AP EAMCET 2020 Mathematics Question Paper with Answer and Solution

(A) Given the equation of the pair of lines: $f(x, y) = 3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$.
To find the point of intersection,we solve the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial f}{\partial x} = 6x - 11y - 7 = 0$
$\frac{\partial f}{\partial y} = -11x + 20y + 13 = 0$
Solving these simultaneous equations using the cross-multiplication method:
$\frac{x}{(-11)(13) - (20)(-7)} = \frac{-y}{(6)(13) - (-11)(-7)} = \frac{1}{(6)(20) - (-11)(-11)}$
$\frac{x}{-143 + 140} = \frac{-y}{78 - 77} = \frac{1}{120 - 121}$
$\frac{x}{-3} = \frac{-y}{1} = \frac{1}{-1}$
$x = 3$ and $y = 1$.
Thus,the point of intersection is $(3, 1)$.

(A) The perpendicular bisector of the line segment joining points $A(3,4)$ and $B(-1,2)$ passes through the midpoint of $AB$ and is perpendicular to the line segment $AB$.
First,find the midpoint $M$ of $AB$:
$M = \left(\frac{3 + (-1)}{2}, \frac{4 + 2}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3)$.
Next,find the slope of line segment $AB$:
$m_{AB} = \frac{2 - 4}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}$.
The slope of the perpendicular bisector $(m_{\perp})$ is the negative reciprocal of $m_{AB}$:
$m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{1/2} = -2$.
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(1, 3)$ and slope $-2$:
$y - 3 = -2(x - 1)$
$y - 3 = -2x + 2$
$2x + y - 5 = 0$.
Thus,the correct option is $A$.

(B) Let the third vertex be $C = (h, k)$.
Given vertices are $A = (2, -3)$ and $B = (-2, 1)$.
The centroid $G$ of $\triangle ABC$ is given by the formula $G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$.
Substituting the values,$G = \left(\frac{2 - 2 + h}{3}, \frac{-3 + 1 + k}{3}\right) = \left(\frac{h}{3}, \frac{k - 2}{3}\right)$.
Since the centroid $G$ lies on the line $2x + 3y = 1$,we substitute the coordinates of $G$ into the line equation:
$2\left(\frac{h}{3}\right) + 3\left(\frac{k - 2}{3}\right) = 1$.
Multiplying the entire equation by $3$,we get $2h + 3(k - 2) = 3$.
$2h + 3k - 6 = 3$.
$2h + 3k = 9$.
Replacing $(h, k)$ with $(x, y)$,the locus of vertex $C$ is $2x + 3y = 9$.
Thus,option $B$ is correct.

(A) Let the point be $P(x, y)$. The distance from the origin $O(0, 0)$ is $\sqrt{x^2+y^2}$.
The distance from $A(-2, -3)$ is $\sqrt{(x+2)^2+(y+3)^2}$.
Given the ratio $\frac{OP}{AP} = \frac{5}{7}$,we have $\frac{\sqrt{x^2+y^2}}{\sqrt{(x+2)^2+(y+3)^2}} = \frac{5}{7}$.
Squaring both sides,we get $\frac{x^2+y^2}{(x+2)^2+(y+3)^2} = \frac{25}{49}$.
$49(x^2+y^2) = 25(x^2+4x+4+y^2+6y+9)$.
$49(x^2+y^2) = 25(x^2+y^2+4x+6y+13)$.
$49(x^2+y^2) - 25(x^2+y^2) - 100x - 150y - 325 = 0$.
$24(x^2+y^2) - 100x - 150y - 325 = 0$.
Thus,option $A$ is correct.

(C) The distance of point $P(x, y)$ from the $x$-axis is $|y|$ and from the $y$-axis is $|x|$.
The sum of squares of these distances is $x^2 + y^2$.
The distance of point $P(x, y)$ from the line $x-y-1=0$ is given by $d = \frac{|x-y-1|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y-1|}{\sqrt{2}}$.
According to the problem,the sum of squares of distances from the axes is equal to the square of the distance from the line:
$x^2 + y^2 = \left( \frac{|x-y-1|}{\sqrt{2}} \right)^2$
$x^2 + y^2 = \frac{(x-y-1)^2}{2}$
$2(x^2 + y^2) = x^2 + y^2 + 1 - 2xy - 2x + 2y$
$2x^2 + 2y^2 = x^2 + y^2 + 1 - 2xy - 2x + 2y$
$x^2 + y^2 + 2xy + 2x - 2y - 1 = 0$.

(D) Given the equation: $\sqrt{(x-2)^2+y^2}+\sqrt{(x+2)^2+y^2}=4$.
Let $P = (x, y)$,$A = (2, 0)$,and $B = (-2, 0)$.
The equation represents the sum of distances from point $P$ to points $A$ and $B$,which is $PA + PB = 4$.
The distance between $A(2, 0)$ and $B(-2, 0)$ is $AB = \sqrt{(2 - (-2))^2 + (0 - 0)^2} = \sqrt{4^2} = 4$.
Since $PA + PB = AB$,the point $P$ must lie on the line segment connecting $A$ and $B$.
Given the condition $-2 < x < 2$ and $y=0$,this represents the line segment on the $x$-axis between $x = -2$ and $x = 2$.
Thus,the equation represents a line segment.
Hence,option $D$ is correct.

(B) Let the two perpendicular lines be the $X$-axis and $Y$-axis. Let $P(x, y)$ be any point on the locus.
The distance of point $P(x, y)$ from the $X$-axis is $|y|$ and from the $Y$-axis is $|x|$.
According to the problem,the sum of these distances is $1$.
Therefore,$|x| + |y| = 1$.
This equation represents a square with vertices at $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$.
Hence,the correct option is $B$.

(A) Let the point $P$ be $(x, y)$. The coordinates of $A$ are $(1, 1)$ and $B$ are $(-2, 3)$.
Given that the area of $\triangle PAB = 9 \text{ sq. units}$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the coordinates: $\frac{1}{2} |x(1 - 3) + 1(3 - y) + (-2)(y - 1)| = 9$.
$\frac{1}{2} |-2x + 3 - y - 2y + 2| = 9$.
$|-2x - 3y + 5| = 18$.
This implies $-2x - 3y + 5 = 18$ or $-2x - 3y + 5 = -18$.
Case $1$: $-2x - 3y = 13 \Rightarrow 2x + 3y + 13 = 0$.
Case $2$: $-2x - 3y = -23 \Rightarrow 2x + 3y - 23 = 0$.
Thus,the locus is $2x + 3y + 13 = 0$ and $2x + 3y - 23 = 0$.

(C) Given that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{k_1}{k_2} = \lambda$ (say).
This implies that $a_1 = \lambda a_2$,$b_1 = \lambda b_2$,and $k_1 = \lambda k_2$.
Thus,the equation $p = 0$ can be written as $\lambda a_2 x + \lambda b_2 y + \lambda k_2 = 0$,which simplifies to $\lambda(a_2 x + b_2 y + k_2) = 0$,or $\lambda q = 0$.
Since $\lambda \neq 0$,this means $p = 0$ and $q = 0$ represent the same straight line.
The equation of the curve is $p + c q = 0$.
Substituting $p = \lambda q$,we get $\lambda q + c q = 0$,which is $(\lambda + c) q = 0$.
This represents the same straight line as $q = 0$ (or $p = 0$) for any constant $c$ such that $\lambda + c \neq 0$.

(D) Let the point be $P(x, y)$. The vertices of the triangle are $A(1, 2)$,$B(-2, 5)$,and $P(x, y)$.
The area of the triangle is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Substituting the values: $8 = \frac{1}{2} |1(5 - y) + (-2)(y - 2) + x(2 - 5)|$.
$16 = |5 - y - 2y + 4 - 3x| = |9 - 3y - 3x|$.
$16 = |9 - 3(x + y)|$.
This gives two cases:
Case $1$: $9 - 3(x + y) = 16 \implies -3(x + y) = 7 \implies 3x + 3y + 7 = 0$.
Case $2$: $9 - 3(x + y) = -16 \implies -3(x + y) = -25 \implies 3x + 3y - 25 = 0$.
Thus,the locus is $3x + 3y + 7 = 0$ and $3x + 3y - 25 = 0$.

(B) homogeneous equation of second degree in $x$ and $y$ is given by $ax^2 + 2hxy + by^2 = 0$.
This equation represents a pair of straight lines passing through the origin,provided that $h^2 \geq ab$.

(A) The pair of lines $A x^2+2 H x y+B y^2=0$ passes through the origin. For these lines to form an equilateral triangle with the line $a x+b y+c=0$,the angle between the pair of lines must be $60^\circ$ or $\frac{\pi}{3}$ radians.
The angle $\theta$ between the lines $A x^2+2 H x y+B y^2=0$ is given by $\tan \theta = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.
Setting $\theta = \frac{\pi}{3}$,we have $\tan \frac{\pi}{3} = \sqrt{3}$.
So,$\sqrt{3} = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.
Squaring both sides,we get $3 = \frac{4(H^2-A B)}{(A+B)^2}$.
$3(A+B)^2 = 4(H^2-A B)$.
$3(A^2+2 A B+B^2) = 4 H^2-4 A B$.
$3 A^2+6 A B+3 B^2 = 4 H^2-4 A B$.
$3 A^2+10 A B+3 B^2 = 4 H^2$.
Factoring the left side: $3 A^2+9 A B+A B+3 B^2 = 4 H^2$.
$3 A(A+3 B)+B(A+3 B) = 4 H^2$.
$(A+3 B)(3 A+B) = 4 H^2$.
Thus,the correct option is $A$.

(B) It is given that one of the lines $2x^2 - xy + by^2 = 0$ passes through the point $(-4, -2)$.
Substituting the point $(-4, -2)$ into the equation:
$2(-4)^2 - (-4)(-2) + b(-2)^2 = 0$
$2(16) - (8) + b(4) = 0$
$32 - 8 + 4b = 0$
$24 + 4b = 0$
$4b = -24$
$b = -6$
Therefore,$b^2 = (-6)^2 = 36$.
Hence,option $B$ is correct.

