AP EAMCET 2009 Mathematics Question Paper with Answer and Solution

(C) Given equation: $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$. \\ Let $t = x-a$. Then the equation becomes: \\ $t(t-1) + (t-1)(t-2) + t(t-2) = 0$ \\ $t^2 - t + t^2 - 3t + 2 + t^2 - 2t = 0$ \\ $3t^2 - 6t + 2 = 0$ \\ The discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$. \\ Since $D > 0$,the roots for $t$ are real and distinct. \\ Consequently,$x = a + t$ will also have real and distinct roots.

(B) Given,$f(x) = x^2 + ax + b$ has imaginary roots.
Therefore,the discriminant $D < 0$,which implies $a^2 - 4b < 0$.
Now,we calculate the derivatives:
$f'(x) = 2x + a$
$f''(x) = 2$
Substituting these into the equation $f(x) + f'(x) + f''(x) = 0$:
$(x^2 + ax + b) + (2x + a) + 2 = 0$
$x^2 + (a + 2)x + (b + a + 2) = 0$
The discriminant $D'$ of this new quadratic equation is:
$D' = (a + 2)^2 - 4(1)(b + a + 2)$
$D' = a^2 + 4a + 4 - 4b - 4a - 8$
$D' = a^2 - 4b - 4$
Since $a^2 - 4b < 0$,it follows that $a^2 - 4b - 4 < -4$.
Thus,$D' < 0$.
Since the discriminant is negative,the roots of the equation $f(x) + f'(x) + f''(x) = 0$ are imaginary.

(C) Given,$\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$.
From the properties of roots,we have:
$\alpha+\beta = 2$ ...$(i)$
$\alpha\beta = 4$ ...$(ii)$
Now,we can write $\alpha^6+\beta^6 = (\alpha^2)^3 + (\beta^2)^3$.
Alternatively,note that the roots of $x^2-2x+4=0$ are $x = \frac{2 \pm \sqrt{4-16}}{2} = 1 \pm \sqrt{3}i$.
In polar form,$1 \pm \sqrt{3}i = 2(\frac{1}{2} \pm i\frac{\sqrt{3}}{2}) = 2e^{\pm i\pi/3}$.
Thus,$\alpha = 2e^{i\pi/3}$ and $\beta = 2e^{-i\pi/3}$.
Then,$\alpha^6 = (2e^{i\pi/3})^6 = 2^6 e^{i2\pi} = 64(1) = 64$.
Similarly,$\beta^6 = (2e^{-i\pi/3})^6 = 2^6 e^{-i2\pi} = 64(1) = 64$.
Therefore,$\alpha^6+\beta^6 = 64+64 = 128$.

(C) Given $n = 3r + 1$ for some integer $r$.
We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1-i\sqrt{3}}{2}$.
Thus,$1+i\sqrt{3} = -2\omega^2$ and $1-i\sqrt{3} = -2\omega$.
Substituting these into the expression:
$(1+i\sqrt{3})^n + (1-i\sqrt{3})^n = (-2\omega^2)^n + (-2\omega)^n$
$= (-2)^n (\omega^{2n} + \omega^n)$
Since $n = 3r+1$,$\omega^n = \omega^{3r+1} = \omega$ and $\omega^{2n} = \omega^{6r+2} = \omega^2$.
Therefore,the expression becomes $(-2)^n (\omega^2 + \omega)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.
Thus,the result is $(-2)^n (-1) = -(-2)^n$.

(A) Let $S$ be the set of all $n$-digit binary sequences. The total number of such sequences is $2^n$.
Let $E$ be the number of sequences with an even number of $0$'s and $O$ be the number of sequences with an odd number of $0$'s.
We know that the sum of binomial coefficients $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + . . . + \binom{n}{n} = 2^n$.
The number of sequences with an even number of $0$'s is given by the sum of even-indexed binomial coefficients: $E = \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + . . .$.
Using the property of binomial coefficients,$\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + . . . = 2^{n-1}$.
Thus,the number of $n$-digit binary sequences containing an even number of $0$'s is $2^{n-1}$.

(D) Total number of points is $3p$.
To form a triangle,we need to select $3$ points out of $3p$.
The total number of ways to select $3$ points is $^{3p}C_3$.
However,if $3$ points are collinear,they do not form a triangle.
There are $3$ lines,each containing $p$ points.
For each line,the number of ways to select $3$ collinear points is $^pC_3$.
Thus,the number of triangles is $^{3p}C_3 - 3 \cdot ^pC_3$.
$= \frac{3p(3p-1)(3p-2)}{6} - 3 \cdot \frac{p(p-1)(p-2)}{6}$
$= \frac{p}{2} [ (3p-1)(3p-2) - (p-1)(p-2) ]$
$= \frac{p}{2} [ (9p^2 - 9p + 2) - (p^2 - 3p + 2) ]$
$= \frac{p}{2} [ 8p^2 - 6p ]$
$= p^2(4p-3)$.

(B) Given,$a_0 = 1$ and $a_{n+1} = 3n^2 + n + a_n$.
For $n = 0$: $a_1 = 3(0)^2 + 0 + a_0 = 0 + 0 + 1 = 1$.
For $n = 1$: $a_2 = 3(1)^2 + 1 + a_1 = 3 + 1 + 1 = 5$.
Testing the options:
For option $(B)$,$P(n) = n^3 - n^2 + 1$:
$P(0) = 0^3 - 0^2 + 1 = 1 = a_0$.
$P(1) = 1^3 - 1^2 + 1 = 1 = a_1$.
$P(2) = 2^3 - 2^2 + 1 = 8 - 4 + 1 = 5 = a_2$.
Since the values match for the initial terms,the correct expression is $a_n = n^3 - n^2 + 1$.

(B) We are given $\lambda = \frac{\cos x}{\cos (x-2 y)}$.
Consider the expression $\tan (x-y) \tan y = \frac{\sin (x-y) \sin y}{\cos (x-y) \cos y}$.
Multiplying the numerator and denominator by $2$,we get $\frac{2 \sin (x-y) \sin y}{2 \cos (x-y) \cos y}$.
Using the product-to-sum formulas $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \cos A \cos B = \cos(A-B) + \cos(A+B)$,we have:
$\tan (x-y) \tan y = \frac{\cos(x-2y) - \cos x}{\cos(x-2y) + \cos x}$.
Dividing the numerator and denominator by $\cos(x-2y)$,we get:
$\frac{1 - \frac{\cos x}{\cos(x-2y)}}{1 + \frac{\cos x}{\cos(x-2y)}} = \frac{1-\lambda}{1+\lambda}$.

(C) Given,$\sinh ^{-1} 2 + \sinh ^{-1} 3 = x$
Taking $\cosh$ on both sides,$\cosh(\sinh ^{-1} 2 + \sinh ^{-1} 3) = \cosh x$
Using the identity $\cosh(A + B) = \cosh A \cosh B + \sinh A \sinh B$,we get:
$\cosh(\sinh ^{-1} 2) \cosh(\sinh ^{-1} 3) + \sinh(\sinh ^{-1} 2) \sinh(\sinh ^{-1} 3) = \cosh x$
Since $\cosh(\sinh ^{-1} y) = \sqrt{1 + y^2}$ and $\sinh(\sinh ^{-1} y) = y$:
$\cosh x = \sqrt{1 + 2^2} \cdot \sqrt{1 + 3^2} + 2 \cdot 3$
$\cosh x = \sqrt{5} \cdot \sqrt{10} + 6$
$\cosh x = \sqrt{50} + 6 = \frac{2 \sqrt{50} + 12}{2} = \frac{1}{2}(12 + 2 \sqrt{50})$

(C) Given equation: $\sin^2 x - \cos 2x = 2 - \sin 2x$
Using identities $\cos 2x = 1 - 2\sin^2 x$ and $\sin 2x = 2\sin x \cos x$:
$\sin^2 x - (1 - 2\sin^2 x) = 2 - 2\sin x \cos x$
$3\sin^2 x - 1 = 2 - 2\sin x \cos x$
Alternatively,using $\cos 2x = 2\cos^2 x - 1$:
$\sin^2 x - (2\cos^2 x - 1) = 2 - 2\sin x \cos x$
$(1 - \cos^2 x) - 2\cos^2 x + 1 = 2 - 2\sin x \cos x$
$2 - 3\cos^2 x = 2 - 2\sin x \cos x$
$-3\cos^2 x + 2\sin x \cos x = 0$
$\cos x (2\sin x - 3\cos x) = 0$
Since $3\cos x \neq 2\sin x$,we must have $\cos x = 0$.
The general solution for $\cos x = 0$ is $x = (2n + 1) \frac{\pi}{2}$,which can be written as $x = (4n \pm 1) \frac{\pi}{2}$ for $n \in \mathbb{Z}$.

(A) The given equation is $x^2+y^2=r^2$.
When the axes are rotated through an angle $\theta = 36^{\circ}$,the transformation equations are:
$x = X \cos \theta - Y \sin \theta$
$y = X \sin \theta + Y \cos \theta$
Substituting these into the original equation:
$(X \cos \theta - Y \sin \theta)^2 + (X \sin \theta + Y \cos \theta)^2 = r^2$
Expanding the terms:
$(X^2 \cos^2 \theta + Y^2 \sin^2 \theta - 2XY \sin \theta \cos \theta) + (X^2 \sin^2 \theta + Y^2 \cos^2 \theta + 2XY \sin \theta \cos \theta) = r^2$
Simplifying:
$X^2(\cos^2 \theta + \sin^2 \theta) + Y^2(\sin^2 \theta + \cos^2 \theta) = r^2$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we get:
$X^2 + Y^2 = r^2$.

(C) Given equation: $(x-a)(x-a-1)+(x-a-1)(x-a-2)+(x-a)(x-a-2)=0$.
Let $t = x-a$. Then the equation becomes $t(t-1)+(t-1)(t-2)+t(t-2)=0$.
Expanding the terms: $(t^2-t) + (t^2-3t+2) + (t^2-2t) = 0$.
Combining like terms: $3t^2 - 6t + 2 = 0$.
The discriminant $D = b^2 - 4ac = (-6)^2 - 4(3)(2) = 36 - 24 = 12$.
Since $D > 0$,the roots for $t$ are real and distinct.
Consequently,$x = a + t$ will also be real and distinct for any $a \in R$.

