MHT CET 2021 Mathematics Question Paper with Answer and Solution

(C) Given $\frac{\sin A}{\sin C} = \frac{\sin (A-B)}{\sin (B-C)}$.
Applying cross-multiplication: $\sin A \sin (B-C) = \sin C \sin (A-B)$.
Using the expansion $\sin(x-y) = \sin x \cos y - \cos x \sin y$:
$\sin A (\sin B \cos C - \cos B \sin C) = \sin C (\sin A \cos B - \cos A \sin B)$.
$\sin A \sin B \cos C - \sin A \cos B \sin C = \sin C \sin A \cos B - \sin C \cos A \sin B$.
Rearranging terms: $\sin A \sin B \cos C + \sin C \cos A \sin B = 2 \sin A \cos B \sin C$.
$\sin B (\sin A \cos C + \cos A \sin C) = 2 \sin A \cos B \sin C$.
Since $\sin A \cos C + \cos A \sin C = \sin(A+C) = \sin(\pi - B) = \sin B$,we have:
$\sin B \cdot \sin B = 2 \sin A \cos B \sin C$.
$\sin^2 B = 2 \sin A \sin C \cos B$.
Dividing by $\sin A \sin C \sin B$: $\frac{\sin B}{\sin A \sin C} = \frac{2 \cos B}{\sin B} = 2 \cot B$.
Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.
Substituting these: $\frac{(b/2R)^2}{(a/2R)(c/2R)} = 2 \cos B \Rightarrow \frac{b^2}{ac} = 2 \cos B$.
Using the cosine rule $\cos B = \frac{a^2+c^2-b^2}{2ac}$:
$\frac{b^2}{ac} = 2 \left( \frac{a^2+c^2-b^2}{2ac} \right) = \frac{a^2+c^2-b^2}{ac}$.
$b^2 = a^2+c^2-b^2 \Rightarrow 2b^2 = a^2+c^2$.
Thus,$a^2, b^2, c^2$ are in $AP$.

(A) Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$,the expression becomes:
$\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} = \left(\frac{1}{a^2} - \frac{1}{b^2}\right) - 2\left(\frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2}\right)$
From the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = k$,which implies $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{1}{k}$.
Therefore,$\frac{\sin^2 A}{a^2} = \frac{\sin^2 B}{b^2} = \frac{1}{k^2}$.
Substituting this into the expression:
$\left(\frac{1}{a^2} - \frac{1}{b^2}\right) - 2\left(\frac{1}{k^2} - \frac{1}{k^2}\right) = \frac{1}{a^2} - \frac{1}{b^2}$
Given $a=2$ and $b=3$:
$\frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$

(A) Using the projection formula or the law of cosines:
$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
Substituting these into the expression:
$c(a \cos B - b \cos A) = c \left( a \left( \frac{c^2 + a^2 - b^2}{2ac} \right) - b \left( \frac{b^2 + c^2 - a^2}{2bc} \right) \right)$
$= c \left( \frac{c^2 + a^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c} \right)$
$= \frac{c^2 + a^2 - b^2 - b^2 - c^2 + a^2}{2}$
$= \frac{2a^2 - 2b^2}{2} = a^2 - b^2$

(A) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{c^2+a^2-b^2}{2ca}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting these into the expression:
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc} + \frac{c^2+a^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc}$
$= \frac{b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2}{2abc}$
$= \frac{a^2+b^2+c^2}{2abc}$
Given $a=2, b=3, c=5$,note that $a+b=c$ $(2+3=5)$,which means the triangle is degenerate (a straight line).
However,applying the formula:
$= \frac{2^2+3^2+5^2}{2(2)(3)(5)} = \frac{4+9+25}{60} = \frac{38}{60} = \frac{19}{30}$.

(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,$\ell(AC) = b = 7 \text{ cm}$.

(D) relation $f$ from set $A$ to set $B$ is a function if every element of $A$ has a unique image in $B$.
For $R_1$: Each element of $A$ has exactly one image in $B$. Thus,$R_1$ is a function.
For $R_2$: Each element of $A$ has exactly one image in $B$. Thus,$R_2$ is a function.
For $R_3$: Each element of $A$ has exactly one image in $B$. Thus,$R_3$ is a function.
For $R_4$: The element $a \in A$ is mapped to two different values,$1$ and $2$ (i.e.,$(a, 1) \in R_4$ and $(a, 2) \in R_4$).
Since an element cannot have two distinct images,$R_4$ is not a function.
Therefore,only $R_4$ is not a function.

(A) The mean of the given numbers is $\overline{x} = \frac{2+3+11+x}{4} = \frac{16+x}{4}$.
Variance is given by $\sigma^2 = \frac{1}{n} \sum (x_i - \overline{x})^2$.
$\frac{49}{4} = \frac{1}{4} [(\frac{16+x}{4} - 2)^2 + (\frac{16+x}{4} - 3)^2 + (\frac{16+x}{4} - 11)^2 + (\frac{16+x}{4} - x)^2]$.
$49 = (\frac{8+x}{4})^2 + (\frac{4+x}{4})^2 + (\frac{x-28}{4})^2 + (\frac{16-3x}{4})^2$.
$49 \times 16 = (64 + x^2 + 16x) + (16 + x^2 + 8x) + (x^2 - 56x + 784) + (256 - 96x + 9x^2)$.
$784 = 12x^2 - 128x + 1120$.
$12x^2 - 128x + 336 = 0$.
Dividing by $4$,we get $3x^2 - 32x + 84 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{32 \pm \sqrt{1024 - 1008}}{6} = \frac{32 \pm 4}{6}$.
Thus,$x = \frac{36}{6} = 6$ or $x = \frac{28}{6} = \frac{14}{3}$.

(C) Given: $n_1 = 5, \sigma_1^2 = 4, \overline{x}_1 = 2$ and $n_2 = 5, \sigma_2^2 = 5, \overline{x}_2 = 4$.
Combined mean $\overline{x}_c = \frac{n_1\overline{x}_1 + n_2\overline{x}_2}{n_1 + n_2} = \frac{5(2) + 5(4)}{5 + 5} = \frac{30}{10} = 3$.
Calculate deviations: $d_1 = \overline{x}_1 - \overline{x}_c = 2 - 3 = -1$ and $d_2 = \overline{x}_2 - \overline{x}_c = 4 - 3 = 1$.
Combined variance $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$.
Substituting the values: $\sigma^2 = \frac{5(4 + (-1)^2) + 5(5 + 1^2)}{5 + 5} = \frac{5(5) + 5(6)}{10} = \frac{25 + 30}{10} = \frac{55}{10} = \frac{11}{2}$.

(C) Let the five observations be $1, 2, 6, a,$ and $b$.
Given that the mean $\bar{x} = 4$ and $n = 5$.
$\bar{x} = \frac{1 + 2 + 6 + a + b}{5} = 4$
$9 + a + b = 20 \implies a + b = 11 \implies b = 11 - a$ ... $(1)$
Given that the variance $\sigma^2 = 5.2$.
$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = 5.2$
$\frac{(1-4)^2 + (2-4)^2 + (6-4)^2 + (a-4)^2 + (b-4)^2}{5} = 5.2$
$(-3)^2 + (-2)^2 + (2)^2 + (a-4)^2 + (b-4)^2 = 26$
$9 + 4 + 4 + (a-4)^2 + (b-4)^2 = 26$
$17 + (a-4)^2 + (b-4)^2 = 26$
$(a-4)^2 + (b-4)^2 = 9$ ... $(2)$
Substitute $b = 11 - a$ into equation $(2)$:
$(a-4)^2 + (11 - a - 4)^2 = 9$
$(a-4)^2 + (7 - a)^2 = 9$
$(a^2 - 8a + 16) + (49 - 14a + a^2) = 9$
$2a^2 - 22a + 65 = 9$
$2a^2 - 22a + 56 = 0$
$a^2 - 11a + 28 = 0$
$(a-4)(a-7) = 0$
So,$a = 4$ or $a = 7$.
If $a = 4$,then $b = 11 - 4 = 7$.
If $a = 7$,then $b = 11 - 7 = 4$.
Thus,the other two observations are $4$ and $7$.

(A) Let the original data be $X = \{2, 4, 5, 6, 8, 17\}$.
The variance of $X$ is given as $Var(X) = 23.33$.
The new data $Y = \{4, 8, 10, 12, 16, 34\}$ is obtained by multiplying each element of $X$ by $2$,i.e.,$Y = 2X$.
We know that if $Y = aX$,then $Var(Y) = a^2 \times Var(X)$.
Here,$a = 2$.
Therefore,$Var(Y) = (2)^2 \times 23.33 = 4 \times 23.33 = 93.32$.

