KCET 2010 Mathematics Question Paper with Answer and Solution

(D) Let the number of vertices be $n$.
Given that the degree of each vertex is $3$.
Then,the sum of the degrees of all vertices in the simple graph is $3n$.
According to the Handshaking Lemma,the sum of the degrees of all vertices is equal to twice the number of edges:
$\sum \text{deg}(v) = 2 \times |E|$
$3n = 2 \times 24$
$3n = 48$
$n = \frac{48}{3}$
$n = 16$
Therefore,the number of vertices is $16$.

(C) Given the cubic equation $x^{3}-5x^{2}-x+5=0$.
Let the roots be $a, -a, b$.
From the relationship between roots and coefficients,the sum of the roots is given by:
$a + (-a) + b = -(-5)/1 = 5$.
This simplifies to $b = 5$.
Now,we check which of the given quadratic equations is satisfied by $x = 5$:
For option $C$,$x^{2}-3x-10=0$:
Substituting $x = 5$,we get $(5)^{2}-3(5)-10 = 25-15-10 = 0$.
Since the equation is satisfied,$b=5$ is a root of $x^{2}-3x-10=0$.

(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$
$= \frac{(1+i)^{n}}{(1-i)^{n}} \times (1-i)^{2}$
$= \left(\frac{1+i}{1-i}\right)^{n} \times (1 + i^{2} - 2i)$
$= \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{n} \times (1 - 1 - 2i)$
$= \left(\frac{1 + i^{2} + 2i}{1 - i^{2}}\right)^{n} \times (-2i)$
$= \left(\frac{2i}{2}\right)^{n} \times (-2i)$
$= i^{n} \times (-2i) = -2i^{n+1}$
For the expression to be a positive real number,$-2i^{n+1}$ must be positive.
If $n=1$,$-2i^{1+1} = -2i^{2} = -2(-1) = 2$,which is positive.
Thus,the least positive integer $n$ is $1$.

(B) Given $x+iy=(-1+i\sqrt{3})^{2010}$.
We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ is the cube root of unity,so $-1+i\sqrt{3} = 2\omega$.
Substituting this into the expression:
$x+iy = (2\omega)^{2010} = 2^{2010} \cdot \omega^{2010}$.
Since $\omega^3 = 1$,we have $\omega^{2010} = (\omega^3)^{670} = 1^{670} = 1$.
Therefore,$x+iy = 2^{2010} \cdot 1 = 2^{2010} + i(0)$.
Comparing the real parts,we get $x = 2^{2010}$.

(A) Let the $n$th term of the series be $t_n$. The series is $1, 3, 7, 13, 21, \ldots$.
Taking the differences between consecutive terms: $3-1=2, 7-3=4, 13-7=6, 21-13=8, \ldots$.
This is an arithmetic progression of differences: $2, 4, 6, 8, \ldots$.
The $n$th term $t_n$ can be expressed as $t_n = t_1 + \sum_{k=1}^{n-1} (2k)$.
$t_n = 1 + 2 \times \frac{(n-1)n}{2} = 1 + n(n-1) = n^2 - n + 1$.
Given $t_n = 9901$,we have $n^2 - n + 1 = 9901$.
$n^2 - n - 9900 = 0$.
Factoring the quadratic equation: $n^2 - 100n + 99n - 9900 = 0$.
$n(n-100) + 99(n-100) = 0$.
$(n-100)(n+99) = 0$.
Since $n$ must be a positive integer,$n = 100$.

(D) Given,the binomial expansion is $(1+x)^{15}$.
The coefficient of $x^{r}$ is $^{15}C_{r}$ and the coefficient of $x^{r+3}$ is $^{15}C_{r+3}$.
According to the problem,the coefficients are equal:
$^{15}C_{r} = ^{15}C_{r+3}$
Using the property of binomial coefficients,if $^{n}C_{x} = ^{n}C_{y}$,then either $x = y$ or $x + y = n$.
Here,$r \neq r+3$,so we must have $r + (r+3) = 15$.
$2r + 3 = 15$
$2r = 12$
$r = 6$

(D) Given,$(49^{2}-4)(49^{3}-49)$
$= [(49)^{2}-(2)^{2}][49(49^{2}-1)]$
$= (49+2)(49-2) \cdot 49(49+1)(49-1)$
$= 51 \cdot 47 \cdot 49 \cdot 50 \cdot 48$
$= 47 \cdot 48 \cdot 49 \cdot 50 \cdot 51$
This is the product of $5$ consecutive integers,which is always divisible by $5!$.

(C) Given expression: $(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)$
$= (\sin \theta + \cos \theta) \left( \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \right)$
$= (\sin \theta + \cos \theta) \left( \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \right)$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$= (\sin \theta + \cos \theta) \left( \frac{1}{\sin \theta \cos \theta} \right)$
$= \frac{\sin \theta}{\sin \theta \cos \theta} + \frac{\cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\cos \theta} + \frac{1}{\sin \theta}$
$= \sec \theta + \operatorname{cosec} \theta$

(B) Let the expression be $E = \cot 12^{\circ} \cot 102^{\circ} + \cot 102^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$.
We know that $\cot 102^{\circ} = \cot(90^{\circ} + 12^{\circ}) = -\tan 12^{\circ}$.
Substituting this into the expression:
$E = \cot 12^{\circ} (-\tan 12^{\circ}) + (-\tan 12^{\circ}) \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$
$E = -1 - \tan 12^{\circ} \cot 66^{\circ} + \cot 66^{\circ} \cot 12^{\circ}$
$E = -1 + \cot 66^{\circ} (\cot 12^{\circ} - \tan 12^{\circ})$
Using the identity $\cot \theta - \tan \theta = 2 \cot 2\theta$:
$E = -1 + \cot 66^{\circ} (2 \cot 24^{\circ})$
Since $\cot 24^{\circ} = \cot(90^{\circ} - 66^{\circ}) = \tan 66^{\circ}$:
$E = -1 + 2 \cot 66^{\circ} \tan 66^{\circ}$
$E = -1 + 2(1) = 1$.

(D) Given equation: $1+\sin ^{2} x=3 \sin x \cdot \cos x$,with $\tan x \neq \frac{1}{2}$.
Dividing both sides by $\cos ^{2} x$ (assuming $\cos x \neq 0$):
$\frac{1}{\cos ^{2} x} + \frac{\sin ^{2} x}{\cos ^{2} x} = 3 \frac{\sin x \cdot \cos x}{\cos ^{2} x}$
$\sec ^{2} x + \tan ^{2} x = 3 \tan x$
Since $\sec ^{2} x = 1 + \tan ^{2} x$,we have:
$1 + \tan ^{2} x + \tan ^{2} x = 3 \tan x$
$2 \tan ^{2} x - 3 \tan x + 1 = 0$
Factoring the quadratic equation:
$2 \tan ^{2} x - 2 \tan x - \tan x + 1 = 0$
$2 \tan x(\tan x - 1) - 1(\tan x - 1) = 0$
$(\tan x - 1)(2 \tan x - 1) = 0$
So,$\tan x = 1$ or $\tan x = \frac{1}{2}$.
Given the condition $\tan x \neq \frac{1}{2}$,we must have $\tan x = 1$.
The general solution for $\tan x = 1$ is $x = n \pi + \frac{\pi}{4}$,where $n \in Z$.

