MHT CET 2026 Mathematics Question Paper with Answer and Solution

(D) Given $a + b + c = 0$, we have $a = -(b + c)$.
Squaring both sides: $|a|^2 = |-(b + c)|^2 = |b + c|^2$.
$|a|^2 = |b|^2 + |c|^2 + 2|b||c| \cos \theta$.
Since $|a| = |b| = |c| = 3$, we have $3^2 = 3^2 + 3^2 + 2(3)(3) \cos \theta$.
$9 = 9 + 9 + 18 \cos \theta$.
$18 \cos \theta = -9$, so $\cos \theta = -1/2$.
Since $\cos \theta = -1/2$, $\theta = 120^\circ$.
Then $\tan \theta = \tan(120^\circ) = -\sqrt{3}$ and $\cot \theta = \cot(120^\circ) = -1/\sqrt{3}$.
$\tan^2 \theta = (-\sqrt{3})^2 = 3$.
$\cot^2 \theta = (-1/\sqrt{3})^2 = 1/3$.
$\tan^2 \theta + \cot^2 \theta = 3 + 1/3 = 10/3$.

(A) Let $\vec{u} = 5\vec{a} + 2\vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$.
The diagonals are $\vec{d_1} = \vec{u} + \vec{v} = 6\vec{a} - \vec{b}$ and $\vec{d_2} = \vec{u} - \vec{v} = 4\vec{a} + 5\vec{b}$.
Given $|\vec{a}|^2 = 8$, $|\vec{b}|^2 = 9$, and $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\frac{\pi}{4}) = (2\sqrt{2})(3)(\frac{1}{\sqrt{2}}) = 6$.
$|\vec{d_1}|^2 = (6\vec{a} - \vec{b}) \cdot (6\vec{a} - \vec{b}) = 36|\vec{a}|^2 + |\vec{b}|^2 - 12(\vec{a} \cdot \vec{b}) = 36(8) + 9 - 12(6) = 288 + 9 - 72 = 225$.
So, $|\vec{d_1}| = \sqrt{225} = 15$.
$|\vec{d_2}|^2 = (4\vec{a} + 5\vec{b}) \cdot (4\vec{a} + 5\vec{b}) = 16|\vec{a}|^2 + 25|\vec{b}|^2 + 40(\vec{a} \cdot \vec{b}) = 16(8) + 25(9) + 40(6) = 128 + 225 + 240 = 593$.
So, $|\vec{d_2}| = \sqrt{593}$.
The lengths are $15$ and $\sqrt{593}$.

(A) The vectors representing the sides are:
$\vec{AB} = \vec{b} - \vec{a} = (4 - (-2))\hat{i} + (0 - (-1))\hat{j} = 6\hat{i} + \hat{j}$
$\vec{BC} = \vec{c} - \vec{b} = (3 - 4)\hat{i} + (3 - 0)\hat{j} = -\hat{i} + 3\hat{j}$
$\vec{CD} = \vec{d} - \vec{c} = (-3 - 3)\hat{i} + (2 - 3)\hat{j} = -6\hat{i} - \hat{j}$
$\vec{DA} = \vec{a} - \vec{d} = (-2 - (-3))\hat{i} + (-1 - 2)\hat{j} = \hat{i} - 3\hat{j}$
Since $\vec{AB} = -\vec{CD}$ and $\vec{BC} = -\vec{DA}$, $ABCD$ is a parallelogram.
Calculate dot product of adjacent sides: $\vec{AB} \cdot \vec{BC} = (6)(-1) + (1)(3) = -6 + 3 = -3 \neq 0$. Thus, it is not a rectangle.
Calculate magnitudes: $|\vec{AB}| = \sqrt{6^2 + 1^2} = \sqrt{37}$ and $|\vec{BC}| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$.
Since $|\vec{AB}| \neq |\vec{BC}|$, it is not a rhombus.
Therefore, $ABCD$ is a parallelogram which is neither a rhombus nor a rectangle.

(D) The magnitude of a vector remains invariant under rotation. Therefore, $|\vec{v}|^2 = |\vec{v}'|^2$.
$|\vec{v}|^2 = 3^2 + 4^2 + (-5)^2 = 9 + 16 + 25 = 50$.
$|\vec{v}'|^2 = (a + 1)^2 + (-3)^2 + 5^2 = (a + 1)^2 + 9 + 25 = (a + 1)^2 + 34$.
Equating the two: $(a + 1)^2 + 34 = 50$.
$(a + 1)^2 = 16$.
$a + 1 = \pm 4$.
Case $1$: $a + 1 = 4 \implies a = 3$.
Case $2$: $a + 1 = -4 \implies a = -5$.
Thus, the possible values of $a$ are $3$ or $-5$.

(C) Step $1$: Calculate $(\vec{a} + \vec{b}) = (4\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + \hat{j} + 2\hat{k}) = 6\hat{i} + 2\hat{j} + 3\hat{k}$.
Step $2$: Calculate $(\vec{b} + \vec{c}) = (2\hat{i} + \hat{j} + 2\hat{k}) + (3\hat{i} + 4\hat{j} + 5\hat{k}) = 5\hat{i} + 5\hat{j} + 7\hat{k}$.
Step $3$: Calculate the dot product $(\vec{a} + \vec{b}) \cdot (\vec{b} + \vec{c}) = (6\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (5\hat{i} + 5\hat{j} + 7\hat{k})$.
Step $4$: Perform the scalar multiplication: $(6 \times 5) + (2 \times 5) + (3 \times 7) = 30 + 10 + 21 = 61$.

