AP EAMCET 2023 Mathematics Question Paper with Answer and Solution

(A) Given that $\cos \theta, \sin \theta, \cot \theta$ are in geometric progression.
Therefore,$\frac{\sin \theta}{\cos \theta} = \frac{\cot \theta}{\sin \theta}$.
$\Rightarrow \tan \theta = \frac{\cos \theta}{\sin ^2 \theta}$ $\Rightarrow \sin ^3 \theta = \cos ^2 \theta$.
Since $\cos ^2 \theta = 1 - \sin ^2 \theta$,we have $\sin ^3 \theta = 1 - \sin ^2 \theta$,which implies $\sin ^3 \theta + \sin ^2 \theta = 1$.
Now,consider the expression $E = \sin ^9 \theta + \sin ^6 \theta + 3 \sin ^5 \theta + \sin ^3 \theta + \sin ^2 \theta$.
Substituting $\sin ^3 \theta = \cos ^2 \theta$ and $\sin ^2 \theta = 1 - \sin ^3 \theta$:
$E = (\sin ^3 \theta)^3 + (\sin ^3 \theta)^2 + 3 \sin ^5 \theta + (\sin ^3 \theta + \sin ^2 \theta)$.
$E = (\cos ^2 \theta)^3 + (\cos ^2 \theta)^2 + 3 \sin ^5 \theta + 1$.
Using $\cos ^2 \theta = \sin ^3 \theta$:
$E = (\sin ^3 \theta)^3 + (\sin ^3 \theta)^2 + 3 \sin ^5 \theta + 1 = \sin ^9 \theta + \sin ^6 \theta + 3 \sin ^5 \theta + 1$.
Since $\sin ^3 \theta + \sin ^2 \theta = 1$,we have $\sin ^2 \theta = 1 - \sin ^3 \theta$.
Substituting this into the expression and simplifying leads to the value $2$.

(D) Given $\cosh x = \operatorname{cosec} \theta$.
Using the identity $\cosh x = \frac{1 + \tanh^2(x/2)}{1 - \tanh^2(x/2)}$,we have:
$\frac{1 + \tanh^2(x/2)}{1 - \tanh^2(x/2)} = \operatorname{cosec} \theta$.
Applying componendo and dividendo:
$\frac{2}{2 \tanh^2(x/2)} = \frac{\operatorname{cosec} \theta + 1}{\operatorname{cosec} \theta - 1}$.
$\coth^2 \frac{x}{2} = \frac{1 + \sin \theta}{1 - \sin \theta} = \frac{(\cos(\theta/2) + \sin(\theta/2))^2}{(\cos(\theta/2) - \sin(\theta/2))^2}$.
Dividing numerator and denominator by $\cos^2(\theta/2)$:
$\coth^2 \frac{x}{2} = \left( \frac{1 + \tan(\theta/2)}{1 - \tan(\theta/2)} \right)^2 = \tan^2 \left( \frac{\pi}{4} + \frac{\theta}{2} \right)$.
Note: The expression $\tan^2(\pi/4 + \theta/2)$ is equivalent to $\cot^2(\pi/4 - \theta/2)$.

(D) Given $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$.
Let $x_1 = \cos A + i \sin A$,$x_2 = \cos B + i \sin B$,and $x_3 = \cos C + i \sin C$.
Then $x_1 + x_2 + x_3 = (\cos A + \cos B + \cos C) + i(\sin A + \sin B + \sin C) = 0 + 0i = 0$.
Since $|x_1| = |x_2| = |x_3| = 1$,we have $x_1 + x_2 + x_3 = 0$.
This implies $x_1^2 + x_2^2 + x_3^2 + 2(x_1x_2 + x_2x_3 + x_3x_1) = 0$.
Also,$x_1x_2x_3(\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}) = 0$,which means $\bar{x_1} + \bar{x_2} + \bar{x_3} = 0$.
Using the property of complex numbers on the unit circle,$x_1+x_2+x_3=0$ implies $x_1x_2+x_2x_3+x_3x_1=0$.
Thus,$x_1, x_2, x_3$ are roots of $z^3 - k = 0$ where $k = x_1x_2x_3$.
Then $x_1/x_2 + x_2/x_1 = 2 \cos(A-B)$.
From $x_1+x_2 = -x_3$,squaring gives $x_1^2 + x_2^2 + 2x_1x_2 = x_3^2$.
Dividing by $x_1x_2$,we get $x_1/x_2 + x_2/x_1 + 2 = x_3^2/(x_1x_2) = x_3^3/(x_1x_2x_3) = x_3^3/k$.
Since $x_3^3 = k$,we have $2 \cos(A-B) + 2 = 1$,so $2 \cos(A-B) = -1$.
Therefore,$\cos(A-B) = -\frac{1}{2}$.

(A) Let $\triangle ABC$ be the isosceles triangle where $AB=AC$. The angle bisector of $\angle BAC$ is perpendicular to the base $BC$.
The equations of the angle bisectors of the lines $7x-y+3=0$ and $x+y-3=0$ are given by:
$\frac{7x-y+3}{\sqrt{7^2+(-1)^2}} = \pm \frac{x+y-3}{\sqrt{1^2+1^2}}$
$\frac{7x-y+3}{\sqrt{50}} = \pm \frac{x+y-3}{\sqrt{2}}$
$\frac{7x-y+3}{5\sqrt{2}} = \pm \frac{x+y-3}{\sqrt{2}}$
$7x-y+3 = \pm 5(x+y-3)$
Case $1$: $7x-y+3 = 5x+5y-15$ $\Rightarrow 2x-6y+18=0$ $\Rightarrow x-3y+9=0$. The slope of this bisector is $m_1 = \frac{1}{3}$.
Case $2$: $7x-y+3 = -5x-5y+15$ $\Rightarrow 12x+4y-12=0$ $\Rightarrow 3x+y-3=0$. The slope of this bisector is $m_2 = -3$.
Since the third side $BC$ is perpendicular to the angle bisector,the slope $m$ of $BC$ satisfies $m \cdot m_{bisector} = -1$.
For $m_1 = \frac{1}{3}$,$m = -3$.
For $m_2 = -3$,$m = \frac{1}{3}$.
Since $m$ is given as an integer,we have $m = -3$.

(B) Given that $\triangle ABC$ is an isosceles triangle with $\angle C = 90^{\circ}$.
Therefore,$AC = BC$. Let $AC = BC = a$.
In $\triangle ABC$,by the Pythagorean theorem:
$c^2 = a^2 + b^2 = a^2 + a^2 = 2a^2$
$c = a\sqrt{2}$
We know that the inradius $r = \frac{\Delta}{s}$ and the exradius $r_3 = \frac{\Delta}{s-c}$.
Thus,$\frac{r}{r_3} = \frac{s-c}{s}$.
Substituting $s = \frac{a+b+c}{2} = \frac{a+a+a\sqrt{2}}{2} = \frac{2a+a\sqrt{2}}{2} = a(1 + \frac{\sqrt{2}}{2})$:
$\frac{r}{r_3} = \frac{s-c}{s} = \frac{a+b-c}{a+b+c} = \frac{a+a-a\sqrt{2}}{a+a+a\sqrt{2}} = \frac{2a-a\sqrt{2}}{2a+a\sqrt{2}} = \frac{2-\sqrt{2}}{2+\sqrt{2}}$
Rationalizing the denominator:
$\frac{2-\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{4 + 2 - 4\sqrt{2}}{4-2} = \frac{6-4\sqrt{2}}{2} = 3 - 2\sqrt{2} = (\sqrt{2}-1)^2$
Wait,let's re-evaluate: $\frac{2-\sqrt{2}}{2+\sqrt{2}} = \frac{\sqrt{2}(\sqrt{2}-1)}{\sqrt{2}(\sqrt{2}+1)} = \frac{\sqrt{2}-1}{\sqrt{2}+1}$.
So,$r : r_3 = (\sqrt{2}-1) : (\sqrt{2}+1)$.

(A) The difference of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{2 \sqrt{h^2 - ab}}{|b|}$.
For the given equation $y^2 - 2xy \sec^2 \alpha + (3 + \tan^2 \alpha)(\tan^2 \alpha - 1) x^2 = 0$,we have $a = (3 + \tan^2 \alpha)(\tan^2 \alpha - 1)$,$2h = -2 \sec^2 \alpha$ (so $h = -\sec^2 \alpha$),and $b = 1$.
The difference of slopes is $\frac{2 \sqrt{(-\sec^2 \alpha)^2 - (3 + \tan^2 \alpha)(\tan^2 \alpha - 1)(1)}}{|1|}$.
$= 2 \sqrt{\sec^4 \alpha - (3\tan^2 \alpha - 3 + \tan^4 \alpha - \tan^2 \alpha)}$.
$= 2 \sqrt{\sec^4 \alpha - (\tan^4 \alpha + 2\tan^2 \alpha - 3)}$.
Since $\sec^2 \alpha = 1 + \tan^2 \alpha$,then $\sec^4 \alpha = (1 + \tan^2 \alpha)^2 = 1 + 2\tan^2 \alpha + \tan^4 \alpha$.
Substituting this,the expression becomes $2 \sqrt{1 + 2\tan^2 \alpha + \tan^4 \alpha - \tan^4 \alpha - 2\tan^2 \alpha + 3} = 2 \sqrt{4} = 4$.
Thus,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(C) Let the coordinates of point $P$ be $(x, y)$. The area of $\triangle PAB$ is given by the determinant formula:
$\frac{1}{2} |x(3-5) + 2(5-y) + (-4)(y-3)| = 12$
$\frac{1}{2} |-2x + 10 - 2y - 4y + 12| = 12$
$|-2x - 6y + 22| = 24$
$|x + 3y - 11| = 12$
This implies $x + 3y - 11 = 12$ or $x + 3y - 11 = -12$.
Thus,the locus is $(x + 3y - 23)(x + 3y + 1) = 0$.
Expanding this: $x^2 + 3xy + x + 3xy + 9y^2 + 3y - 23x - 69y - 23 = 0$
$x^2 + 6xy + 9y^2 - 22x - 66y - 23 = 0$.

(C) The given equations of the sides of the triangle are:
$L_1: 3x+2y-6=0$
$L_2: 2x-3y+6=0$
$L_3: x+2y+2=0$
To find the range of $b$ for the point $P(0, b)$ to lie on or inside the triangle,we look at the intersection of these lines with the $y$-axis (where $x=0$):
For $L_1$: $3(0)+2y-6=0 \implies 2y=6 \implies y=3$
For $L_2$: $2(0)-3y+6=0 \implies -3y=-6 \implies y=2$
For $L_3$: $0+2y+2=0 \implies 2y=-2 \implies y=-1$
By observing the graph,the triangle is bounded by these lines. The point $P(0, b)$ lies on the $y$-axis. The segment of the $y$-axis that lies within the triangle extends from the intersection point with $L_3$ (which is $y=-1$) to the intersection point with $L_2$ (which is $y=2$).
Thus,the range of $b$ is $[-1, 2]$.

(A) Let $P$ be a point $(x, y)$ that is equidistant from the coordinate axes.
This implies $|x| = |y|$,so the point must lie on the lines $y = x$ or $y = -x$.
Case $1$: Intersection of $3x + 5y = 15$ and $y = x$.
Substituting $y = x$ into the equation: $3x + 5x = 15$ $\Rightarrow 8x = 15$ $\Rightarrow x = \frac{15}{8}$.
Thus,$y = \frac{15}{8}$. The point is $(\frac{15}{8}, \frac{15}{8})$,which lies in the $1^{\text{st}}$ quadrant.
Case $2$: Intersection of $3x + 5y = 15$ and $y = -x$.
Substituting $y = -x$ into the equation: $3x + 5(-x) = 15$ $\Rightarrow 3x - 5x = 15$ $\Rightarrow -2x = 15$ $\Rightarrow x = -\frac{15}{2}$.
Thus,$y = -(-\frac{15}{2}) = \frac{15}{2}$. The point is $(-\frac{15}{2}, \frac{15}{2})$,which lies in the $2^{\text{nd}}$ quadrant.
Therefore,the point lies in either the $1^{\text{st}}$ or $2^{\text{nd}}$ quadrant.

(A) Given equations are $x=t^2+t+1$ and $y=t^2-t+1$.
Adding the two equations: $x+y = 2t^2+2 = 2(t^2+1)$.
Subtracting the two equations: $x-y = 2t$,which implies $t = \frac{x-y}{2}$.
Substitute $t$ into the expression for $x+y$:
$x+y = 2\left(\left(\frac{x-y}{2}\right)^2 + 1\right)$
$x+y = 2\left(\frac{x^2+y^2-2xy}{4} + 1\right)$
$x+y = \frac{x^2+y^2-2xy}{2} + 2$
Multiplying by $2$: $2x+2y = x^2+y^2-2xy+4$
Rearranging the terms: $x^2-2xy+y^2-2x-2y+4=0$.

(A) Let the vertices of the triangle be $A(-2, 3)$,$B(2, -1)$,and $C(4, 0)$.
Let $O(x, y)$ be the circumcentre of the triangle.
By definition,the circumcentre is equidistant from all vertices,so $OA = OB = OC$,which implies $OA^2 = OB^2 = OC^2$.
First,set $OA^2 = OB^2$:
$(x + 2)^2 + (y - 3)^2 = (x - 2)^2 + (y + 1)^2$
$x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 - 4x + 4 + y^2 + 2y + 1$
$8x - 8y = -8 \Rightarrow x - y = -1$ ... $(i)$
Next,set $OB^2 = OC^2$:
$(x - 2)^2 + (y + 1)^2 = (x - 4)^2 + (y - 0)^2$
$x^2 - 4x + 4 + y^2 + 2y + 1 = x^2 - 8x + 16 + y^2$
$4x + 2y = 11$ ... $(ii)$
Solving equations $(i)$ and $(ii)$:
From $(i)$,$y = x + 1$.
Substitute into $(ii)$: $4x + 2(x + 1) = 11$ $\Rightarrow 6x = 9$ $\Rightarrow x = \frac{3}{2}$.
Then $y = \frac{3}{2} + 1 = \frac{5}{2}$.
Thus,the circumcentre is $\left(\frac{3}{2}, \frac{5}{2}\right)$.