(C) Given,$f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2 = (x-2)(x-1)$.
Since $f(x)$ is divisible by $(x-2)$ and $(x-1)$,we must have $f(2)=0$ and $f(1)=0$.
For $f(2)=0$:
$2(2)^4-13(2)^2+a(2)+b=0$
$2(16)-13(4)+2a+b=0$
$32-52+2a+b=0$
$2a+b=20$ ... $(i)$
For $f(1)=0$:
$2(1)^4-13(1)^2+a(1)+b=0$
$2-13+a+b=0$
$a+b=11$ ... $(ii)$
Subtracting Eq. $(ii)$ from Eq. $(i)$:
$(2a+b)-(a+b)=20-11$
$a=9$
Substituting $a=9$ in Eq. $(ii)$:
$9+b=11$
$b=2$
Thus,$(a, b) = (9, 2)$.

(C) Given,$\alpha, \beta, \gamma$ are the roots of $x^3+4x+1=0$.
From Vieta's formulas,we have $\alpha+\beta+\gamma=0$,$\alpha\beta+\beta\gamma+\gamma\alpha=4$,and $\alpha\beta\gamma=-1$.
Since $\alpha+\beta+\gamma=0$,we have $\beta+\gamma=-\alpha$,$\gamma+\alpha=-\beta$,and $\alpha+\beta=-\gamma$.
The roots of the new equation are $y_1 = \frac{\alpha^2}{-\alpha} = -\alpha$,$y_2 = \frac{\beta^2}{-\beta} = -\beta$,and $y_3 = \frac{\gamma^2}{-\gamma} = -\gamma$.
Let $y = -x$. Then $x = -y$.
Substituting $x = -y$ into the original equation $x^3+4x+1=0$:
$(-y)^3 + 4(-y) + 1 = 0$
$-y^3 - 4y + 1 = 0$
$y^3 + 4y - 1 = 0$.
Thus,the required equation is $x^3+4x-1=0$.

(C) Given $n = 3r + 1$ for some integer $r$.
We know that $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.
Thus,$1 + i\sqrt{3} = 2\left(\frac{1 + i\sqrt{3}}{2}\right) = -2\omega^2$ and $1 - i\sqrt{3} = 2\left(\frac{1 - i\sqrt{3}}{2}\right) = -2\omega$.
Substituting these into the expression:
$(1 + i\sqrt{3})^n + (1 - i\sqrt{3})^n = (-2\omega^2)^n + (-2\omega)^n$
$= (-2)^n (\omega^{2n} + \omega^n)$
Since $n = 3r + 1$,$\omega^n = \omega^{3r+1} = \omega$ and $\omega^{2n} = \omega^{6r+2} = \omega^2$.
Therefore,the expression becomes $(-2)^n (\omega^2 + \omega)$.
Since $1 + \omega + \omega^2 = 0$,we have $\omega^2 + \omega = -1$.
Thus,the result is $(-2)^n (-1) = -(-2)^n$.

(C) Let $z = x + iy$.
Given,$\left|\frac{z + 2i}{2z + i}\right| < 1$.
This implies $|z + 2i| < |2z + i|$.
Substituting $z = x + iy$,we get $|x + i(y + 2)| < |2x + i(2y + 1)|$.
Squaring both sides,we get $x^2 + (y + 2)^2 < (2x)^2 + (2y + 1)^2$.
$x^2 + y^2 + 4y + 4 < 4x^2 + 4y^2 + 4y + 1$.
$3 < 3x^2 + 3y^2$.
Dividing by $3$,we get $x^2 + y^2 > 1$.

(C) The total number of subsets of the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is $2^n$,where $n=9$.
Total subsets $= 2^9 = 512$.
We want to find the number of subsets that contain at least one odd number.
It is easier to calculate the complement: the number of subsets that contain $NO$ odd numbers.
$A$ subset contains no odd numbers if and only if all its elements are even.
The even numbers in the set are $\{2, 4, 6, 8\}$.
The number of subsets formed using only these even numbers is $2^4 = 16$.
These $16$ subsets include the empty set $\emptyset$.
Therefore,the number of subsets containing at least one odd number is Total subsets $-$ Subsets with only even numbers.
Required number $= 512 - 16 = 496$.

(B) Given,$a_0=1$ and $a_{n+1}=3n^2+n+a_n$.
We can find the first few terms:
$a_1 = 3(0)^2 + 0 + a_0 = 0 + 0 + 1 = 1$.
$a_2 = 3(1)^2 + 1 + a_1 = 3 + 1 + 1 = 5$.
$a_3 = 3(2)^2 + 2 + a_2 = 12 + 2 + 5 = 19$.
Now,check the options:
For $n=0$: $a_0 = 0^3 - 0^2 + 1 = 1$ (Matches).
For $n=1$: $a_1 = 1^3 - 1^2 + 1 = 1$ (Matches).
For $n=2$: $a_2 = 2^3 - 2^2 + 1 = 8 - 4 + 1 = 5$ (Matches).
For $n=3$: $a_3 = 3^3 - 3^2 + 1 = 27 - 9 + 1 = 19$ (Matches).
Thus,$a_n = n^3 - n^2 + 1$ is the correct expression.

(D) We have the expression $\frac{1}{(x-1)^2(x-2)}$.
First,rewrite the expression as $\frac{1}{(-1)^2(1-x)^2(-2)(1-\frac{x}{2})} = -\frac{1}{2}(1-x)^{-2}(1-\frac{x}{2})^{-1}$.
Using the binomial expansion $(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!}x^2 + \dots$ and $(1-y)^{-1} = 1 + y + y^2 + \dots$ for $|x| < 1$ and $|x/2| < 1$:
$(1-x)^{-2} = 1 + 2x + 3x^2 + \dots$
$(1-\frac{x}{2})^{-1} = 1 + \frac{x}{2} + \frac{x^2}{4} + \dots$
Multiplying these series: $-\frac{1}{2}(1 + 2x + \dots)(1 + \frac{x}{2} + \dots) = -\frac{1}{2}(1 + (\frac{1}{2} + 2)x + \dots) = -\frac{1}{2}(1 + \frac{5}{2}x + \dots)$.
The constant term is $-\frac{1}{2} \times 1 = -\frac{1}{2}$.

(B) The given expression is $(1+x^2)^{12}(1+x^{12})(1+x^{24})$.
First,simplify the product $(1+x^{12})(1+x^{24}) = 1 + x^{12} + x^{24} + x^{36}$.
Now,the expression becomes $(1+x^2)^{12}(1 + x^{12} + x^{24} + x^{36})$.
Using the binomial expansion,$(1+x^2)^{12} = \sum_{r=0}^{12} {}^{12}C_r (x^2)^r = \sum_{r=0}^{12} {}^{12}C_r x^{2r}$.
We need the coefficient of $x^{24}$ in the product $(\sum_{r=0}^{12} {}^{12}C_r x^{2r})(1 + x^{12} + x^{24} + x^{36})$.
This is obtained by:
$1$. The term from the sum where $2r = 24$,which is ${}^{12}C_{12} x^{24} \times 1 = {}^{12}C_{12} x^{24}$.
$2$. The term from the sum where $2r = 12$,which is ${}^{12}C_6 x^{12} \times x^{12} = {}^{12}C_6 x^{24}$.
$3$. The term from the sum where $2r = 0$,which is ${}^{12}C_0 x^0 \times x^{24} = 1 \times x^{24} = 1 x^{24}$.
Summing these coefficients: ${}^{12}C_{12} + {}^{12}C_6 + 1 = 1 + {}^{12}C_6 + 1 = {}^{12}C_6 + 2$.

(A) Given expression: $E = (1 + \frac{2x}{3})^{3/2} \cdot (32 + 5x)^{-1/5}$
Using the binomial approximation $(1+u)^n \approx 1+nu$ for small $u$:
$E \approx (1 + \frac{3}{2} \cdot \frac{2x}{3}) \cdot (32)^{-1/5} (1 + \frac{5x}{32})^{-1/5}$
$E \approx (1 + x) \cdot \frac{1}{2} \cdot (1 - \frac{1}{5} \cdot \frac{5x}{32})$
$E \approx \frac{1}{2} (1 + x) (1 - \frac{x}{32})$
Neglecting $x^2$ terms:
$E \approx \frac{1}{2} (1 + x - \frac{x}{32}) = \frac{1}{2} (1 + \frac{31x}{32}) = \frac{32 + 31x}{64}$

(A) To evaluate the product $P = \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$:
$P = \frac{1}{2 \sin A} (2 \sin A \cos A) \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$P = \frac{1}{2 \sin A} (\sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A$
Multiply and divide by $2$:
$P = \frac{1}{2^2 \sin A} (2 \sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A = \frac{\sin 4 A}{2^2 \sin A} \cos 4 A \ldots \cos 2^{n-1} A$
Continuing this process $n$ times,we obtain:
$P = \frac{\sin 2^n A}{2^n \sin A}$

(D) First,find the point of intersection of the lines $x + 3y - 1 = 0$ and $x - 2y + 4 = 0$.
Subtracting the second equation from the first: $(x + 3y - 1) - (x - 2y + 4) = 0$ $\Rightarrow 5y - 5 = 0$ $\Rightarrow y = 1$.
Substituting $y = 1$ into $x + 3y - 1 = 0$,we get $x + 3(1) - 1 = 0 \Rightarrow x = -2$.
The point of intersection is $(-2, 1)$.
The equation of a line perpendicular to $3x + 2y = 0$ is of the form $2x - 3y + \lambda = 0$.
Since the line passes through $(-2, 1)$,we substitute these coordinates into the equation:
$2(-2) - 3(1) + \lambda = 0$ $\Rightarrow -4 - 3 + \lambda = 0$ $\Rightarrow \lambda = 7$.
Thus,the required equation is $2x - 3y + 7 = 0$.

(B) Let the point be $P(x, y)$ on the line $3x + 4y = 5$.
Since $P$ is equidistant from $A(1, 2)$ and $B(3, 4)$,we have $PA^2 = PB^2$.
$(x - 1)^2 + (y - 2)^2 = (x - 3)^2 + (y - 4)^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16$
$-2x - 4y + 5 = -6x - 8y + 25$
$4x + 4y = 20$
$x + y = 5$
Now,solve the system of equations:
$3x + 4y = 5$
$x + y = 5 \Rightarrow x = 5 - y$
Substitute $x$ in the first equation:
$3(5 - y) + 4y = 5$
$15 - 3y + 4y = 5$
$y = -10$
$x = 5 - (-10) = 15$
Thus,the point is $(15, -10)$.

