AP EAMCET 2021 Mathematics Question Paper with Answer and Solution

(B) Let the coordinates of the moving point be $(x, y)$.
According to the definition of an ellipse,the sum of the distances of a point from two fixed points (foci) is constant $(2a)$.
The given condition is $\sqrt{(x - ae)^2 + y^2} + \sqrt{(x + ae)^2 + y^2} = 2a$.
Squaring and simplifying this equation:
$\sqrt{(x - ae)^2 + y^2} = 2a - \sqrt{(x + ae)^2 + y^2}$.
Squaring both sides:
$(x - ae)^2 + y^2 = 4a^2 + (x + ae)^2 + y^2 - 4a\sqrt{(x + ae)^2 + y^2}$.
$x^2 - 2aex + a^2e^2 + y^2 = 4a^2 + x^2 + 2aex + a^2e^2 + y^2 - 4a\sqrt{(x + ae)^2 + y^2}$.
$-4aex - 4a^2 = -4a\sqrt{(x + ae)^2 + y^2}$.
$ex + a = \sqrt{(x + ae)^2 + y^2}$.
Squaring again:
$e^2x^2 + 2aex + a^2 = x^2 + 2aex + a^2e^2 + y^2$.
$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$.
Dividing by $a^2(1 - e^2)$:
$\frac{x^2(1 - e^2)}{a^2(1 - e^2)} + \frac{y^2}{a^2(1 - e^2)} = 1$.
Since $b^2 = a^2(1 - e^2)$,we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

(C) Let the coordinates of point $P$ be $(x, y)$.
Given that $\operatorname{ar}(\triangle POB) = 2 \cdot \operatorname{ar}(\triangle POA)$.
The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.
For $\triangle POB$ with vertices $P(x, y), O(0, 0), B(0, 4)$:
$\operatorname{ar}(\triangle POB) = \frac{1}{2} |x(0-4) + 0(4-y) + 0(y-0)| = \frac{1}{2} |-4x| = 2|x|$.
For $\triangle POA$ with vertices $P(x, y), O(0, 0), A(6, 0)$:
$\operatorname{ar}(\triangle POA) = \frac{1}{2} |x(0-0) + 0(0-y) + 6(y-0)| = \frac{1}{2} |6y| = 3|y|$.
Substituting these into the given condition:
$2|x| = 2 \cdot 3|y| \Rightarrow |x| = 3|y|$.
Squaring both sides,we get $x^2 = 9y^2$.
Thus,the locus is $x^2 - 9y^2 = 0$.

(A) Let the two perpendicular lines be the coordinate axes $OX$ and $OY$. Let the ends of the rod be $A(0, a)$ on the $y$-axis and $B(b, 0)$ on the $x$-axis.
The length of the rod is $AB = 2l$. By the distance formula,$\sqrt{a^2+b^2} = 2l$,which implies $a^2+b^2 = 4l^2$.
Let $P(x, y)$ be the mid-point of the rod $AB$. Then $x = \frac{0+b}{2} = \frac{b}{2}$ and $y = \frac{a+0}{2} = \frac{a}{2}$.
This gives $b = 2x$ and $a = 2y$.
Substituting these values into the equation $a^2+b^2 = 4l^2$,we get $(2y)^2 + (2x)^2 = 4l^2$.
$4x^2 + 4y^2 = 4l^2$,which simplifies to $x^2+y^2 = l^2$.
Thus,the locus of the mid-point is $x^2+y^2 = l^2$.

(C) The given equation is $5x^2 - 8xy + 3y^2 = 0$.
Dividing by $x^2$,we get $3(\frac{y}{x})^2 - 8(\frac{y}{x}) + 5 = 0$.
Let $m = \frac{y}{x}$ be the slope of the lines. Then $3m^2 - 8m + 5 = 0$.
Solving the quadratic equation $3m^2 - 8m + 5 = 0$ using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$m = \frac{8 \pm \sqrt{64 - 4(3)(5)}}{2(3)} = \frac{8 \pm \sqrt{64 - 60}}{6} = \frac{8 \pm \sqrt{4}}{6} = \frac{8 \pm 2}{6}$.
The two slopes are $m_1 = \frac{8+2}{6} = \frac{10}{6} = \frac{5}{3}$ and $m_2 = \frac{8-2}{6} = \frac{6}{6} = 1$.
Since $m_1 > m_2$,we have $m_1 = \frac{5}{3}$ and $m_2 = 1$.
Therefore,the ratio $m_1 : m_2 = \frac{5}{3} : 1 = 5 : 3$.

(C) The given equation of the pair of lines is $4x^2-5xy+y^2=0$.
Dividing by $x^2$,we get the quadratic equation in terms of $m = \frac{y}{x}$:
$m^2-5m+4=0$.
Here,$m_1$ and $m_2$ are the roots of this equation.
Thus,the sum of slopes is $m_1+m_2 = 5$ and the product of slopes is $m_1m_2 = 4$.
We know that $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2}$.
Substituting the values,we get:
$|m_1-m_2| = \sqrt{5^2-4(4)} = \sqrt{25-16} = \sqrt{9} = 3$.

(D) Given the equation: $y^3-4x^2y=0$
Factorizing the equation: $y(y^2-4x^2)=0$
Using the identity $a^2-b^2=(a-b)(a+b)$,we get: $y(y-2x)(y+2x)=0$
This gives the three lines: $L_1: y=0$,$L_2: y=2x$,and $L_3: y=-2x$.
All three lines pass through the origin $(0,0)$.
Since all three lines intersect at a single common point $(0,0)$,they are concurrent lines.

(B) Given the equation of lines: $x^2+2 h x y+2 y^2=0 \dots (i)$
Comparing with $a x^2+2 h x y+b y^2=0$,we get $a=1$ and $b=2$.
Let the slopes of the lines be $m_1$ and $m_2$.
Then,$m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h \dots (ii)$
And $m_1 m_2 = \frac{a}{b} = \frac{1}{2} \dots (iii)$
Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$.
Substituting $m_2 = 2m_1$ into $(iii)$: $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$.
If $m_1 = \frac{1}{2}$,then $m_2 = 1$. From $(ii)$,$m_1+m_2 = -h$ $\Rightarrow \frac{1}{2} + 1 = -h$ $\Rightarrow h = -\frac{3}{2}$.
If $m_1 = -\frac{1}{2}$,then $m_2 = -1$. From $(ii)$,$m_1+m_2 = -h$ $\Rightarrow -\frac{1}{2} - 1 = -h$ $\Rightarrow h = \frac{3}{2}$.
Thus,$h = \pm \frac{3}{2}$.

(D) The given equation is $a x^2+2 h x y+b y^2=0$.
Let the slopes of the lines be $m$ and $m^2$.
From the properties of the homogeneous equation,the sum of slopes is $m+m^2 = -\frac{2h}{b}$ and the product of slopes is $m \cdot m^2 = m^3 = \frac{a}{b}$.
We have $m(1+m) = -\frac{2h}{b}$.
Cubing both sides,we get $m^3(1+m)^3 = -\frac{8h^3}{b^3}$.
Expanding this,$m^3(1+m^3+3m(1+m)) = -\frac{8h^3}{b^3}$.
Substituting $m^3 = \frac{a}{b}$ and $m(1+m) = -\frac{2h}{b}$,we get $\frac{a}{b}(1+\frac{a}{b}+3(-\frac{2h}{b})) = -\frac{8h^3}{b^3}$.
Multiplying by $b^3$,we get $a(b+a-6h) = -8h^3$.
$ab + a^2 - 6ah = -8h^3$.
Dividing by $abh$,we get $\frac{a+b}{h} + \frac{8h^2}{ab} = 6$.

(B) The equation $x^2+kxy+y^2=0$ represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m_1$ and $m_2$. The equation of the third line is $x+y=1$,which can be written as $y=-x+1$,so its slope is $m_3=-1$.
Since the lines form an equilateral triangle,the angle between any two lines is $60^{\circ}$.
The angle between the line $y=mx$ and the line $x+y=1$ (slope $-1$) is given by:
$\tan 60^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} \right|$
$\sqrt{3} = \left| \frac{m+1}{1-m} \right|$
Squaring both sides:
$3 = \frac{(m+1)^2}{(1-m)^2}$
$3(1-2m+m^2) = m^2+2m+1$
$3-6m+3m^2 = m^2+2m+1$
$2m^2-8m+2 = 0$
$m^2-4m+1 = 0$
Since $m = y/x$,we substitute this into the quadratic equation:
$(y/x)^2 - 4(y/x) + 1 = 0$
$y^2 - 4xy + x^2 = 0$
Comparing this with $x^2+kxy+y^2=0$,we get $k=-4$.
Therefore,$k^2 = (-4)^2 = 16$.

(D) The equation of the pair of lines is $ax^2+2hxy+by^2=0$. Let the slopes be $m$ and $m^2$.
From the properties of the quadratic equation in $y/x$,we have:
Sum of slopes: $m+m^2 = -\frac{2h}{b} \quad (1)$
Product of slopes: $m \cdot m^2 = m^3 = \frac{a}{b} \quad (2)$
From $(1)$,$m(1+m) = -\frac{2h}{b}$. Cubing both sides:
$m^3(1+m)^3 = -\frac{8h^3}{b^3}$
$m^3(1+m^3+3m(1+m)) = -\frac{8h^3}{b^3}$
Substitute $m^3 = \frac{a}{b}$ and $m(1+m) = -\frac{2h}{b}$:
$\frac{a}{b} \left(1 + \frac{a}{b} + 3\left(-\frac{2h}{b}\right)\right) = -\frac{8h^3}{b^3}$
$\frac{a}{b} \left(\frac{b+a-6h}{b}\right) = -\frac{8h^3}{b^3}$
$a(a+b-6h) = -8h^3$
$a^2+ab-6ah = -8h^3$
Dividing by $abh$ (assuming $a, b, h \neq 0$):
$\frac{a}{bh} + \frac{1}{h} - \frac{6}{b} = -\frac{8h^2}{ab}$
Rearranging terms gives $\frac{a+b}{h} + \frac{8h^2}{ab} = 6$.

(C) The given equation of the pair of straight lines is $ax^2+2hxy+by^2=0$,where $a=1$,$2h=4$ (so $h=2$),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 $\tan \theta = \left| \frac{2\sqrt{2^2-(1)(1)}}{1+1} \right|$.
$\tan \theta = \frac{2\sqrt{4-1}}{2} = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.

(B) The angle $\theta$ between the lines represented by the homogeneous equation $ax^2+2hxy+by^2=0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Given $\theta = \frac{\pi}{4}$,we have $\tan \frac{\pi}{4} = 1$.
Thus,$1 = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Squaring both sides,we get $1 = \frac{4(h^2-ab)}{(a+b)^2}$.
$(a+b)^2 = 4h^2 - 4ab$.
$a^2 + 2ab + b^2 = 4h^2 - 4ab$.
$4h^2 = a^2 + 6ab + b^2$.