(C) Step $1$: Since $\vec{a} \parallel \vec{b}$, their components are proportional: $\frac{\lambda}{2} = \frac{1}{4} = \frac{1}{4}$.
Step $2$: From $\frac{\lambda}{2} = \frac{1}{4}$, we get $\lambda = \frac{2}{4} = 0.5$.
Step $3$: Since $\vec{b} \perp \vec{c}$, their dot product is zero: $\vec{b} \cdot \vec{c} = 0$.
Step $4$: $(2\hat{i} + 4\hat{j} + 4\hat{k}) \cdot (\hat{i} + \mu \hat{j} + \hat{k}) = 0 \implies (2)(1) + (4)(\mu) + (4)(1) = 0$.
Step $5$: $2 + 4\mu + 4 = 0 \implies 4\mu = -6 \implies \mu = -1.5$.
Step $6$: Calculate $\lambda - \mu = 0.5 - (-1.5) = 0.5 + 1.5 = 2$.

(D) Let $\vec{s} = \vec{u} + \vec{v} = (2+b)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k} = (2+b)\hat{i} + 6\hat{j} - 2\hat{k}$.
The unit vector parallel to $\vec{s}$ is $\hat{s} = \frac{\vec{s}}{|\vec{s}|} = \frac{(2+b)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(2+b)^2 + 6^2 + (-2)^2}}$.
The scalar product of $\vec{a}$ and $\hat{s}$ is $1$, so $\vec{a} \cdot \hat{s} = 1$.
$\frac{(1)(2+b) + (1)(6) + (1)(-2)}{\sqrt{(2+b)^2 + 36 + 4}} = 1$.
$\frac{b+6}{\sqrt{(2+b)^2 + 40}} = 1$.
Squaring both sides: $(b+6)^2 = (2+b)^2 + 40$.
$b^2 + 12b + 36 = b^2 + 4b + 4 + 40$.
$8b = 8$, so $b = 1$.

(A) Let the required vector be $\vec{v}$. Since $\vec{v}$ is coplanar with $\vec{b}$ and $\vec{c}$, it can be written as $\vec{v} = \vec{b} \times (\vec{b} \times \vec{c})$ or simply as a linear combination $\vec{v} = \vec{b} \times (\vec{a} \times \vec{c})$ is not correct here. The vector coplanar to $\vec{b}$ and $\vec{c}$ and orthogonal to $\vec{a}$ is given by $\vec{v} = \vec{a} \times (\vec{b} \times \vec{c})$.
First, calculate $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(6-0) - \hat{j}(9-0) + \hat{k}(3-4) = 6\hat{i} - 9\hat{j} - \hat{k}$.
Now, calculate $\vec{v} = \vec{a} \times (6\hat{i} - 9\hat{j} - \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 6 & -9 & -1 \end{vmatrix}$.
$= \hat{i}(-2 - (-27)) - \hat{j}(-1 - 18) + \hat{k}(-9 - 12) = 25\hat{i} + 19\hat{j} - 21\hat{k}$.

(D) Given $|a| = \sqrt{3}$, $|b| = 3$, and $|a - b| = 4$. Squaring $|a - b| = 4$ gives $|a|^2 + |b|^2 - 2(a \cdot b) = 16$, so $3 + 9 - 2(a \cdot b) = 16$, which implies $a \cdot b = -2$.
Let $v = 2i + 2j - k$. The equation $a \times v = v \times b$ can be written as $a \times v + b \times v = 0$, so $(a + b) \times v = 0$.
This implies $(a + b)$ is parallel to $v = 2i + 2j - k$. Thus, $a + b = k(2i + 2j - k)$ for some scalar $k$.
Calculate $|a + b|^2 = |a|^2 + |b|^2 + 2(a \cdot b) = 3 + 9 + 2(-2) = 8$.
Since $|a + b|^2 = k^2(2^2 + 2^2 + (-1)^2) = 9k^2$, we have $9k^2 = 8$, so $k^2 = 8/9$ and $|k| = \frac{2\sqrt{2}}{3}$.
We need $|(a + b) \cdot (3i + 4j + 2k)| = |k(2i + 2j - k) \cdot (3i + 4j + 2k)| = |k(6 + 8 - 2)| = |12k| = 12 \cdot \frac{2\sqrt{2}}{3} = 8\sqrt{2}$.
Given this equals $\sqrt{\lambda}$, so $\sqrt{\lambda} = 8\sqrt{2} = \sqrt{64 \cdot 2} = \sqrt{128}$. Thus $\lambda = 128$.

(C) Since $\triangle ABC$ is right-angled at $A$, we have $\vec{AB} \cdot \vec{AC} = 0$.
Given expression: $E = \vec{AB} \cdot \vec{AC} + \vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$.
Since $\vec{AB} \cdot \vec{AC} = 0$, the expression becomes $E = \vec{BA} \cdot \vec{BC} + \vec{CA} \cdot \vec{CB}$.
Note that $\vec{BA} \cdot \vec{BC} = |\vec{BA}| |\vec{BC}| \cos B = |\vec{BA}|^2$ (projection of $\vec{BA}$ on $\vec{BC}$ is $\vec{BA}$ itself in right triangle).
Actually, in right triangle $ABC$ with hypotenuse $BC$, $\vec{BA} \cdot \vec{BC} = |\vec{BA}|^2$ and $\vec{CA} \cdot \vec{CB} = |\vec{CA}|^2$.
Thus, $E = |\vec{BA}|^2 + |\vec{CA}|^2 = |\vec{BC}|^2$.
Calculate $|\vec{BC}|^2 = |\vec{c} - \vec{b}|^2 = |(5-3)\hat{i} + (1-(-2))\hat{j} + (-3-1)\hat{k}|^2 = |2\hat{i} + 3\hat{j} - 4\hat{k}|^2$.
$|\vec{BC}|^2 = 2^2 + 3^2 + (-4)^2 = 4 + 9 + 16 = 29$.

(A) Given that each vector is perpendicular to the sum of the other two:
$\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 0$
$\vec{b} \cdot (\vec{a} + \vec{c}) = 0 \implies \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{c} = 0$
$\vec{c} \cdot (\vec{a} + \vec{b}) = 0 \implies \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} = 0$
Adding these equations, we get $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$, so $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = 0$.
Now, $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.
$|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 2(0) = 9 + 16 + 25 = 50$.
Therefore, $|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}$.