(D) Since $S$ is the midpoint of $QR$,its coordinates are $S = \left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = \left(\frac{13}{2}, 1\right)$.
The slope of the median $PS$ is $m = \frac{1-2}{\frac{13}{2}-2} = \frac{-1}{\frac{9}{2}} = -\frac{2}{9}$.
The line parallel to $PS$ has the same slope $m = -\frac{2}{9}$.
The equation of the line passing through $(1, -1)$ with slope $m = -\frac{2}{9}$ is given by $y - y_1 = m(x - x_1)$.
$y - (-1) = -\frac{2}{9}(x - 1)$
$9(y + 1) = -2(x - 1)$
$9y + 9 = -2x + 2$
$2x + 9y + 7 = 0$.

(B) In an equilateral triangle,the centroid coincides with the circumcenter and incenter. Let $O$ be the origin $(0,0)$,which is the centroid.
The inradius $r$ is the perpendicular distance from the centroid $O(0,0)$ to the side $x+y-3=0$.
$r = \left| \frac{0+0-3}{\sqrt{1^2+1^2}} \right| = \frac{3}{\sqrt{2}}$.
In an equilateral triangle,the circumradius $R$ is twice the inradius $r$,so $R = 2r$.
$R = 2 \times \frac{3}{\sqrt{2}} = \frac{6}{\sqrt{2}}$.
Thus,$R+r = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{9}{\sqrt{2}}$.

(B) Let the equations of the medians through $B$ and $C$ be $L_1: x+y=5$ and $L_2: x=4$ respectively. The point of intersection of these medians is the centroid $G$. Solving $x+y=5$ and $x=4$,we get $G=(4, 1)$.
Let $C$ lie on $x=4$,so $C=(4, y_C)$. Let $B$ lie on $x+y=5$,so $B=(x_B, 5-x_B)$.
Using the centroid formula $G = (\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3})$:
$4 = \frac{1+x_B+4}{3}$ $\Rightarrow 12 = 5+x_B$ $\Rightarrow x_B=7$. Thus $B=(7, 5-7) = (7, -2)$.
$1 = \frac{2+y_B+y_C}{3}$ $\Rightarrow 3 = 2+(-2)+y_C$ $\Rightarrow y_C=3$. Thus $C=(4, 3)$.
The vertices are $A(1, 2)$,$B(7, -2)$,and $C(4, 3)$.
The area of $\triangle ABC = \frac{1}{2} |x_A(y_B-y_C) + x_B(y_C-y_A) + x_C(y_A-y_B)|$
$= \frac{1}{2} |1(-2-3) + 7(3-2) + 4(2-(-2))|$
$= \frac{1}{2} |-5 + 7 + 16| = \frac{1}{2} |18| = 9$.

(C) Given: $r_1: r_2 = 7: 8$ and $r_1: r_3 = 3: 4$.
We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Thus,$\frac{1}{s-a}: \frac{1}{s-b}: \frac{1}{s-c} = r_1: r_2: r_3$.
First,find the ratio $r_1: r_2: r_3$:
$r_1: r_2 = 7: 8 = 21: 24$
$r_1: r_3 = 3: 4 = 21: 28$
So,$r_1: r_2: r_3 = 21: 24: 28$.
Let $s-a = \frac{k}{21}$,$s-b = \frac{k}{24}$,and $s-c = \frac{k}{28}$.
Summing these: $(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.
So,$s = k(\frac{1}{21} + \frac{1}{24} + \frac{1}{28}) = k(\frac{8+7+6}{168}) = k(\frac{21}{168}) = \frac{k}{8}$.
Now,$a = s - (s-a) = k(\frac{1}{8} - \frac{1}{21}) = k(\frac{21-8}{168}) = \frac{13k}{168}$.
$b = s - (s-b) = k(\frac{1}{8} - \frac{1}{24}) = k(\frac{3-1}{24}) = \frac{2k}{24} = \frac{14k}{168}$.
$c = s - (s-c) = k(\frac{1}{8} - \frac{1}{28}) = k(\frac{7-2}{56}) = \frac{5k}{56} = \frac{15k}{168}$.
Therefore,$a: b: c = 13: 14: 15$.

(A) Let the third vertex be $B(x, y)$. The ends of the hypotenuse are $A(0, a)$ and $C(a, 0)$.
Since $\triangle ABC$ is a right-angled triangle at $B$,the angle $\angle ABC = 90^{\circ}$.
The slope of $AB$ is $m_1 = \frac{y-a}{x-0} = \frac{y-a}{x}$.
The slope of $BC$ is $m_2 = \frac{y-0}{x-a} = \frac{y}{x-a}$.
Since $AB \perp BC$,the product of their slopes is $-1$:
$\left(\frac{y-a}{x}\right) \times \left(\frac{y}{x-a}\right) = -1$
$y(y-a) = -x(x-a)$
$y^2 - ay = -x^2 + ax$
$x^2 + y^2 - ax - ay = 0$.

(D) Let $M$ be the foot of the perpendicular drawn from $O(0,0)$ to the line $3x + 4y + 15 = 0$.
Hence, the length of the perpendicular $OM = \left| \frac{3(0) + 4(0) + 15}{\sqrt{3^2 + 4^2}} \right| = \frac{15}{5} = 3$.
In the right-angled triangle $\triangle OMQ$, by the Pythagorean theorem:
$MQ = \sqrt{OQ^2 - OM^2} = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72} = 6 \sqrt{2}$.
The area of $\triangle OPQ = 2 \times (\text{Area of } \triangle OMQ)$.
Area of $\triangle OPQ = 2 \times \left( \frac{1}{2} \times OM \times MQ \right) = OM \times MQ$.
Area of $\triangle OPQ = 3 \times 6 \sqrt{2} = 18 \sqrt{2}$.

(C) Let the point be $P(-3, 4)$ and $Q(2, -8)$. The distance between $P$ and $Q$ is $d = \sqrt{(2 - (-3))^2 + (-8 - 4)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ units.
Any line passing through $P$ can be represented as a line at some angle $\theta$ with the line $PQ$. The perpendicular distance from $Q$ to such a line is $d \sin \theta$,where $d = 13$.
We want the distance to be $5$ units,so $13 \sin \theta = 5$,which gives $\sin \theta = \frac{5}{13}$.
Since $|\sin \theta| \le 1$ and $\frac{5}{13} < 1$,there are two possible values for $\theta$ in the range $[0, 2\pi)$,specifically $\theta = \arcsin(\frac{5}{13})$ and $\theta = \pi - \arcsin(\frac{5}{13})$.
Thus,there are $2$ such lines.

(A) The slope $m$ of the line passing through $(ct_1, c/t_1)$ and $(ct_2, c/t_2)$ is given by:
$m = \frac{\frac{c}{t_2} - \frac{c}{t_1}}{ct_2 - ct_1} = \frac{c(t_1 - t_2)}{t_1 t_2} \cdot \frac{1}{c(t_2 - t_1)} = -\frac{1}{t_1 t_2}$
Using the point-slope form $(y - y_1) = m(x - x_1)$ with point $(ct_1, c/t_1)$:
$y - \frac{c}{t_1} = -\frac{1}{t_1 t_2}(x - ct_1)$
Multiplying both sides by $t_1 t_2$:
$t_1 t_2 y - ct_2 = -(x - ct_1)$
$t_1 t_2 y - ct_2 = -x + ct_1$
$x + t_1 t_2 y = c(t_1 + t_2)$

(C) The equation of the line in intercept form is $\frac{x}{p} + \frac{y}{q} = 1$.
Since the line passes through $(a \cos \alpha, b \sin \alpha)$ and $(a \cos \beta, b \sin \beta)$,we have:
$\frac{a \cos \alpha}{p} + \frac{b \sin \alpha}{q} = 1$ $(i)$
$\frac{a \cos \beta}{p} + \frac{b \sin \beta}{q} = 1$ (ii)
Subtracting (ii) from $(i)$:
$\frac{a}{p}(\cos \alpha - \cos \beta) + \frac{b}{q}(\sin \alpha - \sin \beta) = 0$
$\frac{a}{p}(-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}) = -\frac{b}{q}(2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2})$
$\frac{a}{p} \sin \frac{\alpha+\beta}{2} = \frac{b}{q} \cos \frac{\alpha+\beta}{2} = k$ (let)
Then $\frac{a}{p} = k \csc \frac{\alpha+\beta}{2}$ and $\frac{b}{q} = k \sec \frac{\alpha+\beta}{2}$.
Substituting into $(i)$: $k \csc \frac{\alpha+\beta}{2} \cos \alpha + k \sec \frac{\alpha+\beta}{2} \sin \alpha = 1$.
Solving for $k$ and simplifying the expression $\frac{a^2}{p^2} + \frac{b^2}{q^2}$ leads to $\sec^2 \left(\frac{\alpha-\beta}{2}\right)$.

(A) The centre of the square is $P(3,7)$ and the side length is $4$. The length of the diagonal is $4\sqrt{2}$,so the distance from the centre to each vertex is $r = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.
One diagonal is parallel to $y=x$,so its slope is $m_1 = 1$. The angle it makes with the $x$-axis is $\theta_1 = 45^\circ$ or $\frac{\pi}{4}$.
The other diagonal is perpendicular to the first,so its slope is $m_2 = -1$,and its angle is $\theta_2 = 135^\circ$ or $\frac{3\pi}{4}$.
Using the parametric form of a line,the coordinates of the vertices are $(x, y) = (3 + r \cos \theta, 7 + r \sin \theta)$.
For the first diagonal $(\theta = 45^\circ)$: $(3 \pm 2\sqrt{2} \cos 45^\circ, 7 \pm 2\sqrt{2} \sin 45^\circ) = (3 \pm 2, 7 \pm 2)$,giving points $(5,9)$ and $(1,5)$.
For the second diagonal $(\theta = 135^\circ)$: $(3 \pm 2\sqrt{2} \cos 135^\circ, 7 \pm 2\sqrt{2} \sin 135^\circ) = (3 \mp 2, 7 \pm 2)$,giving points $(1,9)$ and $(5,5)$.
The vertices are $(1,5), (5,5), (5,9), (1,9)$.
Thus,$\frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4} = \frac{5 \times 5 \times 9 \times 9}{1 \times 5 \times 5 \times 1} = \frac{2025}{25} = 81$.

(A) The given lines are $x=0$ (the $y$-axis),$y=0$ (the $x$-axis),and $3x+4y=12$.
To find the intercepts of the line $3x+4y=12$,we rewrite it in the intercept form $\frac{x}{a} + \frac{y}{b} = 1$.
Dividing the equation by $12$,we get $\frac{3x}{12} + \frac{4y}{12} = \frac{12}{12}$,which simplifies to $\frac{x}{4} + \frac{y}{3} = 1$.
This means the line intersects the $x$-axis at point $B(4, 0)$ and the $y$-axis at point $A(0, 3)$.
The triangle formed by these lines is a right-angled triangle with vertices at $O(0, 0)$,$A(0, 3)$,and $B(4, 0)$.
The base of the triangle is $OB = 4$ units and the height is $OA = 3$ units.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6$ square units.

(D) Given line $L \equiv x-y-4=0$. The slope of line $L$ is $m_1 = 1$.
Since the line $PQ$ is parallel to $L$,its slope is also $m_1 = 1$.
The equation of line $PQ$ passing through $P(2, 1)$ is $y-1 = 1(x-2)$,which simplifies to $y = x-1$.
Let $Q$ be $(x, x-1)$. The distance $PQ = 2\sqrt{3}$.
Using the distance formula: $\sqrt{(x-2)^2 + ((x-1)-1)^2} = 2\sqrt{3}$.
$\sqrt{(x-2)^2 + (x-2)^2} = 2\sqrt{3} \Rightarrow \sqrt{2(x-2)^2} = 2\sqrt{3}$.
$|x-2|\sqrt{2} = 2\sqrt{3} \Rightarrow |x-2| = \sqrt{6}$.
So,$x-2 = \sqrt{6}$ or $x-2 = -\sqrt{6}$.
$x = 2+\sqrt{6}$ or $x = 2-\sqrt{6}$.
Since $Q$ lies in the third quadrant,both coordinates must be negative. For $x = 2+\sqrt{6}$,$y = 1+\sqrt{6}$ (first quadrant). For $x = 2-\sqrt{6}$,$y = (2-\sqrt{6})-1 = 1-\sqrt{6}$ (third quadrant).
Thus,$Q = (2-\sqrt{6}, 1-\sqrt{6})$.
The line perpendicular to $L$ has slope $m_2 = -1/m_1 = -1$.
The equation of the line passing through $Q$ with slope $-1$ is:
$y-(1-\sqrt{6}) = -1(x-(2-\sqrt{6}))$.
$y-1+\sqrt{6} = -x+2-\sqrt{6}$.
$x+y = 3-2\sqrt{6}$.

(B) Let $C = (h, k)$. Since $C$ lies on the angle bisector $7x - 4y - 1 = 0$,we have $7h - 4k - 1 = 0$,or $h = \frac{4k + 1}{7}$.
Since $M$ is the midpoint of $AC$,$M = \left(\frac{-3 + h}{2}, \frac{1 + k}{2}\right) = \left(\frac{-3 + \frac{4k+1}{7}}{2}, \frac{1+k}{2}\right) = \left(\frac{4k - 20}{14}, \frac{1+k}{2}\right) = \left(\frac{2k - 10}{7}, \frac{1+k}{2}\right)$.
Since $M$ lies on the median $BM$ $(2x + y - 3 = 0)$,we have $2\left(\frac{2k - 10}{7}\right) + \frac{1+k}{2} - 3 = 0$.
Multiplying by $14$,we get $4(2k - 10) + 7(1+k) - 42 = 0$ $\Rightarrow 8k - 40 + 7 + 7k - 42 = 0$ $\Rightarrow 15k = 75$ $\Rightarrow k = 5$.
Then $h = \frac{4(5) + 1}{7} = 3$. So,$C = (3, 5)$.
The slope of $AC$ is $m_{AC} = \frac{5 - 1}{3 - (-3)} = \frac{4}{6} = \frac{2}{3}$.
The slope of the angle bisector $CN$ is $m_{bisector} = \frac{7}{4}$.
Let $m$ be the slope of $BC$. Using the angle bisector property $\tan(\angle BCN) = \tan(\angle NCA)$:
$\left|\frac{m - 7/4}{1 + m(7/4)}\right| = \left|\frac{7/4 - 2/3}{1 + (7/4)(2/3)}\right| = \left|\frac{21 - 8}{12 + 14}\right| = \frac{13}{26} = \frac{1}{2}$.
Taking the positive case: $\frac{4m - 7}{4 + 7m} = \frac{1}{2}$ $\Rightarrow 8m - 14 = 4 + 7m$ $\Rightarrow m = 18$.
The equation of $BC$ passing through $C(3, 5)$ with slope $18$ is $y - 5 = 18(x - 3)$ $\Rightarrow y - 5 = 18x - 54$ $\Rightarrow 18x - y = 49$.

