AP EAMCET 2024 Mathematics Question Paper with Answer and Solution

(D) The equation of the circle is $x^2+y^2-8x-12y+\alpha=0$. The center is $(4, 6)$ and the radius $r = \sqrt{4^2+6^2-\alpha} = \sqrt{52-\alpha}$.
Since the circle lies in the first quadrant without touching the axes,the distance from the center to the axes must be greater than the radius: $r < 4$ and $r < 6$. Thus,$r < 4$,which implies $\sqrt{52-\alpha} < 4$ $\Rightarrow 52-\alpha < 16$ $\Rightarrow \alpha > 36$.
Since $(6, 6)$ is an interior point,substituting it into the circle equation gives $6^2+6^2-8(6)-12(6)+\alpha < 0$.
$36+36-48-72+\alpha < 0$ $\Rightarrow \alpha - 48 < 0$ $\Rightarrow \alpha < 48$.
Combining these,we get $36 < \alpha < 48$.

(D) The point $A(4,2)$ lies on the circle $x^2+y^2-2x-4y+\alpha=0$.
Substituting the coordinates of $A$ into the equation:
$16+4-8-8+\alpha=0 \Rightarrow \alpha=-4$.
The equation of the circle is $x^2+y^2-2x-4y-4=0$.
The power of point $P(5,3)$ with respect to the circle is given by $PA \cdot PB$.
The power of a point $(x_0, y_0)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is $x_0^2+y_0^2+2gx_0+2fy_0+c$.
Substituting $P(5,3)$ into the circle equation:
$PA \cdot PB = 5^2+3^2-2(5)-4(3)-4 = 25+9-10-12-4 = 8$.

(B) To find the angle subtended by the chord $x+y-1=0$ at the center of the circle $x^2+y^2=1$,we follow these steps:
Step $1$: Identify the circle and chord.
The circle $x^2+y^2=1$ has its center at the origin $O(0,0)$ and radius $r=1$.
The chord is given by the line $x+y-1=0$.
Step $2$: Find the perpendicular distance $d$ from the origin to the chord.
The distance $d$ from $(0,0)$ to $x+y-1=0$ is given by $d = \frac{|1(0)+1(0)-1|}{\sqrt{1^2+1^2}} = \frac{1}{\sqrt{2}}$.
Step $3$: Calculate the angle.
Let the chord intersect the circle at points $A$ and $B$. In the isosceles triangle $OAB$,the perpendicular from $O$ to $AB$ bisects the angle $\angle AOB$. Let $\angle AOB = 2\theta$.
In the right-angled triangle formed by the origin,the midpoint of the chord,and one endpoint of the chord,we have $\cos(\theta) = \frac{d}{r} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$.
Thus,$\theta = 45^{\circ}$ or $\frac{\pi}{4}$ radians.
The total angle subtended at the center is $\angle AOB = 2\theta = 2 \times 45^{\circ} = 90^{\circ}$ or $\frac{\pi}{2}$ radians.

(A) The equation of the circle is $x^2 + y^2 - 4 x - 6 y + 9 = 0$.
Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2, f = -3, c = 9$.
Centre $= (-g, -f) = (2, 3)$ and Radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - 9} = 2$.
The line $6 x + 8 y + \alpha = 0$ intersects the circle at two distinct points if the perpendicular distance from the centre $(2, 3)$ to the line is less than the radius.
Distance $d = \frac{|6(2) + 8(3) + \alpha|}{\sqrt{6^2 + 8^2}} = \frac{|12 + 24 + \alpha|}{10} = \frac{|36 + \alpha|}{10}$.
Condition: $d < r$ $\Rightarrow \frac{|36 + \alpha|}{10} < 2$ $\Rightarrow |36 + \alpha| < 20$.
This implies $-20 < 36 + \alpha < 20$,which simplifies to $-56 < \alpha < -16$.
Given $\alpha$ is an integer multiple of $8$,let $\alpha = 8k$ for some integer $k$.
$-56 < 8k < -16 \Rightarrow -7 < k < -2$.
The possible integer values for $k$ are $-6, -5, -4, -3$.
Thus,there are $4$ possible values for $\alpha$.

(B) The given equation of the circle is $x^2 + y^2 + 8x - 6y - 24 = 0$.
Completing the square,we get $(x + 4)^2 + (y - 3)^2 = 24 + 16 + 9 = 49$.
Thus,the center is $C(-4, 3)$ and the radius $r = 7$.
Two lines $l_1: a_1x + b_1y + c_1 = 0$ and $l_2: a_2x + b_2y + c_2 = 0$ are conjugate with respect to a circle with center $(h, k)$ and radius $r$ if $r^2(a_1a_2 + b_1b_2) = (a_1h + b_1k + c_1)(a_2h + b_2k + c_2)$.
Here,$a_1 = 2, b_1 = -3, c_1 = 3$ and $a_2 = 1, b_2 = 2, c_2 = k$.
Substituting the values: $49(2(1) + (-3)(2)) = (2(-4) - 3(3) + 3)(1(-4) + 2(3) + k)$.
$49(2 - 6) = (-8 - 9 + 3)(-4 + 6 + k)$.
$49(-4) = (-14)(2 + k)$.
$196 = 14(2 + k)$ $\Rightarrow 14 = 2 + k$ $\Rightarrow k = 12$.
The point is $\left(\frac{12}{4}, \frac{12}{3}\right) = (3, 4)$.
The length of the tangent from a point $(x_1, y_1)$ to the circle $S = 0$ is $\sqrt{S(x_1, y_1)}$.
$L = \sqrt{3^2 + 4^2 + 8(3) - 6(4) - 24} = \sqrt{9 + 16 + 24 - 24 - 24} = \sqrt{1} = 1$.

(B) The equations of the diameters of the circle are:
$2x - 3y + 1 = 0$ ...$(i)$
$4x - 5y - 1 = 0$ ...(ii)
Solving equations $(i)$ and (ii),we find the centre of the circle $C = (-g, -f) = (3, 4)$,which implies $g = -3$ and $f = -4$.
The equation of the circle is $x^2 + y^2 - 6x - 8y - 11 = 0$.
The radius $r$ of the circle is $CQ = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (-4)^2 - (-11)} = \sqrt{9 + 16 + 11} = \sqrt{36} = 6$.
The distance $CP$ between the centre $C(3, 4)$ and point $P(-2, -2)$ is $CP = \sqrt{(3 - (-2))^2 + (4 - (-2))^2} = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}$.
In the right-angled triangle $\triangle CQP$,the length of the tangent $PQ$ is $PQ = \sqrt{CP^2 - CQ^2} = \sqrt{61 - 36} = \sqrt{25} = 5$.
The area of the quadrilateral $PQCR$ is the sum of the areas of $\triangle CQP$ and $\triangle CRP$.
Area of $PQCR = 2 \times \text{Area}(\triangle CQP) = 2 \times (\frac{1}{2} \times CQ \times PQ) = 6 \times 5 = 30$ square units.

(B) The equation of the circle is $x^2+y^2-4x+6y-4=0$.
Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$ and $f=3$.
The center of the circle is $(-g, -f) = (2, -3)$.
The normal to a circle at any point $(x_1, y_1)$ always passes through the center of the circle.
Thus,the normal is the line passing through $(1, 1)$ and $(2, -3)$.
The slope of this line is $m = \frac{-3-1}{2-1} = \frac{-4}{1} = -4$.
The equation of the line is $(y-1) = -4(x-1)$.
$y-1 = -4x+4$.
$4x+y = 5$.

(D) The equation of the circle is $x^2+y^2-6x+4y+12=0$.
Completing the square,we get $(x-3)^2+(y+2)^2 = 3^2+(-2)^2-12 = 9+4-12 = 1$.
So,the center $O$ is $(3,-2)$ and the radius $r$ is $1$.
The distance $d$ from the point $A(2,3)$ to the center $O(3,-2)$ is $d = \sqrt{(3-2)^2+(-2-3)^2} = \sqrt{1^2+(-5)^2} = \sqrt{26}$.
Let $\alpha$ be the half-angle between the tangents. In the right-angled triangle formed by the point $A$,the center $O$,and the point of tangency $P$,we have $\sin(\alpha) = \frac{r}{d} = \frac{1}{\sqrt{26}}$.
Then $\cos(\alpha) = \sqrt{1-\sin^2(\alpha)} = \sqrt{1-\frac{1}{26}} = \sqrt{\frac{25}{26}} = \frac{5}{\sqrt{26}}$.
Thus,$\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{1/\sqrt{26}}{5/\sqrt{26}} = \frac{1}{5}$.
The total angle between the tangents is $\theta = 2\alpha$.
Using the formula $\tan(2\alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$,we get $\tan(\theta) = \frac{2(1/5)}{1-(1/5)^2} = \frac{2/5}{24/25} = \frac{2}{5} \times \frac{25}{24} = \frac{5}{12}$.
Therefore,$\theta = \tan^{-1}\left(\frac{5}{12}\right)$.

(A) The locus of the point of intersection of perpendicular tangents to a circle is called the director circle.
For the circle $x^2+y^2=a^2$,the equation of the director circle is $x^2+y^2=2a^2$.
Since the point $(10,4)$ lies on the director circle,we have:
$10^2+4^2 = 2a^2$
$100+16 = 2a^2$
$116 = 2a^2$
$a^2 = 58$
$a = \sqrt{58}$

(B) The locus of the point of intersection of perpendicular tangents to a circle is known as its director circle.
For a circle given by the equation $x^2+y^2=r^2$,the equation of the director circle is $x^2+y^2=2r^2$.
Given the circle $x^2+y^2=10$,we have $r^2=10$.
Substituting this into the director circle formula,we get $x^2+y^2=2(10) = 20$.
Thus,the required locus is $x^2+y^2=20$.

(B) The equation of the circle is $x^2+y^2=25$,so the radius $r=5$.
Let $C(x_1, y_1)$ be the midpoint of a chord $AB$ that subtends a right angle at the origin $O(0, 0)$.
In $\triangle OAB$,$OA=OB=5$ and $\angle AOB = 90^\circ$.
Since $OC$ is the median to the hypotenuse $AB$ in the right-angled $\triangle OAB$,$OC = \frac{1}{2} AB$.
Alternatively,in $\triangle OCB$,$\angle COB = 45^\circ$ and $\angle OCB = 90^\circ$.
Thus,$OC = OB \cos(45^\circ) = 5 \times \frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}}$.
The distance of the midpoint $C(x_1, y_1)$ from the origin is $\sqrt{x_1^2+y_1^2}$.
Therefore,$\sqrt{x_1^2+y_1^2} = \frac{5}{\sqrt{2}}$,which implies $x_1^2+y_1^2 = \frac{25}{2}$.
Dividing by $\frac{25}{2}$,we get $\frac{x_1^2}{25/2} + \frac{y_1^2}{25/2} = 1$.
Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1$,we have $a^2 = \frac{25}{2}$,so $|a| = \frac{5}{\sqrt{2}}$.

(C) The equation of the chord is $y = 2x + 3$.
Substituting this into the circle equation $x^2 + y^2 + 6x - 4y + 4 = 0$:
$x^2 + (2x + 3)^2 + 6x - 4(2x + 3) + 4 = 0$
$x^2 + 4x^2 + 12x + 9 + 6x - 8x - 12 + 4 = 0$
$5x^2 + 10x + 1 = 0$.
The $x$-coordinate of the midpoint $a$ is the average of the roots: $a = \frac{x_1 + x_2}{2} = \frac{-10/5}{2} = -1$.
Since $(a, b)$ lies on the chord $y = 2x + 3$,we have $b = 2a + 3 = 2(-1) + 3 = 1$.
Thus,$2a + 3b = 2(-1) + 3(1) = -2 + 3 = 1$.

(B) The equation of the circle is $x^2+y^2-4x-8y+16=0$.
Given the midpoint of the chord is $(x_1, y_1) = (1,3)$.
The equation of the chord with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T = xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c$ and $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.
Substituting the values: $x(1) + y(3) - 2(x+1) - 4(y+3) + 16 = 1^2 + 3^2 - 4(1) - 8(3) + 16$.
$x + 3y - 2x - 2 - 4y - 12 + 16 = 1 + 9 - 4 - 24 + 16$.
$-x - y + 2 = -2$.
$-x - y = -4$,which simplifies to $x + y = 4$.
The chord intersects the coordinate axes at $A(0,4)$ and $B(4,0)$.
The triangle formed by the chord and the coordinate axes is a right-angled triangle with base $OB = 4$ and height $OA = 4$.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8$.

(B) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $S_1: x^2+y^2-2x+6y=0$,we have $g_1=-1, f_1=3, c_1=0$. Thus,$2g(-1) + 2f(3) = c + 0 \Rightarrow -2g + 6f = c$ $(i)$.
For $S_2: x^2+y^2-4x-2y+6=0$,we have $g_2=-2, f_2=-1, c_2=6$. Thus,$2g(-2) + 2f(-1) = c + 6 \Rightarrow -4g - 2f = c + 6$ $(ii)$.
For $S_3: x^2+y^2-12x+2y+3=0$,we have $g_3=-6, f_3=1, c_3=3$. Thus,$2g(-6) + 2f(1) = c + 3 \Rightarrow -12g + 2f = c + 3$ $(iii)$.
Subtracting $(ii)$ from $(i)$: $2g + 8f = -6 \Rightarrow g + 4f = -3$ $(iv)$.
Subtracting $(iii)$ from $(ii)$: $8g - 4f = 3$ $(v)$.
Adding $(iv)$ and $(v)$: $9g = 0 \Rightarrow g = 0$. Then $4f = -3 \Rightarrow f = -3/4$.
Substituting into $(i)$: $c = -2(0) + 6(-3/4) = -18/4 = -9/2$.
The circle $S$ is $x^2+y^2 - \frac{3}{2}y - \frac{9}{2} = 0$.
The tangent at $(0,3)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Substituting $(0,3)$: $x(0) + y(3) + 0(x+0) - \frac{3}{4}(y+3) - \frac{9}{2} = 0$.
$3y - \frac{3}{4}y - \frac{9}{4} - \frac{18}{4} = 0 \Rightarrow \frac{9}{4}y - \frac{27}{4} = 0 \Rightarrow 9y = 27 \Rightarrow y = 3$.