(D) Step $1$: Find the position vector of point $D$ which divides $BC$ in ratio $2:1$. Using section formula, $\vec{d} = \frac{2\vec{c} + 1\vec{b}}{2+1} = \frac{\vec{b} + 2\vec{c}}{3}$.
Step $2$: Point $E$ divides $AD$ in ratio $1:2$. Using section formula, $\vec{e} = \frac{1(\vec{d}) + 2(\vec{a})}{1+2} = \frac{\vec{d} + 2\vec{a}}{3}$.
Step $3$: Substitute $\vec{d}$ into the expression for $\vec{e}$: $\vec{e} = \frac{\frac{\vec{b} + 2\vec{c}}{3} + 2\vec{a}}{3} = \frac{\vec{b} + 2\vec{c} + 6\vec{a}}{9} = \frac{6\vec{a} + \vec{b} + 2\vec{c}}{9}$.

(D) Given that $a \cdot (b + c) = 0$, $b \cdot (c + a) = 0$, and $c \cdot (a + b) = 0$.
This implies $a \cdot b + a \cdot c = 0$, $b \cdot c + b \cdot a = 0$, and $c \cdot a + c \cdot b = 0$.
Adding these three equations: $2(a \cdot b + b \cdot c + c \cdot a) = 0$, so $a \cdot b + b \cdot c + c \cdot a = 0$.
Since $a \cdot b + a \cdot c = 0$, it follows that $a \cdot b = 0$, $b \cdot c = 0$, and $c \cdot a = 0$.
Now, $|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$.
$|a + b + c|^2 = 1^2 + 2^2 + 3^2 + 2(0) = 1 + 4 + 9 = 14$.
Therefore, $|a + b + c| = \sqrt{14}$.

(D) Given $|a+b+c|^2 = 1^2 = 1$. Since $c = \lambda(a \times b)$, $c$ is perpendicular to both $a$ and $b$, so $a \cdot c = 0$ and $b \cdot c = 0$.
Expanding $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) = 1$.
Substituting the values: $(\frac{1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{6}})^2 + 2(a \cdot b + 0 + 0) = 1$.
$\frac{1}{3} + \frac{1}{2} + \frac{1}{6} + 2(a \cdot b) = 1$.
$\frac{2+3+1}{6} + 2(a \cdot b) = 1 \implies 1 + 2(a \cdot b) = 1 \implies a \cdot b = 0$.
Since $a \cdot b = |a||b| \cos \theta = 0$ and $|a|, |b| \neq 0$, we have $\cos \theta = 0$, so $\theta = \frac{\pi}{2}$.

(C) In parallelogram $OBDC$, diagonals are $\vec{d_1} = \vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{d_2} = \vec{CB} = \vec{OB} - \vec{OC} = -3\hat{i} - 2\hat{k}$.
We know $\vec{OD} = \vec{OB} + \vec{OC}$.
Adding the two equations: $2\vec{OB} = (\hat{i} + 2\hat{j} + 6\hat{k}) + (-3\hat{i} - 2\hat{k}) = -2\hat{i} + 2\hat{j} + 4\hat{k} \implies \vec{OB} = -\hat{i} + \hat{j} + 2\hat{k}$.
Subtracting the two equations: $2\vec{OC} = (\hat{i} + 2\hat{j} + 6\hat{k}) - (-3\hat{i} - 2\hat{k}) = 4\hat{i} + 2\hat{j} + 8\hat{k} \implies \vec{OC} = 2\hat{i} + \hat{j} + 4\hat{k}$.
The volume of the parallelepiped is given by the scalar triple product $|\vec{OA} \cdot (\vec{OB} \times \vec{OC})|$.
$\vec{OA} \cdot (\vec{OB} \times \vec{OC}) = \begin{vmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 2 & 1 & 4 \end{vmatrix}$.
$= 1(4 - 2) - 2(-4 - 4) + 3(-1 - 2) = 1(2) - 2(-8) + 3(-3) = 2 + 16 - 9 = 9$.
Thus, the volume is $9$ cubic units.

(C) Given $\vec{p} \times \vec{a} = \vec{b} \times \vec{a}$, we can write $\vec{p} \times \vec{a} - \vec{b} \times \vec{a} = 0$, which implies $(\vec{p} - \vec{b}) \times \vec{a} = 0$.
This means $(\vec{p} - \vec{b})$ is parallel to $\vec{a}$, so $\vec{p} - \vec{b} = t\vec{a}$ for some scalar $t$.
Thus, $\vec{p} = \vec{b} + t\vec{a} = (\hat{i} - 3\hat{j} + 2\hat{k}) + t(\hat{i} + \hat{j} + \hat{k}) = (1+t)\hat{i} + (t-3)\hat{j} + (t+2)\hat{k}$.
Given $\vec{p} \cdot \vec{c} = 0$, where $\vec{c} = 3\hat{i} - 2\hat{k}$, we have $((1+t)\hat{i} + (t-3)\hat{j} + (t+2)\hat{k}) \cdot (3\hat{i} - 2\hat{k}) = 0$.
$3(1+t) - 2(t+2) = 0 \implies 3 + 3t - 2t - 4 = 0 \implies t - 1 = 0 \implies t = 1$.
Substituting $t=1$ into $\vec{p}$, we get $\vec{p} = (1+1)\hat{i} + (1-3)\hat{j} + (1+2)\hat{k} = 2\hat{i} - 2\hat{j} + 3\hat{k}$.
Finally, $|\vec{p}| = \sqrt{2^2 + (-2)^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$.

