TS EAMCET 2002 Mathematics Question Paper with Answer and Solution

(A) Given the partial fraction decomposition: $\frac{1-x+6x^2}{x(1-x)(1+x)} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}$.
Multiplying both sides by $x(1-x)(1+x)$,we get:
$1-x+6x^2 = A(1-x^2) + Bx(1+x) + Cx(1-x)$.
To find $A$,we set $x = 0$:
$1 - 0 + 6(0)^2 = A(1 - 0^2) + B(0)(1+0) + C(0)(1-0)$.
$1 = A(1)$.
Therefore,$A = 1$.

(B) First,we calculate $l$:
$l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} \right) = \lim_{\theta \rightarrow 0} \left( 3 \frac{\sin \theta}{\theta} - 4 \sin \theta \cdot \frac{\sin \theta}{\theta} \right)$
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have $l = 3(1) - 4(0)(1) = 3$.
Next,we calculate $m$:
$m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{\theta}$
Using the limit $\lim_{x \rightarrow 0} \frac{\tan(kx)}{x} = k$,we get $m = 2$.
The quadratic equation with roots $l=3$ and $m=2$ is given by $x^2 - (l+m)x + lm = 0$.
Substituting the values,we get $x^2 - (3+2)x + (3 \times 2) = 0$,which simplifies to $x^2 - 5x + 6 = 0$.

(B) Given that $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+a x^2+b x+c=0$.
By Vieta's formulas,the sum of the roots taken two at a time is $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{b}{1} = b$,and the product of the roots is $\alpha \beta \gamma = -\frac{c}{1} = -c$.
We need to find the value of $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$.
$\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta \gamma + \alpha \gamma + \alpha \beta}{\alpha \beta \gamma}$.
Substituting the values,we get $\frac{b}{-c} = -\frac{b}{c}$.

(B) The given equation is $x^4-8x^3+x^2-x+3=0$.
To remove the second term of an equation of the form $a_0x^n + a_1x^{n-1} + \dots = 0$,we diminish the roots by $h = -\frac{a_1}{n \cdot a_0}$.
Here,$a_0 = 1$,$a_1 = -8$,and $n = 4$.
Thus,$h = -\frac{-8}{4 \times 1} = \frac{8}{4} = 2$.
Therefore,the roots should be diminished by $2$.

(B) Let $f(x) = x^5 - 6x^2 - 4x + 5$.
By Descartes' Rule of Signs,the number of positive real roots is at most the number of sign changes in the coefficients of $f(x)$.
The coefficients are $(1, 0, 0, -6, -4, 5)$.
The sign changes are: $(1$ to $-6)$ and $(-4$ to $5)$.
Thus,there are $2$ sign changes,so there are at most $2$ positive real roots.
Now,consider $f(-x) = (-x)^5 - 6(-x)^2 - 4(-x) + 5 = -x^5 - 6x^2 + 4x + 5$.
The coefficients are $(-1, 0, 0, -6, 4, 5)$.
The sign change is: $(-6$ to $4)$.
Thus,there is $1$ sign change,so there is at most $1$ negative real root.
Therefore,the maximum possible number of real roots is $2 + 1 = 3$.

(C) Given the cubic equation $x^3+ax^2+bx+c=0$ with roots $\alpha, \beta, \gamma$.
By Vieta's formulas:
$\alpha+\beta+\gamma = -a$
$\alpha\beta+\beta\gamma+\gamma\alpha = b$
$\alpha\beta\gamma = -c$
We need to find $\alpha^{-1}+\beta^{-1}+\gamma^{-1} = \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}$.
Taking the common denominator:
$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}$.
Substituting the values from Vieta's formulas:
$\frac{b}{-c} = -\frac{b}{c}$.

(B) Given the cubic equation $2x^3 + 0x^2 - 2x - 1 = 0$.
Let the roots be $\alpha, \beta, \gamma$.
According to Vieta's formulas for a cubic equation $ax^3 + bx^2 + cx + d = 0$,the sum of the product of roots taken two at a time is given by $\Sigma \alpha \beta = \frac{c}{a}$.
Here,$a = 2$,$b = 0$,$c = -2$,and $d = -1$.
Therefore,$\Sigma \alpha \beta = \frac{-2}{2} = -1$.
We need to find $(\Sigma \alpha \beta)^2$.
$(\Sigma \alpha \beta)^2 = (-1)^2 = 1$.

(D) Since the coefficients of the quadratic equation $x^2+ax+b=0$ are real,the complex roots must occur in conjugate pairs.
Given that $1-i$ is a root,its conjugate $1+i$ must also be a root of the equation.
For a quadratic equation $x^2+ax+b=0$,the product of the roots is equal to the constant term $b$.
Therefore,$b = (1-i)(1+i)$.
Using the identity $(x-y)(x+y) = x^2-y^2$:
$b = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2$.
Thus,$b = 2$.

(A) Given $x_n = \cos \left(\frac{\pi}{4^n}\right) + i \sin \left(\frac{\pi}{4^n}\right) = e^{i \frac{\pi}{4^n}}$.
The product $P = x_1 x_2 x_3 \ldots \infty$ is given by:
$P = e^{i \frac{\pi}{4^1}} \cdot e^{i \frac{\pi}{4^2}} \cdot e^{i \frac{\pi}{4^3}} \ldots = e^{i \pi \left(\frac{1}{4} + \frac{1}{4^2} + \frac{1}{4^3} + \ldots \right)}$.
The exponent is a geometric series with first term $a = \frac{1}{4}$ and common ratio $r = \frac{1}{4}$.
The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{1/4}{1 - 1/4} = \frac{1/4}{3/4} = \frac{1}{3}$.
Therefore,$P = e^{i \pi \left(\frac{1}{3}\right)} = e^{i \frac{\pi}{3}}$.
$P = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1 + i \sqrt{3}}{2}$.

(D) Given $z = 3+5i$,then the conjugate is $\bar{z} = 3-5i$.
First,calculate $z^3$:
$z^2 = (3+5i)^2 = 9 + 25i^2 + 30i = 9 - 25 + 30i = -16 + 30i$.
$z^3 = z^2 \cdot z = (-16 + 30i)(3 + 5i) = -48 - 80i + 90i + 150i^2$.
Since $i^2 = -1$,$z^3 = -48 + 10i - 150 = -198 + 10i$.
Now,substitute these into the expression:
$z^3 + \bar{z} + 198 = (-198 + 10i) + (3 - 5i) + 198$.
$= (-198 + 198) + (10i - 5i) + 3 = 0 + 5i + 3 = 3 + 5i$.

(A) We have,$\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$.
Substituting $z=x+iy$:
$\left|x+i\left(y+\frac{1}{2}\right)\right|^2 = \left|x+i\left(y-\frac{1}{2}\right)\right|^2$.
Using the property $|a+ib|^2 = a^2+b^2$:
$x^2 + \left(y+\frac{1}{2}\right)^2 = x^2 + \left(y-\frac{1}{2}\right)^2$.
$x^2 + y^2 + y + \frac{1}{4} = x^2 + y^2 - y + \frac{1}{4}$.
Subtracting $x^2 + y^2 + \frac{1}{4}$ from both sides:
$y = -y$ $\Rightarrow 2y = 0$ $\Rightarrow y = 0$.
The equation $y=0$ represents the $x$-axis.

(B) Using the Pascal's identity,${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$,we have ${}^nC_5 + {}^nC_6 = {}^{n+1}C_6$.
Given inequality: ${}^{n+1}C_6 > {}^{n+1}C_5$.
Expanding the combinations: $\frac{(n+1)!}{6!(n-5)!} > \frac{(n+1)!}{5!(n-4)!}$.
Dividing both sides by $(n+1)!$ and simplifying: $\frac{1}{6!(n-5)!} > \frac{1}{5!(n-4)!}$.
Since $6! = 6 \times 5!$ and $(n-4)! = (n-4) \times (n-5)!$,we get: $\frac{1}{6 \times 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$.
Canceling common terms: $\frac{1}{6} > \frac{1}{n-4}$.
This implies $n-4 > 6$,so $n > 10$.
The least natural number $n$ satisfying this is $n = 11$.

(A) Let $T_n$ be the first term of the $n^{th}$ set. The first terms are $1, 2, 4, 7, 11, \ldots$
This is a sequence where the differences are $1, 2, 3, 4, \ldots$
The $n^{th}$ term is given by $T_n = T_1 + \sum_{k=1}^{n-1} k = 1 + \frac{(n-1)n}{2}$.
For the $50^{th}$ set,$n=50$,so $T_{50} = 1 + \frac{49 \times 50}{2} = 1 + 1225 = 1226$.
The $50^{th}$ set contains $50$ consecutive integers starting from $1226$.
The sum of $n$ terms in an arithmetic progression is $S_n = \frac{n}{2}[2a + (n-1)d]$.
Here $n=50$,$a=1226$,and $d=1$,so $S_{50} = \frac{50}{2}[2(1226) + 49(1)] = 25[2452 + 49] = 25[2501] = 62525$.

(B) In $\triangle ABC$,let $a, b, c$ be the lengths of the sides and $p_1, p_2, p_3$ be the corresponding altitudes.
The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
This implies $p_1 = \frac{2\Delta}{a}$,$p_2 = \frac{2\Delta}{b}$,and $p_3 = \frac{2\Delta}{c}$.
Given that $p_1, p_2, p_3$ are in $AP$,we have $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $AP$.
Dividing by $2\Delta$,we get $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $AP$.
By the definition of harmonic progression,if the reciprocals of terms are in $AP$,then the terms themselves are in $HP$.
Therefore,$a, b, c$ are in $HP$.