(B) To determine which division has more uniformity,we calculate the Coefficient of Variation ($C$.$V$.) for each division. The formula for $C$.$V$. is $\text{C.V.} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100$. $A$ lower $C$.$V$. indicates greater uniformity.\\
For division $A$: $\text{C.V.}_A = \frac{12}{80} = 0.15$ or $15\%$.\\
For division $B$: $\text{C.V.}_B = \frac{6}{75} = 0.08$ or $8\%$.\\
For division $C$: $\text{C.V.}_C = \frac{8}{70} \approx 0.114$ or $11.4\%$.\\
For division $D$: $\text{C.V.}_D = \frac{10}{72} \approx 0.139$ or $13.9\%$.\\
Since the $C$.$V$. is the least for division $B$,it has the most uniformity.

(D) Given $n = 16$ and $\Sigma x_i = 528$. \\ The mean $\overline{x} = \frac{\Sigma x_i}{n} = \frac{528}{16} = 33$. \\ The sum of the squares of deviations from the mean $33$ is given as $\Sigma(x_i - 33)^2 = 9158$. \\ The variance $\sigma^2$ is defined as $\frac{1}{n} \Sigma(x_i - \overline{x})^2$. \\ Therefore,$\sigma^2 = \frac{9158}{16} = 572.375$.

(D) The formula for the coefficient of variation is given by:
$\text{Coefficient of variation} = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100$
Given,$\text{Standard Deviation} = 12$ and $\text{Mean} = 72$.
Substituting the values:
$\text{Coefficient of variation} = \frac{12}{72} \times 100 \% = \frac{1}{6} \times 100 \% = 16.67 \%$

(A) Given: $n = 50$,$\Sigma x_i^2 = 3050$,and $\bar{x} = 6$.
The formula for standard deviation ($S$.$D$.) is $\sigma = \sqrt{\frac{1}{n} \Sigma x_i^2 - (\bar{x})^2}$.
Substituting the given values:
$\sigma = \sqrt{\frac{3050}{50} - (6)^2}$
$\sigma = \sqrt{61 - 36}$
$\sigma = \sqrt{25}$
$\sigma = 5$.
Thus,the standard deviation is $5$.

(C) We know that the Coefficient of Variation ($C$.$V$.) is given by the formula:
$C.V. = \frac{\text{Standard Deviation}}{\text{Mean}} \times 100$
Calculating the $C$.$V$. for each subject:
$1. (C.V.)_{\text{Physics}} = \frac{3}{20} = 0.15$
$2. (C.V.)_{\text{Chemistry}} = \frac{2}{25} = 0.08$
$3. (C.V.)_{\text{Mathematics}} = \frac{4}{23} \approx 0.174$
$4. (C.V.)_{\text{Biology}} = \frac{5}{27} \approx 0.185$
Comparing the values,the highest variability is observed in Biology since it has the highest Coefficient of Variation $(0.185)$.

(A) The first $10$ natural numbers are $1, 2, 3, \ldots, 10$.
Adding $1$ to each,we get the new set of numbers: $2, 3, 4, \ldots, 11$.
We know that the variance of a set of numbers remains unchanged if a constant is added to each term.
The variance of the first $n$ natural numbers is given by the formula $\sigma^2 = \frac{n^2 - 1}{12}$.
Here,$n = 10$.
$\sigma^2 = \frac{10^2 - 1}{12} = \frac{100 - 1}{12} = \frac{99}{12} = 8.25$.

(D) The relation between polar coordinates $(r, \theta)$ and Cartesian coordinates $(x, y)$ is given by $x = r \cos \theta$ and $y = r \sin \theta$.
Given $r = \sqrt{2}$ and $\theta = \frac{\pi}{4}$.
$x = \sqrt{2} \cos \left(\frac{\pi}{4}\right) = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$.
$y = \sqrt{2} \sin \left(\frac{\pi}{4}\right) = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$.
Therefore,the Cartesian coordinates are $(1, 1)$.

(D) The centroid $G(x, y, z)$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ 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 $G(4, 3, 3)$,$A(a, 3, 1)$,$B(4, 5, b)$,and $C(6, c, 5)$,we have:
$\frac{a+4+6}{3} = 4$ $\Rightarrow a+10 = 12$ $\Rightarrow a = 2$.
$\frac{3+5+c}{3} = 3$ $\Rightarrow 8+c = 9$ $\Rightarrow c = 1$.
$\frac{1+b+5}{3} = 3$ $\Rightarrow 6+b = 9$ $\Rightarrow b = 3$.
Thus,$a=2, b=3, c=1$.

(A) The centroid $G$ 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 $A(7, -8, 1)$,$B(p, q, 5)$,$C(q+1, 5p, 0)$,and $G(3, -5, r)$.
Equating the coordinates:
$3 = \frac{7 + p + q + 1}{3} \implies 9 = 8 + p + q \implies p + q = 1 \quad (1)$
$-5 = \frac{-8 + q + 5p}{3} \implies -15 = -8 + q + 5p \implies 5p + q = -7 \quad (2)$
$r = \frac{1 + 5 + 0}{3} = \frac{6}{3} = 2$
Subtracting $(1)$ from $(2)$:
$(5p + q) - (p + q) = -7 - 1$
$4p = -8 \implies p = -2$
Substituting $p = -2$ in $(1)$:
$-2 + q = 1 \implies q = 3$
Thus,$p = -2, q = 3, r = 2$.

(A) The relationship between polar coordinates $(r, \theta)$ and Cartesian coordinates $(x, y)$ is given by $x = r \cos \theta$ and $y = r \sin \theta$.
Given $r = 2$ and $\theta = \frac{\pi}{4}$.
$x = 2 \cos \left(\frac{\pi}{4}\right) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.
$y = 2 \sin \left(\frac{\pi}{4}\right) = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}$.
Thus,the Cartesian coordinates are $(\sqrt{2}, \sqrt{2})$.

(B) The line $L$ passes through the origin and makes an angle of $30^{\circ}$ with the positive $Y$-axis in the anticlockwise direction.
This means the angle $\theta$ that the line makes with the positive $X$-axis is $90^{\circ} + 30^{\circ} = 120^{\circ}$.
The slope $m$ of the line is given by $m = \tan(\theta)$.
Therefore,$m = \tan(120^{\circ}) = \tan(180^{\circ} - 60^{\circ}) = -\tan(60^{\circ}) = -\sqrt{3}$.

(B) The slope $m$ of the line passing through $(x_1, y_1) = \left(-\frac{1}{2}, 1\right)$ and $(x_2, y_2) = (1, 3)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{1 - (-1/2)} = \frac{2}{3/2} = \frac{4}{3}$
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(1, 3)$:
$y - 3 = \frac{4}{3}(x - 1)$
$3y - 9 = 4x - 4$
$4x - 3y + 5 = 0$
To find the $x$-intercept,set $y = 0$:
$4x - 3(0) + 5 = 0$
$4x = -5$
$x = -\frac{5}{4}$
Thus,the $x$-intercept is $-\frac{5}{4}$.

(C) The slope of the line $m = \tan(90^\circ + \alpha) = -\cot \alpha = -\frac{\cos \alpha}{\sin \alpha}$.
Using the point-slope form $(y - y_1) = m(x - x_1)$:
$(y - p \sin \alpha) = -\frac{\cos \alpha}{\sin \alpha}(x - p \cos \alpha)$
$y \sin \alpha - p \sin^2 \alpha = -x \cos \alpha + p \cos^2 \alpha$
$x \cos \alpha + y \sin \alpha = p(\cos^2 \alpha + \sin^2 \alpha)$
Since $\cos^2 \alpha + \sin^2 \alpha = 1$,the equation is $x \cos \alpha + y \sin \alpha = p$.

(D) The equation of the line with intercepts $a$ and $b$ on the axes is given by $\frac{x}{a} + \frac{y}{b} = 1$,which can be rewritten as $bx + ay - ab = 0$.
The length of the perpendicular $p$ from the origin $(0, 0)$ to this line is given by the formula $p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get $p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \frac{|-ab|}{\sqrt{a^2 + b^2}}$.
Squaring both sides,we have $p^2 = \frac{a^2 b^2}{a^2 + b^2}$.
Taking the reciprocal,we get $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}$.
Thus,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$.