(A) The given equation is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos \theta(\cos \theta + 1)$,$2H = -(2 \cos \theta + \sin^2 \theta)$,and $B = 1 - \cos \theta$.
The angle $\alpha$ between the lines is given by $\tan \alpha = \left| \frac{2\sqrt{H^2 - AB}}{A + B} \right|$.
First,calculate $A + B = \cos^2 \theta + \cos \theta + 1 - \cos \theta = \cos^2 \theta + 1$.
Next,calculate $H^2 - AB = \frac{(2 \cos \theta + \sin^2 \theta)^2}{4} - \cos \theta(\cos \theta + 1)(1 - \cos \theta)$.
$= \frac{4 \cos^2 \theta + 4 \cos \theta \sin^2 \theta + \sin^4 \theta}{4} - \cos \theta(1 - \cos^2 \theta) = \cos^2 \theta + \cos \theta \sin^2 \theta + \frac{\sin^4 \theta}{4} - \cos \theta + \cos^3 \theta$.
$= \cos^2 \theta + \cos \theta(1 - \cos^2 \theta) + \frac{\sin^4 \theta}{4} - \cos \theta + \cos^3 \theta = \cos^2 \theta + \frac{\sin^4 \theta}{4}$.
Thus,$\tan \alpha = \frac{2 \sqrt{\cos^2 \theta + \frac{\sin^4 \theta}{4}}}{\cos^2 \theta + 1} = \frac{\sqrt{4 \cos^2 \theta + \sin^4 \theta}}{\cos^2 \theta + 1} = \frac{\sqrt{4 \cos^2 \theta + (1 - \cos^2 \theta)^2}}{\cos^2 \theta + 1}$.
$= \frac{\sqrt{4 \cos^2 \theta + 1 + \cos^4 \theta - 2 \cos^2 \theta}}{\cos^2 \theta + 1} = \frac{\sqrt{\cos^4 \theta + 2 \cos^2 \theta + 1}}{\cos^2 \theta + 1} = \frac{\sqrt{(\cos^2 \theta + 1)^2}}{\cos^2 \theta + 1} = 1$.
Therefore,$\alpha = \tan^{-1}(1) = \frac{\pi}{4}$.

(B) The acute angle $\theta$ between the pair of straight lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
Comparing the given equation $6x^2 + 11xy - 10y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$,we get $a = 6$,$b = -10$,and $2h = 11$,so $h = \frac{11}{2}$.
Substituting these values into the formula:
$\tan \theta = \left| \frac{2\sqrt{(\frac{11}{2})^2 - (6)(-10)}}{6 + (-10)} \right| = \left| \frac{2\sqrt{\frac{121}{4} + 60}}{-4} \right| = \left| \frac{2\sqrt{\frac{121 + 240}{4}}}{-4} \right| = \left| \frac{2\sqrt{\frac{361}{4}}}{-4} \right| = \left| \frac{2 \cdot \frac{\sqrt{361}}{2}}{-4} \right| = \left| \frac{\sqrt{361}}{-4} \right| = \frac{\sqrt{361}}{4}$.
Therefore,$\theta = \tan^{-1}\left(\frac{\sqrt{361}}{4}\right)$.

(B) The given equation of the pair of lines is $ax^2+6xy+by^2-10x+10y-6=0$.
Comparing this with the general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=a, H=3, B=b, G=-5, F=5, C=-6$.
For the lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a+b=0$,which implies $b=-a$.
Also,the condition for the general second-degree equation to represent a pair of lines is $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Substituting the values: $a(b)(-6) + 2(5)(3)(-5) - a(5)^2 - b(-5)^2 - (-6)(3)^2 = 0$.
$-6ab - 150 - 25a - 25b + 54 = 0$.
$-6ab - 25(a+b) - 96 = 0$.
Since $a+b=0$,we have $-6a(-a) - 25(0) - 96 = 0$.
$6a^2 = 96 \Rightarrow a^2 = 16$.
Therefore,$|a| = \sqrt{16} = 4$.

(A) The angle $\theta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
For the equation $x^2 + 2xy \sec \theta + y^2 = 0$,we have $a = 1$,$h = \sec \theta$,and $b = 1$.
Let $\phi$ be the angle between these lines.
Then $\tan \phi = \left| \frac{2\sqrt{(\sec \theta)^2 - (1)(1)}}{1+1} \right|$.
$\tan \phi = \left| \frac{2\sqrt{\sec^2 \theta - 1}}{2} \right|$.
Since $\sec^2 \theta - 1 = \tan^2 \theta$,we have $\tan \phi = \sqrt{\tan^2 \theta} = \tan \theta$.
Therefore,$\phi = \theta$.

(C) The given equation is $(\sin ^2 \alpha) y^2 - 2xy(\cos ^2 \alpha) + (\cos ^2 \alpha - 1) x^2 = 0$.
Comparing this with the standard form $ax^2 + 2hxy + by^2 = 0$,we get:
$a = \cos ^2 \alpha - 1 = -\sin ^2 \alpha$
$h = -\cos ^2 \alpha$
$b = \sin ^2 \alpha$
The angle $\theta$ between the pair of lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Calculating the denominator: $a + b = -\sin ^2 \alpha + \sin ^2 \alpha = 0$.
Since the denominator is $0$,the lines are perpendicular to each other.
Therefore,$\tan \theta = \infty$,which implies $\theta = 90^{\circ}$.

(A) The given equation is $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$.
Expanding this,we get $abx^2 + (a^2 - b^2)xy - aby^2 = 0$.
This is a homogeneous equation of the second degree in $x$ and $y$ of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = ab$,$2H = a^2 - b^2$,and $B = -ab$.
The condition for the lines to be perpendicular is $A + B = 0$.
Here,$A + B = ab + (-ab) = 0$.
Since the sum of the coefficients of $x^2$ and $y^2$ is zero,the lines represented by the equation are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(A) Let the equation be $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4 = 0$.
Assume the equation can be factored as $(a x^2+p x y-a y^2)(x^2+q x y+y^2) = 0$.
Comparing the coefficients of similar terms:
$b = aq - p$,$c = -pq$,$d = aq + p$,and $e = -a$.
Then,$b + d = 2aq$ and $e - a = -2a$.
Also,$ad + be = 2ap$ and $a + c + e = -pq$.
Now,consider the expression $(b+d)(ad+be) = (2aq)(2ap) = 4a^2pq$.
Also,$-(e-a)^2(a+c+e) = -(-2a)^2(-pq) = 4a^2pq$.
Therefore,$(b+d)(ad+be) = -(e-a)^2(a+c+e)$.
This simplifies to $(b+d)(ad+be) + (e-a)^2(a+c+e) = 0$.

(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,$p q = -1$.

(D) The given equation is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $b(\frac{y}{x})^2+2h(\frac{y}{x})+a=0$. Let $m = \frac{y}{x}$,so $bm^2+2hm+a=0$. The roots $m_1$ and $m_2$ represent the slopes of the lines. The quadratic formula gives $m = \frac{-2h \pm \sqrt{4h^2-4ab}}{2b}$. Since $h^2=ab$,the discriminant $4h^2-4ab = 0$. Thus,$m_1 = m_2 = -\frac{2h}{2b} = -\frac{h}{b}$. The ratio of the slopes is $m_1:m_2 = 1:1$.

(B) Given,the pair of lines is $x^2-y^2+2y-1=0$.
$x^2 = y^2-2y+1$
$\Rightarrow x^2 = (y-1)^2$
$\Rightarrow x^2 - (y-1)^2 = 0$
$\Rightarrow (x+y-1)(x-y+1) = 0$
So,the lines are $l_1: x+y-1=0$ and $l_2: x-y+1=0$.
The angle bisectors of these lines are found by $\frac{x+y-1}{\sqrt{1^2+1^2}} = \pm \frac{x-y+1}{\sqrt{1^2+(-1)^2}}$.
$\Rightarrow x+y-1 = \pm(x-y+1)$.
Case $1$: $x+y-1 = x-y+1$ $\Rightarrow 2y = 2$ $\Rightarrow y=1$.
Case $2$: $x+y-1 = -(x-y+1)$ $\Rightarrow x+y-1 = -x+y-1$ $\Rightarrow 2x = 0$ $\Rightarrow x=0$.
The angle bisectors are the lines $x=0$ ($Y$-axis) and $y=1$.
The triangle is formed by the lines $x+y=4$,$x=0$,and $y=1$.
The vertices of the triangle are:
$1$. Intersection of $x=0$ and $y=1$ is $(0, 1)$.
$2$. Intersection of $x=0$ and $x+y=4$ is $(0, 4)$.
$3$. Intersection of $y=1$ and $x+y=4$ is $(3, 1)$.
The base of the triangle along $x=0$ is $|4-1| = 3$.
The height of the triangle from $x=0$ to $x=3$ is $3$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = 4.5$ sq. units.

(B) The given equation is $3 x^2-5 x y+4 y^2=0$.
Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=3$,$2 h=-5$,and $b=4$.
The equation of the bisectors of the angle between the pair of lines is given by $\frac{x^2-y^2}{a-b} = \frac{x y}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{3-4} = \frac{x y}{-5/2}$.
This simplifies to $\frac{x^2-y^2}{-1} = \frac{2 x y}{-5}$.
Multiplying both sides by $-5$,we get $5(x^2-y^2) = 2 x y$.

(A) The given equation is $x^2-2 m x y-y^2=0$.
Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$.
The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{x y}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{1-(-1)} = \frac{x y}{-m}$.
This simplifies to $\frac{x^2-y^2}{2} = \frac{x y}{-m}$,which implies $-m(x^2-y^2) = 2 x y$.
Rearranging the terms,we get $m x^2 + 2 x y - m y^2 = 0$.
Dividing by $m$ (assuming $m \neq 0$),we get $x^2 + \frac{2}{m} x y - y^2 = 0$.
Comparing this with the given bisector equation $x^2-2 n x y-y^2=0$,we have $-2n = \frac{2}{m}$.
Therefore,$-n = \frac{1}{m}$,which gives $mn = -1$ or $mn+1=0$.

(D) The given equation is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $b(\frac{y}{x})^2+2h(\frac{y}{x})+a=0$. Let $m = \frac{y}{x}$,then $bm^2+2hm+a=0$. The roots of this quadratic equation are the slopes $m_1$ and $m_2$ of the lines. Given $h^2=ab$,the discriminant $D = (2h)^2 - 4ab = 4h^2 - 4ab = 4(h^2-ab) = 0$. Since the discriminant is zero,the roots are equal,i.e.,$m_1 = m_2$. Therefore,the ratio of the slopes is $m_1:m_2 = 1:1$.

(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 $p q = -1$.

(C) The given pair of lines is $2y^2+5xy-3x^2=0$,which can be rewritten as $3x^2-5xy-2y^2=0$.
Factoring the quadratic equation: $3x^2-6xy+xy-2y^2=0$ $\Rightarrow 3x(x-2y)+y(x-2y)=0$ $\Rightarrow (x-2y)(3x+y)=0$.
Thus,the two lines are $L_1: x-2y=0$ and $L_2: 3x+y=0$.
The third line is $L_3: x+y=k$.
The vertices of the triangle are the intersection points of these lines:
$O(0,0)$ is the intersection of $L_1$ and $L_2$.
Intersection of $L_1$ and $L_3$: $x-2y=0$ and $x+y=k$ $\Rightarrow 3y=k$ $\Rightarrow y=\frac{k}{3}, x=\frac{2k}{3}$. So,$A(\frac{2k}{3}, \frac{k}{3})$.
Intersection of $L_2$ and $L_3$: $3x+y=0$ and $x+y=k$ $\Rightarrow 2x=-k$ $\Rightarrow x=-\frac{k}{2}, y=\frac{3k}{2}$. So,$B(-\frac{k}{2}, \frac{3k}{2})$.
The centroid of $\triangle OAB$ is $(\frac{0+\frac{2k}{3}-\frac{k}{2}}{3}, \frac{0+\frac{k}{3}+\frac{3k}{2}}{3}) = (\frac{\frac{k}{6}}{3}, \frac{\frac{11k}{6}}{3}) = (\frac{k}{18}, \frac{11k}{18})$.
Given the centroid is $(\frac{1}{18}, \frac{11}{18})$,we have $\frac{k}{18} = \frac{1}{18}$,which implies $k=1$.