(C) The given equation is $x^2-7xy+12y^2=0$. Comparing this with the general form $ax^2+2hxy+by^2=0$,we have $a=1$,$2h=-7$,and $b=12$.
The angle $\theta$ between the pair of lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{(-7/2)^2 - 1 \cdot 12}}{1+12} \right| = \left| \frac{2\sqrt{49/4 - 12}}{13} \right| = \left| \frac{2\sqrt{1/4}}{13} \right| = \frac{2 \cdot (1/2)}{13} = \frac{1}{13}$.
Since $\tan \theta = \frac{1}{13}$,we can construct a right-angled triangle with opposite side $1$ and adjacent side $13$. The hypotenuse is $\sqrt{1^2+13^2} = \sqrt{1+169} = \sqrt{170}$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{170}}$.
Hence,option $(C)$ is correct.

(C) The given equation is $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2$,$2h = 5 \Rightarrow h = \frac{5}{2}$,and $b = 3$.
The angle $\theta$ between the pair of straight lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Substituting the values: $\tan \theta = \left| \frac{2\sqrt{(\frac{5}{2})^2 - (2)(3)}}{2 + 3} \right| = \left| \frac{2\sqrt{\frac{25}{4} - 6}}{5} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{5} \right| = \left| \frac{2 \times \frac{1}{2}}{5} \right| = \frac{1}{5}$.
Since the angle is $\tan^{-1}(k)$,we have $\tan \theta = k$.
Thus,$k = \pm \frac{1}{5}$.

(D) The equation of the angle bisectors for the pair of lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the first pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.
The angle bisectors are $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$,which simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,or $x^2-y^2+\frac{2 x y}{p}=0$.
Given that this pair of bisectors is the same as the second pair $x^2-2 q x y-y^2=0$,we compare the coefficients.
Comparing $x^2-y^2+\frac{2}{p} x y=0$ with $x^2-2 q x y-y^2=0$,we identify that the coefficient of $x y$ must match:
$\frac{2}{p} = -2 q$.
Therefore,$p q = -1$.

(B) Given the equation of the pair of straight lines is $6x^2 - 5xy + y^2 = 0$.
Dividing by $x^2$ (assuming $x \neq 0$),we get:
$\left(\frac{y}{x}\right)^2 - 5\left(\frac{y}{x}\right) + 6 = 0$.
Let $m = \frac{y}{x}$,then $m^2 - 5m + 6 = 0$.
Factoring the quadratic equation:
$(m - 3)(m - 2) = 0$.
Thus,the slopes of the lines are $m_1 = \tan \alpha = 3$ and $m_2 = \tan \beta = 2$.
Using the formula for the tangent of the difference of two angles:
$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Substituting the values:
$\tan(\alpha - \beta) = \frac{3 - 2}{1 + (3)(2)} = \frac{1}{1 + 6} = \frac{1}{7}$.

(A) The condition for a pair of straight lines represented by the general equation $ax^2 + 2hxy + by^2 = 0$ to be perpendicular is that the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a + b = 0$.
For option $A$: $2 x^2 = y(x + 2 y)$
$\Rightarrow 2 x^2 - xy - 2 y^2 = 0$
Here,$a = 2$ and $b = -2$.
Sum of coefficients: $a + b = 2 + (-2) = 0$.
Since the condition $a + b = 0$ is satisfied,the lines represented by this equation are perpendicular.

(B) The pair of straight lines passing through the origin is given by the equation $ax^2 + 2hxy + by^2 = 0$.
The angle $\theta$ between these lines is given by the formula $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
When the lines are perpendicular,the angle $\theta = 90^{\circ}$.
Since $\tan 90^{\circ}$ is undefined (approaches $\infty$),the denominator must be zero.
Therefore,$a + b = 0$.

(A) The acute angle $(\theta)$ between the pair of lines represented by the homogeneous equation $ax^2+2hxy+by^2=0$ is given by the formula:
$\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$
If the lines are at right angles (perpendicular),then $\theta = 90^\circ$.
Since $\tan 90^\circ$ is undefined,the denominator must be zero.
Therefore,$a+b=0$.

(C) The given equation of the pair of straight lines is $x^2+4xy+y^2=0$.
Comparing this with the general equation $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=4 \Rightarrow h=2$,and $b=1$.
The formula for the angle $\theta$ between the pair of lines is $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{2^2-(1)(1)}}{1+1} \right|$.
$\tan \theta = \frac{2\sqrt{4-1}}{2} = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = 60^{\circ}$.

(B) The given equation of the pair of straight lines is $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$.
For a general equation of a pair of lines $Ax^2 + 2Hxy + By^2 = 0$,the lines are perpendicular if $A + B = 0$.
Here,$A = 3d$ and $B = d^2 - 2$.
Setting $A + B = 0$,we get $3d + d^2 - 2 = 0$,which is $d^2 + 3d - 2 = 0$.
Using the quadratic formula $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we find $d = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2}$.
Since there are two distinct real values for $d$,the condition is satisfied for $2$ values of $d$.
Thus,option $B$ is correct.

(C) The equation of the pair of angle bisectors of the pair of lines $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
For the pair of lines $x^2-2pxy-y^2=0$,we have $a=1, h=-p, b=-1$.
The equation of the angle bisectors is $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-p}$,which simplifies to $\frac{x^2-y^2}{2} = \frac{xy}{-p}$,or $x^2-y^2 = -\frac{2}{p}xy$,which is $x^2 + \frac{2}{p}xy - y^2 = 0$.
It is given that this pair of bisectors is the same as the pair of lines $x^2-2qxy-y^2=0$.
Comparing the coefficients of $xy$,we get $\frac{2}{p} = -2q$.
This implies $pq = -1$,or $pq+1=0$.
Thus,option $(C)$ is correct.

(A) Let the two fixed lines be $L_1$ and $L_2$ intersecting at a point $P$.
Any line $L$ passing through $P$ that is equally inclined to $L_1$ and $L_2$ must be an angle bisector of the angle formed by $L_1$ and $L_2$.
There are two such angle bisectors (internal and external),which are always perpendicular to each other.
Since the two lines described in Statement $-I$ are these two angle bisectors,they must be perpendicular.
Thus,Statement $-I$ is true,Statement $-II$ is true,and Statement $-II$ correctly explains Statement $-I$.

(C) The general equation of a pair of lines is $ax^2+2hxy+by^2+2gx+2fy+c=0$.
For the lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a+b=0$.
Given $l+ (-2) = 0$,we get $l=2$.
Now the equation becomes $2x^2+3xy-2y^2-5x+5y+k=0$.
For this to represent a pair of lines,the condition $abc+2fgh-af^2-bg^2-ch^2=0$ must be satisfied.
Here $a=2, b=-2, c=k, h=\frac{3}{2}, g=-\frac{5}{2}, f=\frac{5}{2}$.
Substituting these values: $(2)(-2)(k) + 2(\frac{5}{2})(-\frac{5}{2})(\frac{3}{2}) - 2(\frac{5}{2})^2 - (-2)(-\frac{5}{2})^2 - k(\frac{3}{2})^2 = 0$.
$-4k - \frac{75}{4} - \frac{50}{4} + \frac{50}{4} - \frac{9k}{4} = 0$.
$-4k - \frac{9k}{4} - \frac{75}{4} = 0$.
$-\frac{25k}{4} = \frac{75}{4}$.
$k = -3$.

(B) The equation of the pair of bisectors of the angle between the pair of straight lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.
The equation of the bisectors is $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$.
This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,which implies $-p(x^2-y^2)=2xy$,or $p x^2+2 x y-p y^2=0$.
Dividing by $p$,we get $x^2+\frac{2}{p} x y-y^2=0$.
We are given that this pair is $x^2-2 q x y-y^2=0$.
Comparing the coefficients of $xy$,we have $-2 q = \frac{2}{p}$.
Therefore,$-p q = 1$,which gives $p q = -1$.

(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.
The equation of the pair of angle bisectors for these lines is given by the formula:
$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
This formula is derived by considering the lines $y = m_1x$ and $y = m_2x$ and finding the bisectors of the angles between them.
Thus,the correct option is $D$.

(A) Given the equation of the pair of straight lines: $x^2+4xy-y^2=0$.
Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=1$,$b=-1$,and $2h=4$,which implies $h=2$.
The equation of the bisectors of the angle between the lines is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
Substituting the values: $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{2} \Rightarrow \frac{x^2-y^2}{2} = \frac{xy}{2}$.
This simplifies to $x^2-xy-y^2=0$.
Since $y=mx$ is one of the bisectors,we substitute $y=mx$ into the equation: $x^2-x(mx)-(mx)^2=0$.
Dividing by $x^2$ (assuming $x \neq 0$),we get $1-m-m^2=0$,or $m^2+m-1=0$.
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $m = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Therefore,$2m = -1 \pm \sqrt{5}$.

(D) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$.
We can rewrite the quadratic part as $(3x - y)^2 + 6(3x - y) + 8 = 0$.
Let $t = 3x - y$. Then the equation becomes $t^2 + 6t + 8 = 0$.
Factoring the quadratic,we get $(t + 4)(t + 2) = 0$.
Thus,the two parallel lines are $3x - y + 4 = 0$ and $3x - y + 2 = 0$.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 3$,$b = -1$,$c_1 = 4$,and $c_2 = 2$.
$d = \frac{|4 - 2|}{\sqrt{3^2 + (-1)^2}} = \frac{2}{\sqrt{9 + 1}} = \frac{2}{\sqrt{10}}$.

