MHT CET 2020 Mathematics Question Paper with Answer and Solution

(B) Using the sine rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,which implies $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Given the condition $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,we substitute the values of $a, b, c$:
$\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$.
This simplifies to $\cot A = \cot B = \cot C$.
Since $A, B, C$ are angles of a triangle,$\cot A = \cot B = \cot C$ implies $A = B = C$.
Therefore,the triangle is an equilateral triangle.

(A) Given that $\tan A, \tan B, \tan C$ are in $H.P.$
$\frac{2}{\tan B} = \frac{1}{\tan A} + \frac{1}{\tan C}$
$\frac{2 \cos B}{\sin B} = \frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}$
Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.
Also,$\cos B = \frac{a^{2}+c^{2}-b^{2}}{2ac}, \cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}, \cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$.
Substituting these values:
$2 \left( \frac{a^{2}+c^{2}-b^{2}}{2ac} \right) \cdot \frac{2R}{b} = \frac{b^{2}+c^{2}-a^{2}}{2bc} \cdot \frac{2R}{a} + \frac{a^{2}+b^{2}-c^{2}}{2ab} \cdot \frac{2R}{c}$
Multiplying both sides by $\frac{abc}{2R}$:
$2(a^{2}+c^{2}-b^{2}) = (b^{2}+c^{2}-a^{2}) + (a^{2}+b^{2}-c^{2})$
$2a^{2} + 2c^{2} - 2b^{2} = 2b^{2}$
$2a^{2} + 2c^{2} = 4b^{2}$
$a^{2} + c^{2} = 2b^{2}$
Thus,$a^{2}, b^{2}, c^{2}$ are in $A.P.$

(D) Given $a=\sqrt{3}+1$,$b=\sqrt{3}-1$,$m \angle C=60^{\circ}$.
Using the tangent rule (Napier's Analogy): $\tan \left( \frac{A-B}{2} \right) = \frac{a-b}{a+b} \cot \left( \frac{C}{2} \right)$.
Substitute the values: $\frac{a-b}{a+b} = \frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$.
$\tan \left( \frac{A-B}{2} \right) = \frac{1}{\sqrt{3}} \cot \left( \frac{60^{\circ}}{2} \right) = \frac{1}{\sqrt{3}} \cot(30^{\circ}) = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$.
$\frac{A-B}{2} = 45^{\circ} \Rightarrow A-B = 90^{\circ}$.

(B) We know from the Sine Rule that $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$.
Given that $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.
Dividing the Sine Rule equation by the given equation,we get $\tan A = \tan B = \tan C$.
Since $A, B, C$ are angles of a triangle,this implies $A = B = C = 60^{\circ}$,so the triangle is equilateral.
The area of an equilateral triangle with side $a$ is given by $\text{Area} = \frac{\sqrt{3}}{4} a^2$.
Given $a = \sqrt{6}$,the area is $\frac{\sqrt{3}}{4} (\sqrt{6})^2 = \frac{\sqrt{3}}{4} \times 6 = \frac{3 \sqrt{3}}{2}$ sq. units.

(C) Given the equation: $2 \cos C = \sin B \cdot \operatorname{cosec} A$
Since $\operatorname{cosec} A = \frac{1}{\sin A}$,we have $2 \cos C = \frac{\sin B}{\sin A}$.
Using the Sine Rule,$\frac{\sin B}{\sin A} = \frac{b}{a}$.
Thus,$2 \cos C = \frac{b}{a}$.
Using the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting this into the equation: $2 \left( \frac{a^2 + b^2 - c^2}{2ab} \right) = \frac{b}{a}$.
$\frac{a^2 + b^2 - c^2}{ab} = \frac{b}{a}$.
Multiplying both sides by $ab$,we get $a^2 + b^2 - c^2 = b^2$.
$a^2 - c^2 = 0 \Rightarrow a^2 = c^2$.
Since $a$ and $c$ are side lengths,$a = c$.

(D) Given that $A, B, C$ are in $A$.$P$.,we have $2B = A + C$.
Since $A + B + C = 180^{\circ}$,we substitute $A + C = 2B$ to get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
Using the Sine Rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,we have $\sin C = \frac{c}{b} \sin B$.
Given $b:c = \sqrt{3}:\sqrt{2}$,we have $\frac{c}{b} = \frac{\sqrt{2}}{\sqrt{3}}$.
Thus,$\sin C = \frac{\sqrt{2}}{\sqrt{3}} \times \sin 60^{\circ} = \frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$C = 45^{\circ}$.
Finally,$A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}$.

(B) Given that in $\triangle ABC, \angle C = 90^{\circ}$,so $A+B = 90^{\circ} \Rightarrow B = 90^{\circ}-A$.
Using the sine rule,$a = k \sin A$ and $b = k \sin B$.
Substituting these into the expression:
$\frac{a^{2}+b^{2}}{a^{2}-b^{2}} \sin (A-B) = \frac{\sin^{2} A + \sin^{2} B}{\sin^{2} A - \sin^{2} B} \sin (A-B)$.
Since $B = 90^{\circ}-A$,$\sin B = \cos A$ and $\cos B = \sin A$.
Thus,$\sin^{2} A + \sin^{2} B = \sin^{2} A + \cos^{2} A = 1$.
And $\sin^{2} A - \sin^{2} B = \sin^{2} A - \cos^{2} A = -\cos 2A$.
Also,$\sin (A-B) = \sin (A - (90^{\circ}-A)) = \sin (2A - 90^{\circ}) = -\cos 2A$.
Substituting these values:
$\frac{1}{-\cos 2A} \cdot (-\cos 2A) = 1$.

(D) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting these into the given expression:
$\frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc} = \frac{k+7}{30}$
$\frac{a^2+b^2+c^2}{2abc} = \frac{k+7}{30}$
Given $a=2, b=3, c=5$,we have $a^2+b^2+c^2 = 4+9+25 = 38$ and $2abc = 2 \times 2 \times 3 \times 5 = 60$.
So,$\frac{38}{60} = \frac{k+7}{30}$
$\frac{38}{60} = \frac{2(k+7)}{60}$
$38 = 2k + 14$
$2k = 24 \Rightarrow k = 12$.

(D) Using the cosine rule,we have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$.
Thus,$b^2 + c^2 - a^2 = 2bc \cos A$ and $a^2 + c^2 - b^2 = 2ac \cos B$.
Substituting these into the given expression:
$\frac{c^2 - a^2 + b^2}{a^2 - b^2 + c^2} = \frac{2bc \cos A}{2ac \cos B} = \frac{b \cos A}{a \cos B}$.
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = 2R$,so $a = 2R \sin A$ and $b = 2R \sin B$.
Substituting these values:
$= \frac{(2R \sin B) \cos A}{(2R \sin A) \cos B} = \frac{\sin B \cos A}{\sin A \cos B} = \frac{\tan B}{\tan A}$.

(A) Using the Law of Cosines: $\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}$
Substituting the given values: $\cos 60^{\circ} = \frac{3^{2}+c^{2}-4^{2}}{2(3)(c)}$
$\frac{1}{2} = \frac{9+c^{2}-16}{6c}$
$\frac{1}{2} = \frac{c^{2}-7}{6c}$
$3c = c^{2}-7$
$c^{2}-3c-7 = 0$

(C) We are given the expression $\frac{\cos A-\cos C}{a-c}+\frac{\cos B}{b}$.
By taking the common denominator,we get $\frac{b(\cos A-\cos C) + (a-c)\cos B}{b(a-c)}$.
Expanding the numerator,we have $\frac{b \cos A - b \cos C + a \cos B - c \cos B}{b(a-c)}$.
Rearranging the terms,we get $\frac{(a \cos B + b \cos A) - (b \cos C + c \cos B)}{b(a-c)}$.
Using the projection formula $c = a \cos B + b \cos A$ and $a = b \cos C + c \cos B$,the expression becomes $\frac{c - a}{b(a-c)}$.
Since $c - a = -(a - c)$,the expression simplifies to $\frac{-(a - c)}{b(a - c)} = \frac{-1}{b}$.

(D) Given: $\text{Area} = 10\sqrt{3} \text{ cm}^2$,$\angle B = 60^{\circ}$,and $a+b+c = 20 \text{ cm}$.
Using the area formula: $\text{Area} = \frac{1}{2}ac \sin B$.
$10\sqrt{3} = \frac{1}{2}ac \sin 60^{\circ} \Rightarrow 10\sqrt{3} = \frac{1}{2}ac \left(\frac{\sqrt{3}}{2}\right)$.
$10\sqrt{3} = \frac{ac\sqrt{3}}{4} \Rightarrow ac = 40$.
Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos B$.
$b^2 = (a+c)^2 - 2ac - 2ac \cos 60^{\circ}$.
Since $a+c = 20-b$,we have $b^2 = (20-b)^2 - 2(40) - 2(40)(0.5)$.
$b^2 = 400 + b^2 - 40b - 80 - 40$.
$0 = 280 - 40b$.
$40b = 280 \Rightarrow b = 7 \text{ cm}$.
Thus,$\ell(AC) = b = 7 \text{ cm}$.

(A) Given $b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}=\frac{3 a}{2}$.
Using the identity $\cos ^{2} \theta = \frac{1+\cos 2\theta}{2}$,we have:
$b \left( \frac{1+\cos C}{2} \right) + c \left( \frac{1+\cos B}{2} \right) = \frac{3 a}{2}$.
Multiplying by $2$ on both sides:
$b(1+\cos C) + c(1+\cos B) = 3a$.
$b + b \cos C + c + c \cos B = 3a$.
Rearranging terms:
$(b \cos C + c \cos B) + b + c = 3a$.
By the projection rule,$b \cos C + c \cos B = a$,so:
$a + b + c = 3a$.
$b + c = 2a$.
This implies that $b, a, c$ are in $A$.$P$.