(B) The given equation is $8 x^2-24 x y+18 y^2-6 x+9 y-5=0$.
We can rewrite the quadratic part as $(2 \sqrt{2} x - 3 \sqrt{2} y)^2 = 2(4 x^2 - 12 x y + 9 y^2) = 2(2 x - 3 y)^2$.
Let $2 x - 3 y = t$. Then the equation becomes $2 t^2 - 3 t - 5 = 0$.
Factoring the quadratic: $2 t^2 - 5 t + 2 t - 5 = 0 \implies t(2 t - 5) + 1(2 t - 5) = 0 \implies (t + 1)(2 t - 5) = 0$.
Substituting $t = 2 x - 3 y$ back,we get $(2 x - 3 y + 1)(2(2 x - 3 y) - 5) = 0$.
This simplifies to $(2 x - 3 y + 1)(4 x - 6 y - 5) = 0$.
Since the equation represents two linear factors of the form $(a_1 x + b_1 y + c_1)(a_2 x + b_2 y + c_2) = 0$,it represents a pair of lines.
Comparing the slopes,$m_1 = \frac{2}{3}$ and $m_2 = \frac{4}{6} = \frac{2}{3}$.
Since the slopes are equal,the lines are parallel.

(B) The equation $ax^2+2hxy+by^2=0$ represents a pair of straight lines passing through the origin $(0,0)$.
Line $1$ passes through $(0,0)$ and $(2,3)$. Its equation is $y - 0 = \frac{3-0}{2-0}(x - 0)$ $\Rightarrow y = \frac{3}{2}x$ $\Rightarrow 3x - 2y = 0$.
Line $2$ passes through $(0,0)$ and $(4,5)$. Its equation is $y - 0 = \frac{5-0}{4-0}(x - 0)$ $\Rightarrow y = \frac{5}{4}x$ $\Rightarrow 5x - 4y = 0$.
The combined equation is $(3x - 2y)(5x - 4y) = 0$.
Expanding this,we get $15x^2 - 12xy - 10xy + 8y^2 = 0 \Rightarrow 15x^2 - 22xy + 8y^2 = 0$.
Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a = 15$,$2h = -22$ (so $h = -11$),and $b = 8$.
Therefore,$a + 2h + b = 15 - 22 + 8 = 1$.

(C) Given the equation of the pair of lines: $2x^2 + 4xy - 4y^2 - 6x - 8y + 7 = 0$.
To find the length of the intercept on the $Y$-axis,we set $x = 0$.
Substituting $x = 0$ into the equation,we get: $-4y^2 - 8y + 7 = 0$,which simplifies to $4y^2 + 8y - 7 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have $y = \frac{-8 \pm \sqrt{64 - 4(4)(-7)}}{2(4)} = \frac{-8 \pm \sqrt{64 + 112}}{8} = \frac{-8 \pm \sqrt{176}}{8}$.
Since $\sqrt{176} = \sqrt{16 \times 11} = 4\sqrt{11}$,we get $y = \frac{-8 \pm 4\sqrt{11}}{8} = -1 \pm \frac{\sqrt{11}}{2}$.
The length of the intercept is the absolute difference between the two roots $y_1$ and $y_2$: $|y_1 - y_2| = |(-1 + \frac{\sqrt{11}}{2}) - (-1 - \frac{\sqrt{11}}{2})| = |\frac{\sqrt{11}}{2} + \frac{\sqrt{11}}{2}| = \sqrt{11}$.

(B) Given equation is $(x-2y)^2 + k(x-2y) = 0$.
Factoring the equation,we get $(x-2y)(x-2y+k) = 0$.
This represents two parallel lines: $L_1: x-2y = 0$ and $L_2: x-2y+k = 0$.
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,$A=1, B=-2, C_1=0, C_2=k$.
So,$d = \frac{|k-0|}{\sqrt{1^2+(-2)^2}} = \frac{|k|}{\sqrt{5}}$.
Given $d = 3$,we have $\frac{|k|}{\sqrt{5}} = 3$.
Therefore,$|k| = 3\sqrt{5}$,which implies $k = \pm 3\sqrt{5}$.

(A) The general equation of the second degree $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines if $\Delta = abc+2fgh-af^2-bg^2-ch^2 = 0$.
Comparing the given equation $x^2+pxy+y^2-5x-7y+6=0$ with the general form:
$a=1, b=1, c=6, h=\frac{p}{2}, g=-\frac{5}{2}, f=-\frac{7}{2}$.
Substituting these values into the condition $\Delta=0$:
$(1)(1)(6) + 2(-\frac{7}{2})(-\frac{5}{2})(\frac{p}{2}) - 1(-\frac{7}{2})^2 - 1(-\frac{5}{2})^2 - 6(\frac{p}{2})^2 = 0$
$6 + \frac{35p}{4} - \frac{49}{4} - \frac{25}{4} - \frac{6p^2}{4} = 0$
Multiply by $4$:
$24 + 35p - 49 - 25 - 6p^2 = 0$
$-6p^2 + 35p - 50 = 0$
$6p^2 - 35p + 50 = 0$
$(2p-5)(3p-10) = 0$
Thus,$p = \frac{5}{2}$ or $p = \frac{10}{3}$.
Given the options,the correct value is $\frac{5}{2}$.

(A) The general equation of a second-degree curve $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines if the determinant $\Delta = abc+2fgh-af^2-bg^2-ch^2 = 0$.
Comparing the given equation $x^2+pxy+y^2-5x-7y+6=0$ with the general form:
$a=1, b=1, c=6, h=\frac{p}{2}, g=-\frac{5}{2}, f=-\frac{7}{2}$.
Substituting these values into the condition $\Delta = 0$:
$(1)(1)(6) + 2(-\frac{7}{2})(-\frac{5}{2})(\frac{p}{2}) - (1)(-\frac{7}{2})^2 - (1)(-\frac{5}{2})^2 - (6)(\frac{p}{2})^2 = 0$.
$6 + \frac{35p}{4} - \frac{49}{4} - \frac{25}{4} - \frac{6p^2}{4} = 0$.
Multiplying by $4$:
$24 + 35p - 49 - 25 - 6p^2 = 0$.
$-6p^2 + 35p - 50 = 0$.
$6p^2 - 35p + 50 = 0$.
$(2p-5)(3p-10) = 0$.
Thus,$p = \frac{5}{2}$ or $p = \frac{10}{3}$.

(A) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$.
For these lines to be real,the condition is $h^2 - ab \geq 0$.
Comparing $2x^2 - pxy + 2y^2 = 0$ with the general equation,we have $a = 2$,$2h = -p$ (so $h = -p/2$),and $b = 2$.
Substituting these into the condition:
$(-p/2)^2 - (2)(2) \geq 0$
$\frac{p^2}{4} - 4 \geq 0$
$p^2 - 16 \geq 0$
$(p - 4)(p + 4) \geq 0$
This inequality holds when $p \leq -4$ or $p \geq 4$.
Therefore,$p \in (-\infty, -4] \cup [4, \infty)$.

(D) The given curve is $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$.
The line is $x + 2y = k$,which implies $\frac{x + 2y}{k} = 1$.
Homogenizing the curve equation using the line equation:
$2x^2 - 2xy + 3y^2 + (2x - y)(1) - 1(1)^2 = 0$.
Substituting $1 = \frac{x + 2y}{k}$:
$2x^2 - 2xy + 3y^2 + (2x - y)\left(\frac{x + 2y}{k}\right) - \left(\frac{x + 2y}{k}\right)^2 = 0$.
Multiplying by $k^2$:
$k^2(2x^2 - 2xy + 3y^2) + k(2x^2 + 4xy - xy - 2y^2) - (x^2 + 4xy + 4y^2) = 0$.
$x^2(2k^2 + 2k - 1) + xy(-2k^2 + 3k - 4) + y^2(3k^2 - 2k - 4) = 0$.
Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(2k^2 + 2k - 1) + (3k^2 - 2k - 4) = 0$.
$5k^2 - 5 = 0$.
$5k^2 = 5 \implies k^2 = 1$.

(C) The general equation of a pair of straight lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if the determinant $\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
Comparing with $2x^2 + kxy - 6y^2 + 3x + y + 1 = 0$,we have $a=2, h=k/2, b=-6, g=3/2, f=1/2, c=1$.
The condition $\Delta = 0$ gives $\begin{vmatrix} 2 & k/2 & 3/2 \\ k/2 & -6 & 1/2 \\ 3/2 & 1/2 & 1 \end{vmatrix} = 0$.
Multiplying rows by $2$,we get $\frac{1}{8} \begin{vmatrix} 4 & k & 3 \\ k & -12 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 0$.
Expanding the determinant: $4(-24 - 1) - k(2k - 3) + 3(k + 36) = 0$.
$-100 - 2k^2 + 3k + 3k + 108 = 0$ $\Rightarrow -2k^2 + 6k + 8 = 0$ $\Rightarrow k^2 - 3k - 4 = 0$.
$(k - 4)(k + 1) = 0$. Since $k > 0$,we have $k = 4$.
The equation becomes $2x^2 + 4xy - 6y^2 + 3x + y + 1 = 0$.
Factoring the expression: $(2x - 2y + 1)(x + 3y + 1) = 0$.
Solving the system $2x - 2y + 1 = 0$ and $x + 3y + 1 = 0$:
From the second equation,$x = -3y - 1$. Substituting into the first: $2(-3y - 1) - 2y + 1 = 0$ $\Rightarrow -6y - 2 - 2y + 1 = 0$ $\Rightarrow -8y = 1$ $\Rightarrow y = -1/8$.
Then $x = -3(-1/8) - 1 = 3/8 - 8/8 = -5/8$.
The point of intersection is $\left(-\frac{5}{8}, -\frac{1}{8}\right)$.

(C) The given pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x, y)$ of the pair of lines is given by the formula $\left(\frac{hf-bg}{ab-h^2}, \frac{gh-af}{ab-h^2}\right)$.
Substituting the values: $x = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$.
$y = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$.
The point of intersection is $(\frac{4}{3}, \frac{2}{3})$.
Since the lines are concurrent,this point must satisfy the equation $5x+ky-8=0$.
$5(\frac{4}{3}) + k(\frac{2}{3}) - 8 = 0$.
$\frac{20}{3} + \frac{2k}{3} = 8$.
$20 + 2k = 24$.
$2k = 4 \Rightarrow k = 2$.

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x_0, y_0)$ of the pair of lines is given by the formula:
$x_0 = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$.
$y_0 = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$.
Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the line $5x+\lambda y-8=0$.
$5(\frac{4}{3}) + \lambda(\frac{2}{3}) - 8 = 0$.
$\frac{20}{3} + \frac{2\lambda}{3} - 8 = 0$.
Multiply by $3$: $20 + 2\lambda - 24 = 0$.
$2\lambda - 4 = 0$.
$2\lambda = 4$.
$\lambda = 2$.

(B) The given equation of the pair of straight lines is $2 x^2+3 x y+2 y^2+10 x+5 y=0$.
This is in the general form $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$.
Comparing the coefficients,we get $a=2$,$2h=3$ (so $h=\frac{3}{2}$),and $b=2$.
The equation of the pair of lines passing through the origin and perpendicular to the given pair of lines is given by $b x^2-2 h x y+a y^2=0$.
Substituting the values,we get $2 x^2-3 x y+2 y^2=0$.

(A) The given equation is $2x^2 - xy - y^2 = 0$.
Factoring the quadratic expression:
$2x^2 - 2xy + xy - y^2 = 0$
$2x(x - y) + y(x - y) = 0$
$(2x + y)(x - y) = 0$.
The lines are parallel to $2x + y = 0$ and $x - y = 0$.
Let the required lines be $(2x + y + k_1) = 0$ and $(x - y + k_2) = 0$.
Since these lines pass through $(1, 0)$,we substitute the point:
For the first line: $2(1) + 0 + k_1 = 0 \Rightarrow k_1 = -2$.
For the second line: $1 - 0 + k_2 = 0 \Rightarrow k_2 = -1$.
The combined equation is $(2x + y - 2)(x - y - 1) = 0$.
Expanding the product:
$2x(x - y - 1) + y(x - y - 1) - 2(x - y - 1) = 0$
$2x^2 - 2xy - 2x + xy - y^2 - y - 2x + 2y + 2 = 0$
$2x^2 - xy - y^2 - 4x + y + 2 = 0$.