(D) The given lines are $4x + 3y - 15 = 0$ and $4x + 3y - 5 = 0$. Since the coefficients of $x$ and $y$ are the same,the lines are parallel.
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,$d = \frac{|-15 - (-5)|}{\sqrt{4^2 + 3^2}} = \frac{|-10|}{\sqrt{16 + 9}} = \frac{10}{5} = 2$.
Since the circle touches both lines,its diameter is equal to the distance between the lines,so $2r = 2$,which implies $r = 1$.
The area of the circle is $\pi r^2 = \pi(1)^2 = \pi \text{ square units}$.

(C) The given pair of straight lines is $x^2-3xy+2y^2=0$.
Factoring the equation: $x^2-2xy-xy+2y^2=0$ $\Rightarrow x(x-2y)-y(x-2y)=0$ $\Rightarrow (x-y)(x-2y)=0$.
Thus,the two lines are $L_1: x-y=0$ and $L_2: x-2y=0$.
The third line is $L_3: x+y+1=0$.
The vertices of the triangle are the intersection points of these lines:
$1$. Intersection of $L_1$ and $L_2$: $(0,0)$.
$2$. Intersection of $L_1$ and $L_3$: $x-y=0$ and $x+y=-1$ $\Rightarrow 2x=-1$ $\Rightarrow x=-\frac{1}{2}, y=-\frac{1}{2}$. Point is $(-\frac{1}{2}, -\frac{1}{2})$.
$3$. Intersection of $L_2$ and $L_3$: $x=2y$ and $2y+y=-1$ $\Rightarrow 3y=-1$ $\Rightarrow y=-\frac{1}{3}, x=-\frac{2}{3}$. Point is $(-\frac{2}{3}, -\frac{1}{3})$.
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 $= \frac{1}{2} |0(-\frac{1}{2} - (-\frac{1}{3})) + (-\frac{1}{2})(-\frac{1}{3} - 0) + (-\frac{2}{3})(0 - (-\frac{1}{2}))|$
Area $= \frac{1}{2} |0 + \frac{1}{6} - \frac{1}{3}| = \frac{1}{2} |-\frac{1}{6}| = \frac{1}{12}$ square units.

(C) The given pairs of lines are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$.
Factorizing the first equation: $(x-2y)(x-y)=0$,which gives lines $L_1: x-2y=0$ and $L_2: x-y=0$.
Factorizing the second equation: $(x-2y+2)(x-y-1)=0$,which gives lines $L_3: x-2y+2=0$ and $L_4: x-y-1=0$.
Comparing the equations,we see that $L_1 \parallel L_3$ and $L_2 \parallel L_4$.
Since the opposite sides are parallel,the figure is a parallelogram.
To check if it is a rectangle,we find the angle between $L_1$ and $L_2$. The slopes are $m_1 = 1/2$ and $m_2 = 1$. Since $m_1 \times m_2 \neq -1$,the angle is not $90^{\circ}$.
Thus,the figure is a parallelogram.

(B) The given equation is $2 x^2-10 x y+12 y^2+5 x+\lambda y-3=0$.
Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=2, h=-5, b=12, g=\frac{5}{2}, f=\frac{\lambda}{2}, c=-3$.
For the equation to represent a pair of straight lines,the determinant of the matrix must be zero: $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
Substituting the values: $\begin{vmatrix} 2 & -5 & 5/2 \\ -5 & 12 & \lambda/2 \\ 5/2 & \lambda/2 & -3 \end{vmatrix} = 0$.
Expanding the determinant: $2(-36 - \frac{\lambda^2}{4}) + 5(15 - \frac{5\lambda}{4}) + \frac{5}{2}(\frac{-5\lambda}{2} - 30) = 0$.
$-72 - \frac{\lambda^2}{2} + 75 - \frac{25\lambda}{4} - \frac{25\lambda}{4} - 75 = 0$.
$-\frac{\lambda^2}{2} - \frac{50\lambda}{4} - 72 = 0$.
Multiplying by $-2$: $\lambda^2 + 25\lambda + 144 = 0$.
Factoring the quadratic: $(\lambda + 9)(\lambda + 16) = 0$.
Thus,$\lambda = -9$ or $\lambda = -16$.
Since the condition is $|\lambda| < 16$,the only valid solution is $\lambda = -9$.

(C) The intersection point of the diameter lines is the center of the circle. Solving the equations $2x + y = 7$ and $x + 3y = 11$:
From the first equation,$y = 7 - 2x$.
Substituting into the second: $x + 3(7 - 2x) = 11$ $\Rightarrow x + 21 - 6x = 11$ $\Rightarrow -5x = -10$ $\Rightarrow x = 2$.
Then $y = 7 - 2(2) = 3$. So,the center $(h, k) = (2, 3)$.
The circle passes through $(5, 7)$,so the radius $r$ is the distance between $(2, 3)$ and $(5, 7)$:
$r = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.
The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$:
$(x - 2)^2 + (y - 3)^2 = 5^2$
$x^2 - 4x + 4 + y^2 - 6y + 9 = 25$
$x^2 + y^2 - 4x - 6y + 13 - 25 = 0$
$x^2 + y^2 - 4x - 6y - 12 = 0$.

(A) Let the equation of the circle be $x^2+y^2+2gx+2fy=0$ as it passes through the origin $(0,0)$.
The $x$-intercept is $2|g|=4$ $\Rightarrow |g|=2$ $\Rightarrow g = \pm 2$.
The $y$-intercept is $2|f|=8$ $\Rightarrow |f|=4$ $\Rightarrow f = \pm 4$.
Substituting these values into the general equation,we get $x^2+y^2 \pm 2(2)x \pm 2(4)y=0$.
Thus,the required equations are $x^2+y^2 \pm 4x \pm 8y=0$.

(C) Let the required equation of the circle be $x^2+y^2+2gx+2fy=0$ (since it passes through the origin,$c=0$).
Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2+2f_1f_2=c_1+c_2$.
For the first circle $x^2+y^2-6x+8=0$,$g_1=-3, f_1=0, c_1=8$. Thus,$2g(-3)+2f(0)=0+8$ $\Rightarrow -6g=8$ $\Rightarrow g=-\frac{4}{3}$.
For the second circle $x^2+y^2-2x-2y-7=0$,$g_2=-1, f_2=-1, c_2=-7$. Thus,$2g(-1)+2f(-1)=0-7 \Rightarrow -2g-2f=-7$.
Substituting $g=-\frac{4}{3}$,we get $-2(-\frac{4}{3})-2f=-7$ $\Rightarrow \frac{8}{3}+7=2f$ $\Rightarrow 2f=\frac{29}{3}$.
Substituting $g$ and $f$ into the general equation: $x^2+y^2+2(-\frac{4}{3})x+2(\frac{29}{6})y=0 \Rightarrow x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0$.
Multiplying by $3$,we get $3x^2+3y^2-8x+29y=0$.

(D) Given circles are $S_1: x^2 + y^2 - 2x + 8y + 13 = 0$ and $S_2: x^2 + y^2 - 4x + 6y + 11 = 0$.
For $S_1$,the center $C_1 = (1, -4)$ and radius $r_1 = \sqrt{1^2 + (-4)^2 - 13} = \sqrt{1 + 16 - 13} = 2$.
For $S_2$,the center $C_2 = (2, -3)$ and radius $r_2 = \sqrt{2^2 + (-3)^2 - 11} = \sqrt{4 + 9 - 11} = \sqrt{2}$.
The distance between the centers $d = C_1C_2 = \sqrt{(2 - 1)^2 + (-3 - (-4))^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
The angle $\theta$ between the circles is given by $\cos \theta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2}$.
Substituting the values: $\cos \theta = \frac{(\sqrt{2})^2 - 2^2 - (\sqrt{2})^2}{2 \times 2 \times \sqrt{2}} = \frac{2 - 4 - 2}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = -\frac{1}{\sqrt{2}}$.
Thus,$\theta = 135^{\circ}$.

(B) Let the centre of the circle be $C(h, k)$. Since the circle passes through the origin $(0, 0)$,its radius $r$ is given by $r^2 = h^2 + k^2$.
The perpendicular distance $d$ from the centre $C(h, k)$ to the line $x=3$ is $d = |h-3|$.
The length of the chord cut off by the line $x=3$ is $4$ units. Thus,half the length of the chord is $2$ units.
In the right-angled triangle formed by the radius,the perpendicular distance,and half the chord length,we have:
$r^2 = d^2 + 2^2$
$h^2 + k^2 = (h-3)^2 + 4$
$h^2 + k^2 = h^2 - 6h + 9 + 4$
$k^2 = -6h + 13$
Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = -6x + 13$,or $y^2 + 6x = 13$.

(B) The equation of the parabola is $y^2=4ax$,where $a=1$. The point $(1,0)$ is the focus of the parabola.
Any normal to the parabola $y^2=4ax$ at point $(at^2, 2at)$ is given by the equation $y = -tx + 2at + at^3$.
If this normal passes through the point $(h, k) = (1, 0)$,then $0 = -t(1) + 2(1)t + (1)t^3$.
This simplifies to $0 = t + t^3$,or $t(1 + t^2) = 0$.
The real solution for $t$ is $t=0$.
For $t=0$,the normal is $y = 0$,which is the $x$-axis.
Thus,only $1$ normal can be drawn from the point $(1,0)$ to the parabola $y^2=4x$.

(C) Given that the distance between the foci is $2ae = 6$,so $ae = 3$.
Given that the length of the minor axis is $2b = 8$,so $b = 4$.
We know the relation $b^2 = a^2(1 - e^2) = a^2 - a^2e^2$.
Substituting $b = 4$ and $ae = 3$,we get $16 = a^2 - (3)^2$.
$16 = a^2 - 9$ $\Rightarrow a^2 = 25$ $\Rightarrow a = 5$.
Now,the eccentricity $e = \frac{ae}{a} = \frac{3}{5}$.

(D) Given the equation of the conic: $\frac{5}{r}=2+3 \cos \theta+4 \sin \theta$.
Divide the equation by $2$: $\frac{5/2}{r} = 1 + \frac{3}{2} \cos \theta + 2 \sin \theta$.
We can write $\frac{3}{2} \cos \theta + 2 \sin \theta$ in the form $e \cos(\theta - \phi)$,where $e = \sqrt{(\frac{3}{2})^2 + 2^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.
Thus,the equation becomes $\frac{5/2}{r} = 1 + \frac{5}{2} \cos(\theta - \phi)$.
Comparing this with the standard polar form $\frac{l}{r} = 1 + e \cos(\theta - \phi)$,we identify the eccentricity $e = \frac{5}{2}$.