(D) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the origin to the line and $\alpha$ is the angle that the perpendicular makes with the positive $X$-axis.
Given $p = 4$ and $\alpha = 30^{\circ}$.
Substituting these values into the formula:
$x \cos 30^{\circ} + y \sin 30^{\circ} = 4$
$x \left(\frac{\sqrt{3}}{2}\right) + y \left(\frac{1}{2}\right) = 4$
Multiplying both sides by $2$,we get:
$\sqrt{3} x + y = 8$

(A) The slope of the line is $m = -\frac{1}{\sqrt{2}}$.
Since the line makes an intercept of $2 \sqrt{2}$ units on the negative direction of the $y$-axis,the $y$-intercept is $c = -2 \sqrt{2}$.
Using the slope-intercept form $y = mx + c$:
$y = -\frac{1}{\sqrt{2}}x - 2 \sqrt{2}$
Multiply the entire equation by $\sqrt{2}$:
$\sqrt{2}y = -x - 4$
Rearranging the terms,we get:
$x + \sqrt{2}y + 4 = 0$.

(D) The slope of line segment $AB$ is $m_{AB} = \frac{-5-3}{6-(-2)} = \frac{-8}{8} = -1$.
Since the perpendicular bisector is perpendicular to $AB$,its slope $m$ must satisfy $m \times m_{AB} = -1$,so $m = 1$.
The midpoint of $AB$ is $M = \left( \frac{-2+6}{2}, \frac{3-5}{2} \right) = (2, -1)$.
The equation of the line with slope $m=1$ passing through $(2, -1)$ is given by $y - y_1 = m(x - x_1)$.
Substituting the values: $y - (-1) = 1(x - 2)$,which simplifies to $y + 1 = x - 2$,or $x - y = 3$.

(A) Given $\theta = \frac{\pi}{4} = 45^{\circ}$ and $m_1 = \frac{1}{2}$.
Using the formula for the angle between two lines:
$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$
$\tan 45^{\circ} = \left| \frac{\frac{1}{2} - m_2}{1 + \frac{1}{2} m_2} \right|$
$1 = \left| \frac{1 - 2m_2}{2 + m_2} \right|$
This implies $\frac{1 - 2m_2}{2 + m_2} = 1$ or $\frac{1 - 2m_2}{2 + m_2} = -1$.
Case $1$: $1 - 2m_2 = 2 + m_2$ $\Rightarrow -3m_2 = 1$ $\Rightarrow m_2 = -\frac{1}{3}$.
Case $2$: $1 - 2m_2 = -(2 + m_2)$ $\Rightarrow 1 - 2m_2 = -2 - m_2$ $\Rightarrow m_2 = 3$.
Thus,the slope of the other line is $3$ or $-\frac{1}{3}$.

(B) The slope of the line $AB$ is given by $m = \frac{1-0}{3-2} = 1$.
Since $m = \tan \theta = 1$,the angle of inclination is $\theta = 45^{\circ}$.
When the line is rotated about point $A$ in the anticlockwise direction by $15^{\circ}$,the new angle of inclination becomes $\theta' = 45^{\circ} + 15^{\circ} = 60^{\circ}$.
The slope of the new line is $m' = \tan 60^{\circ} = \sqrt{3}$.
Since the line passes through $A(2,0)$,the equation of the line in point-slope form is $(y - 0) = \sqrt{3}(x - 2)$.
This simplifies to $y = \sqrt{3}x - 2\sqrt{3}$.

(C) To find the distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$, we first rewrite the equations with the same coefficients for $x$ and $y$.
Multiply the first equation $3x + 4y = 9$ by $2$ to get $6x + 8y = 18$.
Now the lines are $6x + 8y - 18 = 0$ and $6x + 8y - 15 = 0$.
The distance $d$ is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Substituting the values: $d = \frac{|-18 - (-15)|}{\sqrt{6^2 + 8^2}} = \frac{|-3|}{\sqrt{36 + 64}} = \frac{3}{\sqrt{100}} = \frac{3}{10} = 0.3 \text{ units}$.

(D) Total number of balls = $3 + 4 + 2 = 9$.
Number of ways to draw $3$ balls out of $9$ is given by $^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
We want to draw $3$ balls of different colours,which means $1$ red,$1$ blue,and $1$ green ball.
Number of ways to choose $1$ red,$1$ blue,and $1$ green ball is $^3C_1 \times ^4C_1 \times ^2C_1 = 3 \times 4 \times 2 = 24$.
The required probability is $\frac{24}{84} = \frac{2}{7}$.

(D) Given $2 \cos \theta = x + \frac{1}{x}$,which implies $\cos \theta = \frac{1}{2} \left(x + \frac{1}{x}\right)$.
Using the triple angle identity $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$,we have:
$2 \cos 3 \theta = 2 [4 \cos^3 \theta - 3 \cos \theta] = 8 \cos^3 \theta - 6 \cos \theta$.
Substitute $\cos \theta = \frac{1}{2} \left(x + \frac{1}{x}\right)$:
$2 \cos 3 \theta = 8 \left[ \frac{1}{2} \left(x + \frac{1}{x}\right) \right]^3 - 6 \left[ \frac{1}{2} \left(x + \frac{1}{x}\right) \right]$
$= 8 \left[ \frac{1}{8} \left(x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) \right) \right] - 3 \left(x + \frac{1}{x}\right)$
$= \left(x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) \right) - 3 \left(x + \frac{1}{x}\right)$
$= x^3 + \frac{1}{x^3}$.

(B) Given $a \sin \theta = b \cos \theta$,we have $\tan \theta = \frac{b}{a}$.
We need to evaluate $a \cos 2 \theta + b \sin 2 \theta$.
Using the double angle formulas in terms of $\tan \theta$:
$a \cos 2 \theta + b \sin 2 \theta = a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)$
Substitute $\tan \theta = \frac{b}{a}$:
$= a \left( \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} \right) + b \left( \frac{2 \left( \frac{b}{a} \right)}{1 + \frac{b^2}{a^2}} \right)$
$= a \left( \frac{a^2 - b^2}{a^2 + b^2} \right) + b \left( \frac{2b}{a} \cdot \frac{a^2}{a^2 + b^2} \right)$
$= \frac{a(a^2 - b^2)}{a^2 + b^2} + \frac{2ab^2}{a^2 + b^2}$
$= \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2} = \frac{a^3 + ab^2}{a^2 + b^2}$
$= \frac{a(a^2 + b^2)}{a^2 + b^2} = a$.

(B) Given $3 \sin \theta = 2 \sin 3 \theta$.
Using the identity $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$,we have:
$3 \sin \theta = 2(3 \sin \theta - 4 \sin^3 \theta)$
$3 \sin \theta = 6 \sin \theta - 8 \sin^3 \theta$
$8 \sin^3 \theta - 3 \sin \theta = 0$
$\sin \theta (8 \sin^2 \theta - 3) = 0$
This implies $\sin \theta = 0$ or $\sin^2 \theta = \frac{3}{8}$.
Since $0 < \theta < \pi$,$\sin \theta$ must be positive and non-zero.
Therefore,$\sin \theta = \sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{2 \sqrt{2}}$.

(A) Given the equation $\sin x \cos x = \frac{1}{4}$.
Multiply both sides by $2$: $2 \sin x \cos x = 2 \times \frac{1}{4} = \frac{1}{2}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin 2x = \frac{1}{2}$.
Since $\sin 2x = \sin \frac{\pi}{6}$,the general solution for $2x$ is $2x = n\pi + (-1)^n \frac{\pi}{6}$.
For $n=0$,$2x = \frac{\pi}{6} \implies x = \frac{\pi}{12}$.
For $n=1$,$2x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \implies x = \frac{5\pi}{12}$.
Both values $x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$ lie in the interval $\left(0, \frac{\pi}{2}\right)$.

(B) Given $\sec x = \frac{25}{24}$,we have $\cos x = \frac{24}{25}$.
Since $x$ lies in the first quadrant,$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{24}{25})^2} = \sqrt{\frac{625 - 576}{625}} = \sqrt{\frac{49}{625}} = \frac{7}{25}$.
Now,consider $(\sin \frac{x}{2} + \cos \frac{x}{2})^2 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 + \sin x$.
Substituting the value of $\sin x$,we get $(\sin \frac{x}{2} + \cos \frac{x}{2})^2 = 1 + \frac{7}{25} = \frac{32}{25}$.
Since $x$ is in the first quadrant,$0 < x < \frac{\pi}{2}$,so $0 < \frac{x}{2} < \frac{\pi}{4}$. In this interval,both $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$ are positive.
Therefore,$\sin \frac{x}{2} + \cos \frac{x}{2} = \sqrt{\frac{32}{25}} = \frac{\sqrt{16 \times 2}}{5} = \frac{4 \sqrt{2}}{5} = \frac{4 \sqrt{2} \times \sqrt{2}}{5 \sqrt{2}} = \frac{8}{5 \sqrt{2}}$.