(D) Given the homogeneous equation $ax^{2}+2hxy+by^{2}=0 \quad \dots(i)$
This represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m$ and $m_{1}$.
Thus,the lines are $y=mx$ and $y=m_{1}x$.
Their product is $(y-mx)(y-m_{1}x) = 0$,which simplifies to $y^{2}-(m+m_{1})xy+mm_{1}x^{2}=0$.
Dividing the original equation by $b$,we get $y^{2}+\frac{2h}{b}xy+\frac{a}{b}x^{2}=0 \quad \dots(ii)$.
Comparing $(i)$ and $(ii)$,we have:
$m+m_{1} = -\frac{2h}{b}$ and $mm_{1} = \frac{a}{b}$.
Since $m$ is a root of the quadratic equation $bm^{2}+2hm+a=0$ (obtained by substituting $y=mx$ in the original equation):
$bm^{2}+2hm+a=0$
$bm^{2}+hm = -a-hm$
$m(bm+h) = -(a+hm)$
This does not lead directly to the target. Let us use the relation $m_{1} = -\frac{2h}{b}-m$.
Substituting into $mm_{1} = \frac{a}{b}$:
$m(-\frac{2h}{b}-m) = \frac{a}{b}$
$-2hm - bm^{2} = a$
$bm^{2} + 2hm + a = 0$
Adding $h^{2}$ to both sides:
$bm^{2} + 2hm + h^{2} = h^{2} - a$
This is not quite right. Let's re-evaluate: $bm^{2}+2hm+a=0 \implies bm^{2}+hm = -a-hm$. Actually,from $bm^{2}+2hm+a=0$,we have $bm^{2}+hm = -a-hm$.
Wait,the expression is $(h+bm)^{2} = h^{2} + 2hbm + b^{2}m^{2}$.
Since $bm^{2}+2hm+a=0$,then $b^{2}m^{2}+2hbm+ab=0$.
Thus $b^{2}m^{2}+2hbm = -ab$.
Adding $h^{2}$ to both sides: $h^{2}+2hbm+b^{2}m^{2} = h^{2}-ab$.
Therefore,$(h+bm)^{2} = h^{2}-ab$.

(A) The points $P, Q, R$ divide the line segment $AB$ into $4$ equal parts. Thus,$AP = PQ = QR = RB = k$ (let).
$P$ divides $AB$ in the ratio $1:3$,$Q$ divides $AB$ in the ratio $2:2$ (or $1:1$),and $R$ divides $AB$ in the ratio $3:1$.
Since $Q$ is the midpoint of $AB$,its coordinates are $\left( \frac{2+6}{2}, \frac{-7+5}{2} \right) = (4, -1)$.
$P$ is the midpoint of $AQ$. The coordinates of $A$ are $(2, -7)$ and $Q$ are $(4, -1)$.
Coordinates of $P = \left( \frac{2+4}{2}, \frac{-7-1}{2} \right) = (3, -4)$.
$R$ is the midpoint of $QB$. The coordinates of $Q$ are $(4, -1)$ and $B$ are $(6, 5)$.
Coordinates of $R = \left( \frac{4+6}{2}, \frac{-1+5}{2} \right) = (5, 2)$.
The midpoint of $PR$ is $\left( \frac{3+5}{2}, \frac{-4+2}{2} \right) = (4, -1)$.

(D) Given points are $P = (-1, 0)$,$Q = (0, 0)$,and $R = (3, 3\sqrt{3})$.
Line $QP$ lies along the negative $x$-axis,so its angle of inclination is $\pi$ radians.
Line $QR$ passes through $(0, 0)$ and $(3, 3\sqrt{3})$. Its slope is $m = \frac{3\sqrt{3} - 0}{3 - 0} = \sqrt{3}$.
The angle of inclination $\phi$ of line $QR$ satisfies $\tan \phi = \sqrt{3}$,so $\phi = \frac{\pi}{3}$.
The angle $\angle PQR$ is the angle between the negative $x$-axis and the line $QR$,which is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
The angle bisector of $\angle PQR$ makes an angle $\alpha$ with the positive $x$-axis. Since the bisector divides $\angle PQR$ into two equal parts of $\frac{\pi}{3}$ each,the angle of the bisector with the positive $x$-axis is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
The slope of the bisector is $m = \tan(\frac{2\pi}{3}) = -\sqrt{3}$.
The equation of the line passing through $(0, 0)$ with slope $-\sqrt{3}$ is $y = -\sqrt{3}x$,which simplifies to $\sqrt{3}x + y = 0$.

(A) Given the equation of the line: $2x + 3y - k = 0, k > 0$.
Rewriting in intercept form: $\frac{x}{k/2} + \frac{y}{k/3} = 1$.
Thus,the coordinates of $A$ and $B$ are $(\frac{k}{2}, 0)$ and $(0, \frac{k}{3})$ respectively.
The area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{k}{2} \times \frac{k}{3} = \frac{k^{2}}{12}$.
Given area $= 12$,so $\frac{k^{2}}{12} = 12$ $\Rightarrow k^{2} = 144$ $\Rightarrow k = 12$ (since $k > 0$).
Therefore,$A = (6, 0)$ and $B = (0, 4)$.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
Substituting the points $(6, 0)$ and $(0, 4)$:
$(x - 6)(x - 0) + (y - 0)(y - 4) = 0$
$x(x - 6) + y(y - 4) = 0$
$x^{2} - 6x + y^{2} - 4y = 0$
$x^{2} + y^{2} - 6x - 4y = 0$.

(D) The equation of the circle passing through the points $(1,0), (0,1)$ and $(0,0)$ is given by the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.
Substituting $(0,0)$,we get $c = 0$.
Substituting $(1,0)$,we get $1 + 2g = 0 \Rightarrow g = -\frac{1}{2}$.
Substituting $(0,1)$,we get $1 + 2f = 0 \Rightarrow f = -\frac{1}{2}$.
Thus,the equation of the circle is $x^2 + y^2 - x - y = 0$.
Since the point $(2k, 3k)$ lies on this circle,it must satisfy the equation:
$(2k)^2 + (3k)^2 - (2k) - (3k) = 0$
$4k^2 + 9k^2 - 5k = 0$
$13k^2 - 5k = 0$
$k(13k - 5) = 0$
Given $k \neq 0$,we have $13k - 5 = 0$,which implies $k = \frac{5}{13}$.

(D) The equation of the circle is $x^{2}+y^{2}-4x=0$. The center of the circle is $C(2,0)$.
Let the chord be bisected at point $M(1,0)$.
The line segment connecting the center $C(2,0)$ and the midpoint $M(1,0)$ is perpendicular to the chord.
The slope of the line $CM$ is $m_{CM} = \frac{0-0}{1-2} = 0$.
Since the chord is perpendicular to $CM$,and $CM$ is a horizontal line (the x-axis),the chord must be a vertical line.
$A$ vertical line passing through $(1,0)$ has the equation $x=1$.
The slope of this chord is undefined (vertical).
We check the options for a line perpendicular to the chord (a vertical line). $A$ line perpendicular to a vertical line must be a horizontal line.
Among the options,$y=1$ is a horizontal line.
Therefore,the chord is perpendicular to the line $y=1$.