(C) We know the half-angle formulas for a triangle:
$\sin^2 \frac{C}{2} = \frac{(s-a)(s-b)}{ab}$ and $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$.
Given the condition $(s-a)(s-b) = s(s-c)$,we can substitute these into the expressions:
$ab \sin^2 \frac{C}{2} = ab \cos^2 \frac{C}{2}$.
Dividing both sides by $ab \cos^2 \frac{C}{2}$ (assuming $ab \neq 0$ and $\cos \frac{C}{2} \neq 0$):
$\tan^2 \frac{C}{2} = 1$.
Since $\frac{C}{2}$ must be an acute angle in a triangle,$\tan \frac{C}{2} = 1$ implies $\frac{C}{2} = 45^{\circ}$.
Therefore,$C = 90^{\circ}$.
Thus,$\triangle ABC$ is a right-angled triangle.

(B) The given equation is of the form $Ax^{2} + 2Hxy + By^{2} = 0$,where $A = \sec^{2} \theta - \sin^{2} \theta$,$2H = -2 \tan \theta$,and $B = \sin^{2} \theta$.
For a pair of lines $Ax^{2} + 2Hxy + By^{2} = 0$,the sum of slopes is $m_{1} + m_{2} = -\frac{2H}{B}$ and the product of slopes is $m_{1}m_{2} = \frac{A}{B}$.
Here,$m_{1} + m_{2} = \frac{2 \tan \theta}{\sin^{2} \theta}$ and $m_{1}m_{2} = \frac{\sec^{2} \theta - \sin^{2} \theta}{\sin^{2} \theta}$.
We know that $|m_{1} - m_{2}| = \sqrt{(m_{1} + m_{2})^{2} - 4m_{1}m_{2}}$.
Substituting the values:
$|m_{1} - m_{2}| = \sqrt{\left(\frac{2 \tan \theta}{\sin^{2} \theta}\right)^{2} - 4\left(\frac{\sec^{2} \theta - \sin^{2} \theta}{\sin^{2} \theta}\right)}$
$|m_{1} - m_{2}| = \sqrt{\frac{4 \tan^{2} \theta - 4(\sec^{2} \theta - \sin^{2} \theta) \sin^{2} \theta}{\sin^{4} \theta}}$
Using $\tan^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta}$ and $\sec^{2} \theta = \frac{1}{\cos^{2} \theta}$:
$|m_{1} - m_{2}| = \sqrt{\frac{4 \frac{\sin^{2} \theta}{\cos^{2} \theta} - 4(\frac{1}{\cos^{2} \theta} - \sin^{2} \theta) \sin^{2} \theta}{\sin^{4} \theta}} = \sqrt{\frac{4 \frac{\sin^{2} \theta}{\cos^{2} \theta} - 4 \frac{\sin^{2} \theta}{\cos^{2} \theta} + 4 \sin^{4} \theta}{\sin^{4} \theta}} = \sqrt{\frac{4 \sin^{4} \theta}{\sin^{4} \theta}} = \sqrt{4} = 2$.

(D) Let the roots of the quadratic equation be $\alpha$ and $\beta$.
Given that the Arithmetic Mean ($A$.$M$.) of the roots is $p$,we have $\frac{\alpha+\beta}{2} = p$,which implies $\alpha+\beta = 2p$.
Given that the Geometric Mean ($G$.$M$.) of the roots is $q$,we have $\sqrt{\alpha\beta} = q$,which implies $\alpha\beta = q^{2}$.
$A$ quadratic equation with roots $\alpha$ and $\beta$ is given by $x^{2} - (\text{sum of roots})x + (\text{product of roots}) = 0$.
Substituting the values,we get $x^{2} - (2p)x + q^{2} = 0$,which simplifies to $x^{2} - 2px + q^{2} = 0$.

(C) Let the roots of the quadratic equation be $a$ and $b$.
Given that the arithmetic mean of $a$ and $b$ is $34$,we have $\frac{a+b}{2} = 34$,which implies $a+b = 68$.
Given that the geometric mean of $a$ and $b$ is $16$,we have $\sqrt{ab} = 16$,which implies $ab = 16^{2} = 256$.
$A$ quadratic equation with roots $a$ and $b$ is given by $x^{2} - (a+b)x + ab = 0$.
Substituting the values,we get $x^{2} - 68x + 256 = 0$.

(D) Given $f(x) = ax^{2} + bx + 2$.
For $f(1) = 4$:
$a(1)^{2} + b(1) + 2 = 4 \implies a + b = 2$ ... $(1)$
For $f(3) = 38$:
$a(3)^{2} + b(3) + 2 = 38 \implies 9a + 3b = 36 \implies 3a + b = 12$ ... $(2)$
Subtracting $(1)$ from $(2)$:
$(3a + b) - (a + b) = 12 - 2
2a = 10 \implies a = 5$
Substituting $a = 5$ in $(1)$:
$5 + b = 2 \implies b = -3$
Therefore,$a - b = 5 - (-3) = 5 + 3 = 8$.

(A) Given the sum of the first $n$ terms $S_{n} = \frac{5^{n} - 2^{n}}{2^{n}} = (\frac{5}{2})^{n} - 1$.
The $n^{th}$ term $T_{n}$ is given by $T_{n} = S_{n} - S_{n-1}$ for $n > 1$.
For $n = 4$,$T_{4} = S_{4} - S_{3}$.
$S_{4} = (\frac{5}{2})^{4} - 1 = \frac{625}{16} - 1 = \frac{625 - 16}{16} = \frac{609}{16}$.
$S_{3} = (\frac{5}{2})^{3} - 1 = \frac{125}{8} - 1 = \frac{125 - 8}{8} = \frac{117}{8} = \frac{234}{16}$.
$T_{4} = \frac{609}{16} - \frac{234}{16} = \frac{375}{16}$.

(A) Let the first term be $a$ and the common difference be $d$ for the given $AP$.
The $n^{th}$ term of an $AP$ is given by $a_n = a + (n-1)d$.
Given that $9 \times a_9 = 13 \times a_{13}$.
Substituting the formula for the terms:
$9[a + (9-1)d] = 13[a + (13-1)d]$
$9[a + 8d] = 13[a + 12d]$
$9a + 72d = 13a + 156d$
Rearranging the terms:
$13a - 9a + 156d - 72d = 0$
$4a + 84d = 0$
Dividing by $4$:
$a + 21d = 0$
Since the $22^{nd}$ term $a_{22} = a + (22-1)d = a + 21d$,
Therefore,$a_{22} = 0$.

(C) Given the sum of the first $n$ terms is $s_{n} = 7(3^{n} - 1)$.
We know that the $n^{th}$ term $t_{n} = s_{n} - s_{n-1}$ for $n > 1$.
$s_{n-1} = 7(3^{n-1} - 1)$.
$t_{n} = 7(3^{n} - 1) - 7(3^{n-1} - 1)$.
$t_{n} = 7(3^{n} - 1 - 3^{n-1} + 1)$.
$t_{n} = 7(3^{n} - 3^{n-1})$.
$t_{n} = 7 \cdot 3^{n-1}(3 - 1)$.
$t_{n} = 7 \cdot 3^{n-1} \cdot 2$.
$t_{n} = 14 \cdot 3^{n-1}$.

(C) Let $x = 1.\overline{41} = 1.414141...$ $(i)$
Multiply both sides by $100$ to shift the decimal point:
$100x = 141.414141...$ (ii)
Subtract equation $(i)$ from equation (ii):
$100x - x = 141.414141... - 1.414141...$
$99x = 140$
$x = \frac{140}{99}$
Therefore,the rational form is $\frac{140}{99}$.

(C) Let the first four terms of the $G.P.$ be $a, ar, ar^2, ar^3$.
Given,common ratio $r = 3$ and the sum of the first four terms $S_4 = 160$.
The formula for the sum of the first $n$ terms is $S_n = \frac{a(r^n - 1)}{r - 1}$.
Substituting the values: $160 = \frac{a(3^4 - 1)}{3 - 1}$.
$160 = \frac{a(81 - 1)}{2} = \frac{a(80)}{2} = 40a$.
Thus,$a = \frac{160}{40} = 4$.
The $4^{th}$ term is $T_4 = ar^3$.
$T_4 = 4 \times (3)^3 = 4 \times 27 = 108$.

(C) In a harmonic progression $(HP)$,the reciprocals of the terms form an arithmetic progression $(AP)$.
Let the corresponding $AP$ have the first term $A$ and common difference $D$.
Given $t_{7} = \frac{1}{10} \Rightarrow T_{7} = 10$,where $T_{n}$ is the $n^{\text{th}}$ term of the $AP$.
Given $t_{12} = \frac{1}{25} \Rightarrow T_{12} = 25$.
Using the formula $T_{n} = A + (n-1)D$:
$A + 6D = 10$ (Equation $1$)
$A + 11D = 25$ (Equation $2$)
Subtracting Equation $1$ from Equation $2$: $5D = 15 \Rightarrow D = 3$.
Substituting $D = 3$ into Equation $1$: $A + 6(3) = 10$ $\Rightarrow A + 18 = 10$ $\Rightarrow A = -8$.
Now,find the $20^{\text{th}}$ term of the $AP$: $T_{20} = A + 19D = -8 + 19(3) = -8 + 57 = 49$.
Therefore,the $20^{\text{th}}$ term of the $HP$ is $t_{20} = \frac{1}{T_{20}} = \frac{1}{49}$.

(A) Given that $\frac{1}{4}, a, b, \frac{1}{19}$ are in $H.P.$
Therefore,their reciprocals $4, \frac{1}{a}, \frac{1}{b}, 19$ are in $A.P.$
Let the $A.P.$ be $4, 4+d, 4+2d, 4+3d$.
Here,$4+3d = 19 \implies 3d = 15 \implies d = 5$.
Thus,the terms are $4, 4+5, 4+10, 19$,which are $4, 9, 14, 19$.
Comparing the terms,$\frac{1}{a} = 9 \implies a = \frac{1}{9}$ and $\frac{1}{b} = 14 \implies b = \frac{1}{14}$.

(A) The sum of the first $n$ even natural numbers is given by $S_e = n(n+1)$.
The sum of the first $n$ odd natural numbers is given by $S_o = n^2$.
Given the ratio: $\frac{n(n+1)}{n^2} = \frac{37}{36}$.
Simplifying the expression: $\frac{n+1}{n} = \frac{37}{36}$.
By cross-multiplication: $36(n+1) = 37n$.
$36n + 36 = 37n$.
$37n - 36n = 36$.
Therefore,$n = 36$.