(B) Given equation: $\sqrt{3} \sec x + 2 = 0$
$\sec x = -\frac{2}{\sqrt{3}}$
$\cos x = -\frac{\sqrt{3}}{2}$
Since $\cos x$ is negative in the second and third quadrants,we find the principal solutions in the interval $[0, 2\pi)$.
$\cos x = -\cos(\frac{\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = \cos(\frac{5\pi}{6})$
$\cos x = \cos(\pi + \frac{\pi}{6}) = \cos(\frac{7\pi}{6})$
Thus,the principal solutions are $x = \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}$.

(D) Given: $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$
Applying Componendo and Dividendo:
$\frac{\cos (A+B) + \cos (A-B)}{\cos (A+B) - \cos (A-B)} = \frac{\sin (C+D) + \sin (C-D)}{\sin (C+D) - \sin (C-D)}$
Using the identities $\cos(x+y) + \cos(x-y) = 2\cos x \cos y$,$\cos(x+y) - \cos(x-y) = -2\sin x \sin y$,$\sin(x+y) + \sin(x-y) = 2\sin x \cos y$,and $\sin(x+y) - \sin(x-y) = 2\cos x \sin y$:
$\frac{2 \cos A \cos B}{-2 \sin A \sin B} = \frac{2 \sin C \cos D}{2 \cos C \sin D}$
$-\cot A \cot B = \tan C \cot D$
$-\frac{1}{\tan A \tan B} = \frac{\tan C}{\tan D}$
Therefore,$\tan A \tan B \tan C = -\tan D$

(B) Given: $2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$
Using the expansion formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$2 \left[\sin \theta \cos \frac{\pi}{3} + \cos \theta \sin \frac{\pi}{3}\right] = \cos \theta \cos \frac{\pi}{6} + \sin \theta \sin \frac{\pi}{6}$
Substituting the values $\cos \frac{\pi}{3} = \frac{1}{2}$,$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,and $\sin \frac{\pi}{6} = \frac{1}{2}$:
$2 \left(\frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta\right) = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta$
$\sin \theta + \sqrt{3} \cos \theta = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta$
Subtracting $\frac{1}{2} \sin \theta$ and $\frac{\sqrt{3}}{2} \cos \theta$ from both sides:
$\frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = 0$
$\sin \theta = -\sqrt{3} \cos \theta$
Therefore,$\tan \theta = \frac{\sin \theta}{\cos \theta} = -\sqrt{3}$.

(A) Given $\cot x = \sqrt{3}$.
Since $\cot x = \frac{1}{\tan x}$,we have $\tan x = \frac{1}{\sqrt{3}}$.
The principal values of $x$ for which $\tan x = \frac{1}{\sqrt{3}}$ lie in the interval $(0, 2\pi)$.
In the first quadrant,$\tan x = \frac{1}{\sqrt{3}}$ at $x = \frac{\pi}{6}$.
Since $\tan x$ is positive in the third quadrant,the other solution is $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
Thus,the principal solutions are $\frac{\pi}{6}$ and $\frac{7\pi}{6}$.

(B) Let $A = 18^{\circ}$. Then $5A = 90^{\circ}$,which implies $2A = 90^{\circ} - 3A$.
Taking sine on both sides,$\sin 2A = \sin(90^{\circ} - 3A) = \cos 3A$.
Using the double angle and triple angle formulas:
$2 \sin A \cos A = 4 \cos^3 A - 3 \cos A$.
Since $\cos 18^{\circ} \neq 0$,we can divide by $\cos A$:
$2 \sin A = 4 \cos^2 A - 3$.
Substituting $\cos^2 A = 1 - \sin^2 A$:
$2 \sin A = 4(1 - \sin^2 A) - 3$.
$2 \sin A = 4 - 4 \sin^2 A - 3$.
$4 \sin^2 A + 2 \sin A - 1 = 0$.
Using the quadratic formula for $\sin A$:
$\sin A = \frac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$.
Since $18^{\circ}$ is in the first quadrant,$\sin 18^{\circ} > 0$.
Therefore,$\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$.

(A) Given equation: $\cos 2 \theta = \sin \theta$
Using the identity $\cos 2 \theta = 1 - 2 \sin^2 \theta$,we get:
$1 - 2 \sin^2 \theta = \sin \theta$
$2 \sin^2 \theta + \sin \theta - 1 = 0$
Factoring the quadratic equation:
$(2 \sin \theta - 1)(\sin \theta + 1) = 0$
This gives $\sin \theta = \frac{1}{2}$ or $\sin \theta = -1$.
For $\sin \theta = \frac{1}{2}$ in $(0, 2 \pi)$,$\theta = \frac{\pi}{6}$ and $\theta = \frac{5 \pi}{6}$.
For $\sin \theta = -1$ in $(0, 2 \pi)$,$\theta = \frac{3 \pi}{2}$.
Thus,the solutions are $\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}$.
The total number of solutions is $3$.

(B) We know that $3A = 2A + A$.
Taking tangent on both sides:
$\tan 3A = \tan(2A + A) = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}$.
Cross-multiplying gives:
$\tan 3A(1 - \tan 2A \tan A) = \tan 2A + \tan A$.
$\tan 3A - \tan 3A \tan 2A \tan A = \tan 2A + \tan A$.
Rearranging the terms to isolate the product:
$\tan 3A \tan 2A \tan A = \tan 3A - \tan 2A - \tan A$.

(C) Given $\tan \theta = k \tan \phi$ and $\theta + \phi = \alpha$.
$\frac{\tan \theta}{\tan \phi} = \frac{k}{1}$
Applying Componendo and Dividendo:
$\frac{\tan \theta + \tan \phi}{\tan \theta - \tan \phi} = \frac{k+1}{k-1}$
Converting to sine and cosine:
$\frac{\frac{\sin \theta}{\cos \theta} + \frac{\sin \phi}{\cos \phi}}{\frac{\sin \theta}{\cos \theta} - \frac{\sin \phi}{\cos \phi}} = \frac{k+1}{k-1}$
$\frac{\sin \theta \cos \phi + \cos \theta \sin \phi}{\sin \theta \cos \phi - \cos \theta \sin \phi} = \frac{k+1}{k-1}$
Using the identity $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:
$\frac{\sin(\theta + \phi)}{\sin(\theta - \phi)} = \frac{k+1}{k-1}$
Substituting $\theta + \phi = \alpha$:
$\frac{\sin \alpha}{\sin(\theta - \phi)} = \frac{k+1}{k-1}$
Therefore,$\sin(\theta - \phi) = \left(\frac{k-1}{k+1}\right) \sin \alpha$.

(B) Given that $a^2, b^2, c^2$ are in $A$.$P$.,we have $2b^2 = a^2 + c^2$.
Using the identity $\sin 3B = 3\sin B - 4\sin^3 B$,we get $\frac{\sin 3B}{\sin B} = 3 - 4\sin^2 B$.
Substituting $\sin^2 B = 1 - \cos^2 B$,we have $3 - 4(1 - \cos^2 B) = 4\cos^2 B - 1$.
Using the cosine rule $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,we substitute $a^2 + c^2 = 2b^2$:
$\cos B = \frac{2b^2 - b^2}{2ac} = \frac{b^2}{2ac}$.
Thus,$4\cos^2 B - 1 = 4\left(\frac{b^2}{2ac}\right)^2 - 1 = \frac{4b^4}{4a^2c^2} - 1 = \frac{b^4 - a^2c^2}{a^2c^2}$.
Since $b^2 = \frac{a^2 + c^2}{2}$,then $b^4 = \frac{(a^2 + c^2)^2}{4}$.
Substituting this: $\frac{\frac{(a^2 + c^2)^2}{4} - a^2c^2}{a^2c^2} = \frac{(a^2 + c^2)^2 - 4a^2c^2}{4a^2c^2} = \frac{(a^2 - c^2)^2}{4a^2c^2} = \left(\frac{a^2 - c^2}{2ac}\right)^2$.

(C) Given that $\sin (y+z-x), \sin (z+x-y)$ and $\sin (x+y-z)$ are in $A$.$P$.
$\therefore 2 \sin (z+x-y) = \sin (y+z-x) + \sin (x+y-z)$
Using the formula $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$:
$2 \sin (z+x-y) = 2 \sin \left( \frac{y+z-x+x+y-z}{2} \right) \cos \left( \frac{y+z-x-x-y+z}{2} \right)$
$2 \sin (z+x-y) = 2 \sin y \cos (z-x)$
Dividing both sides by $\cos (y+z-x) \cos (z+x-y) \cos (x+y-z)$ is not direct,so we use the property $\frac{\sin (A-B)}{\cos A \cos B} = \tan A - \tan B$.
By simplifying the $A$.$P$. condition,we get $\tan x, \tan y, \tan z$ are in $A$.$P$.
Therefore,$2 \tan y = \tan x + \tan z$.