(B) The given equation is $x^2 + 2 \sqrt{2} xy + 2y^2 + 4x + 4 \sqrt{2}y + 1 = 0$.
This can be rewritten as $(x + \sqrt{2}y)^2 + 4(x + \sqrt{2}y) + 1 = 0$.
Let $t = x + \sqrt{2}y$. Then the equation becomes $t^2 + 4t + 1 = 0$.
Solving for $t$ using the quadratic formula,$t = \frac{-4 \pm \sqrt{16 - 4}}{2} = -2 \pm \sqrt{3}$.
Thus,the two parallel lines are $x + \sqrt{2}y + 2 - \sqrt{3} = 0$ and $x + \sqrt{2}y + 2 + \sqrt{3} = 0$.
The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = 1$,$B = \sqrt{2}$,$C_1 = 2 - \sqrt{3}$,and $C_2 = 2 + \sqrt{3}$.
$d = \frac{|(2 - \sqrt{3}) - (2 + \sqrt{3})|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|-2 \sqrt{3}|}{\sqrt{3}} = \frac{2 \sqrt{3}}{\sqrt{3}} = 2$ units.
Therefore,the correct option is $B$.

(D) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$.
This can be written as $(3x - y)^2 + 6(3x - y) + 8 = 0$.
Let $t = 3x - y$. Then the equation becomes $t^2 + 6t + 8 = 0$.
Factoring the quadratic,we get $(t + 4)(t + 2) = 0$.
So,$t = -4$ or $t = -2$.
This gives two parallel lines: $3x - y + 4 = 0$ and $3x - y + 2 = 0$.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 3$,$b = -1$,$c_1 = 4$,and $c_2 = 2$.
$d = \frac{|4 - 2|}{\sqrt{3^2 + (-1)^2}} = \frac{2}{\sqrt{9 + 1}} = \frac{2}{\sqrt{10}}$.

(C) The given equation is $x^2(\sec^2 \theta - \sin^2 \theta) - 2xy \tan \theta + y^2 \sin^2 \theta = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we have:
$a = \sec^2 \theta - \sin^2 \theta$,$h = -\tan \theta$,and $b = \sin^2 \theta$.
Let $m_1$ and $m_2$ be the slopes of the lines. Then $m_1 + m_2 = -\frac{2h}{b} = \frac{2 \tan \theta}{\sin^2 \theta}$ and $m_1 m_2 = \frac{a}{b} = \frac{\sec^2 \theta - \sin^2 \theta}{\sin^2 \theta} = \sec^2 \theta \csc^2 \theta - 1$.
The square of the difference of the slopes is $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$.
$(m_1 - m_2)^2 = \left(\frac{2 \tan \theta}{\sin^2 \theta}\right)^2 - 4\left(\frac{\sec^2 \theta - \sin^2 \theta}{\sin^2 \theta}\right)$.
$(m_1 - m_2)^2 = \frac{4 \sin^2 \theta}{\cos^2 \theta \sin^4 \theta} - \frac{4(\sec^2 \theta - \sin^2 \theta)}{\sin^2 \theta} = \frac{4}{\cos^2 \theta \sin^2 \theta} - \frac{4}{\cos^2 \theta \sin^2 \theta} + 4 = 4$.
Thus,the square of the difference of the slopes is $4$.

(C) The general equation of a second-degree curve is $ax^2+2hxy+by^2+2gx+2fy+c=0$.
Comparing this with the given equation $x^2-4xy-y^2+6x+2y+k=0$,we have:
$a=1, h=-2, b=-1, g=3, f=1, c=k$.
The condition for the equation to represent a pair of straight lines is $\Delta = abc+2fgh-af^2-bg^2-ch^2=0$.
Substituting the values:
$(1)(-1)(k) + 2(1)(3)(-2) - 1(1)^2 - (-1)(3)^2 - k(-2)^2 = 0$
$-k - 12 - 1 + 9 - 4k = 0$
$-5k - 4 = 0$
$5k = -4$
$k = -\frac{4}{5}$.
Thus,the correct option is $C$.

(D) The first pair of lines is given by $xy+x+y+1=0$. Factoring this,we get $x(y+1)+1(y+1)=0$,which implies $(x+1)(y+1)=0$. Thus,the lines are $x=-1$ and $y=-1$.
The second pair of lines is given by $xy+3x+3y+9=0$. Factoring this,we get $x(y+3)+3(y+3)=0$,which implies $(x+3)(y+3)=0$. Thus,the lines are $x=-3$ and $y=-3$.
The four lines forming the quadrilateral are $x=-1, x=-3, y=-1,$ and $y=-3$.
The distance between the vertical lines $x=-1$ and $x=-3$ is $|-1 - (-3)| = 2$.
The distance between the horizontal lines $y=-1$ and $y=-3$ is $|-1 - (-3)| = 2$.
Since the opposite sides are parallel and the distance between the vertical lines is equal to the distance between the horizontal lines,the quadrilateral is a square. Therefore,the correct option is $D$.

(C) The equations of the lines passing through the origin with slopes $m_1 = \frac{2}{3}$ and $m_2 = -\frac{2}{3}$ are given by $y = m_1 x$ and $y = m_2 x$.
Substituting the slopes,we get $y = \frac{2}{3}x \Rightarrow 2x - 3y = 0$ and $y = -\frac{2}{3}x \Rightarrow 2x + 3y = 0$.
The combined equation is the product of these two linear equations:
$(2x - 3y)(2x + 3y) = 0$.
Using the identity $(a-b)(a+b) = a^2 - b^2$,we get:
$(2x)^2 - (3y)^2 = 0$.
Therefore,$4x^2 - 9y^2 = 0$.

(D) The point of intersection of the pair of lines $f(x, y) = x^2+xy+2y^2-3x+2y+4=0$ can be found by solving the partial derivatives of $f(x, y)$ with respect to $x$ and $y$ equal to zero.
Taking partial derivative with respect to $x$:
$\frac{\partial f}{\partial x} = 2x + y - 3 = 0 \quad \dots (i)$
Taking partial derivative with respect to $y$:
$\frac{\partial f}{\partial y} = x + 4y + 2 = 0 \quad \dots (ii)$
Solving the system of linear equations $(i)$ and $(ii)$:
From $(i)$,$y = 3 - 2x$.
Substituting into $(ii)$: $x + 4(3 - 2x) + 2 = 0$
$x + 12 - 8x + 2 = 0$
$-7x + 14 = 0 \implies x = 2$
Substituting $x = 2$ into $y = 3 - 2x$:
$y = 3 - 2(2) = 3 - 4 = -1$
Thus,the point of intersection is $(2, -1)$.
Hence,option $(D)$ is correct.

(C) The equation $x^2+4xy+y^2=0$ represents a pair of straight lines passing through the origin. Let these lines be $L_1$ and $L_2$.
The equation $x-y=4$ represents a third line $L_3$.
The lines $L_1$ and $L_2$ are given by $y = m_1x$ and $y = m_2x$,where $m_1, m_2$ are roots of $m^2+4m+1=0$.
Thus,$m_1+m_2 = -4$ and $m_1m_2 = 1$.
The slopes are $m = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}$.
The angle between $L_1$ and $L_2$ is $\tan \theta = |\frac{2\sqrt{h^2-ab}}{a+b}| = |\frac{2\sqrt{4-1}}{1+1}| = \sqrt{3}$,so $\theta = 60^\circ$.
Since the lines $L_1$ and $L_2$ pass through the origin and the angle between them is $60^\circ$,and the third line $L_3$ intersects them,we check the slopes of the lines formed by the intersection.
By calculating the slopes of the sides of the triangle formed by these three lines,it is found that all sides are equal in length or the angles are $60^\circ$,confirming it is an Equilateral Triangle.

(C) Let the slopes of the lines be $m$ and $3m$ respectively,which are represented by the equation $ax^2+4xy+y^2=0$.
Comparing with $Ax^2+2Hxy+By^2=0$,we have $A=a$,$2H=4 \Rightarrow H=2$,and $B=1$.
The sum of the slopes is $m+3m = -\frac{2H}{B} = -\frac{4}{1} = -4$.
$4m = -4 \Rightarrow m = -1$.
The product of the slopes is $m(3m) = \frac{A}{B} = \frac{a}{1} = a$.
$3m^2 = a$.
Substituting $m = -1$,we get $3(-1)^2 = a \Rightarrow a = 3$.
Hence,option $C$ is correct.

(C) The given equation of the pair of lines is $4x^2 + 2\lambda xy - 7y^2 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = 2\lambda$,and $b = -7$.
Let $m_1$ and $m_2$ be the slopes of the lines.
The sum of the slopes is $m_1 + m_2 = \frac{-2h}{b} = \frac{-2\lambda}{-7} = \frac{2\lambda}{7}$.
The product of the slopes is $m_1m_2 = \frac{a}{b} = \frac{4}{-7} = -\frac{4}{7}$.
Given that the sum of the slopes is equal to the product of the slopes:
$\frac{2\lambda}{7} = -\frac{4}{7}$.
Multiplying both sides by $7$,we get $2\lambda = -4$.
Therefore,$\lambda = -2$.
Hence,option $C$ is correct.

(C) The given equation is a homogeneous equation of the second degree in $x$ and $y$ of the form $ax^2 + 2hxy + by^2 = 0$.
Comparing $4x^2 - 24xy + 11y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = -24$ (so $h = -12$),and $b = 11$.
For a homogeneous equation of the second degree,the discriminant is $h^2 - ab = (-12)^2 - (4)(11) = 144 - 44 = 100$.
Since $h^2 - ab > 0$,the equation represents two distinct real lines passing through the origin.
Therefore,the correct option is $C$.

(C) Let the radius of the required circle be $r$. Given that the circumference is $10 \pi$.
$2 \pi r = 10 \pi$
$\Rightarrow r = 5$
The centre of the circle is $(h, k) = (2, -3)$.
The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
Substituting the values:
$(x - 2)^2 + (y - (-3))^2 = 5^2$
$(x - 2)^2 + (y + 3)^2 = 25$
$x^2 - 4x + 4 + y^2 + 6y + 9 = 25$
$x^2 + y^2 - 4x + 6y + 13 - 25 = 0$
$x^2 + y^2 - 4x + 6y - 12 = 0$
Thus,option $C$ is correct.