(A) Step $1$: Find the intersection points of $x^2 + y^2 = 9$ and $y^2 = 8x$. Substituting $y^2 = 8x$ into the circle equation: $x^2 + 8x - 9 = 0 \implies (x+9)(x-1) = 0$. Since $x \geq 0$, $x = 1$. Then $y^2 = 8$, so $y = \pm 2\sqrt{2}$.
Step $2$: The area is symmetric about the $x$-axis. Area $= 2 \left[ \int_{0}^{1} \sqrt{8x} \, dx + \int_{1}^{3} \sqrt{9-x^2} \, dx \right]$.
Step $3$: Calculate the first integral: $2 \int_{0}^{1} 2\sqrt{2} x^{1/2} \, dx = 4\sqrt{2} [\frac{2}{3} x^{3/2}]_0^1 = \frac{8\sqrt{2}}{3}$.
Step $4$: Calculate the second integral: $2 [\frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}(\frac{x}{3})]_1^3 = 2 [(\frac{3}{2}(0) + \frac{9}{2}\sin^{-1}(1)) - (\frac{1}{2}\sqrt{8} + \frac{9}{2}\sin^{-1}(\frac{1}{3}))] = 2 [\frac{9\pi}{4} - \sqrt{2} - \frac{9}{2}\sin^{-1}(\frac{1}{3})] = \frac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}(\frac{1}{3})$.
Step $5$: Total Area $= \frac{8\sqrt{2}}{3} + \frac{9\pi}{2} - 2\sqrt{2} - 9\sin^{-1}(\frac{1}{3})$.

(B) Step $1$: Identify the curves. The curve $y = |x| - 1$ represents a $V$-shape with vertex at $(0, -1)$. The curve $y = -|x| + 1$ represents an inverted $V$-shape with vertex at $(0, 1)$.
Step $2$: Find the intersection points. Setting $|x| - 1 = -|x| + 1$, we get $2|x| = 2$, so $|x| = 1$, which means $x = 1$ or $x = -1$. At $x = 1$, $y = 0$. At $x = -1$, $y = 0$. The intersection points are $(1, 0)$ and $(-1, 0)$.
Step $3$: The bounded region is a square (or rhombus) with vertices at $(0, 1), (1, 0), (0, -1),$ and $(-1, 0)$.
Step $4$: The area of a rhombus with diagonals $d_1$ and $d_2$ is $\frac{1}{2} \times d_1 \times d_2$. Here, $d_1$ (vertical) $= 1 - (-1) = 2$ and $d_2$ (horizontal) $= 1 - (-1) = 2$.
Step $5$: Area $= \frac{1}{2} \times 2 \times 2 = 2$ sq. units.

(B) Let $I = \int_{0}^{3} \sqrt{3^2 - x^2} dx$.
Using the standard integral formula $\int \sqrt{a^2 - x^2} dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + C$, where $a = 3$.
Applying the limits from $0$ to $3$:
$I = [\frac{x}{2} \sqrt{9 - x^2} + \frac{9}{2} \sin^{-1}(\frac{x}{3})]_{0}^{3}$
$I = (\frac{3}{2} \sqrt{9 - 9} + \frac{9}{2} \sin^{-1}(\frac{3}{3})) - (\frac{0}{2} \sqrt{9 - 0} + \frac{9}{2} \sin^{-1}(\frac{0}{3}))$
$I = (0 + \frac{9}{2} \sin^{-1}(1)) - (0 + 0)$
$I = \frac{9}{2} \times \frac{\pi}{2} = \frac{9\pi}{4}$.

(A) The shaded region is bounded by $y = \sin x$ from $x = 2\pi$ to $x = \frac{9\pi}{4}$ and by $y = \cos x$ from $x = \frac{9\pi}{4}$ to $x = \frac{5\pi}{2}$.
The intersection point is where $\sin x = \cos x$, which is $\tan x = 1$, so $x = \frac{9\pi}{4}$.
The area $A$ is given by:
$A = \int_{2\pi}^{9\pi/4} \sin x \, dx + \int_{9\pi/4}^{5\pi/2} \cos x \, dx$
$A = [-\cos x]_{2\pi}^{9\pi/4} + [\sin x]_{9\pi/4}^{5\pi/2}$
$A = -(\cos(9\pi/4) - \cos(2\pi)) + (\sin(5\pi/2) - \sin(9\pi/4))$
$A = -(\frac{1}{\sqrt{2}} - 1) + (1 - \frac{1}{\sqrt{2}})$
$A = 1 - \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2}$ sq. units.

(D) The area $A$ is given by $\int_{0}^{4\pi} x |\sin(\frac{x}{2})| dx$.
Since $\sin(\frac{x}{2}) \ge 0$ for $x \in [0, 2\pi]$ and $\sin(\frac{x}{2}) \le 0$ for $x \in [2\pi, 4\pi]$, we have:
$A = \int_{0}^{2\pi} x \sin(\frac{x}{2}) dx + \int_{2\pi}^{4\pi} x (-\sin(\frac{x}{2})) dx$.
Using integration by parts $\int u dv = uv - \int v du$ with $u=x, dv=\sin(\frac{x}{2})dx$:
$\int x \sin(\frac{x}{2}) dx = -2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})$.
Evaluating the first integral: $[-2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})]_{0}^{2\pi} = (-2(2\pi)(-1) + 0) - (0) = 4\pi$.
Evaluating the second integral: $-[-2x \cos(\frac{x}{2}) + 4 \sin(\frac{x}{2})]_{2\pi}^{4\pi} = -[(-2(4\pi)(1) + 0) - (-2(2\pi)(-1) + 0)] = -[-8\pi - 4\pi] = 12\pi$.
Total area $A = 4\pi + 12\pi = 16\pi$ square units.

(C) The given equation is $y = 2\sqrt{1 - x^2}$.
Squaring both sides, we get $y^2 = 4(1 - x^2)$, which implies $x^2 + \frac{y^2}{4} = 1$.
This is the equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a = 1$ and $b = 2$.
The curve $y = 2\sqrt{1 - x^2}$ represents the upper half of the ellipse.
The area of the full ellipse is $A = \pi ab = \pi(1)(2) = 2\pi$.
The area bounded by the curve and the $X$-axis is the area of the upper half of the ellipse.
Area $= \frac{1}{2} \times 2\pi = \pi \text{ sq. units}$.