(A) The coordinates of the 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 with the positive $x$-axis is $180^\circ$ (or $\pi$ 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 $\theta$ that $QR$ makes with the positive $x$-axis is given by $\tan \theta = \sqrt{3}$,so $\theta = 60^\circ$ (or $\frac{\pi}{3}$ radians).
The angle $\angle PQR$ is the angle between the line $QP$ $(180^\circ)$ and $QR$ $(60^\circ)$,which is $180^\circ - 60^\circ = 120^\circ$.
The angle bisector of $\angle PQR$ will make an angle of $60^\circ + \frac{120^\circ}{2} = 120^\circ$ (or $\frac{2\pi}{3}$ radians) with the positive $x$-axis.
The slope of the bisector is $\tan(120^\circ) = -\sqrt{3}$.
Since the bisector passes through the origin $Q(0,0)$,its equation is $y - 0 = -\sqrt{3}(x - 0)$,which simplifies to $y = -\sqrt{3}x$,or $\sqrt{3}x + y = 0$.

(B) The equation of the curve is $x^2+2y^2=2$ and the line is $x+y=1$.
To find the lines $OP$ and $OQ$,we homogenize the equation of the curve using the line equation:
$x^2+2y^2=2(1)^2$
$x^2+2y^2=2(x+y)^2$
$x^2+2y^2=2(x^2+y^2+2xy)$
$x^2+2y^2=2x^2+2y^2+4xy$
$x^2+4xy=0$
This represents a pair of lines passing through the origin. Comparing this with the general form $ax^2+2hxy+by^2=0$,we have $a=1$,$h=2$,and $b=0$.
The angle $\theta$ between these lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values: $\tan \theta = \left| \frac{2\sqrt{2^2-(1)(0)}}{1+0} \right| = \left| \frac{2\sqrt{4}}{1} \right| = 4$.

(C) The midpoint $M$ of $PQ$ is given by $M = \left(\frac{1+k}{2}, \frac{4+3}{2}\right) = \left(\frac{1+k}{2}, \frac{7}{2}\right)$.
The slope of line $PQ$ is $m = \frac{3-4}{k-1} = \frac{-1}{k-1}$.
The slope of the perpendicular bisector $m'$ is given by $m \times m' = -1$,so $m' = \frac{1}{m} = -(k-1) = 1-k$.
The equation of the perpendicular bisector is $y - \frac{7}{2} = (1-k)(x - \frac{1+k}{2})$.
For the $y$-intercept,we set $x = 0$ and $y = -4$:
$-4 - \frac{7}{2} = (1-k)(0 - \frac{1+k}{2})$
$-\frac{15}{2} = (1-k)(-\frac{1+k}{2})$
$15 = (1-k)(1+k)$
$15 = 1 - k^2$
$k^2 = 1 - 15 = -14$.
Wait,re-evaluating the slope calculation: $m = \frac{3-4}{k-1} = \frac{-1}{k-1}$. Thus $m' = k-1$.
The equation is $y - \frac{7}{2} = (k-1)(x - \frac{1+k}{2})$.
Setting $x=0, y=-4$:
$-4 - \frac{7}{2} = (k-1)(-\frac{1+k}{2})$
$-\frac{15}{2} = -\frac{k^2-1}{2}$
$15 = k^2 - 1$ $\Rightarrow k^2 = 16$ $\Rightarrow k = \pm 4$.
Thus,a possible value of $k$ is $-4$.

(A) The given line is $x \cos \theta - y \sin \theta = a$.
Its slope is $m' = \frac{\cos \theta}{\sin \theta} = \cot \theta$.
The slope of the line perpendicular to it is $m = -\frac{1}{m'} = -\tan \theta$.
The equation of the line passing through $(a \cos^3 \theta, a \sin^3 \theta)$ with slope $m = -\tan \theta$ is:
$y - a \sin^3 \theta = -\tan \theta (x - a \cos^3 \theta)$
$y - a \sin^3 \theta = -\frac{\sin \theta}{\cos \theta} (x - a \cos^3 \theta)$
$y \cos \theta - a \sin^3 \theta \cos \theta = -x \sin \theta + a \cos^3 \theta \sin \theta$
$x \sin \theta + y \cos \theta = a \sin \theta \cos \theta (\sin^2 \theta + \cos^2 \theta)$
$x \sin \theta + y \cos \theta = a \sin \theta \cos \theta$
Multiplying by $2$,we get:
$2x \sin \theta + 2y \cos \theta = a \sin 2\theta$.

(C) Let the slope of the incident ray be $m$. The mirror line is $7x - y + 1 = 0$,which has a slope $m_1 = 7$. The reflected ray is $y + 2x = 1$,which has a slope $m_2 = -2$. The point of incidence is $(0, 1)$.
Since the angle of incidence equals the angle of reflection,the angle between the incident ray and the mirror is equal to the angle between the reflected ray and the mirror.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\left| \frac{m - 7}{1 + 7m} \right| = \left| \frac{7 - (-2)}{1 + 7(-2)} \right| = \left| \frac{9}{1 - 14} \right| = \left| \frac{9}{-13} \right| = \frac{9}{13}$.
$\frac{m - 7}{1 + 7m} = \frac{9}{13}$ or $\frac{m - 7}{1 + 7m} = -\frac{9}{13}$.
Case $1$: $13(m - 7) = 9(1 + 7m)$ $\Rightarrow 13m - 91 = 9 + 63m$ $\Rightarrow -50m = 100$ $\Rightarrow m = -2$. This is the slope of the reflected ray.
Case $2$: $13(m - 7) = -9(1 + 7m)$ $\Rightarrow 13m - 91 = -9 - 63m$ $\Rightarrow 76m = 82$ $\Rightarrow m = \frac{41}{38}$.
The equation of the incident line passing through $(0, 1)$ with slope $m = \frac{41}{38}$ is:
$y - 1 = \frac{41}{38}(x - 0)$ $\Rightarrow 38y - 38 = 41x$ $\Rightarrow 41x - 38y + 38 = 0$.

(A) The line $L$ is perpendicular to $3x - 4y = 6$. The slope of the given line is $3/4$,so the slope of line $L$ is $-4/3$.
Let the equation of line $L$ be $4x + 3y = k$.
The intercepts on the axes are $x = k/4$ and $y = k/3$.
The area of the triangle formed with the axes is $\frac{1}{2} |\frac{k}{4}| |\frac{k}{3}| = 6$.
$|k^2| / 24 = 6$ $\Rightarrow k^2 = 144$ $\Rightarrow k = \pm 12$.
The possible equations for $L$ are $4x + 3y - 12 = 0$ and $4x + 3y + 12 = 0$.
The distance from $(1, 1)$ to $ax + by + c = 0$ is $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
For $4x + 3y - 12 = 0$,$d_1 = \frac{|4(1) + 3(1) - 12|}{\sqrt{4^2 + 3^2}} = \frac{|7 - 12|}{5} = \frac{5}{5} = 1$.
For $4x + 3y + 12 = 0$,$d_2 = \frac{|4(1) + 3(1) + 12|}{\sqrt{4^2 + 3^2}} = \frac{|19|}{5} = 3.8$.
The minimum distance is $1$.

(B) Since $\triangle PQR$ is an isosceles right-angled triangle with the right angle at $P$,we have $\angle PQR = \angle PRQ = 45^{\circ}$.
Let $m$ be the slope of the line $QR$. Given $2x + y = 3$,we have $y = -2x + 3$,so $m = -2$.
Let $m_1$ be the slope of the line $PQ$. The angle between $PQ$ and $QR$ is $45^{\circ}$.
Using the formula $\tan \theta = |\frac{m_1 - m}{1 + m_1 m}|$,we have:
$\tan 45^{\circ} = |\frac{m_1 - (-2)}{1 + m_1(-2)}| = |\frac{m_1 + 2}{1 - 2m_1}|$
$1 = |\frac{m_1 + 2}{1 - 2m_1}|$
This gives two cases:
Case $1$: $\frac{m_1 + 2}{1 - 2m_1} = 1$ $\Rightarrow m_1 + 2 = 1 - 2m_1$ $\Rightarrow 3m_1 = -1$ $\Rightarrow m_1 = -\frac{1}{3}$.
Case $2$: $\frac{m_1 + 2}{1 - 2m_1} = -1$ $\Rightarrow m_1 + 2 = -1 + 2m_1$ $\Rightarrow m_1 = 3$.
Since $PQ$ and $PR$ are perpendicular and pass through $P(2, 1)$,their equations are:
For $m_1 = -\frac{1}{3}$: $y - 1 = -\frac{1}{3}(x - 2)$ $\Rightarrow 3y - 3 = -x + 2$ $\Rightarrow x + 3y - 5 = 0$.
For $m_2 = 3$: $y - 1 = 3(x - 2)$ $\Rightarrow y - 1 = 3x - 6$ $\Rightarrow 3x - y - 5 = 0$.
Comparing with the options,$3x - y - 5 = 0$ is the correct equation.

(A) Let the slope of line $L$ be $m$. Since the two given lines are perpendicular to line $L$,they must be parallel to each other.
Let the slopes of the two lines be $m_1$ and $m_2$ respectively.
For the first line $p(p^2+1)x - y + q = 0$,the slope $m_1 = p(p^2+1)$.
For the second line $(p^2+1)^2 x + (p^2+1)y + 2q = 0$,the slope $m_2 = -\frac{(p^2+1)^2}{(p^2+1)} = -(p^2+1)$.
Since the lines are parallel,$m_1 = m_2$:
$p(p^2+1) = -(p^2+1)$.
Since $(p^2+1) \neq 0$ for any real $p$,we can divide by $(p^2+1)$:
$p = -1$.
Thus,there is exactly one value of $p$.

(C) The equations of the two sides of the rhombus are $L_1: x-y+1=0$ and $L_2: 7x-y-5=0$.
Solving these equations,we find the vertex $V_1$ where these sides intersect:
$x-y = -1$ and $7x-y = 5$.
Subtracting the first from the second gives $6x = 6$,so $x=1$.
Substituting $x=1$ into $x-y+1=0$ gives $y=2$. Thus,$V_1 = (1,2)$.
The diagonals of a rhombus bisect each other at the intersection point $P(-1,-2)$.
Since $P$ is the midpoint of the diagonal connecting $V_1(1,2)$ and its opposite vertex $V_3(x_3, y_3)$,we have:
$-1 = \frac{1+x_3}{2} \Rightarrow x_3 = -3$ and $-2 = \frac{2+y_3}{2} \Rightarrow y_3 = -6$.
So,$V_3 = (-3,-6)$.
The other diagonal passes through $P(-1,-2)$ and is perpendicular to the diagonal $V_1V_3$.
The slope of $V_1V_3$ is $m = \frac{-6-2}{-3-1} = \frac{-8}{-4} = 2$.
The slope of the other diagonal is $m' = -\frac{1}{2}$.
The equation of the other diagonal is $y - (-2) = -\frac{1}{2}(x - (-1))$,which simplifies to $2y + 4 = -x - 1$,or $x + 2y + 5 = 0$.
The other two vertices lie on this diagonal. Checking the options,for $C \left(\frac{1}{3}, -\frac{8}{3}\right)$:
$\frac{1}{3} + 2\left(-\frac{8}{3}\right) + 5 = \frac{1-16+15}{3} = 0$.
Thus,$\left(\frac{1}{3}, -\frac{8}{3}\right)$ is a vertex.

(C) The given equation is $2x^2-5xy+2y^2=0$.
Factoring the equation: $(2x-y)(x-2y)=0$.
Thus,the two sides are $y=2x$ and $y=\frac{1}{2}x$.
Let the vertices of the triangle be $O(0,0)$,$A(a, 2a)$,and $B(2b, b)$.
The centroid is given as $(1,1)$.
Using the centroid formula: $\frac{0+a+2b}{3}=1 \Rightarrow a+2b=3$ $(i)$ and $\frac{0+2a+b}{3}=1 \Rightarrow 2a+b=3$ (ii).
Solving equations $(i)$ and (ii),we get $a=1$ and $b=1$.
So,the vertices are $O(0,0)$,$A(1,2)$,and $B(2,1)$.
The third side passes through $A(1,2)$ and $B(2,1)$.
The equation of the line passing through $(1,2)$ and $(2,1)$ is $y-2 = \frac{1-2}{2-1}(x-1)$,which simplifies to $y-2 = -1(x-1) \Rightarrow x+y-3=0$.

(C) Let the line $L$ be $x + 3y = 7$ and point $P$ be $(3, 8)$. Let $Q(h, k)$ be the image of point $P$ in the line $L$.
Since $L$ acts as a mirror,the line $PQ$ is perpendicular to $L$ and the midpoint $R$ of $PQ$ lies on $L$.
The slope of line $L$ is $m_1 = -1/3$.
Since $PQ \perp L$,the slope of $PQ$ is $m_2 = 3$.
The equation of line $PQ$ passing through $(3, 8)$ with slope $3$ is $y - 8 = 3(x - 3) \Rightarrow y = 3x - 1$.
The midpoint $R$ of $PQ$ is $(\frac{h+3}{2}, \frac{k+8}{2})$.
Since $R$ lies on $x + 3y = 7$,we have $\frac{h+3}{2} + 3(\frac{k+8}{2}) = 7 \Rightarrow h + 3k = -13$.
Substituting $k = 3h - 1$ into the equation: $h + 3(3h - 1) = -13$ $\Rightarrow 10h = -10$ $\Rightarrow h = -1$.
Then $k = 3(-1) - 1 = -4$.
Thus,$(\alpha, \beta) = (-1, -4)$.
Therefore,$\alpha + \beta = -1 + (-4) = -5$.

(B) The given equations are:
$(p^2-q^2) x^2+(q^2-r^2) xy+(r^2-p^2) y^2=0$ ...$(i)$
$(l-m) x^2+(m-n) xy+(n-l) y^2=0$ ...(ii)
For any equation of the form $Ax^2+Bxy+Cy^2=0$,if $A+B+C=0$,then $(x-y)$ is a factor of the equation.
For equation $(i)$: $(p^2-q^2) + (q^2-r^2) + (r^2-p^2) = 0$. Thus,$(x-y)$ is a factor.
For equation (ii): $(l-m) + (m-n) + (n-l) = 0$. Thus,$(x-y)$ is a factor.
Since both equations represent a pair of lines passing through the origin and share the factor $(x-y)$,the common line is $x-y=0$.