(C) Let the radius of both circles be $r$.
The equation of the circle with center at $(2,3)$ is:
$S_{1} \equiv (x-2)^{2} + (y-3)^{2} = r^{2} \quad \dots(i)$
The equation of the circle with center at $(5,6)$ is:
$S_{2} \equiv (x-5)^{2} + (y-6)^{2} = r^{2} \quad \dots(ii)$
The equation of the common chord is given by the radical axis $S_{1} - S_{2} = 0$.
$(x-2)^{2} + (y-3)^{2} - ((x-5)^{2} + (y-6)^{2}) = 0$
$(x^{2} - 4x + 4 + y^{2} - 6y + 9) - (x^{2} - 10x + 25 + y^{2} - 12y + 36) = 0$
$(x^{2} + y^{2} - 4x - 6y + 13) - (x^{2} + y^{2} - 10x - 12y + 61) = 0$
$6x + 6y - 48 = 0$
Dividing by $6$,we get:
$x + y - 8 = 0$

(A) Given the equation of the circle $x^{2}+y^{2}=1$,the center is $O(0,0)$ and the radius $r_{1} = 1$.
Let the center of the required circle be $C(4,3)$ and its radius be $r_{2}$.
The distance between the centers $O$ and $C$ is $OC = \sqrt{(4-0)^{2} + (3-0)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$.
Since the circles touch externally,the distance between their centers is equal to the sum of their radii:
$OC = r_{1} + r_{2}$
$5 = 1 + r_{2}$
$r_{2} = 4$.
The equation of the circle with center $(h,k) = (4,3)$ and radius $r = 4$ is:
$(x-4)^{2} + (y-3)^{2} = 4^{2}$
$x^{2} - 8x + 16 + y^{2} - 6y + 9 = 16$
$x^{2} + y^{2} - 8x - 6y + 9 = 0$.

(C) The equation of the parabola is $x^{2}=-8y$. Comparing this with $x^{2}=4ay$,we get $4a=-8$,so $a=-2$. The focus of the parabola is $(0, a) = (0, -2)$.
Let the ends of a focal chord be $P(x_1, y_1)$ and $P'(x_2, y_2)$. The tangents at the ends of a focal chord of a parabola intersect at the directrix.
The directrix of the parabola $x^{2}=4ay$ is $y=-a$.
Here,$a=-2$,so the directrix is $y=-(-2) = 2$.
Therefore,the locus of the point of intersection of the tangents is the line $y=2$.

(A) Given the equation of the parabola $y^{2}=4ax$. Let the parametric point on the parabola be $(at^{2}, 2at)$.
The slope of the tangent at this point is $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
The slope of the normal is $-t$. Let the slope of the normal be $m$,so $m = -t$,which implies $t = -m$.
The equation of the normal at point $(at^{2}, 2at)$ is $y - 2at = -t(x - at^{2})$.
Substituting $t = -m$,we get $y - 2a(-m) = -(-m)(x - a(-m)^{2})$.
$y + 2am = m(x - am^{2})$.
$y + 2am = mx - am^{3}$.
$y = mx - 2am - am^{3}$.
Comparing this with the given line $y = mx + c$,we get $c = -2am - am^{3}$.

(C) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$. Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a^{2}=16$ and $b^{2}=4$,so $a=4$ and $b=2$.
Let $\theta$ be the eccentric angle of the point $P(2, \sqrt{3})$.
The parametric coordinates of any point on the ellipse are given by $x = a \cos \theta$ and $y = b \sin \theta$.
Substituting the values,we have $x = 4 \cos \theta$ and $y = 2 \sin \theta$.
Given the point $(2, \sqrt{3})$,we equate:
$2 = 4 \cos \theta \Rightarrow \cos \theta = \frac{2}{4} = \frac{1}{2}$
$\sqrt{3} = 2 \sin \theta \Rightarrow \sin \theta = \frac{\sqrt{3}}{2}$
Since both $\sin \theta = \frac{\sqrt{3}}{2}$ and $\cos \theta = \frac{1}{2}$ are satisfied at $\theta = \frac{\pi}{3}$,the eccentric angle is $\frac{\pi}{3}$.

(D) The given equation of the hyperbola is $x^{2}-y^{2}=4$.
Dividing by $4$,we get $\frac{x^{2}}{4}-\frac{y^{2}}{4}=1$.
Here,$a^{2}=4$ and $b^{2}=4$.
For a hyperbola,$b^{2}=a^{2}(e^{2}-1)$,so $4=4(e^{2}-1)$,which implies $e^{2}-1=1$,so $e^{2}=2$ and $e=\sqrt{2}$.
The coordinates of the foci are $(\pm ae, 0) = (\pm 2\sqrt{2}, 0)$.
The equations of the directrices are $x = \pm \frac{a}{e} = \pm \frac{2}{\sqrt{2}} = \pm \sqrt{2}$.
Consider the focus $(2\sqrt{2}, 0)$ and the nearer directrix $x = \sqrt{2}$.
The distance between them is $|2\sqrt{2} - \sqrt{2}| = \sqrt{2}$.

(B) Let $L = \lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$.
We know that $\sin(2\theta) = 2 \sin \theta \cos \theta$,so $\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)$.
Here,$\theta = \frac{2 \pi}{3 n}$.
Thus,$L = \lim _{n \rightarrow \infty} n \cdot \frac{1}{2} \sin \left( 2 \cdot \frac{2 \pi}{3 n} \right) = \lim _{n \rightarrow \infty} \frac{n}{2} \sin \left( \frac{4 \pi}{3 n} \right)$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we rewrite the expression:
$L = \lim _{n \rightarrow \infty} \frac{n}{2} \cdot \left( \frac{4 \pi}{3 n} \right) \cdot \frac{\sin \left( \frac{4 \pi}{3 n} \right)}{\left( \frac{4 \pi}{3 n} \right)}$.
As $n \rightarrow \infty$,$\frac{4 \pi}{3 n} \rightarrow 0$,so $\frac{\sin \left( \frac{4 \pi}{3 n} \right)}{\left( \frac{4 \pi}{3 n} \right)} \rightarrow 1$.
Therefore,$L = \lim _{n \rightarrow \infty} \frac{n}{2} \cdot \frac{4 \pi}{3 n} \cdot 1 = \frac{4 \pi}{6} = \frac{2 \pi}{3}$.

(C) The condition $a \equiv b \pmod{x}$ implies that $x$ is a divisor of $(a - b)$.
For the first congruence: $21 \equiv 385 \pmod{x}$,we have $x$ divides $(385 - 21) = 364$.
The divisors of $364$ are $1, 2, 4, 7, 13, 14, 26, 28, 52, 91, 182, 364$.
For the second congruence: $587 \equiv 167 \pmod{x}$,we have $x$ divides $(587 - 167) = 420$.
The divisors of $420$ are $1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, 420$.
The greatest common divisor of $364$ and $420$ is $gcd(364, 420) = 28$.
Thus,the greatest value of $x$ is $28$.

(A) We know that $2^2 = 4 \equiv -1 \pmod{5}$.
Now,$2^{2010} = (2^2)^{1005} \equiv (-1)^{1005} \pmod{5}$.
Since $1005$ is an odd number,$(-1)^{1005} = -1$.
So,$2^{2010} \equiv -1 \equiv 4 \pmod{5}$.
Given the congruence $2^{2010} \equiv 3x \pmod{5}$,we substitute $2^{2010} \equiv 4 \pmod{5}$:
$4 \equiv 3x \pmod{5}$.
To solve for $x$,multiply both sides by the modular inverse of $3$ modulo $5$,which is $2$ (since $3 \times 2 = 6 \equiv 1 \pmod{5}$):
$2 \times 4 \equiv 2 \times 3x \pmod{5}$
$8 \equiv 6x \pmod{5}$
$3 \equiv x \pmod{5}$.
Thus,the least positive integer $x$ is $3$.

