AP EAMCET 2021 Mathematics Question Paper with Answer and Solution

(A) Let the points be $A(3,-2,2)$ and $B(6,-17,-4)$. Let $P(2,3,4)$ divide the line segment $AB$ in the ratio $\lambda : 1$.
Using the section formula for the $x$-coordinate:
$2 = \frac{6\lambda + 3}{\lambda + 1}$
$2\lambda + 2 = 6\lambda + 3$
$-4\lambda = 1 \implies \lambda = -\frac{1}{4}$.
The harmonic conjugate $Q$ of $P$ with respect to $A$ and $B$ divides $AB$ externally in the same ratio $\lambda : 1$,which is equivalent to dividing $AB$ internally in the ratio $-\lambda : 1$.
Thus,$Q$ divides $AB$ in the ratio $\frac{1}{4} : 1$,or $1 : 4$.
Using the section formula for $Q$ with ratio $m:n = 1:4$:
$Q = \left( \frac{1(6) + 4(3)}{1+4}, \frac{1(-17) + 4(-2)}{1+4}, \frac{1(-4) + 4(2)}{1+4} \right)$
$Q = \left( \frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5} \right)$
$Q = \left( \frac{18}{5}, -\frac{25}{5}, \frac{4}{5} \right) = \left( \frac{18}{5}, -5, \frac{4}{5} \right)$.

(D) Let the coordinates of the vertices of $\triangle OAB$ be $O(0,0)$,$A(a, 0)$,and $B(0, b)$.
Since the length of the rod $AB$ is $4$,we have $a^2 + b^2 = 4^2 = 16$.
Let $(x, y)$ be the centroid of $\triangle OAB$.
Then,$x = \frac{0+a+0}{3} = \frac{a}{3} \implies a = 3x$.
And $y = \frac{0+0+b}{3} = \frac{b}{3} \implies b = 3y$.
Substituting these into the equation $a^2 + b^2 = 16$,we get $(3x)^2 + (3y)^2 = 16$.
$9x^2 + 9y^2 = 16$.
$x^2 + y^2 = \frac{16}{9}$.
Thus,the locus of the centroid is $x^2 + y^2 = \frac{16}{9}$.

(D) The given equation is $x^2+2 \sqrt{3} xy - y^2 = 8$.
When the axes are rotated through an angle $\theta = \frac{\pi}{3}$,the coordinates $(x, y)$ are replaced by $(X, Y)$ where:
$x = X \cos \frac{\pi}{3} - Y \sin \frac{\pi}{3} = \frac{X - \sqrt{3}Y}{2}$
$y = X \sin \frac{\pi}{3} + Y \cos \frac{\pi}{3} = \frac{\sqrt{3}X + Y}{2}$
Substituting these into the equation:
$\left(\frac{X - \sqrt{3}Y}{2}\right)^2 + 2 \sqrt{3} \left(\frac{X - \sqrt{3}Y}{2}\right) \left(\frac{\sqrt{3}X + Y}{2}\right) - \left(\frac{\sqrt{3}X + Y}{2}\right)^2 = 8$
$\frac{1}{4} [ (X^2 - 2\sqrt{3}XY + 3Y^2) + 2\sqrt{3}(\sqrt{3}X^2 + XY - 3XY - \sqrt{3}Y^2) - (3X^2 + 2\sqrt{3}XY + Y^2) ] = 8$
$\frac{1}{4} [ X^2 - 2\sqrt{3}XY + 3Y^2 + 6X^2 - 4\sqrt{3}XY - 6Y^2 - 3X^2 - 2\sqrt{3}XY - Y^2 ] = 8$
$\frac{1}{4} [ 4X^2 - 8\sqrt{3}XY - 4Y^2 ] = 8$
$X^2 - 2\sqrt{3}XY - Y^2 = 8$
Thus,the transformed equation is $x^2 - 2\sqrt{3}xy - y^2 = 8$.

(B) Let the given point be $P(4, -5)$ and the fixed point be $Q(1, 3)$.
The distance $d$ between $P$ and $Q$ is given by the distance formula:
$d = \sqrt{(4 - 1)^2 + (-5 - 3)^2} = \sqrt{3^2 + (-8)^2} = \sqrt{9 + 64} = \sqrt{73}$.
Since $\sqrt{73} \approx 8.54$,the distance between the point $P$ and the point $Q$ is less than $10$ units.
For a line to be at a distance of $10$ units from point $Q$,the point $P$ must be at a distance of at least $10$ units from $Q$ (if $P$ lies on the line) or the line must be tangent to a circle of radius $10$ centered at $Q$.
Since the distance $PQ < 10$,the point $P$ lies inside the circle of radius $10$ centered at $Q$.
Any line passing through a point inside a circle must intersect the circle at two points,meaning the distance from the center $Q$ to any line passing through $P$ will always be less than the radius $10$.
Therefore,it is impossible to draw a line through $P$ such that its distance from $Q$ is $10$ units.
The number of such lines is $0$.

(D) Let the vertices be $A(1, \sqrt{3})$,$B(-1, -\sqrt{3})$,and $C(3, -\sqrt{3})$.
Calculate the side lengths:
$AB = \sqrt{(-1-1)^2 + (-\sqrt{3}-\sqrt{3})^2} = \sqrt{(-2)^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$.
$BC = \sqrt{(3 - (-1))^2 + (-\sqrt{3} - (-\sqrt{3}))^2} = \sqrt{4^2 + 0^2} = 4$.
$AC = \sqrt{(3-1)^2 + (-\sqrt{3}-\sqrt{3})^2} = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = 4$.
Since $AB = BC = AC = 4$,the triangle is an equilateral triangle.
In an equilateral triangle,the circumcentre coincides with the centroid.
The centroid $(G)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.
$G = \left(\frac{1-1+3}{3}, \frac{\sqrt{3}-\sqrt{3}-\sqrt{3}}{3}\right) = \left(\frac{3}{3}, \frac{-\sqrt{3}}{3}\right) = \left(1, -\frac{1}{\sqrt{3}}\right)$.

(D) Let $O$ be the orthocenter $(3, -4, 2)$ and $C$ be the circumcenter $(2, 1, 3)$.
We know that the centroid $G$ divides the line segment joining the orthocenter and the circumcenter in the ratio $2:1$.
Using the section formula,the coordinates of the centroid $G$ are given by:
$G = \left(\frac{1(3) + 2(2)}{1+2}, \frac{1(-4) + 2(1)}{1+2}, \frac{1(2) + 2(3)}{1+2}\right)$
$G = \left(\frac{3+4}{3}, \frac{-4+2}{3}, \frac{2+6}{3}\right)$
$G = \left(\frac{7}{3}, \frac{-2}{3}, \frac{8}{3}\right)$

(C) Given the equation: $\tan A + \tan B + \tan C = \cot A + \cot B + \cot C$.
We know that for any triangle,$A + B + C = 180^{\circ}$.
If we assume the triangle is equilateral,$A = B = C = 60^{\circ}$,then $\tan 60^{\circ} = \sqrt{3}$ and $\cot 60^{\circ} = \frac{1}{\sqrt{3}}$.
$3\sqrt{3} \neq \frac{3}{\sqrt{3}} = \sqrt{3}$.
For a general triangle,the condition $\tan A + \tan B + \tan C = \cot A + \cot B + \cot C$ implies $\tan A + \tan B + \tan C = \frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C}$.
This equation is satisfied if $A=B=C=60^{\circ}$ is not true,but rather if the angles satisfy specific constraints.
However,in a triangle,$\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
Also,$\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$.
It can be shown that for any triangle,the given condition leads to a contradiction or is impossible for real angles $A, B, C$ summing to $180^{\circ}$ except in degenerate cases.
Thus,the number of such triangles is $0$.

(C) Let the point be $P(x, y)$ which is equidistant from the points $A(2, 3)$ and $B(4, 5)$.
By the distance formula,$PA = PB$,so $PA^2 = PB^2$.
$(x-2)^2 + (y-3)^2 = (x-4)^2 + (y-5)^2$
Expanding both sides:
$x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 10y + 25$
Canceling $x^2$ and $y^2$ from both sides:
$-4x - 6y + 13 = -8x - 10y + 41$
Rearranging the terms:
$8x - 4x + 10y - 6y = 41 - 13$
$4x + 4y = 28$
Dividing by $4$:
$x + y = 7$

(D) Given equation is $x^2+y^2-6x+10y-2=0$.
Completing the square for $x$ and $y$ terms:
$(x^2-6x+9) + (y^2+10y+25) - 9 - 25 - 2 = 0$.
$(x-3)^2 + (y+5)^2 = 36$.
To transform the origin to $(3, -5)$,we substitute $x = X+3$ and $y = Y-5$,which implies $X = x-3$ and $Y = y+5$.
Substituting these into the equation,we get $X^2 + Y^2 = 36$.
Thus,the new equation is $x^2+y^2=36$.

(B) When the origin is shifted to the point $(h, k) = (-1, 1)$,the new coordinates $(X, Y)$ are related to the old coordinates $(x, y)$ by the equations:
$x = X + h = X - 1$
$y = Y + k = Y + 1$
Substituting these into the given equation $3x^2 + y^2 + 2x + 4y + 15 = 0$:
$3(X - 1)^2 + (Y + 1)^2 + 2(X - 1) + 4(Y + 1) + 15 = 0$
$3(X^2 - 2X + 1) + (Y^2 + 2Y + 1) + 2X - 2 + 4Y + 4 + 15 = 0$
$3X^2 - 6X + 3 + Y^2 + 2Y + 1 + 2X - 2 + 4Y + 4 + 15 = 0$
$3X^2 + Y^2 - 4X + 6Y + 21 = 0$
Thus,the transformed equation is $3x^2 + y^2 - 4x + 6y + 21 = 0$.

(A) The given line is $5x - 2y = 10$.
Dividing both sides by $10$,we get $\frac{5x}{10} - \frac{2y}{10} = 1$,which simplifies to $\frac{x}{2} + \frac{y}{-5} = 1$.
Comparing this with the intercept form $\frac{x}{a} + \frac{y}{b} = 1$,the intercepts are $a = 2$ and $b = -5$.
The sum of the squares of the intercepts is $a^2 + b^2 = 2^2 + (-5)^2$.
$= 4 + 25 = 29$.

(D) The equation of a line cutting off equal intercepts on the coordinate axes is given by $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$.
Since the line passes through $(-5, 6)$,we substitute these coordinates into the equation:
$-5 + 6 = a$
$a = 1$
Substituting the value of $a$ back into the equation,we get $x + y = 1$.

(C) The equation of a line with slope $m$ and $y$-intercept $c$ is given by $y = mx + c$.
Given $c = 4$,the equation of the line is $y = mx + 4$,which can be rewritten as $mx - y + 4 = 0$.
The distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here,the point is the origin $(0, 0)$,so $x_1 = 0$ and $y_1 = 0$.
Substituting these values into the distance formula:
$d = \frac{|m(0) - (0) + 4|}{\sqrt{m^2 + (-1)^2}}$
$d = \frac{|4|}{\sqrt{m^2 + 1}}$
$d = \frac{4}{\sqrt{m^2 + 1}}$.

(B) The given line is $x \sec \theta + y \operatorname{cosec} \theta = a$,which can be written as $\frac{x}{\cos \theta} + \frac{y}{\sin \theta} = a$.
Any line perpendicular to $Ax + By + C = 0$ is of the form $Bx - Ay + k = 0$.
Thus,the line perpendicular to the given line is $x \operatorname{cosec} \theta - y \sec \theta + k = 0$,or $\frac{x}{\sin \theta} - \frac{y}{\cos \theta} = -k$.
Since this line passes through $(a \cos^3 \theta, a \sin^3 \theta)$,we substitute these coordinates:
$\frac{a \cos^3 \theta}{\sin \theta} - \frac{a \sin^3 \theta}{\cos \theta} = -k$
$\frac{a(\cos^4 \theta - \sin^4 \theta)}{\sin \theta \cos \theta} = -k$
Using $\cos^4 \theta - \sin^4 \theta = (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) = \cos 2 \theta$,we get:
$\frac{a \cos 2 \theta}{\sin \theta \cos \theta} = -k$
Substituting $-k$ back into the equation $\frac{x}{\sin \theta} - \frac{y}{\cos \theta} = -k$:
$\frac{x}{\sin \theta} - \frac{y}{\cos \theta} = \frac{a \cos 2 \theta}{\sin \theta \cos \theta}$
Multiplying by $\sin \theta \cos \theta$ gives:
$x \cos \theta - y \sin \theta = a \cos 2 \theta$.