(C) Let $E = \cos^2 76^{\circ} + \cos^2 16^{\circ} - \cos 76^{\circ} \cos 16^{\circ}$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have:
$E = \frac{1 + \cos 152^{\circ}}{2} + \frac{1 + \cos 32^{\circ}}{2} - \cos 76^{\circ} \cos 16^{\circ}$
$E = 1 + \frac{1}{2} (\cos 152^{\circ} + \cos 32^{\circ}) - \cos 76^{\circ} \cos 16^{\circ}$
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$\cos 152^{\circ} + \cos 32^{\circ} = 2 \cos 92^{\circ} \cos 60^{\circ} = 2 \cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$\cos 76^{\circ} \cos 16^{\circ} = \frac{1}{2} (\cos 92^{\circ} + \cos 60^{\circ}) = \frac{1}{2} \cos 92^{\circ} + \frac{1}{4}$
Substituting these back:
$E = 1 + \frac{1}{2} \cos 92^{\circ} - (\frac{1}{2} \cos 92^{\circ} + \frac{1}{4})$
$E = 1 - \frac{1}{4} = \frac{3}{4}$

(D) Given that,$x = \tan \theta + \sin \theta$ and $y = \tan \theta - \sin \theta$.
Adding and subtracting the equations,we get:
$\tan \theta = \frac{x + y}{2}$ and $\sin \theta = \frac{x - y}{2}$.
Now,consider the product $xy$:
$xy = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta$.
Using $\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$,we have:
$xy = \sin^2 \theta \left(\frac{1}{\cos^2 \theta} - 1\right) = \sin^2 \theta \left(\frac{1 - \cos^2 \theta}{\cos^2 \theta}\right) = \sin^2 \theta \cdot \tan^2 \theta$.
Also,$x^2 - y^2 = (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2 = 4 \tan \theta \sin \theta$.
Squaring both sides:
$(x^2 - y^2)^2 = 16 \tan^2 \theta \sin^2 \theta$.
Since $xy = \tan^2 \theta \sin^2 \theta$,we get:
$(x^2 - y^2)^2 = 16xy$.

(D) We use the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$.
Here,$A = \frac{\pi}{8} + \frac{x}{2}$ and $B = \frac{\pi}{8} - \frac{x}{2}$.
Then $A + B = \frac{\pi}{8} + \frac{x}{2} + \frac{\pi}{8} - \frac{x}{2} = \frac{\pi}{4}$.
And $A - B = \frac{\pi}{8} + \frac{x}{2} - \left(\frac{\pi}{8} - \frac{x}{2}\right) = x$.
So,$f(x) = \sin\left(\frac{\pi}{4}\right) \sin(x) = \frac{1}{\sqrt{2}} \sin(x)$.
The period of $\sin(x)$ is $2\pi$.
Therefore,the period of $f(x)$ is $2\pi$.

(A) Given that $\cos (\alpha+\beta) = \frac{4}{5}$. Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $0 < \alpha+\beta < \frac{\pi}{2}$,so $\tan (\alpha+\beta) = \frac{3}{4}$.
Given that $\sin (\alpha-\beta) = \frac{5}{13}$. Since $0 < \alpha, \beta < \frac{\pi}{4}$,we have $-\frac{\pi}{4} < \alpha-\beta < \frac{\pi}{4}$,so $\tan (\alpha-\beta) = \frac{5}{12}$.
Now,$\tan 2\alpha = \tan [(\alpha+\beta)+(\alpha-\beta)]$.
Using the formula $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan 2\alpha = \frac{\frac{3}{4} + \frac{5}{12}}{1 - (\frac{3}{4} \cdot \frac{5}{12})}$
$\tan 2\alpha = \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{48-15}{48}} = \frac{14}{12} \cdot \frac{48}{33} = \frac{14 \cdot 4}{33} = \frac{56}{33}$.

(B) We need to evaluate the sum $S = \sum_{k=1}^3 \cos^2 \left((2k-1) \frac{\pi}{12}\right)$.
Expanding the sum for $k=1, 2, 3$:
$S = \cos^2 \left(\frac{\pi}{12}\right) + \cos^2 \left(\frac{3\pi}{12}\right) + \cos^2 \left(\frac{5\pi}{12}\right)$.
Simplifying the angles:
$S = \cos^2 \left(\frac{\pi}{12}\right) + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \left(\frac{5\pi}{12}\right)$.
Since $\cos \left(\frac{5\pi}{12}\right) = \cos \left(\frac{\pi}{2} - \frac{\pi}{12}\right) = \sin \left(\frac{\pi}{12}\right)$,we have $\cos^2 \left(\frac{5\pi}{12}\right) = \sin^2 \left(\frac{\pi}{12}\right)$.
Substituting this into the sum:
$S = \cos^2 \left(\frac{\pi}{12}\right) + \sin^2 \left(\frac{\pi}{12}\right) + \cos^2 \left(\frac{\pi}{4}\right)$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:
$S = 1 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 + \frac{1}{2} = \frac{3}{2}$.

(C) Let the points be $D(4, 2)$ and $C(-2, 6)$. The line $L=0$ is the perpendicular bisector of the segment $CD$.
Slope of the line $CD = \frac{6-2}{-2-4} = \frac{4}{-6} = -\frac{2}{3}$.
Since the line $L$ is perpendicular to $CD$,its slope $m = -\frac{1}{(-2/3)} = \frac{3}{2}$.
The midpoint $O$ of $CD$ is $\left(\frac{4-2}{2}, \frac{2+6}{2}\right) = (1, 4)$.
The equation of the line $L$ passing through $(1, 4)$ with slope $\frac{3}{2}$ is:
$y - 4 = \frac{3}{2}(x - 1)$
$2y - 8 = 3x - 3$
$3x - 2y + 5 = 0$.

(B) The equation of the pair of angle bisectors of the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Since the coordinate axes are the bisectors,their equations are $x = 0$ and $y = 0$,which implies their combined equation is $xy = 0$.
Comparing $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ with $xy = 0$,we see that the coefficient of $x^2$ and $y^2$ in the bisector equation must be zero.
This implies $a - b = 0$ is not possible as per the given condition $a \neq b$,or rather,for the bisectors to be the axes,the $x^2$ and $y^2$ terms must vanish in the bisector equation.
Actually,the condition for the coordinate axes to be the bisectors is $a + b = 0$ and $h \neq 0$ is not required,but rather $h$ must be such that the equation reduces to $xy = 0$.
Given the standard form,if $a + b = 0$,the bisectors are $x^2 - y^2 = 0$,which are $y = \pm x$. If the axes are the bisectors,then $h$ must be $0$ and $a+b=0$ is not the condition; rather,the condition is $a+b=0$ is for perpendicular lines. For axes to be bisectors,$h=0$ is the condition.

(D) The given equation is $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos \theta - \sin \theta$,$H = \cos \theta$,and $B = \cos \theta + \sin \theta$.
The acute angle $\alpha$ between the pair of straight lines is given by $\tan \alpha = \left| \frac{2 \sqrt{H^2 - AB}}{A + B} \right|$.
Substituting the values:
$H^2 - AB = \cos^2 \theta - (\cos \theta - \sin \theta)(\cos \theta + \sin \theta) = \cos^2 \theta - (\cos^2 \theta - \sin^2 \theta) = \sin^2 \theta$.
$A + B = (\cos \theta - \sin \theta) + (\cos \theta + \sin \theta) = 2 \cos \theta$.
Thus,$\tan \alpha = \left| \frac{2 \sqrt{\sin^2 \theta}}{2 \cos \theta} \right| = \left| \frac{2 \sin \theta}{2 \cos \theta} \right| = |\tan \theta|$.
Since $2 \theta$ is acute,$\theta$ is also acute,so $\alpha = \theta$.

(A) The equation of the circle is $x^2+y^2=r^2$ with center $(0,0)$ and radius $r$.
The equation of the line is $mx-y+c=0$.
The line intersects the circle at two distinct points if the perpendicular distance from the center $(0,0)$ to the line is less than the radius $r$.
The perpendicular distance $d$ is given by:
$d = \left| \frac{m(0) - (0) + c}{\sqrt{m^2 + (-1)^2}} \right| = \frac{|c|}{\sqrt{m^2+1}}$.
For two distinct points,we must have $d < r$:
$\frac{|c|}{\sqrt{m^2+1}} < r$
$|c| < r \sqrt{m^2+1}$
$-r \sqrt{m^2+1} < c < r \sqrt{m^2+1}$.

(B) Given that the perimeter $2s = 16 \text{ cm}$,so the semi-perimeter $s = 8 \text{ cm}$.
Let the sides be $a, b, c$. Given $a = 6 \text{ cm}$ and area $\Delta = 12 \text{ cm}^2$.
Using Heron's formula: $\Delta^2 = s(s-a)(s-b)(s-c)$.
$144 = 8(8-6)(8-b)(8-c)$.
$144 = 8(2)(8-b)(8-c)$.
$144 = 16(8-b)(8-c) \Rightarrow 9 = (8-b)(8-c)$.
Since $a+b+c = 16$ and $a=6$,we have $b+c = 10$,so $c = 10-b$.
Substituting $c$ into the equation: $9 = (8-b)(8-(10-b))$.
$9 = (8-b)(b-2)$.
$9 = 8b - 16 - b^2 + 2b$.
$b^2 - 10b + 25 = 0$.
$(b-5)^2 = 0 \Rightarrow b = 5$.
Since $b=5$,then $c = 10-5 = 5$.
As two sides are equal $(b=c=5)$,the triangle is isosceles.

(A) Given the function $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.
We know that $\sin^2 x = 1 - \cos^2 x$ and $\cos^2 x = 1 - \sin^2 x$.
Substituting these into the numerator:
Numerator $= \cos^2 x + \sin^4 x = \cos^2 x + \sin^2 x (\sin^2 x) = \cos^2 x + \sin^2 x (1 - \cos^2 x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x$.
Similarly,for the denominator:
Denominator $= \sin^2 x + \cos^4 x = \sin^2 x + \cos^2 x (\cos^2 x) = \sin^2 x + \cos^2 x (1 - \sin^2 x) = \sin^2 x + \cos^2 x - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x$.
Thus,$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \sin^2 x \cos^2 x} = 1$ for all $x \in R$.
Therefore,$f(2002) = 1$.

(C) Given that $3$ is a root of the equation $x^2+kx-24=0$.
Substituting $x=3$ into the equation:
$(3)^2 + k(3) - 24 = 0$
$9 + 3k - 24 = 0$
$3k - 15 = 0$
$3k = 15 \Rightarrow k = 5$.
Now,we check the options by substituting $k=5$ and $x=3$:
For option $C$: $x^2 - kx + 6 = 0$
Substituting $x=3$ and $k=5$:
$(3)^2 - (5)(3) + 6 = 9 - 15 + 6 = 0$.
Since the equation is satisfied,$3$ is a root of $x^2-kx+6=0$.