(B) The equation of the chord of the hyperbola $S: 2x^2 - 3y^2 - 12 = 0$ with midpoint $M(h, k)$ is given by $T = S_1$,where $T = 2xh - 3yk - 12$ and $S_1 = 2h^2 - 3k^2 - 12$.
Thus,the equation of the chord is $2xh - 3yk = 2h^2 - 3k^2$.
Comparing this with the given chord equation $4x - 3y = 5$,we get:
$\frac{2h}{4} = \frac{-3k}{-3} = \frac{2h^2 - 3k^2}{5}$.
From $\frac{2h}{4} = k$,we have $k = \frac{h}{2}$.
Substituting $k = \frac{h}{2}$ into $\frac{2h}{4} = \frac{2h^2 - 3(h/2)^2}{5}$:
$\frac{h}{2} = \frac{2h^2 - \frac{3h^2}{4}}{5} = \frac{5h^2}{20} = \frac{h^2}{4}$.
So,$\frac{h}{2} = \frac{h^2}{4}$ $\Rightarrow 2h = h^2$ $\Rightarrow h(h - 2) = 0$.
Since the chord exists,$h \neq 0$ (as $h=0$ gives $k=0$,which does not satisfy the chord equation $4(0)-3(0)=5$).
Thus,$h = 2$.
Then $k = \frac{2}{2} = 1$.
The midpoint is $(2, 1)$.

(A) Given equations are $x^2+y^2=a^2$ and $xy=c^2$.
From the second equation,$x = \frac{c^2}{y}$.
Substituting $x$ in the first equation:
$\left(\frac{c^2}{y}\right)^2 + y^2 = a^2$
$\frac{c^4}{y^2} + y^2 = a^2$
$c^4 + y^4 = a^2 y^2$
$y^4 - a^2 y^2 + c^4 = 0$
This is a biquadratic equation in $y$. Let the roots be $y_1, y_2, y_3, y_4$.
The equation is of the form $Ay^4 + By^3 + Cy^2 + Dy + E = 0$,where $B=0$.
By Vieta's formulas,the sum of the roots is $-\frac{B}{A} = -\frac{0}{1} = 0$.
Therefore,$y_1+y_2+y_3+y_4 = 0$.

(C) We know that $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$.
Given expression: $\lim _{x \rightarrow \infty} (\frac{x+5}{x+2})^{x+3} = \lim _{x \rightarrow \infty} (1 + \frac{3}{x+2})^{x+3}$.
Let $t = x+2$,then as $x \rightarrow \infty$,$t \rightarrow \infty$.
$= \lim _{t \rightarrow \infty} (1 + \frac{3}{t})^{t+1} = \lim _{t \rightarrow \infty} (1 + \frac{3}{t})^t \cdot (1 + \frac{3}{t})^1$.
$= e^3 \cdot 1 = e^3$.

(B) Using the projection formula or the cosine rule:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Substituting these into the expression:
$a(b \cos C - c \cos B) = ab \cos C - ac \cos B$
$= ab \left( \frac{a^2 + b^2 - c^2}{2ab} \right) - ac \left( \frac{a^2 + c^2 - b^2}{2ac} \right)$
$= \frac{a^2 + b^2 - c^2}{2} - \frac{a^2 + c^2 - b^2}{2}$
$= \frac{a^2 + b^2 - c^2 - a^2 - c^2 + b^2}{2}$
$= \frac{2b^2 - 2c^2}{2}$
$= b^2 - c^2$.

(C) Let $2s = a+b+c$. Then $b+c-a = 2s-2a$,$c+a-b = 2s-2b$,and $a+b-c = 2s-2c$.
Substituting these into the expression:
$\frac{2s(2s-2a)(2s-2b)(2s-2c)}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2} = 4 \frac{s(s-a)}{bc} \cdot \frac{(s-b)(s-c)}{bc}$.
Using the half-angle formulas $\cos^2(\frac{A}{2}) = \frac{s(s-a)}{bc}$ and $\sin^2(\frac{A}{2}) = \frac{(s-b)(s-c)}{bc}$,we get:
$4 \cos^2(\frac{A}{2}) \sin^2(\frac{A}{2}) = (2 \sin(\frac{A}{2}) \cos(\frac{A}{2}))^2 = \sin^2 A$.

(D) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter of $\triangle ABC$. Then $a+b+c = 2s$,$b+c-a = 2(s-a)$,$c+a-b = 2(s-b)$,and $a+b-c = 2(s-c)$.
Substituting these into the expression:
$\frac{(2s)(2(s-a))(2(s-b))(2(s-c))}{4b^2c^2} = \frac{16s(s-a)(s-b)(s-c)}{4b^2c^2}$.
Using Heron's formula,$\Delta^2 = s(s-a)(s-b)(s-c)$,so the expression becomes $\frac{16\Delta^2}{4b^2c^2} = \frac{4\Delta^2}{b^2c^2}$.
Since the area $\Delta = \frac{1}{2}bc \sin A$,we have $\sin A = \frac{2\Delta}{bc}$.
Thus,$\frac{4\Delta^2}{b^2c^2} = (\frac{2\Delta}{bc})^2 = \sin^2 A$.

(D) Let $f(x) = \sin ^4 x + \cos ^4 x$.
We know that $\sin ^4 x + \cos ^4 x = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x$.
Since $\sin ^2 x + \cos ^2 x = 1$,we have $f(x) = 1 - 2 \sin ^2 x \cos ^2 x$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we can write $2 \sin ^2 x \cos ^2 x = \frac{1}{2} (2 \sin x \cos x)^2 = \frac{1}{2} \sin ^2 2x$.
Thus,$f(x) = 1 - \frac{1}{2} \sin ^2 2x$.
Using the identity $\sin ^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} \right) = 1 - \frac{1}{4} + \frac{\cos 4x}{4} = \frac{3}{4} + \frac{1}{4} \cos 4x$.
The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$.
Therefore,the period of $f(x) = \frac{3}{4} + \frac{1}{4} \cos 4x$ is $\frac{2\pi}{4} = \frac{\pi}{2}$.

(C) Let the two poles be $AD$ and $BC$ with heights $a$ and $b$ respectively,standing on a horizontal line $AB$.
$P$ is a point on $AB$ such that $\angle DPA = 45^{\circ}$ and $\angle CPB = 45^{\circ}$.
In $\triangle APD$,$\tan 45^{\circ} = \frac{AD}{AP} = \frac{a}{AP} \Rightarrow AP = a$.
In $\triangle BPC$,$\tan 45^{\circ} = \frac{BC}{BP} = \frac{b}{BP} \Rightarrow BP = b$.
The horizontal distance between the poles is $AB = AP + PB = a + b$.
Draw a line $DE$ parallel to $AB$ such that $E$ lies on $BC$. Then $DE = AB = a + b$ and $CE = BC - BE = BC - AD = b - a$.
In the right-angled triangle $\triangle DEC$,by Pythagoras theorem:
$DC^2 = DE^2 + CE^2$
$DC^2 = (a + b)^2 + (b - a)^2$
$DC^2 = (a^2 + 2ab + b^2) + (b^2 - 2ab + a^2)$
$DC^2 = 2(a^2 + b^2)$.

(D) Let the vertices of the triangle be $A = (1, 0, 0)$,$B = (0, 1, 0)$,and $C = (0, 0, 1)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$,we calculate the lengths of the sides:
$AB = \sqrt{(0-1)^2 + (1-0)^2 + (0-0)^2} = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1+1} = \sqrt{2}$.
$BC = \sqrt{(0-0)^2 + (0-1)^2 + (1-0)^2} = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.
$CA = \sqrt{(1-0)^2 + (0-0)^2 + (0-1)^2} = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
The perimeter of the triangle is the sum of the lengths of its sides:
$\text{Perimeter} = AB + BC + CA = \sqrt{2} + \sqrt{2} + \sqrt{2} = 3\sqrt{2}$.

(C) Given,$P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B}) = \frac{7}{10}$.
Since $P(A \cap B) + P(\overline{A \cap B}) = 1$,we have:
$P(A \cap B) = 1 - \frac{7}{10} = \frac{3}{10}$.
Also,the formula for the union of two events is:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values:
$\frac{4}{5} = P(A) + \frac{2}{5} - \frac{3}{10}$.
$P(A) = \frac{4}{5} - \frac{2}{5} + \frac{3}{10}$.
$P(A) = \frac{2}{5} + \frac{3}{10} = \frac{4+3}{10} = \frac{7}{10}$.

(D) The given quadratic equation is $x^2 + 4x + c = 0$.
For the roots to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = b^2 - 4ac \geq 0$
Substituting the values $a = 1, b = 4$,we get:
$4^2 - 4(1)(c) \geq 0$
$16 - 4c \geq 0$
$16 \geq 4c$
$c \leq 4$.
Since $c$ is chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,the possible values for $c$ are $\{1, 2, 3, 4\}$.
There are $4$ favorable outcomes out of $9$ total possible outcomes.
Therefore,the required probability is $\frac{4}{9}$.

(D) matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$|A| = 0$.
Given $A = \left[\begin{array}{ccc}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$.
Calculating the determinant along the first row:
$|A| = 1(x - (-1)) - (-1)(1 - x) + x(-1 - x^2) = 0$
$|A| = 1(x + 1) + 1(1 - x) + x(-1 - x^2) = 0$
$|A| = x + 1 + 1 - x - x - x^3 = 0$
$-x^3 - x + 2 = 0$
$x^3 + x - 2 = 0$
By inspection,$x = 1$ is a root because $1^3 + 1 - 2 = 0$.
Dividing $x^3 + x - 2$ by $(x - 1)$,we get $(x - 1)(x^2 + x + 2) = 0$.
For $x^2 + x + 2 = 0$,the discriminant $D = b^2 - 4ac = 1^2 - 4(1)(2) = 1 - 8 = -7$.
Since $D < 0$,there are no real roots for the quadratic part.
Thus,the only real value is $x = 1$.

(C) Given,$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$.
Let $t = \tan \theta$,then $\theta = \tan^{-1} t$.
For $x$,we have $x = \cos^{-1}\left(\frac{1}{\sqrt{1+\tan^2 \theta}}\right) = \cos^{-1}\left(\frac{1}{\sec \theta}\right) = \cos^{-1}(\cos \theta) = \theta = \tan^{-1} t$.
For $y$,we have $y = \sin^{-1}\left(\frac{\tan \theta}{\sqrt{1+\tan^2 \theta}}\right) = \sin^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) = \sin^{-1}(\sin \theta) = \theta = \tan^{-1} t$.
Since $x = \tan^{-1} t$ and $y = \tan^{-1} t$,it follows that $y = x$.
Therefore,differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$.