(A) The given line is $x \cos \theta - y \sin \theta = 2 \cos 2 \theta$.
Comparing this with $y = mx + c$,the slope $m_1 = \frac{\cos \theta}{\sin \theta}$.
The slope of the line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{\sin \theta}{\cos \theta}$.
The equation of the line passing through $(2 \cos^3 \theta, 2 \sin^3 \theta)$ with slope $m_2$ is:
$y - 2 \sin^3 \theta = -\frac{\sin \theta}{\cos \theta} (x - 2 \cos^3 \theta)$
$y \cos \theta - 2 \sin^3 \theta \cos \theta = -x \sin \theta + 2 \cos^3 \theta \sin \theta$
$x \sin \theta + y \cos \theta = 2 \sin \theta \cos \theta (\cos^2 \theta + \sin^2 \theta)$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$x \sin \theta + y \cos \theta = 2 \sin \theta \cos \theta$
Dividing both sides by $\sin \theta \cos \theta$:
$\frac{x \sin \theta}{\sin \theta \cos \theta} + \frac{y \cos \theta}{\sin \theta \cos \theta} = \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$x \sec \theta + y \operatorname{cosec} \theta = 2$.

(B) The slope $m$ of the line is given by $m = \tan \theta$,where $\theta = 150^{\circ}$.
$m = \tan 150^{\circ} = \tan(180^{\circ} - 30^{\circ}) = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}}$.
The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
Substituting the point $(-1, -1)$ and $m = -\frac{1}{\sqrt{3}}$:
$y - (-1) = -\frac{1}{\sqrt{3}}(x - (-1))$
$y + 1 = -\frac{1}{\sqrt{3}}(x + 1)$
$\sqrt{3}(y + 1) = -(x + 1)$
$\sqrt{3}y + \sqrt{3} = -x - 1$
$x + \sqrt{3}y + \sqrt{3} + 1 = 0$
Thus,the equation is $x + \sqrt{3}y + (\sqrt{3} + 1) = 0$.

(C) The given equation of the line is $(a+2b)x + (a-3b)y = a-b$.
Rearranging the terms to group coefficients of $a$ and $b$:
$ax + 2bx + ay - 3by = a - b$
$a(x + y - 1) + b(2x - 3y + 1) = 0$
For this equation to hold for all values of $a$ and $b$,the coefficients must be zero:
$x + y - 1 = 0$ $(i)$
$2x - 3y + 1 = 0$ (ii)
From $(i)$,$y = 1 - x$. Substituting into (ii):
$2x - 3(1 - x) + 1 = 0$
$2x - 3 + 3x + 1 = 0$
$5x - 2 = 0 \Rightarrow x = \frac{2}{5}$
Substituting $x = \frac{2}{5}$ into $(i)$:
$\frac{2}{5} + y - 1 = 0 \Rightarrow y = 1 - \frac{2}{5} = \frac{3}{5}$
Thus,the fixed point is $\left(\frac{2}{5}, \frac{3}{5}\right)$.

(A) Let the equation of the line passing through $A(-5,-4)$ be $\frac{x+5}{\cos \theta} = \frac{y+4}{\sin \theta} = r$.
Any point on this line is $(-5+r\cos \theta, -4+r\sin \theta)$.
For point $B$ on $x+3y+2=0$:
$(-5+r_1\cos \theta) + 3(-4+r_1\sin \theta) + 2 = 0$ $\Rightarrow r_1(\cos \theta + 3\sin \theta) = 15$ $\Rightarrow \frac{15}{r_1} = \cos \theta + 3\sin \theta \dots (i)$.
For point $C$ on $2x+y+4=0$:
$2(-5+r_2\cos \theta) + (-4+r_2\sin \theta) + 4 = 0$ $\Rightarrow r_2(2\cos \theta + \sin \theta) = 10$ $\Rightarrow \frac{10}{r_2} = 2\cos \theta + \sin \theta \dots (ii)$.
For point $D$ on $x-y-5=0$:
$(-5+r_3\cos \theta) - (-4+r_3\sin \theta) - 5 = 0$ $\Rightarrow r_3(\cos \theta - \sin \theta) = 6$ $\Rightarrow \frac{6}{r_3} = \cos \theta - \sin \theta \dots (iii)$.
Given $\left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right)^2$,we have:
$(\cos \theta + 3\sin \theta)^2 + (2\cos \theta + \sin \theta)^2 = (\cos \theta - \sin \theta)^2$.
$(\cos^2 \theta + 9\sin^2 \theta + 6\sin \theta \cos \theta) + (4\cos^2 \theta + \sin^2 \theta + 4\cos \theta \sin \theta) = \cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta$.
$5\cos^2 \theta + 10\sin^2 \theta + 10\sin \theta \cos \theta = \cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta$.
$4\cos^2 \theta + 9\sin^2 \theta + 12\sin \theta \cos \theta = 0$.
$(2\cos \theta + 3\sin \theta)^2 = 0 \Rightarrow 2\cos \theta + 3\sin \theta = 0$.
$\tan \theta = -\frac{2}{3}$.
The equation of line $L$ is $y - (-4) = -\frac{2}{3}(x - (-5))$ $\Rightarrow 3y + 12 = -2x - 10$ $\Rightarrow 2x + 3y + 22 = 0$.

(B) The line $x/a - y/b = 1$ passes through $(8, 6)$,so $8/a - 6/b = 1$ . . . $(i)$.
The intercepts are $(a, 0)$ and $(0, -b)$. The area of the triangle formed with the axes is $1/2 \times |a| \times |-b| = 12$,so $|ab| = 24$,which means $ab = 24$ or $ab = -24$.
Case $1$: $b = 24/a$. Substituting into $(i)$: $8/a - 6/(24/a) = 1 \implies 8/a - a/4 = 1 \implies 32 - a^2 = 4a \implies a^2 + 4a - 32 = 0 \implies (a+8)(a-4) = 0$.
If $a = 4$,$b = 6$. The line is $x/4 - y/6 = 1 \implies 3x - 2y = 12 \implies 3x - 2y - 12 = 0$.
If $a = -8$,$b = -3$. The line is $x/(-8) - y/(-3) = 1 \implies -x/8 + y/3 = 1 \implies -3x + 8y = 24 \implies 3x - 8y + 24 = 0$.
Thus,the equations are $3x - 2y - 12 = 0$ and $3x - 8y + 24 = 0$.

(A) The equation of the line passing through $P(1, 2)$ with slope $m = \tan(45^{\circ}) = 1$ is given by $(y - 2) = 1(x - 1)$,which simplifies to $x - y + 1 = 0$.
To find the point $Q$,we solve the system of equations:
$x - y + 1 = 0 \Rightarrow y = x + 1$
$3x + 4y + 5 = 0$
Substituting $y = x + 1$ into the second equation:
$3x + 4(x + 1) + 5 = 0$
$3x + 4x + 4 + 5 = 0$
$7x + 9 = 0 \Rightarrow x = -\frac{9}{7}$
Then $y = -\frac{9}{7} + 1 = -\frac{2}{7}$.
So,$Q = \left(-\frac{9}{7}, -\frac{2}{7}\right)$.
The distance $PQ$ is calculated using the distance formula:
$PQ = \sqrt{\left(1 - (-\frac{9}{7})\right)^2 + \left(2 - (-\frac{2}{7})\right)^2}$
$PQ = \sqrt{\left(\frac{16}{7}\right)^2 + \left(\frac{16}{7}\right)^2}$
$PQ = \sqrt{2 \times \left(\frac{16}{7}\right)^2} = \frac{16\sqrt{2}}{7}$ units.

(B) The given equation of the line is $px - qy = r$.
Since the line intersects the $x$-axis at $(a, 0)$,we substitute $x = a$ and $y = 0$ into the equation:
$p(a) - q(0) = r$ $\Rightarrow pa = r$ $\Rightarrow a = \frac{r}{p}$.
Since the line intersects the $y$-axis at $(0, b)$,we substitute $x = 0$ and $y = b$ into the equation:
$p(0) - q(b) = r$ $\Rightarrow -qb = r$ $\Rightarrow b = -\frac{r}{q}$.
Therefore,the value of $(a + b)$ is:
$a + b = \frac{r}{p} - \frac{r}{q} = \frac{rq - rp}{pq} = \frac{r(q - p)}{pq}$.

(B) Let the point $P(h, k)$ lie on the line $3x + 5y = 15$.
Since $P$ is equidistant from the coordinate axes,we have $|h| = |k|$,which implies $h = k$ or $h = -k$.
Case $1$: If $h = k$,then $3h + 5h = 15$ $\Rightarrow 8h = 15$ $\Rightarrow h = \frac{15}{8}$. Thus,$P = (\frac{15}{8}, \frac{15}{8})$,which lies in the first quadrant.
Case $2$: If $h = -k$,then $3h + 5(-h) = 15$ $\Rightarrow -2h = 15$ $\Rightarrow h = -\frac{15}{2}$. Thus,$k = \frac{15}{2}$,and $P = (-\frac{15}{2}, \frac{15}{2})$,which lies in the second quadrant.
Therefore,$P$ lies either in the first or in the second quadrant.

(B) The person is standing at the intersection of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$.
Solving these equations,we multiply the first by $4$ and the second by $3$:
$8x - 12y + 16 = 0$
$9x + 12y - 15 = 0$
Adding these gives $17x + 1 = 0$,so $x = -\frac{1}{17}$.
Substituting $x$ into the first equation: $2(-\frac{1}{17}) - 3y + 4 = 0$ $\Rightarrow -\frac{2}{17} + 4 = 3y$ $\Rightarrow 3y = \frac{66}{17}$ $\Rightarrow y = \frac{22}{17}$.
The junction point is $A = (-\frac{1}{17}, \frac{22}{17})$.
To reach the path $6x - 7y + 8 = 0$ in the least time,the person must walk along the perpendicular line.
The slope of the given path is $m_1 = \frac{6}{7}$.
The slope of the perpendicular line is $m_2 = -\frac{1}{m_1} = -\frac{7}{6}$.
The equation of the line passing through $(-\frac{1}{17}, \frac{22}{17})$ with slope $-\frac{7}{6}$ is:
$y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$
$6(17y - 22) = -7(17x + 1)$
$102y - 132 = -119x - 7$
$119x + 102y - 125 = 0$.

(A) Given,the equations of the straight lines are:
$l_1: 3x + 4y + 9 = 0$ ...$(i)$
$l_2: x - 7y - 22 = 0$ ...(ii)
The slope of line $l_1$ is $m_1 = -\frac{3}{4}$.
The slope of line $l_2$ is $m_2 = -\frac{1}{-7} = \frac{1}{7}$.
Let $\theta$ be the angle between the lines $l_1$ and $l_2$. The formula for the angle between two lines is:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
Substituting the values of $m_1$ and $m_2$:
$\tan \theta = \left| \frac{-\frac{3}{4} - \frac{1}{7}}{1 + (-\frac{3}{4})(\frac{1}{7})} \right|$
$\tan \theta = \left| \frac{-\frac{21+4}{28}}{1 - \frac{3}{28}} \right| = \left| \frac{-\frac{25}{28}}{\frac{25}{28}} \right|$
$\tan \theta = |-1| = 1$
Since $\tan \theta = 1$,we have $\theta = \frac{\pi}{4}$.

(A) The given line is $2x + 3y = 6$,which has a slope $m_1 = -2/3$.
Since the pair of lines passing through the origin forms an isosceles right-angled triangle with this line,the lines must make an angle of $45^{\circ}$ with the given line.
Let the slope of the required lines be $m$. The angle between the lines is given by $\tan \theta = |(m - m_1) / (1 + m \cdot m_1)|$.
Setting $\theta = 45^{\circ}$ and $m_1 = -2/3$:
$1 = |(m - (-2/3)) / (1 + m(-2/3))| = |(3m + 2) / (3 - 2m)|$.
This gives two cases:
Case $1$: $(3m + 2) / (3 - 2m) = 1$ $\Rightarrow 3m + 2 = 3 - 2m$ $\Rightarrow 5m = 1$ $\Rightarrow m = 1/5$.
The equation of the line is $y = (1/5)x \Rightarrow x - 5y = 0$.
Case $2$: $(3m + 2) / (3 - 2m) = -1$ $\Rightarrow 3m + 2 = -3 + 2m$ $\Rightarrow m = -5$.
The equation of the line is $y = -5x \Rightarrow 5x + y = 0$.
Thus,the lines are $x - 5y = 0$ and $5x + y = 0$.

(C) Let the points be $A(2, k)$,$B(3, 7)$,$C(-2, 1)$,and $D(3, 0)$.
Since the line $AB$ is parallel to the line $CD$,their slopes must be equal,i.e.,$m_{AB} = m_{CD}$.
The slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.
$m_{AB} = \frac{7 - k}{3 - 2} = \frac{7 - k}{1} = 7 - k$.
$m_{CD} = \frac{0 - 1}{3 - (-2)} = \frac{-1}{3 + 2} = \frac{-1}{5}$.
Equating the slopes: $7 - k = \frac{-1}{5}$.
$35 - 5k = -1$.
$5k = 36$.
$k = \frac{36}{5}$.