(A) Let the equation of the required circle be $x^2+y^2+2gx+2fy+c=0$.
Since it cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $x^2+y^2-4x-4y+7=0$: $2g(-2) + 2f(-2) = c+7$ $\Rightarrow -4g-4f-c=7$ $\Rightarrow 4g+4f+c=-7$ $(i)$.
For $x^2+y^2+4x-4y+6=0$: $2g(2) + 2f(-2) = c+6 \Rightarrow 4g-4f-c=6$ $(ii)$.
For $x^2+y^2+4x+4y+5=0$: $2g(2) + 2f(2) = c+5 \Rightarrow 4g+4f-c=5$ $(iii)$.
Subtracting $(iii)$ from $(i)$: $(4g+4f+c) - (4g+4f-c) = -7-5$ $\Rightarrow 2c = -12$ $\Rightarrow c = -6$.
Substituting $c=-6$ into $(i)$ and $(ii)$: $4g+4f = -1$ and $4g-4f = 0$.
Adding these: $8g = -1 \Rightarrow g = -\frac{1}{8}$.
Subtracting: $8f = -1 \Rightarrow f = -\frac{1}{8}$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-\frac{1}{8})^2 + (-\frac{1}{8})^2 - (-6)} = \sqrt{\frac{1}{64} + \frac{1}{64} + 6} = \sqrt{\frac{2}{64} + 6} = \sqrt{\frac{1}{32} + 6} = \sqrt{\frac{193}{32}} = \frac{\sqrt{193}}{4\sqrt{2}}$.

(A) The given circles are $S_1: x^2+y^2-4x-6y-3=0$ and $S_2: x^2+y^2+8x-4y+11=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get:
For $S_1$: $g_1 = -2, f_1 = -3, c_1 = -3$.
For $S_2$: $g_2 = 4, f_2 = -2, c_2 = 11$.
The angle $\theta$ between two circles is given by $\cos \theta = \frac{|2g_1g_2 + 2f_1f_2 - c_1 - c_2|}{2\sqrt{g_1^2+f_1^2-c_1}\sqrt{g_2^2+f_2^2-c_2}}$.
Substituting the values:
$2g_1g_2 = 2(-2)(4) = -16$.
$2f_1f_2 = 2(-3)(-2) = 12$.
Numerator: $|-16 + 12 - (-3) - 11| = |-16 + 12 + 3 - 11| = |-12| = 12$.
Denominator: $2\sqrt{(-2)^2+(-3)^2-(-3)} \sqrt{4^2+(-2)^2-11} = 2\sqrt{4+9+3}\sqrt{16+4-11} = 2\sqrt{16}\sqrt{9} = 2(4)(3) = 24$.
Thus,$\cos \theta = \frac{12}{24} = \frac{1}{2}$.
Therefore,$\theta = \frac{\pi}{3}$.

(C) The given equations of the circles are $S_1: x^2+y^2+2 \alpha x+2 y-8=0$ and $S_2: x^2+y^2-2 x+\alpha y-14=0$.
Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $(g_1, f_1, c_1) = (\alpha, 1, -8)$ and $(g_2, f_2, c_2) = (-1, \frac{\alpha}{2}, -14)$.
Since the circles intersect orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Substituting the values: $2(\alpha)(-1) + 2(1)(\frac{\alpha}{2}) = -8 - 14$.
$-2\alpha + \alpha = -22$,which gives $\alpha = 22$.
The centre of $S_1$ is $C_1 = (-g_1, -f_1) = (-22, -1)$.
The centre of $S_2$ is $C_2 = (-g_2, -f_2) = (1, -\frac{22}{2}) = (1, -11)$.
The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-22))^2 + (-11 - (-1))^2} = \sqrt{(23)^2 + (-10)^2} = \sqrt{529 + 100} = \sqrt{629}$.

(D) Given circle $S \equiv x^2+y^2-14x+6y+33=0$.
Completing the square: $(x-7)^2 + (y+3)^2 = 49+9-33 = 25$.
So,center $C = (7, -3)$ and radius $r = 5$.
For circle $S' \equiv x^2+y^2=a^2$,center $C' = (0, 0)$ and radius $r' = a$.
For $4$ common tangents,the circles must be separated,which means the distance between centers $CC' > r + r'$.
$CC' = \sqrt{(7-0)^2 + (-3-0)^2} = \sqrt{49+9} = \sqrt{58} \approx 7.616$.
Condition: $7.616 > 5 + a$.
$a < 2.616$.
Since $a \in \mathbb{N}$,the possible values for $a$ are $1$ and $2$.
Thus,there are $2$ possible values for $a$.

(C) For the first circle $S_1: x^2+y^2-8x-8y+28=0$,the center $C_1 = (4, 4)$ and radius $r_1 = \sqrt{4^2+4^2-28} = \sqrt{16+16-28} = \sqrt{4} = 2$.
For the second circle $S_2: x^2+y^2-8x-6y+25-\alpha^2=0$,the center $C_2 = (4, 3)$ and radius $r_2 = \sqrt{4^2+3^2-(25-\alpha^2)} = \sqrt{16+9-25+\alpha^2} = \sqrt{\alpha^2} = |\alpha|$.
Two circles have only one common tangent if they touch each other internally or externally.
The distance between centers $C_1 C_2 = \sqrt{(4-4)^2 + (4-3)^2} = \sqrt{0^2 + 1^2} = 1$.
For internal touch,$|r_1 - r_2| = C_1 C_2 \Rightarrow |2 - |\alpha|| = 1$.
This gives $2 - |\alpha| = 1$ $\Rightarrow |\alpha| = 1$ $\Rightarrow \alpha = \pm 1$ or $2 - |\alpha| = -1$ $\Rightarrow |\alpha| = 3$ $\Rightarrow \alpha = \pm 3$.
Given the options,$\alpha = 1$ is the correct value.

(D) The given equations of the circles are:
$x^2 + y^2 - 6x + 4y + 9 = 0 \Rightarrow (x-3)^2 + (y+2)^2 = 2^2 \quad \dots(i)$
$x^2 + y^2 + 2x - 2y + 1 = 0 \Rightarrow (x+1)^2 + (y-1)^2 = 1^2 \quad \dots(ii)$
From $(i)$ and $(ii)$,the centers and radii are:
$C_1 = (3, -2), r_1 = 2$
$C_2 = (-1, 1), r_2 = 1$
Calculate the distance between the centers $C_1$ and $C_2$:
$C_1C_2 = \sqrt{(3 - (-1))^2 + (-2 - 1)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$
The length of the direct common tangent $AB$ is given by the formula:
$AB = \sqrt{(C_1C_2)^2 - (r_1 - r_2)^2}$
$AB = \sqrt{5^2 - (2 - 1)^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$

(A) The equation of the curve is $2x^2 - y^2 + 3x + 2y = 0$.
Let the equation of the chord be $y = mx + c$,which can be written as $\frac{y - mx}{c} = 1$.
Substituting this into the homogeneous form of the curve equation:
$2x^2 - y^2 + (3x + 2y)(\frac{y - mx}{c}) = 0$.
Multiplying by $c$:
$2cx^2 - cy^2 + 3xy - 3mx^2 + 2y^2 - 2mxy = 0$.
Grouping the terms:
$(2c - 3m)x^2 + (2 - c)y^2 + (3 - 2m)xy = 0$.
Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(2c - 3m) + (2 - c) = 0$.
$c - 3m + 2 = 0$.
Substituting $c = 3m - 2$ into the chord equation $y = mx + c$:
$y = mx + 3m - 2$.
$y + 2 = m(x + 3)$.
This equation represents a family of lines passing through the fixed point $(-3, -2)$.
Thus,$(\alpha, \beta) = (-3, -2)$.

(B) The equation of circle $C_1$ is $x^2+y^2=16$.
The equation of circle $C_2$ is $(x-\alpha)^2+(y-\beta)^2=25$,which expands to $x^2-2\alpha x+\alpha^2+y^2-2\beta y+\beta^2=25$.
The equation of the common chord is given by $C_2-C_1=0$,which is $-2\alpha x-2\beta y+\alpha^2+\beta^2=9$.
The slope of the common chord is $m = -\frac{-2\alpha}{-2\beta} = -\frac{\alpha}{\beta} = \frac{3}{4}$.
Let $\alpha = -3\lambda$ and $\beta = 4\lambda$.
For the common chord to have maximum length,it must be the diameter of the smaller circle $C_1$,meaning the chord must pass through the centre of $C_1$,which is $(0,0)$.
Substituting $(0,0)$ into the chord equation: $-2\alpha(0)-2\beta(0)+\alpha^2+\beta^2=9$,so $\alpha^2+\beta^2=9$.
Substituting $\alpha$ and $\beta$ in terms of $\lambda$: $(-3\lambda)^2+(4\lambda)^2=9$ $\Rightarrow 25\lambda^2=9$ $\Rightarrow \lambda = \pm \frac{3}{5}$.
If $\lambda = \frac{3}{5}$,then $\alpha = -\frac{9}{5}$ and $\beta = \frac{12}{5}$,so $\alpha+\beta = \frac{3}{5}$.
If $\lambda = -\frac{3}{5}$,then $\alpha = \frac{9}{5}$ and $\beta = -\frac{12}{5}$,so $\alpha+\beta = -\frac{3}{5}$.

(D) The equation of the family of circles passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + kS_2 = 0$.
$(x^2+y^2-2x+2y-2) + k(x^2+y^2+2x-2y+1) = 0$
$(1+k)x^2 + (1+k)y^2 + (2k-2)x + (2-2k)y + (k-2) = 0$
Dividing by $(1+k)$,we get $x^2 + y^2 + \frac{2(k-1)}{k+1}x + \frac{2(1-k)}{k+1}y + \frac{k-2}{k+1} = 0$.
The centre of this circle is $\left(-\frac{k-1}{k+1}, -\frac{1-k}{k+1}\right) = \left(\frac{1-k}{k+1}, \frac{k-1}{k+1}\right)$.
Since the centre lies on the line $x-y+6=0$,we substitute the coordinates:
$\frac{1-k}{k+1} - \frac{k-1}{k+1} + 6 = 0$
$\frac{1-k-k+1}{k+1} = -6 \Rightarrow 2-2k = -6k-6$
$4k = -8 \Rightarrow k = -2$.
Substituting $k=-2$ into the equation: $x^2+y^2 + \frac{2(-3)}{-1}x + \frac{2(3)}{-1}y + \frac{-4}{-1} = 0 \Rightarrow x^2+y^2+6x-6y+4=0$.
The radius $r = \sqrt{g^2+f^2-c} = \sqrt{3^2+(-3)^2-4} = \sqrt{9+9-4} = \sqrt{14}$.

(C) The equation of the circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda S_2 = 0$.
$x^2 - 2x + y^2 - 4y - 4 + \lambda(x^2 + 2x + y^2 + 4y - 4) = 0$.
$(1 + \lambda)x^2 + (1 + \lambda)y^2 + 2(\lambda - 1)x + 4(\lambda - 1)y - 4(1 + \lambda) = 0$.
Dividing by $(1 + \lambda)$,we get $x^2 + y^2 + \frac{2(\lambda - 1)}{1 + \lambda}x + \frac{4(\lambda - 1)}{1 + \lambda}y - 4 = 0$.
Since it passes through $(3, 3)$,substitute $x = 3, y = 3$:
$9 + 9 + \frac{6(\lambda - 1)}{1 + \lambda} + \frac{12(\lambda - 1)}{1 + \lambda} - 4 = 0$.
$14 + \frac{18(\lambda - 1)}{1 + \lambda} = 0 \Rightarrow 14(1 + \lambda) + 18(\lambda - 1) = 0$.
$14 + 14\lambda + 18\lambda - 18 = 0$ $\Rightarrow 32\lambda = 4$ $\Rightarrow \lambda = \frac{1}{8}$.
Now,$\alpha = \frac{2(\frac{1}{8} - 1)}{1 + \frac{1}{8}} = \frac{2(-\frac{7}{8})}{\frac{9}{8}} = -\frac{14}{9}$.
$\beta = \frac{4(\frac{1}{8} - 1)}{1 + \frac{1}{8}} = \frac{4(-\frac{7}{8})}{\frac{9}{8}} = -\frac{28}{9}$.
$\gamma = -4$.
$3(\alpha + \beta + \gamma) = 3(-\frac{14}{9} - \frac{28}{9} - 4) = 3(-\frac{42}{9} - 4) = 3(-\frac{14}{3} - 4) = -14 - 12 = -26$.