(A) Given,$z=\tan (y+a x)+\sqrt{y-a x}$
First,find the partial derivatives with respect to $x$:
$z_x = \frac{\partial z}{\partial x} = a \sec^2(y+ax) - \frac{a}{2\sqrt{y-ax}}$
$z_{xx} = \frac{\partial^2 z}{\partial x^2} = 2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}$
Next,find the partial derivatives with respect to $y$:
$z_y = \frac{\partial z}{\partial y} = \sec^2(y+ax) + \frac{1}{2\sqrt{y-ax}}$
$z_{yy} = \frac{\partial^2 z}{\partial y^2} = 2 \sec^2(y+ax) \tan(y+ax) - \frac{1}{4(y-ax)^{3/2}}$
Now,calculate $z_{xx} - a^2 z_{yy}$:
$z_{xx} - a^2 z_{yy} = [2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}}] - a^2 [2 \sec^2(y+ax) \tan(y+ax) - \frac{1}{4(y-ax)^{3/2}}]$
$z_{xx} - a^2 z_{yy} = 2a^2 \sec^2(y+ax) \tan(y+ax) - \frac{a^2}{4(y-ax)^{3/2}} - 2a^2 \sec^2(y+ax) \tan(y+ax) + \frac{a^2}{4(y-ax)^{3/2}} = 0$

(C) Given,$y=e^{a \sin ^{-1} x}$.
On differentiating with respect to $x$,we get:
$y_1 = e^{a \sin ^{-1} x} \cdot \frac{a}{\sqrt{1-x^2}}$
$\Rightarrow y_1 \sqrt{1-x^2} = ay$
Squaring both sides:
$(1-x^2) y_1^2 = a^2 y^2$
Differentiating again with respect to $x$:
$(1-x^2) 2 y_1 y_2 - 2x y_1^2 = a^2 2y y_1$
Dividing by $2y_1$ (assuming $y_1 \neq 0$):
$(1-x^2) y_2 - x y_1 - a^2 y = 0$
Applying Leibnitz's theorem for the $n$-th derivative:
$[(1-x^2) y_{n+2} + n(-2x) y_{n+1} + \frac{n(n-1)}{2}(-2) y_n] - [x y_{n+1} + n(1) y_n] - a^2 y_n = 0$
$(1-x^2) y_{n+2} - 2nx y_{n+1} - n(n-1) y_n - x y_{n+1} - n y_n - a^2 y_n = 0$
$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} - (n^2 - n + n + a^2) y_n = 0$
$(1-x^2) y_{n+2} - (2n+1) x y_{n+1} = (n^2 + a^2) y_n$

(B) Given,$f(x)=x^3+3x-2$.
On differentiating with respect to $x$,we get $f'(x)=3x^2+3$.
Since $x^2 \ge 0$ for all $x \in [2,3]$,$f'(x) = 3x^2+3 \ge 3 > 0$.
Thus,$f(x)$ is a strictly increasing function on the interval $[2,3]$.
For an increasing function,the range is $[f(2), f(3)]$.
At $x=2$,$f(2) = 2^3 + 3(2) - 2 = 8 + 6 - 2 = 12$.
At $x=3$,$f(3) = 3^3 + 3(3) - 2 = 27 + 9 - 2 = 34$.
Therefore,the range of $f(x)$ is $[12, 34]$.

(B) Given the function $f(x)=x^3+a x^2+b x+c$.
To find the extreme values,we differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x)=3 x^2+2 a x+b$.
For extreme values,we set $f^{\prime}(x)=0$:
$3 x^2+2 a x+b=0$.
The roots of this quadratic equation are given by the quadratic formula:
$x = \frac{-2 a \pm \sqrt{(2 a)^2 - 4(3)(b)}}{2(3)} = \frac{-2 a \pm \sqrt{4 a^2 - 12 b}}{6} = \frac{-2 a \pm 2 \sqrt{a^2 - 3 b}}{6} = \frac{-a \pm \sqrt{a^2 - 3 b}}{3}$.
Given the condition $a^2 \leq 3 b$,the term $a^2 - 3 b \leq 0$.
If $a^2 - 3 b < 0$,the roots are imaginary,meaning $f^{\prime}(x)$ is never zero for any real $x$.
If $a^2 = 3 b$,then $f^{\prime}(x) = 3(x + \frac{a}{3})^2$,which is always $\geq 0$,so the function is monotonically increasing and has no local extrema.
Thus,the function has no extreme value.

(D) Let $y = \frac{\log x}{x}$.
Find the derivative with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For critical points,set $\frac{dy}{dx} = 0$:
$1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
To verify the maximum,find the second derivative:
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - \log x)(2x)}{x^4} = \frac{-x - 2x + 2x \log x}{x^4} = \frac{2 \log x - 3}{x^3}$.
At $x = e$,$\frac{d^2y}{dx^2} = \frac{2(1) - 3}{e^3} = -\frac{1}{e^3} < 0$,so $x = e$ is a point of local maxima.
The maximum value is $y(e) = \frac{\log e}{e} = \frac{1}{e} = e^{-1}$.

(A) Let $I = \int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$.
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 - \cos 2x = 2 \sin^2 x$,we get:
$I = \int \left( \frac{2 - 2 \sin x \cos x}{2 \sin^2 x} \right) e^x \, dx$
$I = \int \left( \frac{1}{\sin^2 x} - \frac{\sin x \cos x}{\sin^2 x} \right) e^x \, dx$
$I = \int (\operatorname{cosec}^2 x - \cot x) e^x \, dx$
$I = \int e^x \operatorname{cosec}^2 x \, dx - \int e^x \cot x \, dx$.
Applying integration by parts to the first integral $\int e^x \operatorname{cosec}^2 x \, dx$:
Let $u = \cot x$,then $du = -\operatorname{cosec}^2 x \, dx$.
$\int e^x \operatorname{cosec}^2 x \, dx = e^x(-\cot x) - \int e^x(-\cot x) \, dx = -e^x \cot x + \int e^x \cot x \, dx$.
Substituting this back into the expression for $I$:
$I = (-e^x \cot x + \int e^x \cot x \, dx) - \int e^x \cot x \, dx + c$
$I = -e^x \cot x + c$.

(D) Given the family of curves: $y = (a + bx + cx^2) e^x$
This can be written as $y e^{-x} = a + bx + cx^2$.
Differentiating with respect to $x$ three times:
First derivative: $y' e^{-x} - y e^{-x} = b + 2cx \implies (y' - y) e^{-x} = b + 2cx$
Second derivative: $(y'' - y') e^{-x} - (y' - y) e^{-x} = 2c \implies (y'' - 2y' + y) e^{-x} = 2c$
Third derivative: $(y''' - 2y'' + y') e^{-x} - (y'' - 2y' + y) e^{-x} = 0$
Since $e^{-x} \neq 0$,we have $y''' - 3y'' + 3y' - y = 0$.

(B) Given,$\frac{dy}{dx} = \sin(x+y) \tan(x+y) - 1$.
Let $x+y = z$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dz}{dx}$,so $\frac{dy}{dx} = \frac{dz}{dx} - 1$.
Substituting this into the differential equation:
$\frac{dz}{dx} - 1 = \sin z \tan z - 1$
$\frac{dz}{dx} = \sin z \tan z = \sin z \cdot \frac{\sin z}{\cos z} = \frac{\sin^2 z}{\cos z}$.
Rearranging the variables:
$\int \frac{\cos z}{\sin^2 z} dz = \int dx$.
Let $\sin z = t$,then $\cos z dz = dt$.
$\int \frac{1}{t^2} dt = x + c$
$- \frac{1}{t} = x + c$
$- \operatorname{cosec} z = x + c$
Substituting $z = x+y$ back:
$- \operatorname{cosec}(x+y) = x + c$
$x + \operatorname{cosec}(x+y) = c$ (where $c$ is an arbitrary constant).

(A) The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
Calculating the magnitudes:
$M_1 = |\vec{a}_1| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.45$
$M_2 = |\vec{a}_2| = \sqrt{(-3)^2 + (-4)^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41} \approx 6.40$
$M_3 = |\vec{a}_3| = \sqrt{(-1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.73$
$M_4 = |\vec{a}_4| = \sqrt{(-1)^2 + (3)^2 + (1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \approx 3.32$
Comparing the values: $\sqrt{3} < \sqrt{6} < \sqrt{11} < \sqrt{41}$,which implies $M_3 < M_1 < M_4 < M_2$.

(D) Given equations are $2l + m + 2n = 0$ $(1)$ and $3l^2 + 5m^2 - 11n^2 = 0$ $(2)$.
From $(1)$,$m = -2l - 2n$.
Substitute $m$ into $(2)$: $3l^2 + 5(-2l - 2n)^2 - 11n^2 = 0$.
$3l^2 + 5(4l^2 + 8ln + 4n^2) - 11n^2 = 0$.
$3l^2 + 20l^2 + 40ln + 20n^2 - 11n^2 = 0$.
$23l^2 + 40ln + 9n^2 = 0$.
Divide by $n^2$: $23(\frac{l}{n})^2 + 40(\frac{l}{n}) + 9 = 0$.
Let $x = \frac{l}{n}$. Then $23x^2 + 40x + 9 = 0$.
Let the roots be $x_1 = \frac{l_1}{n_1}$ and $x_2 = \frac{l_2}{n_2}$.
Then $x_1 x_2 = \frac{l_1 l_2}{n_1 n_2} = \frac{9}{23}$.
Similarly,substituting $l = -\frac{m+2n}{2}$ into $(2)$ gives $3(\frac{m+2n}{2})^2 + 5m^2 - 11n^2 = 0$,which leads to $3(m^2 + 4mn + 4n^2) + 20m^2 - 44n^2 = 0$,so $23m^2 + 12mn - 32n^2 = 0$.
Dividing by $n^2$,$23(\frac{m}{n})^2 + 12(\frac{m}{n}) - 32 = 0$.
Let $y_1 = \frac{m_1}{n_1}$ and $y_2 = \frac{m_2}{n_2}$. Then $y_1 y_2 = \frac{m_1 m_2}{n_1 n_2} = -\frac{32}{23}$.
For two lines with direction ratios $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$,$\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
Since $l_i^2 + m_i^2 + n_i^2 = 1$,we use the relation $l_1 l_2 + m_1 m_2 + n_1 n_2 = n_1 n_2 (x_1 x_2 + y_1 y_2 + 1) = n_1 n_2 (\frac{9}{23} - \frac{32}{23} + 1) = n_1 n_2 (\frac{-23}{23} + 1) = 0$.
Thus,$\cos \theta = 0$,which means $\theta = \frac{\pi}{2}$.