(D) Let $ABCD$ be a square where $A \equiv (1,3)$ and $C \equiv (7,5)$.
Diagonal $AC$ has slope $m_{AC} = \frac{5-3}{7-1} = \frac{2}{6} = \frac{1}{3}$.
Let the slope of the side $AB$ be $m$.
Since the diagonal of a square makes an angle of $45^{\circ}$ with the sides,we have $\left| \frac{m - 1/3}{1 + m(1/3)} \right| = \tan 45^{\circ} = 1$.
$\left| \frac{3m-1}{3+m} \right| = 1$.
Case $1$: $\frac{3m-1}{3+m} = 1$ $\Rightarrow 3m-1 = 3+m$ $\Rightarrow 2m = 4$ $\Rightarrow m = 2$.
The equation of the side passing through $A(1,3)$ with slope $m=2$ is $y-3 = 2(x-1)$ $\Rightarrow y-3 = 2x-2$ $\Rightarrow 2x-y+1 = 0$.
Case $2$: $\frac{3m-1}{3+m} = -1$ $\Rightarrow 3m-1 = -3-m$ $\Rightarrow 4m = -2$ $\Rightarrow m = -1/2$.
The equation of the side passing through $A(1,3)$ with slope $m=-1/2$ is $y-3 = -1/2(x-1)$ $\Rightarrow 2y-6 = -x+1$ $\Rightarrow x+2y-7 = 0$.
Comparing with the options,$2x-y+1=0$ is present.

(A) Let $m_1$ be the slope of the line passing through $(3,2)$ and $m_2$ be the slope of $x-2y=3$.
$m_2 = \frac{-1}{-2} = \frac{1}{2}$.
The angle between the lines is $45^{\circ}$,so $\tan 45^{\circ} = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
$1 = \left| \frac{m_1 - 1/2}{1 + m_1/2} \right| = \left| \frac{2m_1 - 1}{2 + m_1} \right|$.
Case $1$: $\frac{2m_1 - 1}{2 + m_1} = 1$ $\Rightarrow 2m_1 - 1 = 2 + m_1$ $\Rightarrow m_1 = 3$.
The equation is $y - 2 = 3(x - 3)$ $\Rightarrow y - 2 = 3x - 9$ $\Rightarrow 3x - y = 7$.
Case $2$: $\frac{2m_1 - 1}{2 + m_1} = -1$ $\Rightarrow 2m_1 - 1 = -2 - m_1$ $\Rightarrow 3m_1 = -1$ $\Rightarrow m_1 = -1/3$.
The equation is $y - 2 = -\frac{1}{3}(x - 3)$ $\Rightarrow 3y - 6 = -x + 3$ $\Rightarrow x + 3y = 9$.
Thus,the equations are $3x - y = 7$ and $x + 3y = 9$.

(C) The product of the perpendiculars from the origin $(0,0)$ to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $p = \left| \frac{c}{\sqrt{(a-b)^2 + 4h^2}} \right|$.
For $xy+x+y+1=0$:
$a=0, b=0, h=1/2, c=1$.
$p_1 = \left| \frac{1}{\sqrt{(0-0)^2 + 4(1/2)^2}} \right| = \left| \frac{1}{\sqrt{1}} \right| = 1$.
For $x^2-y^2+2x+1=0$:
$a=1, b=-1, h=0, c=1$.
$p_2 = \left| \frac{1}{\sqrt{(1-(-1))^2 + 4(0)^2}} \right| = \left| \frac{1}{\sqrt{4}} \right| = 1/2$.
For $2x^2+3xy-2y^2+2x+1=0$:
$a=2, b=-2, h=3/2, c=1$.
$p_3 = \left| \frac{1}{\sqrt{(2-(-2))^2 + 4(3/2)^2}} \right| = \left| \frac{1}{\sqrt{16 + 9}} \right| = \frac{1}{\sqrt{25}} = 1/5$.
Comparing the values: $1/5 < 1/2 < 1$,which means $p_3 < p_2 < p_1$.

(B) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y + 1 = 0$,and $L_3: ax + by - 1 = 0$. The orthocenter is at $(0, 0)$.
Since the orthocenter is the intersection of altitudes,the altitude from vertex $B$ (intersection of $L_1$ and $L_2$) passes through $(0, 0)$ and is perpendicular to $L_3$. The family of lines through $B$ is $(2x + 3y - 1) + \lambda(x + 2y + 1) = 0$. Since it passes through $(0, 0)$,we have $-1 + \lambda = 0 \Rightarrow \lambda = 1$. The altitude is $3x + 5y = 0$,with slope $m = -\frac{3}{5}$. Since this is perpendicular to $L_3$ (slope $-\frac{a}{b}$),we have $(-\frac{a}{b}) \times (-\frac{3}{5}) = -1 \Rightarrow 3a = -5b$.
Similarly,the altitude from vertex $A$ (intersection of $L_1$ and $L_3$) passes through $(0, 0)$ and is perpendicular to $L_2$ (slope $-\frac{1}{2}$). The family of lines through $A$ is $(2x + 3y - 1) + \mu(ax + by - 1) = 0$. Since it passes through $(0, 0)$,we have $-1 - \mu = 0 \Rightarrow \mu = -1$. The altitude is $(2-a)x + (3-b)y = 0$,with slope $-\frac{2-a}{3-b}$. Since this is perpendicular to $L_2$,we have $(-\frac{2-a}{3-b}) \times (-\frac{1}{2}) = -1$ $\Rightarrow 2-a = -2(3-b)$ $\Rightarrow a + 2b = 8$.
Solving $3a = -5b$ and $a + 2b = 8$,we get $a = -40$ and $b = 24$. Wait,checking the calculation: $a = -40, b = 24 \Rightarrow 3(-40) = -120, -5(24) = -120$ (Correct). $a + 2b = -40 + 48 = 8$ (Correct). Thus,$\frac{1}{a} + \frac{1}{b} = -\frac{1}{40} + \frac{1}{24} = \frac{-3 + 5}{120} = \frac{2}{120} = \frac{1}{60}$.

(A) Let the points be $A(2, 3, 4)$,$B(-1, -2, 1)$,and $C(5, 8, 7)$.
We check if the points are collinear by calculating the distance between them.
$AB = \sqrt{(-1-2)^2 + (-2-3)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-5)^2 + (-3)^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$.
$BC = \sqrt{(5-(-1))^2 + (8-(-2))^2 + (7-1)^2} = \sqrt{6^2 + 10^2 + 6^2} = \sqrt{36 + 100 + 36} = \sqrt{172} = 2\sqrt{43}$.
$AC = \sqrt{(5-2)^2 + (8-3)^2 + (7-4)^2} = \sqrt{3^2 + 5^2 + 3^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$.
Since $AB + AC = \sqrt{43} + \sqrt{43} = 2\sqrt{43} = BC$,the points $A, B, C$ are collinear.

(C) The equation $|x+y|=4$ represents two parallel lines: $x+y=4$ and $x+y=-4$.
For a point $(a, a)$ to lie between these two lines,the expression $(x+y)$ evaluated at $(a, a)$ must lie between the values $-4$ and $4$.
Substituting $(a, a)$ into the expression $x+y$,we get $a+a = 2a$.
Thus,the condition is $-4 < 2a < 4$.
Dividing the inequality by $2$,we get $-2 < a < 2$.
This is equivalent to $|a| < 2$.

(C) Let $\triangle ABC$ be the equilateral triangle with base $BC$ given by $x+y-2=0$ and vertex $A$ as $(2,-1)$.
The length of the altitude $h$ from vertex $A$ to the base $BC$ is the perpendicular distance from $(2,-1)$ to $x+y-2=0$.
$h = \frac{|(1)(2) + (1)(-1) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|2 - 1 - 2|}{\sqrt{2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
In an equilateral triangle with side length $s$,the altitude $h$ is given by $h = \frac{\sqrt{3}}{2}s$.
Therefore,$\frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2}s$.
$s = \frac{2}{\sqrt{2} \times \sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}$.

(C) Let the equation of the line parallel to $3x - y = 7$ be $3x - y = \lambda$.
Since it passes through $(1, 2)$,this point must satisfy the equation $3x - y = \lambda$.
$\therefore 3(1) - 2 = \lambda \Rightarrow \lambda = 1$.
So,the line is $3x - y = 1$.
Now,we find the intersection of $x + y + 5 = 0$ and $3x - y = 1$.
Adding the two equations: $(x + y + 5) + (3x - y) = 0 + 1$ $\Rightarrow 4x + 5 = 1$ $\Rightarrow 4x = -4$ $\Rightarrow x = -1$.
Substituting $x = -1$ into $x + y + 5 = 0$: $-1 + y + 5 = 0 \Rightarrow y = -4$.
So,the point of intersection is $(-1, -4)$.
The required distance is the distance between $(1, 2)$ and $(-1, -4)$.
Using the distance formula: $d = \sqrt{(1 - (-1))^2 + (2 - (-4))^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40}$.

(C) Let the image of the point $(3, -4)$ with respect to the line $4x - y - 1 = 0$ be $P(\alpha, \beta)$.
Using the formula for the image of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$ is $\frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$.
Substituting the values,we get:
$\frac{\alpha - 3}{4} = \frac{\beta + 4}{-1} = \frac{-2(4(3) - (-4) - 1)}{4^2 + (-1)^2}$.
$\frac{\alpha - 3}{4} = \frac{\beta + 4}{-1} = \frac{-2(12 + 4 - 1)}{16 + 1} = \frac{-2(15)}{17} = \frac{-30}{17}$.
Now,solving for $\alpha$:
$\frac{\alpha - 3}{4} = \frac{-30}{17}$ $\Rightarrow \alpha - 3 = \frac{-120}{17}$ $\Rightarrow \alpha = 3 - \frac{120}{17} = \frac{51 - 120}{17} = \frac{-69}{17}$.
Solving for $\beta$:
$\frac{\beta + 4}{-1} = \frac{-30}{17}$ $\Rightarrow \beta + 4 = \frac{30}{17}$ $\Rightarrow \beta = \frac{30}{17} - 4 = \frac{30 - 68}{17} = \frac{-38}{17}$.
Finally,calculating $\beta - \alpha$:
$\beta - \alpha = \frac{-38}{17} - (\frac{-69}{17}) = \frac{-38 + 69}{17} = \frac{31}{17}$.

(C) The given equation of the pair of lines is $15 x^2 + 31 x y + 14 y^2 = 0$.
Factoring the quadratic expression: $15 x^2 + 10 x y + 21 x y + 14 y^2 = 0$ $\Rightarrow 5 x(3 x + 2 y) + 7 y(3 x + 2 y) = 0$ $\Rightarrow (3 x + 2 y)(5 x + 7 y) = 0$.
Thus,the two lines are $L_1: 3 x + 2 y = 0$ and $L_2: 5 x + 7 y = 0$.
The perpendicular distance from $(2,3)$ to $3 x + 2 y = 0$ is $d_1 = \frac{|3(2) + 2(3)|}{\sqrt{3^2 + 2^2}} = \frac{12}{\sqrt{13}}$.
The perpendicular distance from $(2,3)$ to $5 x + 7 y = 0$ is $d_2 = \frac{|5(2) + 7(3)|}{\sqrt{5^2 + 7^2}} = \frac{31}{\sqrt{74}}$.
Since $p_1 > p_2$,we have $p_1 = \frac{31}{\sqrt{74}}$ and $p_2 = \frac{12}{\sqrt{13}}$.
Now,calculating $p_1^2 + \frac{1}{74} - p_2^2 + \frac{1}{13} = \frac{31^2}{74} + \frac{1}{74} - \frac{12^2}{13} + \frac{1}{13} = \frac{961+1}{74} - \frac{144-1}{13} = \frac{962}{74} - \frac{143}{13} = 13 - 11 = 2$.

(C) Let $LM$ be the perpendicular bisector of $PQ$ at $R$.
The midpoint $R$ of $PQ$ is given by:
$R = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(\frac{k+1}{2}, \frac{7}{2}\right)$
The slope of $PQ$ is:
$m_{PQ} = \frac{3-4}{k-1} = \frac{-1}{k-1}$
Since $LM$ is perpendicular to $PQ$,the slope of $LM$ $(m_{LM})$ is:
$m_{LM} = -\frac{1}{m_{PQ}} = -\frac{1}{-1/(k-1)} = k-1$
The equation of the line $LM$ with slope $(k-1)$ and $y$-intercept $-4$ is:
$y = (k-1)x - 4$
Since the perpendicular bisector passes through the midpoint $R\left(\frac{k+1}{2}, \frac{7}{2}\right)$,it must satisfy the equation:
$\frac{7}{2} = (k-1)\left(\frac{k+1}{2}\right) - 4$
Multiply by $2$ on both sides:
$7 = (k-1)(k+1) - 8$
$7 = k^2 - 1 - 8$
$7 = k^2 - 9$
$k^2 = 16$
$k = \pm 4$
Thus,a possible value of $k$ is $-4$.