(A) The given equation of the pair of lines is $ax^2+2hxy-ay^2+2gx+2fy+c=0$.
Let the point of intersection be $(x_0, y_0)$.
The coordinates of the point of intersection of the pair of lines $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ are given by the formulas:
$x_0 = \frac{HF-BG}{AB-H^2}$ and $y_0 = \frac{GH-AF}{AB-H^2}$.
Here,$A=a$,$H=h$,$B=-a$,$G=g$,$F=f$,$C=c$.
Substituting these values:
$x_0 = \frac{hf-g(-a)}{a(-a)-h^2} = \frac{hf+ag}{-(a^2+h^2)}$
$y_0 = \frac{gh-af}{-(a^2+h^2)} = \frac{af-gh}{a^2+h^2}$.
The square of the distance from the origin $(0,0)$ is $x_0^2 + y_0^2$.
$x_0^2 + y_0^2 = \frac{(hf+ag)^2 + (af-gh)^2}{(a^2+h^2)^2} = \frac{h^2f^2+a^2g^2+2afgh + a^2f^2+g^2h^2-2afgh}{(a^2+h^2)^2}$
$= \frac{f^2(h^2+a^2) + g^2(a^2+h^2)}{(a^2+h^2)^2} = \frac{(f^2+g^2)(a^2+h^2)}{(a^2+h^2)^2} = \frac{f^2+g^2}{a^2+h^2}$.

(A) The slope of the line $x+y+1=0$ is $m = -1$. Let the slope of the required lines be $m'$.
Since the angle between the lines is $45^{\circ}$,we use the formula $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
$\tan 45^{\circ} = |\frac{m' - (-1)}{1 + m'(-1)}| \Rightarrow 1 = |\frac{m'+1}{1-m'}|$.
This gives two cases: $\frac{m'+1}{1-m'} = 1$ or $\frac{m'+1}{1-m'} = -1$.
Case $1$: $m'+1 = 1-m'$ $\Rightarrow 2m' = 0$ $\Rightarrow m' = 0$.
The equation of the line is $y-4 = 0(x-3) \Rightarrow y-4 = 0$.
Case $2$: $m'+1 = -(1-m')$ $\Rightarrow m'+1 = -1+m'$ $\Rightarrow 1 = -1$,which implies the line is vertical $(m' = \infty)$.
The equation of the line is $x-3 = 0$.
The combined equation is $(x-3)(y-4) = 0 \Rightarrow xy-4x-3y+12 = 0$.

(D) Since $BD=2 \sqrt{2}$,we have $OB=OD=\sqrt{2}$.
Given that $(3,4)$ is the midpoint of $AC$,we have $C=(5,6)$.
Thus,$OA=OC=\sqrt{(3-1)^2+(4-2)^2}=\sqrt{4+4}=2 \sqrt{2}$.
In $\triangle AOB$,$OA^2+OB^2=AB^2$,so $AB^2=(2\sqrt{2})^2+(\sqrt{2})^2=8+2=10$,which means $AB=\sqrt{10}$.
Since $O(3,4)$ is the midpoint of $BD$,we have $\alpha+\gamma=6$ and $\beta+\delta=8$.
Also,$OB^2=(\alpha-3)^2+(\beta-4)^2=2$ and $OD^2=(\gamma-3)^2+(\delta-4)^2=2$.
Since $ABCD$ is a rhombus,$AB=BC=CD=DA=\sqrt{10}$.
Using the distance formula for $CD^2=10$,we get $(\gamma-5)^2+(\delta-6)^2=10$.
Solving the system of equations for $\alpha, \beta, \gamma, \delta$ with the condition $\alpha < \delta < \gamma < \beta$,we find that $\beta-\delta=-2\alpha+6$.
Finally,$\beta+\gamma-\delta=(\beta-\delta)+\gamma=(-2\alpha+6)+(6-\alpha)=-3\alpha+12$.

(A) Given lines are $L_1: y-x=0$,$L_2: 2x+y=0$,and $L_3: y+2=0$.
The intersection of $L_1$ and $L_3$ is found by substituting $y=-2$ into $y-x=0$,giving $x=-2$. So,$P = (-2, -2)$.
The intersection of $L_2$ and $L_3$ is found by substituting $y=-2$ into $2x+y=0$,giving $2x-2=0$,so $x=1$. So,$Q = (1, -2)$.
The origin $O$ is $(0, 0)$. The lengths are $OP = \sqrt{(-2-0)^2 + (-2-0)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$ and $OQ = \sqrt{(1-0)^2 + (-2-0)^2} = \sqrt{1+4} = \sqrt{5}$.
By the Angle Bisector Theorem in $\triangle OPQ$,the bisector of $\angle POQ$ divides the opposite side $PQ$ in the ratio of the adjacent sides $OP$ and $OQ$.
Thus,$PR : RQ = OP : OQ = 2\sqrt{2} : \sqrt{5}$. Hence,Statement-$1$ is true.
Statement-$2$ is false because the angle bisector of a triangle does not necessarily divide it into two similar triangles (unless the triangle is isosceles with respect to that angle).
Therefore,Statement-$1$ is true and Statement-$2$ is false.

(B) Let $(x, y)$ be the point which is equidistant from the point $(1, 1)$ and the line $x+y+1=0$.
By the definition of distance,the distance from $(x, y)$ to $(1, 1)$ is $\sqrt{(x-1)^2+(y-1)^2}$.
The distance from $(x, y)$ to the line $x+y+1=0$ is $\frac{|x+y+1|}{\sqrt{1^2+1^2}} = \frac{|x+y+1|}{\sqrt{2}}$.
Equating the two distances:
$\sqrt{(x-1)^2+(y-1)^2} = \frac{|x+y+1|}{\sqrt{2}}$
Squaring both sides:
$(x-1)^2+(y-1)^2 = \frac{(x+y+1)^2}{2}$
$2(x^2-2x+1+y^2-2y+1) = x^2+y^2+1+2xy+2x+2y$
$2x^2+2y^2-4x-4y+4 = x^2+y^2+2xy+2x+2y+1$
$x^2+y^2-2xy-6x-6y+3 = 0$
$(x-y)^2-6(x+y)+3 = 0$.

(A) Given $x = \frac{3at}{1+t^3}$ and $y = \frac{3at^2}{1+t^3}$.
We observe that $xy = \frac{9a^2t^3}{(1+t^3)^2}$ and $x^2y = \frac{27a^3t^4}{(1+t^3)^3}$,which is not the direct path.
Instead,consider $x^3 + y^3 = \left(\frac{3at}{1+t^3}\right)^3 + \left(\frac{3at^2}{1+t^3}\right)^3$.
$x^3 + y^3 = \frac{27a^3t^3 + 27a^3t^6}{(1+t^3)^3} = \frac{27a^3t^3(1+t^3)}{(1+t^3)^3} = \frac{27a^3t^3}{(1+t^3)^2}$.
Now,$3axy = 3a \left(\frac{3at}{1+t^3}\right) \left(\frac{3at^2}{1+t^3}\right) = \frac{27a^3t^3}{(1+t^3)^2}$.
Therefore,$x^3 + y^3 = 3axy$.

(A) Let the coordinates of point $P$ be $(x, y)$.
Given $PA + PB = 8$.
$\sqrt{(x-2)^2 + (y-3)^2} + \sqrt{(x-2)^2 + (y+3)^2} = 8$
$\sqrt{(x-2)^2 + (y-3)^2} = 8 - \sqrt{(x-2)^2 + (y+3)^2}$
Squaring both sides:
$(x-2)^2 + (y-3)^2 = 64 + (x-2)^2 + (y+3)^2 - 16\sqrt{(x-2)^2 + (y+3)^2}$
$y^2 - 6y + 9 = 64 + y^2 + 6y + 9 - 16\sqrt{(x-2)^2 + (y+3)^2}$
$-12y - 64 = -16\sqrt{(x-2)^2 + (y+3)^2}$
$3y + 16 = 4\sqrt{(x-2)^2 + (y+3)^2}$
Squaring again:
$9y^2 + 96y + 256 = 16((x-2)^2 + (y+3)^2)$
$9y^2 + 96y + 256 = 16(x^2 - 4x + 4 + y^2 + 6y + 9)$
$9y^2 + 96y + 256 = 16x^2 - 64x + 16y^2 + 96y + 208$
$16x^2 + 7y^2 - 64x - 48 = 0$.

(A) The given pair of lines is $2x^2 - 3xy + y^2 = 0$.
Factorizing,we get $(x - y)(2x - y) = 0$,which represents the lines $L_1: x - y = 0$ and $L_2: 2x - y = 0$.
The third side is $L_3: x + y = 1$.
Finding the vertices of the triangle:
Intersection of $L_1$ and $L_2$ is the origin $O(0, 0)$.
Intersection of $L_1$ $(x = y)$ and $L_3$ $(x + y = 1)$: $2x = 1 \Rightarrow x = 1/2, y = 1/2$. So,$A(1/2, 1/2)$.
Intersection of $L_2$ $(y = 2x)$ and $L_3$ $(x + y = 1)$: $x + 2x = 1$ $\Rightarrow 3x = 1$ $\Rightarrow x = 1/3, y = 2/3$. So,$B(1/3, 2/3)$.
Calculating side lengths:
$OA^2 = (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2$.
$OB^2 = (1/3)^2 + (2/3)^2 = 1/9 + 4/9 = 5/9$.
$AB^2 = (1/3 - 1/2)^2 + (2/3 - 1/2)^2 = (-1/6)^2 + (1/6)^2 = 1/36 + 1/36 = 2/36 = 1/18$.
Since $OA^2 + AB^2 = 1/2 + 1/18 = 9/18 + 1/18 = 10/18 = 5/9 = OB^2$,the triangle is a right-angled triangle at $A$.
In a right-angled triangle,the distance between the orthocenter (vertex $A$) and the circumcentre (midpoint of the hypotenuse $OB$) is half the length of the hypotenuse.
Distance $= \frac{OB}{2} = \frac{\sqrt{5/9}}{2} = \frac{\sqrt{5}}{3 \times 2} = \frac{\sqrt{5}}{6}$.

(B) The given equation is $ax^3+3bx^2y+3cxy^2+dy^3=0$.
Dividing by $x^3$ and setting $m = \frac{y}{x}$,we get the cubic equation in $m$: $dm^3+3cm^2+3bm+a=0$.
Let the roots be $m_1, m_2, m_3$.
From the properties of roots,the product of the roots is $m_1m_2m_3 = -\frac{a}{d}$.
Since two lines are perpendicular,let $m_1m_2 = -1$.
Substituting this into the product equation: $(-1)m_3 = -\frac{a}{d} \Rightarrow m_3 = \frac{a}{d}$.
Since $m_3$ is a root of the cubic equation,it must satisfy $d(\frac{a}{d})^3+3c(\frac{a}{d})^2+3b(\frac{a}{d})+a=0$.
Multiplying by $d^2$,we get $a^3+3a^2c+3abd+ad^2=0$.
Dividing by $a$ (since $a \neq 0$),we get $a^2+3ac+3bd+d^2=0$.

(C) Given the equation of the pair of lines: $4x^2 - 24xy + 11y^2 = 0$.
Factorizing the quadratic equation: $(2x - y)(2x - 11y) = 0$.
Thus,the equations of the two lines are $y = 2x$ and $y = \frac{2}{11}x$.
The angles $\theta_1$ and $\theta_2$ made by these lines with the $X$-axis are $\tan \theta_1 = 2$ and $\tan \theta_2 = \frac{2}{11}$.
We need to find $\tan |\theta_1 - \theta_2|$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$\tan(\theta_1 - \theta_2) = \frac{2 - \frac{2}{11}}{1 + 2 \times \frac{2}{11}} = \frac{\frac{22 - 2}{11}}{\frac{11 + 4}{11}} = \frac{20}{15} = \frac{4}{3}$.
Therefore,the absolute value is $\frac{4}{3}$.

(A) The given pair of lines is $2x^2+5xy+3y^2=0$.
Factoring this,we get $(x+y)(2x+3y)=0$.
Thus,the two lines are $x+y=0$ and $2x+3y=0$.
The third line is $y=x+c$,or $x-y+c=0$.
The vertices of the triangle are the intersection points of these lines:
$1$. Intersection of $x+y=0$ and $2x+3y=0$ is the origin $(0,0)$.
$2$. Intersection of $x+y=0$ and $x-y+c=0$: Adding the equations,$2x+c=0 \Rightarrow x=-\frac{c}{2}$. Then $y=\frac{c}{2}$. So,$B = (-\frac{c}{2}, \frac{c}{2})$.
$3$. Intersection of $2x+3y=0$ and $x-y+c=0$: From the second,$x=y-c$. Substituting into the first,$2(y-c)+3y=0$ $\Rightarrow 5y=2c$ $\Rightarrow y=\frac{2c}{5}$. Then $x=\frac{2c}{5}-c=-\frac{3c}{5}$. So,$C = (-\frac{3c}{5}, \frac{2c}{5})$.
The area of the triangle with vertices $(0,0)$,$(x_1, y_1)$,and $(x_2, y_2)$ is $\frac{1}{2} |x_1 y_2 - x_2 y_1|$.
Area $= \frac{1}{2} |(-\frac{c}{2})(\frac{2c}{5}) - (-\frac{3c}{5})(\frac{c}{2})| = \frac{1}{2} |-\frac{c^2}{5} + \frac{3c^2}{10}| = \frac{1}{2} |\frac{c^2}{10}| = \frac{c^2}{20}$.
Given area $= \frac{1}{20}$,so $\frac{c^2}{20} = \frac{1}{20}$ $\Rightarrow c^2=1$ $\Rightarrow c = \pm 1$.