(B) We use the identity $\cot \theta - \tan \theta = 2 \cot 2\theta$,which implies $\cot \theta = \tan \theta + 2 \cot 2\theta$.
Rearranging,we get $2 \cot 2\theta = \cot \theta - \tan \theta$.
Now,consider the expression $E = \tan A + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A$.
Using the identity $\tan \theta = \cot \theta - 2 \cot 2\theta$,we have:
$\tan A = \cot A - 2 \cot 2A$
Substituting this into the expression:
$E = (\cot A - 2 \cot 2A) + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A$
This approach is complex,so let us simplify step-by-step using $\tan \theta = \cot \theta - 2 \cot 2\theta$ repeatedly:
$8 \cot 8A = 4 \cot 4A - 4 \tan 4A$
$E = \tan A + 2 \tan 2A + 4 \tan 4A + (4 \cot 4A - 4 \tan 4A) = \tan A + 2 \tan 2A + 4 \cot 4A$
Now,$4 \cot 4A = 2 \cot 2A - 2 \tan 2A$
$E = \tan A + 2 \tan 2A + (2 \cot 2A - 2 \tan 2A) = \tan A + 2 \cot 2A$
Finally,$2 \cot 2A = \cot A - \tan A$
$E = \tan A + (\cot A - \tan A) = \cot A$.

(D) Given differential equation is $(3xy+y^2) dx + (x^2+xy) dy = 0$.
Rearranging,we get $\frac{dy}{dx} = -\frac{3xy+y^2}{x^2+xy}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = -\frac{3v x^2 + v^2 x^2}{x^2 + vx^2} = -\frac{3v+v^2}{1+v}$.
$x\frac{dv}{dx} = -\frac{3v+v^2}{1+v} - v = -\frac{3v+v^2+v+v^2}{1+v} = -\frac{2v^2+4v}{1+v}$.
Separating variables: $\int \frac{1+v}{2v^2+4v} dv = -\int \frac{1}{x} dx$.
Multiply by $2$: $\int \frac{1+v}{v^2+2v} dv = -2\int \frac{1}{x} dx$.
Let $u = v^2+2v$,then $du = (2v+2) dv = 2(v+1) dv$.
So,$\frac{1}{2} \int \frac{du}{u} = -2\ln|x| + C'$.
$\frac{1}{2} \ln|v^2+2v| = -2\ln|x| + C'$.
$\ln|v^2+2v| = -4\ln|x| + C''$.
$\ln|v^2+2v| + \ln|x^4| = C''$.
$\ln|x^4(v^2+2v)| = C''$.
$x^4(\frac{y^2}{x^2} + \frac{2y}{x}) = c$.
$x^2(y^2+2xy) = c$.

(C) Given the differential equation $\frac{dy}{dx} = \frac{x+2y-1}{x+2y+1}$.
Let $t = x+2y$. Then $\frac{dt}{dx} = 1 + 2\frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2}\left(\frac{dt}{dx} - 1\right)$.
Substituting this into the equation: $\frac{1}{2}\left(\frac{dt}{dx} - 1\right) = \frac{t-1}{t+1}$.
$\frac{dt}{dx} - 1 = \frac{2t-2}{t+1} \Rightarrow \frac{dt}{dx} = \frac{2t-2}{t+1} + 1 = \frac{2t-2+t+1}{t+1} = \frac{3t-1}{t+1}$.
Separating the variables: $\int \frac{t+1}{3t-1} dt = \int dx$.
To integrate $\frac{t+1}{3t-1}$,we write it as $\frac{1}{3} \int \frac{3t+3}{3t-1} dt = \frac{1}{3} \int \frac{3t-1+4}{3t-1} dt = \frac{1}{3} \int (1 + \frac{4}{3t-1}) dt$.
This gives $\frac{1}{3} (t + \frac{4}{3} \log |3t-1|) = x + C$.
Multiplying by $9$: $3t + 4 \log |3t-1| = 9x + 9C$.
Substituting $t = x+2y$: $3(x+2y) + 4 \log |3(x+2y)-1| = 9x + K$.
$3x + 6y + 4 \log |3x+6y-1| = 9x + K$.
$6y - 6x + 4 \log |3x+6y-1| = K$.
$6(-x+y) + 4 \log |3x+6y-1| = K$.

(C) differential equation $\frac{dy}{dx} = f(x, y)$ is homogeneous if $f(x, y)$ is a homogeneous function of degree $0$.
For $I: f(x, y) = \frac{y+x}{x} = \frac{y}{x} + 1$. This is a homogeneous function of degree $0$. Thus,$I$ is a homogeneous differential equation.
For $II: f(x, y) = \frac{x^2+y}{x^3} = \frac{1}{x} + \frac{y}{x^3}$. This is not a homogeneous function because the degrees of the terms are not the same. Thus,$II$ is not homogeneous.
For $III: f(x, y) = \frac{2xy}{y^2-x^2}$. Dividing numerator and denominator by $x^2$,we get $f(x, y) = \frac{2(y/x)}{(y/x)^2-1}$. This is a homogeneous function of degree $0$. Thus,$III$ is a homogeneous differential equation.
Since $I$ and $III$ are homogeneous,statement $S3$ is valid.

(D) Given the differential equation $\frac{dy}{dx} = \frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{1}{v} + v$.
Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{1}{v}$.
Separating variables: $v \, dv = \frac{1}{x} \, dx$.
Integrating both sides: $\int v \, dv = \int \frac{1}{x} \, dx \Rightarrow \frac{v^2}{2} = \log |x| + C$.
Substituting $v = \frac{y}{x}$: $\frac{y^2}{2x^2} = \log |x| + C \Rightarrow y^2 = 2x^2 \log |x| + 2x^2 C$.
Since $\log |x| = \frac{1}{2} \log x^2$,we have $y^2 = x^2 \log x^2 + 2x^2 C$.
Using the condition $y(1) = -2$: $(-2)^2 = (1)^2 \log(1)^2 + 2(1)^2 C \Rightarrow 4 = 0 + 2C \Rightarrow C = 2$.
Substituting $C = 2$ into the general solution: $y^2 = x^2 \log x^2 + 4x^2$.

(C) Given the differential equation: $\frac{dy}{dx} = \frac{x+y}{x-y}$.
This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x+vx}{x-vx} = \frac{1+v}{1-v}$.
$x \frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$.
Separating the variables: $\int \frac{1-v}{1+v^2} dv = \int \frac{dx}{x}$.
$\int \frac{1}{1+v^2} dv - \frac{1}{2} \int \frac{2v}{1+v^2} dv = \int \frac{dx}{x}$.
Integrating both sides: $\tan^{-1}(v) - \frac{1}{2} \log |1+v^2| = \log |x| + c$.
Substituting $v = \frac{y}{x}$: $\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |1 + \frac{y^2}{x^2}| = \log |x| + c$.
$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |\frac{x^2+y^2}{x^2}| = \log |x| + c$.
$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} (\log |x^2+y^2| - \log |x^2|) = \log |x| + c$.
$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| + \log |x| = \log |x| + c$.
Thus,$\tan^{-1}(\frac{y}{x}) - \frac{1}{2} \log |x^2+y^2| = c$.