(B) The region is bounded by $y = x^3$, $y = 8$, and $x = 0$.
Expressing $x$ in terms of $y$, we get $x = y^{1/3}$.
The intersection of $y = x^3$ and $y = 8$ occurs at $x^3 = 8$, which gives $x = 2$.
The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 0$ to $y = 8$:
$A = \int_{0}^{8} y^{1/3} \, dy$
$A = \left[ \frac{y^{4/3}}{4/3} \right]_{0}^{8}$
$A = \frac{3}{4} [8^{4/3} - 0^{4/3}]$
$A = \frac{3}{4} [(2^3)^{4/3}] = \frac{3}{4} [2^4] = \frac{3}{4} \times 16 = 12$ square units.

(C) The area $A$ is given by the integral $\int_{0}^{2} 2^{kx} dx = \frac{2}{\ln 2}$.
Using the formula $\int a^x dx = \frac{a^x}{\ln a}$, we have $\int_{0}^{2} 2^{kx} dx = \left[ \frac{2^{kx}}{k \ln 2} \right]_{0}^{2}$.
Substituting the limits: $\frac{1}{k \ln 2} (2^{2k} - 2^0) = \frac{2}{\ln 2}$.
$\frac{2^{2k} - 1}{k} = 2$.
$2^{2k} - 1 = 2k$.
By inspection, if $k = 1$, $2^2 - 1 = 3 \neq 2(1)$.
If $k = 1/2$, $2^{2(1/2)} - 1 = 2^1 - 1 = 1$, and $2k = 2(1/2) = 1$.
Since $1 = 1$, the value of $k$ is $\frac{1}{2}$.

(A) The equation of the curve is $x^2 = by$, which implies $x = \sqrt{b} \sqrt{y}$ (since it is in the first quadrant).
The area $A$ is given by the integral of $x$ with respect to $y$ from $y = 1$ to $y = 4$:
$A = \int_{1}^{4} \sqrt{b} \sqrt{y} \, dy = 28$
$\sqrt{b} \int_{1}^{4} y^{1/2} \, dy = 28$
$\sqrt{b} \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4} = 28$
$\sqrt{b} \cdot \frac{2}{3} [4^{3/2} - 1^{3/2}] = 28$
$\sqrt{b} \cdot \frac{2}{3} [8 - 1] = 28$
$\sqrt{b} \cdot \frac{2}{3} \cdot 7 = 28$
$\sqrt{b} = \frac{28 \cdot 3}{14} = 2 \cdot 3 = 6$
$b = 6^2 = 36$.

(D) The area $A$ is given by the integral $\int_{3}^{5} |x - 4| \, dx$.
Since $|x - 4| = -(x - 4)$ for $x < 4$ and $|x - 4| = (x - 4)$ for $x \ge 4$, we split the integral:
$A = \int_{3}^{4} -(x - 4) \, dx + \int_{4}^{5} (x - 4) \, dx$.
Evaluating the first part: $\int_{3}^{4} (-x + 4) \, dx = [-\frac{x^2}{2} + 4x]_{3}^{4} = (-8 + 16) - (-4.5 + 12) = 8 - 7.5 = 0.5$.
Evaluating the second part: $\int_{4}^{5} (x - 4) \, dx = [\frac{x^2}{2} - 4x]_{4}^{5} = (12.5 - 20) - (8 - 16) = -7.5 - (-8) = 0.5$.
Total area $A = 0.5 + 0.5 = 1$ sq. unit.

(A) Step $1$: Express $x$ in terms of $y$ from the equation $2x - y + 1 = 0$. We get $2x = y - 1$, so $x = \frac{y - 1}{2}$.
Step $2$: The area $A$ bounded by the curve $x = f(y)$, the $y$-axis, and the lines $y = c$ and $y = d$ is given by $A = \int_{c}^{d} |x| \, dy$.
Step $3$: Substitute the limits $y = -1$ to $y = 3$: $A = \int_{-1}^{3} |\frac{y - 1}{2}| \, dy$.
Step $4$: The expression $\frac{y - 1}{2}$ changes sign at $y = 1$. Thus, $A = \int_{-1}^{1} -(\frac{y - 1}{2}) \, dy + \int_{1}^{3} (\frac{y - 1}{2}) \, dy$.
Step $5$: Calculate the integrals: $\frac{1}{2} [-( \frac{y^2}{2} - y )]_{-1}^{1} + \frac{1}{2} [ \frac{y^2}{2} - y ]_{1}^{3} = \frac{1}{2} [-( (\frac{1}{2} - 1) - (\frac{1}{2} + 1) )] + \frac{1}{2} [ (\frac{9}{2} - 3) - (\frac{1}{2} - 1) ] = \frac{1}{2} [ -(-\frac{1}{2} - \frac{3}{2}) ] + \frac{1}{2} [ \frac{3}{2} - (-\frac{1}{2}) ] = \frac{1}{2} [2] + \frac{1}{2} [2] = 1 + 1 = 2$ square units.

(D) The area $A$ is given by the integral of $|y|$ with respect to $x$ from $x = -2$ to $x = 4$.
$A = \int_{-2}^{4} |x| \, dx$
Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$, we split the integral:
$A = \int_{-2}^{0} (-x) \, dx + \int_{0}^{4} x \, dx$
$A = \left[ -\frac{x^2}{2} \right]_{-2}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{4}$
$A = (0 - (-\frac{(-2)^2}{2})) + (\frac{4^2}{2} - 0)$
$A = (0 - (-2)) + (8 - 0)$
$A = 2 + 8 = 10 \text{ square units}$.