(B) To find the distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$,we first rewrite them in the form $ax + by + c = 0$.
Dividing the second equation by $2$,we get $3x + 4y = 7.5$ or $3x + 4y - 7.5 = 0$.
The first equation is $3x + 4y - 9 = 0$.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \left| \frac{c_1 - c_2}{\sqrt{a^2 + b^2}} \right|$.
Here,$a = 3$,$b = 4$,$c_1 = -9$,and $c_2 = -7.5$.
$d = \left| \frac{-9 - (-7.5)}{\sqrt{3^2 + 4^2}} \right| = \left| \frac{-1.5}{\sqrt{9 + 16}} \right| = \left| \frac{-1.5}{5} \right| = \frac{1.5}{5} = \frac{3}{10} \text{ units}$.

(B) The given equation is $8x^2+8xy+2y^2+26x+13y+15=0$ ...$(i)$
We can rewrite the equation as $2(4x^2+4xy+y^2)+13(2x+y)+15=0$.
Let $2x+y=t$. Then the equation becomes $2t^2+13t+15=0$.
Factoring the quadratic: $2t^2+10t+3t+15=0 \Rightarrow 2t(t+5)+3(t+5)=0$.
So,$(2t+3)(t+5)=0$.
Substituting $t=2x+y$ back,we get $(2(2x+y)+3)(2x+y+5)=0$.
This simplifies to $(4x+2y+3)(2x+y+5)=0$.
Thus,the two lines are $4x+2y+3=0$ and $2x+y+5=0$.
Dividing the first line by $2$,we get $2x+y+1.5=0$.
The distance $d$ between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$A=2, B=1, C_1=5, C_2=1.5$.
$d = \frac{|5-1.5|}{\sqrt{2^2+1^2}} = \frac{3.5}{\sqrt{5}} = \frac{7}{2\sqrt{5}}$.

(B) The given equation is $x^2+2xy+y^2-8ax-8ay-9a^2=0$.
Comparing this with the general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=1, B=1, H=1, G=-4a, F=-4a, C=-9a^2$.
The distance $d$ between two parallel lines represented by $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ is given by the formula $d = 2 \sqrt{\frac{G^2-AC}{A(A+B)}}$.
Substituting the values:
$d = 2 \sqrt{\frac{(-4a)^2 - (1)(-9a^2)}{1(1+1)}}$
$d = 2 \sqrt{\frac{16a^2 + 9a^2}{2}}$
$d = 2 \sqrt{\frac{25a^2}{2}} = 2 \times \frac{5a}{\sqrt{2}} = 5\sqrt{2}a$.
Thus,the distance is $5\sqrt{2}a$ units.

(A) Given that $\triangle ABC$ is an isosceles triangle with base $BC$ where $B \equiv (3, 2)$ and $C \equiv (2, 3)$.
Note that the points $B(3, 2)$ and $C(2, 3)$ are symmetric with respect to the line $y = x$.
Since the triangle is isosceles with base $BC$,the vertex $A$ must lie on the perpendicular bisector of $BC$. The perpendicular bisector of the segment joining $(3, 2)$ and $(2, 3)$ is the line $y = x$.
Because the entire configuration is symmetric about the line $y = x$,the line $AC$ is the reflection of the line $AB$ about the line $y = x$.
To find the reflection of the line $3y = 2x$ about $y = x$,we swap $x$ and $y$ in the equation.
Replacing $x$ with $y$ and $y$ with $x$,we get $3x = 2y$,which is $2y = 3x$.
Thus,the equation of the line $AC$ is $2y = 3x$.

(A) Let the given point be $P(1, 2)$ and the external point be $Q(3, 1)$.
We want to find the equation of a line passing through $P$ such that its perpendicular distance from $Q$ is maximized.
The perpendicular distance from a point $Q$ to a line passing through $P$ is always less than or equal to the distance $PQ$ itself.
The maximum distance is achieved when the line is perpendicular to the segment $PQ$ at point $P$.
Let the line be $L$. Since $L \perp PQ$,the slope of $L$ is the negative reciprocal of the slope of $PQ$.
Slope of $PQ = \frac{1 - 2}{3 - 1} = \frac{-1}{2}$.
Therefore,the slope of line $L$ is $m = -(\frac{1}{-1/2}) = 2$.
The equation of line $L$ passing through $(1, 2)$ with slope $2$ is:
$y - 2 = 2(x - 1)$
$y - 2 = 2x - 2$
$y = 2x$.

(C) Let $P(\alpha, \beta)$ be the point on the line $2x + 3y + 7 = 0$.
Then $2\alpha + 3\beta + 7 = 0$,so $\beta = \frac{-7-2\alpha}{3}$.
The point $P$ is $\left(\alpha, \frac{-7-2\alpha}{3}\right)$.
The distance between $P(\alpha, \beta)$ and $A(1, -3)$ is $3$.
Using the distance formula: $(\alpha - 1)^2 + \left(\frac{-7-2\alpha}{3} + 3\right)^2 = 3^2$.
$(\alpha - 1)^2 + \left(\frac{-7-2\alpha+9}{3}\right)^2 = 9$.
$(\alpha - 1)^2 + \left(\frac{2-2\alpha}{3}\right)^2 = 9$.
$(\alpha - 1)^2 + \frac{4(1-\alpha)^2}{9} = 9$.
$(\alpha - 1)^2 \left(1 + \frac{4}{9}\right) = 9$.
$(\alpha - 1)^2 \left(\frac{13}{9}\right) = 9$.
$(\alpha - 1)^2 = \frac{81}{13}$.
$\alpha - 1 = \pm \frac{9}{\sqrt{13}}$.
$\alpha = 1 \pm \frac{9}{\sqrt{13}} = \frac{\sqrt{13} \pm 9}{\sqrt{13}}$.
For $\alpha = \frac{\sqrt{13}-9}{\sqrt{13}}$,$\beta = \frac{-7 - 2(\frac{\sqrt{13}-9}{\sqrt{13}})}{3} = \frac{-7\sqrt{13} - 2\sqrt{13} + 18}{3\sqrt{13}} = \frac{-9\sqrt{13} + 18}{3\sqrt{13}} = \frac{-3\sqrt{13} + 6}{\sqrt{13}}$.
Thus,the point is $\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3\sqrt{13}+6}{\sqrt{13}}\right)$.

(C) Let the origin be shifted to $(h, k)$.
Let $(x', y')$ be the new coordinates.
Then $x = x' + h$ and $y = y' + k$.
Substituting these into the given equation $9x^2 + 4y^2 + 10x + 12y + 1 = 0$:
$9(x' + h)^2 + 4(y' + k)^2 + 10(x' + h) + 12(y' + k) + 1 = 0$
Expanding the terms:
$9(x'^2 + 2x'h + h^2) + 4(y'^2 + 2y'k + k^2) + 10x' + 10h + 12y' + 12k + 1 = 0$
Grouping the $x'$ and $y'$ terms:
$9x'^2 + 4y'^2 + x'(18h + 10) + y'(8k + 12) + (9h^2 + 4k^2 + 10h + 12k + 1) = 0$
To eliminate the $x'$ and $y'$ terms,we set their coefficients to zero:
$18h + 10 = 0 \implies h = -\frac{10}{18} = -\frac{5}{9}$
$8k + 12 = 0 \implies k = -\frac{12}{8} = -\frac{3}{2}$
Thus,the origin should be shifted to $\left(-\frac{5}{9}, -\frac{3}{2}\right)$.

(D) Let the slopes of the lines represented by $2x^2 + axy + 3y^2 = 0$ be $m$ and $m_1$. Then $m + m_1 = -a/3$ and $mm_1 = 2/3$.
Let the slopes of the lines represented by $2x^2 + bxy - 3y^2 = 0$ be $m$ and $m_2$. Then $m + m_2 = -b/(-3) = b/3$ and $mm_2 = 2/(-3) = -2/3$.
Since the other two lines are perpendicular,$m_1m_2 = -1$.
Dividing the product of slopes: $(mm_1) / (mm_2) = (2/3) / (-2/3) = -1$.
Thus,$m_1 / m_2 = -1$,which implies $m_1 = -m_2$ or $m_1 + m_2 = 0$.
From $mm_1 = 2/3$ and $mm_2 = -2/3$,we have $m_1 = 2/(3m)$ and $m_2 = -2/(3m)$.
Substituting into $m_1m_2 = -1$: $(2/(3m)) \times (-2/(3m)) = -1$ $\Rightarrow -4/(9m^2) = -1$ $\Rightarrow m^2 = 4/9$ $\Rightarrow m = 2/3$ (taking $m = 2/3$).
Then $m_1 = (2/3) / (2/3) = 1$ and $m_2 = (-2/3) / (2/3) = -1$.
Using $m + m_1 = -a/3$: $2/3 + 1 = -a/3$ $\Rightarrow 5/3 = -a/3$ $\Rightarrow a = -5$.
Using $m + m_2 = b/3$: $2/3 - 1 = b/3$ $\Rightarrow -1/3 = b/3$ $\Rightarrow b = -1$.

(D) For the lines to be concurrent,the determinant of their coefficients must be zero:
$\left| \begin{array}{ccc} 2 & -1 & 1 \\ 4 & 1 & 2 \\ 1 & 1 & -k \end{array} \right| = 0$
Expanding along the first row:
$2(-k - 2) - (-1)(-4k - 2) + 1(4 - 1) = 0$
$-2k - 4 - 4k - 2 + 3 = 0$
$-6k - 3 = 0$
$-6k = 3$
$k = -\frac{1}{2}$

(A) The line $2x + 3y + 4 = 0$ is the perpendicular bisector of $AB$.
First,the midpoint $M$ of $AB$ lies on the line:
$M = (\frac{1+\alpha}{2}, \frac{2+\beta}{2})$
$2(\frac{1+\alpha}{2}) + 3(\frac{2+\beta}{2}) + 4 = 0$
$1 + \alpha + 3 + \frac{3\beta}{2} + 4 = 0$
$2\alpha + 3\beta + 16 = 0$ ... $(i)$
Second,the slope of $AB$ is perpendicular to the slope of the given line $(-2/3)$:
$\frac{\beta - 2}{\alpha - 1} \times (-\frac{2}{3}) = -1$
$2(\beta - 2) = 3(\alpha - 1)$
$2\beta - 4 = 3\alpha - 3$
$3\alpha - 2\beta + 1 = 0$ ... $(ii)$
Solving $(i)$ and $(ii)$:
Multiply $(i)$ by $2$ and $(ii)$ by $3$:
$4\alpha + 6\beta = -32$
$9\alpha - 6\beta = -3$
Adding them: $13\alpha = -35 \Rightarrow \alpha = -\frac{35}{13}$
Substituting $\alpha$ in $(ii)$: $3(-\frac{35}{13}) - 2\beta + 1 = 0
-\frac{105}{13} + 1 = 2\beta
-\frac{92}{13} = 2\beta$ $\Rightarrow \beta = -\frac{46}{13}$
Finally,$13\alpha + 13\beta = 13(-\frac{35}{13}) + 13(-\frac{46}{13}) = -35 - 46 = -81$.

(C) The given line is $ax + by + c = 0$.
Since $a, b,$ and $c$ are three consecutive odd integers,they are in arithmetic progression.
Thus,$b - a = c - b$,which implies $c = 2b - a$.
Substituting $c$ into the line equation: $ax + by + (2b - a) = 0$.
Rearranging the terms: $a(x - 1) + b(y + 2) = 0$.
For this line to pass through the fixed point $(\alpha, \beta)$ for all such $a$ and $b$,the coefficients must satisfy:
$\alpha - 1 = 0 \Rightarrow \alpha = 1$ and $\beta + 2 = 0 \Rightarrow \beta = -2$.
Therefore,the value of $\alpha^2 + \beta^2 = 1^2 + (-2)^2 = 1 + 4 = 5$.

(C) The slope of the side $AB$ is $m_{AB} = \frac{8-7}{-7-4} = \frac{1}{-11} = -\frac{1}{11}$.
Since the perpendicular bisector is perpendicular to $AB$,its slope $m$ is given by $m \times m_{AB} = -1$,so $m = 11$.
The midpoint $D$ of side $AB$ is $\left(\frac{4+(-7)}{2}, \frac{7+8}{2}\right) = \left(-\frac{3}{2}, \frac{15}{2}\right)$.
The equation of the line passing through $D\left(-\frac{3}{2}, \frac{15}{2}\right)$ with slope $m=11$ is:
$y - \frac{15}{2} = 11\left(x - (-\frac{3}{2}\right))$
$y - \frac{15}{2} = 11\left(x + \frac{3}{2}\right)$
$y - \frac{15}{2} = 11x + \frac{33}{2}$
$11x - y + \frac{33}{2} + \frac{15}{2} = 0$
$11x - y + \frac{48}{2} = 0$
$11x - y + 24 = 0$.