(A) Let $\alpha$ be the common root of the two equations. Then:
$\alpha^2 + a\alpha + b = 0$ $(1)$
$\alpha^2 + b\alpha + a = 0$ $(2)$
Subtracting equation $(2)$ from $(1)$:
$(\alpha^2 - \alpha^2) + (a\alpha - b\alpha) + (b - a) = 0$
$(a - b)\alpha - (a - b) = 0$
$(a - b)(\alpha - 1) = 0$
Since $a \neq b$,we must have $\alpha - 1 = 0$,which implies $\alpha = 1$.
Substituting $\alpha = 1$ into equation $(1)$:
$(1)^2 + a(1) + b = 0$
$1 + a + b = 0$
$a + b = -1$

(B) First,we calculate $l$:
$l = \lim_{\theta \rightarrow 0} \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} = \lim_{\theta \rightarrow 0} \left( 3 \frac{\sin \theta}{\theta} - 4 \sin \theta \cdot \frac{\sin \theta}{\theta} \right) = 3(1) - 4(0)(1) = 3$.
Next,we calculate $m$:
$m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} = \lim_{\theta \rightarrow 0} \left( \frac{\tan \theta}{\theta} \cdot \frac{2}{1 - \tan^2 \theta} \right) = 1 \cdot \frac{2}{1 - 0} = 2$.
The quadratic equation with roots $l=3$ and $m=2$ is given by $x^2 - (l+m)x + lm = 0$.
Substituting the values,we get $x^2 - (3+2)x + (3 \times 2) = 0$,which simplifies to $x^2 - 5x + 6 = 0$.

(D) Given equation is $x^4-x^2+x-1=0$.
Let $\alpha = \frac{1+\sqrt{3}i}{2}$. Since the coefficients are real,the conjugate $\beta = \frac{1-\sqrt{3}i}{2}$ must also be a root.
Sum of roots $\alpha + \beta = \frac{1+\sqrt{3}i + 1-\sqrt{3}i}{2} = 1$.
Product of roots $\alpha \beta = \frac{1^2 - (\sqrt{3}i)^2}{4} = \frac{1+3}{4} = 1$.
The quadratic factor corresponding to these roots is $x^2 - (\alpha+\beta)x + \alpha\beta = x^2 - x + 1 = 0$.
Dividing $x^4-x^2+x-1$ by $x^2-x+1$,we get $x^4-x^2+x-1 = (x^2-x+1)(x^2+x-1) = 0$.
The real roots are obtained from $x^2+x-1=0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $x = \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$.

(A) Let $z = \frac{3+2i \sin \theta}{1-2i \sin \theta}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1+2i \sin \theta)$:
$z = \frac{(3+2i \sin \theta)(1+2i \sin \theta)}{(1-2i \sin \theta)(1+2i \sin \theta)}$
$z = \frac{3 + 6i \sin \theta + 2i \sin \theta + 4i^2 \sin^2 \theta}{1^2 + (2 \sin \theta)^2}$
Since $i^2 = -1$,we have:
$z = \frac{3 - 4 \sin^2 \theta + 8i \sin \theta}{1 + 4 \sin^2 \theta}$
$z = \frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta} + i \left( \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} \right)$
For $z$ to be a real number,the imaginary part must be zero:
$\frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0$
This implies $\sin \theta = 0$.
Given $0 < \theta < 2\pi$,the only solution is $\theta = \pi$.

(D) Since the coefficients $a$ and $b$ are real,the complex roots of the quadratic equation $x^2+ax+b=0$ must occur in conjugate pairs.
Given that $1-i$ is a root,its conjugate $1+i$ must also be a root of the equation.
For a quadratic equation $x^2+ax+b=0$,the product of the roots is given by the constant term $b$.
Therefore,$b = (1-i)(1+i)$.
Using the identity $(x-y)(x+y) = x^2-y^2$:
$b = 1^2 - i^2$.
Since $i^2 = -1$,we have:
$b = 1 - (-1) = 1 + 1 = 2$.

(A) We know that the definition of the hyperbolic sine function is $\sinh(z) = \frac{e^z - e^{-z}}{2}$.
Substituting $z = ix$,we get:
$\sinh(ix) = \frac{e^{ix} - e^{-ix}}{2}$.
Using Euler's formula,$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$,which implies $\frac{e^{ix} - e^{-ix}}{2} = i \sin x$.
Therefore,$\sinh(ix) = i \sin x$.

(C) The available odd digits are $\{1, 3, 5, 7, 9\}$.
Total number of $5$-digit numbers formed using these $5$ distinct digits is $5! = 120$.
$A$ number is divisible by $5$ if its last digit is $0$ or $5$. Since we only have odd digits,the number is divisible by $5$ only if the last digit is $5$.
If the last digit is fixed as $5$,the remaining $4$ positions can be filled by the remaining $4$ digits $\{1, 3, 7, 9\}$ in $4!$ ways.
Number of $5$-digit numbers divisible by $5 = 4! = 24$.
Therefore,the number of $5$-digit numbers not divisible by $5 = 5! - 4! = 120 - 24 = 96$.

(A) The number of circular permutations of $n$ distinct objects is $(n-1)!$.
For a necklace,the clockwise and anti-clockwise arrangements are considered identical,so the number of ways is $\frac{(n-1)!}{2}$.
Here,$n = 8$.
Number of ways = $\frac{(8-1)!}{2} = \frac{7!}{2} = \frac{5040}{2} = 2520$.

(C) The general term in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is given by:
$T_{r+1} = { }^5 C_r (x^2)^{5-r} (\frac{k}{x})^r$
$T_{r+1} = { }^5 C_r x^{10-2r} \cdot k^r \cdot x^{-r} = { }^5 C_r k^r x^{10-3r}$
For the coefficient of $x$,we set the exponent of $x$ to $1$:
$10 - 3r = 1$
$3r = 9 \Rightarrow r = 3$
Substituting $r = 3$ into the expression for the coefficient:
Coefficient $= { }^5 C_3 k^3 = 10 k^3$
Given that the coefficient is $270$:
$10 k^3 = 270$
$k^3 = 27$
$k = 3$

(B) The $p^{th}$ term in the expansion of $(1+x)^n$ is $T_p = { }^n C_{p-1} x^{p-1}$,so its coefficient is $p = { }^n C_{p-1}$.
The $(p+1)^{th}$ term is $T_{p+1} = { }^n C_p x^p$,so its coefficient is $q = { }^n C_p$.
We know the property of binomial coefficients: $\frac{{ }^n C_r}{{ }^n C_{r-1}} = \frac{n-r+1}{r}$.
Taking the ratio $\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
This implies $p \cdot q = p \cdot (n-p+1)$ is not the correct approach; rather,we use the given values directly.
Actually,the problem states the coefficients are $p$ and $q$. Let's re-evaluate: $p = { }^n C_{p-1}$ and $q = { }^n C_p$.
Using the identity ${ }^n C_{p-1} + { }^n C_p = { }^{n+1} C_p$,this does not directly yield $n+1$ unless specific conditions are met.
However,based on the standard problem type: $\frac{q}{p} = \frac{{ }^n C_p}{{ }^n C_{p-1}} = \frac{n-p+1}{p}$.
Thus,$q = \frac{n-p+1}{p} \cdot p = n-p+1$.
Therefore,$p+q = p + (n-p+1) = n+1$.

(C) To find the sum of the coefficients in the expansion of a polynomial $P(x)$,we substitute $x = 1$ into the expression.
Given the expression $(1+x+x^2)^n$.
Substituting $x = 1$:
$(1 + 1 + 1^2)^n = (1 + 1 + 1)^n = 3^n$.
Thus,the sum of the coefficients is $3^n$.

(C) Let $E = \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}$.
Using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$,we have:
$E = \frac{1+\cos 152^{\circ}}{2} + \frac{1+\cos 32^{\circ}}{2} - \cos 76^{\circ} \cos 16^{\circ}$
$E = 1 + \frac{1}{2}(\cos 152^{\circ} + \cos 32^{\circ}) - \cos 76^{\circ} \cos 16^{\circ}$
Using $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$:
$\cos 152^{\circ} + \cos 32^{\circ} = 2\cos 92^{\circ}\cos 60^{\circ} = 2\cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$
Also,$\cos 92^{\circ} = \cos(180^{\circ} - 88^{\circ}) = -\cos 88^{\circ} = -\sin 2^{\circ}$ (not helpful here).
Alternatively,use $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\frac{1}{2}(\cos 92^{\circ} + \cos 60^{\circ})$
$E = \frac{1+\cos 152^{\circ}}{2} + \frac{1+\cos 32^{\circ}}{2} - \frac{1}{2}\cos 92^{\circ} - \frac{1}{4}$
$E = 1 + \frac{1}{2}(\cos 152^{\circ} + \cos 32^{\circ} - \cos 92^{\circ}) - \frac{1}{4}$
Since $\cos 152^{\circ} + \cos 32^{\circ} = \cos 92^{\circ}$,the expression simplifies to:
$E = 1 + \frac{1}{2}(\cos 92^{\circ} - \cos 92^{\circ}) - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4}$.

(C) We have $\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right) = \cos ^2 \frac{\pi}{12} + \cos ^2 \frac{3 \pi}{12} + \cos ^2 \frac{5 \pi}{12}$.
Since $\frac{3 \pi}{12} = \frac{\pi}{4}$,we have $\cos ^2 \frac{\pi}{4} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
Thus,the expression becomes $\cos ^2 \frac{\pi}{12} + \frac{1}{2} + \cos ^2 \frac{5 \pi}{12}$.
Using the identity $\cos(\frac{\pi}{2} - \theta) = \sin \theta$,we have $\cos \frac{5 \pi}{12} = \cos(\frac{\pi}{2} - \frac{\pi}{12}) = \sin \frac{\pi}{12}$.
Substituting this,we get $\frac{1}{2} + \cos ^2 \frac{\pi}{12} + \sin ^2 \frac{\pi}{12}$.
Since $\cos ^2 \theta + \sin ^2 \theta = 1$,the expression simplifies to $\frac{1}{2} + 1 = \frac{3}{2}$.