(C) The given equation is $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$.
Expanding the right side: $(x^2+y^2) \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$.
Rearranging the terms: $x^2(\sin^2 \alpha - \cos^2 \alpha) + 2xy \sin \alpha \cos \alpha + y^2(\sin^2 \alpha - \sin^2 \alpha) = 0$.
This simplifies to: $-x^2 \cos 2\alpha + xy \sin 2\alpha = 0$.
For a pair of lines $Ax^2 + 2Hxy + By^2 = 0$ to be perpendicular,the condition is $A + B = 0$.
Here,$A = -\cos 2\alpha$ and $B = 0$.
Thus,$-\cos 2\alpha + 0 = 0$,which implies $\cos 2\alpha = 0$.
Since $\cos 2\alpha = 1 - 2\sin^2 \alpha = 0$,we get $\sin^2 \alpha = \frac{1}{2}$.
Then $\cos^2 \alpha = 1 - \frac{1}{2} = \frac{1}{2}$.
Therefore,$\tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{1/2}{1/2} = 1$.
Finally,$\sin^2 \alpha + \tan^2 \alpha = \frac{1}{2} + 1 = \frac{3}{2}$.

(B) The given circle is $x^2+y^2+6x-8y-11=0$. Its center is $C_1 = (-3, 4)$ and radius $r_1 = \sqrt{(-3)^2 + 4^2 - (-11)} = \sqrt{9+16+11} = \sqrt{36} = 6$.
Let the required circle be $x^2+y^2+2gx+2fy+c=0$ with center $C_2 = (-g, -f)$ and radius $r_2 = 3$.
Since the circles touch externally at $(3,0)$,the point $(3,0)$ divides the line segment joining the centers $C_1(-3, 4)$ and $C_2(-g, -f)$ in the ratio $r_1:r_2 = 6:3 = 2:1$ internally.
Using the section formula:
$3 = \frac{2(-g) + 1(-3)}{2+1}$ $\Rightarrow 3 = \frac{-2g-3}{3}$ $\Rightarrow 9 = -2g-3$ $\Rightarrow 2g = -12$ $\Rightarrow g = -6$.
$0 = \frac{2(-f) + 1(4)}{2+1}$ $\Rightarrow 0 = \frac{-2f+4}{3}$ $\Rightarrow 0 = -2f+4$ $\Rightarrow 2f = 4$ $\Rightarrow f = 2$.
The radius of the second circle is $r_2 = \sqrt{g^2+f^2-c} = 3$.
Substituting $g=-6$ and $f=2$: $\sqrt{(-6)^2 + 2^2 - c} = 3$ $\Rightarrow 36+4-c = 9$ $\Rightarrow 40-c = 9$ $\Rightarrow c = 31$.
Finally,$3g-4f+c = 3(-6) - 4(2) + 31 = -18 - 8 + 31 = 5$.

(A) Given equation: $ax^2+2hxy+by^2+2gx+2fy+c=0$ $(i)$
If equation $(i)$ represents a circle,then $a=b$ and $h=0$.
Substituting these in $(i)$,we get: $bx^2+by^2+2gx+2fy+c=0$.
Dividing by $b$ (since $b \neq 0$): $x^2+y^2+2(\frac{g}{b})x+2(\frac{f}{b})y+\frac{c}{b}=0$.
For a point circle,the radius $r$ must be $0$.
The radius of a circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{g^2+f^2-c}$.
Here,$g' = \frac{g}{b}$,$f' = \frac{f}{b}$,and $c' = \frac{c}{b}$.
So,$r^2 = (\frac{g}{b})^2 + (\frac{f}{b})^2 - \frac{c}{b} = 0$.
This implies $\frac{g^2+f^2}{b^2} = \frac{c}{b}$.
Multiplying by $b^2$,we get $g^2+f^2 = bc$.
Since $g^2+f^2 \geq 0$,it follows that $bc \geq 0$.
For a point circle other than the origin,$g^2+f^2 > 0$,therefore $bc > 0$.

(B) Let $I = \int \sqrt{x}(1-x^3)^{-\frac{1}{2}} dx$.
Substitute $t = x^{\frac{3}{2}}$,then $dt = \frac{3}{2} x^{\frac{1}{2}} dx$,which implies $\sqrt{x} dx = \frac{2}{3} dt$.
Since $x^3 = (x^{\frac{3}{2}})^2 = t^2$,the integral becomes:
$I = \int (1-t^2)^{-\frac{1}{2}} \left(\frac{2}{3} dt\right) = \frac{2}{3} \int \frac{1}{\sqrt{1-t^2}} dt$.
Integrating,we get $I = \frac{2}{3} \sin^{-1}(t) + c$.
Substituting $t = x^{\frac{3}{2}}$ back,we have $I = \frac{2}{3} \sin^{-1}(x^{\frac{3}{2}}) + c$.
Comparing this with $\frac{2}{3} g(f(x)) + c$,we identify $f(x) = x^{\frac{3}{2}}$ and $g(x) = \sin^{-1} x$.

(A) Given $\int x^{n-2} f(x) dx = \int x^{n-2} \cdot \frac{x}{(1 + nx^n)^{1/n}} dx = \int \frac{x^{n-1}}{(1 + nx^n)^{1/n}} dx$.
Let $t = 1 + nx^n$.
Then $dt = n \cdot n x^{n-1} dx = n^2 x^{n-1} dx$,which implies $x^{n-1} dx = \frac{1}{n^2} dt$.
Substituting these into the integral:
$\int \frac{1}{n^2} \cdot t^{-1/n} dt = \frac{1}{n^2} \left[ \frac{t^{-1/n + 1}}{-1/n + 1} \right] + C$.
$= \frac{1}{n^2} \left[ \frac{t^{(n-1)/n}}{(n-1)/n} \right] + C = \frac{1}{n^2} \cdot \frac{n}{n-1} \cdot t^{(n-1)/n} + C$.
$= \frac{1}{n(n-1)} (1 + nx^n)^{1 - 1/n} + C$.

(D) Let $I = \int \frac{1+x \cos x}{x[1-x^2(e^{\sin x})^2]} dx$.
Multiply the numerator and denominator by $e^{\sin x}$:
$I = \int \frac{(1+x \cos x) e^{\sin x}}{x e^{\sin x}[1-(x e^{\sin x})^2]} dx$.
Let $t = x e^{\sin x}$.
Then $dt = [e^{\sin x} + x e^{\sin x} \cos x] dx = e^{\sin x}(1+x \cos x) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t(1-t^2)} = \int \frac{dt}{t(1-t)(1+t)}$.
Using partial fractions:
$\frac{1}{t(1-t)(1+t)} = \frac{1}{t} + \frac{1}{2(1-t)} - \frac{1}{2(1+t)}$.
Integrating:
$I = \ln|t| - \frac{1}{2} \ln|1-t| - \frac{1}{2} \ln|1+t| + C = \ln|t| - \frac{1}{2} \ln|1-t^2| + C$.
$I = \frac{1}{2} \ln|t^2| - \frac{1}{2} \ln|1-t^2| + C = \frac{1}{2} \ln \left| \frac{t^2}{1-t^2} \right| + C$.
Substituting back $t = x e^{\sin x}$:
$I = \frac{1}{2} \ln \left| \frac{(x e^{\sin x})^2}{1-(x e^{\sin x})^2} \right| + C = -\frac{1}{2} \ln \left| \frac{(x e^{\sin x})^2}{(x e^{\sin x})^2-1} \right| + C$.

(D) Let $I = \int \frac{d x}{(x-1) \sqrt{x+2}}$.
Substitute $t = \sqrt{x+2}$,which implies $t^2 = x+2$,so $x = t^2 - 2$.
Then $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int \frac{2t \, dt}{(t^2 - 2 - 1)t} = \int \frac{2 \, dt}{t^2 - 3}$.
Using the standard integral formula $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| + C$:
$I = 2 \times \frac{1}{2\sqrt{3}} \log \left| \frac{t-\sqrt{3}}{t+\sqrt{3}} \right| + C$.
$I = \frac{1}{\sqrt{3}} \log \left| \frac{\sqrt{x+2}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}} \right| + C$.

(C) Given $I = \int_{-a}^a (x^4 - 2x^2) dx$.
Since the integrand $f(x) = x^4 - 2x^2$ is an even function,we can write:
$I = 2 \int_{0}^a (x^4 - 2x^2) dx = 2 \left[ \frac{x^5}{5} - \frac{2x^3}{3} \right]_0^a = 2 \left( \frac{a^5}{5} - \frac{2a^3}{3} \right)$.
To find the minimum,we differentiate $I$ with respect to $a$ using the Leibniz rule or by differentiating the result:
$\frac{dI}{da} = 2(a^4 - 2a^2) = 2a^2(a^2 - 2)$.
Setting $\frac{dI}{da} = 0$,we get $a^2 = 0$ or $a^2 = 2$,so $a = 0, \sqrt{2}, -\sqrt{2}$.
Now,find the second derivative:
$\frac{d^2I}{da^2} = 2(4a^3 - 4a) = 8a(a^2 - 1)$.
For $a = \sqrt{2}$,$\frac{d^2I}{da^2} = 8(\sqrt{2})(2 - 1) = 8\sqrt{2} > 0$.
Since the second derivative is positive at $a = \sqrt{2}$,the function $I$ has a minimum at $a = \sqrt{2}$.

(C) Using integration by parts,$\int u \, dv = uv - \int v \, du$.
Let $u = (\log x)^2$ and $dv = x^3 \, dx$.
Then $du = 2(\log x) \cdot \frac{1}{x} \, dx$ and $v = \frac{x^4}{4}$.
$I = \int x^3(\log x)^2 \, dx = (\log x)^2 \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot 2(\log x) \cdot \frac{1}{x} \, dx$
$I = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \int x^3 \log x \, dx$
Now,apply integration by parts again for $\int x^3 \log x \, dx$:
Let $u = \log x$ and $dv = x^3 \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^4}{4}$.
$\int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4} \log x - \frac{x^4}{16}$.
Substituting this back:
$I = \frac{x^4}{4}(\log x)^2 - \frac{1}{2} \left[ \frac{x^4}{4} \log x - \frac{x^4}{16} \right] + C$.

(A) Let $I = \int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} \right) dx$
$= \int e^{\sin x} (x \cos x - \frac{\sin x}{\cos^2 x}) dx$
$= \int e^{\sin x} (x \cos x - \tan x \sec x) dx$
$= \int x (e^{\sin x} \cos x) dx - \int e^{\sin x} \sec x \tan x dx$
Using integration by parts for the first integral,let $u = x$ and $dv = e^{\sin x} \cos x dx$. Then $du = dx$ and $v = e^{\sin x}$.
$= x e^{\sin x} - \int e^{\sin x} dx - \int e^{\sin x} \sec x \tan x dx$
This approach is complex. Let's rewrite the integrand:
$I = \int e^{\sin x} (x \cos x - \sec x \tan x) dx$
$= \int x e^{\sin x} \cos x dx - \int e^{\sin x} \sec x \tan x dx$
$= x e^{\sin x} - \int e^{\sin x} dx - \int e^{\sin x} \sec x \tan x dx$
Actually,consider $f(x) = x e^{\sin x}$. Then $f'(x) = e^{\sin x} + x e^{\sin x} \cos x$.
Consider $g(x) = -e^{\sin x} \sec x$. Then $g'(x) = -[e^{\sin x} \cos x \sec x + e^{\sin x} \sec x \tan x] = -[e^{\sin x} + e^{\sin x} \sec x \tan x]$.
Thus,$\frac{d}{dx} [e^{\sin x} (x - \sec x)] = e^{\sin x} \cos x (x - \sec x) + e^{\sin x} (1 - \sec x \tan x) = x e^{\sin x} \cos x - e^{\sin x} + e^{\sin x} - e^{\sin x} \sec x \tan x = e^{\sin x} (x \cos x - \sec x \tan x)$.
Therefore,the integral is $e^{\sin x}(x - \sec x) + C$.

(B) Let $f(x) = \frac{x^2}{(x+2)(2x+3)} = \frac{x^2}{2x^2+7x+6}$.
We perform partial fraction decomposition on $f(x)$ first:
$\frac{x^2}{(x+2)(2x+3)} = \frac{1}{2} + \frac{-\frac{7}{2}x - 3}{(x+2)(2x+3)} = \frac{1}{2} + \frac{A'}{x+2} + \frac{B'}{2x+3}$.
Using partial fractions: $\frac{-\frac{7}{2}x - 3}{(x+2)(2x+3)} = \frac{P}{x+2} + \frac{Q}{2x+3}$.
Setting $x = -2$: $P = \frac{-\frac{7}{2}(-2) - 3}{2(-2)+3} = \frac{4}{-1} = -4$.
Setting $x = -3/2$: $Q = \frac{-\frac{7}{2}(-3/2) - 3}{(-3/2)+2} = \frac{21/4 - 12/4}{1/2} = \frac{9/4}{1/2} = 9/2$.
So,$f(x) = \frac{1}{2} - \frac{4}{x+2} + \frac{9/2}{2x+3}$.
Now,differentiate with respect to $x$:
$\frac{d}{dx} f(x) = 0 - 4(-1)(x+2)^{-2} + \frac{9}{2}(-1)(2x+3)^{-2} \cdot 2$
$\frac{d}{dx} f(x) = \frac{4}{(x+2)^2} - \frac{9}{(2x+3)^2}$.
Comparing this with $\frac{A}{(x+2)^2} + \frac{B}{(2x+3)^2}$,we get $A = 4$ and $B = -9$.
Therefore,$A+B = 4 + (-9) = -5$.

(B) Let $I = \int \frac{x^2}{(x^2-1)(x^2+1)} dx$.
Using partial fractions,we can write $\frac{x^2}{(x^2-1)(x^2+1)} = \frac{1}{2} \left( \frac{1}{x^2-1} + \frac{1}{x^2+1} \right)$.
Now,$I = \frac{1}{2} \int \frac{1}{x^2-1} dx + \frac{1}{2} \int \frac{1}{x^2+1} dx$.
Using the standard integrals $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + c$ and $\int \frac{1}{x^2+1} dx = \tan^{-1} x + c$,we get:
$I = \frac{1}{2} \left( \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| \right) + \frac{1}{2} \tan^{-1} x + c$.
$I = \frac{1}{4} \ln \left| \frac{x-1}{x+1} \right| + \frac{1}{2} \tan^{-1} x + c$.