(A) Let the two circles be $S_1 \equiv x^2+y^2-6x-7=0$ and $S_2 \equiv x^2+y^2-10x+16=0$.
The common chord is given by $S_1 - S_2 = 0$.
$(x^2+y^2-6x-7) - (x^2+y^2-10x+16) = 0$
$4x - 23 = 0 \Rightarrow x = \frac{23}{4}$.
The points of intersection are found by substituting $x = \frac{23}{4}$ into $S_1$:
$(\frac{23}{4})^2 + y^2 - 6(\frac{23}{4}) - 7 = 0$
$\frac{529}{16} + y^2 - \frac{138}{4} - 7 = 0$
$y^2 = \frac{138}{4} + 7 - \frac{529}{16} = \frac{552 + 112 - 529}{16} = \frac{135}{16}$.
So,$y = \pm \frac{3\sqrt{15}}{4}$.
The diameter endpoints are $(\frac{23}{4}, \frac{3\sqrt{15}}{4})$ and $(\frac{23}{4}, -\frac{3\sqrt{15}}{4})$.
The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.
$(x-\frac{23}{4})(x-\frac{23}{4}) + (y-\frac{3\sqrt{15}}{4})(y+\frac{3\sqrt{15}}{4}) = 0$
$(x-\frac{23}{4})^2 + y^2 - \frac{135}{16} = 0$
$x^2 - \frac{23}{2}x + \frac{529}{16} + y^2 - \frac{135}{16} = 0$
$x^2 + y^2 - \frac{23}{2}x + \frac{394}{16} = 0$
$x^2 + y^2 - \frac{23}{2}x + \frac{197}{8} = 0$
Multiplying by $8$: $8x^2 + 8y^2 - 92x + 197 = 0$.

(C) Let $P(r, s)$ be a point on the circle $x^2+y^2=25$,so $r^2+s^2=25$ $(i)$.
The equation of the chord of contact $L$ of point $P$ with respect to the circle $x^2+y^2=9$ is given by $xr+ys=9$ $(ii)$.
Let $(h, k)$ be the pole of the line $L$ with respect to the circle $x^2+y^2=36$. The equation of the polar of $(h, k)$ with respect to $x^2+y^2=36$ is $xh+yk=36$ $(iii)$.
Comparing equations $(ii)$ and $(iii)$,we have $\frac{r}{h} = \frac{s}{k} = \frac{9}{36} = \frac{1}{4}$.
Thus,$r = \frac{h}{4}$ and $s = \frac{k}{4}$.
Substituting these into equation $(i)$,we get $(\frac{h}{4})^2 + (\frac{k}{4})^2 = 25$.
$\frac{h^2+k^2}{16} = 25 \Rightarrow h^2+k^2 = 400$.
Therefore,the locus of the pole $(h, k)$ is $x^2+y^2=400$.

(B) The equation of the polar of the point $(-1, 1)$ with respect to the circle $x^2+y^2-2x+2y-1=0$ is given by $x(-1) + y(1) - (x-1) + (y+1) - 1 = 0$.
This simplifies to $-x + y - x + 1 + y + 1 - 1 = 0$,which is $-2x + 2y + 1 = 0$ or $2x - 2y - 1 = 0$.
The inverse point $(p, q)$ is the foot of the perpendicular from the center of the circle to the polar line.
The center of the circle $x^2+y^2-2x+2y-1=0$ is $(1, -1)$.
The foot of the perpendicular $(p, q)$ from $(x_0, y_0) = (1, -1)$ to the line $ax + by + c = 0$ is given by $\frac{p-x_0}{a} = \frac{q-y_0}{b} = -\frac{ax_0+by_0+c}{a^2+b^2}$.
Here $a=2, b=-2, c=-1$.
$\frac{p-1}{2} = \frac{q-(-1)}{-2} = -\frac{2(1)-2(-1)-1}{2^2+(-2)^2} = -\frac{2+2-1}{8} = -\frac{3}{8}$.
$p-1 = 2(-\frac{3}{8}) = -\frac{3}{4} \Rightarrow p = 1 - \frac{3}{4} = \frac{1}{4}$.
$q+1 = -2(-\frac{3}{8}) = \frac{3}{4} \Rightarrow q = \frac{3}{4} - 1 = -\frac{1}{4}$.
Thus,$p^2+q^2 = (\frac{1}{4})^2 + (-\frac{1}{4})^2 = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$.

(C) Let $(h, k)$ be the pole of the line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$.
The equation of the polar of a point $(h, k)$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is given by $T = 0$.
The equation is $2hx + 2ky - \frac{3(x + h)}{2} + \frac{5(y + k)}{2} - 7 = 0$.
Multiplying by $2$,we get $4hx + 4ky - 3x - 3h + 5y + 5k - 14 = 0$.
Rearranging terms,we get $(4h - 3)x + (4k + 5)y + (5k - 3h - 14) = 0$.
Comparing this with the given line $9x + y - 28 = 0$,we have:
$\frac{4h - 3}{9} = \frac{4k + 5}{1} = \frac{5k - 3h - 14}{-28}$.
From $\frac{4h - 3}{9} = 4k + 5$,we get $4h - 3 = 36k + 45$ $\Rightarrow 4h - 36k = 48$ $\Rightarrow h - 9k = 12$.
From $\frac{4k + 5}{1} = \frac{5k - 3h - 14}{-28}$,we get $-112k - 140 = 5k - 3h - 14$ $\Rightarrow 3h - 117k = 126$ $\Rightarrow h - 39k = 42$.
Solving $h - 9k = 12$ and $h - 39k = 42$,we subtract the equations: $(h - 9k) - (h - 39k) = 12 - 42$ $\Rightarrow 30k = -30$ $\Rightarrow k = -1$.
Substituting $k = -1$ into $h - 9k = 12$,we get $h - 9(-1) = 12$ $\Rightarrow h + 9 = 12$ $\Rightarrow h = 3$.
Thus,the pole is $(3, -1)$.

(A) The equation of the polar of point $(3,2)$ with respect to the circle $x^2+y^2=4$ is given by $T=0$,which is $3x+2y=4$.
Since $P(\alpha, \beta)$ and $(3,2)$ are conjugate points,the polar of $(3,2)$ must pass through $P(\alpha, \beta)$.
Thus,$3\alpha+2\beta=4$ ...$(i)$.
Given that $P(\alpha, \beta)$ lies on the line $2x+y=1$,we have $2\alpha+\beta=1$ ...(ii).
Solving equations $(i)$ and (ii):
From (ii),$\beta = 1-2\alpha$.
Substituting into $(i)$: $3\alpha + 2(1-2\alpha) = 4$ $\Rightarrow 3\alpha + 2 - 4\alpha = 4$ $\Rightarrow -\alpha = 2$ $\Rightarrow \alpha = -2$.
Then $\beta = 1 - 2(-2) = 5$.
Therefore,$\alpha+\beta = -2+5 = 3$.

(C) Given equations of the two circles are:
$x^2+y^2+4x-5=0 \Rightarrow (x+2)^2+y^2=3^2$
$x^2+y^2-6y+8=0 \Rightarrow x^2+(y-3)^2=1$
Thus,$C_1=(-2, 0), r_1=3$ and $C_2=(0, 3), r_2=1$.
The internal centre of similitude is given by $\left(\frac{r_2x_1+r_1x_2}{r_1+r_2}, \frac{r_2y_1+r_1y_2}{r_1+r_2}\right)$:
$(a, b) = \left(\frac{1(-2)+3(0)}{3+1}, \frac{1(0)+3(3)}{3+1}\right) = \left(-\frac{2}{4}, \frac{9}{4}\right) = \left(-\frac{1}{2}, \frac{9}{4}\right)$.
The external centre of similitude is given by $\left(\frac{r_1x_2-r_2x_1}{r_1-r_2}, \frac{r_1y_2-r_2y_1}{r_1-r_2}\right)$:
$(c, d) = \left(\frac{3(0)-1(-2)}{3-1}, \frac{3(3)-1(0)}{3-1}\right) = \left(\frac{2}{2}, \frac{9}{2}\right) = \left(1, \frac{9}{2}\right)$.
Now,calculating $(a+d)(b+c)$:
$(a+d)(b+c) = \left(-\frac{1}{2} + \frac{9}{2}\right) \left(\frac{9}{4} + 1\right) = \left(\frac{8}{2}\right) \left(\frac{13}{4}\right) = 4 \times \frac{13}{4} = 13$.

(A) Given,$S \equiv x^2+y^2-2x-4y+1=0$.
For the $y$-axis,set $x=0$,which gives $y^2-4y+1=0$.
Solving for $y$,we get $y = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$.
Since $OA > OB$,we have $A(0, 2+\sqrt{3})$ and $B(0, 2-\sqrt{3})$.
The radical axis of $S=0$ and $S^{\prime}=0$ is given by $S-S^{\prime}=0$.
$(x^2+y^2-2x-4y+1) - (x^2+y^2-4x-2y+4) = 0 \Rightarrow 2x-2y-3=0$.
For the $y$-axis,set $x=0$,which gives $-2y-3=0 \Rightarrow y = -\frac{3}{2}$.
So,$C$ is $(0, -\frac{3}{2})$.
Let the ratio in which $C$ divides $AB$ be $k:1$.
Using the section formula for the $y$-coordinate: $-\frac{3}{2} = \frac{k(2-\sqrt{3}) + 1(2+\sqrt{3})}{k+1}$.
$-3(k+1) = 2(k(2-\sqrt{3}) + 2+\sqrt{3})$.
$-3k-3 = 4k - 2k\sqrt{3} + 4 + 2\sqrt{3}$.
$-3k-4k + 2k\sqrt{3} = 4+3+2\sqrt{3}$.
$k(2\sqrt{3}-7) = 7+2\sqrt{3}$.
$k = \frac{7+2\sqrt{3}}{2\sqrt{3}-7} = \frac{7+2\sqrt{3}}{-(7-2\sqrt{3})}$.
Thus,the ratio $k:1$ is $(7+2\sqrt{3}) : (-7+2\sqrt{3})$.

(C) Let the equations of the circles be:
$S_1: x^2+y^2+2x+3y+1=0$
$S_2: x^2+y^2+x-y+3=0$
$S_3: x^2+y^2-3x+2y+5=0$
The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:
$(x^2+y^2+2x+3y+1) - (x^2+y^2+x-y+3) = 0$
$x + 4y - 2 = 0 \quad \dots (i)$
The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:
$(x^2+y^2+x-y+3) - (x^2+y^2-3x+2y+5) = 0$
$4x - 3y - 2 = 0 \quad \dots (ii)$
Solving equations $(i)$ and $(ii)$ for $x$ and $y$:
From $(i)$,$x = 2 - 4y$. Substituting into $(ii)$:
$4(2 - 4y) - 3y - 2 = 0$
$8 - 16y - 3y - 2 = 0$
$6 - 19y = 0 \Rightarrow y = \frac{6}{19}$
Substituting $y = \frac{6}{19}$ into $x = 2 - 4y$:
$x = 2 - 4\left(\frac{6}{19}\right) = 2 - \frac{24}{19} = \frac{38-24}{19} = \frac{14}{19}$
Thus,the radical centre is $\left(\frac{14}{19}, \frac{6}{19}\right)$.

(D) The equation of the first circle is $x^2+y^2+2gx+2fy+c=0$.
Dividing the second circle equation $2x^2+2y^2+3x+8y+2c=0$ by $2$,we get $x^2+y^2+\frac{3}{2}x+4y+c=0$.
The radical axis is obtained by subtracting the two equations: $(2g-\frac{3}{2})x + (2f-4)y = 0$.
This line passes through the origin $(0,0)$.
The circle $x^2+y^2+2x+2y+1=0$ can be written as $(x+1)^2+(y+1)^2=1$,which has center $(-1,-1)$ and radius $r=1$.
For the line $Ax+By=0$ to touch this circle,the perpendicular distance from the center $(-1,-1)$ to the line must equal the radius $1$.
$\frac{|(2g-\frac{3}{2})(-1) + (2f-4)(-1)|}{\sqrt{(2g-\frac{3}{2})^2 + (2f-4)^2}} = 1$.
Squaring both sides: $(-(2g-\frac{3}{2}) - (2f-4))^2 = (2g-\frac{3}{2})^2 + (2f-4)^2$.
Let $A = 2g-\frac{3}{2}$ and $B = 2f-4$. Then $(-A-B)^2 = A^2+B^2$,which simplifies to $A^2+B^2+2AB = A^2+B^2$.
Thus,$2AB = 0$,implying $A=0$ or $B=0$.
If $A=0$,$2g-\frac{3}{2}=0 \Rightarrow g=\frac{3}{4}$.
If $B=0$,$2f-4=0 \Rightarrow f=2$.

(B) Let the equation of the first circle be $S_1 \equiv x^2+y^2-36=0$.
Let the equation of the second circle be $S_2 \equiv x^2+y^2+2gx+2fy+c=0$.
The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$.
Thus,$(x^2+y^2-36) - (x^2+y^2+2gx+2fy+c) = 0$,which simplifies to $-2gx-2fy-36-c=0$.
Given the radical axis is $x-4=0$,we can write it as $x+0y-4=0$.
Comparing the coefficients,we get $\frac{-2g}{1} = \frac{-2f}{0} = \frac{-36-c}{-4} = k$.
This gives $f=0$ and $2g = -k$,so $g = -k/2$. Also,$36+c = 4k$,so $c = 4k-36$.
Since the circles are orthogonal,the condition is $2g_1g_2 + 2f_1f_2 = c_1+c_2$.
Here $g_1=0, f_1=0, c_1=-36$ and $g_2=g, f_2=f, c_2=c$.
Substituting these,$2(0)(g) + 2(0)(f) = -36 + c$.
So,$0 = -36 + c$,which means $c = 36$.
From $c = 4k-36$,we have $36 = 4k-36$,so $4k = 72$,which gives $k = 18$.
Then $g = -k/2 = -18/2 = -9$.
The centre of the second circle is $(-g, -f) = (9, 0)$.