(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.
Since the circle passes through $(0,0)$,we have $0 + 0 + 0 + 0 + c = 0$,so $c = 0$.
Since it passes through $(2,0)$,we have $4 + 0 + 4g + 0 + 0 = 0$,which gives $g = -1$.
Since it passes through $(0,-2)$,we have $0 + 4 + 0 - 4f + 0 = 0$,which gives $f = 1$.
The equation of the circle is $x^2 + y^2 - 2x + 2y = 0$.
For the point $(k,-2)$ to lie on this circle,we substitute the coordinates into the equation:
$k^2 + (-2)^2 - 2(k) + 2(-2) = 0$
$k^2 + 4 - 2k - 4 = 0$
$k^2 - 2k = 0$
$k(k - 2) = 0$
Since the points must be distinct,$k$ cannot be $0$ (as $(0,-2)$ is already a point).
Therefore,$k = 2$.

(A) Let the old coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The transformation equations for rotation of axes by an angle $\theta$ are:
$x = x' \cos \theta - y' \sin \theta$
$y = x' \sin \theta + y' \cos \theta$
Given $\theta = 45^{\circ}$,$x' = \sqrt{2}$,and $y' = 4$.
Substituting these values:
$x = \sqrt{2} \cos 45^{\circ} - 4 \sin 45^{\circ} = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) - 4 \left( \frac{1}{\sqrt{2}} \right) = 1 - \frac{4}{\sqrt{2}} = 1 - 2\sqrt{2}$
$y = \sqrt{2} \sin 45^{\circ} + 4 \cos 45^{\circ} = \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) + 4 \left( \frac{1}{\sqrt{2}} \right) = 1 + \frac{4}{\sqrt{2}} = 1 + 2\sqrt{2}$
Thus,the coordinates in the old system are $(1 - 2\sqrt{2}, 1 + 2\sqrt{2})$.

(B) The given line is $2x - 3y + 7 = 0$.
Any line perpendicular to the given line is of the form $3x + 2y + k = 0$.
To find the intercepts,set $y = 0$: $3x + k = 0 \Rightarrow x = -\frac{k}{3}$.
Set $x = 0$: $2y + k = 0 \Rightarrow y = -\frac{k}{2}$.
The area of the triangle formed with the coordinate axes is given by $\frac{1}{2} |x_{intercept} \cdot y_{intercept}| = 3$.
$\frac{1}{2} |(-\frac{k}{3}) \cdot (-\frac{k}{2})| = 3$.
$\frac{1}{2} |\frac{k^2}{6}| = 3$.
$|k^2| = 36 \Rightarrow k = \pm 6$.
Substituting $k$ into the equation $3x + 2y + k = 0$,we get $3x + 2y = \pm 6$.

(B) The given equation of the pair of straight lines is $xy-x-y+1=0$.
Factoring the expression: $x(y-1)-1(y-1)=0$,which gives $(x-1)(y-1)=0$.
This represents two lines: $x=1$ and $y=1$.
The point of intersection of these two lines is $(1, 1)$.
Since the lines are concurrent with the line $ax+2y-3a=0$,the point $(1, 1)$ must satisfy the equation of the line $ax+2y-3a=0$.
Substituting $x=1$ and $y=1$ into the equation: $a(1)+2(1)-3a=0$.
$a+2-3a=0$.
$-2a+2=0$.
$2a=2$.
$a=1$.

(D) Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. Since it is at a constant distance $c$ from the origin $(0,0)$,we have $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = c$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$.
Points $A$ and $B$ are $(a, 0)$ and $(0, b)$ respectively.
The circle passing through $O(0,0)$,$A(a,0)$,and $B(0,b)$ has the equation $x^2 + y^2 - ax - by = 0$.
The centre of this circle is $(h, k) = (\frac{a}{2}, \frac{b}{2})$.
Thus,$a = 2h$ and $b = 2k$.
Substituting these into the line equation condition: $\frac{1}{(2h)^2} + \frac{1}{(2k)^2} = \frac{1}{c^2}$.
$\frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{c^2} \Rightarrow \frac{h^2 + k^2}{4h^2k^2} = \frac{1}{c^2}$.
This approach is complex; alternatively,since the triangle $OAB$ is a right-angled triangle at $O$,the hypotenuse $AB$ is the diameter of the circle.
The centre is the midpoint of $AB$,which is $(\frac{a}{2}, \frac{b}{2})$.
The distance from origin to line $AB$ is $c = \frac{|ab|}{\sqrt{a^2+b^2}}$.
Let the centre be $(x, y) = (\frac{a}{2}, \frac{b}{2})$,so $a=2x, b=2y$.
$c = \frac{|(2x)(2y)|}{\sqrt{(2x)^2+(2y)^2}} = \frac{4|xy|}{2\sqrt{x^2+y^2}} = \frac{2|xy|}{\sqrt{x^2+y^2}}$.
Squaring both sides: $c^2 = \frac{4x^2y^2}{x^2+y^2}$.
Actually,the locus of the midpoint of the hypotenuse of a right triangle with fixed distance $c$ from the vertex is $x^{-2} + y^{-2} = c^{-2}$ if the axes are the legs. However,given the options,let's re-evaluate: The distance from origin to line $x/a + y/b = 1$ is $c = \frac{1}{\sqrt{1/a^2 + 1/b^2}}$. The centre is $(a/2, b/2)$.
If the question implies the circle passes through $O, A, B$,the diameter is $AB$. The distance from origin to $AB$ is $c$. The locus of the midpoint of $AB$ where $AB$ is a line at distance $c$ from origin is $x^{-2} + y^{-2} = c^{-2}$.

(B) Let the radius of the circle be $r$. Since the circle lies in the first quadrant and touches both coordinate axes,its center must be $(r, r)$.
The equation of the circle is $(x - r)^2 + (y - r)^2 = r^2$.
The circle touches the line $4x + 3y - 12 = 0$. The perpendicular distance from the center $(r, r)$ to the line must be equal to the radius $r$:
$\frac{|4r + 3r - 12|}{\sqrt{4^2 + 3^2}} = r$
$\frac{|7r - 12|}{5} = r$
This gives two cases:
$1) 7r - 12 = 5r$ $\Rightarrow 2r = 12$ $\Rightarrow r = 6$
$2) 7r - 12 = -5r$ $\Rightarrow 12r = 12$ $\Rightarrow r = 1$
Since we are looking for the radius of the larger circle,the radius is $6$.

(B) Since the circle touches both coordinate axes in the third quadrant,its center must be at a distance of $5$ units from both axes in the negative direction.
Therefore,the center of the circle is $(-5, -5)$.
The radius of the circle is given as $r = 5$.
The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting $h = -5$,$k = -5$,and $r = 5$,we get:
$(x - (-5))^2 + (y - (-5))^2 = 5^2$
$(x+5)^2 + (y+5)^2 = 25$.

(A) The equation of the circle is $x^2+y^2=r^2$ with center $(0,0)$ and radius $r$.
The equation of the line is $mx-y+c=0$.
The line intersects the circle at two distinct points if the perpendicular distance from the center $(0,0)$ to the line is less than the radius $r$.
The perpendicular distance $d$ is given by:
$d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{m^2+1}}$.
Setting $d < r$,we get:
$\frac{|c|}{\sqrt{m^2+1}} < r$
$|c| < r \sqrt{m^2+1}$
This inequality implies:
$-r \sqrt{m^2+1} < c < r \sqrt{m^2+1}$.

(C) Given that the focus is $S(3,0)$,let $P(x, y)$ be any point on the parabola.
By the definition of a parabola,the distance from $P$ to the focus $S$ is equal to the perpendicular distance from $P$ to the directrix.
$SP^2 = PM^2$
$(x-3)^2 + (y-0)^2 = (x+3)^2$
$y^2 = (x+3)^2 - (x-3)^2$
Using the identity $a^2 - b^2 = (a+b)(a-b)$:
$y^2 = (x+3+x-3)(x+3-x+3)$
$y^2 = (2x)(6)$
$y^2 = 12x$

(C) Let the parabola be $y^2 = 4ax$. Let $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ be two points on the parabola such that $PQ$ is a focal chord passing through the focus $S(a, 0)$.
The equation of the chord $PQ$ is $y(t_1 + t_2) = 2x + 2at_1t_2$.
Since it passes through $(a, 0)$,we have $0 = 2a + 2at_1t_2$,which implies $t_1t_2 = -1$.
Let $(x_1, y_1)$ be the pole of the chord $PQ$ with respect to the parabola $y^2 = 4ax$.
The equation of the polar of $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$,which can be rewritten as $yy_1 - 2ax = 2ax_1$.
Comparing this with the equation of the chord $y(t_1 + t_2) - 2x = 2at_1t_2$,we get:
$\frac{y_1}{t_1 + t_2} = \frac{-2a}{-2} = \frac{2ax_1}{2at_1t_2}$
From $\frac{y_1}{t_1 + t_2} = a$,we have $y_1 = a(t_1 + t_2)$.
From $a = \frac{x_1}{t_1t_2}$,we have $x_1 = at_1t_2$.
Since $t_1t_2 = -1$,we get $x_1 = a(-1) = -a$.
Thus,the locus of the pole $(x_1, y_1)$ is $x = -a$,which is the directrix of the parabola.

(A) The equation of the line is $x+4y-4=0$. Comparing this with $lx+my+n=0$,we get $l=1, m=4, n=-4$.
The equation of the ellipse is $x^2+4y^2=4$,which can be written as $\frac{x^2}{4}+\frac{y^2}{1}=1$.
Here,$a^2=4$ and $b^2=1$.
The formula for the pole $(x_1, y_1)$ of the line $lx+my+n=0$ with respect to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by:
$x_1 = -\frac{a^2l}{n}$ and $y_1 = -\frac{b^2m}{n}$.
Substituting the values:
$x_1 = -\frac{4 \times 1}{-4} = 1$
$y_1 = -\frac{1 \times 4}{-4} = 1$
Thus,the pole is $(1,1)$.

(A) Given that $A$ is a non-singular matrix,its determinant $|A| \neq 0$,which implies that the inverse matrix $A^{-1}$ exists.
Given the equation $AB = O$,where $O$ is the zero matrix.
Multiplying both sides by $A^{-1}$ on the left:
$A^{-1}(AB) = A^{-1}O$
$(A^{-1}A)B = O$
$IB = O$
$B = O$
Therefore,$B$ must be a null matrix.