(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. \\
Since the circle touches the $x$-axis at $(3, 0)$,the center is $(h, k)$ and the radius $r = |k|$. \\
Thus,the equation becomes $(x-h)^2 + (y-k)^2 = k^2$. \\
Since it passes through $(3, 0)$,we have $(3-h)^2 + (0-k)^2 = k^2$ $\Rightarrow (3-h)^2 = 0$ $\Rightarrow h = 3$. \\
Since it also passes through $(1, -2)$,we substitute $h=3$ into the equation: $(1-3)^2 + (-2-k)^2 = k^2$. \\
$(-2)^2 + 4 + 4k + k^2 = k^2$ \\
$4 + 4 + 4k = 0$ $\Rightarrow 4k = -8$ $\Rightarrow k = -2$. \\
Substituting $h=3$ and $k=-2$ into $(x-h)^2 + (y-k)^2 = k^2$: \\
$(x-3)^2 + (y+2)^2 = (-2)^2$ \\
$x^2 - 6x + 9 + y^2 + 4y + 4 = 4$ \\
$x^2 + y^2 - 6x + 4y + 9 = 0$.

(B) The equation of the circle is $x^2+y^2-x+3y=0$. The center is $C = (\frac{1}{2}, -\frac{3}{2})$.
Let the line $L_1$ passing through the origin be $y = mx$,or $mx - y = 0$.
The line $L_2$ is $x + y - 1 = 0$.
Since the intercepts made by the circle on $L_1$ and $L_2$ are equal,the perpendicular distances from the center $C$ to these lines must be equal.
The perpendicular distance from $C(\frac{1}{2}, -\frac{3}{2})$ to $L_1$ is $d_1 = \frac{|m(\frac{1}{2}) - (-\frac{3}{2})|}{\sqrt{m^2 + (-1)^2}} = \frac{|\frac{m+3}{2}|}{\sqrt{m^2+1}}$.
The perpendicular distance from $C(\frac{1}{2}, -\frac{3}{2})$ to $L_2$ is $d_2 = \frac{|\frac{1}{2} - \frac{3}{2} - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$.
Equating $d_1^2 = d_2^2$,we get $\frac{(m+3)^2}{4(m^2+1)} = 2$.
$(m+3)^2 = 8(m^2+1) \Rightarrow m^2 + 6m + 9 = 8m^2 + 8$.
$7m^2 - 6m - 1 = 0$.
$(7m+1)(m-1) = 0$,so $m = 1$ or $m = -\frac{1}{7}$.
For $m = 1$,$L_1$ is $y = x$ or $x - y = 0$.
For $m = -\frac{1}{7}$,$L_1$ is $y = -\frac{1}{7}x$ or $x + 7y = 0$.

(A) The circle passes through points $(4, 0)$ and $(-4, 0)$ on the $x$-axis.
Since the points are symmetric about the $y$-axis,the center of the circle must lie on the $y$-axis.
Let the center be $(0, c)$.
The distance from the center $(0, c)$ to the point $(4, 0)$ is the radius $r = 5$.
Using the distance formula: $\sqrt{(4-0)^2 + (0-c)^2} = 5$.
$16 + c^2 = 25$ $\Rightarrow c^2 = 9$ $\Rightarrow c = \pm 3$.
Thus,the centers are $(0, 3)$ and $(0, -3)$.
For center $(0, 3)$,the equation is $x^2 + (y-3)^2 = 5^2$ $\Rightarrow x^2 + y^2 - 6y + 9 = 25$ $\Rightarrow x^2 + y^2 - 6y - 16 = 0$.
For center $(0, -3)$,the equation is $x^2 + (y+3)^2 = 5^2$ $\Rightarrow x^2 + y^2 + 6y + 9 = 25$ $\Rightarrow x^2 + y^2 + 6y - 16 = 0$.
Both $x^2 + y^2 - 6y - 16 = 0$ and $x^2 + y^2 + 6y - 16 = 0$ are valid equations.

(D) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$.
Since the circle passes through the origin $(0,0)$,we have $c=0$.
The circle cuts off an intercept of $-2$ on the $x$-axis,meaning it passes through $(-2,0)$. Substituting this into the equation: $(-2)^2 + 0^2 + 2g(-2) + 2f(0) + 0 = 0$ $\Rightarrow 4 - 4g = 0$ $\Rightarrow g = 1$.
The circle cuts off an intercept of $3$ on the $y$-axis,meaning it passes through $(0,3)$. Substituting this into the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + 0 = 0$ $\Rightarrow 9 + 6f = 0$ $\Rightarrow f = -\frac{3}{2}$.
Substituting $g=1$,$f=-\frac{3}{2}$,and $c=0$ into the general equation: $x^2+y^2+2(1)x+2(-\frac{3}{2})y+0=0$.
This simplifies to $x^2+y^2+2x-3y=0$.

(C) Let $S_1 = x^2+y^2-8x-6y+21=0$ and $S_2 = x^2+y^2-2x-15=0$.
Using the family of circles equation $S_1 + \lambda S_2 = 0$,we have:
$(x^2+y^2-8x-6y+21) + \lambda(x^2+y^2-2x-15) = 0$.
Since the circle passes through the point $(1, 2)$,substitute $x=1$ and $y=2$ into the equation:
$(1^2+2^2-8(1)-6(2)+21) + \lambda(1^2+2^2-2(1)-15) = 0$.
$(1+4-8-12+21) + \lambda(1+4-2-15) = 0$.
$6 + \lambda(-12) = 0$.
$12\lambda = 6 \Rightarrow \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ back into the family equation:
$(x^2+y^2-8x-6y+21) + \frac{1}{2}(x^2+y^2-2x-15) = 0$.
Multiply the entire equation by $2$:
$2(x^2+y^2-8x-6y+21) + (x^2+y^2-2x-15) = 0$.
$2x^2+2y^2-16x-12y+42 + x^2+y^2-2x-15 = 0$.
$3x^2+3y^2-18x-12y+27 = 0$.
$3(x^2+y^2)-18x-12y+27 = 0$.

(D) Given the equation of the circle: $x^2+y^2-3x-4y+2=0$.
Since the circle cuts the $x$-axis,the $y$-coordinate at these points must be $0$.
Substituting $y=0$ into the equation:
$x^2 + (0)^2 - 3x - 4(0) + 2 = 0$
$x^2 - 3x + 2 = 0$
Factoring the quadratic equation:
$(x-1)(x-2) = 0$
This gives $x=1$ and $x=2$.
Therefore,the points of intersection are $(1,0)$ and $(2,0)$.

(A) The given equation of the circle is $x^2+y^2+8x+10y-8=0$.
Comparing this with the general equation of a circle $x^2+y^2+2gx+2fy+c=0$,we get:
$2g = 8 \implies g = 4$
$2f = 10 \implies f = 5$
$c = -8$
The centre of the circle is $(-g, -f) = (-4, -5)$.
The radius $r$ is given by $\sqrt{g^2+f^2-c}$.
$r = \sqrt{4^2+5^2-(-8)} = \sqrt{16+25+8} = \sqrt{49} = 7$.
Thus,the centre is $(-4, -5)$ and the radius is $7$.

(D) The end points of the diameter are $(4,7)$ and $(-2,-1)$.
The equation of the circle is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
Substituting the values,we get $(x-4)(x+2) + (y-7)(y+1) = 0$.
Expanding this,we have $x^2 - 2x - 8 + y^2 - 6y - 7 = 0$,which simplifies to $x^2 + y^2 - 2x - 6y - 15 = 0$.
The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing the equations,we get $g = -1$,$f = -3$,and $c = -15$.
The length of the $X$-intercept is given by the formula $2\sqrt{g^2 - c}$.
Substituting the values,$AB = 2\sqrt{(-1)^2 - (-15)} = 2\sqrt{1 + 15} = 2\sqrt{16} = 2 \times 4 = 8$.
Thus,$AB = 8$.

(C) Given circle equation is $x^2+y^2+4x-4y+4=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=2$,$f=-2$,and $c=4$.
The center of the circle is $(-g, -f) = (-2, 2)$.
The radius of the circle is $r = \sqrt{g^2+f^2-c} = \sqrt{2^2+(-2)^2-4} = \sqrt{4+4-4} = \sqrt{4} = 2$.
Since the absolute value of the $x$-coordinate of the center is $|-2| = 2$ (which equals the radius) and the absolute value of the $y$-coordinate of the center is $|2| = 2$ (which also equals the radius),the circle touches both the $X$-axis and the $Y$-axis.

(B) The circumference of the circle is $10 \pi$. Given $2 \pi r = 10 \pi$,we find the radius $r = 5$.
Since the diameters of a circle intersect at its center,we solve the system of equations:
$2x + 3y + 1 = 0$ $(i)$
$3x - y - 4 = 0$ $(ii)$
From $(ii)$,$y = 3x - 4$. Substituting this into $(i)$:
$2x + 3(3x - 4) + 1 = 0$
$2x + 9x - 12 + 1 = 0$
$11x - 11 = 0 \Rightarrow x = 1$.
Then $y = 3(1) - 4 = -1$.
The center of the circle is $(1, -1)$.
The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
$(x - 1)^2 + (y + 1)^2 = 5^2$
$x^2 - 2x + 1 + y^2 + 2y + 1 = 25$
$x^2 + y^2 - 2x + 2y - 23 = 0$.

(C) We have,$\int \frac{2x+3}{x(x+1)(x+2)(x+3)+1} dx = \frac{-1}{ax^2+bx+c} + \alpha$.
Rearranging the denominator: $x(x+3)(x+1)(x+2)+1 = (x^2+3x)(x^2+3x+2)+1$.
Let $t = x^2+3x$,then $dt = (2x+3)dx$.
Substituting into the integral: $\int \frac{dt}{t(t+2)+1} = \int \frac{dt}{t^2+2t+1} = \int \frac{dt}{(t+1)^2}$.
Integrating gives: $-\frac{1}{t+1} + \alpha = \frac{-1}{x^2+3x+1} + \alpha$.
Comparing this with $\frac{-1}{ax^2+bx+c} + \alpha$,we get $a=1, b=3, c=1$.
Therefore,$a+b+c = 1+3+1 = 5$.

(B) To solve the integral $I = \int \frac{x^2+1}{x^4+1} dx$,we divide the numerator and denominator by $x^2$:
$I = \int \frac{1 + \frac{1}{x^2}}{x^2 + \frac{1}{x^2}} dx$
We can rewrite the denominator as $(x - \frac{1}{x})^2 + 2$:
$I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + (\sqrt{2})^2} dx$
Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + (\sqrt{2})^2} = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + c$
Substituting $t = x - \frac{1}{x} = \frac{x^2-1}{x}$ back:
$I = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right) + c$
Thus,$f(x) = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)$.

(D) Let $I = \int \frac{1+\cos (4 x)}{\cot x-\tan x} d x$.
Using the identity $1+\cos(2\theta) = 2\cos^2(\theta)$,we have $1+\cos(4x) = 2\cos^2(2x)$.
Also,$\cot x - \tan x = \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} = \frac{\cos(2x)}{\frac{1}{2}\sin(2x)} = 2\frac{\cos(2x)}{\sin(2x)}$.
Substituting these into the integral:
$I = \int \frac{2\cos^2(2x)}{2\frac{\cos(2x)}{\sin(2x)}} dx = \int \cos(2x) \sin(2x) dx$.
Using the identity $\sin(2\theta) = 2\sin\theta \cos\theta$,we have $\sin(2x)\cos(2x) = \frac{1}{2}\sin(4x)$.
$I = \int \frac{1}{2}\sin(4x) dx = \frac{1}{2} \left( \frac{-\cos(4x)}{4} \right) + C = -\frac{1}{8}\cos(4x) + C$.
Comparing this with $A\cos(4x)+B$,we get $A = -\frac{1}{8}$.