(A) Let the point on the circle $x^2+y^2-2x+2y+1=0$ be $P(1,0)$.
Any chord passing through $P(1,0)$ can be represented as a line $L$ passing through $(1,0)$.
The pole of a chord with respect to the circle $x^2+y^2=4$ is a point $Q(h,k)$ such that the chord is the polar of $Q$ with respect to $x^2+y^2=4$.
The equation of the polar of $Q(h,k)$ with respect to $x^2+y^2=4$ is $hx+ky=4$.
Since this polar passes through $P(1,0)$,we have $h(1)+k(0)=4$,which implies $h=4$.
Thus,the locus of the poles $(h,k)$ is $x=4$.

(A) Given the rotation angle $\theta = 45^{\circ}$.
Let the new coordinates be $(x', y')$ and the old coordinates be $(x, y)$.
The transformation equations are:
$x = x' \cos 45^{\circ} - y' \sin 45^{\circ} = \frac{x'-y'}{\sqrt{2}}$
$y = x' \sin 45^{\circ} + y' \cos 45^{\circ} = \frac{x'+y'}{\sqrt{2}}$
Substituting these into the equation $y^2 = 4ax$:
$(\frac{x'+y'}{\sqrt{2}})^2 = 4a(\frac{x'-y'}{\sqrt{2}})$
$\frac{(x'+y')^2}{2} = \frac{4a(x'-y')}{\sqrt{2}}$
$(x'+y')^2 = \frac{8a}{\sqrt{2}}(x'-y') = 4\sqrt{2}a(x'-y')$
Replacing $(x', y')$ with $(x, y)$,the transformed equation is $(x+y)^2 = 4\sqrt{2}a(x-y)$.

(A) The standard form of the parabola is $y^2=4ax$. The focus is given as $(a, 0) = (\frac{1}{4}, k)$.
Comparing these,we get $a = \frac{1}{4}$ and $k = 0$.
So,the parabola is $y^2 = 4(\frac{1}{4})x$,which simplifies to $y^2 = x$.
The line is $x - 2y - 3 = 0$,so $x = 2y + 3$.
Substituting $x$ in the parabola equation: $y^2 = 2y + 3 \Rightarrow y^2 - 2y - 3 = 0$.
Factoring the quadratic: $(y-3)(y+1) = 0$,so $y = 3$ or $y = -1$.
If $y = 3$,$x = 2(3) + 3 = 9$. So $Q = (9, 3)$.
If $y = -1$,$x = 2(-1) + 3 = 1$. So $P = (1, -1)$.
The distance $PQ = \sqrt{(9-1)^2 + (3-(-1))^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5}$.
Since $a = \frac{1}{4}$,we have $16a = 16(\frac{1}{4}) = 4$.
Therefore,$PQ = 16a\sqrt{5}$.

(A) The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.
Here,$4a = 8$,so $a = 2$.
The equation of the tangent is $y = mx + \frac{2}{m}$.
Since it passes through the point $(1, 3)$,we substitute $x = 1$ and $y = 3$:
$3 = m(1) + \frac{2}{m}$
$3 = m + \frac{2}{m}$
$m^2 - 3m + 2 = 0$
$(m - 1)(m - 2) = 0$
Thus,$m = 1$ or $m = 2$.
For $m = 1$,the tangent is $y = 1x + \frac{2}{1} \Rightarrow y = x + 2$.
For $m = 2$,the tangent is $y = 2x + \frac{2}{2} \Rightarrow y = 2x + 1$.
Comparing with the given options,$y = 2x + 1$ is present.

(B) Let $y=mx+c$ be the equation of a common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=9$.
Condition for $y=mx+c$ to be a tangent to the parabola $y^2=4ax$ is $c=\frac{a}{m}$. Here $4a=8$,so $a=2$. Thus,$c=\frac{2}{m}$ $(i)$.
The line $y=mx+c$ is also tangent to the circle $x^2+y^2=9$,so the perpendicular distance from the center $(0,0)$ to the line $mx-y+c=0$ must equal the radius $r=3$.
$\frac{|c|}{\sqrt{m^2+1}}=3 \Rightarrow c^2=9(m^2+1)$.
Substituting $c=\frac{2}{m}$ into the equation: $\frac{4}{m^2}=9(m^2+1)$ $\Rightarrow 4=9m^2(m^2+1)$ $\Rightarrow 9m^4+9m^2-4=0$.
Let $m^2=t$,then $9t^2+9t-4=0 \Rightarrow (3t-1)(3t+4)=0$.
Since $m^2=t > 0$,we have $t=\frac{1}{3}$,so $m^2=\frac{1}{3}$ and $m=\pm\frac{1}{\sqrt{3}}$.
For $m=\frac{1}{\sqrt{3}}$,$c=\frac{2}{1/\sqrt{3}}=2\sqrt{3}$.
The equation is $y=\frac{1}{\sqrt{3}}x+2\sqrt{3}$ $\Rightarrow \sqrt{3}y=x+6$ $\Rightarrow x-\sqrt{3}y+6=0$.

(D) Let the equation of the tangent to the parabola $y^2=4x$ be $y=mx+\frac{1}{m}$.
For the parabola $x^2=-32y$,the equation of the tangent with slope $m$ is $y=mx-a m^2$,where $x^2=4ay$ gives $a=-8$.
Thus,the tangent is $y=mx-(-8)m^2$,which simplifies to $y=mx+8m^2$.
Comparing the two tangent equations,we have $\frac{1}{m}=8m^2$.
This implies $m^3=\frac{1}{8}$,so $m=\frac{1}{2}$.
Substituting $m=\frac{1}{2}$ into the first tangent equation: $y=\frac{1}{2}x+\frac{1}{1/2} = \frac{1}{2}x+2$.
Multiplying by $2$,we get $2y=x+4$,or $x-2y+4=0$.

(D) Given the parabola $y^2 = 8x$,we have $4a = 8$,so $a = 2$.
For a point $(at^2, 2at)$ on the parabola,we have $(2, -4) = (2t^2, 4t)$,which gives $t = -1$.
The normal at parameter $t$ meets the parabola again at parameter $t_2 = -t - \frac{2}{t}$.
Substituting $t = -1$,we get $t_2 = -(-1) - \frac{2}{-1} = 1 + 2 = 3$.
The coordinates of the point $(\alpha, \beta)$ are $(at_2^2, 2at_2) = (2(3)^2, 2(2)(3)) = (18, 12)$.
Thus,$\alpha + \beta = 18 + 12 = 30$.

(B) The equation of the parabola is $y^2=6x$. Comparing this with $y^2=4ax$,we get $4a=6$,so $a=1.5$.
The equation of the normal to the parabola $y^2=4ax$ at the point $(x_1, y_1)$ is given by $y-y_1 = -\frac{y_1}{2a}(x-x_1)$.
Substituting the values $x_1=24$,$y_1=12$,and $a=1.5$:
$y-12 = -\frac{12}{2(1.5)}(x-24)$
$y-12 = -\frac{12}{3}(x-24)$
$y-12 = -4(x-24)$
$y-12 = -4x+96$
$4x+y = 108$.

(B) The parabola is $y^2 = 4ax$. The point $P$ is $(2a, 2a\sqrt{2})$.
Comparing $P(2a, 2a\sqrt{2})$ with $(at^2, 2at)$,we get $t = \sqrt{2}$.
The normal at $t$ meets the parabola at $t_1 = -t - \frac{2}{t} = -\sqrt{2} - \frac{2}{\sqrt{2}} = -2\sqrt{2}$.
The coordinates of $Q$ are $(at_1^2, 2at_1) = (a(-2\sqrt{2})^2, 2a(-2\sqrt{2})) = (8a, -4a\sqrt{2})$.
The vertex is $O(0, 0)$.
The slope of $OP$ is $m_1 = \frac{2a\sqrt{2}}{2a} = \sqrt{2}$.
The slope of $OQ$ is $m_2 = \frac{-4a\sqrt{2}}{8a} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.
Since $m_1 \times m_2 = \sqrt{2} \times (-\frac{1}{\sqrt{2}}) = -1$,the lines $OP$ and $OQ$ are perpendicular.
Therefore,$\theta = 90^{\circ}$.

(A) Given the parabola $y^2=12x$,we have $4a=12$,so $a=3$. Let the parameters of points $P$ and $Q$ be $t_1$ and $t_2$ respectively. The ordinates are $y_1=2at_1$ and $y_2=2at_2$. Given $y_1:y_2=1:2$,we have $t_1:t_2=1:2$,so $t_2=2t_1$. Let $t_1=t$,then $t_2=2t$.
The point of intersection $(x, y)$ of the normals at $t_1$ and $t_2$ is given by:
$x = 2a + a(t_1^2 + t_2^2 + t_1t_2) = 2(3) + 3(t^2 + 4t^2 + 2t^2) = 6 + 21t^2$
$y = -at_1t_2(t_1+t_2) = -3(t)(2t)(t+2t) = -6t^2(3t) = -18t^3$
From $x=6+21t^2$,we get $t^2 = \frac{x-6}{21}$,so $t = \pm \left(\frac{x-6}{21}\right)^{1/2}$.
Substituting $t$ into $y=-18t^3$,we get $y = -18 \left(\frac{x-6}{21}\right)^{3/2}$,which simplifies to $y+18\left(\frac{x-6}{21}\right)^{3/2}=0$.

(C) The given parabola is $y^2=12x$. Comparing with $y^2=4ax$,we get $a=3$. The focus is $(3, 0)$.
Let $Q(3t^2, 6t)$ be a point on the parabola.
Point $P(x, y)$ divides the segment joining $(3, 0)$ and $(3t^2, 6t)$ in the ratio $1:2$.
Using the section formula,$P(x, y) = \left(\frac{1(3t^2) + 2(3)}{1+2}, \frac{1(6t) + 2(0)}{1+2}\right) = \left(\frac{3t^2+6}{3}, \frac{6t}{3}\right) = (t^2+2, 2t)$.
Thus,$x = t^2+2$ and $y = 2t$.
From $y = 2t$,we get $t = \frac{y}{2}$.
Substituting $t$ into the equation for $x$: $x = (\frac{y}{2})^2 + 2 = \frac{y^2}{4} + 2$.
$x - 2 = \frac{y^2}{4} \Rightarrow y^2 = 4(x-2)$.

(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{2x^2}{5} + \left(\frac{5}{x}\right)^{1/2}\right)^{10}$ is given by:
$T_{r+1} = {}^{10}C_r \left(\frac{2x^2}{5}\right)^{10-r} \left(\frac{5^{1/2}}{x^{1/2}}\right)^r$
$T_{r+1} = {}^{10}C_r \cdot \frac{2^{10-r}}{5^{10-r}} \cdot x^{2(10-r)} \cdot \frac{5^{r/2}}{x^{r/2}}$
$T_{r+1} = {}^{10}C_r \cdot \frac{2^{10-r}}{5^{10-r-r/2}} \cdot x^{20-2r-r/2}$
For the term to be independent of $x$,the exponent of $x$ must be zero:
$20 - 2r - \frac{r}{2} = 0$ $\Rightarrow 20 = \frac{5r}{2}$ $\Rightarrow r = 8$
Substituting $r=8$ into the expression:
$T_9 = {}^{10}C_8 \cdot \frac{2^{10-8}}{5^{10-8-4}} = {}^{10}C_2 \cdot \frac{2^2}{5^{-2}} = 45 \cdot 4 \cdot 25 = 4500$
The square root of the independent term is $\sqrt{4500} = \sqrt{900 \times 5} = 30\sqrt{5}$.

(B) Let $T_{r+1}$ be the greatest term,so $\frac{T_{r+1}}{T_r} \geq 1$.
For the expansion $(a+b)^n$,the condition is $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \times |\frac{b}{a}| \geq 1$.
Here $n=6$,$a=5$,$b=3x$. At $x=1$,$b=3$.
$\frac{6-r+1}{r} \times \frac{3}{5} \geq 1$
$\Rightarrow \frac{7-r}{r} \times \frac{3}{5} \geq 1$
$\Rightarrow 21 - 3r \geq 5r$
$\Rightarrow 8r \leq 21$
$\Rightarrow r \leq 2.625$.
Since $r$ must be an integer,the greatest term occurs at $r=2$,which is $T_{2+1} = T_3$.
$T_3 = {}^6C_2 \times 5^{6-2} \times (3 \times 1)^2$
$T_3 = 15 \times 5^4 \times 3^2$
$T_3 = (3 \times 5) \times 5^4 \times 3^2 = 3^3 \times 5^5$.

(B) Given the expression $(3x - 2)^{2/3}$. For the binomial expansion,we rewrite it as:
$(3x - 2)^{2/3} = [-2(1 - \frac{3x}{2})]^{2/3} = (-2)^{2/3} (1 - \frac{3x}{2})^{2/3} = \sqrt[3]{4} (1 - \frac{3x}{2})^{2/3}$.
The general term $T_{r+1}$ in the expansion of $(1+y)^n$ is $\binom{n}{r} y^r$.
Here $n = \frac{2}{3}$ and $y = -\frac{3x}{2}$.
The $4^{th}$ term $(T_4)$ corresponds to $r = 3$:
$T_4 = \sqrt[3]{4} \times \frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{3!} (-\frac{3x}{2})^3$
$T_4 = \sqrt[3]{4} \times \frac{\frac{2}{3} \times (-\frac{1}{3}) \times (-\frac{4}{3})}{6} \times (-\frac{27x^3}{8})$
$T_4 = \sqrt[3]{4} \times \frac{8/27}{6} \times (-\frac{27x^3}{8})$
$T_4 = \sqrt[3]{4} \times \frac{8}{162} \times (-\frac{27x^3}{8}) = -\frac{\sqrt[3]{4}}{6} x^3$.

