. The median through $A$ is the vector $\vec{AD}$.
Since $D$ is the midpoint of $BC$,the position vector of $D$ relative to $A$ is given by the average of the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$
$\vec{AD} = \frac{1}{2}((-3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k}))$
$\vec{AD} = \frac{1}{2}(2\hat{i} - 2\hat{j} + 8\hat{k})$
$\vec{AD} = \hat{i} - \hat{j} + 4\hat{k}$
The length of the median $\vec{AD}$ is the magnitude of the vector $\vec{AD}$:
$|\vec{AD}| = \sqrt{(1)^2 + (-1)^2 + (4)^2}$
$|\vec{AD}| = \sqrt{1 + 1 + 16}$
$|\vec{AD}| = \sqrt{18}$

(D) Given: $|a|=1, |b|=2$ and the angle $\theta = 120^{\circ}$.
We need to evaluate the expression ${(a+3b) \times (3a-b)}^2$.
First,expand the cross product:
$(a+3b) \times (3a-b) = a \times (3a) - a \times b + (3b) \times (3a) - (3b) \times b$
Since $a \times a = 0$ and $b \times b = 0$,we have:
$= 0 - (a \times b) + 9(b \times a) - 0$
Since $b \times a = -(a \times b)$,the expression becomes:
$= -(a \times b) - 9(a \times b) = -10(a \times b)$
Now,square the magnitude:
${(-10)(a \times b)}^2 = 100 |a \times b|^2$
Using the formula $|a \times b| = |a||b| \sin \theta$:
$|a \times b| = 1 \times 2 \times \sin 120^{\circ} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$
Therefore,$100 |a \times b|^2 = 100 \times (\sqrt{3})^2 = 100 \times 3 = 300$.

(A) Two vectors are orthogonal if their dot product is equal to $0$.
Given vectors are $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$.
Taking the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (-2x)(3x) + (-3y)(2y) = 0$.
This simplifies to $1 - 6x^2 - 6y^2 = 0$.
Rearranging the terms,we get $6x^2 + 6y^2 = 1$,which implies $x^2 + y^2 = \frac{1}{6}$.
This is the equation of a circle with center at the origin $(0, 0)$ and radius $\frac{1}{\sqrt{6}}$.
Thus,the locus of the point $(x, y)$ is a circle.

(C) The scalar triple product is defined as $[u v w] = u \cdot (v \times w)$.
Since $u$ is a unit vector,$|u| = 1$.
Using the property of the dot product,$|u \cdot (v \times w)| \leq |u| |v \times w| = |v \times w|$.
First,calculate the cross product $v \times w$:
$v \times w = \begin{vmatrix} i & j & k \\ 2 & 1 & -1 \\ 1 & 0 & 3 \end{vmatrix} = i(3 - 0) - j(6 - (-1)) + k(0 - 1) = 3i - 7j - k$.
Now,find the magnitude of the resulting vector:
$|v \times w| = \sqrt{3^2 + (-7)^2 + (-1)^2} = \sqrt{9 + 49 + 1} = \sqrt{59}$.
Therefore,the maximum value of $[u v w]$ is $\sqrt{59}$.

(B) We use the vector triple product formula: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$.
Applying this to the first term: $i \times (r \times i) = (i \cdot i)r - (i \cdot r)i = r - r_x i$,where $r = r_x i + r_y j + r_z k$.
Similarly,for the other terms:
$j \times (r \times j) = (j \cdot j)r - (j \cdot r)j = r - r_y j$
$k \times (r \times k) = (k \cdot k)r - (k \cdot r)k = r - r_z k$.
Summing these three expressions:
$(i \times (r \times i)) + (j \times (r \times j)) + (k \times (r \times k)) = (r - r_x i) + (r - r_y j) + (r - r_z k)$
$= 3r - (r_x i + r_y j + r_z k)$
$= 3r - r = 2r$.