(C) Given functions are $f(x) = 3x - 4$ and $g(x) = 3x + 2$.
To find $f^{-1}(y)$,let $f(x) = y$.
$3x - 4 = y \implies 3x = y + 4 \implies x = \frac{y + 4}{3}$.
Thus,$f^{-1}(y) = \frac{y + 4}{3}$.
Now,$f^{-1}(5) = \frac{5 + 4}{3} = \frac{9}{3} = 3$.
To find $g^{-1}(z)$,let $g(x) = z$.
$3x + 2 = z \implies 3x = z - 2 \implies x = \frac{z - 2}{3}$.
Thus,$g^{-1}(z) = \frac{z - 2}{3}$.
Finally,$g^{-1}(f^{-1}(5)) = g^{-1}(3) = \frac{3 - 2}{3} = \frac{1}{3}$.

(B) Given curves are $x=y^2$ $(i)$ and $xy=a^3$ (ii).
For curve $(i)$,differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.
For curve (ii),differentiating with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.
To find the point of intersection,substitute $x=y^2$ into $xy=a^3$: $y^2 \cdot y = a^3 \Rightarrow y^3 = a^3 \Rightarrow y = a$.
Then $x = a^2$. So the point of intersection is $(a^2, a)$.
Let $m_1$ and $m_2$ be the slopes of the tangents at $(a^2, a)$.
$m_1 = \left(\frac{1}{2y}\right)_{(a^2, a)} = \frac{1}{2a}$.
$m_2 = \left(-\frac{y}{x}\right)_{(a^2, a)} = -\frac{a}{a^2} = -\frac{1}{a}$.
Since the curves cut orthogonally,$m_1 m_2 = -1$.
$\left(\frac{1}{2a}\right) \left(-\frac{1}{a}\right) = -1$.
$-\frac{1}{2a^2} = -1$.
$2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$.

(A) Let $f(x) = \log(1+x) - \frac{2x}{2+x}$.
For the function to be defined,we must have $1+x > 0$,which implies $x > -1$.
Now,find the derivative $f'(x)$:
$f'(x) = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2}$
$f'(x) = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^2} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$
For the function to be increasing,$f'(x) > 0$:
$\frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} > 0$
$\frac{4+x^2+4x - 4 - 4x}{(1+x)(2+x)^2} > 0$
$\frac{x^2}{(1+x)(2+x)^2} > 0$
Since $x^2 \ge 0$ and $(2+x)^2 > 0$ for $x \neq -2$,the inequality holds when $1+x > 0$ and $x \neq 0$.
Thus,$x > -1$ and $x \neq 0$.

(A) Given the function $f(x) = x e^{-x}$.
To find the critical points,we calculate the first derivative $f'(x)$ using the product rule:
$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.
Setting $f'(x) = 0$ for maxima or minima:
$e^{-x}(1 - x) = 0$.
Since $e^{-x} \neq 0$ for any $x \in R$,we have $1 - x = 0$,which gives $x = 1$.
Now,we find the second derivative $f''(x)$ to check for maxima:
$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - [e^{-x} - x e^{-x}] = e^{-x}(x - 2)$.
Evaluating at $x = 1$:
$f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0$.
Since the second derivative is negative at $x = 1$,the function attains a maximum value at $x = 1$.

(C) We have,$I = \int \frac{dx}{1-\cos x-\sin x}$.
Using the half-angle formulas $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and $\sin x = \frac{2\tan(x/2)}{1+\tan^2(x/2)}$,we get:
$I = \int \frac{dx}{1 - \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} - \frac{2\tan(x/2)}{1+\tan^2(x/2)}}$
$I = \int \frac{(1+\tan^2(x/2)) dx}{1+\tan^2(x/2) - 1 + \tan^2(x/2) - 2\tan(x/2)}$
$I = \int \frac{\sec^2(x/2) dx}{2\tan^2(x/2) - 2\tan(x/2)} = \frac{1}{2} \int \frac{\sec^2(x/2) dx}{\tan(x/2)(\tan(x/2)-1)}$.
Let $z = \tan(x/2)$,then $dz = \frac{1}{2}\sec^2(x/2) dx$,so $\sec^2(x/2) dx = 2dz$.
$I = \int \frac{2dz}{2z(z-1)} = \int \frac{dz}{z(z-1)}$.
Using partial fractions: $\frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$.
$I = \int \left(\frac{1}{z-1} - \frac{1}{z}\right) dz = \log|z-1| - \log|z| + C = \log\left|\frac{z-1}{z}\right| + C$.
Substituting $z = \tan(x/2)$ back:
$I = \log\left|\frac{\tan(x/2)-1}{\tan(x/2)}\right| + C = \log|1 - \cot(x/2)| + C$.

(B) Let $I = \int \frac{dx}{7+5 \cos x}$.
Using the identity $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$ and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$,we get:
$I = \int \frac{dx}{7(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) + 5(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})}$
$I = \int \frac{dx}{12 \cos^2 \frac{x}{2} + 2 \sin^2 \frac{x}{2}}$
Divide numerator and denominator by $\cos^2 \frac{x}{2}$:
$I = \int \frac{\sec^2 \frac{x}{2} dx}{12 + 2 \tan^2 \frac{x}{2}} = \frac{1}{2} \int \frac{\sec^2 \frac{x}{2} dx}{6 + \tan^2 \frac{x}{2}}$
Let $\tan \frac{x}{2} = z$,then $\frac{1}{2} \sec^2 \frac{x}{2} dx = dz$.
$I = \int \frac{dz}{6 + z^2} = \int \frac{dz}{(\sqrt{6})^2 + z^2}$
Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + c$:
$I = \frac{1}{\sqrt{6}} \tan^{-1} \frac{z}{\sqrt{6}} + c = \frac{1}{\sqrt{6}} \tan^{-1} \left(\frac{1}{\sqrt{6}} \tan \frac{x}{2}\right) + c$.

(A) Let $I = \int \frac{3^x \, dx}{\sqrt{9^x-1}} = \int \frac{3^x \, dx}{\sqrt{(3^x)^2-1}}$.
Substitute $3^x = z$. Then,differentiating both sides with respect to $x$,we get $3^x \log 3 \, dx = dz$,which implies $3^x \, dx = \frac{dz}{\log 3}$.
Substituting these into the integral,we get $I = \frac{1}{\log 3} \int \frac{dz}{\sqrt{z^2-1}}$.
Using the standard integration formula $\int \frac{dt}{\sqrt{t^2-a^2}} = \log |t + \sqrt{t^2-a^2}| + c$,we obtain:
$I = \frac{1}{\log 3} \log |z + \sqrt{z^2-1}| + c$.
Substituting $z = 3^x$ back into the expression,we get:
$I = \frac{1}{\log 3} \log |3^x + \sqrt{9^x-1}| + c$.

(B) Since the integrand $f(x) = \sin^4 x \cos^6 x$ is an even function,we have:
$\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx = 2 \int_0^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$
Using Wallis' formula $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$:
$I = 2 \times \frac{\Gamma(\frac{4+1}{2}) \Gamma(\frac{6+1}{2})}{2 \Gamma(\frac{4+6+2}{2})} = \frac{\Gamma(\frac{5}{2}) \Gamma(\frac{7}{2})}{\Gamma(6)}$
Using $\Gamma(n+1) = n\Gamma(n)$ and $\Gamma(\frac{1}{2}) = \sqrt{\pi}$:
$\Gamma(\frac{5}{2}) = \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{3}{4} \sqrt{\pi}$
$\Gamma(\frac{7}{2}) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}$
$\Gamma(6) = 5! = 120$
$I = \frac{(\frac{3}{4} \sqrt{\pi}) (\frac{15}{8} \sqrt{\pi})}{120} = \frac{45 \pi}{32 \times 120} = \frac{45 \pi}{3840} = \frac{3 \pi}{256}$

(B) We have,
$\int_2^3 \frac{d x}{x^2-x} = \int_2^3 \frac{1}{x(x-1)} d x$
Using partial fractions,$\frac{1}{x(x-1)} = \frac{1}{x-1} - \frac{1}{x}$
So,$\int_2^3 \left( \frac{1}{x-1} - \frac{1}{x} \right) d x = [\log |x-1| - \log |x|]_2^3$
$= [\log |\frac{x-1}{x}|]_2^3$
$= \log |\frac{3-1}{3}| - \log |\frac{2-1}{2}|$
$= \log \frac{2}{3} - \log \frac{1}{2}$
$= \log \left( \frac{2/3}{1/2} \right) = \log \frac{4}{3}$

(A) Given the integral $\int_{1}^{9} x^2 dx$ with $n = 4$ intervals.
The width of each interval is $h = \frac{9-1}{4} = 2$.
The values of $x$ are $x_0=1, x_1=3, x_2=5, x_3=7, x_4=9$.
The corresponding values of $y = f(x) = x^2$ are:
$y_0 = f(1) = 1^2 = 1$
$y_1 = f(3) = 3^2 = 9$
$y_2 = f(5) = 5^2 = 25$
$y_3 = f(7) = 7^2 = 49$
$y_4 = f(9) = 9^2 = 81$
Using the trapezoidal rule formula:
$\int_{a}^{b} f(x) dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$
Substituting the values:
$\int_{1}^{9} x^2 dx \approx \frac{2}{2} [1 + 2(9 + 25 + 49) + 81]$
$\int_{1}^{9} x^2 dx \approx 1 [1 + 2(83) + 81]$
$\int_{1}^{9} x^2 dx \approx 1 + 166 + 81 = 248$.

(A) The equation of the family of all concentric circles centered at $(h, k)$ is given by $(x-h)^2 + (y-k)^2 = r^2$.
Here,$(h, k)$ are fixed constants (the center),and $r$ is the radius,which is the only arbitrary parameter.
Since there is only one arbitrary parameter $(r)$,the order of the differential equation is equal to the number of arbitrary parameters.
Therefore,the order of the differential equation is $1$.