(C) The given equation of the circle is $x^2+y^2-4x-6y+11=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $2g=-4 \Rightarrow g=-2$ and $2f=-6 \Rightarrow f=-3$.
The center of the circle is $(-g, -f) = (2, 3)$.
Let the other end of the diameter be $(h, k)$.
Since the center of the circle is the midpoint of the diameter,we have:
$\frac{h+3}{2} = 2$ $\Rightarrow h+3 = 4$ $\Rightarrow h = 1$
$\frac{k+4}{2} = 3$ $\Rightarrow k+4 = 6$ $\Rightarrow k = 2$
Thus,the other end of the diameter is $(1, 2)$.

(D) The centre of the circle is $(h, k) = (5, 4)$.
Since the circle touches the $Y$-axis,the radius $r$ is equal to the absolute value of the $x$-coordinate of the centre.
Thus,$r = |5| = 5$.
The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$.
Substituting the values,we get $(x-5)^2 + (y-4)^2 = 5^2$.
Expanding this,we have $(x^2 - 10x + 25) + (y^2 - 8y + 16) = 25$.
Simplifying,$x^2 + y^2 - 10x - 8y + 41 = 25$.
Therefore,$x^2 + y^2 - 10x - 8y + 16 = 0$.

(C) The given equation of the circle is $x^2+y^2+6x+2ky+25=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=3$,$f=k$,and $c=25$.
The center of the circle is $(-g, -f) = (-3, -k)$ and the radius $r$ is given by $\sqrt{g^2+f^2-c} = \sqrt{3^2+k^2-25} = \sqrt{k^2-16}$.
Since the circle touches the $Y$-axis,the radius must be equal to the absolute value of the $x$-coordinate of the center,i.e.,$r = |-g| = |-3| = 3$.
Equating the two expressions for the radius: $\sqrt{k^2-16} = 3$.
Squaring both sides: $k^2-16 = 9$.
$k^2 = 25$,which gives $k = \pm 5$.

(D) The circle passes through the origin $(0,0)$ and makes intercepts $a$ and $b$ on the $x$ and $y$ axes respectively.
Therefore,the circle passes through the points $(a,0)$ and $(0,b)$.
Since the angle between the coordinate axes is $90^{\circ}$,the line segment joining $(a,0)$ and $(0,b)$ is a diameter of the circle.
The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting $(x_1, y_1) = (a,0)$ and $(x_2, y_2) = (0,b)$:
$(x-a)(x-0) + (y-0)(y-b) = 0$
$x(x-a) + y(y-b) = 0$
$x^2 - ax + y^2 - by = 0$
$x^2 + y^2 - ax - by = 0$

(B) The given equation of the circle is $x^2+y^2-6x-8y=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$,$f=-4$,and $c=0$.
The radius $r$ of the circle is given by the formula $r = \sqrt{g^2+f^2-c}$.
Substituting the values,$r = \sqrt{(-3)^2+(-4)^2-0} = \sqrt{9+16} = \sqrt{25} = 5 \text{ units}$.
The diameter of the circle is $d = 2r = 2 \times 5 = 10 \text{ units}$.

(B) The family of circles passing through the intersection of the circle $x^2+y^2=9$ and the line $x+y=1$ is given by $(x^2+y^2-9) + \lambda(x+y-1) = 0$.
This simplifies to $x^2+y^2+\lambda x+\lambda y - (\lambda+9) = 0$.
The center of this circle is $(-\frac{\lambda}{2}, -\frac{\lambda}{2})$.
For the smallest circle,the line $x+y=1$ must be the diameter of the circle.
Substituting the center into the line equation: $-\frac{\lambda}{2} - \frac{\lambda}{2} = 1$,which gives $-\lambda = 1$,so $\lambda = -1$.
Substituting $\lambda = -1$ into the family equation,we get $(x^2+y^2-9) - (x+y-1) = 0$.

(D) The point of reflection of $(m, n)$ on the line $y-x=0$ is $(n, m)$.
Let the centre of the circle be $(n, m)$ and the radius be $r$.
The equation of the circle is $(x-n)^2 + (y-m)^2 = r^2$ ... $(i)$.
Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre,so $r = |m|$.
Thus,$r^2 = m^2$.
Substituting $r^2$ into equation $(i)$:
$(x-n)^2 + (y-m)^2 = m^2$
$x^2 - 2nx + n^2 + y^2 - 2my + m^2 = m^2$
$x^2 + y^2 - 2nx - 2my + n^2 = 0$.

(D) Centre $C = (2,3)$.
Radius $r$ is the perpendicular distance from the centre $(2,3)$ to the line $3x-4y+1=0$.
$r = \frac{|3(2)-4(3)+1|}{\sqrt{3^2+(-4)^2}} = \frac{|6-12+1|}{\sqrt{9+16}} = \frac{|-5|}{5} = 1$.
The equation of the circle is $(x-h)^2+(y-k)^2 = r^2$.
$(x-2)^2+(y-3)^2 = 1^2$.
$x^2-4x+4+y^2-6y+9 = 1$.
$x^2+y^2-4x-6y+12 = 0$.
Thus,option $D$ is correct.

(A) The normals to a circle always intersect at its center. Given the normals $(x-1)(y-2)=0$,the center of the circle is $(1, 2)$.
Since $3x+4y=6$ is a tangent to the circle,the radius $r$ is the perpendicular distance from the center $(1, 2)$ to the line $3x+4y-6=0$.
$r = \frac{|3(1) + 4(2) - 6|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 6|}{\sqrt{9 + 16}} = \frac{5}{5} = 1$.
The equation of the circle with center $(h, k) = (1, 2)$ and radius $r = 1$ is $(x-h)^2 + (y-k)^2 = r^2$.
$(x-1)^2 + (y-2)^2 = 1^2$,which simplifies to $(x-1)^2 + (y-2)^2 = 1$.
Thus,option $A$ is correct.

(A) Let $I = \int_0^1 \frac{8 \log (1+x)}{1+x^2} dx$.
Substitute $x = \tan \theta$,so $dx = \sec^2 \theta d\theta$.
When $x = 0$,$\theta = 0$,and when $x = 1$,$\theta = \frac{\pi}{4}$.
Then $I = 8 \int_0^{\pi/4} \frac{\log(1+\tan \theta)}{1+\tan^2 \theta} \sec^2 \theta d\theta = 8 \int_0^{\pi/4} \log(1+\tan \theta) d\theta$.
Using the property $\int_0^a f(\theta) d\theta = \int_0^a f(a-\theta) d\theta$:
$I = 8 \int_0^{\pi/4} \log(1+\tan(\frac{\pi}{4}-\theta)) d\theta$.
Since $\tan(\frac{\pi}{4}-\theta) = \frac{1-\tan \theta}{1+\tan \theta}$,we have:
$I = 8 \int_0^{\pi/4} \log(1 + \frac{1-\tan \theta}{1+\tan \theta}) d\theta = 8 \int_0^{\pi/4} \log(\frac{2}{1+\tan \theta}) d\theta$.
$I = 8 \int_0^{\pi/4} (\log 2 - \log(1+\tan \theta)) d\theta = 8 \int_0^{\pi/4} \log 2 d\theta - I$.
$2I = 8 \log 2 [\theta]_0^{\pi/4} = 8 \log 2 (\frac{\pi}{4}) = 2\pi \log 2$.
Therefore,$I = \pi \log 2$.

(C) Let $I = \int_0^{\pi / 2} e^{\sin x} \cos x \, dx$.
Substitute $\sin x = t$. Then,differentiating both sides with respect to $x$,we get $\cos x \, dx = dt$.
Now,change the limits of integration:
When $x = 0$,$t = \sin(0) = 0$.
When $x = \frac{\pi}{2}$,$t = \sin(\frac{\pi}{2}) = 1$.
Substituting these into the integral,we get:
$I = \int_0^1 e^t \, dt$.
The integral of $e^t$ is $e^t$.
$I = [e^t]_0^1 = e^1 - e^0 = e - 1$.

(D) Let $I = \int_0^{\pi/4} (\tan^2 x - \tan^4 x) dx$.
We can factor the integrand as:
$\tan^2 x - \tan^4 x = \tan^2 x (1 - \tan^2 x)$.
Using the identity $\tan^2 x = \sec^2 x - 1$,we have:
$I = \int_0^{\pi/4} (\sec^2 x - 1)(1 - (\sec^2 x - 1)) dx = \int_0^{\pi/4} (\sec^2 x - 1)(2 - \sec^2 x) dx$.
Expanding the product:
$I = \int_0^{\pi/4} (2\sec^2 x - \sec^4 x - 2 + \sec^2 x) dx = \int_0^{\pi/4} (3\sec^2 x - \sec^4 x - 2) dx$.
Using $\sec^4 x = \sec^2 x (1 + \tan^2 x)$:
$I = \int_0^{\pi/4} (3\sec^2 x - \sec^2 x(1 + \tan^2 x) - 2) dx = \int_0^{\pi/4} (2\sec^2 x - \sec^2 x \tan^2 x - 2) dx$.
Integrating term by term:
$I = [2\tan x - \frac{\tan^3 x}{3} - 2x]_0^{\pi/4}$.
Evaluating at the limits:
$I = (2(1) - \frac{1^3}{3} - 2(\frac{\pi}{4})) - (0) = 2 - \frac{1}{3} - \frac{\pi}{2} = \frac{5}{3} - \frac{\pi}{2}$.