(D) Let the point be $(x, y)$.
The distance between $(x, y)$ and $(3, -2)$ is given as $4$ units.
Using the distance formula,$\sqrt{(x-3)^2 + (y-(-2))^2} = 4$.
Squaring both sides,we get $(x-3)^2 + (y+2)^2 = 16$.
Expanding the terms,$x^2 - 6x + 9 + y^2 + 4y + 4 = 16$.
Simplifying the equation,$x^2 + y^2 - 6x + 4y + 13 = 16$.
Thus,the locus is $x^2 + y^2 - 6x + 4y - 3 = 0$.

(A) Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.
Given, $r = 9 \ cm$ and $\Delta r = 0.03 \ cm$.
The surface area $S$ of a sphere is given by $S = 4 \pi r^2$.
To find the approximate error in the surface area, we differentiate $S$ with respect to $r$:
$\frac{dS}{dr} = 8 \pi r$.
Using the differential approximation, $\Delta S \approx \frac{dS}{dr} \times \Delta r$.
Substituting the values:
$\Delta S = 8 \pi \times 9 \times 0.03$.
$\Delta S = 72 \pi \times 0.03 = 2.16 \pi \ cm^2$.
Thus, the approximate error in calculating the surface area is $2.16 \pi \ cm^2$.

(D) Let $D$ be the diameter,$r$ be the radius,and $h$ be the altitude of the right circular cone.
Given $D = 10 \ cm$,so $r = 5 \ cm$.
Given $h = 20 \ cm$.
The rate of change of diameter is $\frac{dD}{dt} = 2 \ cm/s$.
Since $D = 2r$,we have $\frac{dr}{dt} = \frac{1}{2} \frac{dD}{dt} = \frac{1}{2} \times 2 = 1 \ cm/s$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h$.
To keep the volume constant,$\frac{dV}{dt} = 0$.
Using the product rule,$\frac{dV}{dt} = \frac{1}{3} \pi (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) = 0$.
Substituting the values: $2(5)(20)(1) + (5)^2 \frac{dh}{dt} = 0$.
$200 + 25 \frac{dh}{dt} = 0$.
$25 \frac{dh}{dt} = -200$.
$\frac{dh}{dt} = -8 \ cm/s$.
Thus,the altitude must change at a rate of $-8 \ cm/s$.

(D) Let the radius of the iron ball be $r_0 = 10 \ cm$ and the thickness of the ice layer be $x \ cm$. The total radius of the sphere (iron ball + ice) is $R = 10 + x \ cm$.
The volume of the ice layer is $V = \frac{4}{3}\pi (10+x)^3 - \frac{4}{3}\pi (10)^3$.
Since the iron ball is constant,the rate of change of the total volume is equal to the rate of change of the ice volume.
$V = \frac{4}{3}\pi (10+x)^3$.
Differentiating with respect to time $t$:
$\frac{dV}{dt} = 4\pi (10+x)^2 \frac{dx}{dt}$.
Given $\frac{dV}{dt} = -50 \ cm^3/min$ (since it melts) and $x = 5 \ cm$:
$-50 = 4\pi (10+5)^2 \frac{dx}{dt}$.
$-50 = 4\pi (15)^2 \frac{dx}{dt}$.
$-50 = 4\pi (225) \frac{dx}{dt}$.
$-50 = 900\pi \frac{dx}{dt}$.
$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \ cm/min$.
The negative sign indicates a decrease in thickness.
Therefore,the rate at which the thickness decreases is $\frac{1}{18\pi} \ cm/min$.
Thus,option $(D)$ is correct.

(C) Let $v$ be the volume and $s$ be the surface area of the spherical balloon with radius $r$.
Given: $\frac{dv}{dt} = 30 \ cm^3/min$.
The volume of a sphere is $v = \frac{4}{3} \pi r^3$.
Differentiating with respect to $t$: $\frac{dv}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given values: $30 = 4 \pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{30}{4 \pi r^2} = \frac{15}{2 \pi r^2}$.
The surface area of a sphere is $s = 4 \pi r^2$.
Differentiating with respect to $t$: $\frac{ds}{dt} = 8 \pi r \frac{dr}{dt}$.
Substituting $\frac{dr}{dt}$: $\frac{ds}{dt} = 8 \pi r \left( \frac{15}{2 \pi r^2} \right) = \frac{60}{r}$.
For $r = 6 \ cm$: $\frac{ds}{dt} = \frac{60}{6} = 10 \ cm^2/min$.

(A) Given,$T=2 \pi \sqrt{\frac{l}{g}}$.
Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.
Differentiating both sides with respect to $l$,we get $\frac{1}{T} \frac{dT}{dl} = \frac{1}{2l}$.
Thus,the relative change is $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.
Given that the length is increased by $1 \%$,we have $\frac{dl}{l} \times 100 = 1 \%$.
Therefore,the percentage change in the time period is $\frac{dT}{T} \times 100 = \frac{1}{2} \times \left( \frac{dl}{l} \times 100 \right) = \frac{1}{2} \times 1 \% = 0.5 \%$.
Hence,the approximate change in the time period is $0.5 \%$.

(B) Given,$\frac{dV}{dt} = 4 \pi \text{ cm}^3/\text{s}$ ...$(i)$
We need to find $\frac{dr}{dt}$ when $V = 288 \pi \text{ cm}^3$.
Let $r$ be the radius of the spherical ball.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Differentiating both sides with respect to $t$,we get:
$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Substituting the given rate $\frac{dV}{dt} = 4 \pi$:
$4 \pi = 4 \pi r^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{r^2}$ ...(ii)
When $V = 288 \pi$,we have:
$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow r^3 = 288 \times \frac{3}{4} = 216$.
Thus,$r = \sqrt[3]{216} = 6 \text{ cm}$.
Substituting $r = 6$ into equation (ii):
$\frac{dr}{dt} = \frac{1}{6^2} = \frac{1}{36} \text{ cm/s}$.

(C) Given the curve $y = 5x - 2x^3$.
The slope of the curve is given by $m = \frac{dy}{dx} = 5 - 6x^2$.
We need to find the rate of change of the slope with respect to time $t$,which is $\frac{dm}{dt}$.
Differentiating $m$ with respect to $t$ using the chain rule:
$\frac{dm}{dt} = \frac{d}{dt}(5 - 6x^2) = -12x \cdot \frac{dx}{dt}$.
Given that $\frac{dx}{dt} = 2 \text{ units/sec}$ and we need to evaluate this at $x = 3$:
$\left(\frac{dm}{dt}\right)_{x=3} = -12(3) \times 2 = -72$.
Thus,the rate of change of the slope at $x = 3$ is $-72 \text{ units/sec}^2$.

(D) Given the distance function $s = 60t - 5t^2$.
The velocity $v$ is the rate of change of distance with respect to time $t$,given by $v = \frac{ds}{dt}$.
$\frac{ds}{dt} = \frac{d}{dt}(60t - 5t^2) = 60 - 10t$.
The particle comes to rest when its velocity $v = 0$.
Setting $60 - 10t = 0$,we get $10t = 60$,which implies $t = 6 \text{ s}$.
Now,we calculate the distance covered at $t = 6 \text{ s}$:
$s(6) = 60(6) - 5(6)^2$
$s(6) = 360 - 5(36)$
$s(6) = 360 - 180 = 180 \text{ units}$.
Thus,the distance covered before coming to rest is $180 \text{ units}$.

(B) Given $y = f(x) = 5x^2 + 6x + 6$,$x = 2$,and $\Delta x = 0.001$.
First,calculate the differential $dy$:
$\frac{dy}{dx} = 10x + 6$
$dy = \left(\frac{dy}{dx}\right) \Delta x = (10x + 6) \Delta x$
Substituting $x = 2$ and $\Delta x = 0.001$:
$dy = (10(2) + 6)(0.001) = (26)(0.001) = 0.026$.
Next,calculate the increment $\Delta y$:
$\Delta y = f(x + \Delta x) - f(x)$
$\Delta y = [5(x + \Delta x)^2 + 6(x + \Delta x) + 6] - [5x^2 + 6x + 6]$
$\Delta y = 5(x^2 + 2x\Delta x + (\Delta x)^2) + 6x + 6\Delta x + 6 - 5x^2 - 6x - 6$
$\Delta y = 10x\Delta x + 5(\Delta x)^2 + 6\Delta x$
Substituting $x = 2$ and $\Delta x = 0.001$:
$\Delta y = 10(2)(0.001) + 5(0.001)^2 + 6(0.001)$
$\Delta y = 0.020 + 0.000005 + 0.006 = 0.026005$.
Thus,$\Delta y = 0.026005$ and $dy = 0.026$.

(A) Given curve is $y = 8x^2 - x^4 - 4$.
To find the stationary points,we differentiate the function with respect to $x$:
$\frac{dy}{dx} = 16x - 4x^3$.
Stationary points occur where $\frac{dy}{dx} = 0$:
$16x - 4x^3 = 0$
$4x(4 - x^2) = 0$
This gives $x = 0$ or $x^2 = 4$,which implies $x = 0, 2, -2$.
Now,we find the corresponding $y$-values:
For $x = 0$: $y = 8(0)^2 - (0)^4 - 4 = -4$.
For $x = 2$: $y = 8(2)^2 - (2)^4 - 4 = 8(4) - 16 - 4 = 32 - 20 = 12$.
For $x = -2$: $y = 8(-2)^2 - (-2)^4 - 4 = 8(4) - 16 - 4 = 32 - 20 = 12$.
Thus,the stationary points are $(0, -4), (2, 12), (-2, 12)$.

(A) $(i)$ Given $f(x) = x|x|$. For $x > 0$,$f(x) = x^2$,so $f'(x) = 2x > 0$. For $x < 0$,$f(x) = -x^2$,so $f'(x) = -2x > 0$. Thus,$f'(x) > 0$ for all $x \in R - \{0\}$,meaning $f(x)$ is strictly increasing. This statement is true.
$(ii)$ Given $f(x) = \log_{1/4} (x)$. Since the base $1/4 < 1$,the logarithmic function is strictly decreasing on $(0, \infty)$. This statement is false.
$(iii)$ $A$ one-one function can be strictly increasing,strictly decreasing,or neither (e.g.,$f(x) = 1/x$ is one-one but not monotonic). This statement is false.
$(iv)$ Given $f(x) = x^{1/3}$. $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} > 0$ for all $x \neq 0$. Thus,it is strictly increasing on $R$. This statement is false.

(D) Given the function $f(x) = -x^3 + 4ax^2 + 2x - 5$.
To find where the function is decreasing,we calculate the derivative $f'(x) = -3x^2 + 8ax + 2$.
For the function to be decreasing for every $x$,we must have $f'(x) < 0$ for all $x \in R$.
This implies $-3x^2 + 8ax + 2 < 0$ for all $x$.
For a quadratic expression $Ax^2 + Bx + C$ to be negative for all $x$,the conditions are $A < 0$ and the discriminant $\Delta = B^2 - 4AC < 0$.
Here,$A = -3$,which is less than $0$.
The discriminant is $\Delta = (8a)^2 - 4(-3)(2) = 64a^2 + 24$.
Since $64a^2 + 24$ is always positive for any real value of $a$,the condition $\Delta < 0$ can never be satisfied.
Therefore,there is no value of $a$ for which the function is decreasing for every $x$.

(D) Given that $f''(x) > 0$ for all $x \in R$,it implies that $f'(x)$ is a strictly increasing function for all $x \in R$.
Since $f'(3) = 0$,we have $f'(x) < 0$ for $x < 3$ and $f'(x) > 0$ for $x > 3$.
Now,consider $g(x) = f(\tan^2 x - 2 \tan x + 4)$.
Differentiating with respect to $x$,we get $g'(x) = f'(\tan^2 x - 2 \tan x + 4) \cdot \frac{d}{dx}(\tan^2 x - 2 \tan x + 4)$.
$g'(x) = f'(\tan^2 x - 2 \tan x + 4) \cdot (2 \tan x \sec^2 x - 2 \sec^2 x) = f'(\tan^2 x - 2 \tan x + 4) \cdot 2 \sec^2 x (\tan x - 1)$.
Let $u = \tan^2 x - 2 \tan x + 4 = (\tan x - 1)^2 + 3$. Since $(\tan x - 1)^2 \ge 0$,we have $u \ge 3$.
For $u > 3$,we know $f'(u) > 0$ because $f'(x)$ is increasing and $f'(3) = 0$.
For $g(x)$ to be increasing,we require $g'(x) > 0$.
Since $f'(u) > 0$ and $2 \sec^2 x > 0$ for $x \in (0, \frac{\pi}{2})$,the sign of $g'(x)$ depends on $(\tan x - 1)$.
Thus,$g'(x) > 0$ when $\tan x - 1 > 0$,which means $\tan x > 1$.
For $x \in (0, \frac{\pi}{2})$,$\tan x > 1$ implies $x \in (\frac{\pi}{4}, \frac{\pi}{2})$.