(B) We need to find the sum $S = 5^{2} + 6^{2} + 7^{2} + \ldots + 20^{2}$.
This can be expressed as the difference of two sums of squares:
$S = \sum_{k=1}^{20} k^{2} - \sum_{k=1}^{4} k^{2}$.
Using the formula $\sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$:
For $n=20$: $\sum_{k=1}^{20} k^{2} = \frac{20(21)(41)}{6} = 10 \times 7 \times 41 = 2870$.
For $n=4$: $\sum_{k=1}^{4} k^{2} = \frac{4(5)(9)}{6} = 2 \times 5 \times 3 = 30$.
Therefore,$S = 2870 - 30 = 2840$.

(B) The $n^{th}$ term of the series is given by $T_n = \frac{\sum_{k=1}^{n} k^2}{n+1}$.
Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:
$T_n = \frac{n(n+1)(2n+1)}{6(n+1)} = \frac{n(2n+1)}{6} = \frac{2n^2 + n}{6}$.
We need to find the sum of the first $8$ terms,$S_8 = \sum_{n=1}^{8} T_n = \sum_{n=1}^{8} \frac{2n^2 + n}{6}$.
$S_8 = \frac{1}{6} \left[ 2 \sum_{n=1}^{8} n^2 + \sum_{n=1}^{8} n \right]$.
Using $\sum_{n=1}^{8} n^2 = \frac{8(9)(17)}{6} = 204$ and $\sum_{n=1}^{8} n = \frac{8(9)}{2} = 36$.
$S_8 = \frac{1}{6} [2(204) + 36] = \frac{1}{6} [408 + 36] = \frac{444}{6} = 74$.

(D) The $n^{th}$ term of the series is given by $T_n = n(2n+1)^2$.
Expanding this,we get $T_n = n(4n^2 + 4n + 1) = 4n^3 + 4n^2 + n$.
The sum of $10$ terms is $S_{10} = \sum_{n=1}^{10} (4n^3 + 4n^2 + n)$.
Using the standard summation formulas:
$\sum_{n=1}^{10} n^3 = \left(\frac{10 \times 11}{2}\right)^2 = 55^2 = 3025$.
$\sum_{n=1}^{10} n^2 = \frac{10 \times 11 \times 21}{6} = 385$.
$\sum_{n=1}^{10} n = \frac{10 \times 11}{2} = 55$.
Therefore,$S_{10} = 4(3025) + 4(385) + 55$.
$S_{10} = 12100 + 1540 + 55 = 13695$.

(B) Given that $n(X)=200, n(A)=90, n(B)=80$ and $n(A' \cap B')=40$.
By De Morgan's Law,$n(A' \cap B') = n((A \cup B)') = n(X) - n(A \cup B)$.
So,$40 = 200 - n(A \cup B)$,which implies $n(A \cup B) = 160$.
We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
$160 = 90 + 80 - n(A \cap B) \implies 160 = 170 - n(A \cap B) \implies n(A \cap B) = 10$.
Now,$n(A \cap B') = n(A) - n(A \cap B) = 90 - 10 = 80$.

(C) $A \cap B = \{4\}$
$A \cup B = \{2, 3, 4, 5\}$
Therefore,the Cartesian product is:
$(A \cap B) \times (A \cup B) = \{4\} \times \{2, 3, 4, 5\}$
$= \{(4, 2), (4, 3), (4, 4), (4, 5)\}$

(C) The set $A$ consists of prime numbers between $0$ and $9$.
$A = \{2, 3, 5, 7\}$.
The number of elements in set $A$ is $n(A) = 4$.
The number of elements in the power set of $A$ is given by $2^{n(A)}$.
$2^{4} = 16$.

(A) The relation $R$ is defined from $A$ to $B$ as $R = \{(x, y) : x \in A, y \in B, x > y\}$.
We check each element $x \in A$ against $y \in B$ where $x > y$:
For $x = 2$,$y = 1$ $(2 > 1)$,so $(2, 1) \in R$.
For $x = 3$,$y = 1$ $(3 > 1)$,so $(3, 1) \in R$.
For $x = 4$,$y = 1$ $(4 > 1)$ and $y = 1$ is not the only one,$y = 1$ is valid. Also $4 > 1$,so $(4, 1) \in R$.
For $x = 5$,$y = 1$ $(5 > 1)$ and $y = 4$ $(5 > 4)$,so $(5, 1) \in R$ and $(5, 4) \in R$.
The set of all ordered pairs in $R$ is $\{(2, 1), (3, 1), (4, 1), (5, 1), (5, 4)\}$.
The range is the set of all second elements in the ordered pairs of $R$.
Range $R = \{1, 4\}$.

(A) Given sets are $A = \{x, y, z\}$ and $B = \{1, 2\}$.
Number of elements in $A$ is $n(A) = 3$.
Number of elements in $B$ is $n(B) = 2$.
The Cartesian product $A \times B$ contains $n(A) \times n(B) = 3 \times 2 = 6$ elements.
$A$ relation from $A$ to $B$ is a subset of $A \times B$.
The total number of subsets of a set with $n$ elements is $2^n$.
Therefore,the total number of relations from $A$ to $B$ is $2^6 = 64$.

(D) Given polar coordinates are $P(r, \theta) = \left(\frac{1}{2}, 120^{\circ}\right)$,where $r = \frac{1}{2}$ and $\theta = 120^{\circ}$.
We use the conversion formulas $x = r \cos \theta$ and $y = r \sin \theta$.
For $x$: $x = \frac{1}{2} \cos 120^{\circ} = \frac{1}{2} \left(-\frac{1}{2}\right) = -\frac{1}{4}$.
For $y$: $y = \frac{1}{2} \sin 120^{\circ} = \frac{1}{2} \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}$.
Therefore,the Cartesian coordinates are $\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$.

(C) The vertices are $A(0,4,0)$,$B(0,0,3)$,and $C(0,4,3)$.
Calculate the side lengths:
$a = BC = \sqrt{(0-0)^2 + (4-0)^2 + (3-3)^2} = \sqrt{0+16+0} = 4$.
$b = AC = \sqrt{(0-0)^2 + (4-4)^2 + (3-0)^2} = \sqrt{0+0+9} = 3$.
$c = AB = \sqrt{(0-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{0+16+9} = 5$.
The incentre $I(x, y, z)$ is given by the formula:
$I = \left( \frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}, \frac{az_A + bz_B + cz_C}{a+b+c} \right)$.
Substituting the values:
$x = \frac{4(0) + 3(0) + 5(0)}{4+3+5} = 0$.
$y = \frac{4(4) + 3(0) + 5(4)}{4+3+5} = \frac{16+0+20}{12} = \frac{36}{12} = 3$.
$z = \frac{4(0) + 3(3) + 5(3)}{4+3+5} = \frac{0+9+15}{12} = \frac{24}{12} = 2$.
Thus,the incentre is $(0,3,2)$.

(D) The centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Given that the origin $(0, 0, 0)$ is the centroid:
For the $x$-coordinate: $\frac{2+q-5}{3} = 0$ $\Rightarrow q-3 = 0$ $\Rightarrow q = 3$.
For the $y$-coordinate: $\frac{p-2+1}{3} = 0$ $\Rightarrow p-1 = 0$ $\Rightarrow p = 1$.
For the $z$-coordinate: $\frac{-3+5+r}{3} = 0$ $\Rightarrow r+2 = 0$ $\Rightarrow r = -2$.
Thus,$p=1, q=3, r=-2$.

(A) Since the points $A(5, k)$,$B(-3, 1)$,and $C(-7, -2)$ are collinear,the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{1 - k}{-3 - 5} = \frac{1 - k}{-8}$.
Slope of $BC = \frac{-2 - 1}{-7 - (-3)} = \frac{-3}{-4} = \frac{3}{4}$.
Equating the slopes: $\frac{1 - k}{-8} = \frac{3}{4}$.
Multiplying both sides by $-8$: $1 - k = \frac{3}{4} \times (-8)$.
$1 - k = -6$.
$k = 1 + 6 = 7$.

(A) We know that $r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}$.
Since the point $(-2, -2)$ lies in the $III$ quadrant,the angle $\theta$ is given by $\tan \theta = \frac{y}{x} = \frac{-2}{-2} = 1$.
Since the point is in the $III$ quadrant,$\theta = \pi + \tan^{-1}(1) = \pi + \frac{\pi}{4} = \frac{5 \pi}{4}$.
Therefore,the polar coordinates $(r, \theta)$ are $(2 \sqrt{2}, \frac{5 \pi}{4})$.

(A) To check if the points are collinear,we can check the slopes of the line segments formed by these points.
Slope of $AB = \frac{0 - (-b)}{0 - (-a)} = \frac{b}{a}$.
Slope of $BC = \frac{b - 0}{a - 0} = \frac{b}{a}$.
Since the slope of $AB$ is equal to the slope of $BC$,the points $A, B,$ and $C$ are collinear.
Now,check the slope of $CD = \frac{ab - b}{a^2 - a} = \frac{b(a - 1)}{a(a - 1)} = \frac{b}{a}$ (for $a \neq 1, a \neq 0$).
Since the slope of $BC$ is equal to the slope of $CD$,the points $B, C,$ and $D$ are also collinear.
Since all points lie on the same line with slope $\frac{b}{a}$,the points $A, B, C,$ and $D$ are collinear.

(C) The equation of the line passing through points $(x_1, y_1) = (1, 4)$ and $(x_2, y_2) = (-5, 1)$ is given by the two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$.
Substituting the values: $\frac{y - 4}{1 - 4} = \frac{x - 1}{-5 - 1}$.
$\frac{y - 4}{-3} = \frac{x - 1}{-6}$.
$2(y - 4) = x - 1$ $\Rightarrow 2y - 8 = x - 1$ $\Rightarrow x - 2y + 7 = 0$ ...$(1)$.
We are given the second line: $4x + 3y - 5 = 0$ ...$(2)$.
From $(1)$,$x = 2y - 7$. Substituting this into $(2)$:
$4(2y - 7) + 3y - 5 = 0$.
$8y - 28 + 3y - 5 = 0$.
$11y - 33 = 0 \Rightarrow y = 3$.
Substituting $y = 3$ into $x = 2y - 7$: $x = 2(3) - 7 = 6 - 7 = -1$.
Thus,the point of intersection is $(-1, 3)$.