(A) We evaluate each option using truth tables:
$(a)$ Let $x = (p \wedge \sim q)$ and $y = (p \rightarrow q)$. The truth table shows that $(x \leftrightarrow y)$ is not a tautology (it is a contingency).
$(b)$ This is a standard tautology known as the Law of Hypothetical Syllogism.
$(c)$ This is a standard logical equivalence (Distributive Law of implication over conjunction).
$(d)$ This is a standard logical equivalence for the negation of a biconditional statement.
Therefore,option $(a)$ is not true.

(C) Given that,$a=2$.
In $\triangle ABC$,$B=\tan ^{-1}(\frac{1}{2})$ and $C=\tan ^{-1}(\frac{1}{3})$.
We know that in $\triangle ABC$,$A+B+C=\pi$.
$A = \pi - (B+C) = \pi - (\tan ^{-1}(\frac{1}{2}) + \tan ^{-1}(\frac{1}{3}))$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1}(\frac{x+y}{1-xy})$,we get:
$B+C = \tan ^{-1}(\frac{1/2 + 1/3}{1 - 1/6}) = \tan ^{-1}(\frac{5/6}{5/6}) = \tan ^{-1}(1) = \frac{\pi}{4}$.
Thus,$A = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Now,$\sin A = \sin(\frac{3\pi}{4}) = \sin(135^{\circ}) = \frac{1}{\sqrt{2}}$.
Since $\tan B = \frac{1}{2}$,we have $\sin B = \frac{1}{\sqrt{1^2+2^2}} = \frac{1}{\sqrt{5}}$.
Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B}$.
$b = a \cdot \frac{\sin B}{\sin A} = 2 \cdot \frac{1/\sqrt{5}}{1/\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{5}}$.
Therefore,$(A, b) = (\frac{3\pi}{4}, \frac{2\sqrt{2}}{\sqrt{5}})$.

(A) Given the sides of the triangle are $a = 6+2 \sqrt{3}$,$b = 4 \sqrt{3}$,and $c = \sqrt{24} = 2 \sqrt{6}$.
Comparing the values: $a \approx 6 + 3.464 = 9.464$,$b \approx 4 \times 1.732 = 6.928$,and $c \approx 2 \times 2.449 = 4.898$.
Since $c$ is the smallest side,the smallest angle is $C$.
Using the law of cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
$a^2 = (6+2 \sqrt{3})^2 = 36 + 12 + 24 \sqrt{3} = 48 + 24 \sqrt{3}$.
$b^2 = (4 \sqrt{3})^2 = 48$.
$c^2 = 24$.
$\cos C = \frac{48 + 24 \sqrt{3} + 48 - 24}{2(6+2 \sqrt{3})(4 \sqrt{3})} = \frac{72 + 24 \sqrt{3}}{8 \sqrt{3}(3+\sqrt{3})} = \frac{24(3+\sqrt{3})}{8 \sqrt{3}(3+\sqrt{3})} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
Wait,checking the calculation again: $\cos C = \frac{72+24 \sqrt{3}}{16 \sqrt{3}(3+\sqrt{3})} = \frac{24(3+\sqrt{3})}{16 \sqrt{3}(3+\sqrt{3})} = \frac{24}{16 \sqrt{3}} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$.
Thus,$\cos C = \frac{\sqrt{3}}{2}$,which means $C = 30^{\circ}$.
The tangent of the smallest angle is $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$.

(B) The given curve is $y^{2} = \left|\begin{array}{ll}x+1 & x+2 \\ x+3 & x+5\end{array}\right|$.
Expanding the determinant:
$y^{2} = (x+1)(x+5) - (x+2)(x+3)$
$y^{2} = (x^{2} + 6x + 5) - (x^{2} + 5x + 6)$
$y^{2} = x - 1$,which is a parabola.
Let the parametric coordinates of a point $Q$ on the parabola be $Q(t^{2}+1, t)$.
Given $P(x, y)$ is the midpoint of the line segment joining $A(1, 0)$ and $Q(t^{2}+1, t)$:
$x = \frac{1 + t^{2} + 1}{2} = \frac{t^{2} + 2}{2} \Rightarrow t^{2} = 2x - 2$
$y = \frac{0 + t}{2} = \frac{t}{2} \Rightarrow t = 2y$
Substituting $t$ in the equation for $x$:
$(2y)^{2} = 2x - 2$
$4y^{2} = 2(x - 1)$
$y^{2} = \frac{1}{2}(x - 1)$
This is a parabola with its axis along the $x$-axis. Therefore,the locus of $P$ is symmetrical about the $x$-axis.

(B) Given,$\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$
Multiplying both sides by $(3-5 x)(2+3 x)$,we get:
$1 = A(2+3 x) + B(3-5 x)$
$1 = (2A + 3B) + x(3A - 5B)$
Comparing the coefficients of $x$ and the constant terms on both sides:
$3A - 5B = 0 \implies 3A = 5B \implies A = \frac{5}{3}B$
$2A + 3B = 1$
Substituting $A = \frac{5}{3}B$ into the second equation:
$2(\frac{5}{3}B) + 3B = 1$
$\frac{10}{3}B + 3B = 1$
$\frac{10B + 9B}{3} = 1$
$19B = 3 \implies B = \frac{3}{19}$
Now,$A = \frac{5}{3} \times \frac{3}{19} = \frac{5}{19}$
Therefore,$A : B = \frac{5}{19} : \frac{3}{19} = 5 : 3$

(B) Let the radius of the sector be $r$ and the arc length be $s$. The perimeter of the sector is given by $P = 2r + s = 20 \text{ cm}$.
Thus,$s = 20 - 2r$.
The area $A$ of the sector is given by $A = \frac{1}{2} s r$.
Substituting the value of $s$,we get $A = \frac{1}{2} (20 - 2r) r = 10r - r^2$.
To find the maximum area,we differentiate $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = 10 - 2r = 0 \Rightarrow r = 5 \text{ cm}$.
Substituting $r = 5$ into the area equation:
$A = 10(5) - (5)^2 = 50 - 25 = 25 \text{ cm}^2$.

(B) The binary operation is defined as $A * B = A \cup B$ for all $A, B \in P(X)$.
$1$. Commutative Law: $A * B = A \cup B = B \cup A = B * A$. Thus,it is commutative.
$2$. Associative Law: $(A * B) * C = (A \cup B) \cup C = A \cup (B \cup C) = A * (B * C)$. Thus,it is associative.
$3$. Identity Law: The identity element $E$ must satisfy $A * E = A$,which means $A \cup E = A$. This holds for $E = \phi$. Thus,the identity element exists.
$4$. Inverse Law: For an element $A$ to have an inverse $B$,we must have $A * B = E$,where $E = \phi$. This implies $A \cup B = \phi$. This is only possible if $A = \phi$ and $B = \phi$. For any non-empty set $A \neq \phi$,there is no $B \in P(X)$ such that $A \cup B = \phi$. Therefore,the inverse law is not satisfied.

(D) Let $e$ be the identity element in $Q^{+}$ such that $a * e = a$ for all $a \in Q^{+}$.
$\frac{a \times e}{2010} = a \implies e = 2010$.
Thus,the identity element is $2010$.
Let $x$ be the inverse of $2010$. By definition,$2010 * x = e$.
$\frac{2010 \times x}{2010} = 2010$.
$x = 2010$.
Therefore,the inverse of $2010$ is $2010$.