(B) The area $A$ is given by $\int_{-2}^{1} |x^3 + ax| \, dx$.
Since $y = x^3 + ax = x(x^2 + a)$, and $a > 0$, the curve crosses the $x$-axis only at $x = 0$ in the interval $[-2, 1]$.
For $x \in [-2, 0]$, $x^3 + ax \le 0$, so $|x^3 + ax| = -(x^3 + ax)$.
For $x \in [0, 1]$, $x^3 + ax \ge 0$, so $|x^3 + ax| = x^3 + ax$.
Area $= \int_{-2}^{0} -(x^3 + ax) \, dx + \int_{0}^{1} (x^3 + ax) \, dx = \frac{37}{4}$.
Evaluating the integrals: $-[\frac{x^4}{4} + \frac{ax^2}{2}]_{-2}^{0} + [\frac{x^4}{4} + \frac{ax^2}{2}]_{0}^{1} = \frac{37}{4}$.
$-(0 - (4 + 2a)) + (\frac{1}{4} + \frac{a}{2}) = \frac{37}{4}$.
$4 + 2a + \frac{1}{4} + \frac{a}{2} = \frac{37}{4}$.
$\frac{5a}{2} = \frac{37}{4} - \frac{17}{4} = \frac{20}{4} = 5$.
$a = 2$.

(B) The area $A$ bounded by the parabola $y^2 = 4kx$ and the line $x = k$ is given by $A = 2 \int_{0}^{k} \sqrt{4kx} \, dx$.
$A = 2 \cdot 2\sqrt{k} \int_{0}^{k} x^{1/2} \, dx = 4\sqrt{k} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{k}$.
$A = 4\sqrt{k} \cdot \frac{2}{3} k^{3/2} = \frac{8}{3} k^2$.
Given $A = \frac{128}{3}$, so $\frac{8}{3} k^2 = \frac{128}{3}$.
$8k^2 = 128 \implies k^2 = 16 \implies k = 4$ (since $k > 0$).
We need to find $\sin^{-1}(\frac{2}{k}) = \sin^{-1}(\frac{2}{4}) = \sin^{-1}(\frac{1}{2})$.
$\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$.

(A) Step $1$: Find the points of intersection with the $X$-axis by setting $y = 0$.
$2x - x^2 = 0 \implies x(2 - x) = 0$.
So, $x = 0$ and $x = 2$.
Step $2$: The area $A$ is given by the integral $\int_{0}^{2} (2x - x^2) \, dx$.
Step $3$: Evaluate the integral: $\left[ x^2 - \frac{x^3}{3} \right]_{0}^{2}$.
Step $4$: Substitute the limits: $\left( 2^2 - \frac{2^3}{3} \right) - (0 - 0) = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3} \text{ sq. units}$.

(A) Given the curve $y = 4x^2$, we express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.
The area $A$ in the first quadrant bounded by the $Y$-axis and the lines $y = 2$ and $y = 4$ is given by the integral:
$A = \int_{2}^{4} x \, dy = \int_{2}^{4} \frac{\sqrt{y}}{2} \, dy$.
$A = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} [y^{3/2}]_{2}^{4}$.
$A = \frac{1}{3} [4^{3/2} - 2^{3/2}] = \frac{1}{3} [8 - 2\sqrt{2}]$ sq. units.

(D) The given curve is $y = \sqrt{16 - x^2}$, which represents the upper semi-circle of $x^2 + y^2 = 4^2$ with radius $r = 4$.
We need to find the area bounded by $x = 0$ and $x = 4$ above the $X$-axis.
This region represents one-quarter of the circle with radius $r = 4$.
The area of a full circle is $A = \pi r^2 = \pi(4)^2 = 16\pi$.
The area of the required region is $\frac{1}{4} \times 16\pi = 4\pi$ square units.

(A) Let $V$ be the volume and $S$ be the surface area of the spherical raindrop.
$V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$.
The rate of evaporation is the rate of change of volume, $\frac{dV}{dt}$.
Given $\frac{dV}{dt} \propto -S$, so $\frac{dV}{dt} = -k' S$ (where $k' > 0$ is a constant).
Differentiating $V$ with respect to $t$: $\frac{dV}{dt} = \frac{d}{dt} (\frac{4}{3} \pi r^3) = 4 \pi r^2 \frac{dr}{dt}$.
Equating the two expressions for $\frac{dV}{dt}$:
$4 \pi r^2 \frac{dr}{dt} = -k' (4 \pi r^2)$.
Dividing both sides by $4 \pi r^2$:
$\frac{dr}{dt} = -k'$.
Rearranging gives $\frac{dr}{dt} + k' = 0$. Since $k'$ is a positive constant, we can write it as $k$.
Thus, $\frac{dr}{dt} + k = 0$.

(A) Step $1$: Rewrite the differential equation as $(c + dy) dy = (a + bx) dx$.
Step $2$: Integrate both sides: $\int (c + dy) dy = \int (a + bx) dx$.
Step $3$: This gives $cy + \frac{d}{2} y^2 = ax + \frac{b}{2} x^2 + C$.
Step $4$: Rearrange the terms: $\frac{b}{2} x^2 - \frac{d}{2} y^2 - ax + cy + C = 0$.
Step $5$: For this to represent a circle centered at the origin, the coefficients of $x^2$ and $y^2$ must be equal, and the coefficients of $x$ and $y$ must be zero.
Step $6$: Thus, $a = 0$, $c = 0$, and $\frac{b}{2} = -\frac{d}{2}$, which implies $b = -d$ or $b + d = 0$.