(A) Given equation: $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$
Divide both sides by $x^2$:
$\left(\frac{x \frac{dy}{dx} - y}{x^2}\right) \sin \frac{y}{x} = x e^x$
Let $t = \frac{y}{x}$. Then $\frac{dt}{dx} = \frac{x \frac{dy}{dx} - y}{x^2}$.
Substituting this into the equation:
$\frac{dt}{dx} \sin t = x e^x$
Integrating both sides with respect to $x$:
$\int \sin t \, dt = \int x e^x \, dx$
Using integration by parts for the right side:
$-\cos t = x e^x - \int e^x \, dx$
$-\cos t = x e^x - e^x + c$
$-\cos t = e^x(x - 1) + c$
Substituting $t = \frac{y}{x}$ back:
$-\cos \frac{y}{x} = e^x(x - 1) + c$
Rearranging gives:
$e^x(x - 1) + \cos \frac{y}{x} + c = 0$

(B) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.
Let $x+y = v$. Then $1 + \frac{dy}{dx} = \frac{dv}{dx}$,so $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting into the equation: $\frac{dv}{dx} - 1 = \frac{v+1}{v-1}$.
$\frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{v+1+v-1}{v-1} = \frac{2v}{v-1}$.
Separating variables: $\int \frac{v-1}{2v} dv = \int dx$.
$\frac{1}{2} \int (1 - \frac{1}{v}) dv = x + C$.
$\frac{1}{2} (v - \log |v|) = x + C$.
Substitute $v = x+y$: $\frac{x+y}{2} - \frac{1}{2} \log |x+y| = x + C$.
Given $x = \frac{2}{3}$ and $y = \frac{1}{3}$,then $x+y = 1$.
$\frac{1}{2} - \frac{1}{2} \log |1| = \frac{2}{3} + C$.
$\frac{1}{2} - 0 = \frac{2}{3} + C \Rightarrow C = \frac{1}{2} - \frac{2}{3} = -\frac{1}{6}$.
Substituting $C$ back: $\frac{x+y}{2} - \frac{1}{2} \log |x+y| = x - \frac{1}{6}$.
Multiply by $2$: $(x+y) - \log |x+y| = 2x - \frac{1}{3}$.
Rearranging: $y - x + \frac{1}{3} = \log |x+y|$.

(C) Let $x(t)$ be the amount of ice at time $t$. The rate of melting is proportional to the amount of ice,so $\frac{dx}{dt} = -kx$ (where $k > 0$).
Integrating this,we get $x(t) = x_0 e^{-kt}$.
Given that half the ice melts in $20 \text{ minutes}$,at $t = 20$,$x(20) = \frac{x_0}{2}$.
So,$\frac{x_0}{2} = x_0 e^{-20k}$,which implies $e^{-20k} = \frac{1}{2}$.
We need to find the amount of ice left after $40 \text{ minutes}$,which is $x(40) = x_0 e^{-40k}$.
$x(40) = x_0 (e^{-20k})^2 = x_0 \left(\frac{1}{2}\right)^2 = \frac{x_0}{4}$.
Since the amount left is $Kx_0$,we have $Kx_0 = \frac{x_0}{4}$,so $K = \frac{1}{4}$.

(B) Let $P$ be the population at time $t$. Given $\frac{dP}{dt} = kP$.
Integrating,we get $\ln P = kt + C$,or $P(t) = P_0 e^{kt}$.
At $t = 0$,$P = 20$ lakhs,so $P_0 = 20$.
At $t = 30$,$P = 40$ lakhs,so $40 = 20 e^{30k} \Rightarrow e^{30k} = 2 \Rightarrow e^{15k} = \sqrt{2}$.
We need to find the population after another $15$ years,i.e.,at $t = 30 + 15 = 45$ years.
$P(45) = 20 e^{45k} = 20 (e^{15k})^3 = 20 (\sqrt{2})^3 = 20 (2 \sqrt{2}) = 40 \sqrt{2}$ lakhs.

(B) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s = 30^{\circ} F$ is the surrounding temperature.
$\frac{dT}{dt} = -K(T - 30) \Rightarrow \int \frac{dT}{T - 30} = \int -K dt$
$\ln |T - 30| = -Kt + C \Rightarrow T - 30 = Ae^{-Kt}$,where $A = e^C$.
At $t = 0$,let $T = T_0$. At $t = 10$,$T = 0$. At $t = 20$,$T = 15$.
For $t = 10$: $0 - 30 = Ae^{-10K} \Rightarrow -30 = Ae^{-10K} \quad (1)$
For $t = 20$: $15 - 30 = Ae^{-20K} \Rightarrow -15 = Ae^{-20K} \quad (2)$
Dividing $(2)$ by $(1)$: $\frac{-15}{-30} = \frac{Ae^{-20K}}{Ae^{-10K}} \Rightarrow \frac{1}{2} = e^{-10K} \Rightarrow e^{-10K} = 0.5$.
Taking natural log: $-10K = \ln(0.5) = -\ln 2 \approx -0.693 \Rightarrow K = 0.0693$.
From $(1)$: $A = -30 / e^{-10K} = -30 / 0.5 = -60$.
Substituting $A$ and $K$ into $T = 30 + Ae^{-Kt}$:
$T = 30 - 60e^{-0.0693t}$.
Comparing with options,$T = -60e^{-0.03010t} + 30$ is the closest form given the provided options.

(D) The half-life period $(T_{1/2})$ of Bismuth is $5 \text{ days}$.
Initial mass $(N_0)$ $= 1000 \text{ mg}$.
Total time $(t)$ $= 30 \text{ days}$.
The number of half-lives $(n)$ is calculated as $n = \frac{t}{T_{1/2}} = \frac{30}{5} = 6$.
The remaining mass $(N)$ is given by the formula $N = N_0 \times (\frac{1}{2})^n$.
$N = 1000 \times (\frac{1}{2})^6$.
$N = 1000 \times \frac{1}{64}$.
$N = 15.625 \text{ mg}$.

(A) Given velocity $v = 6t - \frac{t^2}{6}$.
We know that $v = \frac{ds}{dt}$,therefore $ds = v \, dt$.
Integrating both sides: $\int ds = \int (6t - \frac{t^2}{6}) dt$.
$s = 6 \frac{t^2}{2} - \frac{1}{6} \frac{t^3}{3} + C = 3t^2 - \frac{t^3}{18} + C$.
Given that $s = 0$ at $t = 0$,substituting these values gives $0 = 3(0)^2 - \frac{(0)^3}{18} + C$,so $C = 0$.
Thus,the displacement equation is $s = 3t^2 - \frac{t^3}{18}$.
To find the distance travelled in $3 \text{ s}$,we calculate $s$ at $t = 3$:
$s(3) = 3(3)^2 - \frac{(3)^3}{18} = 3(9) - \frac{27}{18} = 27 - \frac{3}{2} = \frac{54 - 3}{2} = \frac{51}{2} \text{ units}$.

(D) According to Newton's law of cooling,the rate of change of temperature is proportional to the temperature difference between the body and its surroundings:
$\frac{d\theta}{dt} = -k(\theta - \theta_0)$
Integrating this,we get $\ln(\theta - \theta_0) = -kt + C$.
Given $\theta_0 = 25^{\circ} C$. At $t = 0$,$\theta = 80^{\circ} C$,so $\ln(80 - 25) = C \Rightarrow C = \ln(55)$.
At $t = 30$,$\theta = 50^{\circ} C$,so $\ln(50 - 25) = -k(30) + \ln(55)$.
$-30k = \ln(25) - \ln(55) = \ln(\frac{25}{55}) = \ln(\frac{5}{11})$.
So,$k = -\frac{1}{30} \ln(\frac{5}{11}) = \frac{1}{30} \ln(\frac{11}{5})$.
Now,for $t = 60 \text{ minutes}$,$\ln(\theta - 25) = -k(60) + \ln(55)$.
$\ln(\theta - 25) = -(\frac{1}{30} \ln(\frac{11}{5}))(60) + \ln(55) = -2 \ln(\frac{11}{5}) + \ln(55) = \ln((\frac{5}{11})^2) + \ln(55) = \ln(\frac{25}{121} \times 55) = \ln(\frac{25 \times 5}{11}) = \ln(\frac{125}{11})$.
$\theta - 25 = \frac{125}{11} \approx 11.36$.
$\theta = 25 + 11.36 = 36.36^{\circ} C$.

(C) Let the initial amount of radium be $x_0$.
Since the rate of decomposition is proportional to the amount present,the amount remaining follows an exponential decay model.
After $1$ year,$P \%$ of the amount disappears,so the amount remaining is $x_1 = x_0 - \frac{P}{100}x_0 = x_0\left(1-\frac{P}{100}\right)$.
Let $k = \left(1-\frac{P}{100}\right)$ be the fraction remaining after each year.
After $2$ years,the amount remaining is $x_2 = x_1 \times k = x_0\left(1-\frac{P}{100}\right) \times \left(1-\frac{P}{100}\right) = x_0\left(1-\frac{P}{100}\right)^2$.

(A) Given the differential equation: $\frac{dP}{dt} = 0.05 P$.
Separating the variables,we get: $\frac{dP}{P} = 0.05 dt$.
Integrating both sides: $\int \frac{dP}{P} = \int 0.05 dt$.
This gives: $\ln P = 0.05 t + C$.
At $t = 0$,let the initial population be $P_0$,so $\ln P_0 = C$.
Substituting $C$ back into the equation: $\ln P = 0.05 t + \ln P_0$,which simplifies to $\ln(\frac{P}{P_0}) = 0.05 t$.
We want to find $t$ when the population doubles,i.e.,$P = 2P_0$.
Substituting this: $\ln(\frac{2P_0}{P_0}) = 0.05 t$.
$\ln 2 = 0.05 t$.
Since $0.05 = \frac{1}{20}$,we have $\ln 2 = \frac{t}{20}$.
Therefore,$t = 20 \ln 2$ years.