(A) Given $g(x) = \frac{1}{6} f(3 x^2 - 1) + \frac{1}{2} f(1 - x^2)$.
Taking the derivative with respect to $x$:
$g'(x) = \frac{1}{6} f'(3 x^2 - 1) \cdot (6x) + \frac{1}{2} f'(1 - x^2) \cdot (-2x)$
$g'(x) = x [f'(3 x^2 - 1) - f'(1 - x^2)]$.
Since $f''(x) > 0$,$f'(x)$ is an increasing function.
For $g(x)$ to be increasing,$g'(x) > 0$.
Case $1$: If $x > 0$,then we need $f'(3 x^2 - 1) - f'(1 - x^2) > 0$,which implies $f'(3 x^2 - 1) > f'(1 - x^2)$.
Since $f'$ is increasing,this means $3 x^2 - 1 > 1 - x^2$,so $4 x^2 > 2$,or $x^2 > \frac{1}{2}$.
Given $x > 0$,this gives $x \in \left( \frac{1}{\sqrt{2}}, \infty \right)$.
Case $2$: If $x < 0$,then we need $f'(3 x^2 - 1) - f'(1 - x^2) < 0$,which implies $f'(3 x^2 - 1) < f'(1 - x^2)$.
Since $f'$ is increasing,this means $3 x^2 - 1 < 1 - x^2$,so $4 x^2 < 2$,or $x^2 < \frac{1}{2}$.
Given $x < 0$,this gives $x \in \left( -\frac{1}{\sqrt{2}}, 0 \right)$.
Thus,$g(x)$ is increasing in $\left( -\frac{1}{\sqrt{2}}, 0 \right) \cup \left( \frac{1}{\sqrt{2}}, \infty \right)$.

(B) Given $f(x)=k x^3-9 x^2+9 x+3$ $(k>0)$ is increasing for all $x$.
Since $f(x)$ is increasing,its derivative $f^{\prime}(x) \geq 0$ for all $x$.
$f^{\prime}(x) = 3 k x^2-18 x+9$.
Setting $f^{\prime}(x) \geq 0$,we get $3 k x^2-18 x+9 \geq 0$,which simplifies to $k x^2-6 x+3 \geq 0$.
For a quadratic expression $a x^2+b x+c$ to be non-negative for all $x$ where $a>0$,the discriminant $D$ must be less than or equal to $0$.
Here $a=k$,$b=-6$,and $c=3$.
$D = b^2-4 a c = (-6)^2-4(k)(3) = 36-12 k$.
Setting $D \leq 0$,we have $36-12 k \leq 0$,which implies $12 k \geq 36$,so $k \geq 3$.
Thus,the correct condition is $k \geq 3$.

(B) Given the displacement function: $s = \frac{1}{3} t^3 - 3 t^2 + 9 t + 17$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{ds}{dt} = t^2 - 6 t + 9$.
To find when the velocity decreases,we calculate the acceleration $a = \frac{dv}{dt}$.
$a = \frac{dv}{dt} = 2 t - 6$.
Velocity decreases when the acceleration is negative,i.e.,$\frac{dv}{dt} < 0$.
$2 t - 6 < 0 \implies 2 t < 6 \implies t < 3$.
Since time $t$ must be greater than $0$,the velocity decreases in the interval $0 < t < 3$.

(B) Given,$a > 0$.
$f(x) = 2 x^3 - 9 a x^2 + 12 a^2 x + 1$.
Find the derivative: $f'(x) = 6 x^2 - 18 a x + 12 a^2$.
Factor the derivative: $f'(x) = 6 (x^2 - 3 a x + 2 a^2) = 6 (x - 2 a) (x - a)$.
Setting $f'(x) = 0$ gives critical points $x = a$ and $x = 2 a$.
Find the second derivative: $f''(x) = 6 (2 x - 3 a) = 12 x - 18 a$.
Check for local maxima and minima:
At $x = a$,$f''(a) = 6 (2 a - 3 a) = -6 a < 0$ (since $a > 0$),so $\alpha = a$ is the point of local maximum.
At $x = 2 a$,$f''(2 a) = 6 (4 a - 3 a) = 6 a > 0$ (since $a > 0$),so $\beta = 2 a$ is the point of local minimum.
Given the condition $2 \alpha + \beta = 8$,substitute $\alpha = a$ and $\beta = 2 a$:
$2(a) + 2 a = 8$.
$4 a = 8$.
$a = 2$.

(B) Given: $xy = 6$
$\implies y = \frac{6}{x}$
Let $f(x) = 2x + 3y = 2x + 3 \left( \frac{6}{x} \right) = 2x + \frac{18}{x}$
To find the minimum value,differentiate $f(x)$ with respect to $x$:
$f'(x) = 2 - \frac{18}{x^2}$
Set $f'(x) = 0$ for critical points:
$2 - \frac{18}{x^2} = 0 \implies x^2 = 9 \implies x = 3$ (assuming $x > 0$ for positive values)
Now,find the second derivative:
$f''(x) = \frac{36}{x^3}$
At $x = 3$,$f''(3) = \frac{36}{27} > 0$,which confirms a local minimum.
The minimum value is $f(3) = 2(3) + \frac{18}{3} = 6 + 6 = 12$.

(C) Given $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$.
Differentiating $f(x)$ with respect to $x$,we get $f'(x) = 6x^2 - 18ax + 12a^2$.
For maximum or minimum,we set $f'(x) = 0$:
$6(x^2 - 3ax + 2a^2) = 0$
$6(x - a)(x - 2a) = 0$
Thus,the critical points are $x = a$ and $x = 2a$.
Now,we find the second derivative $f''(x) = 12x - 18a$.
At $x = a$,$f''(a) = 12a - 18a = -6a$. Since $a > 0$,$f''(a) < 0$,so $f(x)$ has a maximum at $p = a$.
At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a$. Since $a > 0$,$f''(2a) > 0$,so $f(x)$ has a minimum at $q = 2a$.
Given the condition $p^2 = q$,we substitute the values:
$a^2 = 2a$
$a^2 - 2a = 0$
$a(a - 2) = 0$
Since $a > 0$,we must have $a = 2$.

(B) Let $f(x) = x^4-x^2-2x+5$.
Find the derivative: $f'(x) = 4x^3-2x-2$.
Set $f'(x) = 0$ to find critical points: $4x^3-2x-2 = 0 \Rightarrow 2x^3-x-1 = 0$.
Factoring the cubic equation: $2x^3-2x^2+2x^2-2x+x-1 = 0 \Rightarrow 2x^2(x-1) + 2x(x-1) + 1(x-1) = 0 \Rightarrow (x-1)(2x^2+2x+1) = 0$.
The real root is $x = 1$. The quadratic factor $2x^2+2x+1$ has a negative discriminant $(D = 4-8 = -4)$,so it yields no real roots.
Check the second derivative: $f''(x) = 12x^2-2$.
At $x = 1$,$f''(1) = 12(1)^2-2 = 10 > 0$.
Since $f''(1) > 0$,the function has a local minimum at $x = 1$.
As $x \to \pm \infty$,$f(x) \to \infty$,so the local minimum is the absolute minimum.
Calculating the value: $f(1) = (1)^4-(1)^2-2(1)+5 = 1-1-2+5 = 3$.
Thus,the absolute minimum value is $3$.

(B) Given,the perimeter of the circular sector is $P = 60 \ m$.
Let the radius be $r \ m$ and the arc length be $l \ m$.
The perimeter of a circular sector is given by $P = l + 2r$.
Substituting the given perimeter,we have $60 = l + 2r$,which implies $l = 60 - 2r$.
The area $A$ of a circular sector is given by $A = \frac{1}{2}lr$.
Substituting $l = 60 - 2r$ into the area formula,we get:
$A = \frac{1}{2}(60 - 2r)r = 30r - r^2$.
To maximize the area,we find the derivative of $A$ with respect to $r$ and set it to zero:
$\frac{dA}{dr} = \frac{d}{dr}(30r - r^2) = 30 - 2r$.
Setting $\frac{dA}{dr} = 0$ gives $30 - 2r = 0$,so $2r = 30$,which means $r = 15 \ m$.
To confirm this is a maximum,we check the second derivative: $\frac{d^2A}{dr^2} = -2$,which is less than $0$,confirming that $r = 15 \ m$ provides the maximum area.

(D) Given $y = x(\log x)^2$.
To find the critical points,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = x \cdot (2 \log x \cdot \frac{1}{x}) + (\log x)^2 \cdot 1 = 2 \log x + (\log x)^2$.
Setting $\frac{dy}{dx} = 0$:
$\log x (2 + \log x) = 0$.
This gives $\log x = 0$ or $\log x = -2$.
Thus,$x = e^0 = 1$ or $x = e^{-2}$.
Now,we use the second derivative test:
$\frac{d^2y}{dx^2} = \frac{2}{x} + 2 \log x \cdot \frac{1}{x} = \frac{2 + 2 \log x}{x}$.
At $x = 1$,$\frac{d^2y}{dx^2} = \frac{2 + 0}{1} = 2 > 0$,so $x = 1$ is a point of local minima.
At $x = e^{-2}$,$\frac{d^2y}{dx^2} = \frac{2 + 2(-2)}{e^{-2}} = \frac{-2}{e^{-2}} < 0$,so $x = e^{-2}$ is a point of local maxima.
The maximum value is $y(e^{-2}) = e^{-2}(\log e^{-2})^2 = e^{-2}(-2)^2 = 4e^{-2}$.

(C) Given the function $f(x) = x^3 - 6x^2 - 12x - 3$.
First,find the first derivative: $f'(x) = 3x^2 - 12x - 12$.
Next,find the second derivative: $f''(x) = 6x - 12$.
Evaluate the second derivative at $x = 2$: $f''(2) = 6(2) - 12 = 12 - 12 = 0$.
Since $f'(2) = 3(2)^2 - 12(2) - 12 = 12 - 24 - 12 = -24 \neq 0$,$x=2$ is not a critical point.
However,checking the options provided,the question implies evaluating the nature of the point where $f''(x)=0$.
Since $f''(x)$ changes sign at $x=2$ (for $x < 2$,$f''(x) < 0$ and for $x > 2$,$f''(x) > 0$),$x=2$ is a point of inflection.

(B) Let $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$.
To find the absolute maximum,we first find the critical points by setting the derivative $f'(x) = 0$.
$f'(x) = 40x^{39} - 20x^{19} = 20x^{19}(2x^{20} - 1)$.
Setting $f'(x) = 0$,we get $x = 0$ or $2x^{20} = 1$,which implies $x^{20} = \frac{1}{2}$,so $x = (\frac{1}{2})^{1/20}$.
Now,evaluate $f(x)$ at the critical point and the endpoints $x=0$ and $x=1$:
$f(0) = 0^{40} - 0^{20} = 0$.
$f(1) = 1^{40} - 1^{20} = 1 - 1 = 0$.
$f((\frac{1}{2})^{1/20}) = ((\frac{1}{2})^{1/20})^{40} - ((\frac{1}{2})^{1/20})^{20} = (\frac{1}{2})^2 - (\frac{1}{2})^1 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.
Comparing the values $f(0)=0$,$f(1)=0$,and $f((\frac{1}{2})^{1/20}) = -\frac{1}{4}$,the absolute maximum value is $0$.

(B) Let the sides of the rectangle be $x$ and $y$.
Given,the perimeter is $20$ units.
$2(x + y) = 20 \Rightarrow x + y = 10 \Rightarrow y = 10 - x$.
The area $A$ of the rectangle is given by $A = xy$.
Substituting $y$,we get $A = x(10 - x) = 10x - x^2$.
To find the maximum area,we differentiate $A$ with respect to $x$:
$\frac{dA}{dx} = 10 - 2x$.
Setting $\frac{dA}{dx} = 0$,we get $10 - 2x = 0$,which implies $x = 5$.
Now,we check the second derivative: $\frac{d^2A}{dx^2} = -2$.
Since $\frac{d^2A}{dx^2} < 0$,the area is maximum at $x = 5$.
When $x = 5$,$y = 10 - 5 = 5$.
Thus,the maximum area is $A = 5 \times 5 = 25$ sq units.

(A) Given $f(x) = x^5 - 5x^4 + 5x^3 - 10$.
First,find the derivative $f'(x)$:
$f'(x) = 5x^4 - 20x^3 + 15x^2$.
For local maxima and minima,set $f'(x) = 0$:
$5x^2(x^2 - 4x + 3) = 0
\Rightarrow 5x^2(x - 1)(x - 3) = 0$.
The critical points are $x = 0, 1, 3$.
Now,find the second derivative $f''(x)$:
$f''(x) = 20x^3 - 60x^2 + 30x$.
Check the nature of the critical points:
For $x = 1$: $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 < 0$. Thus,$x = 1$ is a point of local maxima. So,$a = 1$.
For $x = 3$: $f''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$. Thus,$x = 3$ is a point of local minima. So,$b = 3$.
For $x = 0$: $f''(0) = 0$,which is a point of inflection.
Finally,calculate $2a + b$:
$2a + b = 2(1) + 3 = 5$.