(B) An equivalence relation $R$ on a set $A$ must satisfy the reflexive,symmetric,and transitive properties.
For a set $A$ with $n$ elements,the reflexive property requires that for every element $a \in A$,the ordered pair $(a, a)$ must be in $R$.
Given that the set contains $6$ elements,we must have at least the pairs $(a_1, a_1), (a_2, a_2), (a_3, a_3), (a_4, a_4), (a_5, a_5),$ and $(a_6, a_6)$ in $R$.
Thus,the minimum number of ordered pairs required is $6$.

(A) Given,$AB = B$ and $BA = A$ ... $(i)$
We need to find $A^{2} + B^{2}$.
Using the given relations:
$A^{2} = A \cdot A = A(BA) = (AB)A = BA = A$
Similarly,$B^{2} = B \cdot B = B(AB) = (BA)B = AB = B$
Therefore,$A^{2} + B^{2} = A + B$.

(A) Given $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$.
The characteristic equation of $A$ is given by $|A - \lambda I| = 0$.
$|A - \lambda I| = \begin{vmatrix} 3 - \lambda & 2 \\ 1 & 1 - \lambda \end{vmatrix} = 0$.
$(3 - \lambda)(1 - \lambda) - (2)(1) = 0$.
$3 - 3\lambda - \lambda + \lambda^{2} - 2 = 0$.
$\lambda^{2} - 4\lambda + 1 = 0$.
By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation. Substituting $\lambda = A$,we get:
$A^{2} - 4A + I = 0$.
Comparing this with the given equation $A^{2} + xA + yI = 0$,we get:
$x = -4$ and $y = 1$.
Thus,$(x, y) = (-4, 1)$.

(D) Given that $A$ is a $3 \times 3$ matrix,so $n=3$.
We are given $|A|=3$.
We know that for any non-singular matrix $A$,$|A^{-1}| = \frac{1}{|A|}$.
Also,for a scalar $k$ and a matrix $A$ of order $n \times n$,$|kA| = k^n |A|$.
Therefore,$|2A| = 2^3 |A| = 8 \times 3 = 24$.
Now,$|(2A)^{-1}| = \frac{1}{|2A|} = \frac{1}{24}$.

(C) Let $f(x) = \left|\begin{array}{ccc} x+3 & x & x+2 \\ x & x+1 & x-1 \\ x+2 & 2x & 3x+1 \end{array}\right|$.
To find the constant term,we can set $x = 0$ in the determinant.
Substituting $x = 0$ into the determinant:
$f(0) = \left|\begin{array}{ccc} 0+3 & 0 & 0+2 \\ 0 & 0+1 & 0-1 \\ 0+2 & 2(0) & 3(0)+1 \end{array}\right|$
$f(0) = \left|\begin{array}{ccc} 3 & 0 & 2 \\ 0 & 1 & -1 \\ 2 & 0 & 1 \end{array}\right|$
Expanding along the second column $(C_2)$:
$f(0) = 0 - 1 \times \left|\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}\right| + 0$
$f(0) = -1 \times (3(1) - 2(2))$
$f(0) = -1 \times (3 - 4)$
$f(0) = -1 \times (-1) = 1$
Wait,let us re-evaluate the expansion of the polynomial $f(x) = 8x^2 + 9x - 1$. The constant term is the value of the polynomial when $x=0$. Evaluating $f(0) = 8(0)^2 + 9(0) - 1 = -1$. The previous expansion was correct,and the constant term is $-1$.

(B) Given $\sec^{-1} \left( \frac{a+b}{a-b} \right) = 2 \sin^{-1} x$.
Taking the reciprocal,we get $\cos^{-1} \left( \frac{a-b}{a+b} \right) = 2 \sin^{-1} x$.
Dividing the numerator and denominator by $a$,we have $\cos^{-1} \left( \frac{1 - b/a}{1 + b/a} \right) = 2 \sin^{-1} x$.
Using the identity $\cos^{-1} \left( \frac{1 - y^2}{1 + y^2} \right) = 2 \tan^{-1} y$,where $y = \sqrt{b/a}$,we get $2 \tan^{-1} \left( \sqrt{\frac{b}{a}} \right) = 2 \sin^{-1} x$.
Dividing by $2$,we have $\tan^{-1} \left( \sqrt{\frac{b}{a}} \right) = \sin^{-1} x$.
Using the identity $\tan^{-1} y = \sin^{-1} \left( \frac{y}{\sqrt{1+y^2}} \right)$,we get $\sin^{-1} \left( \frac{\sqrt{b/a}}{\sqrt{1 + b/a}} \right) = \sin^{-1} x$.
Simplifying the expression inside,we get $\sin^{-1} \left( \sqrt{\frac{b/a}{(a+b)/a}} \right) = \sin^{-1} x$.
Thus,$x = \sqrt{\frac{b}{a+b}}$.

(A) Given expression: $\frac{\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)}{\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)}$
We know that $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$ for $\theta \in [-1, 1]$ and $\tan ^{-1} \theta + \cot ^{-1} \theta = \frac{\pi}{2}$ for $\theta \in \mathbb{R}$.
Using the identities $\cos x = \sin(\frac{\pi}{2} - x)$ and $\sin x = \cos(\frac{\pi}{2} - x)$:
Numerator: $\sin ^{-1}(\sin(\frac{\pi}{2} - x)) + \cos ^{-1}(\cos(\frac{\pi}{2} - x)) = (\frac{\pi}{2} - x) + (\frac{\pi}{2} - x) = \pi - 2x$.
Denominator: $\tan ^{-1}(\tan(\frac{\pi}{2} - x)) + \cot ^{-1}(\cot(\frac{\pi}{2} - x)) = (\frac{\pi}{2} - x) + (\frac{\pi}{2} - x) = \pi - 2x$.
Thus,the expression becomes $\frac{\pi - 2x}{\pi - 2x} = 1$.

(A) The function $f(x) = [x]$ is known as the Greatest Integer Function.
For any integer $n$,the left-hand limit is $\lim_{x \to n^-} f(x) = n-1$ and the right-hand limit is $\lim_{x \to n^+} f(x) = n$.
Since the left-hand limit is not equal to the right-hand limit at any integer $n$,the function is discontinuous at all integral values of $x$.
However,for any non-integral value $c$ (where $c$ is not an integer),the function is locally constant in a small neighborhood around $c$,making it continuous at all such points.
Therefore,the function $f(x) = [x]$ is continuous for all non-integral values of $x$.