(B) Let $A \equiv (a, 0)$ and $B \equiv (0, b)$.
Let $P \equiv (5, 6)$ be the point that divides the line segment $AB$ internally in the ratio $3: 1$.
Using the section formula,the coordinates of $P$ are given by:
$5 = \frac{3 \times 0 + 1 \times a}{3 + 1}$ $\Rightarrow 5 = \frac{a}{4}$ $\Rightarrow a = 20$.
$6 = \frac{3 \times b + 1 \times 0}{3 + 1}$ $\Rightarrow 6 = \frac{3b}{4}$ $\Rightarrow 3b = 24$ $\Rightarrow b = 8$.
Thus,the $X$-intercept is $20$ and the $Y$-intercept is $8$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values,we get $\frac{x}{20} + \frac{y}{8} = 1$.
Multiplying by $40$,we get $2x + 5y = 40$.

(C) Let $a$ and $b$ be the intercepts made by the line on the axes.
We are given $a + b = 8$ and $ab = 15$.
Solving the quadratic equation $t^2 - 8t + 15 = 0$,we get $(t - 3)(t - 5) = 0$,so $(a, b) = (3, 5)$ or $(5, 3)$.
The intercept form of the equation of a line is $\frac{x}{a} + \frac{y}{b} = 1$.
Case $1$: When $a = 3$ and $b = 5$,the equation is $\frac{x}{3} + \frac{y}{5} = 1$,which simplifies to $5x + 3y - 15 = 0$.
Case $2$: When $a = 5$ and $b = 3$,the equation is $\frac{x}{5} + \frac{y}{3} = 1$,which simplifies to $3x + 5y - 15 = 0$.
Thus,the equations are $5x + 3y - 15 = 0$ and $3x + 5y - 15 = 0$.

(B) Given lines are $L_1: x - 2y + 8 = 0$ and $L_2: 3x - y + 4 = 0$.
To find the point of intersection,multiply $L_2$ by $2$: $6x - 2y + 8 = 0$.
Subtract $L_1$ from this: $(6x - 2y + 8) - (x - 2y + 8) = 0$,which gives $5x = 0$,so $x = 0$.
Substituting $x = 0$ in $L_2$: $3(0) - y + 4 = 0$,which gives $y = 4$.
The point of intersection is $(0, 4)$.
$A$ line passing through $(0, 0)$ and $(0, 4)$ is the $y$-axis,which is $x = 0$.
However,checking the options,if the line passes through $(0, 0)$ and the intersection point,the slope $m = \frac{4-0}{0-0}$ is undefined. Re-evaluating the intersection: $x-2y+8=0$ and $3x-y+4=0$. $y=3x+4$. $x-2(3x+4)+8=0 \implies x-6x-8+8=0 \implies -5x=0 \implies x=0, y=4$. The line is $x=0$. Given the options,there might be a typo in the question constants. If the intersection was $(4, 5)$,the line would be $5x-4y=0$.

(B) Let the line intersect the $x$-axis at $(h, 0)$ and the $y$-axis at $(0, k)$.
Since $(a, -2a)$ is the midpoint of the line segment,we have:
$\frac{h + 0}{2} = a \Rightarrow h = 2a$
$\frac{0 + k}{2} = -2a \Rightarrow k = -4a$
Using the intercept form of the line equation,$\frac{x}{h} + \frac{y}{k} = 1$:
$\frac{x}{2a} + \frac{y}{-4a} = 1$
Multiplying by $4a$,we get:
$2x - y = 4a$
Thus,the equation of the line is $2x - y = 4a$.

(A) The slope $m_1$ of the line passing through $(2, 3)$ and $(1, -2)$ is given by $m_1 = \frac{-2-3}{1-2} = \frac{-5}{-1} = 5$.
Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 \times m_2 = -1$.
Thus,$m_2 = -\frac{1}{5}$.
Using the point-slope form $(y - y_1) = m(x - x_1)$ for the point $(7, -4)$:
$y - (-4) = -\frac{1}{5}(x - 7)$
$5(y + 4) = -(x - 7)$
$5y + 20 = -x + 7$
$x + 5y + 13 = 0$.

(A) The equation of the first line is $x \sin \theta - y \cos \theta = 5$. Its slope $m_1 = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
The equation of the second line is $x \sin \alpha - y \cos \alpha + 11 = 0$. Its slope $m_2 = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha$.
Let $\beta$ be the acute angle between the lines.
Then,$\tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha} \right|$.
Using the trigonometric identity $\tan(\theta - \alpha) = \frac{\tan \theta - \tan \alpha}{1 + \tan \theta \tan \alpha}$,we get $\tan \beta = |\tan(\theta - \alpha)|$.
Therefore,$\beta = |\theta - \alpha|$.

(D) The given equations of the lines are $y = \sqrt{3}x - 1$ and $y = \frac{1}{\sqrt{3}}x - \frac{7}{\sqrt{3}}$.
Comparing these with $y = mx + c$,the slopes are $m_{1} = \sqrt{3}$ and $m_{2} = \frac{1}{\sqrt{3}}$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} \right|$.
Substituting the values: $\tan \theta = \left| \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})(\frac{1}{\sqrt{3}})} \right| = \left| \frac{\frac{3-1}{\sqrt{3}}}{1 + 1} \right| = \left| \frac{2/\sqrt{3}}{2} \right| = \frac{1}{\sqrt{3}}$.
Since $\tan \theta = \frac{1}{\sqrt{3}}$,we have $\theta = 30^{\circ}$.

(D) The given lines are $3x + 4y = 9$ and $6x + 8y = 15$.
To find the distance between parallel lines,we first make the coefficients of $x$ and $y$ identical.
Multiply the first equation by $2$: $2(3x + 4y) = 2(9) \Rightarrow 6x + 8y = 18$.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 6$,$b = 8$,$c_1 = -18$,and $c_2 = -15$.
$d = \frac{|-18 - (-15)|}{\sqrt{6^2 + 8^2}} = \frac{|-3|}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3$ units.

(B) The given line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be rewritten as $bx + ay - ab = 0$.
The length of the perpendicular from a point $(x_{1}, y_{1})$ to the line $Ax + By + C = 0$ is given by $d = \left| \frac{Ax_{1} + By_{1} + C}{\sqrt{A^{2} + B^{2}}} \right|$.
Substituting $A = b$,$B = a$,$C = -ab$,$x_{1} = a$,and $y_{1} = b$:
$d = \left| \frac{b(a) + a(b) - ab}{\sqrt{b^{2} + a^{2}}} \right| = \left| \frac{ab + ab - ab}{\sqrt{a^{2} + b^{2}}} \right| = \left| \frac{ab}{\sqrt{a^{2} + b^{2}}} \right|$ units.

(B) Let $I = \int_{-2}^{1} [x+1] \, dx$.
Using the property $[x+n] = [x] + n$ for any integer $n$,we have $[x+1] = [x] + 1$.
So,$I = \int_{-2}^{1} ([x] + 1) \, dx = \int_{-2}^{1} [x] \, dx + \int_{-2}^{1} 1 \, dx$.
Evaluating the second integral: $\int_{-2}^{1} 1 \, dx = [x]_{-2}^{1} = 1 - (-2) = 3$.
Evaluating the first integral: $\int_{-2}^{1} [x] \, dx = \int_{-2}^{-1} -2 \, dx + \int_{-1}^{0} -1 \, dx + \int_{0}^{1} 0 \, dx$.
$= -2[x]_{-2}^{-1} - 1[x]_{-1}^{0} + 0 = -2(-1 - (-2)) - 1(0 - (-1)) = -2(1) - 1(1) = -2 - 1 = -3$.
Therefore,$I = -3 + 3 = 0$.

(D) We use the trigonometric identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$.
$\int_{0}^{\frac{\pi}{2}} \sin^{2} x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx$
$= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos 2x) \, dx$
$= \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}$
$= \frac{1}{2} \left[ \left( \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \right]$
$= \frac{1}{2} \left[ \frac{\pi}{2} - 0 - 0 + 0 \right]$
$= \frac{\pi}{4}$

(D) The integral is $\int_{-2}^{2.24} [x] \, dx$. We split the interval based on the jumps of the greatest integer function $[x]$:
$\int_{-2}^{2.24} [x] \, dx = \int_{-2}^{-1} -2 \, dx + \int_{-1}^{0} -1 \, dx + \int_{0}^{1} 0 \, dx + \int_{1}^{2} 1 \, dx + \int_{2}^{2.24} 2 \, dx$
Evaluating each part:
$\int_{-2}^{-1} -2 \, dx = -2[-1 - (-2)] = -2(1) = -2$
$\int_{-1}^{0} -1 \, dx = -1[0 - (-1)] = -1(1) = -1$
$\int_{0}^{1} 0 \, dx = 0$
$\int_{1}^{2} 1 \, dx = 1[2 - 1] = 1$
$\int_{2}^{2.24} 2 \, dx = 2[2.24 - 2] = 2(0.24) = 0.48$
Summing these values: $-2 - 1 + 0 + 1 + 0.48 = -2 + 0.48 = -1.52$.
Note: Since the provided options do not match the calculated result,the correct value is $-1.52$.

(B) We need to evaluate the definite integral $\int_{0}^{4}|x-2| d x$.
The function $|x-2|$ changes its definition at $x=2$. Specifically,$|x-2| = -(x-2) = 2-x$ for $x < 2$ and $|x-2| = x-2$ for $x \ge 2$.
Thus,we split the integral at $x=2$:
$\int_{0}^{4}|x-2| d x = \int_{0}^{2}(2-x) d x + \int_{2}^{4}(x-2) d x$
Evaluating the first part: $\int_{0}^{2}(2-x) d x = [2x - \frac{x^2}{2}]_{0}^{2} = (4 - 2) - (0 - 0) = 2$.
Evaluating the second part: $\int_{2}^{4}(x-2) d x = [\frac{x^2}{2} - 2x]_{2}^{4} = (8 - 8) - (2 - 4) = 0 - (-2) = 2$.
Adding the two parts: $2 + 2 = 4$.