(C) Let $I = \int \frac{dx}{\sqrt{7-6x-x^2}}$.
First,complete the square for the quadratic expression $7-6x-x^2$:
$7-6x-x^2 = 7 - (x^2 + 6x) = 7 - (x^2 + 6x + 9 - 9) = 7 - ((x+3)^2 - 9) = 16 - (x+3)^2 = 4^2 - (x+3)^2$.
Now,substitute this into the integral:
$I = \int \frac{dx}{\sqrt{4^2 - (x+3)^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C$,where $a = 4$ and $x$ is replaced by $(x+3)$:
$I = \sin^{-1}\left(\frac{x+3}{4}\right) + C$.

(A) We are given the integral $I = \int \frac{e^x(x + 3)}{(x + 5)^3} dx$.
First,we rewrite the numerator $(x + 3)$ as $(x + 5 - 2)$:
$I = \int \frac{e^x(x + 5 - 2)}{(x + 5)^3} dx$
$I = \int e^x \left[ \frac{x + 5}{(x + 5)^3} - \frac{2}{(x + 5)^3} \right] dx$
$I = \int e^x \left[ \frac{1}{(x + 5)^2} - \frac{2}{(x + 5)^3} \right] dx$
We use the standard integration formula $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c$.
Let $f(x) = \frac{1}{(x + 5)^2} = (x + 5)^{-2}$.
Then $f'(x) = -2(x + 5)^{-3} = -\frac{2}{(x + 5)^3}$.
Since the integrand is in the form $e^x(f(x) + f'(x))$,the integral is $e^x f(x) + c$.
Thus,$I = \frac{e^x}{(x + 5)^2} + c$.

(A) Let $I = \int \sqrt{e^{4x} + e^{2x}} \, dx = \int e^x \sqrt{e^{2x} + 1} \, dx$.
Substitute $e^x = v$,then $dv = e^x \, dx$.
$I = \int \sqrt{v^2 + 1} \, dv$.
Using the standard integral formula $\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln|x + \sqrt{x^2 + a^2}| + c$.
Here $a = 1$,so $I = \frac{v}{2} \sqrt{v^2 + 1} + \frac{1}{2} \ln|v + \sqrt{v^2 + 1}| + c$.
Since $\sinh^{-1}(v) = \ln(v + \sqrt{v^2 + 1})$,we have $I = \frac{v}{2} \sqrt{v^2 + 1} + \frac{1}{2} \sinh^{-1}(v) + c$.
Substituting $v = e^x$ back,we get $I = \frac{1}{2} e^x \sqrt{e^{2x} + 1} + \frac{1}{2} \sinh^{-1}(e^x) + c$.

(A) We are given the integral $I = \int [ \cos(x) \cdot \frac{d}{dx}(\csc(x)) ] dx$.
First,we know that $\frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x)$.
Substituting this into the integral,we get:
$I = \int [ \cos(x) \cdot (-\csc(x) \cot(x)) ] dx$
$I = - \int [ \cos(x) \cdot \frac{1}{\sin(x)} \cdot \frac{\cos(x)}{\sin(x)} ] dx$
$I = - \int [ \frac{\cos^2(x)}{\sin^2(x)} ] dx$
$I = - \int \cot^2(x) dx$
Using the identity $\cot^2(x) = \csc^2(x) - 1$,we have:
$I = - \int (\csc^2(x) - 1) dx$
$I = - \int \csc^2(x) dx + \int 1 dx$
Since $\int \csc^2(x) dx = -\cot(x)$,we get:
$I = -(-\cot(x)) + x + c$
$I = \cot(x) + x + c$
Comparing this with $f(x) + g(x) + c$,we have $f(x) = \cot(x)$ and $g(x) = x$ (or vice versa).
Therefore,$f(x) \cdot g(x) = x \cot(x)$.

(B) The primitive means the indefinite integral. Let $I = \int \cos(\log x) dx$.
Using integration by parts,let $u = \cos(\log x)$ and $dv = dx$. Then $du = -\sin(\log x) \cdot \frac{1}{x} dx$ and $v = x$.
$I = x \cos(\log x) - \int x \cdot (-\sin(\log x)) \cdot \frac{1}{x} dx = x \cos(\log x) + \int \sin(\log x) dx$.
Now,integrate $\int \sin(\log x) dx$ by parts: $u = \sin(\log x)$,$dv = dx$.
$I = x \cos(\log x) + [x \sin(\log x) - \int x \cdot \cos(\log x) \cdot \frac{1}{x} dx]$.
$I = x \cos(\log x) + x \sin(\log x) - I$.
$2I = x [\cos(\log x) + \sin(\log x)] + C$.
$I = \frac{x}{2} [\cos(\log x) + \sin(\log x)] + C$.
Comparing this with $f(x)\{\cos(g(x)) + \sin(h(x))\}$,we get $f(x) = \frac{x}{2}$,$g(x) = \log x$,and $h(x) = \log x$.
Thus,$f^{\prime}(x) = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.
Therefore,option $B$ is correct.

(B) Let $x+2=t$,then $dx=dt$.
Substituting these into the integral:
$I = \int(t-1)(t)^4(t+1) \, dt = \int(t^2-1)t^4 \, dt$.
Expanding the integrand:
$I = \int(t^6-t^4) \, dt$.
Integrating term by term:
$I = \frac{t^7}{7} - \frac{t^5}{5} + C$.
Substituting back $t = x+2$:
$I = \frac{(x+2)^7}{7} - \frac{(x+2)^5}{5} + C$.

(A) Let $I = \int \frac{e^{\tan ^{-1} x}}{1+x^2} \left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2 + \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$ for $x > 0$.
Substitute $\tan ^{-1} x = \theta$,so $x = \tan \theta$ and $\frac{1}{1+x^2} d x = d \theta$.
Since $x > 0$,$\theta \in (0, \pi/2)$.
Note that $\sec ^{-1} \sqrt{1+x^2} = \sec ^{-1} \sqrt{1+\tan^2 \theta} = \sec ^{-1} \sec \theta = \theta$.
Also,$\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \cos ^{-1}(\cos 2\theta) = 2\theta$ (since $2\theta \in (0, \pi)$).
Substituting these into the integral:
$I = \int e^{\theta} [\theta^2 + 2\theta] d \theta$.
Using the standard integral formula $\int e^x [f(x) + f'(x)] d x = e^x f(x) + c$,where $f(\theta) = \theta^2$ and $f'(\theta) = 2\theta$:
$I = e^{\theta} \theta^2 + c$.
Substituting back $\theta = \tan ^{-1} x$:
$I = e^{\tan ^{-1} x} (\tan ^{-1} x)^2 + c$.

(C) Let $I = \int 3^x \left(f^{\prime}(x) + f(x) \log 3\right) dx$.
We can rewrite the integrand as:
$I = \int \left(3^x f^{\prime}(x) + 3^x f(x) \log 3\right) dx$.
Recall the product rule for differentiation: $\frac{d}{dx} \left(3^x f(x)\right) = 3^x f^{\prime}(x) + f(x) \cdot \frac{d}{dx}(3^x) = 3^x f^{\prime}(x) + f(x) \cdot 3^x \log 3$.
Thus,the integrand is the derivative of $3^x f(x)$.
Therefore,$\int \frac{d}{dx} \left(3^x f(x)\right) dx = 3^x f(x) + c$.

(D) Let $I = \int \frac{x+\sin x}{1+\cos x} dx$.
Using the identities $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$,we get:
$I = \int \frac{x + 2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} dx$
$I = \int \left( \frac{x}{2 \cos^2(x/2)} + \frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} \right) dx$
$I = \int \left( \frac{1}{2} x \sec^2(x/2) + \tan(x/2) \right) dx$
Now,apply integration by parts to the first term $\int \frac{1}{2} x \sec^2(x/2) dx$:
Let $u = x$ and $dv = \frac{1}{2} \sec^2(x/2) dx$. Then $du = dx$ and $v = \tan(x/2)$.
Using $\int u dv = uv - \int v du$:
$\int \frac{1}{2} x \sec^2(x/2) dx = x \tan(x/2) - \int \tan(x/2) dx$
Substituting this back into the expression for $I$:
$I = x \tan(x/2) - \int \tan(x/2) dx + \int \tan(x/2) dx + c$
$I = x \tan(x/2) + c$

(A) Let $I = \int \cos \sqrt{x} \, dx$.
Substitute $\sqrt{x} = t$,so $x = t^2$.
Differentiating with respect to $x$,we get $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int \cos(t) \cdot 2t \, dt = 2 \int t \cos(t) \, dt$.
Using integration by parts,where $u = t$ and $dv = \cos(t) \, dt$:
$I = 2 \left[ t \sin(t) - \int \sin(t) \, dt \right] = 2 [t \sin(t) + \cos(t)] + c$.
Replacing $t$ with $\sqrt{x}$:
$I = 2 \sqrt{x} \sin \sqrt{x} + 2 \cos \sqrt{x} + c$.

(D) Let $I = \int \frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} d x = A \log |5 \cos x+4 \sin x|+B x+c$.
By differentiating both sides with respect to $x$,we get:
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = A \frac{d}{dx}(\log |5 \cos x+4 \sin x|) + B$
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = A \frac{-5 \sin x+4 \cos x}{5 \cos x+4 \sin x} + B$
$\frac{3 \cos x-2 \sin x}{4 \sin x+5 \cos x} = \frac{A(-5 \sin x+4 \cos x) + B(4 \sin x+5 \cos x)}{4 \sin x+5 \cos x}$
Equating the numerators:
$3 \cos x-2 \sin x = (4A+5B) \cos x + (-5A+4B) \sin x$
Comparing coefficients of $\cos x$ and $\sin x$:
$4A+5B = 3$ --- $(1)$
$-5A+4B = -2$ --- $(2)$
Multiplying $(1)$ by $5$ and $(2)$ by $4$:
$20A+25B = 15$
$-20A+16B = -8$
Adding these equations: $41B = 7 \Rightarrow B = \frac{7}{41}$.
Substituting $B$ in $(1)$: $4A + 5(\frac{7}{41}) = 3 \Rightarrow 4A = 3 - \frac{35}{41} = \frac{123-35}{41} = \frac{88}{41} \Rightarrow A = \frac{22}{41}$.
Thus,$A = \frac{22}{41}$ and $B = \frac{7}{41}$.

(D) Let $I = \int \frac{x-1}{(x-2)(x-3)} \, dx$.
Using partial fractions,we write: $\frac{x-1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.
Equating numerators: $x-1 = A(x-3) + B(x-2)$.
For $x=2$: $2-1 = A(2-3) \implies 1 = -A \implies A = -1$.
For $x=3$: $3-1 = B(3-2) \implies 2 = B$.
So,$\frac{x-1}{(x-2)(x-3)} = \frac{2}{x-3} - \frac{1}{x-2}$.
Integrating both sides: $I = \int \left( \frac{2}{x-3} - \frac{1}{x-2} \right) \, dx$.
$I = 2 \log |x-3| - \log |x-2| + c$.
Using logarithmic properties: $I = \log |x-3|^2 - \log |x-2| + c = \log \left| \frac{(x-3)^2}{x-2} \right| + c$.