(A) The general term in the expansion of $(\sqrt{2} + 3^{1/5})^{10}$ is given by $T_{r+1} = {}^{10}C_r (\sqrt{2})^{10-r} (3^{1/5})^r = {}^{10}C_r (2)^{(10-r)/2} (3)^{r/5}$.
For the term to be rational,the exponents of $2$ and $3$ must be integers.
Thus,$(10-r)/2$ must be an integer (which is true for all even $r$) and $r/5$ must be an integer.
For $0 \le r \le 10$,the values of $r$ that satisfy both conditions are $r = 0$ and $r = 10$.
For $r = 0$: $T_1 = {}^{10}C_0 (2)^5 (3)^0 = 1 \times 32 \times 1 = 32$.
For $r = 10$: $T_{11} = {}^{10}C_{10} (2)^0 (3)^2 = 1 \times 1 \times 9 = 9$.
The sum of the rational terms is $32 + 9 = 41$.

(B) The middle term is the $(\frac{12}{2}+1)^{\text{th}}$ term,i.e.,the $7^{\text{th}}$ term.
The terms equidistant from the middle term are $T_{7-k}$ and $T_{7+k}$.
For $k=2$,the ratio is $\frac{T_5}{T_9} = \frac{{}^{12}C_4 x^4}{{}^{12}C_8 x^8} = \frac{1}{256}$.
Since ${}^{12}C_4 = {}^{12}C_8$,we have $\frac{1}{x^4} = \frac{1}{256}$ $\Rightarrow x^4 = 256$ $\Rightarrow x = 4$.
For $k=4$,the ratio is $\frac{T_3}{T_{11}} = \frac{{}^{12}C_2 x^2}{{}^{12}C_{10} x^{10}} = \frac{1}{256}$.
Since ${}^{12}C_2 = {}^{12}C_{10}$,we have $\frac{1}{x^8} = \frac{1}{256}$ $\Rightarrow x^8 = 256$ $\Rightarrow x = \sqrt{2}$.
However,given $x \in N$,we check $k=1$: $\frac{T_6}{T_8} = \frac{{}^{12}C_5 x^5}{{}^{12}C_7 x^7} = \frac{1}{x^2} = \frac{1}{256} \Rightarrow x = 16$.
The sum of all terms in the expansion $(1+x)^{12}$ is $(1+x)^{12}$.
For $x=4$,the sum is $(1+4)^{12} = 5^{12}$.
For $x=2$,the sum is $(1+2)^{12} = 3^{12}$.

(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{1^{77}+2^{77}+\ldots+n^{77}}{n^{78}}$
This can be rewritten as: $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}\right)^{77}$
Using the definition of a definite integral as the limit of a Riemann sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_{0}^{1} f(x) dx$
Here,$f(x) = x^{77}$.
Therefore,the limit becomes: $\int_{0}^{1} x^{77} dx$
Evaluating the integral: $\left[ \frac{x^{78}}{78} \right]_{0}^{1} = \frac{1^{78}}{78} - \frac{0^{78}}{78} = \frac{1}{78}$

(C) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2-r^2}}$.
We can rewrite the expression by taking $n$ common from the square root in the denominator:
$L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{n \sqrt{1-(\frac{r}{n})^2}}$.
This is a Riemann sum of the form $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(\frac{r}{n})$,which is equivalent to the definite integral $\int_0^1 f(x) dx$.
Here,$f(x) = \frac{1}{\sqrt{1-x^2}}$.
Thus,$L = \int_0^1 \frac{1}{\sqrt{1-x^2}} dx$.
The integral of $\frac{1}{\sqrt{1-x^2}}$ is $\sin^{-1}(x)$.
Evaluating the definite integral from $0$ to $1$:
$L = [\sin^{-1}(x)]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

(C) Let $I = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^3}{n^3}\right)^{\frac{r^2}{n^3}}$.
Taking the natural logarithm on both sides:
$\log I = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3} \log \left(1+\frac{r^3}{n^3}\right)$.
We can rewrite this as:
$\log I = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \left(\frac{r}{n}\right)^2 \log \left(1+\left(\frac{r}{n}\right)^3\right)$.
Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$:
$\log I = \int_0^1 x^2 \log(1+x^3) dx$.
Let $t = 1+x^3$,then $dt = 3x^2 dx$,or $x^2 dx = \frac{1}{3} dt$.
When $x=0, t=1$. When $x=1, t=2$.
$\log I = \frac{1}{3} \int_1^2 \log t dt = \frac{1}{3} [t \log t - t]_1^2$.
$\log I = \frac{1}{3} [(2 \log 2 - 2) - (1 \log 1 - 1)] = \frac{1}{3} [2 \log 2 - 2 + 1] = \frac{2 \log 2 - 1}{3}$.
Therefore,$I = e^{\left(\frac{2 \log 2 - 1}{3}\right)}$.

(A) Let $I = \int_1^2 \frac{x^4-1}{x^6-1} d x$.
We simplify the integrand: $\frac{x^4-1}{x^6-1} = \frac{(x^2-1)(x^2+1)}{(x^2-1)(x^4+x^2+1)} = \frac{x^2+1}{x^4+x^2+1}$.
Note that $x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$.
Using partial fractions: $\frac{x^2+1}{(x^2+x+1)(x^2-x+1)} = \frac{1}{2} \left( \frac{1}{x^2+x+1} + \frac{1}{x^2-x+1} \right)$.
Integrating: $I = \frac{1}{2} \int_1^2 \left( \frac{1}{(x+\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} + \frac{1}{(x-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \right) d x$.
$I = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{x+1/2}{\sqrt{3}/2}\right) + \tan^{-1}\left(\frac{x-1/2}{\sqrt{3}/2}\right) \right]_1^2$.
$I = \frac{1}{\sqrt{3}} \left[ \tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right) \right]_1^2$.
Evaluating at limits: $I = \frac{1}{\sqrt{3}} \left[ (\tan^{-1}(\frac{5}{\sqrt{3}}) + \tan^{-1}(\sqrt{3})) - (\tan^{-1}(\sqrt{3}) + \tan^{-1}(\frac{1}{\sqrt{3}})) \right]$.
$I = \frac{1}{\sqrt{3}} \left[ \tan^{-1}(\frac{5}{\sqrt{3}}) - \tan^{-1}(\frac{1}{\sqrt{3}}) \right]$.
Using $\tan^{-1} A - \tan^{-1} B = \tan^{-1}(\frac{A-B}{1+AB})$: $I = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{5/\sqrt{3} - 1/\sqrt{3}}{1 + 5/3} \right) = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{4/\sqrt{3}}{8/3} \right) = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\sqrt{3}}{2} \right)$.

(D) Let $I = \int_{-1/24}^{1/24} \sec x \log \left(\frac{1-x}{1+x}\right) dx$.
Define $f(x) = \sec x \log \left(\frac{1-x}{1+x}\right)$.
We check if the function is odd or even by evaluating $f(-x)$:
$f(-x) = \sec(-x) \log \left(\frac{1-(-x)}{1+(-x)}\right) = \sec x \log \left(\frac{1+x}{1-x}\right)$.
Using the property $\log(a^{-1}) = -\log a$,we have:
$f(-x) = \sec x \log \left(\frac{1-x}{1+x}\right)^{-1} = -\sec x \log \left(\frac{1-x}{1+x}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
For an odd function,the integral over a symmetric interval $[-a, a]$ is always $0$.
Therefore,$\int_{-1/24}^{1/24} f(x) dx = 0$.

(D) Let $I = \int_0^{\frac{\pi}{4}} \log (1+\tan x) \, dx$.
Using the property $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$,we get:
$I = \int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) \, dx$.
Since $\tan(\frac{\pi}{4}-x) = \frac{\tan(\frac{\pi}{4})-\tan x}{1+\tan(\frac{\pi}{4})\tan x} = \frac{1-\tan x}{1+\tan x}$,we have:
$I = \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) \, dx = \int_0^{\frac{\pi}{4}} \log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right) \, dx$.
$I = \int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) \, dx = \int_0^{\frac{\pi}{4}} (\log 2 - \log(1+\tan x)) \, dx$.
$I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log(1+\tan x) \, dx$.
$I = [x \log 2]_0^{\frac{\pi}{4}} - I$.
$2I = \frac{\pi}{4} \log 2$.
$I = \frac{\pi}{8} \log 2$.

(B) The curve is given by $y=x^3-19x+30$. Factoring the polynomial,we get $y=(x+5)(x-2)(x-3)$.
The roots are $x=-5, 2, 3$. The curve is above the $x$-axis on the interval $[-5, 2]$ and below the $x$-axis on the interval $[2, 3]$.
The total area $A$ is given by:
$A = \int_{-5}^{2} (x^3-19x+30) dx + \left| \int_{2}^{3} (x^3-19x+30) dx \right|$
$A = \int_{-5}^{2} (x^3-19x+30) dx - \int_{2}^{3} (x^3-19x+30) dx$
Evaluating the integral $\int (x^3-19x+30) dx = \frac{x^4}{4} - \frac{19x^2}{2} + 30x + C$.
For the first part: $\left[ \frac{x^4}{4} - \frac{19x^2}{2} + 30x \right]_{-5}^{2} = (4 - 38 + 60) - (\frac{625}{4} - \frac{475}{2} - 150) = 26 - (\frac{625-950-600}{4}) = 26 - (-\frac{925}{4}) = 26 + 231.25 = 257.25 = \frac{1029}{4}$.
For the second part: $\left[ \frac{x^4}{4} - \frac{19x^2}{2} + 30x \right]_{2}^{3} = (\frac{81}{4} - \frac{171}{2} + 90) - (4 - 38 + 60) = (\frac{81-342+360}{4}) - 26 = \frac{99}{4} - 26 = \frac{99-104}{4} = -\frac{5}{4}$.
Total Area $A = \frac{1029}{4} - (-\frac{5}{4}) = \frac{1034}{4} = \frac{517}{2} \text{ sq. units}$.

(A) The given curves are $x=y^2$ and $x=3-2y^2$.
To find the points of intersection, set $y^2 = 3-2y^2$, which gives $3y^2 = 3$, so $y^2 = 1$, implying $y = \pm 1$.
The area is symmetric about the $x$-axis.
$\text{Required Area} = 2 \int_{-1}^{1} (x_{\text{right}} - x_{\text{left}}) dy = 2 \int_{-1}^{1} ((3-2y^2) - y^2) dy$
$= 2 \int_{-1}^{1} (3-3y^2) dy = 6 \int_{-1}^{1} (1-y^2) dy$
$= 6 \left[ y - \frac{y^3}{3} \right]_{-1}^{1} = 6 \left( (1 - \frac{1}{3}) - (-1 + \frac{1}{3}) \right)$
$= 6 \left( \frac{2}{3} - (-\frac{2}{3}) \right) = 6 \left( \frac{4}{3} \right) = 8 \text{ sq. units.}$

(B) Given the curves are $x^2+y^2=2ax$ (which is $(x-a)^2+y^2=a^2$) and $y^2=ax$.
To find the intersection points,substitute $y^2=ax$ into the circle equation: $x^2+ax=2ax \Rightarrow x^2-ax=0 \Rightarrow x(x-a)=0$. Thus,$x=0$ or $x=a$.
For $x=a$,$y^2=a^2 \Rightarrow y=a$ (since we consider the region above the $X$-axis).
The area of the shaded region is the area under the circle minus the area under the parabola from $x=0$ to $x=a$.
Area $= \int_0^a (y_{circle} - y_{parabola}) dx = \int_0^a (\sqrt{a^2-(x-a)^2} - \sqrt{ax}) dx$.
Area $= \int_0^a \sqrt{a^2-(x-a)^2} dx - \int_0^a \sqrt{ax} dx$.
The first integral represents the area of a quarter circle of radius $a$,which is $\frac{\pi a^2}{4}$.
The second integral is $\sqrt{a} \int_0^a x^{1/2} dx = \sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_0^a = \sqrt{a} \cdot \frac{2}{3} a^{3/2} = \frac{2a^2}{3}$.
Therefore,the required area is $\frac{\pi a^2}{4} - \frac{2a^2}{3} = a^2\left(\frac{\pi}{4}-\frac{2}{3}\right)$ sq. units.

(C) The given curves are $y^2=8(x+2)$ and $y^2=4(1-x)$.
First,find the intersection point of the two parabolas:
$8(x+2) = 4(1-x)$
$2(x+2) = 1-x$
$2x+4 = 1-x$
$3x = -3 \implies x = -1$.
At $x=-1$,$y^2 = 8(-1+2) = 8$,so $y = \pm 2\sqrt{2}$.
The region is bounded by the two parabolas and the $Y$-axis $(x=0)$.
The area $A$ is given by $2 \int_{-2}^{0} |y| dx$. Specifically,from $x=-2$ to $x=-1$,the boundary is $y^2=8(x+2)$,and from $x=-1$ to $x=0$,the boundary is $y^2=4(1-x)$.
$A = 2 \left[ \int_{-2}^{-1} \sqrt{8(x+2)} dx + \int_{-1}^{0} \sqrt{4(1-x)} dx \right]$
$A = 2 \left[ 2\sqrt{2} \int_{-2}^{-1} (x+2)^{1/2} dx + 2 \int_{-1}^{0} (1-x)^{1/2} dx \right]$
$A = 2 \left[ 2\sqrt{2} \left[ \frac{2}{3}(x+2)^{3/2} \right]_{-2}^{-1} + 2 \left[ -\frac{2}{3}(1-x)^{3/2} \right]_{-1}^{0} \right]$
$A = 2 \left[ 2\sqrt{2} \cdot \frac{2}{3}(1)^{3/2} - \frac{4}{3} \left( (1-0)^{3/2} - (1-(-1))^{3/2} \right) \right]$
$A = 2 \left[ \frac{4\sqrt{2}}{3} - \frac{4}{3} (1 - 2\sqrt{2}) \right]$
$A = 2 \left[ \frac{4\sqrt{2}}{3} - \frac{4}{3} + \frac{8\sqrt{2}}{3} \right] = 2 \left[ \frac{12\sqrt{2}-4}{3} \right] = \frac{8}{3}(3\sqrt{2}-1)$.