(C) Given function: $f(x) = a \sin(x) + \frac{1}{3} \sin(3x)$.
First,find the derivative $f'(x)$:
$f'(x) = a \cos(x) + \frac{1}{3} \cos(3x) \cdot 3 = a \cos(x) + \cos(3x)$.
Since the function attains its maximum at $x = \frac{\pi}{3}$,the first derivative must be zero at this point:
$f'\left(\frac{\pi}{3}\right) = 0$.
Substituting $x = \frac{\pi}{3}$ into the derivative:
$a \cos\left(\frac{\pi}{3}\right) + \cos\left(3 \cdot \frac{\pi}{3}\right) = 0$.
$a \cdot \frac{1}{2} + \cos(\pi) = 0$.
Since $\cos(\pi) = -1$,we have:
$\frac{a}{2} - 1 = 0$.
$\frac{a}{2} = 1$.
$a = 2$.

(A) $f(x) = ax^3 + bx^2 + 26x - 24$ ...$(i)$ on $[2, 4]$.
Since $f(x)$ satisfies Rolle's theorem,$f(2) = f(4)$.
$f(2) = a(8) + b(4) + 26(2) - 24 = 8a + 4b + 28$.
$f(4) = a(64) + b(16) + 26(4) - 24 = 64a + 16b + 80$.
Equating $f(2) = f(4)$: $8a + 4b + 28 = 64a + 16b + 80 \Rightarrow 56a + 12b + 52 = 0 \Rightarrow 14a + 3b + 13 = 0$ ...(ii).
Differentiating $f(x)$ w.r.t. $x$: $f^{\prime}(x) = 3ax^2 + 2bx + 26$.
Given $f^{\prime}\left(3 + \frac{1}{\sqrt{3}}\right) = 0$:
$3a\left(3 + \frac{1}{\sqrt{3}}\right)^2 + 2b\left(3 + \frac{1}{\sqrt{3}}\right) + 26 = 0$.
$3a\left(9 + \frac{1}{3} + 2\sqrt{3}\right) + 6b + \frac{2b}{\sqrt{3}} + 26 = 0$.
$3a\left(\frac{28}{3} + 2\sqrt{3}\right) + 6b + \frac{2b}{\sqrt{3}} + 26 = 0$.
$28a + 6\sqrt{3}a + 6b + \frac{2b}{\sqrt{3}} + 26 = 0$.
$(28a + 6b + 26) + \frac{1}{\sqrt{3}}(18a + 2b) = 0$.
Comparing rational and irrational parts: $28a + 6b + 26 = 0$ and $18a + 2b = 0$.
From $18a + 2b = 0$,$b = -9a$.
Substitute into $28a + 6(-9a) + 26 = 0 \Rightarrow 28a - 54a + 26 = 0 \Rightarrow -26a = -26 \Rightarrow a = 1$.
Then $b = -9(1) = -9$.
Therefore,$ab = 1 \times (-9) = -9$.

(D) For Rolle's theorem to be applicable on $[0, 1]$,the function $f(x)$ must satisfy three conditions:
$1$. $f(x)$ is continuous on $[0, 1]$.
$2$. $f(x)$ is differentiable on $(0, 1)$.
$3$. $f(0) = f(1)$.
Given $f(x) = x^\alpha \log x$ and $f(0) = 0$.
First,check the condition $f(0) = f(1)$:
$f(1) = 1^\alpha \log(1) = 1 \times 0 = 0$.
Since $f(0) = 0$,the condition $f(0) = f(1)$ is satisfied for any $\alpha$.
Next,check continuity at $x = 0$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^\alpha \log x = 0$.
This limit exists and equals $0$ if $\alpha > 0$.
Using $L$'$H$ôpital's rule:
$\lim_{x \to 0^+} \frac{\log x}{x^{-\alpha}} = \lim_{x \to 0^+} \frac{1/x}{-\alpha x^{-\alpha-1}} = \lim_{x \to 0^+} \frac{x^\alpha}{-\alpha} = 0$ (for $\alpha > 0$).
Checking the options,$\alpha = 1/2$ is the only value greater than $0$.

(B) Given $f(x)=\sqrt{x^2-x}$ on the interval $[1,4]$.
First,calculate the values at the endpoints:
$f(1)=\sqrt{1^2-1}=0$
$f(4)=\sqrt{4^2-4}=\sqrt{12}=2\sqrt{3}$
Next,find the derivative $f^{\prime}(x)$:
$f^{\prime}(x)=\frac{1}{2\sqrt{x^2-x}} \cdot (2x-1) = \frac{2x-1}{2\sqrt{x^2-x}}$
According to Lagrange's mean value theorem,there exists $c \in (1,4)$ such that $f^{\prime}(c) = \frac{f(4)-f(1)}{4-1}$.
Substituting the values:
$\frac{2c-1}{2\sqrt{c^2-c}} = \frac{2\sqrt{3}-0}{3} = \frac{2}{\sqrt{3}}$
$\sqrt{3}(2c-1) = 4\sqrt{c^2-c}$
Squaring both sides:
$3(4c^2-4c+1) = 16(c^2-c)$
$12c^2-12c+3 = 16c^2-16c$
$4c^2-4c-3 = 0$
Solving the quadratic equation using the quadratic formula or factoring:
$(2c-3)(2c+1) = 0$
This gives $c = \frac{3}{2}$ or $c = -\frac{1}{2}$.
Since $c \in (1,4)$,we discard $c = -\frac{1}{2}$.
Thus,$c = \frac{3}{2}$.

(A) We are given the integral $I = \int \frac{\sin \alpha}{\sqrt{1 + \cos \alpha}} d \alpha$.
Using the trigonometric identity $1 + \cos \alpha = 2 \cos^2 (\frac{\alpha}{2})$,we substitute this into the integral:
$I = \int \frac{2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})}{\sqrt{2 \cos^2 (\frac{\alpha}{2})}} d \alpha$
Assuming $\cos (\frac{\alpha}{2}) > 0$,the denominator becomes $\sqrt{2} \cos (\frac{\alpha}{2})$:
$I = \int \frac{2 \sin (\frac{\alpha}{2}) \cos (\frac{\alpha}{2})}{\sqrt{2} \cos (\frac{\alpha}{2})} d \alpha$
$I = \sqrt{2} \int \sin (\frac{\alpha}{2}) d \alpha$
Integrating $\sin (\frac{\alpha}{2})$ with respect to $\alpha$:
$I = \sqrt{2} \times (-2 \cos (\frac{\alpha}{2})) + c$
$I = -2 \sqrt{2} \cos (\frac{\alpha}{2}) + c$.

(C) We are given the integral $I = \int_{1}^{2} \frac{x^{3} - 1}{x^{2}} dx$.
First,simplify the integrand:
$\frac{x^{3} - 1}{x^{2}} = \frac{x^{3}}{x^{2}} - \frac{1}{x^{2}} = x - x^{-2}$.
Now,integrate term by term:
$I = \int_{1}^{2} (x - x^{-2}) dx = \left[ \frac{x^{2}}{2} - \frac{x^{-1}}{-1} \right]_{1}^{2} = \left[ \frac{x^{2}}{2} + \frac{1}{x} \right]_{1}^{2}$.
Substitute the limits:
$I = \left( \frac{2^{2}}{2} + \frac{1}{2} \right) - \left( \frac{1^{2}}{2} + \frac{1}{1} \right) = \left( 2 + \frac{1}{2} \right) - \left( \frac{1}{2} + 1 \right) = 2 + \frac{1}{2} - \frac{1}{2} - 1 = 1$.

(B) We need to evaluate the integral: $\int \left( \frac{x}{a} + \frac{b}{x} + x^a + b^x + ab \right) dx$
Applying the linearity property of integration:
$= \int \frac{x}{a} dx + \int \frac{b}{x} dx + \int x^a dx + \int b^x dx + \int ab dx$
$= \frac{1}{a} \int x dx + b \int \frac{1}{x} dx + \int x^a dx + \int b^x dx + ab \int 1 dx$
Using standard integration formulas:
$= \frac{1}{a} \cdot \frac{x^2}{2} + b \log |x| + \frac{x^{a+1}}{a+1} + \frac{b^x}{\log b} + abx + C$
$= \frac{x^2}{2a} + b \log |x| + \frac{x^{a+1}}{a+1} + \frac{b^x}{\log b} + abx + C$

(B) Given $f^{\prime}(x)=\tan ^2 x+\cot ^2 x$.
Integrating both sides with respect to $x$:
$f(x)=\int(\tan ^2 x+\cot ^2 x) dx$.
Using the identity $\tan^2 x = \sec^2 x - 1$ and $\cot^2 x = \operatorname{cosec}^2 x - 1$:
$f(x)=\int(\sec^2 x - 1 + \operatorname{cosec}^2 x - 1) dx$.
$f(x)=\int(\sec^2 x + \operatorname{cosec}^2 x - 2) dx$.
$f(x)=\tan x - \cot x - 2x + C$.
Given $f\left(\frac{\pi}{4}\right)=0$:
$f\left(\frac{\pi}{4}\right)=\tan\left(\frac{\pi}{4}\right) - \cot\left(\frac{\pi}{4}\right) - 2\left(\frac{\pi}{4}\right) + C = 0$.
$1 - 1 - \frac{\pi}{2} + C = 0$.
$C = \frac{\pi}{2}$.
Therefore,$f(x)=\tan x - \cot x - 2x + \frac{\pi}{2}$.

(C) Given integral is $I = \int \frac{x^3-x^2+x-1}{x-1} dx$.
Factorize the numerator: $x^3-x^2+x-1 = x^2(x-1) + 1(x-1) = (x^2+1)(x-1)$.
Substitute this into the integral: $I = \int \frac{(x^2+1)(x-1)}{x-1} dx$.
Cancel the common term $(x-1)$: $I = \int (x^2+1) dx$.
Integrate term by term: $I = \frac{x^3}{3} + x + C$.

(B) Let $I = \int \frac{dx}{x \sqrt{x^4 - 1}}$.
Multiply the numerator and denominator by $x$:
$I = \int \frac{x \ dx}{x^2 \sqrt{(x^2)^2 - 1}}$.
Let $x^2 = t$,then $2x \ dx = dt$,or $x \ dx = \frac{dt}{2}$.
Substituting these into the integral:
$I = \int \frac{dt/2}{t \sqrt{t^2 - 1}} = \frac{1}{2} \int \frac{dt}{t \sqrt{t^2 - 1}}$.
We know that $\int \frac{dt}{t \sqrt{t^2 - 1}} = \sec^{-1}(t) + C$.
Therefore,$I = \frac{1}{2} \sec^{-1}(t) + C = \frac{1}{2} \sec^{-1}(x^2) + C$.
Comparing this with the given expression $\frac{1}{k} \sec^{-1}(x^k)$,we get $k = 2$.

(B) Given the integral $I = \int \frac{(x - 1)^2}{(x^2 + 1)^2} dx$.
Expanding the numerator,we get $(x - 1)^2 = x^2 - 2x + 1$.
So,$I = \int \frac{x^2 + 1 - 2x}{(x^2 + 1)^2} dx$.
Splitting the integral,we have $I = \int \frac{x^2 + 1}{(x^2 + 1)^2} dx - \int \frac{2x}{(x^2 + 1)^2} dx$.
This simplifies to $I = \int \frac{1}{x^2 + 1} dx - \int 2x(x^2 + 1)^{-2} dx$.
We know that $\int \frac{1}{x^2 + 1} dx = \tan^{-1}(x)$.
For the second part,let $u = x^2 + 1$,then $du = 2x dx$.
Thus,$\int 2x(x^2 + 1)^{-2} dx = \int u^{-2} du = \frac{u^{-1}}{-1} = -\frac{1}{u} = -\frac{1}{x^2 + 1}$.
Substituting these back,$I = \tan^{-1}(x) - (-\frac{1}{x^2 + 1}) + k = \tan^{-1}(x) + \frac{1}{x^2 + 1} + k$.
Comparing this with the given form $\tan^{-1}(x) + g(x) + k$,we find $g(x) = \frac{1}{x^2 + 1}$.