(C) The expression $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$ represents the expansion of the determinant of matrix $A$ along the second row.
By the property of determinants,the sum of the products of elements of any row (or column) with their corresponding cofactors is equal to the value of the determinant $|A|$.
First,we calculate the determinant $|A|$:
$|A| = 1 \begin{vmatrix} 1 & 3 \\ 3 & -5 \end{vmatrix} - 0 \begin{vmatrix} 2 & 3 \\ 0 & -5 \end{vmatrix} + 2 \begin{vmatrix} 2 & 1 \\ 0 & 3 \end{vmatrix}$
$|A| = 1(-5 - 9) - 0 + 2(6 - 0)$
$|A| = 1(-14) + 2(6) = -14 + 12 = -2$.
Thus,$a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = |A| = -2$.

(B) Given the matrix equation $AX=B$,where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 3 & 3 & -4\end{array}\right]$,$B=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$,and $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$.
First,we find the determinant of $A$:
$|A| = 1((-1)(-4) - (0)(3)) - (-1)((2)(-4) - (0)(3)) + 1((2)(3) - (-1)(3))$
$|A| = 1(4) + 1(-8) + 1(6+3) = 4 - 8 + 9 = 5$.
Since $|A| \neq 0$,$A^{-1}$ exists.
The adjoint of $A$,$adj(A)$,is the transpose of the cofactor matrix:
$C_{11} = 4, C_{12} = 8, C_{13} = 9$
$C_{21} = -1, C_{22} = -7, C_{23} = -6$
$C_{31} = 1, C_{32} = 2, C_{33} = 1$
$adj(A) = \left[\begin{array}{ccc}4 & -1 & 1 \\ 8 & -7 & 2 \\ 9 & -6 & 1\end{array}\right]$.
Thus,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{5} \left[\begin{array}{ccc}4 & -1 & 1 \\ 8 & -7 & 2 \\ 9 & -6 & 1\end{array}\right]$.
Now,$X = A^{-1}B = \frac{1}{5} \left[\begin{array}{ccc}4 & -1 & 1 \\ 8 & -7 & 2 \\ 9 & -6 & 1\end{array}\right] \left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right] = \frac{1}{5} \left[\begin{array}{l}4(1) - 1(1) + 1(2) \\ 8(1) - 7(1) + 2(2) \\ 9(1) - 6(1) + 1(2)\end{array}\right] = \frac{1}{5} \left[\begin{array}{l}5 \\ 5 \\ 5\end{array}\right] = \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$.
Therefore,$x=1, y=1, z=1$.
Finally,$x+y+z = 1+1+1 = 3$.

(A) Given the matrix equation $AX=B$:
$\begin{bmatrix} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 12 \\ 15 \\ 13 \end{bmatrix}$
Applying row operations $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$:
$\begin{bmatrix} 1 & 3 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 12 \\ 3 \\ 1 \end{bmatrix}$
This gives the system of equations:
$x + 3y + 3z = 12$
$y + z = 3$
$z = 1$
From $z = 1$,substitute into $y + z = 3$ to get $y + 1 = 3$,so $y = 2$.
Substitute $y = 2$ and $z = 1$ into $x + 3y + 3z = 12$:
$x + 3(2) + 3(1) = 12$
$x + 6 + 3 = 12$
$x + 9 = 12 \Rightarrow x = 3$
Finally,calculate $x^{2} + y^{2} + z^{2} = 3^{2} + 2^{2} + 1^{2} = 9 + 4 + 1 = 14$.

(C) The degree of a differential equation is defined as the power of the highest ordered derivative,provided the equation can be expressed as a polynomial in terms of its derivatives.
In the given equation $e^{\frac{dy}{dx}} + (\frac{dy}{dx})^3 = x$,the term $e^{\frac{dy}{dx}}$ involves the derivative in the exponent,which means it cannot be expressed as a polynomial in terms of $\frac{dy}{dx}$.
Therefore,the degree of this differential equation is not defined.

(B) Given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{\frac{7}{3}}=7 \frac{d^{2}y}{dx^{2}}$.
To find the degree,we must eliminate the fractional exponent by raising both sides to the power of $3$:
$\left(\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{\frac{7}{3}}\right)^{3} = \left(7 \frac{d^{2}y}{dx^{2}}\right)^{3}$.
This simplifies to $\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{7} = 343 \left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.
The highest order derivative present is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $3$,so the degree is $3$.
Thus,the order and degree are $2$ and $3$ respectively.

(A) Given differential equation is $\sqrt{1+\frac{1}{(\frac{dy}{dx})^2}} = (\frac{d^2y}{dx^2})^{3/2}$.
Squaring both sides,we get $1+\frac{1}{(\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^3$.
Multiplying both sides by $(\frac{dy}{dx})^2$,we get $(\frac{dy}{dx})^2 + 1 = (\frac{d^2y}{dx^2})^3 (\frac{dy}{dx})^2$.
The highest order derivative present is $\frac{d^2y}{dx^2}$,so the order is $2$.
The power of the highest order derivative after making the equation free from radicals and fractions is $3$,so the degree is $3$.
Thus,the order and degree are $2$ and $3$ respectively.

(A) Given differential equation is $\left[1+\frac{1}{(\frac{dy}{dx})^{2}}\right]^{\frac{5}{3}}=5 \frac{d^{2}y}{dx^{2}}$.
To find the order and degree,we first eliminate the fractional exponent by raising both sides to the power of $3$:
$\left[1+\frac{1}{(\frac{dy}{dx})^{2}}\right]^{5} = (5 \frac{d^{2}y}{dx^{2}})^{3}$
$\left[\frac{(\frac{dy}{dx})^{2}+1}{(\frac{dy}{dx})^{2}}\right]^{5} = 125 (\frac{d^{2}y}{dx^{2}})^{3}$
$((\frac{dy}{dx})^{2}+1)^{5} = 125 (\frac{d^{2}y}{dx^{2}})^{3} (\frac{dy}{dx})^{10}$
The highest order derivative present is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
The power of the highest order derivative after making the equation a polynomial in derivatives is $3$.
Thus,the order is $2$ and the degree is $3$.

(A) Given the differential equation: $y = px + \sqrt{a^{2}p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$.
Rearranging the terms,we get: $y - px = \sqrt{a^{2}p^{2} + b^{2}}$.
Squaring both sides:
$(y - px)^{2} = a^{2}p^{2} + b^{2}$
$y^{2} - 2pxy + p^{2}x^{2} = a^{2}p^{2} + b^{2}$
Substituting $p = \frac{dy}{dx}$:
$y^{2} - 2xy\left(\frac{dy}{dx}\right) + x^{2}\left(\frac{dy}{dx}\right)^{2} = a^{2}\left(\frac{dy}{dx}\right)^{2} + b^{2}$
The highest derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest derivative is $2$,so the degree is $2$.
Therefore,the order and degree are $1$ and $2$ respectively.

(A) Given the differential equation: $\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{\frac{7}{3}}=7\left(\frac{d^{2}y}{dx^{2}}\right)$.
To find the degree,we must eliminate the fractional exponent. Raise both sides to the power of $3$:
$\left[1+\left(\frac{dy}{dx}\right)^{3}\right]^{7} = 7^{3}\left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.
The order of a differential equation is the highest derivative present,which is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.
The degree is the power of the highest order derivative after the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^{2}y}{dx^{2}}$ is $3$.
Thus,the order is $2$ and the degree is $3$.

(D) The equation of a line with $x$-intercept $a$ and $y$-intercept $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
Multiplying by $ab$,we get $bx + ay = ab$.
Differentiating both sides with respect to $x$,we get $b + a \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{b}{a}$.
Differentiating again with respect to $x$,since $-\frac{b}{a}$ is a constant,we get $\frac{d^{2} y}{d x^{2}} = 0$.

(B) Given equation is $y^{2}=(2 x+c)^{5}$ ...$(i)$
Differentiating both sides with respect to $x$:
$2 y \frac{dy}{dx} = 5(2 x+c)^{4} \times 2$
$y \frac{dy}{dx} = 5(2 x+c)^{4}$
From this,we find $(2 x+c)$:
$(2 x+c)^{4} = \frac{y}{5} \frac{dy}{dx}$
$(2 x+c) = \left(\frac{y}{5} \frac{dy}{dx}\right)^{1/4}$
Substituting this into equation $(i)$:
$y^{2} = \left[\left(\frac{y}{5} \frac{dy}{dx}\right)^{1/4}\right]^{5}$
$y^{2} = \left(\frac{y}{5} \frac{dy}{dx}\right)^{5/4}$
Raising both sides to the power of $4$:
$(y^{2})^{4} = \left(\frac{y}{5} \frac{dy}{dx}\right)^{5}$
$y^{8} = \frac{y^{5}}{5^{5}} \left(\frac{dy}{dx}\right)^{5}$
$y^{8} = \frac{y^{5}}{3125} \left(\frac{dy}{dx}\right)^{5}$
Dividing by $y^{5}$ (assuming $y \neq 0$):
$y^{3} = \frac{1}{3125} \left(\frac{dy}{dx}\right)^{5}$
$3125 y^{3} = \left(\frac{dy}{dx}\right)^{5}$
$\left(\frac{dy}{dx}\right)^{5} - 3125 y^{3} = 0$

(D) Given $y = e^{ax}$.
Taking the natural logarithm on both sides,we get $\log y = \log(e^{ax})$.
Since $\log(e^{ax}) = ax$,we have $\log y = ax$ ...$(1)$.
Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = a$.
Now,substitute the value of $a$ from the derivative into equation $(1)$:
$\log y = \left( \frac{1}{y} \frac{dy}{dx} \right) x$.
Multiplying both sides by $y$,we obtain $y \log y = x \frac{dy}{dx}$,which is $x \frac{dy}{dx} = y \log y$.

(A) Given the equation: $\tan^{-1} x + \tan^{-1} y = c$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\tan^{-1} x) + \frac{d}{dx}(\tan^{-1} y) = \frac{d}{dx}(c)$
$\frac{1}{1+x^2} + \frac{1}{1+y^2} \cdot \frac{dy}{dx} = 0$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{1}{1+y^2} \cdot \frac{dy}{dx} = -\frac{1}{1+x^2}$
$\frac{dy}{dx} = -\frac{1+y^2}{1+x^2}$
Thus,the correct option is $A$.