(C) Let the ratio in which the plane divides the line segment joining $A(2, -4, 3)$ and $B(-4, 5, -6)$ be $\lambda : 1$.
By the section formula,the coordinates of the point of intersection $P$ are:
$P = \left( \frac{-4\lambda + 2}{\lambda + 1}, \frac{5\lambda - 4}{\lambda + 1}, \frac{-6\lambda + 3}{\lambda + 1} \right)$
Since point $P$ lies on the plane $3x + 2y + z - 4 = 0$,we substitute these coordinates into the plane equation:
$3\left( \frac{-4\lambda + 2}{\lambda + 1} \right) + 2\left( \frac{5\lambda - 4}{\lambda + 1} \right) + \left( \frac{-6\lambda + 3}{\lambda + 1} \right) - 4 = 0$
Multiplying by $(\lambda + 1)$:
$3(-4\lambda + 2) + 2(5\lambda - 4) + (-6\lambda + 3) - 4(\lambda + 1) = 0$
$-12\lambda + 6 + 10\lambda - 8 - 6\lambda + 3 - 4\lambda - 4 = 0$
$-12\lambda - 3 = 0$
$-12\lambda = 3$
$\lambda = -\frac{3}{12} = -\frac{1}{4}$
Thus,the ratio is $-1 : 4$.

(C) We know that if $\alpha, \beta, \gamma$ are the angles made by a line with the coordinate axes,then the sum of the squares of their cosines is $1$,i.e.,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given that the angles are $\alpha, \frac{\pi}{2}-\alpha, \beta$.
Substituting these values into the identity,we get:
$\cos^2 \alpha + \cos^2 \left(\frac{\pi}{2}-\alpha\right) + \cos^2 \beta = 1$
Since $\cos \left(\frac{\pi}{2}-\alpha\right) = \sin \alpha$,the equation becomes:
$\cos^2 \alpha + \sin^2 \alpha + \cos^2 \beta = 1$
Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we have:
$1 + \cos^2 \beta = 1$
$\cos^2 \beta = 0$
$\cos \beta = 0$
Since $\cos \beta = 0$,we have $\beta = \frac{\pi}{2}$.

(B) Let the unit vector along the line be $\hat{u} = \cos \alpha \hat{i} + \cos \alpha \hat{j} + \cos \alpha \hat{k}$.
Since $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$,we have $3 \cos^2 \alpha = 1$,which implies $\cos \alpha = \pm \frac{1}{\sqrt{3}}$.
Thus,the unit vector along the line is $\hat{u} = \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})$.
The projection of vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line is given by the magnitude of the dot product $\vec{a} \cdot \hat{u}$.
$|\vec{a} \cdot \hat{u}| = |(4\hat{i} - 3\hat{j} + 2\hat{k}) \cdot \pm \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})|$
$= |\pm \frac{1}{\sqrt{3}}(4 - 3 + 2)|$
$= |\pm \frac{3}{\sqrt{3}}| = \sqrt{3}$.

(B) The equation of the sphere is $x^2+y^2+z^2+2x-2y-4z-19=0$. Comparing this with the general equation $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $u=1, v=-1, w=-2, d=-19$.
The centre of the sphere is $(-u, -v, -w) = (-1, 1, 2)$.
The radius of the sphere $R$ is given by $\sqrt{u^2+v^2+w^2-d} = \sqrt{1^2+(-1)^2+(-2)^2-(-19)} = \sqrt{1+1+4+19} = \sqrt{25} = 5$.
Now,the perpendicular distance $p$ from the centre $(-1, 1, 2)$ to the plane $x+2y+2z+7=0$ is:
$p = \frac{|(-1) + 2(1) + 2(2) + 7|}{\sqrt{1^2+2^2+2^2}} = \frac{|-1+2+4+7|}{\sqrt{1+4+4}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4$.
Let $r$ be the radius of the circle. In the right-angled triangle formed by the centre of the sphere,the centre of the circle,and a point on the circumference of the circle,we have $R^2 = p^2 + r^2$.
$r^2 = R^2 - p^2 = 5^2 - 4^2 = 25 - 16 = 9$.
Therefore,$r = \sqrt{9} = 3$.