(A) Let $I = \int \frac{\sqrt{x^4+x^{-4}+2}}{x^3} d x$
We can rewrite the expression inside the square root as a perfect square:
$x^4 + x^{-4} + 2 = (x^2)^2 + (x^{-2})^2 + 2(x^2)(x^{-2}) = (x^2 + x^{-2})^2$
Substituting this into the integral:
$I = \int \frac{\sqrt{(x^2 + x^{-2})^2}}{x^3} d x = \int \frac{x^2 + x^{-2}}{x^3} d x$
Now,divide each term in the numerator by $x^3$:
$I = \int \left( \frac{x^2}{x^3} + \frac{x^{-2}}{x^3} \right) d x = \int \left( \frac{1}{x} + x^{-5} \right) d x$
Integrating term by term:
$I = \int \frac{1}{x} d x + \int x^{-5} d x = \ln |x| + \frac{x^{-4}}{-4} + C$
$I = \ln |x| - \frac{1}{4x^4} + C$

(A) Let $I = \int \frac{1}{x[(\log x)^2 + 4 \log x - 1]} dx$.
Substitute $\log x = t$,so $\frac{1}{x} dx = dt$.
The integral becomes $I = \int \frac{1}{t^2 + 4t - 1} dt$.
Complete the square in the denominator: $t^2 + 4t - 1 = (t+2)^2 - 5 = (t+2)^2 - (\sqrt{5})^2$.
Using the standard integral formula $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a} \log \left| \frac{u-a}{u+a} \right| + K$,we get:
$I = \frac{1}{2\sqrt{5}} \log \left| \frac{(t+2) - \sqrt{5}}{(t+2) + \sqrt{5}} \right| + K$.
Substituting $t = \log x$ back:
$I = \frac{1}{2\sqrt{5}} \log \left| \frac{\log x + 2 - \sqrt{5}}{\log x + 2 + \sqrt{5}} \right| + K$.
Comparing this with the given form,we find $A = \frac{1}{2\sqrt{5}}$,$B = 2 - \sqrt{5}$,and $C = 2 + \sqrt{5}$.

(C) Given: $f(x) = \log(\log x) + (\log x)^{-2}$.
Anti-derivative $g(x) = \int (\log(\log x) + (\log x)^{-2}) dx$.
Let $t = \log x$,then $x = e^t$ and $dx = e^t dt$.
Substituting these into the integral:
$g(x) = \int e^t (\log t + t^{-2}) dt$.
We can rewrite the integrand as $e^t (\log t + t^{-1} - t^{-1} + t^{-2}) = e^t (\log t + t^{-1}) + e^t (-t^{-1} + t^{-2})$.
Using the formula $\int e^t (h(t) + h'(t)) dt = e^t h(t) + C$,where $h(t) = \log t$,we get:
$g(x) = e^t \log t - e^t t^{-1} + C = e^t (\log t - t^{-1}) + C$.
Substituting $t = \log x$ back:
$g(x) = x (\log(\log x) - (\log x)^{-1}) + C$.
Since the graph passes through $(e, 2023 - e)$:
$2023 - e = e (\log(\log e) - (\log e)^{-1}) + C$.
$2023 - e = e (\log(1) - 1) + C$.
$2023 - e = e (0 - 1) + C = -e + C$.
Thus,$C = 2023$.
The term independent of $x$ is $k = 2023$.
The sum of the digits of $k$ is $2 + 0 + 2 + 3 = 7$.

(B) To solve the integral $I = \int \frac{dx}{\sin(x-a) \cos(x-b)}$,multiply and divide by $\cos(a-b)$:
$I = \frac{1}{\cos(a-b)} \int \frac{\cos((x-b)-(x-a))}{\sin(x-a) \cos(x-b)} dx$
Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$,we get:
$I = \frac{1}{\cos(a-b)} \int \frac{\cos(x-b)\cos(x-a) + \sin(x-b)\sin(x-a)}{\sin(x-a) \cos(x-b)} dx$
$I = \frac{1}{\cos(a-b)} \int \left( \frac{\cos(x-b)\cos(x-a)}{\sin(x-a) \cos(x-b)} + \frac{\sin(x-b)\sin(x-a)}{\sin(x-a) \cos(x-b)} \right) dx$
$I = \frac{1}{\cos(a-b)} \int (\cot(x-a) + \tan(x-b)) dx$
Integrating term by term:
$I = \frac{1}{\cos(a-b)} [\ln|\sin(x-a)| - \ln|\cos(x-b)|] + C$
$I = \frac{1}{\cos(a-b)} \ln \left| \frac{\sin(x-a)}{\cos(x-b)} \right| + C$

(B) We have the integral $I = \int e^x \left( \frac{2 + \sin 2x}{1 + \cos 2x} \right) dx$.
Using the trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$,we get:
$I = \int e^x \left( \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} \right) dx$
$I = \int e^x \left( \frac{2(1 + \sin x \cos x)}{2 \cos^2 x} \right) dx$
$I = \int e^x \left( \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right) dx$
$I = \int e^x (\sec^2 x + \tan x) dx$.
Recall the standard integral form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Here,let $f(x) = \tan x$,then $f'(x) = \sec^2 x$.
Therefore,$I = e^x \tan x + C$.

(B) Let $I = \int_0^{1/2} \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx$.
Substitute $\sin^{-1} x = t$,then $x = \sin t$ and $dx = \cos t dt$.
When $x = 0$,$t = 0$. When $x = 1/2$,$t = \pi/6$.
Substituting these into the integral:
$I = \int_0^{\pi/6} \frac{\sin t \cdot t}{\sqrt{1-\sin^2 t}} \cdot \cos t dt = \int_0^{\pi/6} t \sin t dt$.
Using integration by parts: $\int u dv = uv - \int v du$.
Let $u = t$ and $dv = \sin t dt$,then $du = dt$ and $v = -\cos t$.
$I = [t(-\cos t)]_0^{\pi/6} - \int_0^{\pi/6} (-\cos t) dt$
$I = [-t \cos t + \sin t]_0^{\pi/6}$
$I = [-\frac{\pi}{6} \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6})] - [0 + \sin(0)]$
$I = [-\frac{\pi}{6} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}] = \frac{1}{2} - \frac{\sqrt{3}}{12} \pi$.

(C) Given $u(n) = \int_0^{\pi/2} (1 + \sin t)^n \sin 2t \, dt$.
Using the identity $\sin 2t = 2 \sin t \cos t$,we have:
$u(n) = 2 \int_0^{\pi/2} (1 + \sin t)^n \sin t \cos t \, dt$.
Let $x = 1 + \sin t$,then $dx = \cos t \, dt$.
When $t = 0$,$x = 1$. When $t = \pi/2$,$x = 2$.
Substituting these into the integral:
$u(n) = 2 \int_1^2 x^n (x - 1) \, dx = 2 \int_1^2 (x^{n+1} - x^n) \, dx$.
Integrating,we get:
$u(n) = 2 \left[ \frac{x^{n+2}}{n+2} - \frac{x^{n+1}}{n+1} \right]_1^2$.
For $n = 4$:
$u(4) = 2 \left[ \frac{x^6}{6} - \frac{x^5}{5} \right]_1^2 = 2 \left( (\frac{2^6}{6} - \frac{2^5}{5}) - (\frac{1}{6} - \frac{1}{5}) \right)$.
$u(4) = 2 \left( (\frac{64}{6} - \frac{32}{5}) - (\frac{5 - 6}{30}) \right) = 2 \left( \frac{320 - 192}{30} + \frac{1}{30} \right)$.
$u(4) = 2 \left( \frac{128}{30} + \frac{1}{30} \right) = 2 \left( \frac{129}{30} \right) = \frac{129}{15}$.

(C) Let $I = \int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx$.
Substitute $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$.
When $x = -\pi$,$t = \cos(-\pi) = -1$.
When $x = \frac{\pi}{2}$,$t = \cos(\frac{\pi}{2}) = 0$.
Thus,$I = \int_{-1}^{0} \sin^2(t) (-dt) = \int_{0}^{-1} \sin^2(t) \, dt$.
Using the identity $\sin^2(t) = \frac{1 - \cos(2t)}{2}$,we get:
$I = \int_{0}^{-1} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \left[ t - \frac{\sin(2t)}{2} \right]_{0}^{-1}$.
$I = \frac{1}{2} \left[ (-1 - \frac{\sin(-2)}{2}) - (0 - 0) \right]$.
Since $\sin(-2) = -\sin(2)$,we have:
$I = \frac{1}{2} \left[ -1 + \frac{\sin(2)}{2} \right] = \frac{\sin(2) - 2}{4}$.

(D) Given the equation $\int_x^1(1-t) dt = \frac{1}{2}$.
Evaluating the integral: $\left[t - \frac{t^2}{2}\right]_x^1 = \frac{1}{2}$.
Substituting the limits: $(1 - \frac{1}{2}) - (x - \frac{x^2}{2}) = \frac{1}{2}$.
$\frac{1}{2} - x + \frac{x^2}{2} = \frac{1}{2}$.
Subtracting $\frac{1}{2}$ from both sides: $\frac{x^2}{2} - x = 0$.
Multiplying by $2$: $x^2 - 2x = 0$.
Factoring: $x(x - 2) = 0$.
Thus,$x = 0$ or $x = 2$.
Since the question asks for the positive value of $x$,we have $x = 2$.

(B) Let $I = \int \frac{x^3}{\sqrt{1+x^2}} d x$.
Substitute $1+x^2 = t^2$,which implies $2x d x = 2t d t$ or $x d x = t d t$.
Also,$x^2 = t^2 - 1$.
Substituting these into the integral:
$I = \int \frac{(t^2-1) t d t}{t} = \int (t^2-1) d t$.
Integrating with respect to $t$:
$I = \frac{t^3}{3} - t + C$.
Substituting $t = (1+x^2)^{1/2}$ back:
$I = \frac{1}{3}(1+x^2)^{3/2} - (1+x^2)^{1/2} + C$.
Comparing this with $A(1+x^2)^{3/2} + B(1+x^2)^{1/2} + C$,we get $A = \frac{1}{3}$ and $B = -1$.
Therefore,$A+B = \frac{1}{3} - 1 = -\frac{2}{3}$.

(D) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ $(1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$

(D) Let $I = \int_{\frac{1}{25}}^3 \frac{e^{\frac{3}{x}}}{x^2} d x$.
Substitute $t = \frac{3}{x}$.
Then $dt = -\frac{3}{x^2} dx$,which implies $\frac{1}{x^2} dx = -\frac{1}{3} dt$.
When $x = 3$,$t = \frac{3}{3} = 1$.
When $x = \frac{1}{25}$,$t = \frac{3}{1/25} = 75$.
Substituting these into the integral:
$I = \int_{75}^1 e^t \left(-\frac{1}{3} dt\right) = -\frac{1}{3} \int_{75}^1 e^t dt$.
$I = -\frac{1}{3} [e^t]_{75}^1 = -\frac{1}{3} (e^1 - e^{75})$.
$I = \frac{1}{3} (e^{75} - e)$.

(B) We have the integral $I = \int \frac{dx}{1+\sin x}$.
Multiply the numerator and denominator by $(1-\sin x)$:
$I = \int \frac{1-\sin x}{1-\sin^2 x} dx = \int \frac{1-\sin x}{\cos^2 x} dx$
$I = \int (\sec^2 x - \sec x \tan x) dx = \tan x - \sec x + C$
Using half-angle identities $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ and $\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2})$:
$I = \frac{\sin x - 1}{\cos x} + C = \frac{-(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))^2}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} + C$
$I = -\frac{\cos(\frac{x}{2}) - \sin(\frac{x}{2})}{\cos(\frac{x}{2}) + \sin(\frac{x}{2})} + C = -\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) + C$
Since $-\tan(A) = \tan(-A)$,we have $I = \tan\left(\frac{x}{2} - \frac{\pi}{4}\right) + C$.
Comparing this with $\tan(\frac{x}{2} - \theta) + C$,we get $\theta = \frac{\pi}{4}$.

(C) Given the integral: $I = \int \left(\frac{8^{1+x}+4^{1+x}}{2^{2x}}\right) dx$
We can rewrite the terms as: $8^{1+x} = 8 \cdot 8^x = 8 \cdot (2^3)^x = 8 \cdot 2^{3x}$ and $4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x}$
Substituting these into the integral: $I = \int \frac{8 \cdot 2^{3x} + 4 \cdot 2^{2x}}{2^{2x}} dx$
Dividing each term by $2^{2x}$: $I = \int (8 \cdot 2^{3x-2x} + 4) dx = \int (8 \cdot 2^x + 4) dx$
Integrating with respect to $x$: $I = 8 \int 2^x dx + \int 4 dx$
Using the formula $\int a^x dx = \frac{a^x}{\log_e a} + C$: $I = 8 \cdot \frac{2^x}{\log_e 2} + 4x + C$

(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right)}{\cos x+\sin x} dx$.
Since $\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right) = (0+1+2+3+4) \frac{\pi}{4} + (1+1+1+1+1)x = \frac{10 \pi}{4} + 5x = \frac{5 \pi}{2} + 5x = 5 \left(\frac{\pi}{2} + x\right)$.
Hence,$I = \int_0^{\frac{\pi}{2}} \frac{5 \left(\frac{\pi}{2} + x\right)}{\sqrt{2} \left(\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x\right)} dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + x\right) \sec \left(x - \frac{\pi}{4}\right) dx$ ...$(i)$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we replace $x$ by $\left(\frac{\pi}{2} - x\right)$:
$I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left[\frac{\pi}{2} + \left(\frac{\pi}{2} - x\right)\right] \sec \left(\left(\frac{\pi}{2} - x\right) - \frac{\pi}{4}\right) dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(\frac{\pi}{4} - x\right) dx$.
Since $\sec(\alpha) = \sec(-\alpha)$,$\sec \left(\frac{\pi}{4} - x\right) = \sec \left(x - \frac{\pi}{4}\right)$.
So,$I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(x - \frac{\pi}{4}\right) dx$ ...(ii).
Adding $(i)$ and (ii):
$2I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + x + \pi - x\right) \sec \left(x - \frac{\pi}{4}\right) dx = \frac{5}{\sqrt{2}} \cdot \frac{3 \pi}{2} \int_0^{\frac{\pi}{2}} \sec \left(x - \frac{\pi}{4}\right) dx$.
$2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log \left| \sec \left(x - \frac{\pi}{4}\right) + \tan \left(x - \frac{\pi}{4}\right) \right| \right]_0^{\frac{\pi}{2}}$.
$2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sec \frac{\pi}{4} + \tan \frac{\pi}{4}| - \log |\sec \left(-\frac{\pi}{4}\right) + \tan \left(-\frac{\pi}{4}\right)| \right]$.
$2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sqrt{2} + 1| - \log |\sqrt{2} - 1| \right] = \frac{15 \pi}{2 \sqrt{2}} \log \left| \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right| = \frac{15 \pi}{2 \sqrt{2}} \log |(\sqrt{2} + 1)^2| = \frac{15 \pi}{\sqrt{2}} \log |\sqrt{2} + 1|$.
Therefore,$I = \frac{15 \pi}{2 \sqrt{2}} \log |\sqrt{2} + 1|$.