(B) Step $1$: Divide the equation by $xy^6$ to get $\frac{1}{y^6} \frac{dy}{dx} + \frac{1}{xy^5} = x^2$.
Step $2$: Let $v = y^{-5}$, then $\frac{dv}{dx} = -5y^{-6} \frac{dy}{dx}$, which implies $\frac{1}{y^6} \frac{dy}{dx} = -\frac{1}{5} \frac{dv}{dx}$.
Step $3$: Substitute into the equation: $-\frac{1}{5} \frac{dv}{dx} + \frac{v}{x} = x^2$, which simplifies to $\frac{dv}{dx} - \frac{5}{x}v = -5x^2$.
Step $4$: The integrating factor is $IF = e^{\int -\frac{5}{x} dx} = e^{-5 \ln x} = x^{-5}$.
Step $5$: The solution is $v \cdot x^{-5} = \int (-5x^2) \cdot x^{-5} dx = \int -5x^{-3} dx = -5 \frac{x^{-2}}{-2} + c = \frac{5}{2}x^{-2} + c$.
Step $6$: Multiplying by $x^5$, we get $v = \frac{5}{2}x^3 + cx^5$, so $y^{-5} = \frac{5}{2}x^3 + cx^5$.

(C) Given differential equation: $2x \frac{dy}{dx} - y = 3$
Rearranging the terms: $2x \frac{dy}{dx} = y + 3$
Separating the variables: $\frac{dy}{y + 3} = \frac{dx}{2x}$
Integrating both sides: $\int \frac{dy}{y + 3} = \int \frac{dx}{2x}$
$\ln|y + 3| = \frac{1}{2} \ln|x| + C$
Multiply by $2$: $2 \ln|y + 3| = \ln|x| + 2C$
$\ln(y + 3)^2 = \ln|x| + C'$
$(y + 3)^2 = kx$, where $k = e^{C'}$
This equation is of the form $(y - k_1)^2 = 4a(x - k_2)$, which represents a family of parabolas.

(A) Given: $(x + y) \frac{dy}{dx} = 1$.
Rewrite as: $\frac{dx}{dy} = x + y$.
This is a linear differential equation of the form $\frac{dx}{dy} - x = y$.
Here, $P = -1$ and $Q = y$.
The integrating factor $IF = e^{\int P dy} = e^{\int -1 dy} = e^{-y}$.
The solution is given by $x \cdot IF = \int (Q \cdot IF) dy + c$.
$x e^{-y} = \int y e^{-y} dy + c$.
Using integration by parts: $\int y e^{-y} dy = y(-e^{-y}) - \int 1(-e^{-y}) dy = -y e^{-y} - e^{-y}$.
So, $x e^{-y} = -y e^{-y} - e^{-y} + c$.
Multiply by $e^y$: $x = -y - 1 + ce^y$.
Rearranging gives: $x + y + 1 = ce^y$.

(C) Step $1$: Rewrite the differential equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
Divide by $x^2$: $\frac{dy}{dx} - \frac{1}{x}y = \frac{1}{x^2}$.
Step $2$: Find the integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x}$.
Step $3$: Multiply the equation by $IF$ and integrate: $\frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = \frac{1}{x^3}$.
This is $\frac{d}{dx} (y \cdot \frac{1}{x}) = x^{-3}$.
Step $4$: Integrate both sides: $y \cdot \frac{1}{x} = \int x^{-3} dx = \frac{x^{-2}}{-2} + c = -\frac{1}{2x^2} + c$.
Step $5$: Multiply by $2x^2$: $2xy = -1 + 2cx^2$, which rearranges to $2xy - 2cx^2 + 1 = 0$.

(A) Given equation: $(x + 2y^3) \frac{dy}{dx} - y = 0$
Rearranging the equation: $(x + 2y^3) = y \frac{dx}{dy}$
$\frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y} + 2y^2$
$\frac{dx}{dy} - \frac{1}{y} x = 2y^2$
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = -\frac{1}{y}$ and $Q = 2y^2$.
Integrating factor $IF = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.
The solution is $x \cdot IF = \int (Q \cdot IF) dy + c$.
$x \cdot \frac{1}{y} = \int (2y^2 \cdot \frac{1}{y}) dy + c$
$\frac{x}{y} = \int 2y dy + c$
$\frac{x}{y} = y^2 + c$
$x = y(y^2 + c) = (c + y^2)y$.

(A) The given differential equation is $(1 + t^2) + (x - e^{\tan^{-1} t}) \frac{dt}{dx} = 0$.
Rearranging the terms, we get $(x - e^{\tan^{-1} t}) \frac{dt}{dx} = -(1 + t^2)$.
Taking the reciprocal, $\frac{dx}{dt} = -\frac{x - e^{\tan^{-1} t}}{1 + t^2} = -\frac{x}{1 + t^2} + \frac{e^{\tan^{-1} t}}{1 + t^2}$.
This can be written as $\frac{dx}{dt} + \frac{1}{1 + t^2} x = \frac{e^{\tan^{-1} t}}{1 + t^2}$.
This is a linear differential equation of the form $\frac{dx}{dt} + P(t)x = Q(t)$, where $P(t) = \frac{1}{1 + t^2}$.
The integrating factor $(IF)$ is given by $e^{\int P(t) dt} = e^{\int \frac{1}{1 + t^2} dt} = e^{\tan^{-1} t}$.

(A) Given $y = e^{-mx}$.
First derivative: $\frac{dy}{dx} = -m e^{-mx}$.
Second derivative: $\frac{d^2y}{dx^2} = m^2 e^{-mx}$.
Substitute these into the differential equation: $m^2 e^{-mx} + 4(-m e^{-mx}) + 3(e^{-mx}) = 0$.
Factor out $e^{-mx}$: $e^{-mx}(m^2 - 4m + 3) = 0$.
Since $e^{-mx} \neq 0$, we have $m^2 - 4m + 3 = 0$.
Factor the quadratic: $(m - 1)(m - 3) = 0$.
Thus, $m = 1$ or $m = 3$.