(B) Given $\frac{x}{x-y} = \log a - \log(x-y)$.
Rearranging,we get $\log(x-y) + \frac{x}{x-y} = \log a$.
Differentiating both sides with respect to $x$:
$\frac{1}{x-y} \left(1 - \frac{dy}{dx}\right) + \frac{(x-y)(1) - x(1 - \frac{dy}{dx})}{(x-y)^2} = 0$.
Multiply by $(x-y)^2$:
$(x-y)(1 - \frac{dy}{dx}) + x - y - x + x \frac{dy}{dx} = 0$.
$x - y - (x-y) \frac{dy}{dx} - y + x \frac{dy}{dx} = 0$.
$x - 2y + \frac{dy}{dx} (x - (x-y)) = 0$.
$x - 2y + y \frac{dy}{dx} = 0$.
$y \frac{dy}{dx} = 2y - x$.
Therefore,$\frac{dy}{dx} = \frac{2y - x}{y}$.

(C) Given the equation $y = mx + \frac{2}{m} \dots (1)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = m$.
Substituting $m = \frac{dy}{dx}$ into equation $(1)$:
$y = x \left(\frac{dy}{dx}\right) + \frac{2}{\frac{dy}{dx}}$
Multiplying both sides by $\frac{dy}{dx}$:
$y \left(\frac{dy}{dx}\right) = x \left(\frac{dy}{dx}\right)^{2} + 2$.

(C) Given the differential equation $\cos \left(\frac{dy}{dx}\right) = a$.
Taking $\cos^{-1}$ on both sides,we get $\frac{dy}{dx} = \cos^{-1} a$.
Integrating both sides with respect to $x$,we have $\int dy = \int \cos^{-1} a \, dx$.
This gives $y = x \cos^{-1} a + c$ .... $(1)$.
Given the initial condition $y(0) = 2$,substitute $x = 0$ and $y = 2$ into equation $(1)$:
$2 = 0 \cdot \cos^{-1} a + c \Rightarrow c = 2$.
Substituting $c = 2$ back into equation $(1)$,we get $y = x \cos^{-1} a + 2$.
Rearranging the terms,$y - 2 = x \cos^{-1} a$,which implies $\frac{y - 2}{x} = \cos^{-1} a$.
Taking $\cos$ on both sides,we obtain $\cos \left(\frac{y - 2}{x}\right) = a$.

(A) Given $y=e^{x}(A \cos x+B \sin x)$ ... $(1)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)$
$\frac{dy}{dx} = y + e^{x}(-A \sin x + B \cos x)$ ... $(2)$
Differentiating again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + e^{x}(-A \sin x + B \cos x) + e^{x}(-A \cos x - B \sin x)$
Substitute $e^{x}(-A \sin x + B \cos x) = \frac{dy}{dx} - y$ from $(2)$:
$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^{x}(A \cos x + B \sin x)$
Since $e^{x}(A \cos x + B \sin x) = y$:
$\frac{d^{2}y}{dx^{2}} = 2\frac{dy}{dx} - y - y$
$\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0$

(B) Given the function $y = a(x-a)^{2} \quad ...(1)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2a(x-a) \quad ...(2)$
From $(2)$,$a = \frac{1}{2} \frac{dy/dx}{x-a}$. Substituting $a$ into $(1)$:
$y = \frac{1}{2} \frac{dy/dx}{x-a} (x-a)^2 = \frac{1}{2} (x-a) \frac{dy}{dx}$
Thus,$(x-a) = \frac{2y}{dy/dx}$.
Substituting $(x-a)$ back into $(2)$:
$\frac{dy}{dx} = 2a \left( \frac{2y}{dy/dx} \right) \implies a = \frac{(dy/dx)^2}{4y}$.
Now,substitute $a$ and $(x-a)$ into $y = a(x-a)^2$:
$y = \left( \frac{(dy/dx)^2}{4y} \right) \left( \frac{2y}{dy/dx} \right)^2$
$y = \frac{(dy/dx)^2}{4y} \cdot \frac{4y^2}{(dy/dx)^2} = y$.
To find the differential equation,we eliminate $a$ from $y = a(x-a)^2$ and $y' = 2a(x-a)$.
From $y' = 2a(x-a)$,we have $a = x - \frac{y'}{2a}$ is not correct.
Correct approach: $x-a = \frac{y'}{2a}$. Substitute into $y = a(x-a)^2$:
$y = a \left( \frac{y'}{2a} \right)^2 = \frac{(y')^2}{4a}$.
Since $a = x - (x-a) = x - \frac{y'}{2a}$,we have $2a^2 - 2ax + y' = 0$.
Solving for $a$ and substituting leads to the relation $8y^3 = (y')^2 (2x - y')$. Option $B$ is the correct form.

(B) Let $(h, 8)$ be the centre of the circle.
Since the circle touches the $X$-axis,its radius is $r = 8$.
The equation of the circle is $(x-h)^{2} + (y-8)^{2} = 8^{2} = 64$ ... $(1)$
Differentiating equation $(1)$ with respect to $x$,we get:
$2(x-h) + 2(y-8) \frac{d y}{d x} = 0$
$(x-h) = -(y-8) \frac{d y}{d x}$
Substituting the value of $(x-h)$ into equation $(1)$:
$[-(y-8) \frac{d y}{d x}]^{2} + (y-8)^{2} = 64$
$(y-8)^{2} (\frac{d y}{d x})^{2} + (y-8)^{2} = 64$
$(y-8)^{2} [1 + (\frac{d y}{d x})^{2}] = 64$
Thus,the correct option is $B$.

(A) Given: $x = a \sin t - b \cos t$ and $y = a \cos t + b \sin t$.
Squaring and adding both equations:
$x^{2} + y^{2} = (a \sin t - b \cos t)^{2} + (a \cos t + b \sin t)^{2}$
$x^{2} + y^{2} = a^{2} \sin^{2} t + b^{2} \cos^{2} t - 2ab \sin t \cos t + a^{2} \cos^{2} t + b^{2} \sin^{2} t + 2ab \sin t \cos t$
$x^{2} + y^{2} = a^{2}(\sin^{2} t + \cos^{2} t) + b^{2}(\cos^{2} t + \sin^{2} t) = a^{2} + b^{2}$.
Differentiating $x^{2} + y^{2} = a^{2} + b^{2}$ with respect to $x$:
$2x + 2y \frac{dy}{dx} = 0 \implies y \frac{dy}{dx} = -x$.
Differentiating again with respect to $x$:
$y \frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2} = -1$.
Substitute $\frac{dy}{dx} = -\frac{x}{y}$:
$y \frac{d^{2}y}{dx^{2}} + \left(-\frac{x}{y}\right)^{2} = -1$
$y \frac{d^{2}y}{dx^{2}} + \frac{x^{2}}{y^{2}} = -1$.
Multiplying by $y^{2}$:
$y^{3} \frac{d^{2}y}{dx^{2}} + x^{2} = -y^{2}$
$y^{3} \frac{d^{2}y}{dx^{2}} + x^{2} + y^{2} = 0$.

(A) Given the differential equation: $\sec^{2} x \tan y \, dx + \sec^{2} y \tan x \, dy = 0$
Rearranging the terms,we get: $\sec^{2} x \tan y \, dx = -\sec^{2} y \tan x \, dy$
Separating the variables: $\frac{\sec^{2} x}{\tan x} \, dx = -\frac{\sec^{2} y}{\tan y} \, dy$
Integrating both sides: $\int \frac{\sec^{2} x}{\tan x} \, dx = -\int \frac{\sec^{2} y}{\tan y} \, dy$
Let $u = \tan x$,then $du = \sec^{2} x \, dx$. Similarly,let $v = \tan y$,then $dv = \sec^{2} y \, dy$.
Thus,$\int \frac{1}{u} \, du = -\int \frac{1}{v} \, dv$
$\log |u| = -\log |v| + \log |c|$
$\log |\tan x| + \log |\tan y| = \log |c|$
$\log |\tan x \tan y| = \log |c|$
Therefore,the general solution is $\tan x \tan y = c$.

(B) Given differential equation is $x dy + 2y dx = 0$.
Rearranging the terms,we get $x dy = -2y dx$.
Separating the variables,we have $\frac{dy}{y} = -2 \frac{dx}{x}$.
Integrating both sides,$\int \frac{dy}{y} = -2 \int \frac{dx}{x}$.
This gives $\ln|y| = -2 \ln|x| + C$.
Using properties of logarithms,$\ln|y| + 2 \ln|x| = C$,which simplifies to $\ln|y| + \ln|x^2| = C$.
Thus,$\ln|yx^2| = C$,which implies $yx^2 = e^C = k$.
Given the initial condition $x = 2$ and $y = 1$,substitute these values into $x^2y = k$:
$(2)^2(1) = k \Rightarrow 4(1) = k \Rightarrow k = 4$.
Therefore,the particular solution is $x^2y = 4$.

(C) Given the differential equation: $\frac{dy}{dx} + \frac{1}{\sqrt{1-x^2}} = 0$
Rearranging the terms,we get: $\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$
Integrating both sides with respect to $x$: $\int dy = -\int \frac{1}{\sqrt{1-x^2}} dx$
We know that $\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + c$
Therefore,the solution is: $y = -\sin^{-1} x + c$
This can be rewritten as: $y + \sin^{-1} x = c$

(B) Given the differential equation: $\log \left(\frac{dy}{dx}\right) = 9x - 6y + 6$.
Taking the exponential of both sides,we get: $\frac{dy}{dx} = e^{9x - 6y + 6} = e^{9x+6} \cdot e^{-6y}$.
Separating the variables,we have: $e^{6y} dy = e^{9x+6} dx$.
Integrating both sides: $\int e^{6y} dy = \int e^{9x+6} dx$.
This gives: $\frac{e^{6y}}{6} = \frac{e^{9x+6}}{9} + C$.
Given $y = 1$ when $x = 0$: $\frac{e^{6}}{6} = \frac{e^{6}}{9} + C$.
Solving for $C$: $C = \frac{e^{6}}{6} - \frac{e^{6}}{9} = \frac{3e^{6} - 2e^{6}}{18} = \frac{e^{6}}{18}$.
Substituting $C$ back into the equation: $\frac{e^{6y}}{6} = \frac{e^{9x+6}}{9} + \frac{e^{6}}{18}$.
Multiplying by $18$: $3e^{6y} = 2e^{9x+6} + e^{6}$.