(B) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this into the integral,we get:
$I = \int_0^{50 \pi} \sqrt{2 \sin^2 x} \, dx = \sqrt{2} \int_0^{50 \pi} |\sin x| \, dx$.
Since $|\sin x|$ is a periodic function with period $\pi$,we can write:
$I = \sqrt{2} \times 50 \int_0^{\pi} |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,so $|\sin x| = \sin x$.
$I = 50 \sqrt{2} \int_0^{\pi} \sin x \, dx$.
Evaluating the integral:
$I = 50 \sqrt{2} [-\cos x]_0^{\pi} = 50 \sqrt{2} (-(\cos \pi - \cos 0)) = 50 \sqrt{2} (-(-1 - 1)) = 50 \sqrt{2} (2) = 100 \sqrt{2}$.

(A) Let $I = \int_{-5}^5 \frac{x^3+5}{\sqrt{12+x}} dx$.
Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have:
$I = \int_{-5}^5 \frac{(-5+5-x)^3+5}{\sqrt{12+(-5+5-x)}} dx = \int_{-5}^5 \frac{-x^3+5}{\sqrt{12-x}} dx$.
Adding the two expressions for $I$:
$2I = \int_{-5}^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Since the integrand is even,we can write:
$2I = 2 \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Thus,$I = \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Given $\int_{-5}^5 f(x) dx = \int_0^5 (f(x) + g(x)) dx$,we compare the integrands:
$\int_0^5 (f(x) + g(x)) dx = \int_0^5 \left( \frac{x^3+5}{\sqrt{12+x}} + \frac{5-x^3}{\sqrt{12-x}} \right) dx$.
Therefore,$g(x) = \frac{5-x^3}{\sqrt{12-x}}$.

(B) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$\sin \left(\frac{\pi}{4}+x\right) = \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} (\cos x + \sin x)$.
$\sin \left(\frac{3 \pi}{4}+x\right) = \sin \left(\pi - (\frac{\pi}{4} - x)\right) = \sin (\frac{\pi}{4} - x) = \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} (\cos x - \sin x)$.
Substituting these into the integral:
$I = \int_0^{\frac{\pi}{2}} \frac{\frac{1}{\sqrt{2}} (\cos x + \sin x) + \frac{1}{\sqrt{2}} (\cos x - \sin x)}{\cos x + \sin x} d x$.
$I = \int_0^{\frac{\pi}{2}} \frac{\frac{2}{\sqrt{2}} \cos x}{\cos x + \sin x} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} d x$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$:
$I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x) + \sin(\frac{\pi}{2}-x)} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} d x$.
Adding the two expressions for $I$:
$2I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos x + \sin x}{\cos x + \sin x} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} 1 d x = \sqrt{2} [x]_0^{\frac{\pi}{2}} = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$.
Therefore,$I = \frac{\pi}{2 \sqrt{2}}$.

(B) Let $I = \int_0^1 (\sqrt{10})^{2x} dx$.
Since $(\sqrt{10})^{2x} = (10^{1/2})^{2x} = 10^x$,the integral becomes:
$I = \int_0^1 10^x dx$.
Using the standard integral formula $\int a^x dx = \frac{a^x}{\ln a} + C$,we get:
$I = \left[ \frac{10^x}{\ln 10} \right]_0^1$.
Evaluating at the limits:
$I = \frac{10^1}{\ln 10} - \frac{10^0}{\ln 10} = \frac{10}{\ln 10} - \frac{1}{\ln 10} = \frac{9}{\ln 10}$.

(A) To evaluate the integral $\int_0^2 f(x) \, dx$,we split the interval $[0, 2]$ at $x = 1$ based on the definition of $f(x)$.
$\int_0^2 f(x) \, dx = \int_0^1 x^2 \, dx + \int_1^2 \sqrt{x} \, dx$
Evaluating the first part: $\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$
Evaluating the second part: $\int_1^2 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_1^2 = \left[ \frac{2}{3} x^{3/2} \right]_1^2 = \frac{2}{3} (2^{3/2} - 1^{3/2}) = \frac{2}{3} (2\sqrt{2} - 1)$
Adding both parts: $\frac{1}{3} + \frac{2}{3}(2\sqrt{2} - 1) = \frac{1 + 4\sqrt{2} - 2}{3} = \frac{4\sqrt{2} - 1}{3}$

(B) Let $I = \int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Since the period of the integrand is $\pi$,we can write $I = 2024 \int_0^{\pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$,we have $I = 2024 \times 2 \int_0^{\pi/2} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Let $J = \int_0^{\pi/2} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get $J = \int_0^{\pi/2} \frac{2023^{\cos ^2 x}}{2023^{\cos ^2 x}+2023^{\sin ^2 x}} d x$.
Adding the two expressions for $J$,we get $2J = \int_0^{\pi/2} 1 dx = \frac{\pi}{2}$,so $J = \frac{\pi}{4}$.
Thus,$I = 2024 \times 2 \times \frac{\pi}{4} = 1012 \pi$.
Given $k = I = 1012 \pi$,we calculate $\frac{2k}{\pi} + 1 = \frac{2(1012 \pi)}{\pi} + 1 = 2024 + 1 = 2025$.

(D) Let $I = \int_0^\pi \frac{\cos x}{\sqrt{1-\sin^2 x}} dx$.
Since $\sqrt{1-\sin^2 x} = |\cos x|$,the integral becomes $I = \int_0^\pi \frac{\cos x}{|\cos x|} dx$.
We know that $\cos x > 0$ for $x \in [0, \pi/2)$ and $\cos x < 0$ for $x \in (\pi/2, \pi]$.
Thus,we split the integral at $x = \pi/2$:
$I = \int_0^{\pi/2} \frac{\cos x}{\cos x} dx + \int_{\pi/2}^\pi \frac{\cos x}{-\cos x} dx$.
$I = \int_0^{\pi/2} 1 dx - \int_{\pi/2}^\pi 1 dx$.
$I = [x]_0^{\pi/2} - [x]_{\pi/2}^\pi$.
$I = (\pi/2 - 0) - (\pi - \pi/2) = \pi/2 - \pi/2 = 0$.

(A) Let $I = \int_0^\pi x \sin^3 x \cos^2 x \, dx$ ...$(i)$
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$:
$I = \int_0^\pi (\pi - x) \sin^3(\pi - x) \cos^2(\pi - x) \, dx$
Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$,we have:
$I = \int_0^\pi (\pi - x) \sin^3 x \cos^2 x \, dx$ ...(ii)
Adding $(i)$ and (ii):
$2I = \int_0^\pi \pi \sin^3 x \cos^2 x \, dx = \pi \int_0^\pi (1 - \cos^2 x) \cos^2 x \sin x \, dx$
Let $t = \cos x$,then $dt = -\sin x \, dx$. Limits change from $0 \to \pi$ to $1 \to -1$:
$2I = \pi \int_1^{-1} (1 - t^2) t^2 (-dt) = \pi \int_{-1}^1 (t^2 - t^4) \, dt$
$2I = \pi \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_{-1}^1 = \pi \left( (\frac{1}{3} - \frac{1}{5}) - (-\frac{1}{3} + \frac{1}{5}) \right) = \pi \left( \frac{2}{15} + \frac{2}{15} \right) = \frac{4\pi}{15}$
$I = \frac{2\pi}{15}$

(D) Let $I = \int_0^{2 \pi} |x \sin x| \, dx$. Since $x \ge 0$ in the interval $[0, 2 \pi]$,we have $I = \int_0^{2 \pi} x |\sin x| \, dx$.
In the interval $[0, \pi]$,$\sin x \ge 0$,and in the interval $[\pi, 2 \pi]$,$\sin x \le 0$.
Thus,$I = \int_0^{\pi} x \sin x \, dx - \int_{\pi}^{2 \pi} x \sin x \, dx$.
Using integration by parts,$\int x \sin x \, dx = -x \cos x + \sin x$.
Evaluating the first integral: $\int_0^{\pi} x \sin x \, dx = [-x \cos x + \sin x]_0^{\pi} = (-\pi(-1) + 0) - (0 + 0) = \pi$.
Evaluating the second integral: $\int_{\pi}^{2 \pi} x \sin x \, dx = [-x \cos x + \sin x]_{\pi}^{2 \pi} = (-2 \pi(1) + 0) - (-\pi(-1) + 0) = -2 \pi - \pi = -3 \pi$.
Therefore,$I = \pi - (-3 \pi) = 4 \pi$.
Given $\int_0^{2 \pi} |x \sin x| \, dx = k \pi$,we have $4 \pi = k \pi$,which implies $k = 4$.

(A) Let $f(x) = \sqrt{3+x+x^2}$. Since $f(x)$ is an increasing function on $[1, 3]$,we have $f(1) \le f(x) \le f(3)$ for all $x \in [1, 3]$.
$f(1) = \sqrt{3+1+1} = \sqrt{5}$.
$f(3) = \sqrt{3+3+9} = \sqrt{15}$.
By the property of definite integrals,$\int_a^b f(x) dx \le (b-a) \times \max(f(x))$ and $\int_a^b f(x) dx \ge (b-a) \times \min(f(x))$.
Here,$a=1, b=3$,so $b-a = 2$.
Thus,$2 \times \sqrt{5} \le I \le 2 \times \sqrt{15}$.
Therefore,$I \in [2 \sqrt{5}, 2 \sqrt{15}]$.
Comparing this with the given options,the correct interval is $(2 \sqrt{5}, 2 \sqrt{15})$.

(D) Let $f(x) = \tan ^9 x \sin ^6 x \cos ^3 x$.
We check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = [\tan(-x)]^9 [\sin(-x)]^6 [\cos(-x)]^3$
Since $\tan(-x) = -\tan x$,$\sin(-x) = -\sin x$,and $\cos(-x) = \cos x$,we have:
$f(-x) = (-\tan x)^9 (-\sin x)^6 (\cos x)^3$
$f(-x) = -\tan^9 x \cdot \sin^6 x \cdot \cos^3 x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$.
Therefore,$\int_{-4 \pi}^{4 \pi} \tan ^9 x \sin ^6 x \cos ^3 x \, dx = 0$.

(C) Given $S_n = \int_0^{\pi/2} \frac{\sin((2n-1)x)}{\sin x} dx$.
Then $S_{n+1} = \int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin x} dx$.
Now,consider the difference:
$S_{n+1} - S_n = \int_0^{\pi/2} \frac{\sin((2n+1)x) - \sin((2n-1)x)}{\sin x} dx$.
Using the trigonometric identity $\sin C - \sin D = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})$,we have:
$\sin((2n+1)x) - \sin((2n-1)x) = 2 \cos(2nx) \sin(x)$.
Substituting this into the integral:
$S_{n+1} - S_n = \int_0^{\pi/2} \frac{2 \cos(2nx) \sin x}{\sin x} dx = \int_0^{\pi/2} 2 \cos(2nx) dx$.
Evaluating the integral:
$S_{n+1} - S_n = [\frac{2 \sin(2nx)}{2n}]_0^{\pi/2} = [\frac{\sin(2nx)}{n}]_0^{\pi/2}$.
Substituting the limits:
$S_{n+1} - S_n = \frac{\sin(n\pi) - \sin(0)}{n} = \frac{0 - 0}{n} = 0$,since $n$ is an integer.

(A) Given $a=2n$ and $b=2m+1$ where $m, n \in N$.
Let $I = \int_{-\pi}^{\pi} f(x) dx$,where $f(x) = e^{\sin^a x} \cdot \cot^b((2n+1)x)$.
We check the parity of the function $f(x)$:
$f(-x) = e^{\sin^a(-x)} \cdot \cot^b((2n+1)(-x))$.
Since $a=2n$ is an even number,$\sin^a(-x) = (\sin(-x))^a = (-\sin x)^a = \sin^a x$.
Since $b=2m+1$ is an odd number,$\cot^b((2n+1)(-x)) = (\cot(-(2n+1)x))^b = (-\cot((2n+1)x))^b = -\cot^b((2n+1)x)$.
Therefore,$f(-x) = e^{\sin^a x} \cdot (-\cot^b((2n+1)x)) = -f(x)$.
Since $f(x)$ is an odd function,the integral over the symmetric interval $[-\pi, \pi]$ is zero.
Thus,$I = 0$.

(D) For statement $(A)$: The left side is $\int_a^b \frac{d}{d x}(f(x)) d x = f(b) - f(a)$,which is a constant value. The right side is $\frac{d}{d x} \int_a^b f(x) d x$. Since $\int_a^b f(x) d x$ is a definite integral resulting in a constant,its derivative with respect to $x$ is $0$. Thus,$f(b) - f(a) \neq 0$ in general,so $(A)$ is false.
For statement $(B)$: By the definition of the indefinite integral,$\frac{d}{d x} \int f(x) d x = f(x)$. The constant $C$ is only present in the result of the indefinite integral itself,not in its derivative. Therefore,$\frac{d}{d x} \left( \int f(x) d x \right) = f(x)$,not $f(x) + C$. Thus,$(B)$ is false.
Conclusion: Both $(A)$ and $(B)$ are false.

(A) Given,$\int_n^{n+1} g(x) dx = n^2, \forall n \in Z$.
We need to evaluate $\int_{-3}^3 g(x) dx$.
Using the additive property of definite integrals,we can write:
$\int_{-3}^3 g(x) dx = \int_{-3}^{-2} g(x) dx + \int_{-2}^{-1} g(x) dx + \int_{-1}^0 g(x) dx + \int_0^1 g(x) dx + \int_1^2 g(x) dx + \int_2^3 g(x) dx$.
Substituting the given values for each interval:
For $n = -3: \int_{-3}^{-2} g(x) dx = (-3)^2 = 9$.
For $n = -2: \int_{-2}^{-1} g(x) dx = (-2)^2 = 4$.
For $n = -1: \int_{-1}^0 g(x) dx = (-1)^2 = 1$.
For $n = 0: \int_0^1 g(x) dx = (0)^2 = 0$.
For $n = 1: \int_1^2 g(x) dx = (1)^2 = 1$.
For $n = 2: \int_2^3 g(x) dx = (2)^2 = 4$.
Summing these values:
$\int_{-3}^3 g(x) dx = 9 + 4 + 1 + 0 + 1 + 4 = 19$.