(D) Let $I = \int \frac{1 - (\cot x)^{2021}}{\tan x + (\cot x)^{2022}} dx$.
Multiplying the numerator and denominator by $(\sin x)^{2022} \cos x$,we get:
$I = \int \frac{1 - \frac{(\cos x)^{2021}}{(\sin x)^{2021}}}{\frac{\sin x}{\cos x} + \frac{(\cos x)^{2022}}{(\sin x)^{2022}}} dx = \int \frac{(\sin x)^{2021} - (\cos x)^{2021}}{(\sin x)^{2021}} \cdot \frac{(\sin x)^{2022} \cos x}{(\sin x)^{2023} + (\cos x)^{2023}} dx$.
Simplifying the expression:
$I = \int \frac{(\sin x)^{2021} - (\cos x)^{2021}}{1} \cdot \frac{\sin x \cos x}{(\sin x)^{2023} + (\cos x)^{2023}} dx$.
$I = \int \frac{(\sin x)^{2022} \cos x - (\cos x)^{2022} \sin x}{(\sin x)^{2023} + (\cos x)^{2023}} dx$.
Let $f(x) = (\sin x)^{2023} + (\cos x)^{2023}$.
Then $f'(x) = 2023(\sin x)^{2022} \cos x - 2023(\cos x)^{2022} \sin x = 2023 [(\sin x)^{2022} \cos x - (\cos x)^{2022} \sin x]$.
Thus,$I = \frac{1}{2023} \int \frac{f'(x)}{f(x)} dx = \frac{1}{2023} \ln |f(x)| + c$.
Comparing this with the given form,we get $A = 2023$.

(A) Given,$\int \frac{(2x+1)^6}{(3x+2)^8} dx = P \left( \frac{2x+1}{3x+2} \right)^Q + R$
Let $t = \frac{2x+1}{3x+2}$.
Then,$\frac{dt}{dx} = \frac{(3x+2)(2) - (2x+1)(3)}{(3x+2)^2} = \frac{6x+4-6x-3}{(3x+2)^2} = \frac{1}{(3x+2)^2}$.
Thus,$dt = \frac{dx}{(3x+2)^2}$.
The integral becomes $\int \left( \frac{2x+1}{3x+2} \right)^6 \cdot \frac{dx}{(3x+2)^2} = \int t^6 dt$.
Integrating,we get $\int t^6 dt = \frac{t^7}{7} + C = \frac{1}{7} \left( \frac{2x+1}{3x+2} \right)^7 + C$.
Comparing this with $P \left( \frac{2x+1}{3x+2} \right)^Q + R$,we get $P = \frac{1}{7}$,$Q = 7$,and $R = C$.
Therefore,$\frac{P}{Q} = \frac{1/7}{7} = \frac{1}{49} = \frac{1}{7^2}$.

(B) Let $I = \int \frac{\sqrt{1-x^4}}{x^7} dx$.
Substitute $x^2 = u$,so $2x dx = du$,which implies $dx = \frac{du}{2x} = \frac{du}{2\sqrt{u}}$.
Then $I = \int \frac{\sqrt{1-u^2}}{u^{7/2}} \cdot \frac{du}{2\sqrt{u}} = \frac{1}{2} \int \frac{\sqrt{1-u^2}}{u^4} du$.
Substitute $u = \sin v$,so $du = \cos v dv$.
$I = \frac{1}{2} \int \frac{\cos v \cdot \cos v}{\sin^4 v} dv = \frac{1}{2} \int \cot^2 v \csc^2 v dv$.
Let $\cot v = w$,then $-\csc^2 v dv = dw$.
$I = \frac{1}{2} \int -w^2 dw = -\frac{1}{2} \cdot \frac{w^3}{3} + C = -\frac{1}{6} \cot^3 v + C$.
Since $\cot v = \frac{\cos v}{\sin v} = \frac{\sqrt{1-u^2}}{u} = \frac{\sqrt{1-x^4}}{x^2}$,we have $I = -\frac{1}{6} \left( \frac{\sqrt{1-x^4}}{x^2} \right)^3 + C = -\frac{1}{6x^6} (\sqrt{1-x^4})^3 + C$.
Comparing with $f(x) \{\sqrt{1-x^4}\}^n + C$,we get $f(x) = -\frac{1}{6x^6}$ and $n = 3$.
Thus,$(f(x))^n = (-\frac{1}{6x^6})^3 = -\frac{1}{216x^{18}}$.

(D) We have,$\int \frac{5 \tan x}{\tan x-2} d x = x + a \log |\sin x - 2 \cos x| + K$.
On differentiating both sides with respect to $x$,we get:
$\frac{5 \tan x}{\tan x-2} = 1 + a \frac{d}{dx} (\log |\sin x - 2 \cos x|) $
$\frac{5 \tan x}{\tan x-2} = 1 + a \frac{\cos x + 2 \sin x}{\sin x - 2 \cos x} $
Dividing the numerator and denominator of the fraction by $\cos x$:
$\frac{5 \tan x}{\tan x-2} = 1 + a \frac{1 + 2 \tan x}{\tan x - 2} $
$\frac{5 \tan x}{\tan x-2} = \frac{\tan x - 2 + a + 2a \tan x}{\tan x - 2} $
Comparing the coefficients of $\tan x$ and the constant terms:
$5 = 2a + 1 \Rightarrow 2a = 4 \Rightarrow a = 2$
Also,$a - 2 = 0 \Rightarrow a = 2$.
Thus,$a = 2$.

(B) Let $I = \int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx$.
We express the numerator as:
$2 \cos x + 3 \sin x = A(4 \cos x + 5 \sin x) + B \frac{d}{dx}(4 \cos x + 5 \sin x)$.
$2 \cos x + 3 \sin x = A(4 \cos x + 5 \sin x) + B(-4 \sin x + 5 \cos x)$.
$2 \cos x + 3 \sin x = (4A + 5B) \cos x + (5A - 4B) \sin x$.
Comparing coefficients:
$4A + 5B = 2$ and $5A - 4B = 3$.
Multiplying the first by $4$ and second by $5$:
$16A + 20B = 8$ and $25A - 20B = 15$.
Adding them: $41A = 23 \implies A = \frac{23}{41}$.
Substituting $A$ in $4A + 5B = 2$:
$4(\frac{23}{41}) + 5B = 2 \implies \frac{92}{41} + 5B = 2 \implies 5B = 2 - \frac{92}{41} = \frac{82-92}{41} = \frac{-10}{41} \implies B = \frac{-2}{41}$.
Thus,$\int \frac{2 \cos x+3 \sin x}{4 \cos x+5 \sin x} dx = \int \left( \frac{23}{41} + \frac{-2}{41} \frac{-4 \sin x + 5 \cos x}{4 \cos x + 5 \sin x} \right) dx$.
$= \frac{23}{41} x - \frac{2}{41} \log |4 \cos x + 5 \sin x| + c$.
Comparing with the given form,$K = \frac{-2}{41}$.

(A) Let $I = \int_{0}^{\pi/2} \tan^{n}(x) dx$.
Using the property of definite integration $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $x$ with $\frac{\pi}{2} - x$.
$I = \int_{0}^{\pi/2} \tan^{n}(\frac{\pi}{2} - x) dx$.
Since $\tan(\frac{\pi}{2} - x) = \cot(x)$,we have $I = \int_{0}^{\pi/2} \cot^{n}(x) dx$.
Comparing this with the given equation $\int_{0}^{\pi/2} \tan^{n}(x) dx = k \int_{0}^{\pi/2} \cot^{n}(x) dx$,we see that $I = k \cdot I$.
Therefore,$k = 1$.

(B) $I = \int_0^1 \frac{x e^x}{(x+1)^2} d x$
We can rewrite the integrand as:
$I = \int_0^1 e^x \left[ \frac{x+1-1}{(x+1)^2} \right] d x$
$I = \int_0^1 e^x \left[ \frac{1}{x+1} - \frac{1}{(x+1)^2} \right] d x$
Using the standard integral formula $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$,let $f(x) = \frac{1}{x+1}$.
Then,$f'(x) = -\frac{1}{(x+1)^2}$.
Thus,the integral becomes:
$I = \left[ e^x \cdot \frac{1}{x+1} \right]_0^1$
$I = \left( \frac{e^1}{1+1} \right) - \left( \frac{e^0}{0+1} \right)$
$I = \frac{e}{2} - 1$

(D) Let $I = \int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{1/3}} dx$.
Substitute $t = \log x$,then $dt = \frac{1}{x} dx$.
When $x = \frac{1}{e}$,$t = \log(\frac{1}{e}) = -1$.
When $x = \frac{2}{e}$,$t = \log(\frac{2}{e}) = \log 2 - \log e = \log 2 - 1$.
Thus,$I = \int_{\log 2 - 1}^{-1} t^{-1/3} dt$.
Integrating,we get $I = \left[ \frac{t^{2/3}}{2/3} \right]_{\log 2 - 1}^{-1} = \frac{3}{2} \left[ t^{2/3} \right]_{\log 2 - 1}^{-1}$.
$I = \frac{3}{2} \left[ (-1)^{2/3} - (\log 2 - 1)^{2/3} \right]$.
Since $(-1)^{2/3} = ((-1)^2)^{1/3} = 1^{1/3} = 1$,we have $I = \frac{3}{2} \left[ 1 - (\log 2 - 1)^{2/3} \right]$.

(D) Let $I = \int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}$.
Expanding the denominator: $(x-a)(b-x) = -x^2 + (a+b)x - ab$.
Completing the square: $-[x^2 - (a+b)x + ab] = -[x^2 - (a+b)x + (\frac{a+b}{2})^2 - (\frac{a+b}{2})^2 + ab] = -[(x - \frac{a+b}{2})^2 - (\frac{b-a}{2})^2] = (\frac{b-a}{2})^2 - (x - \frac{a+b}{2})^2$.
Thus,$I = \int_a^b \frac{dx}{\sqrt{(\frac{b-a}{2})^2 - (x - \frac{a+b}{2})^2}}$.
Using the standard integral $\int \frac{dx}{\sqrt{A^2 - u^2}} = \sin^{-1}(\frac{u}{A}) + C$,we get:
$I = [\sin^{-1}(\frac{x - \frac{a+b}{2}}{\frac{b-a}{2}})]_a^b$.
Evaluating at the limits:
Upper limit $(x=b)$: $\sin^{-1}(\frac{b - \frac{a+b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(\frac{\frac{b-a}{2}}{\frac{b-a}{2}}) = \sin^{-1}(1) = \frac{\pi}{2}$.
Lower limit $(x=a)$: $\sin^{-1}(\frac{a - \frac{a+b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(\frac{\frac{a-b}{2}}{\frac{b-a}{2}}) = \sin^{-1}(-1) = -\frac{\pi}{2}$.
Therefore,$I = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

(B) Let $I = \int_1^4 x \sqrt{x^2-1} \, dx$.
Substitute $t = x^2 - 1$,so $dt = 2x \, dx$ or $x \, dx = \frac{dt}{2}$.
When $x = 1$,$t = 1^2 - 1 = 0$.
When $x = 4$,$t = 4^2 - 1 = 15$.
Thus,$I = \int_0^{15} \sqrt{t} \frac{dt}{2} = \frac{1}{2} \int_0^{15} t^{1/2} \, dt$.
$I = \frac{1}{2} \left[ \frac{t^{3/2}}{3/2} \right]_0^{15} = \frac{1}{2} \times \frac{2}{3} [t^{3/2}]_0^{15} = \frac{1}{3} (15)^{3/2}$.
Comparing this with $\alpha(k)^\beta$,we get $\alpha = \frac{1}{3}$,$k = 15$,and $\beta = \frac{3}{2}$.
Therefore,$\alpha \beta = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2}$.

(A) To evaluate the integral $\int_{0}^{2} x e^{x} dx$,we use the method of integration by parts: $\int u dv = uv - \int v du$.
Let $u = x$ and $dv = e^{x} dx$. Then $du = dx$ and $v = e^{x}$.
Applying the formula: $\int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x}$.
Now,applying the limits from $0$ to $2$:
$\left[ x e^{x} - e^{x} \right]_{0}^{2} = (2 e^{2} - e^{2}) - (0 \cdot e^{0} - e^{0})$.
$= (e^{2}) - (0 - 1) = e^{2} + 1$.