(B) Given: $\sec y + (x - e^{\sin y}) \frac{dy}{dx} = 0$
Rearranging the equation: $(e^{\sin y} - x) \frac{dy}{dx} = \sec y$
$\frac{dx}{dy} = \frac{e^{\sin y} - x}{\sec y} = (e^{\sin y} - x) \cos y$
$\frac{dx}{dy} + x \cos y = e^{\sin y} \cos y$
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = \cos y$ and $Q = e^{\sin y} \cos y$.
Integrating factor $IF = e^{\int P dy} = e^{\int \cos y dy} = e^{\sin y}$.
The solution is $x \cdot IF = \int (Q \cdot IF) dy + c$.
$x e^{\sin y} = \int (e^{\sin y} \cos y \cdot e^{\sin y}) dy + c$.
Let $u = \sin y$, then $du = \cos y dy$.
$x e^{\sin y} = \int e^{2u} du + c = \frac{e^{2u}}{2} + c$.
$x e^{\sin y} = \frac{e^{2 \sin y}}{2} + c$.

(D) Step $1$: Divide the equation by $x \sin x$ to get the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
$\frac{dy}{dx} + \frac{x \cos x + \sin x}{x \sin x} y = \frac{1}{x}$.
Step $2$: Identify $P(x) = \frac{x \cos x + \sin x}{x \sin x} = \cot x + \frac{1}{x}$.
Step $3$: Find the integrating factor $IF = e^{\int P(x) dx} = e^{\int (\cot x + \frac{1}{x}) dx} = e^{\ln |\sin x| + \ln |x|} = e^{\ln |x \sin x|} = x \sin x$.
Step $4$: The solution is $y \cdot IF = \int Q(x) \cdot IF dx$.
$y(x \sin x) = \int \frac{1}{x} (x \sin x) dx = \int \sin x dx$.
$xy \sin x = -\cos x + c$.
$xy \sin x + \cos x = c$.

(B) Step $1$: Rewrite the differential equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
Divide the given equation $x \frac{dy}{dx} + 2y = x^2 \log x$ by $x$:
$\frac{dy}{dx} + \frac{2}{x}y = x \log x$.
Step $2$: Identify $P(x)$.
Here, $P(x) = \frac{2}{x}$.
Step $3$: Calculate the integrating factor $(IF)$.
$IF = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \log x} = e^{\log x^2} = x^2$.

(D) Step $1$: Rewrite the equation in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
$\frac{dy}{dx} + \frac{6x^2}{1+x^3}y = \frac{1+x^2}{1+x^3}$.
Step $2$: Find the integrating factor $IF = e^{\int P(x) dx} = e^{\int \frac{6x^2}{1+x^3} dx} = e^{2 \ln(1+x^3)} = (1+x^3)^2$.
Step $3$: Multiply by $IF$ and integrate: $y(1+x^3)^2 = \int (1+x^2)(1+x^3) dx = \int (1 + x^2 + x^3 + x^5) dx = x + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^6}{6} + c$.
Step $4$: Comparing with $y = \frac{1}{(1+x^3)^2} [x + \frac{x^3}{3} + \frac{x^4}{4} + \frac{x^6}{6} + c]$, we get $s=2, p=3, q=4, r=6$.
Step $5$: The $LCM$ of $2, 3, 4, 6$ is $12$.

(D) The integrating factor $(IF)$ of the linear differential equation $\frac{dy}{dx} + Py = Q$ is given by $IF = e^{\int P \, dx}$.
Given $IF = \tan x$, we have $e^{\int P \, dx} = \tan x$.
Taking the natural logarithm on both sides: $\int P \, dx = \ln(\tan x)$.
Differentiating both sides with respect to $x$: $P = \frac{d}{dx}(\ln(\tan x))$.
Using the chain rule: $P = \frac{1}{\tan x} \cdot \sec^2 x$.
$P = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x}$.
Multiply numerator and denominator by $2$: $P = \frac{2}{2 \sin x \cos x} = \frac{2}{\sin 2x} = 2 \csc 2x$.

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = x^2 + 5$.
Step $1$: Find the integrating factor $(IF)$.
$IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.
Step $2$: Multiply the differential equation by $IF$ and integrate.
$x \frac{dy}{dx} + y = x(x^2 + 5) = x^3 + 5x$.
$\frac{d}{dx}(xy) = x^3 + 5x$.
Integrating both sides with respect to $x$:
$xy = \int (x^3 + 5x) dx = \frac{x^4}{4} + \frac{5x^2}{2} + c$.

(B) Given $\frac{dy}{dx} = \frac{x + y}{x - y}$.
Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the equation: $v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$.
$x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$.
Separating variables: $\int \frac{1 - v}{1 + v^2} dv = \int \frac{1}{x} dx$.
$\int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \ln|x| + C$.
$\tan^{-1}(v) - \frac{1}{2} \ln(1 + v^2) = \ln|x| + C$.
$\tan^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(1 + \frac{y^2}{x^2}) + C = \ln|x| + \frac{1}{2} \ln(\frac{x^2 + y^2}{x^2}) + C$.
$\tan^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(x^2 + y^2) - \ln|x| + C = \frac{1}{2} \ln(x^2 + y^2) + C$.
$2\tan^{-1}(\frac{y}{x}) = \ln(x^2 + y^2) + 2C$.
Taking exponential on both sides: $e^{2\tan^{-1}(\frac{y}{x})} = e^{\ln(x^2 + y^2)} \cdot e^{2C} = k(x^2 + y^2)$.
This matches the form $c(x^2 + y^2)^{\frac{1}{2}} = e^{\tan^{-1}(\frac{y}{x})}$.

(D) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x + y}$.
Taking the reciprocal: $\frac{dx}{dy} = \frac{x + y}{y} = \frac{x}{y} + 1$.
This is a linear differential equation of the form $\frac{dx}{dy} - \frac{1}{y}x = 1$.
The integrating factor $IF = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.
Multiplying by $IF$: $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = \frac{1}{y}$.
Integrating both sides with respect to $y$: $\frac{x}{y} = \int \frac{1}{y} dy = \ln |y| + C$.
Since the curve passes through $(0, 1)$, substitute $x=0, y=1$: $\frac{0}{1} = \ln(1) + C \implies C = 0$.
Thus, $\frac{x}{y} = \ln y$, which implies $y = e^{\frac{x}{y}}$.