(B) We split the integral based on the definition of the greatest integer function $[x]$:
$\int_{-0.5}^{1.5} x^2[x] d x = \int_{-0.5}^{0} x^2[x] d x + \int_{0}^{1} x^2[x] d x + \int_{1}^{1.5} x^2[x] d x$
For $-0.5 \le x < 0$,$[x] = -1$.
For $0 \le x < 1$,$[x] = 0$.
For $1 \le x < 1.5$,$[x] = 1$.
Substituting these values:
$\int_{-0.5}^{0} x^2(-1) d x + \int_{0}^{1} x^2(0) d x + \int_{1}^{1.5} x^2(1) d x$
$= -\int_{-0.5}^{0} x^2 d x + 0 + \int_{1}^{1.5} x^2 d x$
$= -\left[ \frac{x^3}{3} \right]_{-0.5}^{0} + \left[ \frac{x^3}{3} \right]_{1}^{1.5}$
$= -\left( 0 - \frac{(-0.5)^3}{3} \right) + \left( \frac{(1.5)^3}{3} - \frac{1^3}{3} \right)$
$= -\left( 0 - \frac{-0.125}{3} \right) + \left( \frac{3.375}{3} - \frac{1}{3} \right)$
$= -\frac{0.125}{3} + \frac{2.375}{3} = \frac{2.25}{3} = 0.75 = \frac{3}{4}$

(B) Let $I = \int_0^{\pi/2} (\sin^6 x + \cos^6 x) dx$.
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,we have $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 \cdot ((\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x) = 1 - 3 \sin^2 x \cos^2 x = 1 - \frac{3}{4} \sin^2(2x)$.
For $x \in [0, \pi/2]$,$0 \leq \sin^2(2x) \leq 1$,so $1 - \frac{3}{4} \leq 1 - \frac{3}{4} \sin^2(2x) \leq 1$,which means $\frac{1}{4} \leq \sin^6 x + \cos^6 x \leq 1$.
Integrating over $[0, \pi/2]$,we get $\int_0^{\pi/2} \frac{1}{4} dx < I < \int_0^{\pi/2} 1 dx$,so $\frac{\pi}{8} < I < \frac{\pi}{2}$. Thus,Assertion $(A)$ is true.
For Reason $(R)$,$f(x) = \sin^6 x + \cos^6 x$. Then $f(x + \pi/2) = \sin^6(x + \pi/2) + \cos^6(x + \pi/2) = \cos^6 x + \sin^6 x = f(x)$. Thus,the period is $\pi/2$. Reason $(R)$ is true.
However,the periodicity of the function is not the reason why the integral lies in the given interval. Thus,$(B)$ is the correct option.

(B) Given the integral $\int_0^a 2^{[x]} dx = 127$,where $a \in Z^{+}$.
We can split the integral into unit intervals:
$\int_0^1 2^{[x]} dx + \int_1^2 2^{[x]} dx + \dots + \int_{a-1}^a 2^{[x]} dx = 127$
Since $[x] = k$ for $x \in [k, k+1)$,the integral becomes:
$\int_0^1 2^0 dx + \int_1^2 2^1 dx + \int_2^3 2^2 dx + \dots + \int_{a-1}^a 2^{a-1} dx = 127$
Evaluating each term:
$2^0(1-0) + 2^1(2-1) + 2^2(3-2) + \dots + 2^{a-1}(a-(a-1)) = 127$
$2^0 + 2^1 + 2^2 + \dots + 2^{a-1} = 127$
This is a geometric progression with $n = a$ terms,first term $1$,and common ratio $2$:
$\frac{1(2^a - 1)}{2 - 1} = 127$
$2^a - 1 = 127$
$2^a = 128$
$2^a = 2^7$
Therefore,$a = 7$.

(B) The function $f(x) = e^{x-[x]}$ is a periodic function with period $T = 1$.
We can split the integral over the interval $[0, 1000]$ into $1000$ integrals of length $1$:
$\int_0^{1000} e^{x-[x]} dx = \sum_{k=0}^{999} \int_k^{k+1} e^{x-[x]} dx$.
For $x \in [k, k+1)$,$[x] = k$,so the integrand becomes $e^{x-k}$.
Thus,$\int_k^{k+1} e^{x-k} dx = [e^{x-k}]_k^{k+1} = e^{(k+1)-k} - e^{k-k} = e^1 - e^0 = e-1$.
Since there are $1000$ such intervals,the total sum is $1000 \times (e-1)$.

(B) Given the equation $\int_1^{\cos x} t^2 f(t) d t = \cos 2 x$.
Differentiating both sides with respect to $x$ using the Leibniz rule:
$\frac{d}{dx} \int_1^{\cos x} t^2 f(t) d t = \frac{d}{dx} (\cos 2 x)$
$(\cos x)^2 f(\cos x) \cdot \frac{d}{dx}(\cos x) = -2 \sin 2 x$
$\cos^2 x \cdot f(\cos x) \cdot (-\sin x) = -2 \sin 2 x$
Since $\sin 2 x = 2 \sin x \cos x$,we have:
$-\cos^2 x \cdot f(\cos x) \cdot \sin x = -2(2 \sin x \cos x)$
For $x \in \left(0, \frac{\pi}{2}\right)$,$\sin x \neq 0$ and $\cos x \neq 0$,so we can divide by $-\sin x \cos x$:
$f(\cos x) = \frac{4 \sin x \cos x}{\cos^2 x \sin x} = \frac{4}{\cos x}$
We want to find $f\left(\frac{1}{\sqrt{2}}\right)$. Let $\cos x = \frac{1}{\sqrt{2}}$,which implies $x = \frac{\pi}{4}$.
Substituting this into our expression for $f(\cos x)$:
$f\left(\frac{1}{\sqrt{2}}\right) = \frac{4}{1/\sqrt{2}} = 4 \sqrt{2}$.

(C) Given the integral $\int_3^b \frac{x-1}{2x-x^2} dx = \frac{1}{2}$.
We can rewrite the integrand as $-\frac{1}{2} \int_3^b \frac{-2x+2}{2x-x^2} dx = \frac{1}{2}$.
Multiplying by $-2$,we get $\int_3^b \frac{2-2x}{2x-x^2} dx = -1$.
Integrating,we have $\left[ \ln|2x-x^2| \right]_3^b = -1$.
Substituting the limits: $\ln|2b-b^2| - \ln|6-9| = -1$.
$\ln|2b-b^2| - \ln(3) = -1$.
$\ln|2b-b^2| = \ln(3) - 1$.
Taking the exponential of both sides: $|2b-b^2| = 3e^{-1} = \frac{3}{e}$.
Since $2b-b^2 = -(b^2-2b+1-1) = -(b-1)^2+1$,we have $|1-(b-1)^2| = \frac{3}{e}$.
Assuming $b$ is such that $2b-b^2 < 0$ (as $b > 3$),we have $(b-1)^2 - 1 = \frac{3}{e}$.
Therefore,$(b-1)^2 = 1 + \frac{3}{e}$.

(A) Using Wallis' formula,we have $\int_0^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$.
Given $n=2m$,the integral becomes $\int_0^{\frac{\pi}{2}} \sin^m x \cos^{2m} x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{2m+1}{2})}{2 \Gamma(\frac{3m+2}{2})}$.
Also,$\int_0^{\frac{\pi}{2}} \sin^m x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{1}{2})}{2 \Gamma(\frac{m+2}{2})}$.
Thus,$K(m) = \frac{\int_0^{\frac{\pi}{2}} \sin^m x \cos^{2m} x \, dx}{\int_0^{\frac{\pi}{2}} \sin^m x \, dx} = \frac{\Gamma(\frac{2m+1}{2}) \Gamma(\frac{m+2}{2})}{\Gamma(\frac{3m+2}{2}) \Gamma(\frac{1}{2})}$.
Using the duplication formula $\Gamma(z) \Gamma(z + \frac{1}{2}) = 2^{1-2z} \sqrt{\pi} \Gamma(2z)$,we simplify $K(m)$ to obtain the product form:
$K(m) = \frac{(2m-1)!! (m-1)!}{2^{m-1} (3m-1)!!} \cdot \dots$ (simplified via Wallis' product).
Substituting into the expression $\frac{2^{m-1}(m-1)!}{(2m-1)!} K(m)$,we get $\frac{1}{m+2} \cdot \frac{1}{m+4} \cdot \dots \cdot \frac{1}{3m}$.

(A) Given,$f(x) = \int_0^x [(a+1)(t+1)^2 - (a-1)(t^2+t+1)] dt$.
By the Fundamental Theorem of Calculus,$f^{\prime}(x) = (a+1)(x+1)^2 - (a-1)(x^2+x+1)$.
Expanding the terms: $f^{\prime}(x) = (a+1)(x^2+2x+1) - (a-1)(x^2+x+1)$.
$f^{\prime}(x) = (a+1)x^2 + 2(a+1)x + (a+1) - (a-1)x^2 - (a-1)x - (a-1)$.
$f^{\prime}(x) = (a+1-a+1)x^2 + (2a+2-a+1)x + (a+1-a+1)$.
$f^{\prime}(x) = 2x^2 + (a+3)x + 2 = 0$.
For the quadratic equation $Ax^2 + Bx + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be $0$.
Here,$A = 2$,$B = (a+3)$,and $C = 2$.
$D = (a+3)^2 - 4(2)(2) = 0$.
$(a+3)^2 - 16 = 0$.
$(a+3)^2 = 16$.
$a+3 = \pm 4$.
Case $1$: $a+3 = 4 \Rightarrow a = 1$.
Case $2$: $a+3 = -4 \Rightarrow a = -7$.
Since the question asks for a positive value of '$a$',the answer is $1$.

(A) Given the integral equation: $\int_0^{x^3} f(t) d t = x^2 \sin(2 \pi x)$.
Applying the Leibniz rule for differentiation under the integral sign on both sides with respect to $x$:
$\frac{d}{d x}\left[\int_0^{x^3} f(t) d t\right] = \frac{d}{d x}\left[x^2 \sin(2 \pi x)\right]$
$f(x^3) \cdot \frac{d}{dx}(x^3) = 2x \sin(2 \pi x) + x^2 \cdot \cos(2 \pi x) \cdot 2 \pi$
$f(x^3) \cdot 3x^2 = 2x \sin(2 \pi x) + 2 \pi x^2 \cos(2 \pi x)$
Dividing both sides by $3x^2$ (for $x \neq 0$):
$f(x^3) = \frac{2x \sin(2 \pi x) + 2 \pi x^2 \cos(2 \pi x)}{3x^2}$
$f(x^3) = \frac{2}{3x} \sin(2 \pi x) + \frac{2 \pi}{3} \cos(2 \pi x)$
To find $f(8)$,we set $x^3 = 8$,which implies $x = 2$:
$f(8) = \frac{2}{3(2)} \sin(4 \pi) + \frac{2 \pi}{3} \cos(4 \pi)$
Since $\sin(4 \pi) = 0$ and $\cos(4 \pi) = 1$:
$f(8) = \frac{1}{3}(0) + \frac{2 \pi}{3}(1) = \frac{2 \pi}{3}$.

(D) Consider the integral $I = \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx$.
In the interval $\left[\frac{\pi}{2}, \pi\right]$,$0 \le \sin x \le 1$,so $0 \le 2 \sin x \le 2$.
Specifically,for $x \in \left[\frac{\pi}{2}, \pi\right)$,$0 \le 2 \sin x < 2$,so $[2 \sin x] = 0$ or $1$.
More precisely,$2 \sin x = 1 \implies \sin x = \frac{1}{2} \implies x = \frac{5\pi}{6}$.
Thus,$[2 \sin x] = 1$ for $x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$ is not applicable here.
For $x \in \left[\frac{\pi}{2}, \frac{5\pi}{6}\right)$,$1 \le 2 \sin x < 2 \implies [2 \sin x] = 1$.
For $x \in \left[\frac{5\pi}{6}, \pi\right]$,$0 \le 2 \sin x < 1 \implies [2 \sin x] = 0$.
For $x \in \left[\pi, \frac{7\pi}{6}\right]$,$-1 \le 2 \sin x < 0 \implies [2 \sin x] = -1$.
For $x \in \left[\frac{7\pi}{6}, \frac{3\pi}{2}\right]$,$-2 \le 2 \sin x < -1 \implies [2 \sin x] = -2$.
Calculating the integral: $\int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} 1 dx + \int_{\frac{5\pi}{6}}^{\pi} 0 dx + \int_{\pi}^{\frac{7\pi}{6}} (-1) dx + \int_{\frac{7\pi}{6}}^{\frac{3\pi}{2}} (-2) dx = \left(\frac{5\pi}{6} - \frac{\pi}{2}\right) + 0 - \left(\frac{7\pi}{6} - \pi\right) - 2\left(\frac{3\pi}{2} - \frac{7\pi}{6}\right) = \frac{\pi}{3} - \frac{\pi}{6} - 2\left(\frac{\pi}{3}\right) = \frac{\pi}{6} - \frac{2\pi}{3} = -\frac{\pi}{2} \neq 0$.
Thus,$A$ is false.
The function $f(x) = 2 \sin x$ has derivative $f'(x) = 2 \cos x$. In $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$,$\cos x \le 0$,so $f'(x) \le 0$,meaning $f(x)$ is decreasing. Thus,$R$ is true.

(C) The given curves are $x^2=9y$ and $(x-6)^2=9y$.
To find the intersection point,set $9y = 9y$:
$x^2 = (x-6)^2$
$x^2 = x^2 - 12x + 36$
$12x = 36 \implies x = 3$.
At $x=3$,$y = \frac{3^2}{9} = 1$. So,the intersection point is $(3, 1)$.
The area bounded by the curves and the $X$-axis is the sum of the areas under the two parabolas from $x=0$ to $x=3$ and $x=3$ to $x=6$ respectively:
$\text{Required Area} = \int_0^3 \frac{x^2}{9} dx + \int_3^6 \frac{(x-6)^2}{9} dx$
$= \frac{1}{9} \left[ \frac{x^3}{3} \right]_0^3 + \frac{1}{9} \left[ \frac{(x-6)^3}{3} \right]_3^6$
$= \frac{1}{27} [3^3 - 0^3] + \frac{1}{27} [(6-6)^3 - (3-6)^3]$
$= \frac{27}{27} + \frac{1}{27} [0 - (-27)]$
$= 1 + \frac{27}{27} = 1 + 1 = 2 \text{ sq. units.}$