(B) Given integral is $I = \int \frac{1 + x + \sqrt{x(1 + x)}}{\sqrt{x} + \sqrt{1 + x}} dx$.
We can rewrite the numerator as:
$1 + x + \sqrt{x} \cdot \sqrt{1 + x} = \sqrt{1 + x} \cdot \sqrt{1 + x} + \sqrt{x} \cdot \sqrt{1 + x}$.
Factoring out $\sqrt{1 + x}$,we get:
$\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})$.
Substituting this back into the integral:
$I = \int \frac{\sqrt{1 + x} (\sqrt{1 + x} + \sqrt{x})}{\sqrt{x} + \sqrt{1 + x}} dx$.
Canceling the common term $(\sqrt{x} + \sqrt{1 + x})$:
$I = \int \sqrt{1 + x} dx = \int (1 + x)^{1/2} dx$.
Using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + c$:
$I = \frac{(1 + x)^{3/2}}{3/2} + c = \frac{2}{3} (1 + x)^{3/2} + c$.

(C) Using integration by parts,let $u = \log \cot (\frac{x}{2})$ and $dv = \cos x dx$.
Then $du = \frac{1}{\cot (\frac{x}{2})} \cdot (-\csc^2 (\frac{x}{2})) \cdot \frac{1}{2} dx = -\frac{1}{2 \sin (\frac{x}{2}) \cos (\frac{x}{2})} dx = -\frac{1}{\sin x} dx$.
And $v = \int \cos x dx = \sin x$.
Using the formula $\int u dv = uv - \int v du$:
$\int \cos x \log \cot (\frac{x}{2}) dx = \sin x \log \cot (\frac{x}{2}) - \int \sin x \cdot (-\frac{1}{\sin x}) dx$
$= \sin x \log \cot (\frac{x}{2}) + \int 1 dx$
$= \sin x \log \cot (\frac{x}{2}) + x + c$.

(B) Given integral: $I = \int \frac{\cos 4x + 1}{\cot x - \tan x} dx$
Using the identity $\cos 4x + 1 = 2 \cos^2 2x$ and $\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2 \cos 2x}{\sin 2x}$.
Substituting these into the integral:
$I = \int \frac{2 \cos^2 2x}{\frac{2 \cos 2x}{\sin 2x}} dx$
$I = \int \cos 2x \sin 2x dx$
$I = \frac{1}{2} \int \sin 4x dx$
$I = \frac{1}{2} \left( \frac{-\cos 4x}{4} \right) + c$
$I = \frac{-1}{8} \cos 4x + c$
Comparing with $k \cos 4x + c$,we get $k = \frac{-1}{8}$.

(B) Given the integral: $\int_{0}^{a} \frac{dx}{4 + x^2} = \frac{\pi}{8}$
We know that $\int \frac{dx}{k^2 + x^2} = \frac{1}{k} \tan^{-1} \left( \frac{x}{k} \right) + C$.
Here,$k^2 = 4$,so $k = 2$.
Applying the limits from $0$ to $a$:
$\left[ \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) \right]_{0}^{a} = \frac{\pi}{8}$
$\frac{1}{2} \tan^{-1} \left( \frac{a}{2} \right) - \frac{1}{2} \tan^{-1} (0) = \frac{\pi}{8}$
Since $\tan^{-1}(0) = 0$,we have:
$\frac{1}{2} \tan^{-1} \left( \frac{a}{2} \right) = \frac{\pi}{8}$
$\tan^{-1} \left( \frac{a}{2} \right) = \frac{\pi}{4}$
Taking the tangent of both sides:
$\frac{a}{2} = \tan \left( \frac{\pi}{4} \right)$
$\frac{a}{2} = 1$
$a = 2$

(C) We are given $f(x) = \frac{1}{\cos^2 x \sqrt{1 + \tan x}} = \sec^2 x (1 + \tan x)^{-1/2}$.
Let $t = 1 + \tan x$. Then $dt = \sec^2 x \ dx$.
Substituting these into the integral:
$F(x) = \int f(x) \ dx = \int (1 + \tan x)^{-1/2} \sec^2 x \ dx = \int t^{-1/2} \ dt$.
Integrating with respect to $t$:
$F(x) = \frac{t^{1/2}}{1/2} + C = 2 \sqrt{t} + C = 2 \sqrt{1 + \tan x} + C$.
Given $F(0) = 4$,we substitute $x = 0$:
$F(0) = 2 \sqrt{1 + \tan 0} + C = 2 \sqrt{1 + 0} + C = 2 + C$.
Since $F(0) = 4$,we have $2 + C = 4$,which implies $C = 2$.
Thus,$F(x) = 2 \sqrt{1 + \tan x} + 2 = 2 (\sqrt{1 + \tan x} + 1)$.

(B) Given $I = \int \frac{\sec x}{\sqrt{\sin (2 x + \theta) + \sin \theta}} d x$
Using the formula $\sin C + \sin D = 2 \sin (\frac{C + D}{2}) \cos (\frac{C - D}{2})$:
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \sin (\frac{2 x + 2 \theta}{2}) \cos (\frac{2 x + \theta - \theta}{2})}} d x$
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \sin (x + \theta) \cos x}} d x$
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 (\sin x \cos \theta + \cos x \sin \theta) \cos x}} d x$
Divide numerator and denominator by $\cos x$ inside the square root:
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2 \cos^2 x (\tan x \cos \theta + \sin \theta)}} d x$
$\Rightarrow I = \int \frac{\sec x}{\sqrt{2} \cos x \sqrt{\tan x \cos \theta + \sin \theta}} d x$
$\Rightarrow I = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x}{\sqrt{\tan x \cos \theta + \sin \theta}} d x$
Let $t = \tan x \cos \theta + \sin \theta$,then $dt = \sec^2 x \cos \theta d x$,so $\sec^2 x d x = \frac{dt}{\cos \theta}$.
$\Rightarrow I = \frac{1}{\sqrt{2} \cos \theta} \int t^{-1/2} dt$
$\Rightarrow I = \frac{1}{\sqrt{2} \cos \theta} \cdot 2 t^{1/2} + C$
$\Rightarrow I = \sqrt{\frac{2}{\cos^2 \theta}} \sqrt{\tan x \cos \theta + \sin \theta} + C$
$\Rightarrow I = \sqrt{2 \sec^2 \theta (\tan x \cos \theta + \sin \theta)} + C$
$\Rightarrow I = \sqrt{2 \sec \theta (\tan x + \tan \theta)} + C$

(A) Given $I = \int \frac{d x}{\cos ^4 x+\sin ^4 x}$.
Dividing numerator and denominator by $\cos^4 x$,we get $I = \int \frac{\sec^4 x}{1+\tan^4 x} d x = \int \frac{\sec^2 x(1+\tan^2 x)}{1+\tan^4 x} d x$.
Let $\tan x = t$,then $\sec^2 x d x = d t$.
$I = \int \frac{1+t^2}{1+t^4} d t = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}} d t$.
$I = \int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2} d t$.
Let $u = t-\frac{1}{t}$,then $d u = (1+\frac{1}{t^2}) d t$.
$I = \int \frac{d u}{u^2+(\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{u}{\sqrt{2}}) + C$.
Substituting $u = t-\frac{1}{t} = \tan x - \cot x$,we get $I = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{\tan x - \cot x}{\sqrt{2}}) + C$.
Comparing this with the given form,$g(x) = \frac{\tan x - \cot x}{\sqrt{2}}$.

(A) Let $I = \int (1 - \cos x) \operatorname{cosec}^2 x \, dx$.
Using the identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$ and $\operatorname{cosec}^2 x = \frac{1}{\sin^2 x}$,we have:
$I = \int 2 \sin^2 \left(\frac{x}{2}\right) \cdot \frac{1}{\sin^2 x} \, dx$.
Since $\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$,then $\sin^2 x = 4 \sin^2 \left(\frac{x}{2}\right) \cos^2 \left(\frac{x}{2}\right)$.
Substituting this into the integral:
$I = \int \frac{2 \sin^2 \left(\frac{x}{2}\right)}{4 \sin^2 \left(\frac{x}{2}\right) \cos^2 \left(\frac{x}{2}\right)} \, dx$.
$I = \int \frac{1}{2 \cos^2 \left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) \, dx$.
Integrating $\sec^2 \left(\frac{x}{2}\right)$,we get $2 \tan \left(\frac{x}{2}\right)$.
$I = \frac{1}{2} \cdot 2 \tan \left(\frac{x}{2}\right) + C = \tan \left(\frac{x}{2}\right) + C$.

(B) Given the integral: $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.
Let $I = \int \cos ^k(x) \sin (x) d x$.
Substitute $t = \cos x$,then $dt = -\sin x d x$,which implies $\sin x d x = -dt$.
Substituting these into the integral,we get:
$I = \int t^k (-dt) = -\int t^k dt$.
Integrating $t^k$ with respect to $t$ gives:
$I = -\frac{t^{k+1}}{k+1} + C$.
Substituting back $t = \cos x$:
$I = -\frac{\cos^{k+1} x}{k+1} + C$.
Comparing this with the given expression $\frac{-1}{4} \cos^4(x) + C$,we have:
$\frac{1}{k+1} = \frac{1}{4}$ and $k+1 = 4$.
Therefore,$k = 3$.

(C) Let $I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx$.
We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 1 - (\sin x - \cos x)^2$.
Alternatively,we can write $\sin 2x = (\sin x + \cos x)^2 - 1$.
So,$I = \int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^2 - 1}} \, dx$.
Let $t = \sin x + \cos x$.
Then $dt = (\cos x - \sin x) \, dx$,which implies $(\sin x - \cos x) \, dx = -dt$.
Substituting these into the integral:
$I = \int \frac{-dt}{\sqrt{t^2 - 1}} = -\int \frac{dt}{\sqrt{t^2 - 1}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + C$,we get:
$I = -\log |t + \sqrt{t^2 - 1}| + C$.
Substituting $t = \sin x + \cos x$ back:
$I = -\log |\sin x + \cos x + \sqrt{(\sin x + \cos x)^2 - 1}| + C$.
Since $(\sin x + \cos x)^2 - 1 = 1 + \sin 2x - 1 = \sin 2x$,we have:
$I = -\log |\sin x + \cos x + \sqrt{\sin 2x}| + C$.

(B) Let $I = \int \sqrt[3]{x}\left(1+\sqrt[3]{x^4}\right)^{\frac{1}{7}} d x = \int x^{\frac{1}{3}}\left(1+x^{\frac{4}{3}}\right)^{\frac{1}{7}} d x$.
Let $1+x^{\frac{4}{3}} = t$.
Then,$\frac{4}{3} x^{\frac{1}{3}} d x = d t$,which implies $x^{\frac{1}{3}} d x = \frac{3}{4} d t$.
Substituting these into the integral,we get:
$I = \int t^{\frac{1}{7}} \cdot \frac{3}{4} d t = \frac{3}{4} \int t^{\frac{1}{7}} d t$.
Integrating,we have:
$I = \frac{3}{4} \cdot \frac{t^{\frac{1}{7}+1}}{\frac{1}{7}+1} + c = \frac{3}{4} \cdot \frac{t^{\frac{8}{7}}}{\frac{8}{7}} + c = \frac{3}{4} \cdot \frac{7}{8} t^{\frac{8}{7}} + c = \frac{21}{32} \left(1+x^{\frac{4}{3}}\right)^{\frac{8}{7}} + c$.
Comparing this with $A\left(1+\sqrt[3]{x^4}\right)^B+c$,we get $A = \frac{21}{32}$ and $B = \frac{8}{7}$.
Therefore,$A B = \frac{21}{32} \times \frac{8}{7} = \frac{3}{4}$.

(D) Let $I = \int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$.
Using the substitution $t = x+\frac{\pi}{4}$,we have $x = t-\frac{\pi}{4}$ and $dx = dt$.
Substituting these into the integral:
$I = \int \frac{\sqrt{2} \sin \left(t-\frac{\pi}{4}\right)}{\sin t} dt$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$I = \int \sqrt{2} \frac{\sin t \cos \frac{\pi}{4} - \cos t \sin \frac{\pi}{4}}{\sin t} dt$.
Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$I = \int \sqrt{2} \frac{\left(\frac{1}{\sqrt{2}} \sin t - \frac{1}{\sqrt{2}} \cos t\right)}{\sin t} dt = \int \frac{\sin t - \cos t}{\sin t} dt$.
$I = \int (1 - \cot t) dt = t - \ln |\sin t| + c$.
Substituting back $t = x+\frac{\pi}{4}$:
$I = x + \frac{\pi}{4} - \ln |\sin (x+\frac{\pi}{4})| + c$.
Since $\frac{\pi}{4}$ is a constant,it can be absorbed into the constant $c$:
$I = x - \log |\sin (x+\frac{\pi}{4})| + c$.

(C) Let $I = \int \frac{dx}{\sin x \cos x}$.
Multiply the numerator and denominator by $\sec^2 x$:
$I = \int \frac{\sec^2 x dx}{\tan x}$.
Substitute $\tan x = t$,then $\sec^2 x dx = dt$.
Therefore,$I = \int \frac{dt}{t} = \log |t| + c$.
Substituting back $t = \tan x$,we get $I = \log |\tan x| + c$.