(D) Given $f(x) = |x-2| + x$.
First,we check for continuity at $x=0$ and $x=2$.
At $x=0$: $f(0) = |0-2| + 0 = 2$. $\lim_{x \to 0^-} f(x) = |0-2| + 0 = 2$ and $\lim_{x \to 0^+} f(x) = |0-2| + 0 = 2$. Since $\lim_{x \to 0} f(x) = f(0)$,the function is continuous at $x=0$.
At $x=2$: $f(2) = |2-2| + 2 = 2$. $\lim_{x \to 2^-} f(x) = |2-2| + 2 = 2$ and $\lim_{x \to 2^+} f(x) = |2-2| + 2 = 2$. Since $\lim_{x \to 2} f(x) = f(2)$,the function is continuous at $x=2$.
Now,we check for differentiability. We can write $f(x)$ as:
$f(x) = \begin{cases} -(x-2) + x, & x < 2 \\ (x-2) + x, & x \geq 2 \end{cases} = \begin{cases} 2, & x < 2 \\ 2x-2, & x \geq 2 \end{cases}$.
For $x < 2$,$f'(x) = 0$. For $x > 2$,$f'(x) = 2$.
At $x=2$,$LHD$ $= \lim_{h \to 0^-} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0^-} \frac{2-2}{h} = 0$.
$RHD$ $= \lim_{h \to 0^+} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0^+} \frac{2(2+h)-2-2}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2$.
Since $LHD$ $\neq$ $RHD$,$f(x)$ is not differentiable at $x=2$.
At $x=0$,$f(x) = 2$ in the neighborhood,so $f'(0) = 0$. Thus,$f(x)$ is differentiable at $x=0$.
Therefore,the function is continuous at both $x=0$ and $x=2$.

(C) Given,$h(x) = f(x) \cdot g(x)$ and $f^{\prime}(x) \cdot g^{\prime}(x) = c$.
First,find the first derivative of $h(x)$ using the product rule:
$h^{\prime}(x) = f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)$.
Next,find the second derivative $h^{\prime \prime}(x)$:
$h^{\prime \prime}(x) = \frac{d}{dx}[f^{\prime}(x) \cdot g(x) + f(x) \cdot g^{\prime}(x)]$
$h^{\prime \prime}(x) = [f^{\prime \prime}(x) \cdot g(x) + f^{\prime}(x) \cdot g^{\prime}(x)] + [f^{\prime}(x) \cdot g^{\prime}(x) + f(x) \cdot g^{\prime \prime}(x)]$
$h^{\prime \prime}(x) = f^{\prime \prime}(x) \cdot g(x) + f(x) \cdot g^{\prime \prime}(x) + 2f^{\prime}(x) \cdot g^{\prime}(x)$.
Since $f^{\prime}(x) \cdot g^{\prime}(x) = c$,we have:
$h^{\prime \prime}(x) = f^{\prime \prime}(x) \cdot g(x) + f(x) \cdot g^{\prime \prime}(x) + 2c \quad \dots(i)$.
Now,evaluate the expression:
$\frac{f^{\prime \prime}(x)}{f(x)} + \frac{g^{\prime \prime}(x)}{g(x)} + \frac{2c}{f(x) \cdot g(x)} = \frac{f^{\prime \prime}(x) \cdot g(x) + g^{\prime \prime}(x) \cdot f(x) + 2c}{f(x) \cdot g(x)}$.
Substituting from equation $(i)$:
$= \frac{h^{\prime \prime}(x)}{h(x)}$.

(D) Given $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$.
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = 3a \cos^{2} \theta (-\sin \theta) = -3a \cos^{2} \theta \sin \theta$.
$\frac{dy}{d\theta} = 3a \sin^{2} \theta (\cos \theta) = 3a \sin^{2} \theta \cos \theta$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3a \sin^{2} \theta \cos \theta}{-3a \cos^{2} \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta$.
From the given equations:
$\frac{y}{a} = \sin^{3} \theta \Rightarrow \sin \theta = (\frac{y}{a})^{1/3}$.
$\frac{x}{a} = \cos^{3} \theta \Rightarrow \cos \theta = (\frac{x}{a})^{1/3}$.
Substituting these into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{(y/a)^{1/3}}{(x/a)^{1/3}} = -(\frac{y}{x})^{1/3} = -\sqrt[3]{\frac{y}{x}}$.

(A) Let $y = e^{ax} \cos bx$.
Using the product rule,the derivative is:
$\frac{dy}{dx} = \frac{d}{dx}(e^{ax}) \cdot \cos bx + e^{ax} \cdot \frac{d}{dx}(\cos bx)$
$\frac{dy}{dx} = ae^{ax} \cos bx - be^{ax} \sin bx$
$\frac{dy}{dx} = e^{ax} (a \cos bx - b \sin bx)$
Let $a = r \cos \alpha$ and $b = r \sin \alpha$.
Then $a^2 + b^2 = r^2(\cos^2 \alpha + \sin^2 \alpha) = r^2$.
Thus,$r = \sqrt{a^2 + b^2}$.
Substituting these into the derivative expression:
$\frac{dy}{dx} = e^{ax} (r \cos \alpha \cos bx - r \sin \alpha \sin bx)$
$\frac{dy}{dx} = re^{ax} (\cos bx \cos \alpha - \sin bx \sin \alpha)$
Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:
$\frac{dy}{dx} = re^{ax} \cos(bx + \alpha)$
Therefore,$r = \sqrt{a^2 + b^2}$.

(B) Given $y = \tan^{-1} \sqrt{x^{2}-1}$.
Let $x = \sec \theta$,then $\sqrt{x^{2}-1} = \tan \theta$.
So,$y = \tan^{-1}(\tan \theta) = \theta = \sec^{-1} x$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sec^{-1} x) = \frac{1}{x \sqrt{x^{2}-1}}$.
Now,find the second derivative $\frac{d^{2}y}{dx^{2}}$:
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left( (x(x^{2}-1)^{1/2})^{-1} \right) = -1 \cdot (x(x^{2}-1)^{1/2})^{-2} \cdot \frac{d}{dx} (x(x^{2}-1)^{1/2})$.
Using the product rule: $\frac{d}{dx} (x(x^{2}-1)^{1/2}) = (x^{2}-1)^{1/2} + x \cdot \frac{1}{2}(x^{2}-1)^{-1/2} \cdot 2x = \sqrt{x^{2}-1} + \frac{x^{2}}{\sqrt{x^{2}-1}} = \frac{x^{2}-1+x^{2}}{\sqrt{x^{2}-1}} = \frac{2x^{2}-1}{\sqrt{x^{2}-1}}$.
Thus,$\frac{d^{2}y}{dx^{2}} = - \frac{1}{x^{2}(x^{2}-1)} \cdot \frac{2x^{2}-1}{\sqrt{x^{2}-1}} = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}}$.
Finally,the ratio is:
$\frac{d^{2}y/dx^{2}}{dy/dx} = \left( - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} \right) \div \left( \frac{1}{x \sqrt{x^{2}-1}} \right) = - \frac{2x^{2}-1}{x^{2}(x^{2}-1)^{3/2}} \cdot x \sqrt{x^{2}-1} = - \frac{2x^{2}-1}{x(x^{2}-1)} = \frac{1-2x^{2}}{x(x^{2}-1)}$.

(D) Given the curve $y = \log_{e} x$ $(i)$.
Let the point of contact be $P(\alpha, \beta)$.
The slope of the tangent is $\frac{dy}{dx} = \frac{1}{x}$.
The equation of the tangent at $P$ is $(y - \beta) = \frac{1}{\alpha}(x - \alpha)$.
Since the tangent passes through the origin $(0,0)$,we have $(0 - \beta) = \frac{1}{\alpha}(0 - \alpha)$,which gives $\beta = 1$.
From equation $(i)$,at $P$,$\beta = \log_{e} \alpha$. Since $\beta = 1$,we have $1 = \log_{e} \alpha$,so $\alpha = e$.
Thus,the point of contact is $P(e, 1)$.
The slope of the tangent at $P$ is $\frac{1}{e}$,so the slope of the normal at $P$ is $-e$.
The equation of the normal at $P(e, 1)$ is $(y - 1) = -e(x - e)$,which simplifies to $ex + y - (e^{2} + 1) = 0$.
The length of the perpendicular from the origin $(0,0)$ to the line $ex + y - (e^{2} + 1) = 0$ is given by $d = \frac{|e(0) + 1(0) - (e^{2} + 1)|}{\sqrt{e^{2} + 1^{2}}} = \frac{e^{2} + 1}{\sqrt{e^{2} + 1}} = \sqrt{e^{2} + 1}$.