(B) Let $I = \int_{0}^{\pi / 4} \log (1 + \tan x) dx$.
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:
$I = \int_{0}^{\pi / 4} \log (1 + \tan (\frac{\pi}{4} - x)) dx$.
Since $\tan (\frac{\pi}{4} - x) = \frac{1 - \tan x}{1 + \tan x}$,we get:
$I = \int_{0}^{\pi / 4} \log (1 + \frac{1 - \tan x}{1 + \tan x}) dx = \int_{0}^{\pi / 4} \log (\frac{2}{1 + \tan x}) dx$.
$I = \int_{0}^{\pi / 4} (\log 2 - \log (1 + \tan x)) dx = \frac{\pi}{4} \log 2 - I$.
$2I = \frac{\pi}{4} \log 2 \implies I = \frac{\pi}{8} \log 2$.
Given $\int_{0}^{\pi} \log (\sin x) dx = 8k$. We know $\int_{0}^{\pi} \log (\sin x) dx = -\pi \log 2$.
So,$8k = -\pi \log 2 \implies \pi \log 2 = -8k$.
Substituting this into $I = \frac{\pi}{8} \log 2$,we get $I = \frac{1}{8} (-8k) = -k$.

(D) We are given the integral $I = \int_{0}^{1} x^{m} (1 - x)^{n} dx$.
Using the property of definite integrals,$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we substitute $x$ with $(1 - x)$:
$I = \int_{0}^{1} (1 - x)^{m} (1 - (1 - x))^{n} dx$
$I = \int_{0}^{1} (1 - x)^{m} (x)^{n} dx$
$I = \int_{0}^{1} x^{n} (1 - x)^{m} dx$.
Comparing this with the given equation $\int_{0}^{1} x^{m} (1 - x)^{n} dx = k \int_{0}^{1} x^{n} (1 - x)^{m} dx$,we see that $I = k \cdot I$.
Since the integral is non-zero,we conclude that $k = 1$.

(C) We are given the integral $I = \int_{-1}^1 \frac{|x|}{x} \, dx$.
First,we define the function $f(x) = \frac{|x|}{x}$.
Based on the definition of the absolute value function,we have:
$f(x) = \begin{cases} -\frac{x}{x} = -1, & x < 0 \\ \frac{x}{x} = 1, & x > 0 \end{cases}$
Since the function is discontinuous at $x = 0$,we split the integral at $x = 0$:
$I = \int_{-1}^0 (-1) \, dx + \int_0^1 (1) \, dx$
Evaluating the integrals:
$I = [-x]_{-1}^0 + [x]_0^1$
$I = (-(0) - (-(-1))) + (1 - 0)$
$I = (-1) + 1 = 0$
Thus,the value of the integral is $0$.

(A) Let $I = \int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x \quad ...(i)$
Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,we get:
$I = \int_0^\pi \frac{(\pi-x) \tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)} d x$
Since $\tan(\pi-x) = -\tan x$ and $\sec(\pi-x) = -\sec x$,we have:
$I = \int_0^\pi \frac{(\pi-x)(-\tan x)}{-\sec x-\tan x} d x = \int_0^\pi \frac{(\pi-x) \tan x}{\sec x+\tan x} d x \quad ...(ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^\pi \frac{x \tan x + (\pi-x) \tan x}{\sec x+\tan x} d x = \int_0^\pi \frac{\pi \tan x}{\sec x+\tan x} d x$
$I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\sin x} d x$
Multiply numerator and denominator by $(1-\sin x)$:
$I = \frac{\pi}{2} \int_0^\pi \frac{\sin x(1-\sin x)}{1-\sin^2 x} d x = \frac{\pi}{2} \int_0^\pi \frac{\sin x - \sin^2 x}{\cos^2 x} d x$
$I = \frac{\pi}{2} \int_0^\pi (\sec x \tan x - \tan^2 x) d x = \frac{\pi}{2} \int_0^\pi (\sec x \tan x - (\sec^2 x - 1)) d x$
$I = \frac{\pi}{2} [\sec x - \tan x + x]_0^\pi$
$I = \frac{\pi}{2} [(\sec \pi - \tan \pi + \pi) - (\sec 0 - \tan 0 + 0)]$
$I = \frac{\pi}{2} [(-1 - 0 + \pi) - (1 - 0 + 0)] = \frac{\pi}{2} (\pi - 2)$

(A) Let $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \tan \left(1+x^2\right) d x$.
Define the integrand as $f(x) = x \tan \left(1+x^2\right)$.
To check if the function is odd or even,we evaluate $f(-x)$:
$f(-x) = (-x) \tan \left(1+(-x)^2\right) = -x \tan \left(1+x^2\right) = -f(x)$.
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 $\int_{-a}^{a} f(x) d x = 0$.
Therefore,$I = 0$.

(B) We have $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$.
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get:
$I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin(\pi - x)) + \cos^2(\cos(\pi - x)) \right\} dx$
$I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(-\cos x) \right\} dx$
$I = \int_0^\pi (\pi - x) \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
Adding the two expressions for $I$:
$2I = \int_0^\pi \pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
$I = \frac{\pi}{2} \int_0^\pi \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$:
$I = \frac{\pi}{2} \cdot 2 \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
$I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$
Now,using $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ again:
$I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin(\pi/2 - x)) + \cos^2(\cos(\pi/2 - x)) \right\} dx$
$I = \pi \int_0^{\pi/2} \left\{ \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$
Adding these two forms of $I$:
$2I = \pi \int_0^{\pi/2} \left\{ \sin^2(\sin x) + \cos^2(\cos x) + \sin^2(\cos x) + \cos^2(\sin x) \right\} dx$
$2I = \pi \int_0^{\pi/2} \left\{ (\sin^2(\sin x) + \cos^2(\sin x)) + (\cos^2(\cos x) + \sin^2(\cos x)) \right\} dx$
$2I = \pi \int_0^{\pi/2} (1 + 1) dx = 2\pi \int_0^{\pi/2} dx = 2\pi \cdot \frac{\pi}{2} = \pi^2$
Thus,$I = \frac{\pi^2}{2}$.
Since $\pi^2 \approx 9.869$,then $I \approx \frac{9.869}{2} = 4.9345$.
Therefore,$[I] = [4.9345] = 4$.

(A) Given,$\int_a^b x^3 dx = 0$ and $\int_a^b x^2 dx = \frac{2}{3}$.
First,evaluate the integral $\int_a^b x^3 dx = 0$:
$\left[\frac{x^4}{4}\right]_a^b = 0 \Rightarrow \frac{b^4 - a^4}{4} = 0 \Rightarrow b^4 = a^4$.
This implies $b = a$ or $b = -a$. Since $a$ and $b$ are limits of integration,we assume $a \neq b$,so $b = -a$.
Next,evaluate the integral $\int_a^b x^2 dx = \frac{2}{3}$:
$\left[\frac{x^3}{3}\right]_a^b = \frac{2}{3} \Rightarrow \frac{b^3 - a^3}{3} = \frac{2}{3} \Rightarrow b^3 - a^3 = 2$.
Substitute $b = -a$ into the equation:
$(-a)^3 - a^3 = 2 \Rightarrow -a^3 - a^3 = 2 \Rightarrow -2a^3 = 2$.
$a^3 = -1 \Rightarrow a = -1$.
Since $b = -a$,we have $b = -(-1) = 1$.
Thus,$a = -1$ and $b = 1$.

(D) We are given the integrals $\int_0^1 f(x) dx = 1$,$\int_0^1 x f(x) dx = a$,and $\int_0^1 x^2 f(x) dx = a^2$.
Expanding the integrand $(x-a)^2 f(x) = (x^2 - 2ax + a^2) f(x) = x^2 f(x) - 2ax f(x) + a^2 f(x)$.
Now,integrate term by term:
$\int_0^1 (x-a)^2 f(x) dx = \int_0^1 x^2 f(x) dx - 2a \int_0^1 x f(x) dx + a^2 \int_0^1 f(x) dx$.
Substituting the given values:
$= a^2 - 2a(a) + a^2(1) = a^2 - 2a^2 + a^2 = 0$.

(C) We need to evaluate the integral $I = \int_{2}^{4} (|x - 2| + |x - 3|) dx$.
Split the integral based on the critical points $x = 2$ and $x = 3$.
For $2 \le x \le 3$,$|x - 2| = x - 2$ and $|x - 3| = 3 - x$.
For $3 \le x \le 4$,$|x - 2| = x - 2$ and $|x - 3| = x - 3$.
Thus,$I = \int_{2}^{3} ((x - 2) + (3 - x)) dx + \int_{3}^{4} ((x - 2) + (x - 3)) dx$.
$I = \int_{2}^{3} (1) dx + \int_{3}^{4} (2x - 5) dx$.
$I = [x]_{2}^{3} + [x^2 - 5x]_{3}^{4}$.
$I = (3 - 2) + ((16 - 20) - (9 - 15))$.
$I = 1 + (-4 - (-6)) = 1 + 2 = 3$.

(C) Let $I = \int_{- 1 / 2}^{1 / 2} \{ [x] + \log (\frac{1 + x}{1 - x}) \} dx$.
We can split the integral as $I = \int_{- 1 / 2}^{1 / 2} [x] dx + \int_{- 1 / 2}^{1 / 2} \log (\frac{1 + x}{1 - x}) dx$.
Let $f(x) = \log (\frac{1 + x}{1 - x})$. Then $f(- x) = \log (\frac{1 - x}{1 + x}) = \log (\frac{1 + x}{1 - x})^{- 1} = - \log (\frac{1 + x}{1 - x}) = - f(x)$.
Since $f(x)$ is an odd function,$\int_{- 1 / 2}^{1 / 2} \log (\frac{1 + x}{1 - x}) dx = 0$.
Now,consider $\int_{- 1 / 2}^{1 / 2} [x] dx$.
For $x \in [- 1 / 2, 0)$,$[x] = - 1$.
For $x \in [0, 1 / 2]$,$[x] = 0$.
Thus,$\int_{- 1 / 2}^{1 / 2} [x] dx = \int_{- 1 / 2}^{0} (- 1) dx + \int_{0}^{1 / 2} 0 dx = [- x]_{- 1 / 2}^{0} = 0 - (1 / 2) = - 1 / 2$.
Therefore,$I = - 1 / 2 + 0 = - 1 / 2$.

(B) The given curve is $y = x|x|$. We can define this as a piecewise function:
$y = \begin{cases} x^2 & \text{if } x \geq 0 \\ -x^2 & \text{if } x < 0 \end{cases}$
We need to find the area enclosed by this curve between $x = -1$ and $x = 1$.
The total area $A$ is given by the integral of the absolute value of $y$ from $-1$ to $1$:
$A = \int_{-1}^{1} |y| dx = \int_{-1}^{0} |-x^2| dx + \int_{0}^{1} |x^2| dx$
$A = \int_{-1}^{0} x^2 dx + \int_{0}^{1} x^2 dx$
Calculating the integrals:
$A = \left[ \frac{x^3}{3} \right]_{-1}^{0} + \left[ \frac{x^3}{3} \right]_{0}^{1}$
$A = (0 - (-\frac{1}{3})) + (\frac{1}{3} - 0)$
$A = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
Thus,the area is $\frac{2}{3}$ sq units.