(C) We know that for a line making angles $\alpha, \beta, \gamma$ with the coordinate axes,the direction cosines satisfy $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Given expression: $E = \cos 2\alpha + \cos 2\beta + \cos 2\gamma + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Using the identity $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$,we have:
$E = (\cos^2 \alpha - \sin^2 \alpha) + (\cos^2 \beta - \sin^2 \beta) + (\cos^2 \gamma - \sin^2 \gamma) + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Simplifying the expression:
$E = \cos^2 \alpha - \sin^2 \alpha + \cos^2 \beta - \sin^2 \beta + \cos^2 \gamma - \sin^2 \gamma + \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$.
Canceling the $\sin^2$ terms:
$E = \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma$.
Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,the value of the expression is $1$.

(C) Given,$P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$,and $P(E_1 / E_2) = \frac{1}{4}$.
$(A)$ Since $P(E_2 / E_1) = \frac{P(E_1 \cap E_2)}{P(E_1)}$,we have $\frac{1}{2} = \frac{P(E_1 \cap E_2)}{1/4}$,so $P(E_1 \cap E_2) = \frac{1}{8}$.
Also,$P(E_1 / E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} = \frac{1}{4}$.
Substituting $P(E_1 \cap E_2) = \frac{1}{8}$,we get $\frac{1/8}{P(E_2)} = \frac{1}{4}$,which implies $P(E_2) = \frac{1}{2}$. Thus,$(A)$ matches with $(iv)$.
$(B)$ $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2+4-1}{8} = \frac{5}{8}$. Thus,$(B)$ matches with $(ii)$.
$(C)$ $P(\bar{E}_1 / \bar{E}_2) = \frac{P(\bar{E}_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{1 - P(E_1 \cup E_2)}{1 - P(E_2)} = \frac{1 - 5/8}{1 - 1/2} = \frac{3/8}{1/2} = \frac{3}{4}$. Thus,$(C)$ matches with $(vi)$.
$(D)$ $P(E_1 / \bar{E}_2) = \frac{P(E_1 \cap \bar{E}_2)}{P(\bar{E}_2)} = \frac{P(E_1) - P(E_1 \cap E_2)}{1 - P(E_2)} = \frac{1/4 - 1/8}{1 - 1/2} = \frac{1/8}{1/2} = \frac{1}{4}$. Thus,$(D)$ matches with $(i)$.

(A) Given that $X$ is a binomial variate with range $\{0, 1, 2, 3, 4, 5, 6\}$,so the number of trials $n = 6$.
The probability mass function of a binomial distribution is given by $P(X=k) = {^nC_k} p^k q^{n-k}$,where $q = 1-p$.
According to the problem,$P(X=2) = 4 P(X=4)$.
Substituting the values:
${^6C_2} p^2 q^4 = 4 \cdot {^6C_4} p^4 q^2$.
Since ${^6C_2} = \frac{6 \times 5}{2 \times 1} = 15$ and ${^6C_4} = {^6C_2} = 15$,we have:
$15 p^2 q^4 = 4 \cdot 15 p^4 q^2$.
Dividing both sides by $15 p^2 q^2$ (assuming $p, q \neq 0$):
$q^2 = 4 p^2$.
Substituting $q = 1-p$:
$(1-p)^2 = 4 p^2$.
Taking the square root on both sides:
$1-p = 2p$ or $1-p = -2p$.
Case $1$: $1 = 3p \Rightarrow p = \frac{1}{3}$.
Case $2$: $1 = -p \Rightarrow p = -1$ (which is impossible as $0 \leq p \leq 1$).
Therefore,the parameter $p = \frac{1}{3}$.

(B) Given,$f(x)=x^2+a x+b$ has imaginary roots.
Since the roots are imaginary,the discriminant $D < 0$,so $a^2-4b < 0$.
Now,we find the derivatives:
$f^{\prime}(x) = 2x+a$
$f^{\prime \prime}(x) = 2$
Substituting these into the equation $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$:
$(x^2+ax+b) + (2x+a) + 2 = 0$
$x^2+(a+2)x+(a+b+2) = 0$
Let $D'$ be the discriminant of this new quadratic equation:
$D' = (a+2)^2 - 4(a+b+2)$
$D' = a^2+4a+4 - 4a-4b-8$
$D' = a^2-4b-4$
Since $a^2-4b < 0$,it follows that $a^2-4b-4 < -4$.
Therefore,$D' < 0$,which means the roots of the equation $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are also imaginary.

(D) Given,$\frac{1}{e^{3x}}(e^x + e^{5x}) = a_0 + a_1x + a_2x^2 + \ldots$
$\Rightarrow e^{-2x} + e^{2x} = a_0 + a_1x + a_2x^2 + \ldots$
Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \ldots$,we have:
$e^{-2x} + e^{2x} = (1 - 2x + \frac{(-2x)^2}{2!} - \frac{(-2x)^3}{3!} + \ldots) + (1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \ldots)$
$= 2(1 + \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + \ldots) = 2 + 4x^2 + \frac{2}{3}x^4 + \ldots$
Comparing coefficients,we see that all odd-indexed coefficients $a_1, a_3, a_5, \ldots$ are $0$.
Therefore,$2a_1 + 2^3a_3 + 2^5a_5 + \ldots = 0$.

(C) Given the probability distribution:
Mean $m = \sum p_i x_i = (0 \times \frac{1}{3}) + (1 \times \frac{1}{2}) + (2 \times 0) + (3 \times \frac{1}{6})$
$m = 0 + \frac{1}{2} + 0 + \frac{1}{2} = 1$
Variance $\sigma^2 = \sum p_i (x_i - m)^2$
$\sigma^2 = \frac{1}{3}(0 - 1)^2 + \frac{1}{2}(1 - 1)^2 + 0(2 - 1)^2 + \frac{1}{6}(3 - 1)^2$
$\sigma^2 = \frac{1}{3}(1) + \frac{1}{2}(0) + 0 + \frac{1}{6}(4)$
$\sigma^2 = \frac{1}{3} + 0 + 0 + \frac{4}{6} = \frac{1}{3} + \frac{2}{3} = 1$
Thus,$m = 1$ and $\sigma^2 = 1$,so $m = \sigma^2 = 1$.

(D) The region is bounded by $y=\sin x$,$y=\cos x$ and the $x$-axis for $0 \leq x \leq \frac{\pi}{2}$.
The area $A_1$ is the area under $y=\sin x$ from $x=0$ to $x=\frac{\pi}{4}$:
$A_1 = \int_0^{\pi/4} \sin x \, dx = [-\cos x]_0^{\pi/4} = -(\cos \frac{\pi}{4} - \cos 0) = -(\frac{1}{\sqrt{2}} - 1) = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.
The area $A_2$ is the area under $y=\cos x$ from $x=\frac{\pi}{4}$ to $x=\frac{\pi}{2}$:
$A_2 = \int_{\pi/4}^{\pi/2} \cos x \, dx = [\sin x]_{\pi/4}^{\pi/2} = \sin \frac{\pi}{2} - \sin \frac{\pi}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.
Thus,the ratio $A_1 : A_2 = \frac{\sqrt{2}-1}{\sqrt{2}} : \frac{\sqrt{2}-1}{\sqrt{2}} = 1 : 1$.

(B) Given that $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
Consider the transpose of the matrix $(AB - BA)$:
$(AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$
Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$,we get:
$(AB - BA)^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$
Substituting $A^{\prime} = A$ and $B^{\prime} = B$:
$(AB - BA)^{\prime} = BA - AB$
$(AB - BA)^{\prime} = -(AB - BA)$
Since the transpose of the matrix $(AB - BA)$ is equal to its negative,the matrix $(AB - BA)$ is a skew-symmetric matrix.

(D) square matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$|A| = 0$.
Given the matrix $A = \begin{bmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{bmatrix}$.
We calculate the determinant:
$|A| = 1(x - (-1)) - (-1)(1 - x) + x(-1 - x^2) = 0$
$|A| = 1(x + 1) + 1(1 - x) + x(-1 - x^2) = 0$
$|A| = x + 1 + 1 - x - x - x^3 = 0$
$|A| = -x^3 - x + 2 = 0$
$x^3 + x - 2 = 0$
By inspection,if $x = 1$,then $1^3 + 1 - 2 = 0$,which satisfies the equation.
Alternatively,if $x = 1$,the matrix becomes $\begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{bmatrix}$.
Since the first and third rows (or columns) are identical,the determinant is $0$.
Thus,the real value of $x$ is $1$.

(A) Given the determinant equation: $\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0$
Expanding along the first row:
$3(3x - 35) - 5(21 - 7x) + x(35 - x^2) = 0$
$9x - 105 - 105 + 35x + 35x - x^3 = 0$
$-x^3 + 79x - 210 = 0$
$x^3 - 79x + 210 = 0$
Since $x = -10$ is a root,$(x + 10)$ is a factor.
Performing polynomial division or synthetic division:
$(x + 10)(x^2 - 10x + 21) = 0$
$(x + 10)(x - 3)(x - 7) = 0$
The roots are $x = -10, 3, 7$.
Thus,the other roots are $3$ and $7$.

(D) Let $a$ be the first term and $R$ be the common ratio of the geometric progression $(GP)$.
Then,the $n$-th term is given by $T_n = a R^{n-1}$.
Given $x = T_p = a R^{p-1}$,$y = T_q = a R^{q-1}$,and $z = T_r = a R^{r-1}$.
Taking the logarithm on both sides:
$\log x = \log a + (p-1) \log R$
$\log y = \log a + (q-1) \log R$
$\log z = \log a + (r-1) \log R$
Now,substitute these into the determinant:
$\Delta = \left|\begin{array}{lll} \log a + (p-1) \log R & p & 1 \\ \log a + (q-1) \log R & q & 1 \\ \log a + (r-1) \log R & r & 1 \end{array}\right|$
Using the property of determinants,we can split this into two determinants:
$\Delta = \left|\begin{array}{lll} \log a & p & 1 \\ \log a & q & 1 \\ \log a & r & 1 \end{array}\right| + \left|\begin{array}{lll} (p-1) \log R & p & 1 \\ (q-1) \log R & q & 1 \\ (r-1) \log R & r & 1 \end{array}\right|$
In the first determinant,the first column is $\log a$ times the third column,so it is $0$.
In the second determinant,perform the operation $C_2 \rightarrow C_2 - C_3$:
$\Delta = 0 + \log R \left|\begin{array}{lll} p-1 & p-1 & 1 \\ q-1 & q-1 & 1 \\ r-1 & r-1 & 1 \end{array}\right|$
Since the first and second columns are identical,the value of the determinant is $0$.
Thus,$\Delta = 0 + 0 = 0$.