(B) Given the differential equation: $y \frac{dx}{dy} = x \log x$
Separate the variables: $\frac{dx}{x \log x} = \frac{dy}{y}$
Integrate both sides: $\int \frac{1}{x \log x} dx = \int \frac{1}{y} dy$
Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1}{u} du = \log |u| = \log |\log x|$.
So,$\log |\log x| = \log |y| + C$.
Given $x = e$ and $y = 1$: $\log |\log e| = \log |1| + C \Rightarrow \log |1| = 0 + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.
Thus,$\log |\log x| = \log |y|$.
Taking the exponential of both sides: $\log x = y$,which implies $x = e^y$.

(A) Given the differential equation: $(1+y^{2})+(x-e^{\tan ^{-1} y}) \frac{dy}{dx}=0$
Rearranging the terms: $(x-e^{\tan ^{-1} y}) \frac{dy}{dx} = -(1+y^{2})$
Taking the reciprocal: $\frac{dx}{dy} = \frac{-(x-e^{\tan ^{-1} y})}{1+y^{2}} = \frac{-x}{1+y^{2}} + \frac{e^{\tan ^{-1} y}}{1+y^{2}}$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^{2}}$ and $Q(y) = \frac{e^{\tan ^{-1} y}}{1+y^{2}}$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^{2}} dy} = e^{\tan ^{-1} y}$.
The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$.
$x \cdot e^{\tan ^{-1} y} = \int \frac{e^{\tan ^{-1} y}}{1+y^{2}} \cdot e^{\tan ^{-1} y} dy + c$.
Let $t = e^{\tan ^{-1} y}$,then $dt = \frac{e^{\tan ^{-1} y}}{1+y^{2}} dy$.
Substituting this into the integral: $x \cdot e^{\tan ^{-1} y} = \int t dt + c = \frac{t^{2}}{2} + c$.
Thus,the general solution is $x \cdot e^{\tan ^{-1} y} = \frac{(e^{\tan ^{-1} y})^{2}}{2} + c$.

(B) Given the differential equation: $\sin ^{-1}\left(\frac{dy}{dx}\right)=x+y$
$\implies \frac{dy}{dx}=\sin (x+y)$
Let $x+y=t$. Then differentiating with respect to $x$,we get $1+\frac{dy}{dx}=\frac{dt}{dx}$,so $\frac{dy}{dx}=\frac{dt}{dx}-1$.
Substituting this into the equation: $\frac{dt}{dx}-1=\sin t$
$\implies \frac{dt}{dx}=1+\sin t$
$\implies \int \frac{dt}{1+\sin t}=\int dx$
To integrate $\frac{1}{1+\sin t}$,multiply numerator and denominator by $(1-\sin t)$:
$\int \frac{1-\sin t}{1-\sin^2 t} dt = \int dx$
$\implies \int \frac{1-\sin t}{\cos^2 t} dt = x+c$
$\implies \int (\sec^2 t - \sec t \tan t) dt = x+c$
$\implies \tan t - \sec t = x+c$
Substituting $t=x+y$ back,we get: $x = \tan (x+y) - \sec (x+y) + c$.

(A) Given the differential equation $x^{2} \frac{dy}{dx} = y^{2} + xy$.
Dividing by $x^{2}$,we get $\frac{dy}{dx} = \frac{y^{2} + xy}{x^{2}} = \left(\frac{y}{x}\right)^{2} + \frac{y}{x}$.
This is a homogeneous differential equation. Let $y = ux$,then $\frac{dy}{dx} = u + x \frac{du}{dx}$.
Substituting these into the equation: $u + x \frac{du}{dx} = u^{2} + u$.
Subtracting $u$ from both sides: $x \frac{du}{dx} = u^{2}$.
Separating the variables: $\frac{du}{u^{2}} = \frac{dx}{x}$.
Integrating both sides: $\int u^{-2} du = \int \frac{1}{x} dx$.
$-u^{-1} = \log |x| + c$.
Since $u = \frac{y}{x}$,we have $-\frac{x}{y} = \log |x| + c$.
Rearranging the terms: $\frac{x}{y} + \log |x| = -c$,which can be written as $\frac{x}{y} + \log |x| = C$ (where $C = -c$).

(D) Given the slope of the tangent is $\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2$.
This is a homogeneous differential equation.
Let $y = ux$,then $\frac{dy}{dx} = u + x\frac{du}{dx}$.
Substituting this into the equation:
$u + x\frac{du}{dx} = 1 + u + u^2$.
Subtracting $u$ from both sides:
$x\frac{du}{dx} = 1 + u^2$.
Separating the variables:
$\int \frac{du}{1 + u^2} = \int \frac{dx}{x}$.
Integrating both sides:
$\tan^{-1}(u) = \log|x| + C$.
Substituting $u = \frac{y}{x}$:
$\tan^{-1}\left(\frac{y}{x}\right) = \log|x| + C$.
Since the curve passes through $(1,0)$,we have $\tan^{-1}(0) = \log|1| + C$,which gives $0 = 0 + C$,so $C = 0$.
Thus,the equation is $\tan^{-1}\left(\frac{y}{x}\right) = \log|x|$.

(B) Given the equation $y=c^{2}+\frac{c}{x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{d y}{d x} = -\frac{c}{x^{2}}$
From this,we can express $c$ in terms of $x$ and $\frac{d y}{d x}$:
$c = -x^{2} \frac{d y}{d x}$
Now,substitute this value of $c$ back into the original equation:
$y = (-x^{2} \frac{d y}{d x})^{2} + \frac{-x^{2} \frac{d y}{d x}}{x}$
$y = x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x}$
Rearranging the terms to form the differential equation:
$x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x} - y = 0$
Thus,the correct option is $B$.

(C) Given the differential equation: $x \sin \left(\frac{y}{x}\right) dy = \left[y \sin \left(\frac{y}{x}\right) - x\right] dx$
Dividing both sides by $dx \cdot x \sin \left(\frac{y}{x}\right)$,we get:
$\frac{dy}{dx} = \frac{y \sin \left(\frac{y}{x}\right) - x}{x \sin \left(\frac{y}{x}\right)} = \frac{y}{x} - \frac{1}{\sin \left(\frac{y}{x}\right)}$
Let $v = \frac{y}{x}$,so $y = vx$. Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation:
$v + x \frac{dv}{dx} = v - \frac{1}{\sin v}$
$x \frac{dv}{dx} = -\frac{1}{\sin v} = -\operatorname{cosec} v$
Separating the variables:
$\sin v \, dv = -\frac{1}{x} dx$
Integrating both sides:
$\int \sin v \, dv = -\int \frac{1}{x} dx$
$-\cos v = -\log |x| + c$
$\cos v = \log |x| + c$
Substituting $v = \frac{y}{x}$ back:
$\cos \left(\frac{y}{x}\right) = \log |x| + c$

(A) Given the differential equation: $\sin^{2} y \frac{dx}{dy} + x = \cot y$.
Dividing by $\sin^{2} y$,we get: $\frac{dx}{dy} + (\operatorname{cosec}^{2} y)x = \cot y \operatorname{cosec}^{2} y$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \operatorname{cosec}^{2} y$ and $Q(y) = \cot y \operatorname{cosec}^{2} y$.
The Integrating Factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \operatorname{cosec}^{2} y dy} = e^{-\cot y}$.
The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$.
$x e^{-\cot y} = \int \cot y \operatorname{cosec}^{2} y e^{-\cot y} dy + C$.
Let $t = -\cot y$,then $dt = \operatorname{cosec}^{2} y dy$.
$x e^{-\cot y} = \int (-t) e^{t} dt + C = -(t e^{t} - e^{t}) + C = e^{t}(1 - t) + C$.
Substituting back $t = -\cot y$: $x e^{-\cot y} = e^{-\cot y}(1 + \cot y) + C$.
Given $x = 0$ at $y = \frac{3\pi}{4}$,we have $0 = e^{-\cot(3\pi/4)}(1 + \cot(3\pi/4)) + C$.
Since $\cot(3\pi/4) = -1$,$0 = e^{1}(1 - 1) + C \implies C = 0$.
Thus,$x e^{-\cot y} = e^{-\cot y}(1 + \cot y)$,which simplifies to $x = 1 + \cot y$.

(C) The given differential equation is $\frac{dy}{dx} + \frac{1}{x}y = x^3 - 3$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = x^3 - 3$.
The integrating factor ($I$.$F$.) is given by the formula $I.F. = e^{\int P dx}$.
Substituting $P = \frac{1}{x}$,we get $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.
Therefore,the integrating factor is $x$.

(D) Given the differential equation $r dx + (x - r^2) dr = 0$.
Rearranging the terms,we get $r dx = -(x - r^2) dr$.
Dividing by $dr$,we have $r \frac{dx}{dr} = r^2 - x$.
Rearranging into the standard linear form $\frac{dx}{dr} + P(r)x = Q(r)$,we get $\frac{dx}{dr} + \frac{1}{r}x = r$.
The Integrating Factor ($I$.$F$.) is $e^{\int \frac{1}{r} dr} = e^{\log r} = r$.
The general solution is given by $x \cdot (I.F.) = \int Q(r) \cdot (I.F.) dr + c$.
Substituting the values,$x \cdot r = \int r \cdot r dr + c$.
Integrating the right side,$xr = \int r^2 dr + c$.
Thus,$xr = \frac{r^3}{3} + c$.

(B) The given differential equation is $\frac{dy}{dx} + 2y = e^{-x}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = e^{-x}$.
The Integrating Factor ($I$.$F$.) is given by $I.F. = e^{\int P dx} = e^{\int 2 dx} = e^{2x}$.
The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.
Substituting the values,we get $y e^{2x} = \int (e^{-x} \cdot e^{2x}) dx + c$.
$y e^{2x} = \int e^{x} dx + c$.
$y e^{2x} = e^{x} + c$.