(A) The area $A$ of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is given by the integral:
$A = \int_2^3 \sqrt{16-x^2} dx$
Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$:
$A = \left[\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_2^3$
Evaluating the definite integral:
$A = \left(\frac{3}{2}\sqrt{16-9} + 8\sin^{-1}\left(\frac{3}{4}\right)\right) - \left(\frac{2}{2}\sqrt{16-4} + 8\sin^{-1}\left(\frac{2}{4}\right)\right)$
$A = \left(\frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}\right)\right) - \left(\sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}\right)\right)$
Since $\sqrt{12} = 2\sqrt{3}$ and $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$:
$A = \frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}\right) - 2\sqrt{3} - 8\left(\frac{\pi}{6}\right)$
$A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}\right)$
Comparing this with the given expression $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$:
Note that the given expression is $2 \times \left(\frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + \frac{k}{2}\right)$.
Actually,evaluating the integral result directly: $A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}\right)$.
To match the form $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$,we multiply the expression by $2$ and divide by $2$:
$A = \frac{1}{2} \left(3\sqrt{7} - 4\sqrt{3} - \frac{8\pi}{3} + 16\sin^{-1}\left(\frac{3}{4}\right)\right)$
Thus,$k = 16\sin^{-1}\left(\frac{3}{4}\right)$.

(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4\frac{dy}{dx} - 7x = 0$.
Rearranging the terms,we get $\sqrt{\frac{dy}{dx}} = 4\frac{dy}{dx} + 7x$.
Squaring both sides to remove the radical sign,we get:
$\frac{dy}{dx} = (4\frac{dy}{dx} + 7x)^2$
$\frac{dy}{dx} = 16(\frac{dy}{dx})^2 + 49x^2 + 56x\frac{dy}{dx}$.
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.
The exponent of the highest order derivative after making the equation a polynomial in derivatives is $2$,so the degree is $2$.
Thus,the order is $1$ and the degree is $2$.

(B) Given the equation $x^2+y^2=1$.
On differentiating both sides with respect to $x$,we get:
$2x + 2y y^{\prime} = 0$
Dividing by $2$,we have $x + y y^{\prime} = 0$.
Now,differentiating again with respect to $x$ using the product rule on $y y^{\prime}$:
$\frac{d}{dx}(x) + \frac{d}{dx}(y y^{\prime}) = 0$
$1 + (y y^{\prime \prime} + (y^{\prime}) \cdot y^{\prime}) = 0$
$1 + y y^{\prime \prime} + (y^{\prime})^2 = 0$
Thus,the correct differential equation is $y y^{\prime \prime} + (y^{\prime})^2 + 1 = 0$.

(A) Given the differential equation: $\frac{d^2 y}{d x^2}+y=0$.
We check if $y=3 \sin x+4 \cos x$ is a solution.
Differentiating with respect to $x$:
$\frac{d y}{d x}=3 \cos x-4 \sin x$.
Differentiating again with respect to $x$:
$\frac{d^2 y}{d x^2}=-3 \sin x-4 \cos x$.
Substituting this into the original equation:
$\frac{d^2 y}{d x^2}+y = (-3 \sin x-4 \cos x) + (3 \sin x+4 \cos x) = 0$.
Since the equation is satisfied,$y=3 \sin x+4 \cos x$ is a solution.
Alternative Method:
The characteristic equation for $\frac{d^2 y}{d x^2}+y=0$ is $m^2+1=0$,which gives $m = \pm i$.
The general solution is $y=c_1 \cos x+c_2 \sin x$.
Option $A$ is of this form with $c_1=4$ and $c_2=3$.

(B) Given differential equation is $2 x \frac{d y}{d x} - y = 4$.
Rearranging the terms,we get $2 x \frac{d y}{d x} = 4 + y$.
Separating the variables,we have $\frac{2}{4 + y} d y = \frac{1}{x} d x$.
Integrating both sides,$\int \frac{2}{4 + y} d y = \int \frac{1}{x} d x$.
This gives $2 \ln |4 + y| = \ln |x| + C$,where $C = \ln |c|$.
Using logarithmic properties,$\ln (4 + y)^2 = \ln |c x|$.
Taking the exponential of both sides,$(4 + y)^2 = c x$.
This equation is of the form $(y - k)^2 = 4 a (x - h)$,which represents a family of parabolas.

(B) Given that $f^{\prime}(x)=x+\frac{1}{x}$.
To find $f(x)$,we integrate both sides with respect to $x$:
$f(x) = \int f^{\prime}(x) dx = \int \left(x + \frac{1}{x}\right) dx$.
Using the linearity property of integration:
$f(x) = \int x dx + \int \frac{1}{x} dx$.
Applying the standard integration formulas $\int x^n dx = \frac{x^{n+1}}{n+1}$ and $\int \frac{1}{x} dx = \log |x| + C$:
$f(x) = \frac{x^2}{2} + \log |x| + C$.

(A) Given differential equation: $e^x \tan y \, dx + (1+e^x) \sec^2 y \, dy = 0$
Rearranging the terms: $e^x \tan y \, dx = -(1+e^x) \sec^2 y \, dy$
Separating the variables: $\frac{e^x}{1+e^x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy$
Integrating both sides: $\int \frac{e^x}{1+e^x} \, dx = -\int \frac{\sec^2 y}{\tan y} \, dy$
Using the formula $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$,we get:
$\ln(1+e^x) = -\ln(\tan y) + \ln C$
$\ln(1+e^x) + \ln(\tan y) = \ln C$
$\ln[(1+e^x) \tan y] = \ln C$
$(1+e^x) \tan y = C$
Since the curve passes through the point $\left(0, \frac{\pi}{4}\right)$,substitute $x=0$ and $y=\frac{\pi}{4}$:
$(1+e^0) \tan(\frac{\pi}{4}) = C$
$(1+1)(1) = C \implies C = 2$
Thus,the equation of the curve is $(1+e^x) \tan y = 2$.

(C) Given the differential equation: $\frac{dy}{dx} = e^{x+y}$
Using the property of exponents,we can write: $\frac{dy}{dx} = e^x \cdot e^y$
Separating the variables,we get: $\frac{dy}{e^y} = e^x dx$
This can be rewritten as: $e^{-y} dy = e^x dx$
Integrating both sides: $\int e^{-y} dy = \int e^x dx$
Performing the integration: $-e^{-y} = e^x + C_1$
Rearranging the terms: $e^x + e^{-y} = -C_1$
Letting $-C_1 = c$,we get the final solution: $e^x + e^{-y} = c$

(NONE) Given differential equation: $(1+x^2) \frac{dy}{dx} = e^{\tan^{-1} x} - y$.
Divide by $(1+x^2)$: $\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{\tan^{-1} x}}{1+x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{1+x^2}$ and $Q = \frac{e^{\tan^{-1} x}}{1+x^2}$.
Integrating Factor $(IF)$ = $e^{\int P dx} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1} x}$.
The solution is $y \cdot IF = \int Q \cdot IF dx + C$.
$y e^{\tan^{-1} x} = \int \frac{e^{\tan^{-1} x}}{1+x^2} \cdot e^{\tan^{-1} x} dx + C$.
Let $t = \tan^{-1} x$,then $dt = \frac{1}{1+x^2} dx$.
$y e^{\tan^{-1} x} = \int e^{2t} dt + C = \frac{e^{2t}}{2} + C = \frac{e^{2 \tan^{-1} x}}{2} + C$.
Given $y=1$ when $x=0$: $1 \cdot e^0 = \frac{e^0}{2} + C \Rightarrow 1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}$.
Thus,$y e^{\tan^{-1} x} = \frac{e^{2 \tan^{-1} x} + 1}{2}$.

(B) Given,the slope of the tangent is $\frac{dy}{dx} = x + xy$.
Rearranging the terms,we have $\frac{dy}{dx} = x(1+y)$.
Separating the variables,we get $\frac{dy}{1+y} = x dx$.
Integrating both sides,we get $\int \frac{dy}{1+y} = \int x dx$,which gives $\ln(y+1) = \frac{x^2}{2} + C$.
Exponentiating both sides,$y+1 = e^{\frac{x^2}{2} + C} = e^C \cdot e^{\frac{x^2}{2}}$.
Let $e^C = A$,so $y = A e^{\frac{x^2}{2}} - 1$.
Since the curve passes through $(0,1)$,we substitute $x=0$ and $y=1$: $1 = A e^0 - 1 \Rightarrow 1 = A - 1 \Rightarrow A = 2$.
Thus,the equation of the curve is $y = 2 e^{\frac{x^2}{2}} - 1$.

(B) Given the differential equation: $\left(x^2+1\right) \frac{dy}{dx} + 4xy = \frac{1}{x^2+1}$
Divide by $(x^2+1)$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:
$\frac{dy}{dx} + \frac{4x}{x^2+1}y = \frac{1}{(x^2+1)^2}$
Here,$P(x) = \frac{4x}{x^2+1}$ and $Q(x) = \frac{1}{(x^2+1)^2}$.
The Integrating Factor $(IF)$ is given by $e^{\int P(x) dx}$:
$IF = e^{\int \frac{4x}{x^2+1} dx} = e^{2 \ln(x^2+1)} = e^{\ln(x^2+1)^2} = (x^2+1)^2$.
The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + c$:
$y(x^2+1)^2 = \int \left( \frac{1}{(x^2+1)^2} \cdot (x^2+1)^2 \right) dx + c$
$y(x^2+1)^2 = \int 1 dx + c$
$y(x^2+1)^2 = x + c$.

(D) The given differential equation is $\frac{dy}{dx}+y \tan x=2x+x^2 \tan x$.
This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=\tan x$ and $Q=2x+x^2 \tan x$.
The integrating factor $IF = e^{\int \tan x dx} = e^{\ln |\sec x|} = \sec x$.
The solution is given by $y \cdot IF = \int Q \cdot IF dx + C$.
$y \sec x = \int (2x+x^2 \tan x) \sec x dx + C$.
$y \sec x = \int 2x \sec x dx + \int x^2 \sec x \tan x dx + C$.
Using integration by parts on the second integral: $\int x^2 \sec x \tan x dx = x^2 \sec x - \int 2x \sec x dx$.
Substituting this back: $y \sec x = \int 2x \sec x dx + x^2 \sec x - \int 2x \sec x dx + C$.
$y \sec x = x^2 \sec x + C$,which simplifies to $y = x^2 + C \cos x$.
Given $Y(0)=1$,we have $1 = 0^2 + C \cos(0) \implies C=1$.
So,$Y(x) = x^2 + \cos x$.
Then $Y'(x) = 2x - \sin x$.
Calculating $Y'(\frac{\pi}{4}) - Y'(-\frac{\pi}{4}) = (2(\frac{\pi}{4}) - \sin(\frac{\pi}{4})) - (2(-\frac{\pi}{4}) - \sin(-\frac{\pi}{4}))$.
$= (\frac{\pi}{2} - \frac{1}{\sqrt{2}}) - (-\frac{\pi}{2} + \frac{1}{\sqrt{2}}) = \pi - \frac{2}{\sqrt{2}} = \pi - \sqrt{2}$.
Thus,option $D$ is correct.

(A) Let $A_1$ be the area of the first square and $A_2$ be the area of the second square.
Given the side of the first square is $x$,its area is $A_1 = x^2$.
Given the side of the second square is $y = x - x^2$,its area is $A_2 = y^2 = (x - x^2)^2$.
Expanding this,we get $A_2 = x^2 - 2 x^3 + x^4$.
We need to find the rate of change of $A_2$ with respect to $A_1$,which is $\frac{d A_2}{d A_1}$.
Using the chain rule,$\frac{d A_2}{d A_1} = \frac{d A_2 / d x}{d A_1 / d x}$.
First,calculate the derivatives with respect to $x$:
$\frac{d A_1}{d x} = \frac{d}{d x}(x^2) = 2 x$.
$\frac{d A_2}{d x} = \frac{d}{d x}(x^2 - 2 x^3 + x^4) = 2 x - 6 x^2 + 4 x^3$.
Now,substitute these into the chain rule formula:
$\frac{d A_2}{d A_1} = \frac{2 x - 6 x^2 + 4 x^3}{2 x} = \frac{2 x(1 - 3 x + 2 x^2)}{2 x} = 1 - 3 x + 2 x^2$.