(B) Given differential equation is $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$.
By simplifying the right side,we get $\frac{dy}{dx} = (\frac{y}{x})^{1/3} = \frac{y^{1/3}}{x^{1/3}}$.
Separating the variables,we have $y^{-1/3} dy = x^{-1/3} dx$.
Integrating both sides,we get $\int y^{-1/3} dy = \int x^{-1/3} dx$.
Applying the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$,we get $\frac{y^{2/3}}{2/3} = \frac{x^{2/3}}{2/3} + C'$.
Multiplying by $\frac{2}{3}$,we get $y^{2/3} = x^{2/3} + \frac{2}{3}C'$.
Let $c = \frac{2}{3}C'$,then $y^{2/3} - x^{2/3} = c$.

(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{3}$ and $Q = 1$.
The integrating factor $(IF)$ is given by $e^{\int P dx} = e^{\int \frac{1}{3} dx} = e^{x/3}$.
Multiplying both sides of the equation by the $IF$,we get:
$e^{x/3} \frac{dy}{dx} + \frac{1}{3} e^{x/3} y = e^{x/3}$
This can be written as:
$\frac{d}{dx} (y \cdot e^{x/3}) = e^{x/3}$
Integrating both sides with respect to $x$:
$y \cdot e^{x/3} = \int e^{x/3} dx + c$
$y \cdot e^{x/3} = 3e^{x/3} + c$
Dividing by $e^{x/3}$:
$y = 3 + ce^{-x/3}$

(A) We have,$\frac{dy}{dx} - y = x^2$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -1$ and $Q = x^2$.
The integrating factor is $IF = e^{\int P dx} = e^{\int -1 dx} = e^{-x}$.
The general solution is given by $y \cdot IF = \int Q \cdot IF dx + c$.
Substituting the values,we get $y e^{-x} = \int x^2 e^{-x} dx + c$.
Using integration by parts,$\int x^2 e^{-x} dx = -x^2 e^{-x} + \int 2x e^{-x} dx = -x^2 e^{-x} - 2x e^{-x} - 2e^{-x} + c$.
Thus,$y e^{-x} = -e^{-x}(x^2 + 2x + 2) + c$.
Multiplying both sides by $e^x$,we get $y = -(x^2 + 2x + 2) + ce^x$.
Rearranging the terms,we obtain $y + x^2 + 2x + 2 = ce^x$.

(A) Given that the position vectors are $A = \hat{i} + x\hat{j} + 3\hat{k}$,$B = 3\hat{i} + 4\hat{j} + 7\hat{k}$,and $C = y\hat{i} - 2\hat{j} - 5\hat{k}$.
Since the points are collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel,meaning $\vec{AB} = t\vec{BC}$ for some scalar $t$.
First,calculate the vectors:
$\vec{AB} = (3-1)\hat{i} + (4-x)\hat{j} + (7-3)\hat{k} = 2\hat{i} + (4-x)\hat{j} + 4\hat{k}$.
$\vec{BC} = (y-3)\hat{i} + (-2-4)\hat{j} + (-5-7)\hat{k} = (y-3)\hat{i} - 6\hat{j} - 12\hat{k}$.
Equating $\vec{AB} = t\vec{BC}$:
$2\hat{i} + (4-x)\hat{j} + 4\hat{k} = t(y-3)\hat{i} - 6t\hat{j} - 12t\hat{k}$.
Comparing the coefficients of $\hat{k}$:
$4 = -12t \Rightarrow t = -\frac{1}{3}$.
Comparing the coefficients of $\hat{j}$:
$4 - x = -6t = -6(-\frac{1}{3}) = 2 \Rightarrow x = 2$.
Comparing the coefficients of $\hat{i}$:
$t(y - 3) = 2 \Rightarrow -\frac{1}{3}(y - 3) = 2 \Rightarrow y - 3 = -6 \Rightarrow y = -3$.
Thus,$(x, y) = (2, -3)$.

(D) Let the vertices of the triangle be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.
The side vectors are:
$\vec{AB} = (1-2)\hat{i} + (-3 - (-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} - 2\hat{j} - 6\hat{k}$
$\vec{BC} = (3-1)\hat{i} + (-4 - (-3))\hat{j} + (-4 - (-5))\hat{k} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{CA} = (2-3)\hat{i} + (-1 - (-4))\hat{j} + (1 - (-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$
The lengths of the sides are:
$c = |\vec{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$
$a = |\vec{BC}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$
$b = |\vec{CA}| = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$
Checking for the right-angled triangle condition $(a^2 + b^2 = c^2)$:
$a^2 + b^2 = 6 + 35 = 41$
$c^2 = 41$
Since $a^2 + b^2 = c^2$,the triangle is a right-angled triangle.

(A) Let the vector $a = x\hat{i} + y\hat{j} + z\hat{k}$.
Given $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$.
From $a \cdot \hat{i} = a \cdot (\hat{i} + \hat{j})$,we get $a \cdot \hat{i} = a \cdot \hat{i} + a \cdot \hat{j}$,which implies $a \cdot \hat{j} = 0$. Thus,$y = 0$.
From $a \cdot (\hat{i} + \hat{j}) = a \cdot (\hat{i} + \hat{j} + \hat{k})$,we get $a \cdot \hat{i} + a \cdot \hat{j} = a \cdot \hat{i} + a \cdot \hat{j} + a \cdot \hat{k}$,which implies $a \cdot \hat{k} = 0$. Thus,$z = 0$.
Since $a \cdot \hat{i} = x$,and the problem implies $a$ is a unit vector or specific vector satisfying these conditions,if we assume $a = \hat{i}$,then $a \cdot \hat{i} = 1$,$a \cdot (\hat{i} + \hat{j}) = 1$,and $a \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$.
Therefore,$a = \hat{i}$ satisfies the given conditions.

(B) The orthogonal projection of a vector $\vec{a}$ on a non-zero vector $\vec{b}$ is defined as the component of $\vec{a}$ along the direction of $\vec{b}$.
The formula for the projection of $\vec{a}$ on $\vec{b}$ is given by:
$\text{Proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \right) \hat{b}$
Since the unit vector $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$,we substitute this into the formula:
$\text{Proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \right) \left( \frac{\vec{b}}{|\vec{b}|} \right)$
Simplifying the expression,we get:
$\text{Proj}_{\vec{b}} \vec{a} = \frac{(\vec{a} \cdot \vec{b}) \vec{b}}{|\vec{b}|^2}$

(B) We are given the expression $(a+b) \cdot ((b+c) \times (a+b+c))$.
First,expand the cross product term: $(b+c) \times (a+b+c) = (b \times a) + (b \times b) + (b \times c) + (c \times a) + (c \times b) + (c \times c)$.
Since the cross product of a vector with itself is zero ($b \times b = 0$ and $c \times c = 0$) and $c \times b = -(b \times c)$,we have:
$(b+c) \times (a+b+c) = (b \times a) + (b \times c) + (c \times a) - (b \times c) = (b \times a) + (c \times a)$.
Now,take the dot product with $(a+b)$:
$(a+b) \cdot ((b \times a) + (c \times a)) = a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times a) + b \cdot (c \times a)$.
Using the properties of the scalar triple product $[x y z] = x \cdot (y \times z)$:
$a \cdot (b \times a) = [a b a] = 0$ (since two vectors are same).
$a \cdot (c \times a) = [a c a] = 0$.
$b \cdot (b \times a) = [b b a] = 0$.
$b \cdot (c \times a) = [b c a]$.
Since $[b c a] = [a b c]$,the final result is $[a b c]$.

(B) The vector representing the line segment $PQ$ is $\vec{PQ} = (0-0, 0-1, 1-0) = (0, -1, 1)$.
The normal vector to the plane $x+y+z=3$ is $\vec{n} = (1, 1, 1)$.
The length of the projection of a vector $\vec{v}$ onto a plane with normal $\vec{n}$ is given by the formula $L = |\vec{v}| \sin(\theta)$,where $\theta$ is the angle between the vector $\vec{v}$ and the normal $\vec{n}$.
First,calculate the magnitude of $\vec{PQ}$:
$|\vec{PQ}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.
Next,calculate the cosine of the angle $\theta$ between $\vec{PQ}$ and $\vec{n}$:
$\cos(\theta) = \frac{|\vec{PQ} \cdot \vec{n}|}{|\vec{PQ}| |\vec{n}|} = \frac{|(0)(1) + (-1)(1) + (1)(1)|}{\sqrt{2} \cdot \sqrt{1^2+1^2+1^2}} = \frac{|0 - 1 + 1|}{\sqrt{2} \cdot \sqrt{3}} = 0$.
Since $\cos(\theta) = 0$,the angle $\theta = 90^\circ$,which means $\sin(\theta) = 1$.
Therefore,the length of the projection is $L = |\vec{PQ}| \sin(90^\circ) = \sqrt{2} \cdot 1 = \sqrt{2}$.

(A) Let the points be $A(0,0,1)$,$B(0,1,2)$,and $C(1,0,3)$.
The vectors lying on the plane are $\vec{AB} = (0-0, 1-0, 2-1) = (0,1,1)$ and $\vec{AC} = (1-0, 0-0, 3-1) = (1,0,2)$.
The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{AB} \times \vec{AC}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{vmatrix}$
$\vec{n} = \hat{i}(1 \times 2 - 1 \times 0) - \hat{j}(0 \times 2 - 1 \times 1) + \hat{k}(0 \times 0 - 1 \times 1)$
$\vec{n} = \hat{i}(2) - \hat{j}(-1) + \hat{k}(-1)$
$\vec{n} = 2\hat{i} + \hat{j} - \hat{k}$
The direction ratios of the normal are $(2,1,-1)$.

(A) The given equation of the plane is $by + cz + d = 0$.
Since the variable $x$ is missing from the equation,the normal vector to the plane is $\vec{n} = 0\hat{i} + b\hat{j} + c\hat{k}$.
The normal vector lies in the $YOZ$-plane.
$A$ plane is perpendicular to another plane if their normal vectors are perpendicular.
The normal to the $YOZ$-plane is the $x$-axis,which is $\hat{i}$.
Since the dot product of the normal vector $\vec{n} = b\hat{j} + c\hat{k}$ and the $x$-axis vector $\hat{i}$ is $0$,the plane $by + cz + d = 0$ is perpendicular to the $YOZ$-plane.