(C) The area is given by the integral $A = \int_0^{\frac{\pi}{2}} |\sin x - \cos x| \, dx$.
Since $\cos x \geq \sin x$ for $0 \leq x \leq \frac{\pi}{4}$ and $\sin x \geq \cos x$ for $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$,we split the integral:
$A = \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx$.
Evaluating the first part: $[\sin x + \cos x]_0^{\frac{\pi}{4}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \sqrt{2} - 1$.
Evaluating the second part: $[-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (-0 - 1) - (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = -1 + \sqrt{2} = \sqrt{2} - 1$.
Adding both parts: $A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)$ square units.

(C) Given the curve $y=2x-x^2$.
For the stationary point,we set $\frac{dy}{dx}=0$.
$\frac{dy}{dx} = 2-2x = 0 \Rightarrow x=1$.
Thus,$\alpha = 1$.
The area is bounded by $y=2^x, y=2x-x^2, x=0$ and $x=1$.
Required Area $= \int_0^1 (2^x - (2x-x^2)) dx$.
$= \int_0^1 2^x dx - \int_0^1 (2x-x^2) dx$.
$= \left[ \frac{2^x}{\log 2} \right]_0^1 - \left[ x^2 - \frac{x^3}{3} \right]_0^1$.
$= \left( \frac{2^1}{\log 2} - \frac{2^0}{\log 2} \right) - \left( (1^2 - \frac{1^3}{3}) - (0) \right)$.
$= \frac{2-1}{\log 2} - (1 - \frac{1}{3}) = \frac{1}{\log 2} - \frac{2}{3}$.
$= \frac{3 - 2 \log 2}{3 \log 2} = \frac{3 - \log 4}{3 \log 2}$.

(D) Given differential equation is $\left(\frac{d^2 y}{d x^2}\right)^{-\frac{7}{2}} \left(\frac{d^3 y}{d x^3}\right)^2 = \left(\frac{d^2 y}{d x^2}\right)^{-\frac{5}{2}} \left(\frac{d^4 y}{d x^4}\right)$.
Multiply both sides by $\left(\frac{d^2 y}{d x^2}\right)^{\frac{7}{2}}$ to eliminate negative exponents:
$\left(\frac{d^3 y}{d x^3}\right)^2 = \left(\frac{d^2 y}{d x^2}\right) \left(\frac{d^4 y}{d x^4}\right)$.
To remove the fractional power,we square both sides:
$\left(\frac{d^3 y}{d x^3}\right)^4 = \left(\frac{d^2 y}{d x^2}\right)^2 \left(\frac{d^4 y}{d x^4}\right)^2$.
The highest order derivative present is $\frac{d^4 y}{d x^4}$,so the order is $4$.
The power of the highest order derivative is $2$,so the degree is $2$.
The difference between order and degree is $4 - 2 = 2$.

(B) Given the differential equation: $\frac{d^4 y}{d x^4} = \{c + (\frac{d y}{d x})^2\}^{\frac{3}{2}}$.
To find the degree,we must eliminate the fractional exponent by squaring both sides:
$(\frac{d^4 y}{d x^4})^2 = \{c + (\frac{d y}{d x})^2\}^3$.
The order of the differential equation is the highest derivative present,which is $4$.
The degree is the power of the highest derivative after the equation is made a polynomial in derivatives,which is $2$.
Therefore,the sum of the order and degree is $4 + 2 = 6$.

(C) Given the differential equation: $\frac{d^3 y}{d x^3} = \left[1 + \left(\frac{d y}{d x}\right)^2\right]^{5/2}$.
To find the degree,we must eliminate the fractional exponent by squaring both sides:
$\left(\frac{d^3 y}{d x^3}\right)^2 = \left[1 + \left(\frac{d y}{d x}\right)^2\right]^5$.
The order of a differential equation is the highest derivative present,which is $3$ (from $\frac{d^3 y}{d x^3}$).
The degree is the power of the highest derivative after the equation is made free from radicals and fractions,which is $2$.
Therefore,the order is $3$ and the degree is $2$.

(A) Given the general solution $y=a^3 e^{b^2 x+c}$.
We can rewrite this as $y = (a^3 e^c) e^{b^2 x}$.
Let $K = a^3 e^c$,where $K$ is an arbitrary constant because $a$ and $c$ are arbitrary constants.
Thus,the equation becomes $y = K e^{b^2 x}$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = K e^{b^2 x} \cdot b^2$.
Since $y = K e^{b^2 x}$,we substitute this back into the derivative:
$\frac{dy}{dx} = b^2 y$.
This is a first-order differential equation because it involves only the first derivative $\frac{dy}{dx}$.
Therefore,the order of the differential equation is $1$.

(A) Given differential equation: $x\left(\frac{d^2 y}{d x^2}\right)^{1/2} = \left(1+\frac{d y}{d x}\right)^{4/3}$.
To find the degree,we must eliminate the fractional exponents.
First,square both sides: $x^2 \left(\frac{d^2 y}{d x^2}\right) = \left(1+\frac{d y}{d x}\right)^{8/3}$.
Next,cube both sides to remove the remaining fraction: $x^6 \left(\frac{d^2 y}{d x^2}\right)^3 = \left(1+\frac{d y}{d x}\right)^8$.
Now,the equation is in polynomial form with respect to the derivatives.
The highest order derivative is $\frac{d^2 y}{d x^2}$,so the order is $2$.
The power of the highest order derivative is $3$,so the degree is $3$.
The sum of the order and degree is $2 + 3 = 5$.

(A) The order of a differential equation is equal to the number of arbitrary constants present in the general equation of the family of curves.
For option $A$,the equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$.
Here,$g$ and $f$ are two independent arbitrary constants.
Since there are $2$ arbitrary constants,the differential equation formed will be of order $2$.
For option $B$,the equation is $y^2 = 4a(x-h)$,which involves two constants,but the condition of passing through the origin and focus on the $x$-axis restricts it.
For option $C$,the equation is $y = mx$,which has only $1$ arbitrary constant.
For option $D$,the equation $x^2 - y^2 = k^2$ has only $1$ arbitrary constant $k$.
Therefore,the correct option is $A$.

(C) The equation of a family of circles with center $(0, b)$ on the $Y$-axis and radius $a$ is given by $x^2 + (y - b)^2 = a^2$.
Differentiating with respect to $x$:
$2x + 2(y - b)y_1 = 0$
$x + (y - b)y_1 = 0$ ... $(i)$
Differentiating again with respect to $x$:
$1 + (y - b)y_2 + y_1^2 = 0$
$(y - b)y_2 = -(1 + y_1^2)$
$y - b = -\frac{1 + y_1^2}{y_2}$ ... $(ii)$
Substituting $(ii)$ into $(i)$:
$x + \left(-\frac{1 + y_1^2}{y_2}\right)y_1 = 0$
$x - \frac{y_1(1 + y_1^2)}{y_2} = 0$
$xy_2 = y_1(1 + y_1^2)$

(D) Given equation: $y = a e^{2x} + b x e^{2x} = e^{2x}(a + bx)$.
First derivative with respect to $x$: $\frac{dy}{dx} = 2e^{2x}(a + bx) + b e^{2x} = 2y + b e^{2x}$.
Rearranging gives $b e^{2x} = \frac{dy}{dx} - 2y$ ... $(i)$.
Second derivative with respect to $x$: $\frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2b e^{2x}$ ... $(ii)$.
Substitute $(i)$ into $(ii)$:
$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2(\frac{dy}{dx} - 2y)$.
$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2\frac{dy}{dx} - 4y$.
$\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0$.
Thus,the differential equation is $y^{\prime \prime} - 4y^{\prime} + 4y = 0$.

(B) Given the equation: $y = A \cos 3x + B \sin 3x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = -3A \sin 3x + 3B \cos 3x$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = -9A \cos 3x - 9B \sin 3x$
Factor out $-9$:
$\frac{d^2y}{dx^2} = -9(A \cos 3x + B \sin 3x)$
Since $y = A \cos 3x + B \sin 3x$,we substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = -9y$
Rearranging gives:
$\frac{d^2y}{dx^2} + 9y = 0$

(A) The equation of the family of hyperbolas with centres at the origin and axes along the coordinate axes is given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (or $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for the general conic form). Differentiating with respect to $x$: $\frac{2x}{a^2} - \frac{2y}{b^2} y_1 = 0 \Rightarrow \frac{x}{a^2} = \frac{y y_1}{b^2} \Rightarrow \frac{y y_1}{x} = \frac{b^2}{a^2} = k$ (constant).
Differentiating again with respect to $x$: $\frac{d}{dx} \left( \frac{y y_1}{x} \right) = 0$.
Using the quotient rule: $\frac{x(y y_2 + y_1^2) - y y_1}{x^2} = 0$.
Since $x \neq 0$,we have $x y y_2 + x y_1^2 - y y_1 = 0$.

(B) Given $y = (\tan^{-1} 2x)^2 + (\cot^{-1} 2x)^2$.
Differentiating with respect to $x$:
$y' = 2(\tan^{-1} 2x) \cdot \frac{2}{1 + 4x^2} + 2(\cot^{-1} 2x) \cdot \frac{-2}{1 + 4x^2}$
$y' = \frac{4(\tan^{-1} 2x - \cot^{-1} 2x)}{1 + 4x^2}$
$(1 + 4x^2) y' = 4(\tan^{-1} 2x - \cot^{-1} 2x)$
Differentiating again with respect to $x$:
$8x y' + (1 + 4x^2) y'' = 4 \left( \frac{2}{1 + 4x^2} - \frac{-2}{1 + 4x^2} \right)$
$8x y' + (1 + 4x^2) y'' = 4 \left( \frac{4}{1 + 4x^2} \right) = \frac{16}{1 + 4x^2}$
Multiply both sides by $(1 + 4x^2)$:
$8x(1 + 4x^2) y' + (1 + 4x^2)^2 y'' = 16$
$(1 + 4x^2)^2 y'' - 16 = -8x(1 + 4x^2) y'$

(D) Given the equation: $ax + by = 1$.
Differentiating with respect to $x$:
$a + b \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{a}{b}$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = 0$.
Thus,the required differential equation is $\frac{d^2y}{dx^2} = 0$.

(A) Given the differential equation $x dy - y dx = \sqrt{x^2 + y^2} dx$.
Rearranging the terms,we get $\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$.
Simplifying,we have $x \frac{dv}{dx} = \sqrt{1 + v^2}$.
Separating the variables: $\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$.
Integrating both sides: $\ln(v + \sqrt{v^2 + 1}) = \ln|x| + C = \ln|cx|$.
Thus,$v + \sqrt{v^2 + 1} = cx$.
Substituting $v = \frac{y}{x}$: $\frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 1} = cx \Rightarrow y + \sqrt{x^2 + y^2} = cx^2$.
Given $y(\sqrt{3}) = 1$,we have $1 + \sqrt{3 + 1} = c(\sqrt{3})^2 \Rightarrow 1 + 2 = 3c \Rightarrow c = 1$.
Therefore,the solution is $y + \sqrt{x^2 + y^2} = x^2$.

(C) Given the differential equation: $(x+y) y dx + (y-x) x dy = 0$.
Rearranging the terms: $(y-x) x dy = -(x+y) y dx$,which gives $\frac{dy}{dx} = \frac{(x+y) y}{(x-y) x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{(x+vx) vx}{(x-vx) x} = \frac{v(1+v)}{1-v} = \frac{v+v^2}{1-v}$.
$x \frac{dv}{dx} = \frac{v+v^2}{1-v} - v = \frac{v+v^2-v+v^2}{1-v} = \frac{2v^2}{1-v}$.
Separating the variables: $\int \frac{1-v}{v^2} dv = \int \frac{2}{x} dx$.
Integrating both sides: $\int (v^{-2} - v^{-1}) dv = 2 \int \frac{1}{x} dx$.
$-v^{-1} - \ln|v| = 2 \ln|x| + C$.
Substituting $v = \frac{y}{x}$: $-\frac{x}{y} - \ln(\frac{y}{x}) = 2 \ln|x| + C$.
$-\frac{x}{y} - \ln|y| + \ln|x| = 2 \ln|x| + C$.
$-\frac{x}{y} = \ln|y| + \ln|x| + C = \ln|xy| + C$.
Multiplying by $-y$: $x = -y \ln|xy| - yC$.
$x + y(\ln|xy| + C) = 0$.
$x + y \ln|cxy| = 0$,where $C = \ln|c|$.