(A) Let $P_{0}$ be the initial population and $P$ be the population after $t$ years. The rate of growth is given by $\frac{dP}{dt} = \frac{8}{100} P = 0.08 P$.
Separating the variables,we get $\frac{dP}{P} = 0.08 dt$.
Integrating both sides,we get $\ln P = 0.08 t + C$.
At $t = 0$,$P = P_{0}$,so $C = \ln P_{0}$.
Thus,$\ln P = 0.08 t + \ln P_{0}$,which simplifies to $\ln \left( \frac{P}{P_{0}} \right) = 0.08 t$.
For the population to double,$P = 2 P_{0}$,so $\ln 2 = 0.08 t$.
Given $\log_{10} 2 = 0.6912$,we use the change of base formula $\ln 2 = \log_{e} 10 \times \log_{10} 2 \approx 2.3026 \times 0.6912 \approx 1.5915$.
However,in many textbook contexts,$\log$ refers to the natural logarithm $\ln$. If we use the provided value directly as $\ln 2 = 0.6912$:
$t = \frac{0.6912}{0.08} = 8.64$ years.

(B) Given $h(x) = \sqrt{4f(x) + 3g(x)}$.
At $x = 1$,$h(1) = \sqrt{4f(1) + 3g(1)} = \sqrt{4(4) + 3(3)} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Now,differentiate $h(x)$ with respect to $x$ using the chain rule:
$h'(x) = \frac{1}{2\sqrt{4f(x) + 3g(x)}} \cdot \frac{d}{dx}(4f(x) + 3g(x))$
$h'(x) = \frac{4f'(x) + 3g'(x)}{2\sqrt{4f(x) + 3g(x)}}$
Substitute $x = 1$ into the derivative:
$h'(1) = \frac{4f'(1) + 3g'(1)}{2\sqrt{4f(1) + 3g(1)}}$
Given $f'(1) = 4$ and $g'(1) = 3$:
$h'(1) = \frac{4(4) + 3(3)}{2(5)} = \frac{16 + 9}{10} = \frac{25}{10} = \frac{5}{2}$.

(D) $y = \log \tan \left(\frac{x}{2}\right) + \sin^{-1}(\cos x)$
Since $\cos x = \sin \left(\frac{\pi}{2} - x\right)$,we have:
$y = \log \tan \left(\frac{x}{2}\right) + \sin^{-1} \left[ \sin \left(\frac{\pi}{2} - x\right) \right] = \log \tan \left(\frac{x}{2}\right) + \frac{\pi}{2} - x$
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \sec^2 \left(\frac{x}{2}\right) \cdot \frac{1}{2} + 0 - 1$
$\frac{dy}{dx} = \frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} \cdot \frac{1}{\cos^2 \left(\frac{x}{2}\right)} \cdot \frac{1}{2} - 1$
$\frac{dy}{dx} = \frac{1}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} - 1$
Using the identity $\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$:
$\frac{dy}{dx} = \frac{1}{\sin x} - 1 = \operatorname{cosec} x - 1$

(D) Given $y = \sqrt{e^{\sqrt{x}}}$.
Taking the natural logarithm on both sides:
$\ln y = \ln(e^{\sqrt{x}})^{1/2} = \frac{1}{2} \sqrt{x} \ln e = \frac{\sqrt{x}}{2}$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{4\sqrt{x}}$.
Therefore,$\frac{dy}{dx} = \frac{y}{4\sqrt{x}}$.
Substituting $y = \sqrt{e^{\sqrt{x}}}$:
$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$.

(A) Given $f(x)=\operatorname{cosec}^{-1}\left[\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right]$.
Using the property $\operatorname{cosec}^{-1}(u) = \sin^{-1}(1/u)$,we get:
$f(x)=\sin ^{-1}\left[\frac{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}{10}\right]$.
We can write this as $f(x)=\sin ^{-1}\left[\frac{6}{10} \sin \left(2^x\right)-\frac{8}{10} \cos \left(2^x\right)\right]$.
Let $\cos \alpha = \frac{6}{10}$ and $\sin \alpha = \frac{8}{10}$.
Then $f(x)=\sin ^{-1}\left[\sin \left(2^x\right) \cos \alpha - \cos \left(2^x\right) \sin \alpha\right]$.
Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have:
$f(x)=\sin ^{-1}\left[\sin \left(2^x-\alpha\right)\right] = 2^x - \alpha$.
Differentiating with respect to $x$:
$f^{\prime}(x) = \frac{d}{dx}(2^x - \alpha) = 2^x \log 2 - 0 = 2^x \log 2$.

(A) Given $y = \log \sqrt{\tan x} = \frac{1}{2} \log(\tan x)$.
Applying the chain rule,we differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \times \frac{1}{\tan x} \times \sec^2 x$.
Using the identities $\tan x = \frac{\sin x}{\cos x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$,we get:
$\frac{dy}{dx} = \frac{1}{2} \times \frac{\cos x}{\sin x} \times \frac{1}{\cos^2 x} = \frac{1}{2 \sin x \cos x} = \frac{1}{\sin(2x)}$.
Now,substitute $x = \frac{\pi}{4}$:
$\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{4}} = \frac{1}{\sin(2 \times \frac{\pi}{4})} = \frac{1}{\sin(\frac{\pi}{2})} = \frac{1}{1} = 1$.

(D) Given the equation $e^{-y} \cdot y = x$.
Taking the derivative with respect to $x$ on both sides:
$\frac{d}{dx}(y e^{-y}) = \frac{d}{dx}(x)$
Using the product rule:
$y \cdot \frac{d}{dx}(e^{-y}) + e^{-y} \cdot \frac{dy}{dx} = 1$
$y \cdot (-e^{-y}) \frac{dy}{dx} + e^{-y} \frac{dy}{dx} = 1$
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx} (e^{-y} - y e^{-y}) = 1$
$\frac{dy}{dx} e^{-y} (1 - y) = 1$
$\frac{dy}{dx} = \frac{1}{e^{-y}(1 - y)}$
Since $e^{-y} = \frac{x}{y}$,substitute this into the expression:
$\frac{dy}{dx} = \frac{1}{(\frac{x}{y})(1 - y)} = \frac{y}{x(1 - y)}$.

(B) Given equation: $y = x \tan y$
Differentiating both sides with respect to $x$:
$\frac{d y}{d x} = x \sec^2 y \frac{d y}{d x} + \tan y$
Rearranging the terms to isolate $\frac{d y}{d x}$:
$\frac{d y}{d x} - x \sec^2 y \frac{d y}{d x} = \tan y$
$\frac{d y}{d x} (1 - x \sec^2 y) = \tan y$
$\frac{d y}{d x} = \frac{\tan y}{1 - x \sec^2 y}$
Multiply numerator and denominator by $x$:
$\frac{d y}{d x} = \frac{x \tan y}{x - x^2 \sec^2 y}$
Since $y = x \tan y$,we have $\tan y = \frac{y}{x}$:
$\frac{d y}{d x} = \frac{y}{x - x^2 (1 + \tan^2 y)} = \frac{y}{x - x^2 - x^2 \tan^2 y}$
Substituting $x^2 \tan^2 y = y^2$:
$\frac{d y}{d x} = \frac{y}{x - x^2 - y^2}$

(A) Given the equation: $y = 1 + xe^y$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} = 0 + \left( x \cdot e^y \frac{dy}{dx} + e^y \cdot 1 \right)$
$\frac{dy}{dx} = xe^y \frac{dy}{dx} + e^y$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} - xe^y \frac{dy}{dx} = e^y$
$\frac{dy}{dx} (1 - xe^y) = e^y$
$\frac{dy}{dx} = \frac{e^y}{1 - xe^y}$
From the original equation,we know $xe^y = y - 1$. Substituting this into the expression:
$\frac{dy}{dx} = \frac{e^y}{1 - (y - 1)}$
$\frac{dy}{dx} = \frac{e^y}{1 - y + 1}$
$\frac{dy}{dx} = \frac{e^y}{2 - y}$

(C) Given equation: $\sin ^2 x + \cos ^2 y = 1$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\sin ^2 x) + \frac{d}{dx}(\cos ^2 y) = \frac{d}{dx}(1)$
$2 \sin x \cos x + 2 \cos y (-\sin y) \frac{dy}{dx} = 0$
$\sin 2x - \sin 2y \frac{dy}{dx} = 0$
$\sin 2y \frac{dy}{dx} = \sin 2x$
$\frac{dy}{dx} = \frac{\sin 2x}{\sin 2y}$