(C) First,we find the prime factorization of $2001$.
$2001 = 3 \times 667 = 3 \times 23 \times 29$.
Given that $a, b, c$ are factors of $2001$ such that $a < b < c$,we have $a = 3$,$b = 23$,and $c = 29$.
The direction ratios of the normal to the plane are given as $b, c, a$,which are $23, 29, 3$.
The equation of a plane with normal vector $(b, c, a)$ passing through $(x_0, y_0, z_0)$ is $b(x - x_0) + c(y - y_0) + a(z - z_0) = 0$.
Substituting the values,we get $23(x - 1) + 29(y - 1) + 3(z - 1) = 0$.
Expanding this,we get $23x - 23 + 29y - 29 + 3z - 3 = 0$.
$23x + 29y + 3z - 55 = 0$.
Therefore,the equation of the plane is $23x + 29y + 3z = 55$.

(A) The given equation of the plane is $7x + 11y + 13z = 3003$.
Dividing the entire equation by $3003$,we get the intercept form of the plane:
$\frac{7x}{3003} + \frac{11y}{3003} + \frac{13z}{3003} = 1$
$\frac{x}{429} + \frac{y}{273} + \frac{z}{231} = 1$
The plane meets the coordinate axes at points $A, B,$ and $C$. These points are the intercepts on the $x, y,$ and $z$ axes respectively:
$A = (429, 0, 0)$
$B = (0, 273, 0)$
$C = (0, 0, 231)$
The centroid of $\triangle ABC$ with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),$ and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Centroid $= \left(\frac{429+0+0}{3}, \frac{0+273+0}{3}, \frac{0+0+231}{3}\right)$
Centroid $= (143, 91, 77)$.

(D) Given that the probability of success $p = \frac{1}{4}$.
Then,the probability of failure $q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$.
The standard deviation $(SD)$ of a binomial distribution is given by $\sqrt{npq} = 3$.
Squaring both sides,we get $npq = 9$.
Substituting the values of $p$ and $q$:
$n \times \frac{1}{4} \times \frac{3}{4} = 9$
$n \times \frac{3}{16} = 9$
$n = 9 \times \frac{16}{3} = 48$.
The mean of a binomial distribution is given by $np$.
Mean $= 48 \times \frac{1}{4} = 12$.

(A) To find the determinant of matrix $A$,we expand along the first row:
$|A| = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{vmatrix}$
$|A| = 1 \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} - 0 \begin{vmatrix} 2 & 0 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix}$
$|A| = 1(1 \times 1 - 0 \times 2) - 0 + 1(2 \times 2 - 1 \times 3)$
$|A| = 1(1 - 0) + 1(4 - 3)$
$|A| = 1(1) + 1(1) = 1 + 1 = 2$
Thus,the value of $\det(A)$ is $2$.

(A) Given the system of equations:
$x - cy - bz = 0$
$-cx + y - az = 0$
$-bx - ay + z = 0$
Since $x^2+y^2+z^2 \neq 0$,the system has a non-trivial solution. Therefore,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 & -c & -b \\ -c & 1 & -a \\ -b & -a & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(1 - a^2) - (-c)(-c - ab) + (-b)(ca + b) = 0$
$1 - a^2 - c^2 - abc - abc - b^2 = 0$
$1 - a^2 - b^2 - c^2 - 2abc = 0$
Rearranging the terms,we get:
$a^2 + b^2 + c^2 + 2abc = 1$

(B) We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.
Given equation: $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$.
Adding the two equations:
$(\sin ^{-1} x + \cos ^{-1} x) + (\sin ^{-1} x - \cos ^{-1} x) = \frac{\pi}{2} + \frac{\pi}{6}$
$2 \sin ^{-1} x = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$
$\sin ^{-1} x = \frac{\pi}{3}$
$x = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

(C) Given $f(x) = \cos^2 x + \sin^4 x$.
Using $\cos^2 x = 1 - \sin^2 x$,we get:
$f(x) = 1 - \sin^2 x + \sin^4 x = 1 - \sin^2 x(1 - \sin^2 x) = 1 - \sin^2 x \cos^2 x$.
Multiplying and dividing by $4$,we get:
$f(x) = 1 - \frac{4 \sin^2 x \cos^2 x}{4} = 1 - \frac{(2 \sin x \cos x)^2}{4} = 1 - \frac{\sin^2 2x}{4}$.
Since $0 \leq \sin^2 2x \leq 1$,we have:
$0 \leq \frac{\sin^2 2x}{4} \leq \frac{1}{4}$.
Multiplying by $-1$ and adding $1$:
$1 - 0 \geq 1 - \frac{\sin^2 2x}{4} \geq 1 - \frac{1}{4}$.
$1 \geq f(x) \geq \frac{3}{4}$.
Thus,the range $f(R) = \left[\frac{3}{4}, 1\right]$.

(A) Given the function $f(x) = \frac{|x|}{x}$ for $x \in A$,where $A = \{x \in R, x \neq 0, -4 \leq x \leq 4\}$.
If $x > 0$,then $|x| = x$,so $f(x) = \frac{x}{x} = 1$.
If $x < 0$,then $|x| = -x$,so $f(x) = \frac{-x}{x} = -1$.
Since $x$ cannot be $0$,the function $f(x)$ only takes the values $1$ and $-1$.
Therefore,the range of $f$ is $\{1, -1\}$.

(A) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit,right-hand limit,and the value of the function at $x=0$ must be equal.
$1$. The value of the function at $x=0$ is $f(0) = a^2 \cos^2(0) + b^2 \sin^2(0) = a^2(1) + b^2(0) = a^2$.
$2$. The right-hand limit $(RHL)$ as $x \rightarrow 0^+$ is $\lim_{x \rightarrow 0^+} e^{ax+b} = e^{a(0)+b} = e^b$.
$3$. The left-hand limit $(LHL)$ as $x \rightarrow 0^-$ is $\lim_{x \rightarrow 0^-} (a^2 \cos^2 x + b^2 \sin^2 x) = a^2(1) + b^2(0) = a^2$.
For continuity,$LHL = RHL = f(0)$,so $e^b = a^2$.
Taking the natural logarithm on both sides,we get $\ln(e^b) = \ln(a^2)$,which simplifies to $b = 2 \ln |a|$ or $b = 2 \log |a|$.

(C) The function $f(x) = x - [x]$ is the fractional part function,denoted as $\{x\}$.
We check the continuity of $f(x)$ at any integer $n \in Z$.
The left-hand limit is:
$\lim_{x \rightarrow n^-} f(x) = \lim_{h \rightarrow 0} f(n-h) = \lim_{h \rightarrow 0} ((n-h) - [n-h]) = \lim_{h \rightarrow 0} ((n-h) - (n-1)) = \lim_{h \rightarrow 0} (1-h) = 1$.
The right-hand limit is:
$\lim_{x \rightarrow n^+} f(x) = \lim_{h \rightarrow 0} f(n+h) = \lim_{h \rightarrow 0} ((n+h) - [n+h]) = \lim_{h \rightarrow 0} ((n+h) - n) = \lim_{h \rightarrow 0} h = 0$.
The value of the function at $x=n$ is:
$f(n) = n - [n] = n - n = 0$.
Since $\lim_{x \rightarrow n^-} f(x) \neq \lim_{x \rightarrow n^+} f(x)$,the function $f(x)$ is discontinuous at every integer $n \in Z$.
Therefore,the set of points of discontinuity is $Z$.

(A) Given function is $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$.
We can rewrite the function as $f(x) = \sqrt{a} \cdot x^{1/2} + a^2 \cdot \sqrt{a}^{-1} \cdot x^{-1/2} = \sqrt{a} \cdot x^{1/2} + a^{3/2} \cdot x^{-1/2}$.
Differentiating with respect to $x$ using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$:
$f^{\prime}(x) = \sqrt{a} \cdot \frac{1}{2} x^{-1/2} + a^{3/2} \cdot \left(-\frac{1}{2}\right) x^{-3/2}$.
$f^{\prime}(x) = \frac{\sqrt{a}}{2\sqrt{x}} - \frac{a^{3/2}}{2x\sqrt{x}}$.
Now,substitute $x = a$ into the derivative:
$f^{\prime}(a) = \frac{\sqrt{a}}{2\sqrt{a}} - \frac{a^{3/2}}{2a\sqrt{a}}$.
$f^{\prime}(a) = \frac{1}{2} - \frac{a^{3/2}}{2a^{3/2}}$.
$f^{\prime}(a) = \frac{1}{2} - \frac{1}{2} = 0$.

(B) Given $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$.
First,calculate $\frac{\partial z}{\partial x}$:
$\frac{\partial z}{\partial x} = \frac{y}{x} \left[ \cos \left( \frac{x}{y} \right) \cdot \frac{1}{y} - \sin \left( 1 + \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right) \right] - \frac{y}{x^2} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$
$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$
$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - z \quad \dots (i)$
Next,calculate $\frac{\partial z}{\partial y}$:
$\frac{\partial z}{\partial y} = \frac{1}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] + \frac{y}{x} \left[ \cos \left( \frac{x}{y} \right) \cdot \left( -\frac{x}{y^2} \right) - \sin \left( 1 + \frac{y}{x} \right) \cdot \frac{1}{x} \right]$
$y \frac{\partial z}{\partial y} = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right)$
$y \frac{\partial z}{\partial y} = z - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right) \quad \dots (ii)$
Comparing $(i)$ and $(ii)$,we observe that $x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$.

(B) Given that $f(x)=e^x$ and $g(x)=\sin^{-1} x$.
Then $h(x) = f(g(x)) = e^{\sin^{-1} x}$.
Taking the natural logarithm on both sides,we get $\ln(h(x)) = \sin^{-1} x$.
Differentiating both sides with respect to $x$,we get $\frac{1}{h(x)} \cdot h'(x) = \frac{d}{dx}(\sin^{-1} x)$.
Thus,$\frac{h'(x)}{h(x)} = \frac{1}{\sqrt{1-x^2}}$.

(C) Given the function: $y = ae^x + be^{-x} + c$
Differentiating with respect to $x$ once:
$y^{\prime} = \frac{d}{dx}(ae^x + be^{-x} + c) = ae^x - be^{-x}$
Differentiating again with respect to $x$:
$y^{\prime \prime} = \frac{d}{dx}(ae^x - be^{-x}) = ae^x + be^{-x}$
Differentiating a third time with respect to $x$:
$y^{\prime \prime \prime} = \frac{d}{dx}(ae^x + be^{-x}) = ae^x - be^{-x}$
Comparing this result with the first derivative,we see that $y^{\prime \prime \prime} = y^{\prime}$.