(D) $(i)$ Let $I_1 = \int_0^{\pi / 2} \sin ^m x \cos x d x$.
Substitute $\sin x = t$,then $\cos x d x = d t$.
When $x = 0, t = 0$ and when $x = \pi / 2, t = 1$.
$I_1 = \int_0^1 t^m d t = \left[ \frac{t^{m+1}}{m+1} \right]_0^1 = \frac{1}{m+1} (1^{m+1} - 0^{m+1}) = \frac{1}{m+1}$.
Thus,statement $(i)$ is true.
(ii) Let $I_2 = \int_0^{\pi / 2} \sin x \cos ^n x d x$.
Substitute $\cos x = t$,then $-\sin x d x = d t$,or $\sin x d x = -d t$.
When $x = 0, t = 1$ and when $x = \pi / 2, t = 0$.
$I_2 = \int_1^0 t^n (-d t) = \int_0^1 t^n d t = \left[ \frac{t^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1} (1^{n+1} - 0^{n+1}) = \frac{1}{n+1}$.
Thus,statement (ii) is true.
Therefore,both $(i)$ and (ii) are true.

(C) Given,$I_1 = \int_0^{\pi / 2} \frac{x}{\sin x} dx$ and $I_2 = \int_0^1 \frac{\tan^{-1} x}{x} dx$.
For $I_2$,let $\tan^{-1} x = t$,then $x = \tan t$ and $dx = \sec^2 t dt$.
When $x = 0, t = 0$ and when $x = 1, t = \pi / 4$.
Substituting these into $I_2$:
$I_2 = \int_0^{\pi / 4} \frac{t}{\tan t} \sec^2 t dt = \int_0^{\pi / 4} \frac{t \cos t}{\sin t} \cdot \frac{1}{\cos^2 t} dt = \int_0^{\pi / 4} \frac{t}{\sin t \cos t} dt$.
Multiplying numerator and denominator by $2$:
$I_2 = \int_0^{\pi / 4} \frac{2t}{\sin 2t} dt$.
Let $2t = u$,then $2dt = du$ or $dt = du / 2$.
When $t = 0, u = 0$ and when $t = \pi / 4, u = \pi / 2$.
$I_2 = \int_0^{\pi / 2} \frac{u}{\sin u} \cdot \frac{du}{2} = \frac{1}{2} \int_0^{\pi / 2} \frac{u}{\sin u} du = \frac{1}{2} I_1$.
Therefore,$I_1 / I_2 = 2 / 1$,so $I_1 : I_2 = 2 : 1$.

(C) Let $I = \int \frac{(1+x) e^x}{\cot \left(x e^x\right)} d x$.
Substitute $t = x e^x$.
Differentiating with respect to $x$,we get $dt = (e^x + x e^x) dx = e^x(1+x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{\cot t} = \int \tan t dt$.
The integral of $\tan t$ is $\log |\sec t| + c$.
Thus,$I = \log |\sec(x e^x)| + c$.
Therefore,the correct option is $C$.

(A) First,evaluate $A$: $A = \int_{1}^{\sin \theta} \frac{t}{1+t^2} dt = \frac{1}{2} [\ln(1+t^2)]_{1}^{\sin \theta} = \frac{1}{2} \ln(1+\sin^2 \theta) - \frac{1}{2} \ln(2) = \frac{1}{2} \ln(\frac{1+\sin^2 \theta}{2})$.
Next,evaluate $B$: $B = \int_{1}^{\operatorname{cosec} \theta} \frac{1}{t(1+t^2)} dt$. Using partial fractions,$\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.
So,$B = [\ln|t| - \frac{1}{2} \ln(1+t^2)]_{1}^{\operatorname{cosec} \theta} = [\ln(\frac{t}{\sqrt{1+t^2}})]_{1}^{\operatorname{cosec} \theta}$.
Since $\frac{t}{\sqrt{1+t^2}} = \frac{\operatorname{cosec} \theta}{\sqrt{1+\operatorname{cosec}^2 \theta}} = \frac{1}{\sqrt{\sin^2 \theta + 1}}$,we have $B = \ln(\frac{1}{\sqrt{1+\sin^2 \theta}}) - \ln(\frac{1}{\sqrt{2}}) = \ln(\sqrt{\frac{2}{1+\sin^2 \theta}}) = -\frac{1}{2} \ln(\frac{1+\sin^2 \theta}{2}) = -A$.
Thus,$A+B = 0$,which implies $e^{A+B} = e^0 = 1$.
The determinant becomes $\left| \begin{array}{ccc} A & A^2 & -A \\ 1 & (-A)^2 & -1 \\ 1 & A^2+(-A)^2 & -1 \end{array} \right| = \left| \begin{array}{ccc} A & A^2 & -A \\ 1 & A^2 & -1 \\ 1 & 2A^2 & -1 \end{array} \right|$.
Since the first and third columns are proportional (column $3$ is $-1$ times column $1$),the determinant is $0$.

(C) Let $f(x) = x^2 \sin x$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = (-x)^2 \sin(-x) = x^2 (-\sin x) = -x^2 \sin x = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x) = x^2 \sin x$ 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,$\int_{-\pi}^{\pi} x^2 \sin x \, dx = 0$.

(A) Let $I = \int_{-\pi / 4}^{\pi / 4} x^3 \sin^4(x) dx$.
We know that for a definite integral,$\int_{-a}^{a} f(x) dx = 0$ if $f(x)$ is an odd function,i.e.,$f(-x) = -f(x)$.
Here,$f(x) = x^3 \sin^4(x)$.
Calculating $f(-x)$:
$f(-x) = (-x)^3 \sin^4(-x) = -x^3 (\sin(x))^4 = -x^3 \sin^4(x) = -f(x)$.
Since $f(x)$ is an odd function,the integral over the symmetric interval $[-\pi/4, \pi/4]$ is $0$.
Therefore,$I = 0$.

(A) Let $I = \int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{2 \sin 2 x}}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{4 \sin x \cos x}} = \int_0^{\pi / 4} \frac{d x}{2 \cos ^4 x \sqrt{\tan x}}$.
$I = \frac{1}{2} \int_0^{\pi / 4} \frac{\sec^2 x (1 + \tan^2 x)}{\sqrt{\tan x}} d x$.
Let $\tan x = t^2$,then $\sec^2 x d x = 2t dt$.
When $x = 0, t = 0$ and when $x = \frac{\pi}{4}, t = 1$.
Substituting these into the integral:
$I = \frac{1}{2} \int_0^1 \frac{(1 + t^4) \cdot 2t dt}{t} = \int_0^1 (1 + t^4) dt$.
$I = [t + \frac{t^5}{5}]_0^1 = 1 + \frac{1}{5} = \frac{6}{5}$.

(C) We know that the absolute value function $|x|$ is defined as:
$|x| = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}$
Therefore,the integral can be split at $x = 0$:
$I = \int_{-1}^2 |x| \, dx = \int_{-1}^0 (-x) \, dx + \int_0^2 (x) \, dx$
Evaluating the first part:
$\int_{-1}^0 (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-1}^0 = 0 - \left( -\frac{(-1)^2}{2} \right) = 0 - (-\frac{1}{2}) = \frac{1}{2}$
Evaluating the second part:
$\int_0^2 (x) \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - 0 = \frac{4}{2} = 2$
Adding both parts:
$I = \frac{1}{2} + 2 = \frac{1+4}{2} = \frac{5}{2}$

(A) Let $I = \int_0^{b-c} f(x+c) dx$.
Substitute $x+c = t$,then $dx = dt$.
When $x = 0$,$t = c$.
When $x = b-c$,$t = b$.
Substituting these into the integral,we get:
$I = \int_c^b f(t) dt$.
Since the variable of integration is a dummy variable,we can write $\int_c^b f(t) dt = \int_c^b f(x) dx$.
Thus,$\int_0^{b-c} f(x+c) dx = 1 \cdot \int_c^b f(x) dx$.
Comparing this with the given expression $\int_0^{b-c} f(x+c) dx = k \int_c^b f(x) dx$,we find that $k = 1$.

(C) Let $I = \int_0^{\pi / 2} \frac{1}{1+\tan ^{2020} x} d x$.
Since $\tan x = \frac{\sin x}{\cos x}$,we can write $I = \int_0^{\pi / 2} \frac{\cos ^{2020} x}{\cos ^{2020} x+\sin ^{2020} x} d x$ $(i)$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get $I = \int_0^{\pi / 2} \frac{\sin ^{2020} x}{\sin ^{2020} x+\cos ^{2020} x} d x$ (ii).
Adding equations $(i)$ and (ii),we get $2I = \int_0^{\pi / 2} \frac{\cos ^{2020} x + \sin ^{2020} x}{\cos ^{2020} x + \sin ^{2020} x} d x = \int_0^{\pi / 2} 1 d x$.
Thus,$2I = [x]_0^{\pi / 2} = \frac{\pi}{2}$,which implies $I = \frac{\pi}{4}$.
Therefore,option $C$ is correct.

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} (2 \sin |x| + \cos |x|) dx$.
Since $f(x) = 2 \sin |x| + \cos |x|$ is an even function because $f(-x) = 2 \sin |-x| + \cos |-x| = 2 \sin |x| + \cos |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$.
$I = 2 \int_{0}^{\pi / 2} (2 \sin x + \cos x) dx$.
Integrating the terms,we get $I = 2 [-2 \cos x + \sin x]_{0}^{\pi / 2}$.
Evaluating at the limits,$I = 2 [(-2 \cos(\pi / 2) + \sin(\pi / 2)) - (-2 \cos(0) + \sin(0))]$.
$I = 2 [(-2 \times 0 + 1) - (-2 \times 1 + 0)]$.
$I = 2 [1 - (-2)] = 2 [1 + 2] = 2 \times 3 = 6$.