(D) We use the properties of inverse trigonometric functions: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,$\sin^{-1}(-x) = -\sin^{-1}(x)$,and $\tan^{-1}(-x) = -\tan^{-1}(x)$.
Given expression: $E = \cos^{-1}\left(-\frac{1}{2}\right) - 2\sin^{-1}\left(\frac{1}{2}\right) + 3\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) - 4\tan^{-1}(-1)$
Step $1$: Apply the properties:
$E = \left(\pi - \cos^{-1}\left(\frac{1}{2}\right)\right) - 2\left(\frac{\pi}{6}\right) + 3\left(\pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) - 4\left(-\frac{\pi}{4}\right)$
Step $2$: Substitute the standard values:
$E = \left(\pi - \frac{\pi}{3}\right) - \frac{\pi}{3} + 3\left(\pi - \frac{\pi}{4}\right) + \pi$
Step $3$: Simplify the expression:
$E = \frac{2\pi}{3} - \frac{\pi}{3} + 3\left(\frac{3\pi}{4}\right) + \pi$
$E = \frac{\pi}{3} + \frac{9\pi}{4} + \pi$
$E = \frac{4\pi + 27\pi + 12\pi}{12} = \frac{43\pi}{12}$

(C) The expression $\frac{2x-1}{x^3+4x^2+3x}$ is defined for all real numbers $x$ except where the denominator is zero.
Set the denominator equal to zero:
$x^3+4x^2+3x = 0$
$x(x^2+4x+3) = 0$
$x(x+1)(x+3) = 0$
Thus,the expression is undefined at $x = 0$,$x = -1$,and $x = -3$.
Therefore,the set is $R - \{0, -1, -3\}$.

(B) The Trapezoidal rule for numerical integration is given by:
$\text{Distance} = \int_{0}^{10} v \ dt \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3 + y_4) + y_5]$
Here,the interval width $h = 2 \ s$.
The values are $y_0=0, y_1=12, y_2=16, y_3=20, y_4=35, y_5=60$.
Substituting these values into the formula:
$\text{Distance} = \frac{2}{2} [0 + 2(12 + 16 + 20 + 35) + 60]$
$= 1 \times [0 + 2(83) + 60]$
$= 166 + 60$
$= 226 \ \text{m}$.

(D) For a function $f(x)$ to be continuous at $x=0$,the condition $\lim_{x \rightarrow 0} f(x) = f(0)$ must hold.
Given $f(x) = \frac{2 \sin x - \sin 2x}{2x \cos x}$ for $x \neq 0$.
We know that $\sin 2x = 2 \sin x \cos x$.
Substituting this into the expression:
$\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{2 \sin x - 2 \sin x \cos x}{2x \cos x}$
$= \lim_{x \rightarrow 0} \frac{2 \sin x (1 - \cos x)}{2x \cos x}$
$= \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right) \cdot \left( \frac{1 - \cos x}{\cos x} \right)$
Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{1 - \cos x}{\cos x} = \frac{1 - 1}{1} = 0$.
Therefore,$\lim_{x \rightarrow 0} f(x) = 1 \cdot 0 = 0$.
Since $f(0) = a$,for continuity,we must have $a = 0$.

(B) Given,$x = \frac{1-\sqrt{y}}{1+\sqrt{y}}$.
Applying componendo and dividendo,we get:
$\frac{1+x}{1-x} = \frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y})-(1-\sqrt{y})}$
$\frac{1+x}{1-x} = \frac{2}{2\sqrt{y}} = \frac{1}{\sqrt{y}}$
Squaring both sides,we get $\sqrt{y} = \frac{1-x}{1+x}$,so $y = \left(\frac{1-x}{1+x}\right)^2$.
Now,differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$
Using the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v-uv'}{v^2}$:
$\frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(-1)(1+x) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}$
Therefore,$\frac{dy}{dx} = 2\left(\frac{1-x}{1+x}\right) \cdot \left(\frac{-2}{(1+x)^2}\right) = \frac{-4(1-x)}{(1+x)^3} = \frac{4(x-1)}{(1+x)^3}$.

(D) Given,error in diameter $\Delta D = \pm 0.04 \text{ cm}$.
Since $D = 2r$,the error in radius is $\Delta r = \frac{\Delta D}{2} = \pm 0.02 \text{ cm}$.
The volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Differentiating with respect to $r$,we get $dV = 4 \pi r^2 dr$.
The relative error in volume is $\frac{dV}{V} = \frac{4 \pi r^2 dr}{\frac{4}{3} \pi r^3} = 3 \frac{dr}{r}$.
Percentage error in volume $= \frac{dV}{V} \times 100 = 3 \times \frac{\Delta r}{r} \times 100$.
Substituting the values $r = 10 \text{ cm}$ and $\Delta r = \pm 0.02 \text{ cm}$:
Percentage error $= 3 \times \frac{\pm 0.02}{10} \times 100 = 3 \times (\pm 0.002) \times 100 = \pm 0.6\%$.

(B) Given the function $f(x) = x^3 + ax^2 + bx + c$.
To find the extreme values,we differentiate $f(x)$ with respect to $x$:
$f'(x) = 3x^2 + 2ax + b$.
For extreme values,we set $f'(x) = 0$:
$3x^2 + 2ax + b = 0$.
The discriminant of this quadratic equation is $D = (2a)^2 - 4(3)(b) = 4a^2 - 12b = 4(a^2 - 3b)$.
Given the condition $a^2 \leq 3b$,it follows that $a^2 - 3b \leq 0$,which implies $D \leq 0$.
If $D < 0$,the quadratic equation $f'(x) = 0$ has no real roots,meaning $f'(x)$ does not change sign and the function is strictly monotonic.
If $D = 0$,$f'(x) = 3(x + a/3)^2$,which is always $\geq 0$,so the function is monotonically increasing and has a point of inflection,not an extreme value.
Therefore,the function has no extreme value.

(B) Given,$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$.
Integrating both sides with respect to $x$,we get:
$a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) = \int \frac{1}{x^4-1} dx$.
We know that $\frac{1}{x^4-1} = \frac{1}{(x^2-1)(x^2+1)} = \frac{1}{2} \left[ \frac{1}{x^2-1} - \frac{1}{x^2+1} \right]$.
So,$\int \frac{1}{x^4-1} dx = \frac{1}{2} \int \frac{1}{x^2-1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$.
Using standard integrals $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right|$ and $\int \frac{1}{x^2+1} dx = \tan^{-1} x$:
$a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) = \frac{1}{2} \left( \frac{1}{2} \log \left| \frac{x-1}{x+1} \right| \right) - \frac{1}{2} \tan^{-1} x$.
$a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right) = -\frac{1}{2} \tan^{-1} x + \frac{1}{4} \log \left| \frac{x-1}{x+1} \right|$.
Comparing the coefficients,we get $a = -\frac{1}{2}$ and $b = \frac{1}{4}$.
Therefore,$a - 2b = -\frac{1}{2} - 2(\frac{1}{4}) = -\frac{1}{2} - \frac{1}{2} = -1$.

(C) Let $I = \int \frac{d x}{(x+1) \sqrt{4 x+3}}$.
Substitute $4x + 3 = t^2$,which implies $4dx = 2tdt$ or $dx = \frac{1}{2}tdt$.
Also,$x = \frac{t^2 - 3}{4}$,so $x + 1 = \frac{t^2 - 3}{4} + 1 = \frac{t^2 + 1}{4}$.
Substituting these into the integral:
$I = \int \frac{\frac{1}{2} t dt}{(\frac{t^2 + 1}{4}) t} = \int \frac{\frac{1}{2} dt}{\frac{t^2 + 1}{4}} = 2 \int \frac{dt}{t^2 + 1}$.
Integrating,we get $I = 2 \tan^{-1}(t) + c$.
Substituting back $t = \sqrt{4x + 3}$,we get $I = 2 \tan^{-1}(\sqrt{4x + 3}) + c$.

(C) We are given $I_n = \int \sin^n x \, dx$.
Using integration by parts,let $u = \sin^{n-1} x$ and $dv = \sin x \, dx$.
Then $du = (n-1) \sin^{n-2} x \cos x \, dx$ and $v = -\cos x$.
Applying the formula $\int u \, dv = uv - \int v \, du$:
$I_n = -\sin^{n-1} x \cos x - \int (-\cos x) (n-1) \sin^{n-2} x \cos x \, dx$
$I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x \cos^2 x \, dx$
$I_n = -\sin^{n-1} x \cos x + (n-1) \int \sin^{n-2} x (1 - \sin^2 x) \, dx$
$I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2} - (n-1) I_n$
Rearranging the terms:
$I_n + (n-1) I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$
$n I_n = -\sin^{n-1} x \cos x + (n-1) I_{n-2}$
$n I_n - (n-1) I_{n-2} = -\sin^{n-1} x \cos x$.

(B) Let $I = \int_0^\pi \frac{1}{1+\sin x} dx$.
Using the identity $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int_0^\pi \frac{1}{1+\frac{2 \tan(x/2)}{1+\tan^2(x/2)}} dx = \int_0^\pi \frac{1+\tan^2(x/2)}{1+\tan^2(x/2)+2 \tan(x/2)} dx$.
Since $1+\tan^2(x/2) = \sec^2(x/2)$,we have:
$I = \int_0^\pi \frac{\sec^2(x/2)}{(1+\tan(x/2))^2} dx$.
Let $t = \tan(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) dx$,which implies $\sec^2(x/2) dx = 2 dt$.
As $x \to 0$,$t \to 0$. As $x \to \pi$,$t \to \infty$.
Thus,$I = \int_0^\infty \frac{2 dt}{(1+t)^2} = 2 \left[ -\frac{1}{1+t} \right]_0^\infty = 2(0 - (-1)) = 2$.