(B) Given $y=a \cos (\log x)+b \sin (\log x)$.
First,differentiate with respect to $x$:
$y^{\prime} = \frac{d}{dx} [a \cos (\log x)+b \sin (\log x)] = -a \sin (\log x) \cdot \frac{1}{x} + b \cos (\log x) \cdot \frac{1}{x} = \frac{-a \sin (\log x) + b \cos (\log x)}{x}$.
So,$x y^{\prime} = -a \sin (\log x) + b \cos (\log x)$.
Now,differentiate again with respect to $x$:
$\frac{d}{dx} (x y^{\prime}) = \frac{d}{dx} [-a \sin (\log x) + b \cos (\log x)]$.
Using the product rule on the left side: $x y^{\prime \prime} + y^{\prime} = -a \cos (\log x) \cdot \frac{1}{x} - b \sin (\log x) \cdot \frac{1}{x}$.
Multiply both sides by $x$:
$x^2 y^{\prime \prime} + x y^{\prime} = -[a \cos (\log x) + b \sin (\log x)]$.
Since $y = a \cos (\log x) + b \sin (\log x)$,we have:
$x^2 y^{\prime \prime} + x y^{\prime} = -y$.

(A) Given $z = \sec(y-ax) + \tan(y+ax)$.
First,find the partial derivatives with respect to $x$:
$\frac{\partial z}{\partial x} = \sec(y-ax)\tan(y-ax)(-a) + \sec^2(y+ax)(a) = -a\sec(y-ax)\tan(y-ax) + a\sec^2(y+ax)$.
$\frac{\partial^2 z}{\partial x^2} = -a[\sec(y-ax)\sec^2(y-ax)(-a) + \tan(y-ax)\sec(y-ax)\tan(y-ax)(-a)] + a[2\sec(y+ax)\sec(y+ax)\tan(y+ax)(a)]$
$= a^2\sec^3(y-ax) + a^2\sec(y-ax)\tan^2(y-ax) + 2a^2\sec^2(y+ax)\tan(y+ax)$.
Next,find the partial derivatives with respect to $y$:
$\frac{\partial z}{\partial y} = \sec(y-ax)\tan(y-ax) + \sec^2(y+ax)$.
$\frac{\partial^2 z}{\partial y^2} = \sec(y-ax)\sec^2(y-ax) + \tan(y-ax)\sec(y-ax)\tan(y-ax) + 2\sec(y+ax)\sec(y+ax)\tan(y+ax)$
$= \sec^3(y-ax) + \sec(y-ax)\tan^2(y-ax) + 2\sec^2(y+ax)\tan(y+ax)$.
Now,calculate $\frac{\partial^2 z}{\partial x^2} - a^2\frac{\partial^2 z}{\partial y^2}$:
$= [a^2\sec^3(y-ax) + a^2\sec(y-ax)\tan^2(y-ax) + 2a^2\sec^2(y+ax)\tan(y+ax)] - a^2[\sec^3(y-ax) + \sec(y-ax)\tan^2(y-ax) + 2\sec^2(y+ax)\tan(y+ax)]$
$= 0$.

(C) Let $y = f(x) = x^{3000}$.
We use the differential approximation formula: $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
Here,let $x = 1$ and $\Delta x = 0.0002$.
Then $f(x) = 1^{3000} = 1$.
The derivative is $f'(x) = 3000 x^{2999}$.
At $x = 1$,$f'(1) = 3000(1)^{2999} = 3000$.
Now,calculate the change $\Delta y \approx f'(x) \Delta x = 3000 \times 0.0002 = 0.6$.
Therefore,$f(1.0002) \approx f(1) + \Delta y = 1 + 0.6 = 1.6$.

(A) Given the integral $\int_2^{10} x^2 dx$ with $n = 4$ intervals.
Here,$a = 2$,$b = 10$,and the step size $h = \frac{b - a}{n} = \frac{10 - 2}{4} = 2$.
The points are $x_0 = 2, x_1 = 4, x_2 = 6, x_3 = 8, x_4 = 10$.
The corresponding values of $f(x) = x^2$ are:
$y_0 = f(2) = 4$
$y_1 = f(4) = 16$
$y_2 = f(6) = 36$
$y_3 = f(8) = 64$
$y_4 = f(10) = 100$
Using the trapezoidal rule:
$\int_a^b f(x) dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$
$\int_2^{10} x^2 dx \approx \frac{2}{2} [4 + 2(16 + 36 + 64) + 100]$
$= 1 \cdot [4 + 2(116) + 100] = 4 + 232 + 100 = 336$.

(A) The equation of the family of all concentric circles centered at $(h, k)$ is $(x-h)^2 + (y-k)^2 = r^2$.
Since $(h, k)$ are fixed constants,the only arbitrary constant (parameter) in this equation is $r$.
The order of a differential equation is equal to the number of arbitrary constants present in the general solution of the family of curves.
Since there is only $1$ arbitrary constant $(r)$,the order of the differential equation is $1$.

(B) The orthogonal projection of a vector $a$ onto a vector $b$ is defined as the component of $a$ along the direction of $b$.
The formula for the projection of $a$ on $b$ is given by:
$\text{Proj}_{b} a = \left( \frac{a \cdot b}{|b|^2} \right) b$
This represents a vector that is parallel to $b$ and has a magnitude equal to the scalar projection of $a$ on $b$.

(D) Given that,$[a, b, c] = 3$.
The volume of the parallelepiped with edges $\vec{u} = 2a+b$,$\vec{v} = 2b+c$,and $\vec{w} = 2c+a$ is given by the scalar triple product $[\vec{u}, \vec{v}, \vec{w}]$.
$[\vec{u}, \vec{v}, \vec{w}] = [2a+b, 2b+c, 2c+a]$.
Using the properties of the scalar triple product:
$[2a+b, 2b+c, 2c+a] = 2a \cdot ((2b+c) \times (2c+a)) + b \cdot ((2b+c) \times (2c+a))$.
Expanding the cross products:
$(2b+c) \times (2c+a) = 4(b \times c) + 2(b \times a) + 2(c \times c) + (c \times a) = 4(b \times c) + 2(b \times a) + (c \times a)$ (since $c \times c = 0$).
Now,$[2a+b, 2b+c, 2c+a] = 2a \cdot (4(b \times c) + 2(b \times a) + (c \times a)) + b \cdot (4(b \times c) + 2(b \times a) + (c \times a))$.
Distributing the dot products:
$= 8[a, b, c] + 4[a, b, a] + 2[a, c, a] + 4[b, b, c] + 2[b, b, a] + [b, c, a]$.
Since any scalar triple product with two identical vectors is $0$:
$[a, b, a] = 0, [a, c, a] = 0, [b, b, c] = 0, [b, b, a] = 0$.
Thus,the expression simplifies to:
$8[a, b, c] + [b, c, a]$.
Since $[b, c, a] = [a, b, c]$,we have:
$8[a, b, c] + [a, b, c] = 9[a, b, c]$.
Given $[a, b, c] = 3$,the volume is $9 \times 3 = 27$ cubic units.

(D) Given the equations for the direction ratios $(l, m, n)$:
$l+m-n=0$ $(i)$
$l^2+m^2-n^2=0$ (ii)
From equation $(i)$,we have $l = n-m$.
Substituting this into equation (ii):
$(n-m)^2 + m^2 - n^2 = 0$
$n^2 + m^2 - 2nm + m^2 - n^2 = 0$
$2m^2 - 2nm = 0$
$2m(m-n) = 0$
This gives two cases: $m=0$ or $m=n$.
Case $1$: If $m=0$,then $l=n$. The direction ratios are $(1, 0, 1)$.
Case $2$: If $m=n$,then $l=0$. The direction ratios are $(0, 1, 1)$.
Let the two lines have direction ratios $\vec{a} = (1, 0, 1)$ and $\vec{b} = (0, 1, 1)$.
The cosine of the angle $\theta$ between them is given by:
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(1)(0) + (0)(1) + (1)(1)|}{\sqrt{1^2+0^2+1^2} \sqrt{0^2+1^2+1^2}}$
$\cos \theta = \frac{1}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,the acute angle is $\theta = \frac{\pi}{3}$.

(D) The sum of probabilities in a probability distribution is always $1$.
Given $P(X=0) = 0.4$.
Since $P(X=0) + P(X=1) + P(X=2) = 1$,we have $0.4 + P(X=1) + P(X=2) = 1$.
$P(X=1) + P(X=2) = 0.6$.
Given $P(X=1) = P(X=2)$,let $P(X=1) = P(X=2) = p$.
Then $p + p = 0.6 \Rightarrow 2p = 0.6 \Rightarrow p = 0.3$.
So,$P(X=1) = 0.3$ and $P(X=2) = 0.3$.
The mean $E(X)$ is given by $\sum x_i P(x_i)$.
$E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))$.
$E(X) = (0 \times 0.4) + (1 \times 0.3) + (2 \times 0.3)$.
$E(X) = 0 + 0.3 + 0.6 = 0.9$.

(C) Given that the mean of the Poisson distribution is $\lambda = \frac{1}{2}$.
The probability mass function of a Poisson distribution is given by $P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}$.
We need to find the ratio $\frac{P(X=3)}{P(X=2)}$.
$P(X=3) = \frac{(\frac{1}{2})^3 e^{-1/2}}{3!}$ and $P(X=2) = \frac{(\frac{1}{2})^2 e^{-1/2}}{2!}$.
Taking the ratio:
$\frac{P(X=3)}{P(X=2)} = \frac{\frac{(\frac{1}{2})^3 e^{-1/2}}{3!}}{\frac{(\frac{1}{2})^2 e^{-1/2}}{2!}}$
$= \frac{(\frac{1}{2})^3}{3!} \times \frac{2!}{(\frac{1}{2})^2}$
$= \frac{1}{2} \times \frac{2!}{3!}$
$= \frac{1}{2} \times \frac{2}{6} = \frac{1}{6}$.
Thus,the ratio is $1:6$.