(D) Let $I = \int_0^{2 \pi} \frac{x \cos x}{1+\cos x} d x$ ... $(i)$
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^{2 \pi} \frac{(2 \pi - x) \cos(2 \pi - x)}{1 + \cos(2 \pi - x)} d x$
Since $\cos(2 \pi - x) = \cos x$,we have:
$I = \int_0^{2 \pi} \frac{(2 \pi - x) \cos x}{1 + \cos x} d x$ ... (ii)
Adding $(i)$ and (ii):
$2I = \int_0^{2 \pi} \frac{2 \pi \cos x}{1 + \cos x} d x = 2 \pi \int_0^{2 \pi} \frac{\cos x}{1 + \cos x} d x$
$I = \pi \int_0^{2 \pi} \frac{\cos x}{1 + \cos x} d x = 2 \pi \int_0^{\pi} \frac{\cos x}{1 + \cos x} d x$
$I = 2 \pi \int_0^{\pi} (1 - \frac{1}{1 + \cos x}) d x = 2 \pi \int_0^{\pi} (1 - \frac{1}{2 \cos^2(x/2)}) d x$
$I = 2 \pi \int_0^{\pi} (1 - \frac{1}{2} \sec^2(x/2)) d x$
$I = 2 \pi [x - \tan(x/2)]_0^{\pi}$
Evaluating at limits: $I = 2 \pi [(\pi - \tan(\pi/2)) - (0 - \tan(0))]$. Since $\tan(\pi/2)$ is undefined,the integral diverges. Thus,the correct option is $D$.

(A) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x$.
We can rewrite the integrand as $I = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ $(i)$.
Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we get:
$I = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}} d x = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$ (ii).
Adding $(i)$ and (ii):
$2I = \int_{\pi / 6}^{\pi / 3} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) d x = \int_{\pi / 6}^{\pi / 3} 1 d x$.
$2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(A) We are given the property of definite integrals: $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$.
Applying this property to the integral $I = \int_{0}^{\pi/2} \cos^{4}(x) \cdot \sin^{2}(x) dx$,we replace $x$ with $(\frac{\pi}{2} - x)$.
Since $\cos(\frac{\pi}{2} - x) = \sin(x)$ and $\sin(\frac{\pi}{2} - x) = \cos(x)$,the integral becomes:
$I = \int_{0}^{\pi/2} \sin^{4}(x) \cdot \cos^{2}(x) dx$.
Given that $\int_{0}^{\pi/2} \sin^{4}(x) \cdot \cos^{2}(x) dx = \frac{\pi}{32}$,it follows that $I = \frac{\pi}{32}$.

(D) Let $I = \int_{-a}^{a} \sqrt{\frac{a - x}{a + x}} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$,where $a = -a$ and $b = a$,we have $f(x) \rightarrow f(-a + a - x) = f(-x)$.
$I = \int_{-a}^{a} \sqrt{\frac{a - (-x)}{a + (-x)}} dx = \int_{-a}^{a} \sqrt{\frac{a + x}{a - x}} dx$.
Adding the two expressions for $I$:
$2I = \int_{-a}^{a} \left( \sqrt{\frac{a - x}{a + x}} + \sqrt{\frac{a + x}{a - x}} \right) dx$.
$2I = \int_{-a}^{a} \frac{(a - x) + (a + x)}{\sqrt{(a + x)(a - x)}} dx = \int_{-a}^{a} \frac{2a}{\sqrt{a^2 - x^2}} dx$.
Since the integrand is an even function,$2I = 2 \int_{0}^{a} \frac{2a}{\sqrt{a^2 - x^2}} dx$.
$I = 2a \int_{0}^{a} \frac{1}{\sqrt{a^2 - x^2}} dx$.
$I = 2a [\sin^{-1}(\frac{x}{a})]_{0}^{a}$.
$I = 2a (\sin^{-1}(1) - \sin^{-1}(0)) = 2a (\frac{\pi}{2} - 0) = a \pi$.

(C) Let $I = \int_0^{\pi / 2} \frac{2 \sin (x)+3 \cos (x)}{\sin (x)+\cos (x)} d x$.
Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get:
$I = \int_0^{\pi / 2} \frac{2 \sin (\pi/2-x)+3 \cos (\pi/2-x)}{\sin (\pi/2-x)+\cos (\pi/2-x)} d x = \int_0^{\pi / 2} \frac{2 \cos (x)+3 \sin (x)}{\cos (x)+\sin (x)} d x$.
Adding the two expressions for $I$:
$2I = \int_0^{\pi / 2} \frac{(2 \sin x + 3 \cos x) + (2 \cos x + 3 \sin x)}{\sin x + \cos x} d x$.
$2I = \int_0^{\pi / 2} \frac{5 \sin x + 5 \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 5 d x$.
$2I = [5x]_0^{\pi / 2} = 5(\frac{\pi}{2} - 0) = \frac{5\pi}{2}$.
Therefore,$I = \frac{5\pi}{4}$.

(A) We are given the expression: $I = \int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx$.
Using the property of definite integrals,$\int_0^{2a} f(x) dx = \int_0^a f(x) dx + \int_a^{2a} f(x) dx$.
Substituting this into the expression,we get: $I = (\int_0^a f(x) dx + \int_a^{2a} f(x) dx) - \int_a^{2a} f(x) dx$.
Simplifying the expression,the term $\int_a^{2a} f(x) dx$ cancels out.
Therefore,$I = \int_0^a f(x) dx$.

(D) We are given the integral equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$.
Recall the standard integral formula $\int \frac{dt}{|t| \sqrt{t^2-1}} = \sec^{-1}(t) + C$.
Applying the limits of integration,we get:
$\left[ \sec^{-1}(t) \right]_{\sqrt{2}}^x = \frac{\pi}{12}$.
Substituting the limits,we have $\sec^{-1}(x) - \sec^{-1}(\sqrt{2}) = \frac{\pi}{12}$.
Since $\sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$,the equation becomes $\sec^{-1}(x) - \frac{\pi}{4} = \frac{\pi}{12}$.
Adding $\frac{\pi}{4}$ to both sides,we get $\sec^{-1}(x) = \frac{\pi}{12} + \frac{\pi}{4} = \frac{\pi + 3\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.
Therefore,$x = \sec\left(\frac{\pi}{3}\right) = 2$.

(D) Given the equation: $\int_0^{x^2} x f(t) dt = x^5 - x^3$.
Since $x$ is independent of $t$,we can write it as $x \int_0^{x^2} f(t) dt = x^5 - x^3$.
Dividing by $x$ (assuming $x \neq 0$),we get $\int_0^{x^2} f(t) dt = x^4 - x^2$.
Differentiating both sides with respect to $x$ using the Leibniz integral rule:
$f(x^2) \cdot \frac{d}{dx}(x^2) = \frac{d}{dx}(x^4 - x^2)$.
$f(x^2) \cdot (2x) = 4x^3 - 2x$.
$f(x^2) = \frac{4x^3 - 2x}{2x} = 2x^2 - 1$.
To find $f(1)$,we set $x^2 = 1$,which implies $x = 1$.
Substituting $x = 1$ into the expression for $f(x^2)$:
$f(1) = 2(1)^2 - 1 = 2 - 1 = 1$.
Thus,the value of $f(1)$ is $1$.

(B) We are given $I_n = \int_0^{\pi / 2} \sin^n(x) dx$.
Using integration by parts,let $u = \sin^{n-1}(x)$ and $dv = \sin(x) dx$.
Then $du = (n-1) \sin^{n-2}(x) \cos(x) dx$ and $v = -\cos(x)$.
$I_n = [-\sin^{n-1}(x) \cos(x)]_0^{\pi / 2} + \int_0^{\pi / 2} (n-1) \sin^{n-2}(x) \cos^2(x) dx$.
Since $\cos(\pi/2) = 0$ and $\sin(0) = 0$,the boundary term is $0$.
$I_n = (n-1) \int_0^{\pi / 2} \sin^{n-2}(x) (1 - \sin^2(x)) dx$.
$I_n = (n-1) \int_0^{\pi / 2} \sin^{n-2}(x) dx - (n-1) \int_0^{\pi / 2} \sin^n(x) dx$.
$I_n = (n-1) I_{n-2} - (n-1) I_n$.
$I_n + (n-1) I_n = (n-1) I_{n-2}$.
$n I_n = (n-1) I_{n-2}$.
$I_n = \frac{n-1}{n} I_{n-2}$.
Comparing this with $I_n = k I_{n-2}$,we get $k = \frac{n-1}{n}$.
Hence,option $B$ is correct.

(B) Let $I = \int_0^{\pi / 2} |\sin t - \cos t| \, dt$.
Since $\sin t - \cos t \le 0$ for $t \in [0, \pi / 4]$ and $\sin t - \cos t \ge 0$ for $t \in [\pi / 4, \pi / 2]$,we split the integral:
$I = \int_0^{\pi / 4} (\cos t - \sin t) \, dt + \int_{\pi / 4}^{\pi / 2} (\sin t - \cos t) \, dt$.
Evaluating the first part: $[\sin t + \cos t]_0^{\pi / 4} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
Evaluating the second part: $[-\cos t - \sin t]_{\pi / 4}^{\pi / 2} = (0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$.
Adding both parts: $I = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$.
Thus,option $B$ is correct.

(C) The given curves are $x^2 = 2 - y$ (which is $y = 2 - x^2$) and $x^2 = y$.
To find the intersection points,set $2 - x^2 = x^2$.
$2x^2 = 2 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 1$.
$A = \int_{-1}^{1} ((2 - x^2) - x^2) dx = \int_{-1}^{1} (2 - 2x^2) dx$.
Since the function is even,$A = 2 \int_{0}^{1} (2 - 2x^2) dx$.
$A = 4 \int_{0}^{1} (1 - x^2) dx = 4 [x - \frac{x^3}{3}]_{0}^{1}$.
$A = 4 (1 - \frac{1}{3}) = 4 (\frac{2}{3}) = \frac{8}{3}$ square units.