(B) Given differential equation: $(x \sin \frac{y}{x}) dy = (y \sin \frac{y}{x} - x) dx$
Rearranging the terms: $\frac{dy}{dx} = \frac{y \sin \frac{y}{x} - x}{x \sin \frac{y}{x}} = \frac{y}{x} - \frac{1}{\sin \frac{y}{x}} = \frac{y}{x} - \operatorname{cosec} \frac{y}{x}$
Let $\frac{y}{x} = v$,then $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$
Substituting these into the equation: $v + x \frac{dv}{dx} = v - \operatorname{cosec} v$
$x \frac{dv}{dx} = -\operatorname{cosec} v$
Separating the variables: $\frac{dv}{\operatorname{cosec} v} = -\frac{dx}{x} \Rightarrow \sin v dv = -\frac{dx}{x}$
Integrating both sides: $\int \sin v dv = -\int \frac{1}{x} dx$
$-\cos v = -\log_e |x| + C$
$\cos v = \log_e |x| + C'$
Substituting $v = \frac{y}{x}$ back: $\cos \frac{y}{x} = \log_e x + C$

(B) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + \tan(\frac{y}{x})$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v + \tan(v)$.
Subtracting $v$ from both sides:
$x \frac{dv}{dx} = \tan(v)$.
Separating the variables:
$\int \cot(v) dv = \int \frac{1}{x} dx$.
Integrating both sides:
$\ln|\sin(v)| = \ln|x| + \ln|C|$.
$\ln|\sin(v)| = \ln|Cx|$.
Taking the exponential of both sides:
$\sin(v) = Cx$.
Substituting $v = \frac{y}{x}$ back:
$\sin(\frac{y}{x}) = Cx$.

(A) Given differential equation is $x dy - y dx = \sqrt{x^2 + y^2} dx$.
Dividing both sides by $x dx$ (assuming $x \neq 0$),we get:
$\frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}$.
Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v + \sqrt{1 + v^2}$.
Subtracting $v$ from both sides:
$x \frac{dv}{dx} = \sqrt{1 + v^2}$.
Separating the variables:
$\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$.
Integrating both sides:
$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$.
$\log |v + \sqrt{1 + v^2}| = \log |x| + \log |c|$.
$v + \sqrt{1 + v^2} = cx$.
Substituting $v = \frac{y}{x}$ back:
$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = cx$.
$\frac{y + \sqrt{x^2 + y^2}}{x} = cx$.
$y + \sqrt{x^2 + y^2} = cx^2$.

(B) Given differential equation: $(xy + y^2) dx - (x^2 - 2xy) dy = 0$
Rearranging the terms: $\frac{dy}{dx} = \frac{xy + y^2}{x^2 - 2xy}$
Divide numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{\frac{y}{x} + (\frac{y}{x})^2}{1 - 2(\frac{y}{x})}$
Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the equation: $v + x\frac{dv}{dx} = \frac{v + v^2}{1 - 2v}$
$x\frac{dv}{dx} = \frac{v + v^2}{1 - 2v} - v = \frac{v + v^2 - v + 2v^2}{1 - 2v} = \frac{3v^2}{1 - 2v}$
Separating variables: $(\frac{1 - 2v}{3v^2}) dv = \frac{dx}{x} \Rightarrow (\frac{1}{3v^2} - \frac{2}{3v}) dv = \frac{dx}{x}$
Integrating both sides: $\int (\frac{1}{3}v^{-2} - \frac{2}{3v}) dv = \int \frac{1}{x} dx$
$-\frac{1}{3v} - \frac{2}{3} \ln|v| = \ln|x| + C_1$
Substitute $v = \frac{y}{x}$: $-\frac{x}{3y} - \frac{2}{3} \ln(\frac{y}{x}) = \ln|x| + C_1$
Multiply by $3$: $-\frac{x}{y} - 2(\ln y - \ln x) = 3\ln x + 3C_1$
$-\frac{x}{y} = 3\ln x + 2\ln y - 2\ln x + C_2 = \ln x + 2\ln y + C_2 = \ln(xy^2) + C_2$
$-\frac{x}{y} = \ln(Cxy^2) \Rightarrow e^{-\frac{x}{y}} = Cxy^2$
Therefore,$Cxy^2 e^{\frac{x}{y}} = 1$.

(D) Given differential equation: $(\sin y \cos^2 y - x \sec^2 y) dy = \tan y dx$
Rearranging the terms: $\tan y \frac{dx}{dy} + x \sec^2 y = \sin y \cos^2 y$
Dividing by $\tan y$: $\frac{dx}{dy} + x \frac{\sec^2 y}{\tan y} = \frac{\sin y \cos^2 y}{\tan y} = \cos^3 y$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{\sec^2 y}{\tan y}$ and $Q(y) = \cos^3 y$.
Integrating factor ($I$.$F$.) $= e^{\int P(y) dy} = e^{\int \frac{\sec^2 y}{\tan y} dy} = e^{\ln(\tan y)} = \tan y$.
The solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$.
$x \tan y = \int \cos^3 y \cdot \tan y dy = \int \cos^3 y \cdot \frac{\sin y}{\cos y} dy = \int \cos^2 y \sin y dy$.
Let $u = \cos y$,then $du = -\sin y dy$.
$x \tan y = -\int u^2 du = -\frac{u^3}{3} + C = -\frac{\cos^3 y}{3} + C$.
Multiplying by $3$: $3x \tan y = -\cos^3 y + 3C$.
Thus,$3x \tan y + \cos^3 y = C$ (where $C$ is a constant).

(D) Given the differential equation $x dy + (y + y^2 x) dx = 0$.
Dividing by $x dx$,we get $\frac{dy}{dx} + \frac{y}{x} + y^2 = 0$.
This is a Bernoulli differential equation. Divide by $y^2$: $y^{-2} \frac{dy}{dx} + \frac{1}{x} y^{-1} = -1$.
Let $v = y^{-1}$,then $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$,so $- \frac{dv}{dx} + \frac{v}{x} = -1$,which simplifies to $\frac{dv}{dx} - \frac{v}{x} = 1$.
This is a linear differential equation with Integrating Factor $I.F. = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$.
The solution is $v \cdot \frac{1}{x} = \int 1 \cdot \frac{1}{x} dx = \log x + C$.
Substituting $v = \frac{1}{y}$,we get $\frac{1}{xy} = \log x + C$.
Given $y = 1$ at $x = 1$,we have $\frac{1}{1 \cdot 1} = \log 1 + C \Rightarrow 1 = 0 + C \Rightarrow C = 1$.
Thus,$\frac{1}{xy} = \log x + 1$,which gives $y = \frac{1}{x(1 + \log x)}$.

(D) Given the differential equation: $(y^2+x+1) dy = (y+1) dx$.
Rearranging the terms to form a linear differential equation in $x$:
$\frac{dx}{dy} = \frac{y^2+x+1}{y+1} = \frac{y^2+1}{y+1} + \frac{x}{y+1}$.
$\frac{dx}{dy} - \frac{1}{y+1} x = \frac{y^2+1}{y+1}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y+1}$ and $Q(y) = \frac{y^2+1}{y+1}$.
The integrating factor $IF = e^{\int P(y) dy} = e^{-\int \frac{1}{y+1} dy} = e^{-\log(y+1)} = \frac{1}{y+1}$.
The solution is $x \cdot IF = \int Q(y) \cdot IF dy + c$.
$x \cdot \frac{1}{y+1} = \int \frac{y^2+1}{(y+1)^2} dy + c$.
Using the substitution $y^2+1 = (y+1)^2 - 2y - 1 + 1 = (y+1)^2 - 2(y+1) + 2$:
$\frac{x}{y+1} = \int \left( 1 - \frac{2}{y+1} + \frac{2}{(y+1)^2} \right) dy + c$.
$\frac{x}{y+1} = y - 2 \log |y+1| - \frac{2}{y+1} + c$.
Multiplying by $(y+1)$: $x = y(y+1) - 2(y+1) \log |y+1| - 2 + c(y+1)$.
Alternatively,rearranging the integral result: $\frac{x+2}{y+1} = y - 2 \log |y+1| + c$.
Thus,$\frac{x+2}{y+1} + 2 \log |y+1| = y + c$,which is $\frac{x+2}{y+1} + \log (y+1)^2 = y + c$.

(D) Given the differential equation: $\sin x \frac{dy}{dx} - y \cos x = 1$.
Divide the entire equation by $\sin x$ to write it in the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} - y \cot x = \operatorname{cosec} x$.
Here,$P = -\cot x$.
The integrating factor $(IF)$ is given by the formula $IF = e^{\int P dx}$.
$IF = e^{\int -\cot x dx} = e^{-\ln|\sin x|} = e^{\ln|\sin x|^{-1}} = \frac{1}{\sin x} = \operatorname{cosec} x$.

(B) Given the differential equation: $(x + 2y^3) \frac{dy}{dx} = y$.
Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y} + 2y^2$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y^2$.
The integrating factor ($I$.$F$.) is given by $e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.
The general solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
Substituting the values: $x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + c$.
$\frac{x}{y} = \int 2y dy + c$.
$\frac{x}{y} = y^2 + c$.
Therefore,$x = y(y^2 + c)$.

(C) Given differential equation is $(1+\tan y)(dx-dy)+2x dy=0$.
Rearranging the terms,we get $(1+\tan y)dx = (1+\tan y - 2x)dy$.
Dividing by $(1+\tan y)dy$,we obtain $\frac{dx}{dy} = 1 - \frac{2x}{1+\tan y}$,which simplifies to $\frac{dx}{dy} + \left(\frac{2}{1+\tan y}\right)x = 1$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{2}{1+\tan y} = \frac{2\cos y}{\sin y + \cos y}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(y)dy} = e^{\int \frac{2\cos y}{\sin y + \cos y} dy}$.
Since $\int \frac{2\cos y}{\sin y + \cos y} dy = \int \frac{(\cos y - \sin y) + (\cos y + \sin y)}{\sin y + \cos y} dy = \int \left(\frac{\cos y - \sin y}{\sin y + \cos y} + 1\right) dy = \ln|\sin y + \cos y| + y$.
Thus,$I.F. = e^{\ln(\sin y + \cos y) + y} = e^y(\sin y + \cos y)$.
The solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
$x e^y(\sin y + \cos y) = \int e^y(\sin y + \cos y) dy + c$.
Using the formula $\int e^y(f(y) + f'(y)) dy = e^y f(y) + c$,where $f(y) = \sin y$ and $f'(y) = \cos y$,we get:
$x e^y(\sin y + \cos y) = e^y \sin y + c$.
Rearranging gives $e^y(x \sin y + x \cos y - \sin y) = c$.

(A) Given differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x-y-1}$.
Let $x = X+h$ and $y = Y+k$,then $\frac{dy}{dx} = \frac{dY}{dX}$.
Substituting these,we get $\frac{dY}{dX} = \frac{X+Y+h+k+1}{X-Y+h-k-1}$.
For the equation to be homogeneous,we set $h+k+1 = 0$ and $h-k-1 = 0$.
Solving these equations,we find $h = 0$ and $k = -1$.
Thus,the equation becomes $\frac{dY}{dX} = \frac{X+Y}{X-Y}$.
Rearranging,we get $(X-Y) dY = (X+Y) dX$,which implies $X dY - Y dX = X dX + Y dY$.
Dividing by $X^2+Y^2$,we have $\frac{X dY - Y dX}{X^2+Y^2} = \frac{X dX + Y dY}{X^2+Y^2}$.
Integrating both sides,$\int d\left(\tan^{-1}\left(\frac{Y}{X}\right)\right) = \frac{1}{2} \int d(\log(X^2+Y^2))$.
This gives $\tan^{-1}\left(\frac{Y}{X}\right) = \frac{1}{2} \log(X^2+Y^2) + C$.
Substituting $X = x$ and $Y = y+1$,we get $\tan^{-1}\left(\frac{y+1}{x}\right) = \frac{1}{2} \log(x^2+(y+1)^2) + C$.
Therefore,$\tan^{-1}\left(\frac{y+1}{x}\right) - \frac{1}{2} \log(x^2+y^2+2y+1) = C$.

(C) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6x$.
This can be written as the derivative of a product: $\frac{d}{dx}(y \cos x) = 6x$.
Integrating both sides with respect to $x$: $y \cos x = \int 6x \, dx = 3x^2 + C$.
Using the initial condition $y(\frac{\pi}{3}) = 0$:
$0 \cdot \cos(\frac{\pi}{3}) = 3(\frac{\pi}{3})^2 + C \Rightarrow 0 = \frac{\pi^2}{3} + C \Rightarrow C = \frac{-\pi^2}{3}$.
Thus,the general solution is $y \cos x = 3x^2 - \frac{\pi^2}{3}$.
Now,to find $y(\frac{\pi}{6})$,substitute $x = \frac{\pi}{6}$:
$y(\frac{\pi}{6}) \cos(\frac{\pi}{6}) = 3(\frac{\pi}{6})^2 - \frac{\pi^2}{3}$.
$y(\frac{\pi}{6}) \cdot \frac{\sqrt{3}}{2} = 3(\frac{\pi^2}{36}) - \frac{\pi^2}{3} = \frac{\pi^2}{12} - \frac{4\pi^2}{12} = \frac{-3\pi^2}{12} = \frac{-\pi^2}{4}$.
Therefore,$y(\frac{\pi}{6}) = \frac{-\pi^2}{4} \cdot \frac{2}{\sqrt{3}} = \frac{-\pi^2}{2 \sqrt{3}}$.

(B) Given differential equation: $e^x y dx + e^x dy + x dx = 0$
We can rewrite this as: $d(ye^x) + x dx = 0$
Integrating both sides: $\int d(ye^x) + \int x dx = \int 0 dx$
$ye^x + \frac{x^2}{2} = C_1$
Multiplying by $2$: $2ye^x + x^2 = 2C_1$
Let $2C_1 = C$,so the solution is $2ye^x + x^2 = C$.