(A) Given the differential equation: $(1-x^{2}) \frac{dy}{dx} + 2xy = x(1-x^{2})^{\frac{1}{2}}$
Divide by $(1-x^{2})$ to get the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{2x}{1-x^{2}}y = \frac{x}{\sqrt{1-x^{2}}}$
Here,$P = \frac{2x}{1-x^{2}}$ and $Q = \frac{x}{\sqrt{1-x^{2}}}$.
Integrating Factor $(I.F.) = e^{\int P dx} = e^{\int \frac{2x}{1-x^{2}} dx} = e^{-\ln(1-x^{2})} = e^{\ln(\frac{1}{1-x^{2}})} = \frac{1}{1-x^{2}}$.
The general solution is $y \times (I.F.) = \int Q \times (I.F.) dx + c$.
$y \times \frac{1}{1-x^{2}} = \int \frac{x}{\sqrt{1-x^{2}}} \times \frac{1}{1-x^{2}} dx + c$
$y \times \frac{1}{1-x^{2}} = \int x(1-x^{2})^{-\frac{3}{2}} dx + c$
Let $u = 1-x^{2}$,then $du = -2x dx$,so $x dx = -\frac{1}{2} du$.
$y \times \frac{1}{1-x^{2}} = -\frac{1}{2} \int u^{-\frac{3}{2}} du + c = -\frac{1}{2} \left( \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} \right) + c = u^{-\frac{1}{2}} + c = \frac{1}{\sqrt{1-x^{2}}} + c$.
Multiplying both sides by $(1-x^{2})$:
$y = \frac{1-x^{2}}{\sqrt{1-x^{2}}} + c(1-x^{2}) = \sqrt{1-x^{2}} + c(1-x^{2})$.

(D) Given the differential equation: $(1+x^{2}) dt = (\tan^{-1} x - t) dx$.
Divide both sides by $(1+x^{2}) dx$:
$\frac{dt}{dx} = \frac{\tan^{-1} x - t}{1+x^{2}}$
Rearrange the equation into the standard linear form $\frac{dt}{dx} + P(x)t = Q(x)$:
$\frac{dt}{dx} + \frac{1}{1+x^{2}}t = \frac{\tan^{-1} x}{1+x^{2}}$
Here,$P(x) = \frac{1}{1+x^{2}}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx}$:
$I.F. = e^{\int \frac{1}{1+x^{2}} dx} = e^{\tan^{-1} x}$.

(A) Given differential equation: $\sin y \frac{d y}{d x} = \cos y (1 - x \cos y)$.
Expanding the right side: $\sin y \frac{d y}{d x} = \cos y - x \cos^2 y$.
Dividing both sides by $\cos^2 y$: $\frac{\sin y}{\cos^2 y} \frac{d y}{d x} = \frac{1}{\cos y} - x$.
This simplifies to: $\sec y \tan y \frac{d y}{d x} = \sec y - x$.
Rearranging the terms: $\sec y \tan y \frac{d y}{d x} - \sec y = -x$.
Let $v = \sec y$. Then $\frac{d v}{d x} = \sec y \tan y \frac{d y}{d x}$.
Substituting into the equation: $\frac{d v}{d x} - v = -x$.
This is a linear differential equation of the form $\frac{d v}{d x} + P(x)v = Q(x)$,where $P(x) = -1$.
The integrating factor ($I$.$F$.) is given by $e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$.

(B) Given differential equation is $(1+y+x^{2}y)dx + (x+x^{3})dy = 0$.
Rearranging the terms,we have $(x+x^{3})dy = -(1+y(1+x^{2}))dx$.
Dividing by $dx(x+x^{3})$,we get $\frac{dy}{dx} = -\frac{1+y(1+x^{2})}{x(1+x^{2})}$.
This simplifies to $\frac{dy}{dx} = -\frac{1}{x(1+x^{2})} - \frac{y(1+x^{2})}{x(1+x^{2})}$.
Thus,$\frac{dy}{dx} + \left(\frac{1}{x}\right)y = -\frac{1}{x(1+x^{2})}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = -\frac{1}{x(1+x^{2})}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\log x} = x$.

(D) The given differential equation is $x \frac{dy}{dx} + y \log x = x^2$.
Dividing both sides by $x$,we get:
$\frac{dy}{dx} + \left(\frac{\log x}{x}\right) y = x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{\log x}{x}$ and $Q = x$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$.
$I$.$F$. $= e^{\int \frac{\log x}{x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
So,$\int \frac{\log x}{x} dx = \int u du = \frac{u^2}{2} = \frac{(\log x)^2}{2}$.
Therefore,$I$.$F$. $= e^{\frac{(\log x)^2}{2}} = (e^{\log x})^{\frac{\log x}{2}} = x^{\frac{\log x}{2}} = x^{\log (\sqrt{x})}$.
Thus,the correct option is $D$.

(B) The given differential equation is $y \log y \left(\frac{dx}{dy}\right) + x = \log y$.
Dividing both sides by $y \log y$,we get:
$\frac{dx}{dy} + \frac{x}{y \log y} = \frac{\log y}{y \log y} = \frac{1}{y}$.
This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{1}{y \log y}$ and $Q = \frac{1}{y}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dy}$.
$I$.$F$. $= e^{\int \frac{1}{y \log y} dy}$.
Let $u = \log y$,then $du = \frac{1}{y} dy$.
So,$\int \frac{1}{y \log y} dy = \int \frac{1}{u} du = \log u = \log(\log y)$.
Therefore,$I$.$F$. $= e^{\log(\log y)} = \log y$.

(D) Given the differential equation: $\frac{dy}{dx}(x \log x) + y = 4 \log x$.
Divide the entire equation by $(x \log x)$ to bring it into the standard form $\frac{dy}{dx} + Py = Q$:
$\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{4 \log x}{x \log x} = \frac{4}{x}$.
Here,$P = \frac{1}{x \log x}$.
The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$:
$I.F. = e^{\int \frac{1}{x \log x} dx}$.
Let $u = \log x$,then $du = \frac{1}{x} dx$.
So,$\int \frac{1}{x \log x} dx = \int \frac{1}{u} du = \log u = \log(\log x)$.
Therefore,$I.F. = e^{\log(\log x)} = \log x$.

(A) Given the differential equation: $(y + x \frac{dy}{dx}) \sin(xy) = \cos x$
Let $u = xy$. Then,differentiating with respect to $x$,we get $\frac{du}{dx} = y + x \frac{dy}{dx}$.
Substituting this into the equation,we have: $\sin(u) \frac{du}{dx} = \cos x$.
Integrating both sides with respect to $x$: $\int \sin(u) du = \int \cos x dx$.
This yields: $-\cos(u) = \sin x + C$.
Substituting $u = xy$ back,we get: $-\cos(xy) = \sin x + C$.
At $x = 0$,the value of $y$ is not explicitly given,but since $\cos(xy) = -\sin x - C$,at $x=0$,$\cos(0) = -\sin(0) - C$,which implies $1 = 0 - C$,so $C = -1$.
Thus,$-\cos(xy) = \sin x - 1$,which simplifies to $\sin x + \cos(xy) = 1$.

(D) The given line is $5x + 2y + 7 = 0$.
Its slope is $m_1 = -\frac{5}{2}$.
Any line perpendicular to this line will have a slope $m_2 = -\frac{1}{m_1} = \frac{2}{5}$.
The equation of a family of lines with slope $\frac{2}{5}$ is $y = \frac{2}{5}x + c$,which can be written as $2x - 5y + 5c = 0$.
Let $5c = k$,so the equation is $2x - 5y + k = 0$.
Differentiating both sides with respect to $x$,we get $2 - 5\frac{dy}{dx} = 0$.
Multiplying by $dx$,we get $2dx - 5dy = 0$,or $5dy - 2dx = 0$.

(C) The rate of growth is proportional to the number present.
$\frac{dN}{dt} = kN \Rightarrow \frac{dN}{N} = k dt$.
Integrating both sides,we get $\ln N = kt + C$.
At $t = 0$,$N = 1000$,so $C = \ln 1000$.
Thus,$\ln N = kt + \ln 1000 \Rightarrow \ln(\frac{N}{1000}) = kt \Rightarrow N = 1000 e^{kt}$.
Given that at $t = 1$,$N = 2000$,we have $2000 = 1000 e^k$,which implies $e^k = 2$.
Substituting this into the equation,$N = 1000 \times (e^k)^t = 1000 \times 2^t$.
For $t = 2 \frac{1}{2} = 2.5$ hours:
$N = 1000 \times 2^{2.5} = 1000 \times 2^2 \times 2^{0.5} = 1000 \times 4 \times \sqrt{2}$.
Given $\sqrt{2} = 1.414$,we get $N = 1000 \times 4 \times 1.414 = 5656$.

(B) Let $\theta$ be the temperature of the body at time $t$. The temperature of the surrounding is $\theta_s = 20^{\circ} C$. According to Newton's law of cooling,$\frac{d\theta}{dt} = -K(\theta - \theta_s)$.
Integrating this,we get $\ln(\theta - 20) = -Kt + C$.
At $t = 0$,$\theta = 100^{\circ} C$,so $\ln(100 - 20) = C \Rightarrow C = \ln(80)$.
Thus,$\ln\left(\frac{\theta - 20}{80}\right) = -Kt$.
At $t = 20 \text{ min}$,$\theta = 60^{\circ} C$,so $\ln\left(\frac{60 - 20}{80}\right) = -K(20) \Rightarrow \ln(0.5) = -20K \Rightarrow K = \frac{-\ln(0.5)}{20}$.
We need to find $\theta$ at $t = 60 \text{ min}$ (one hour).
$\ln\left(\frac{\theta - 20}{80}\right) = -\left(\frac{-\ln(0.5)}{20}\right)(60) = 3 \ln(0.5) = \ln(0.5^3) = \ln(0.125)$.
$\frac{\theta - 20}{80} = 0.125 = \frac{1}{8}$.
$\theta - 20 = 10 \Rightarrow \theta = 30^{\circ} C$.