(C) Given curve: $4x^{5} = 5y^{4}$.
Differentiating with respect to $x$: $20x^{4} = 20y^{3} \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x^{4}}{y^{3}}$.
Length of subtangent $(ST) = \left| y \frac{dx}{dy} \right| = \left| y \frac{y^{3}}{x^{4}} \right| = \frac{y^{4}}{x^{4}}$.
Length of subnormal $(SN) = \left| y \frac{dy}{dx} \right| = \left| y \frac{x^{4}}{y^{3}} \right| = \frac{x^{4}}{y^{2}}$.
We need the ratio of the cube of the subtangent to the square of the subnormal:
$\frac{(ST)^{3}}{(SN)^{2}} = \frac{(y^{4}/x^{4})^{3}}{(x^{4}/y^{2})^{2}} = \frac{y^{12}/x^{12}}{x^{8}/y^{4}} = \frac{y^{16}}{x^{20}} = \left(\frac{y^{4}}{x^{5}}\right)^{4}$.
Since $4x^{5} = 5y^{4}$,we have $\frac{y^{4}}{x^{5}} = \frac{4}{5}$.
Therefore,the ratio is $\left(\frac{4}{5}\right)^{4} = \frac{4^{4}}{5^{4}}$.

(A) Given the function $f(x) = \frac{x}{\log x}$.
First,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(\log x)(1) - (x)(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}$.
For the function to be increasing,we must have $f'(x) > 0$.
Since $(\log x)^2$ is always positive for $x > 0$ and $x \neq 1$,the condition $f'(x) > 0$ implies $\log x - 1 > 0$.
$\log x > 1$.
Since the base of the logarithm is $e$,we have $\log_e x > \log_e e$.
Therefore,$x > e$.
Thus,the set of values is $\{x: x > e\}$.

(B) Given the integral equation: $\int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c$.
Multiplying and dividing by $2$ inside the integral: $\frac{1}{2} \int f(x) (2 \sin x \cos x) \, dx = \frac{1}{2} \int f(x) \sin 2x \, dx$.
Let $f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$.
Substituting this into the integral: $\frac{1}{2} \int \frac{2}{(b^2 - a^2) \cos 2x} \sin 2x \, dx = \frac{1}{b^2 - a^2} \int \tan 2x \, dx$.
Integrating $\tan 2x$: $\frac{1}{b^2 - a^2} \cdot \frac{\log |\sec 2x|}{2} + c = \frac{1}{2(b^2 - a^2)} \log |\sec 2x| + c$.
Since $\sec 2x = \frac{1}{\cos 2x}$,this is $\frac{1}{2(b^2 - a^2)} \log |\frac{1}{\cos 2x}| + c$.
Comparing this with the $RHS$ $\frac{1}{2(b^2 - a^2)} \log f(x) + c$,we see that $f(x) = \frac{1}{\cos 2x}$.
However,to match the constant factor in the provided options,$f(x) = \frac{2}{(b^2 - a^2) \cos 2x}$ is the correct form.

(D) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$.
Substitute $x = \tan^2 \theta$,which implies $dx = 2 \tan \theta \sec^2 \theta d\theta$.
Then,$\sqrt{x} = \tan \theta$.
Substituting these into the integral:
$I = \int \frac{\tan \theta}{\tan^2 \theta (\tan^2 \theta + 1)} \cdot (2 \tan \theta \sec^2 \theta) d\theta$.
Since $\tan^2 \theta + 1 = \sec^2 \theta$,the expression simplifies to:
$I = \int \frac{\tan \theta}{\tan^2 \theta \sec^2 \theta} \cdot (2 \tan \theta \sec^2 \theta) d\theta$.
$I = \int \frac{2 \tan^2 \theta \sec^2 \theta}{\tan^2 \theta \sec^2 \theta} d\theta = \int 2 d\theta$.
$I = 2\theta + C$.
Since $x = \tan^2 \theta$,we have $\theta = \tan^{-1}(\sqrt{x})$.
Therefore,$I = 2 \tan^{-1}(\sqrt{x}) + C$.
Comparing this with $k \tan^{-1} m$,we get $k = 2$ and $m = \sqrt{x}$.
Thus,$(k, m) = (2, \sqrt{x})$.

(B) Let $I = \int_{0}^{1} x(1-x)^{3 / 2} d x$.
Using the Beta function property $\int_{0}^{1} x^{m-1}(1-x)^{n-1} dx = B(m, n) = \frac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}$,
Here $m-1 = 1 \implies m = 2$ and $n-1 = 3/2 \implies n = 5/2$.
$I = B(2, 5/2) = \frac{\Gamma(2) \Gamma(5/2)}{\Gamma(2 + 5/2)} = \frac{\Gamma(2) \Gamma(5/2)}{\Gamma(9/2)}$.
Since $\Gamma(n) = (n-1)!$ for integers and $\Gamma(n+1) = n\Gamma(n)$,
$\Gamma(2) = 1! = 1$.
$\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \Gamma(1/2) = \frac{3}{4} \sqrt{\pi}$.
$\Gamma(9/2) = \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{105}{16} \sqrt{\pi}$.
Therefore,$I = \frac{1 \cdot \frac{3}{4} \sqrt{\pi}}{\frac{105}{16} \sqrt{\pi}} = \frac{3}{4} \cdot \frac{16}{105} = \frac{3 \cdot 4}{105} = \frac{12}{105} = \frac{4}{35}$.

(A) Let $I = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$.
So,the integral becomes $I = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3 + 1 - (\sin x - \cos x)^2} d x = \int_{0}^{\pi / 4} \frac{\sin x+\cos x}{4 - (\sin x - \cos x)^2} d x$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$.
When $x = \pi/4$,$t = \sin(\pi/4) - \cos(\pi/4) = 0$.
Thus,$I = \int_{-1}^{0} \frac{dt}{4 - t^2} = \int_{-1}^{0} \frac{dt}{2^2 - t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,we get:
$I = \left[ \frac{1}{2(2)} \log \left| \frac{2+t}{2-t} \right| \right]_{-1}^{0} = \frac{1}{4} \left[ \log \left| \frac{2+0}{2-0} \right| - \log \left| \frac{2-1}{2-(-1)} \right| \right]$.
$I = \frac{1}{4} [ \log(1) - \log(1/3) ] = \frac{1}{4} [ 0 - (-\log 3) ] = \frac{1}{4} \log 3$.