(A) Given vectors are $\overrightarrow{a}_1=2 \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{a}_2=3 \hat{i}-4 \hat{j}-4 \hat{k}$,$\overrightarrow{a}_3=\hat{i}+\hat{j}-\hat{k}$,and $\overrightarrow{a}_4=-\hat{i}+3 \hat{j}+\hat{k}$.
Calculating the magnitudes $m_1, m_2, m_3, m_4$:
$m_1 = |\overrightarrow{a}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$
$m_2 = |\overrightarrow{a}_2| = \sqrt{3^2 + (-4)^2 + (-4)^2} = \sqrt{9+16+16} = \sqrt{41}$
$m_3 = |\overrightarrow{a}_3| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1+1+1} = \sqrt{3}$
$m_4 = |\overrightarrow{a}_4| = \sqrt{(-1)^2 + 3^2 + 1^2} = \sqrt{1+9+1} = \sqrt{11}$
Comparing the values: $\sqrt{3} < \sqrt{6} < \sqrt{11} < \sqrt{41}$.
Therefore,$m_3 < m_1 < m_4 < m_2$.

(A) The angle $\theta$ between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$.
Since the angle $\theta > 90^{\circ}$,we have $\cos \theta < 0$,which implies $\overrightarrow{a} \cdot \overrightarrow{b} < 0$.
Calculating the dot product: $\overrightarrow{a} \cdot \overrightarrow{b} = (\lambda)(\lambda) + (-7)(1) + (3)(2\lambda) = \lambda^2 - 7 + 6\lambda$.
Setting the dot product to be less than zero: $\lambda^2 + 6\lambda - 7 < 0$.
Factoring the quadratic expression: $(\lambda + 7)(\lambda - 1) < 0$.
Solving the inequality,we find that $\lambda$ must lie between the roots $-7$ and $1$.
Therefore,$-7 < \lambda < 1$.

(A) The given equation of the sphere is $x^2+y^2+z^2-12x-4y-3z=0$.
Comparing this with the general equation of a sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$,we get $2u=-12$,$2v=-4$,and $2w=-3$.
Thus,$u=-6$,$v=-2$,and $w=-\frac{3}{2}$.
The centre of the sphere is $(-u, -v, -w) = (6, 2, \frac{3}{2})$.
The radius $r$ of the sphere is given by the formula $r = \sqrt{u^2+v^2+w^2-d}$.
Substituting the values,we get $r = \sqrt{(-6)^2+(-2)^2+(-\frac{3}{2})^2-0}$.
$r = \sqrt{36+4+\frac{9}{4}} = \sqrt{40+\frac{9}{4}} = \sqrt{\frac{160+9}{4}} = \sqrt{\frac{169}{4}}$.
Therefore,$r = \frac{13}{2}$.

(A) Given the expression: $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}$.
Using the triangle law of vector addition,$\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}$.
So,the expression becomes $\overrightarrow{AC}+2\overrightarrow{AD}-2\overrightarrow{DC}$.
Since $\overrightarrow{AD} = \overrightarrow{AC}+\overrightarrow{CD}$,we have:
$\overrightarrow{AC}+2(\overrightarrow{AC}+\overrightarrow{CD})-2\overrightarrow{DC} = \overrightarrow{AC}+2\overrightarrow{AC}+2\overrightarrow{CD}+2\overrightarrow{CD} = 3\overrightarrow{AC}+4\overrightarrow{CD} = 3\overrightarrow{AC}-4\overrightarrow{DC}$.
Given $Q$ is the midpoint of $AC$,$\overrightarrow{AC} = 2\overrightarrow{QC}$.
Given $P$ divides $DC$ in ratio $1:2$,$\overrightarrow{DP} = \frac{1}{3}\overrightarrow{DC}$ and $\overrightarrow{PC} = \frac{2}{3}\overrightarrow{DC}$,so $\overrightarrow{DC} = \frac{3}{2}\overrightarrow{PC}$.
Substituting these: $3(2\overrightarrow{QC}) - 4(\frac{3}{2}\overrightarrow{PC}) = 6\overrightarrow{QC} - 6\overrightarrow{PC} = 6(\overrightarrow{QC}+\overrightarrow{CP}) = 6\overrightarrow{QP}$.
Since $\overrightarrow{QP} = -\overrightarrow{PQ}$,we have $6\overrightarrow{QP} = -6\overrightarrow{PQ}$.
Comparing with $k\overrightarrow{PQ}$,we get $k = -6$.

(B) Given equations for direction ratios $(l, m, n)$ are $l+m+n=0$ and $l^2=m^2+n^2$.
From $l+m+n=0$,we have $l=-(m+n)$.
Substituting this into $l^2=m^2+n^2$,we get $(-(m+n))^2 = m^2+n^2$.
$m^2+n^2+2mn = m^2+n^2$,which implies $2mn=0$,so $mn=0$.
This gives two cases:
Case $I$: $m=0$. Then $l=-n$. Let the direction ratios be $(k, 0, -k)$. The unit vector is $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
Case $II$: $n=0$. Then $l=-m$. Let the direction ratios be $(k, -k, 0)$. The unit vector is $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
Let $\vec{a} = (\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$ and $\vec{b} = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0)$.
The cosine of the angle $\theta$ between them is given by $\cos \theta = |\vec{a} \cdot \vec{b}|$.
$\cos \theta = |(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (0)(-\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(0)| = |\frac{1}{2} + 0 + 0| = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(C) Given equations for direction cosines $(l, m, n)$ are:
$l+m+n=0$ --- $(i)$
$l^2+m^2-n^2=0$ --- $(ii)$
From $(i)$,$n = -(l+m)$.
Substituting this into $(ii)$:
$l^2 + m^2 - (-(l+m))^2 = 0$
$l^2 + m^2 - (l^2 + m^2 + 2lm) = 0$
$-2lm = 0 \Rightarrow lm = 0$.
This implies either $l=0$ or $m=0$.
Case $1$: If $l=0$,then from $(i)$,$m+n=0 \Rightarrow m=-n$. Let $m=1$,then $n=-1$. The direction ratios are $(0, 1, -1)$.
Case $2$: If $m=0$,then from $(i)$,$l+n=0 \Rightarrow l=-n$. Let $l=1$,then $n=-1$. The direction ratios are $(1, 0, -1)$.
Let the two vectors be $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$ and $\vec{b} = 1\hat{i} + 0\hat{j} - 1\hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (-1)(-1) = 1$.
$|\vec{a}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(D) Given the equations for direction cosines $l, m, n$:
$l+m+n=0 \implies n = -(l+m)$
Substitute $n$ into $l^2-5m^2+n^2=0$:
$l^2-5m^2+(-l-m)^2=0$
$l^2-5m^2+l^2+2lm+m^2=0$
$2l^2+2lm-4m^2=0$
$l^2+lm-2m^2=0$
$(l+2m)(l-m)=0$
Case $1$: $l=m$. Then $n = -(l+m) = -2l$. Direction ratios are $(l, l, -2l)$,i.e.,$(1, 1, -2)$.
Normalized direction cosines $(l_1, m_1, n_1) = (\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}})$.
Case $2$: $l=-2m$. Then $n = -(-2m+m) = m$. Direction ratios are $(-2m, m, m)$,i.e.,$(-2, 1, 1)$.
Normalized direction cosines $(l_2, m_2, n_2) = (-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}})$.
The cosine of the angle $\theta$ is given by $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.
$\cos \theta = |(\frac{1}{\sqrt{6}})(-\frac{2}{\sqrt{6}}) + (\frac{1}{\sqrt{6}})(\frac{1}{\sqrt{6}}) + (-\frac{2}{\sqrt{6}})(\frac{1}{\sqrt{6}})|$
$\cos \theta = |-\frac{2}{6} + \frac{1}{6} - \frac{2}{6}| = |-\frac{3}{6}| = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(B) First,we calculate the vectors $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$.
$2 \overrightarrow{a}-\overrightarrow{c} = 2(-\hat{i}+\hat{j}+2 \hat{k}) - (-2 \hat{i}+\hat{j}+3 \hat{k}) = (-2\hat{i}+2\hat{j}+4\hat{k}) + (2\hat{i}-\hat{j}-3\hat{k}) = \hat{j}+\hat{k}$.
$\overrightarrow{a}+\overrightarrow{b} = (-\hat{i}+\hat{j}+2 \hat{k}) + (2 \hat{i}-\hat{j}-\hat{k}) = \hat{i}+\hat{k}$.
Let $\theta$ be the angle between these two vectors. The cosine of the angle is given by $\cos \theta = \frac{(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k})}{|\hat{j}+\hat{k}| |\hat{i}+\hat{k}|}$.
Calculating the dot product: $(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k}) = (0)(1) + (1)(0) + (1)(1) = 1$.
Calculating the magnitudes: $|\hat{j}+\hat{k}| = \sqrt{0^2+1^2+1^2} = \sqrt{2}$ and $|\hat{i}+\hat{k}| = \sqrt{1^2+0^2+1^2} = \sqrt{2}$.
Thus,$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

(C) Given equations are $l+m+n=0$ and $l^2+m^2-n^2=0$.
From the first equation,$l = -(m+n)$.
Substituting this into the second equation: $(-m-n)^2 + m^2 - n^2 = 0$.
$m^2 + 2mn + n^2 + m^2 - n^2 = 0$.
$2m^2 + 2mn = 0 \Rightarrow 2m(m+n) = 0$.
This gives two cases:
Case $1$: $m=0$. Then $l = -n$. The direction ratios are $(-1, 0, 1)$.
Case $2$: $m = -n$. Then $l = 0$. The direction ratios are $(0, -1, 1)$.
Let the two lines be represented by vectors $\vec{a} = -\hat{i} + \hat{k}$ and $\vec{b} = -\hat{j} + \hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (-1)(0) + (0)(-1) + (1)(1) = 1$.
$|\vec{a}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}$.
$|\vec{b}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
$\cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is given by:
$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$
Here,the point is $(3, 2, 1)$ and the plane equation is $2x - y + 3z - 7 = 0$.
Substituting the values:
$\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(2(3) - 1(2) + 3(1) - 7)}{2^2 + (-1)^2 + 3^2}$
$\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(6 - 2 + 3 - 7)}{4 + 1 + 9}$
$\frac{x - 3}{2} = \frac{y - 2}{-1} = \frac{z - 1}{3} = \frac{-2(0)}{14} = 0$
Setting each part to $0$:
$x - 3 = 0 \Rightarrow x = 3$
$y - 2 = 0 \Rightarrow y = 2$
$z - 1 = 0 \Rightarrow z = 1$
Thus,the image of the point is $(3, 2, 1)$,which means the point lies on the plane.