(A) In a regular hexagon $ABCDEF$,the center $O$ is such that $\overrightarrow{OA}=\overrightarrow{BC}=\vec{b}$ and $\overrightarrow{AB}=\overrightarrow{OC}=\vec{a}$.
Also,$\overrightarrow{CD}=\overrightarrow{AF}$ and $\overrightarrow{DE}=\overrightarrow{BA}=-\vec{a}$.
By the polygon law of vector addition,the sum of vectors along the sides of a closed polygon is zero:
$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FA} = 0$
Since $\overrightarrow{CD} = \overrightarrow{AF} = -\overrightarrow{FA}$,we have:
$\vec{a} + \vec{b} + \overrightarrow{CD} - \vec{a} + \overrightarrow{EF} - \overrightarrow{CD} = 0$
Alternatively,using the property of a regular hexagon,$\overrightarrow{FA} = \overrightarrow{CD} - \overrightarrow{BC} = \vec{a} - \vec{b}$.

(B) Let the points be $A(\alpha, 10, 13)$,$B(6, 11, 11)$,and $C(\frac{9}{2}, \beta, -8)$.
Since the points are collinear,the vectors $\vec{AB}$ and $\vec{BC}$ must be proportional.
$\vec{AB} = (6-\alpha)\hat{i} + (11-10)\hat{j} + (11-13)\hat{k} = (6-\alpha)\hat{i} + 1\hat{j} - 2\hat{k}$.
$\vec{BC} = (\frac{9}{2}-6)\hat{i} + (\beta-11)\hat{j} + (-8-11)\hat{k} = -\frac{3}{2}\hat{i} + (\beta-11)\hat{j} - 19\hat{k}$.
For collinearity,$\frac{6-\alpha}{-3/2} = \frac{1}{\beta-11} = \frac{-2}{-19}$.
From $\frac{1}{\beta-11} = \frac{2}{19}$,we get $2(\beta-11) = 19 \Rightarrow 2\beta - 22 = 19 \Rightarrow 2\beta = 41 \Rightarrow 6\beta = 123$.
From $\frac{6-\alpha}{-3/2} = \frac{2}{19}$,we get $19(6-\alpha) = -3 \Rightarrow 114 - 19\alpha = -3 \Rightarrow 19\alpha = 117$.
Thus,$(19\alpha - 6\beta)^2 = (117 - 123)^2 = (-6)^2 = 36$.

(C) Three vectors $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\left|\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & p & 5 \end{array}\right| = 0$
Expanding the determinant along the first row:
$2(2 \times 5 - (-3) \times p) - (-1)(1 \times 5 - (-3) \times 3) + 1(1 \times p - 2 \times 3) = 0$
$2(10 + 3p) + 1(5 + 9) + 1(p - 6) = 0$
$20 + 6p + 14 + p - 6 = 0$
$7p + 28 = 0$
$7p = -28$
$p = -4$

(A) Given,$|\vec{a}|=|\vec{b}|=1$.
Since $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$,let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$. Then $\vec{a} \cdot \vec{b} = \cos \theta$.
So,$\vec{u} = \vec{a} - \cos \theta \vec{b}$.
Calculating the magnitude squared: $|\vec{u}|^2 = |\vec{a}|^2 + \cos^2 \theta |\vec{b}|^2 - 2 \cos \theta (\vec{a} \cdot \vec{b}) = 1 + \cos^2 \theta - 2 \cos^2 \theta = 1 - \cos^2 \theta = \sin^2 \theta$.
Thus,$|\vec{u}| = \sin \theta$ (since $\sin \theta > 0$ for non-collinear vectors).
Also,$|\vec{v}| = |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = \sin \theta$.
Therefore,$|\vec{v}| = |\vec{u}|$.
Now,$\vec{u} \cdot \vec{v} = (\vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) \cdot (\vec{a} \times \vec{b}) = \vec{a} \cdot (\vec{a} \times \vec{b}) - (\vec{a} \cdot \vec{b}) \vec{b} \cdot (\vec{a} \times \vec{b}) = 0 - 0 = 0$.
Hence,$|\vec{v}| = |\vec{u}| + |\vec{u} \cdot \vec{v}| = |\vec{u}| + 0 = |\vec{u}|$.

(D) Given: $\vec{a}+\vec{b}+\vec{c}=\alpha \vec{d}$ ....$(i)$
Given: $\vec{b}+\vec{c}+\vec{d}=\beta \vec{a}$ ....(ii)
From $(i)$,$\vec{b}+\vec{c}=\alpha \vec{d}-\vec{a}$.
Substituting this into (ii): $(\alpha \vec{d}-\vec{a})+\vec{d}=\beta \vec{a}$.
Rearranging the terms: $(\alpha+1)\vec{d} = (\beta+1)\vec{a}$.
Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$\vec{d}$ must be a linear combination of $\vec{a}, \vec{b}, \vec{c}$. For the equation $(\alpha+1)\vec{d} = (\beta+1)\vec{a}$ to hold for non-coplanar vectors,the coefficients must be zero,implying $\alpha+1=0$ and $\beta+1=0$,so $\alpha=-1$ and $\beta=-1$.
Substituting $\alpha=-1$ into $(i)$: $\vec{a}+\vec{b}+\vec{c} = -\vec{d}$.
Therefore,$\vec{a}+\vec{b}+\vec{c}+\vec{d} = \vec{0}$.
Thus,$|\vec{a}+\vec{b}+\vec{c}+\vec{d}| = |\vec{0}| = 0$.

(A) Let $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ and $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$ be the unit vectors along $\vec{a}$ and $\vec{b}$ respectively.
Then,$|\vec{b}| \vec{a} + |\vec{a}| \vec{b} = |\vec{a}| |\vec{b}| \left( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right) = |\vec{a}| |\vec{b}| (\hat{a} + \hat{b})$.
Since $\hat{a} + \hat{b}$ is the diagonal of the rhombus formed by unit vectors $\hat{a}$ and $\hat{b}$,it represents the angle bisector of the angle between $\vec{a}$ and $\vec{b}$.
Thus,$|\vec{b}| \vec{a} + |\vec{a}| \vec{b}$ is a vector parallel to the angle bisector of $\vec{a}$ and $\vec{b}$.

(C) Let the position vectors of the points be $\overrightarrow{OA} = -\hat{i} + 4\hat{j} - 4\hat{k}$,$\overrightarrow{OB} = 3\hat{i} + 2\hat{j} - 5\hat{k}$,$\overrightarrow{OC} = -3\hat{i} + 8\hat{j} - 5\hat{k}$,and $\overrightarrow{OD} = -3\hat{i} + 2\hat{j} + \lambda\hat{k}$.
For the points to be coplanar,the scalar triple product of the vectors $\overrightarrow{AB}$,$\overrightarrow{AC}$,and $\overrightarrow{AD}$ must be zero.
First,calculate the vectors:
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3 - (-1))\hat{i} + (2 - 4)\hat{j} + (-5 - (-4))\hat{k} = 4\hat{i} - 2\hat{j} - \hat{k}$
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (-3 - (-1))\hat{i} + (8 - 4)\hat{j} + (-5 - (-4))\hat{k} = -2\hat{i} + 4\hat{j} - \hat{k}$
$\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = (-3 - (-1))\hat{i} + (2 - 4)\hat{j} + (\lambda - (-4))\hat{k} = -2\hat{i} - 2\hat{j} + (\lambda + 4)\hat{k}$
Now,set the determinant of these vectors to zero:
$\begin{vmatrix} 4 & -2 & -1 \\ -2 & 4 & -1 \\ -2 & -2 & \lambda + 4 \end{vmatrix} = 0$
Expanding the determinant:
$4(4(\lambda + 4) - 2) - (-2)(-2(\lambda + 4) - 2) - 1(4 - (-8)) = 0$
$4(4\lambda + 16 - 2) + 2(-2\lambda - 8 - 2) - 1(12) = 0$
$4(4\lambda + 14) + 2(-2\lambda - 10) - 12 = 0$
$16\lambda + 56 - 4\lambda - 20 - 12 = 0$
$12\lambda + 24 = 0$
$12\lambda = -24$
$\lambda = -2$

(B) We are given $|\vec{f}|=10$,$|\vec{g}|=14$,and $|\vec{f}-\vec{g}|=15$.
Using the property $|\vec{f}-\vec{g}|^2 = |\vec{f}|^2 + |\vec{g}|^2 - 2(\vec{f} \cdot \vec{g})$,we have:
$15^2 = 10^2 + 14^2 - 2(\vec{f} \cdot \vec{g})$
$225 = 100 + 196 - 2(\vec{f} \cdot \vec{g})$
$225 = 296 - 2(\vec{f} \cdot \vec{g})$
$2(\vec{f} \cdot \vec{g}) = 296 - 225 = 71$.
Now,we need to find $|\vec{f}+\vec{g}|$.
Using the property $|\vec{f}+\vec{g}|^2 = |\vec{f}|^2 + |\vec{g}|^2 + 2(\vec{f} \cdot \vec{g})$,we substitute the known values:
$|\vec{f}+\vec{g}|^2 = 10^2 + 14^2 + 71$
$|\vec{f}+\vec{g}|^2 = 100 + 196 + 71 = 367$
Therefore,$|\vec{f}+\vec{g}| = \sqrt{367}$.

(D) Given that $|\vec{a}|=|\vec{b}|=|\vec{c}|=\sqrt{3}$.
Expanding the given equation $(\vec{a}+\vec{b}+\vec{c})^2+(\vec{b}+\vec{c}-\vec{a})^2+(\vec{c}+\vec{a}-\vec{b})^2=36$:
$3(|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36$.
Substituting $|\vec{a}|^2=|\vec{b}|^2=|\vec{c}|^2=3$:
$3(3+3+3) + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36$.
$27 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 36 \Rightarrow 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.
Now,consider $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3+3+3+9 = 18$.
We need to find $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = |2(\vec{a}+\vec{c})-3 \vec{b}|^2$.
From the expansion,we have $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$.
Using the identity $|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = 4|\vec{a}|^2 + 9|\vec{b}|^2 + 4|\vec{c}|^2 - 12(\vec{a} \cdot \vec{b}) - 12(\vec{b} \cdot \vec{c}) + 8(\vec{a} \cdot \vec{c})$.
Given the symmetry,$|2 \vec{a}-3 \vec{b}+2 \vec{c}|^2 = 4(3) + 9(3) + 4(3) - 12(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c}) + 8(\vec{a} \cdot \vec{c}) = 75$.

(A) Since the vectors $\vec{a}, \vec{b}, \vec{c}$ are linearly dependent,their scalar triple product must be zero:
$\begin{vmatrix} \alpha & \beta & 3 \\ 0 & 1 & 2 \\ 3 & 2 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$\alpha(1 - 4) - \beta(0 - 6) + 3(0 - 3) = 0$
$-3\alpha + 6\beta - 9 = 0$
Dividing by $-3$:
$\alpha - 2\beta + 3 = 0 \Rightarrow \alpha = 2\beta - 3$
Given that the magnitude $|\vec{a}| = \sqrt{14}$,we have:
$\alpha^2 + \beta^2 + 3^2 = 14$
$\alpha^2 + \beta^2 = 5$
Substituting $\alpha = 2\beta - 3$:
$(2\beta - 3)^2 + \beta^2 = 5$
$4\beta^2 - 12\beta + 9 + \beta^2 = 5$
$5\beta^2 - 12\beta + 4 = 0$
Factoring the quadratic equation:
$(5\beta - 2)(\beta - 2) = 0$
Since $\beta$ is an integer,we take $\beta = 2$.
Then $\alpha = 2(2) - 3 = 1$.
Therefore,$\alpha + \beta = 1 + 2 = 3$.

(B) First,find the unit vectors along $\vec{a}$ and $\vec{b}$.
$|\vec{a}| = \sqrt{4^2 + 7^2 + (-4)^2} = \sqrt{16 + 49 + 16} = \sqrt{81} = 9$.
$|\vec{b}| = \sqrt{12^2 + (-3)^2 + 4^2} = \sqrt{144 + 9 + 16} = \sqrt{169} = 13$.
Unit vectors are $\hat{a} = \frac{4 \hat{i} + 7 \hat{j} - 4 \hat{k}}{9}$ and $\hat{b} = \frac{12 \hat{i} - 3 \hat{j} + 4 \hat{k}}{13}$.
The vector along the internal angle bisector is given by $\vec{v} = \lambda(\hat{a} + \hat{b})$.
$\hat{a} + \hat{b} = \frac{13(4 \hat{i} + 7 \hat{j} - 4 \hat{k}) + 9(12 \hat{i} - 3 \hat{j} + 4 \hat{k})}{117} = \frac{(52 + 108) \hat{i} + (91 - 27) \hat{j} + (-52 + 36) \hat{k}}{117} = \frac{160 \hat{i} + 64 \hat{j} - 16 \hat{k}}{117} = \frac{16}{117}(10 \hat{i} + 4 \hat{j} - \hat{k})$.
Let $\vec{c} = k(10 \hat{i} + 4 \hat{j} - \hat{k})$. The magnitude is $|\vec{c}| = |k| \sqrt{10^2 + 4^2 + (-1)^2} = |k| \sqrt{100 + 16 + 1} = |k| \sqrt{117} = |k| \sqrt{9 \times 13} = 3|k| \sqrt{13}$.
Given $|\vec{c}| = 3 \sqrt{13}$,we have $3|k| \sqrt{13} = 3 \sqrt{13}$,so $k = 1$.
Thus,$\vec{c} = 10 \hat{i} + 4 \hat{j} - \hat{k}$.