(C) The curve is defined as $y = x^2$ for $x < 0$ and $y = x$ for $x \geq 0$. The line is $y = 4$.
For $x < 0$,the curve is $y = x^2$,which implies $x = -\sqrt{y}$ (since $x$ is negative). The intersection with $y=4$ is at $x = -2$. The area $A_1$ in the second quadrant is given by the integral of $|x|$ with respect to $y$ from $y=0$ to $y=4$:
$A_1 = \int_{0}^{4} |-\sqrt{y}| dy = \int_{0}^{4} y^{1/2} dy = \left[ \frac{2}{3} y^{3/2} \right]_{0}^{4} = \frac{2}{3}(8) = \frac{16}{3}$.
For $x \geq 0$,the curve is $y = x$. The intersection with $y=4$ is at $x = 4$. The area $A_2$ in the first quadrant is the area of the triangle formed by the vertices $(0,0)$,$(4,0)$,and $(4,4)$,or the integral of $x$ with respect to $y$ from $y=0$ to $y=4$:
$A_2 = \int_{0}^{4} x dy = \int_{0}^{4} y dy = \left[ \frac{y^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8$.
The total area is $A_1 + A_2 = \frac{16}{3} + 8 = \frac{16 + 24}{3} = \frac{40}{3}$ square units.

(A) Given differential equation is $y = x \frac{dp}{dx} + \sqrt{a^{2} p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$.
Substituting $p = \frac{dy}{dx}$,we get $y = x \frac{d}{dx} \left( \frac{dy}{dx} \right) + \sqrt{a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}}$.
This simplifies to $y = x \frac{d^{2}y}{dx^{2}} + \sqrt{a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}}$.
Rearranging the terms,we have $y - x \frac{d^{2}y}{dx^{2}} = \sqrt{a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}}$.
Squaring both sides,we get $\left( y - x \frac{d^{2}y}{dx^{2}} \right)^{2} = a^{2} \left( \frac{dy}{dx} \right)^{2} + b^{2}$.
The highest order derivative present is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
The power of the highest order derivative after rationalizing is $2$,so the degree is $2$.
Thus,the order and degree are $2$ and $2$ respectively.

(B) The given differential equation is $2x \frac{dy}{dx} - y = 3$.
Rearranging the terms,we get $2x \frac{dy}{dx} = y + 3$.
Separating the variables,we have $\frac{dy}{y+3} = \frac{dx}{2x}$.
Integrating both sides,we get $\int \frac{dy}{y+3} = \frac{1}{2} \int \frac{dx}{x}$.
This gives $\ln|y+3| = \frac{1}{2} \ln|x| + C_1$.
Multiplying by $2$,we get $2 \ln|y+3| = \ln|x| + 2C_1$.
Using properties of logarithms,$\ln(y+3)^2 = \ln|x| + \ln|c|$,where $c = e^{2C_1}$.
Thus,$(y+3)^2 = cx$,which is the equation of a family of parabolas.

(C) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are unit vectors,so $|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$.
Given $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$.
Squaring both sides:
$(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) \cdot (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}) = 0$.
$|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0$.
$1+1+1+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0$.
$3+2(\overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}) = 0 \implies \overrightarrow{a} \cdot \overrightarrow{b}+\overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a} = -3/2$.
Since $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,the vectors form an equilateral triangle,so the angle between any two vectors is $120^{\circ}$.
Thus,$\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a} = 1 \times 1 \times \cos(120^{\circ}) = -1/2$.
Substituting these values:
$3(-1/2)+2(-1/2)+(-1/2) = -3/2 - 2/2 - 1/2 = -6/2 = -3$.

(C) Given that $\hat{i}, \hat{j}, \hat{k}$ are unit vectors along the positive $x, y, z$-axes respectively.
$(a)$ $\sum \hat{i} \times(\hat{j}+\hat{k}) = \hat{i} \times(\hat{j}+\hat{k}) + \hat{j} \times(\hat{k}+\hat{i}) + \hat{k} \times(\hat{i}+\hat{j})$
$= (\hat{k} - \hat{j}) + (\hat{i} - \hat{k}) + (\hat{j} - \hat{i}) = \vec{0}$. This is true.
$(b)$ $\sum \hat{i} \times(\hat{j} \times \hat{k}) = \hat{i} \times \hat{i} + \hat{j} \times \hat{j} + \hat{k} \times \hat{k} = \vec{0} + \vec{0} + \vec{0} = \vec{0}$. This is true.
$(c)$ $\sum \hat{i} \cdot(\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{k} = 1 + 1 + 1 = 3$. This is false as it is not equal to $\vec{0}$.
$(d)$ $\sum \hat{i} \cdot(\hat{j}+\hat{k}) = (\hat{i} \cdot \hat{j} + \hat{i} \cdot \hat{k}) + (\hat{j} \cdot \hat{k} + \hat{j} \cdot \hat{i}) + (\hat{k} \cdot \hat{i} + \hat{k} \cdot \hat{j}) = 0 + 0 + 0 = 0$. This is true.

(B) Given that $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are non-zero coplanar vectors,their scalar triple product is zero,i.e.,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$.
We need to evaluate the scalar triple product $[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}]$.
Using the definition of the scalar triple product:
$[2 \overrightarrow{a}-\overrightarrow{b} \quad 3 \overrightarrow{b}-\overrightarrow{c} \quad 4 \overrightarrow{c}-\overrightarrow{a}] = (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [(3 \overrightarrow{b}-\overrightarrow{c}) \times (4 \overrightarrow{c}-\overrightarrow{a})]$
$= (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [12(\overrightarrow{b} \times \overrightarrow{c}) - 3(\overrightarrow{b} \times \overrightarrow{a}) - 4(\overrightarrow{c} \times \overrightarrow{c}) + (\overrightarrow{c} \times \overrightarrow{a})]$
Since $\overrightarrow{c} \times \overrightarrow{c} = 0$,the expression simplifies to:
$= (2 \overrightarrow{a}-\overrightarrow{b}) \cdot [12(\overrightarrow{b} \times \overrightarrow{c}) - 3(\overrightarrow{b} \times \overrightarrow{a}) + (\overrightarrow{c} \times \overrightarrow{a})]$
$= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - 6[\overrightarrow{a} \overrightarrow{b} \overrightarrow{a}] + 2[\overrightarrow{a} \overrightarrow{c} \overrightarrow{a}] - 12[\overrightarrow{b} \overrightarrow{b} \overrightarrow{c}] + 3[\overrightarrow{b} \overrightarrow{b} \overrightarrow{a}] - [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$
Since any scalar triple product with two identical vectors is zero,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{a}] = 0, [\overrightarrow{a} \overrightarrow{c} \overrightarrow{a}] = 0, [\overrightarrow{b} \overrightarrow{b} \overrightarrow{c}] = 0, [\overrightarrow{b} \overrightarrow{b} \overrightarrow{a}] = 0$.
Thus,the expression becomes:
$= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$
Since $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = [\overrightarrow{b} \overrightarrow{c} \overrightarrow{a}]$,we have:
$= 24[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 23[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$
Since $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are coplanar,$[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = 0$.
Therefore,$23 \times 0 = 0$.

(A) We know that for a space vector making angles $\alpha, \beta$,and $\gamma$ with the positive direction of the $x, y$,and $z$-axes respectively,the direction cosines satisfy the relation:
$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$
Given that $\alpha = 150^{\circ}$ and $\beta = 60^{\circ}$.
Substituting these values into the equation:
$\cos^{2} 150^{\circ} + \cos^{2} 60^{\circ} + \cos^{2} \gamma = 1$
Since $\cos 150^{\circ} = -\frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$,we have:
$(-\frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} + \cos^{2} \gamma = 1$
$\frac{3}{4} + \frac{1}{4} + \cos^{2} \gamma = 1$
$1 + \cos^{2} \gamma = 1$
$\cos^{2} \gamma = 0$
$\cos \gamma = 0$
Therefore,$\gamma = 90^{\circ}$.