(A) The given general solution is $y=(c_1+c_2) \sin (x+c_3)+c_4 e^{x+c_5}$.
We can simplify the expression by substituting constants:
Let $A = c_1+c_2$ and $B = c_4 e^{c_5}$.
Then the equation becomes $y = A \sin (x+c_3) + B e^x$.
Using the trigonometric identity $\sin (x+c_3) = \sin x \cos c_3 + \cos x \sin c_3$,we get:
$y = A (\sin x \cos c_3 + \cos x \sin c_3) + B e^x$
$y = (A \cos c_3) \sin x + (A \sin c_3) \cos x + B e^x$.
Let $K_1 = A \cos c_3$,$K_2 = A \sin c_3$,and $K_3 = B$.
Thus,$y = K_1 \sin x + K_2 \cos x + K_3 e^x$.
There are $3$ essential arbitrary constants $(K_1, K_2, K_3)$.
The order of a differential equation is equal to the number of essential arbitrary constants in its general solution.
Therefore,the order of the differential equation is $3$.

(B) The given differential equation is $\left(\frac{d^2 y}{d x^2}\right)^2-\left(\frac{d y}{d x}\right)^3=y^3$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2 y}{d x^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^2 y}{d x^2}$ is $2$,so the degree is $2$.
The product of the degree and the order is $2 \times 2 = 4$.

(B) The given differential equation is $y = x(\frac{dy}{dx})^3 + \frac{d^2y}{dx^2}$.
The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Here,the power of the highest derivative $\frac{d^2y}{dx^2}$ is $1$.
Thus,the order is $2$ and the degree is $1$.
The sum of the order and degree is $2 + 1 = 3$.
Therefore,option $(B)$ is correct.

(B) The given differential equation is $y_3^{2/3} + 2 + 3y_2 + y_1 = 0$.
To find the degree,we must first eliminate the fractional exponent of the highest order derivative.
Rearranging the equation,we get $y_3^{2/3} = -(2 + 3y_2 + y_1)$.
Cubing both sides to remove the fractional power,we obtain $(y_3^{2/3})^3 = (-(2 + 3y_2 + y_1))^3$,which simplifies to $y_3^2 = -(2 + 3y_2 + y_1)^3$.
The highest order derivative present is $y_3$ (the third derivative),and its exponent after rationalizing the equation is $2$.
Therefore,the degree of the differential equation is $2$.
Hence,option $B$ is correct.

(C) The equation of the family of lines passing through the origin is given by $y = mx$,where $m$ is an arbitrary constant.
To find the differential equation,differentiate both sides with respect to $x$:
$\frac{dy}{dx} = m$
Substitute the value of $m = \frac{y}{x}$ into the differentiated equation:
$\frac{dy}{dx} = \frac{y}{x}$
Multiplying both sides by $x$,we get:
$y = x \frac{dy}{dx}$

(A) Given the relation: $y = a \log x + b$
Differentiating both sides with respect to $x$:
$y_1 = \frac{d}{dx}(a \log x + b) = \frac{a}{x}$
Multiplying both sides by $x$:
$x y_1 = a$
Differentiating again with respect to $x$:
$\frac{d}{dx}(x y_1) = \frac{d}{dx}(a)$
$x y_2 + y_1(1) = 0$
$x y_2 + y_1 = 0$

(A) Given the differential equation $\frac{d^2 y}{d x^2} = 0$.
Integrating both sides with respect to $x$,we get $\frac{d y}{d x} = a$,where $a$ is an arbitrary constant.
Integrating again with respect to $x$,we get $y = ax + b$,where $b$ is another arbitrary constant.
This equation $y = ax + b$ is the general form of the equation of a straight line.

(D) Given the differential equation: $\frac{dy}{dx} = \sec y$.
Separating the variables,we get: $\cos y \, dy = dx$.
Integrating both sides: $\int \cos y \, dy = \int dx$.
This gives: $\sin y = x + c$.
Using the initial condition $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $\sin(0) = 0 + c$,which implies $c = 0$.
Substituting $c = 0$ back into the general solution,we get: $\sin y = x$,or $x = \sin y$.
Thus,the correct option is $D$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{y(1+x)}{-x(1+y)}$.
Separate the variables $x$ and $y$:
$\frac{1+y}{y} dy = -\frac{1+x}{x} dx$.
This can be written as:
$(\frac{1}{y} + 1) dy = -(\frac{1}{x} + 1) dx$.
Integrate both sides:
$\int (\frac{1}{y} + 1) dy = -\int (\frac{1}{x} + 1) dx$.
$\log|y| + y = -(\log|x| + x) + C$.
$\log|y| + y = -\log|x| - x + C$.
Rearranging the terms:
$x + y + \log|x| + \log|y| = C$.
Using the property $\log a + \log b = \log(ab)$:
$x + y + \log|xy| = C$.
Thus,the correct option is $C$.

(C) Given the differential equation: $\{x \cos (y/x) + y \sin (y/x)\} y dx = \{y \sin (y/x) - x \cos (y/x)\} x dy$.
Rearranging the terms,we get: $\frac{dy}{dx} = \frac{y \{x \cos (y/x) + y \sin (y/x)\}}{x \{y \sin (y/x) - x \cos (y/x)\}} = \frac{(y/x) \cos (y/x) + (y/x)^2 \sin (y/x)}{(y/x) \sin (y/x) - \cos (y/x)}$.
Let $v = y/x$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v}$.
$x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v(v \sin v - \cos v)}{v \sin v - \cos v} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} = \frac{2v \cos v}{v \sin v - \cos v}$.
Separating the variables: $\frac{v \sin v - \cos v}{v \cos v} dv = 2 \frac{dx}{x}$.
Integrating both sides: $\int (\tan v - \frac{1}{v}) dv = 2 \int \frac{1}{x} dx$.
$-\ln |\cos v| - \ln |v| = 2 \ln |x| + C$.
$-\ln |v \cos v| = \ln |x^2| + C$.
$-\ln |(y/x) \cos (y/x)| = \ln |x^2| + C$.
$\ln |(y/x) \cos (y/x)| + \ln |x^2| = -C$.
$\ln |xy \cos (y/x)| = -C$.
$xy \cos (y/x) = \pm e^{-C}$.

(A) Given differential equation is $\cos(x+y) dy = dx$,which can be written as $\frac{dy}{dx} = \sec(x+y)$.
Let $x+y = t$. Then $1 + \frac{dy}{dx} = \frac{dt}{dx}$,so $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting this into the equation: $\frac{dt}{dx} - 1 = \sec(t) \Rightarrow \frac{dt}{dx} = 1 + \sec(t) = \frac{1+\cos(t)}{\cos(t)}$.
Separating the variables: $\int \frac{\cos(t)}{1+\cos(t)} dt = \int dx$.
Using the identity $1+\cos(t) = 2\cos^2(t/2)$,we get $\int \frac{\cos(t)}{2\cos^2(t/2)} dt = \int dx$.
Since $\cos(t) = 2\cos^2(t/2) - 1$,the integral becomes $\int \frac{2\cos^2(t/2)-1}{2\cos^2(t/2)} dt = \int dx$.
This simplifies to $\int (1 - \frac{1}{2}\sec^2(t/2)) dt = \int dx$.
Integrating both sides: $t - \tan(t/2) = x + C$.
Substituting $t = x+y$: $(x+y) - \tan((x+y)/2) = x + C \Rightarrow y - \tan((x+y)/2) = C$.
Given $y(0) = 0$,we have $0 - \tan(0/2) = C \Rightarrow C = 0$.
Thus,the solution is $y = \tan \left(\frac{x+y}{2}\right)$.

(A) Given differential equation is $x \frac{d y}{d x} = y(\log(\frac{y}{x}) + 1)$.
Dividing by $x$,we get $\frac{d y}{d x} = \frac{y}{x}(\log(\frac{y}{x}) + 1)$.
This is a homogeneous differential equation. Let $y = v x$,then $\frac{d y}{d x} = v + x \frac{d v}{d x}$.
Substituting these into the equation:
$v + x \frac{d v}{d x} = v(\log v + 1)$
$v + x \frac{d v}{d x} = v \log v + v$
$x \frac{d v}{d x} = v \log v$
Separating variables: $\frac{d v}{v \log v} = \frac{d x}{x}$.
Integrating both sides: $\int \frac{1}{v \log v} d v = \int \frac{1}{x} d x$.
Let $\log v = t$,then $\frac{1}{v} d v = d t$.
$\int \frac{1}{t} d t = \int \frac{1}{x} d x \implies \log t = \log x + \log c$.
$\log(\log v) = \log(c x) \implies \log v = c x$.
Since $v = \frac{y}{x}$,we have $\log(\frac{y}{x}) = c x \implies \frac{y}{x} = e^{c x} \implies y = x e^{c x}$.

(A) Given differential equation:
$\frac{x dy}{dx} = y + \sqrt{x^2 + y^2}$
$\Rightarrow \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$
$x \frac{dv}{dx} = \sqrt{1 + v^2}$
Separating the variables:
$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$
Integrating both sides:
$\ln|v + \sqrt{1 + v^2}| = \ln|x| + \ln|c|$
$\ln|v + \sqrt{1 + v^2}| = \ln|cx|$
$v + \sqrt{1 + v^2} = cx$
Substituting $v = \frac{y}{x}$ back:
$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = cx$
$\frac{y + \sqrt{x^2 + y^2}}{x} = cx$
$y + \sqrt{x^2 + y^2} = cx^2$
Or $x^2 = \frac{1}{c} [y + \sqrt{x^2 + y^2}]$,which can be written as $x^2 = C[y + \sqrt{x^2 + y^2}]$.
Thus,option $A$ is correct.

(C) Given differential equation: $(e^{y-x}) dy = (e^x - e^y) dx$
Multiply both sides by $e^x$: $e^y dy = (e^{2x} - e^x e^y) dx$
Rearrange the terms: $e^y \frac{dy}{dx} + e^x e^y = e^{2x}$
Let $z = e^y$,then $\frac{dz}{dx} = e^y \frac{dy}{dx}$.
The equation becomes: $\frac{dz}{dx} + e^x z = e^{2x}$.
This is a linear differential equation of the form $\frac{dz}{dx} + P(x)z = Q(x)$,where $P(x) = e^x$ and $Q(x) = e^{2x}$.
Integrating factor $IF = e^{\int P(x) dx} = e^{\int e^x dx} = e^{e^x}$.
The solution is $z \cdot IF = \int Q(x) \cdot IF dx + c$.
$z \cdot e^{e^x} = \int e^{2x} \cdot e^{e^x} dx + c$.
Let $u = e^x$,then $du = e^x dx$.
$z \cdot e^{e^x} = \int u \cdot e^u du + c$.
Using integration by parts: $\int u e^u du = u e^u - e^u + c$.
Substituting back $z = e^y$ and $u = e^x$: $e^y e^{e^x} = e^x e^{e^x} - e^{e^x} + c$.

(A) The given linear differential equation is $x \frac{dy}{dx} + 3y = x^2$.
Dividing both sides by $x$,we get:
$\frac{dy}{dx} + \frac{3}{x} y = x$.
Comparing this with the standard form $\frac{dy}{dx} + Py = Q$,we identify $P = \frac{3}{x}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P dx}$.
$IF = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = e^{\ln x^3} = x^3$.
Thus,the integrating factor is $x^3$.

(B) The given differential equation is $\frac{dy}{dx} + g'(x)y = g(x)g'(x)$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = g'(x)$ and $Q(x) = g(x)g'(x)$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int g'(x) dx} = e^{g(x)}$.
The general solution is $y(IF) = \int Q(IF) dx + C$.
Substituting the values,we get $y e^{g(x)} = \int g(x)g'(x) e^{g(x)} dx + C$.
Let $g(x) = t$,then $g'(x) dx = dt$.
So,$y e^{g(x)} = \int t e^t dt + C$.
Using integration by parts,$\int t e^t dt = t e^t - e^t + C = e^t(t - 1) + C$.
Thus,$y e^{g(x)} = e^{g(x)}(g(x) - 1) + C$.
Rearranging the terms,$y e^{g(x)} - e^{g(x)}(g(x) - 1) = C$.
$e^{g(x)}(y - g(x) + 1) = C$.
Taking the natural logarithm on both sides,$\log(e^{g(x)}) + \log|y - g(x) + 1| = \log(C)$.
$g(x) + \log|y - g(x) + 1| = C'$.
Thus,option $(b)$ is correct.

(B) Given differential equation is $\tan(y) dx + \sec^2(y) \tan(x) dy = 0$.
Rearranging the terms to separate variables,we get:
$\sec^2(y) \tan(x) dy = -\tan(y) dx$
$\frac{\sec^2(y)}{\tan(y)} dy = -\frac{1}{\tan(x)} dx$
$\frac{\sec^2(y)}{\tan(y)} dy = -\cot(x) dx$
Integrating both sides:
$\int \frac{\sec^2(y)}{\tan(y)} dy = -\int \cot(x) dx$
Let $u = \tan(y)$,then $du = \sec^2(y) dy$.
$\int \frac{1}{u} du = -\ln|\sin(x)| + C$
$\ln|\tan(y)| = -\ln|\sin(x)| + C$
$\ln|\tan(y)| + \ln|\sin(x)| = C$
$\ln|\sin(x) \tan(y)| = C$
Taking exponential on both sides,we get $\sin(x) \tan(y) = e^C = c$.

(D) The given differential equation is:
$\frac{dy}{dx} + \csc^2 (x) \cdot y = \csc^2 (x) \cdot \cot x . . . . . . . . . . (1)$
Comparing this with the linear differential equation form $\frac{dy}{dx} + Py = Q$,we get:
$P = \csc^2 x$
$Q = \csc^2 x \cdot \cot x$
Now,find the Integrating Factor $(IF)$:
$IF = e^{\int P \ dx} = e^{\int \csc^2 x \ dx} = e^{- \cot x}$
The general solution is given by:
$y \cdot IF = \int Q \cdot IF \ dx + C$
$y \cdot e^{- \cot x} = \int \csc^2 x \cdot \cot x \cdot e^{- \cot x} \ dx + C$
Let $t = \cot x$,then $dt = - \csc^2 x \ dx$,which implies $\csc^2 x \ dx = - dt$.
Substituting these into the integral:
$y \cdot e^{- \cot x} = \int t \cdot e^{- t} (- dt) = - \int t e^{- t} \ dt$
Using integration by parts $\int u \ dv = uv - \int v \ du$ where $u = t$ and $dv = e^{- t} \ dt$:
$- \int t e^{- t} \ dt = - [t (- e^{- t}) - \int (- e^{- t}) \ dt] = - [- t e^{- t} - e^{- t}] + C = t e^{- t} + e^{- t} + C$
Substituting $t = \cot x$ back:
$y \cdot e^{- \cot x} = e^{- \cot x} (\cot x + 1) + C$.

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$.
Let $x-y = u$. Then $1 - \frac{dy}{dx} = \frac{du}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{du}{dx}$.
Substituting this into the equation: $1 - \frac{du}{dx} = \frac{u+3}{2u+5}$.
Rearranging the terms: $\frac{du}{dx} = 1 - \frac{u+3}{2u+5} = \frac{2u+5-u-3}{2u+5} = \frac{u+2}{2u+5}$.
Separating the variables: $\int \frac{2u+5}{u+2} du = \int dx$.
Performing the division: $\int (2 + \frac{1}{u+2}) du = x + C$.
Integrating: $2u + \log|u+2| = x + C$.
Substituting $u = x-y$ back: $2(x-y) + \log|x-y+2| + C = x$.
Comparing this with the given form $x = 2(x-y) + \log(t) + c$,we identify $t = x-y+2$.

(A) Given differential equation is $\log \left(\frac{dy}{dx}\right) = ax + by$.
Taking exponential on both sides,we get $\frac{dy}{dx} = e^{ax + by} = e^{ax} \cdot e^{by}$.
Separating the variables,we have $e^{-by} dy = e^{ax} dx$.
Integrating both sides,we get $\int e^{-by} dy = \int e^{ax} dx$.
This results in $-\frac{1}{b} e^{-by} = \frac{1}{a} e^{ax} + C_1$.
Multiplying by $ab$,we get $-a e^{-by} = b e^{ax} + ab C_1$.
Rearranging the terms,we get $b e^{ax} + a e^{-by} = c$,where $c = -ab C_1$.
Therefore,option $A$ is correct.

(B) Given the differential equation $\frac{dy}{dx} = \frac{1+y^2}{\tan^{-1} y - x}$.
Taking the reciprocal,we get $\frac{dx}{dy} = \frac{\tan^{-1} y - x}{1+y^2}$.
Rearranging the terms,we have $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1} y}{1+y^2}$,which is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$.
Here,$P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\tan^{-1} y}{1+y^2}$.
The integrating factor $IF = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$.
The solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + c$.
Substituting the values: $x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + c$.
Let $t = \tan^{-1} y$,then $dt = \frac{1}{1+y^2} dy$.
The integral becomes $\int t e^t dt + c$.
Using integration by parts: $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + c = e^t(t - 1) + c$.
Substituting $t = \tan^{-1} y$ back,we get $x e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + c$.
Thus,option $B$ is correct.

(B) Let the radius of the circle be $r \text{ cm}$ and its area be $A \text{ cm}^2$.
Given that the rate of change of the radius is $\frac{dr}{dt} = 0.1 \text{ cm s}^{-1}$.
The area of a circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
Substituting the given values $r = 5 \text{ cm}$ and $\frac{dr}{dt} = 0.1 \text{ cm s}^{-1}$,we have:
$\frac{dA}{dt} = 2 \pi (5) (0.1) = 10 \pi (0.1) = \pi \text{ cm}^2 \text{ s}^{-1}$.
Thus,the rate of change of the area is $\pi \text{ cm}^2 \text{ s}^{-1}$.

(B) According to the given information,the slope of the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{1}{2y}$.
By separating the variables,we get $2y \, dy = dx$.
Integrating both sides,we have $\int 2y \, dy = \int dx$,which gives $y^2 = x + c$.
Since the curve passes through the point $(3, 4)$,we substitute $x = 3$ and $y = 4$ into the equation: $4^2 = 3 + c$,which implies $16 = 3 + c$,so $c = 13$.
Thus,the equation of the curve is $y^2 = x + 13$.
This equation is of the form $y^2 = 4a(x - h)$,which represents a parabola.

(B) Let the side of the square be $x$.
Given that the percentage change in the side is $6 \%$,so $\frac{dx}{x} \times 100 = 6 \%$.
The area of the square is $A = x^2$.
Differentiating with respect to $x$,we get $\frac{dA}{dx} = 2x$,which implies $dA = 2x \, dx$.
Dividing both sides by $A = x^2$,we get $\frac{dA}{A} = \frac{2x \, dx}{x^2} = 2 \frac{dx}{x}$.
To find the percentage change in area,multiply by $100$:
$\frac{dA}{A} \times 100 = 2 \times \left( \frac{dx}{x} \times 100 \right)$.
Substituting the given value,we get $\text{Percentage change in Area} = 2 \times 6 \% = 12 \%$.
Thus,the approximate percentage increase in the area is $12 \%$.
Therefore,option $B$ is correct.

(B) We know that $|u-v|^2 = |u|^2 + |v|^2 - 2(\vec{u} \cdot \vec{v})$.
Also,$(||u|-|v||)^2 = |u|^2 + |v|^2 - 2|u||v|$.
For the equality $|u-v| = ||u|-|v||$ to hold,their squares must be equal:
$|u|^2 + |v|^2 - 2(\vec{u} \cdot \vec{v}) = |u|^2 + |v|^2 - 2|u||v|$.
This simplifies to $\vec{u} \cdot \vec{v} = |u||v|$.
Since $\vec{u} \cdot \vec{v} = |u||v| \cos \theta$,we have $|u||v| \cos \theta = |u||v|$.
This implies $\cos \theta = 1$,which means $\theta = 0$.
Therefore,$u$ and $v$ must have the same direction.