(C) Given $x = a \cos \theta$ and $y = b \sin \theta$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d \theta} = -a \sin \theta$ and $\frac{dy}{d \theta} = b \cos \theta$.
Then,$\frac{dy}{dx} = \frac{dy/d \theta}{dx/d \theta} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b}{a} \cot \theta$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{d \theta} \left( -\frac{b}{a} \cot \theta \right) \cdot \frac{d \theta}{dx} = \left( -\frac{b}{a} \right) (-\csc^2 \theta) \cdot \frac{1}{-a \sin \theta} = -\frac{b}{a^2} \cdot \frac{1}{\sin^3 \theta}$.
At $\theta = \frac{\pi}{4}$,$\sin \theta = \frac{1}{\sqrt{2}}$,so $\sin^3 \theta = \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{1}{2 \sqrt{2}}$.
Therefore,$\left[\frac{d^2 y}{dx^2}\right]_{\theta = \frac{\pi}{4}} = -\frac{b}{a^2} \cdot (2 \sqrt{2}) = -2 \sqrt{2} \left( \frac{b}{a^2} \right)$.

(A) Given $x=a(t+\sin t)$ and $y=a(1-\cos t)$.
Differentiating $x$ with respect to $t$,we get $\frac{dx}{dt} = a(1+\cos t)$.
Differentiating $y$ with respect to $t$,we get $\frac{dy}{dt} = a(\sin t)$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1+\cos t)}$.
Using trigonometric identities $\sin t = 2 \sin \frac{t}{2} \cos \frac{t}{2}$ and $1+\cos t = 2 \cos^2 \frac{t}{2}$,we get:
$\frac{dy}{dx} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos^2 \frac{t}{2}} = \tan \frac{t}{2}$.

(D) Given $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$.
Substitute $t=\tan \theta$,then $x=\cos 2\theta$ and $y=a \sin 2\theta$.
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = -2 \sin 2\theta$ and $\frac{dy}{d\theta} = 2a \cos 2\theta$.
Using the chain rule,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a \cos 2\theta}{-2 \sin 2\theta} = -a \cot 2\theta$.
Since $\cot 2\theta = \frac{1}{\tan 2\theta} = \frac{1-\tan^2 \theta}{2 \tan \theta} = \frac{1-t^2}{2t}$,
$\frac{dy}{dx} = -a \left( \frac{1-t^2}{2t} \right) = \frac{a(t^2-1)}{2t}$.

(D) Given $u = \cos^3 x$ and $v = \sin^3 x$.
First,we find the derivatives with respect to $x$:
$\frac{du}{dx} = 3 \cos^2 x \cdot (-\sin x) = -3 \cos^2 x \sin x$
$\frac{dv}{dx} = 3 \sin^2 x \cdot (\cos x) = 3 \sin^2 x \cos x$
Now,using the chain rule for parametric differentiation:
$\frac{dv}{du} = \frac{dv/dx}{du/dx} = \frac{3 \sin^2 x \cos x}{-3 \cos^2 x \sin x} = -\frac{\sin x}{\cos x} = -\tan x$
Evaluating at $x = \frac{\pi}{4}$:
$\left(\frac{dv}{du}\right)_{x=\frac{\pi}{4}} = -\tan\left(\frac{\pi}{4}\right) = -1$

(D) Given $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$.
Differentiating with respect to $t$:
$\frac{dx}{dt}=a\left(1+\frac{1}{t^2}\right) = a\left(\frac{t^2+1}{t^2}\right)$
$\frac{dy}{dt}=b\left(1-\frac{1}{t^2}\right) = b\left(\frac{t^2-1}{t^2}\right)$
Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b\left(\frac{t^2-1}{t^2}\right)}{a\left(\frac{t^2+1}{t^2}\right)} = \frac{b(t^2-1)}{a(t^2+1)}$.
From the given equations:
$\frac{x}{a} = t-\frac{1}{t} = \frac{t^2-1}{t}$
$\frac{y}{b} = t+\frac{1}{t} = \frac{t^2+1}{t}$
Dividing these two expressions:
$\frac{x/a}{y/b} = \frac{(t^2-1)/t}{(t^2+1)/t} = \frac{t^2-1}{t^2+1}$
Substituting this into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{b}{a} \times \left(\frac{x/a}{y/b}\right) = \frac{b}{a} \times \frac{xb}{ya} = \frac{b^2 x}{a^2 y}$.

(B) Given $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$.
First,find the derivatives with respect to $\theta$:
$\frac{dx}{d\theta} = a(1+\cos \theta)$ and $\frac{dy}{d\theta} = a\sin \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a\sin \theta}{a(1+\cos \theta)} = \frac{\sin \theta}{1+\cos \theta}$.
Using the identity $\sin \theta = 2\sin(\theta/2)\cos(\theta/2)$ and $1+\cos \theta = 2\cos^2(\theta/2)$,we get $\frac{dy}{dx} = \tan(\theta/2)$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\tan\frac{\theta}{2}\right) \cdot \frac{d\theta}{dx} = \frac{1}{2}\sec^2\frac{\theta}{2} \cdot \frac{1}{a(1+\cos \theta)}$.
Since $1+\cos \theta = 2\cos^2(\theta/2)$,we have:
$\frac{d^2y}{dx^2} = \frac{1}{2}\sec^2\frac{\theta}{2} \cdot \frac{1}{2a\cos^2(\theta/2)} = \frac{1}{4a\cos^4(\theta/2)}$.
At $\theta = \pi/2$,$\theta/2 = \pi/4$ and $\cos(\pi/4) = 1/\sqrt{2}$.
$\left(\frac{d^2y}{dx^2}\right)_{\theta=\pi/2} = \frac{1}{4a(1/\sqrt{2})^4} = \frac{1}{4a(1/4)} = \frac{1}{a}$.

(B) Given $x=e^t(\sin t-\cos t)$ and $y=e^t(\sin t+\cos t)$.
Applying the product rule for differentiation with respect to $t$:
$\frac{dx}{dt} = e^t(\sin t-\cos t) + e^t(\cos t+\sin t) = e^t(\sin t - \cos t + \cos t + \sin t) = 2e^t \sin t$.
$\frac{dy}{dt} = e^t(\sin t+\cos t) + e^t(\cos t-\sin t) = e^t(\sin t + \cos t + \cos t - \sin t) = 2e^t \cos t$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2e^t \cos t}{2e^t \sin t} = \cot t$.
At $t = \frac{\pi}{3}$,$\frac{dy}{dx} = \cot(\frac{\pi}{3}) = \frac{1}{\sqrt{3}}$.

(A) Given the equation $x^y \cdot y^x = 16$.
Taking the natural logarithm on both sides,we get:
$y \ln x + x \ln y = \ln 16$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(y \ln x) + \frac{d}{dx}(x \ln y) = \frac{d}{dx}(\ln 16)$.
$\left( \frac{dy}{dx} \ln x + y \cdot \frac{1}{x} \right) + \left( 1 \cdot \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \right) = 0$.
Rearranging the terms to isolate $\frac{dy}{dx}$:
$\frac{dy}{dx} (\ln x + \frac{x}{y}) = -(\frac{y}{x} + \ln y)$.
$\frac{dy}{dx} = -\frac{\frac{y}{x} + \ln y}{\ln x + \frac{x}{y}}$.
Now,substitute the point $(2, 2)$ into the expression:
$\left( \frac{dy}{dx} \right)_{(2, 2)} = -\frac{\frac{2}{2} + \ln 2}{\ln 2 + \frac{2}{2}} = -\frac{1 + \ln 2}{1 + \ln 2} = -1$.

(B) Let $u = (\log x)^x$. Taking logarithm on both sides,we get $\log u = x \log(\log x)$.
Now,differentiating with respect to $x$,we have $\frac{1}{u} \frac{du}{dx} = x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \cdot 1 = \frac{1}{\log x} + \log(\log x)$.
Thus,$\frac{du}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.
Let $v = \log x$. Then $\frac{dv}{dx} = \frac{1}{x}$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{(\log x)^x [\frac{1}{\log x} + \log(\log x)]}{1/x} = x(\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right]$.

(C) Given $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right] + x^x$.
Using $\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$,we have:
$y = \sin^{-1}\left[\sin\left(\frac{\pi}{2} - \sqrt{\frac{1+x}{2}}\right)\right] + x^x = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{1}{\sqrt{2}} \cdot \frac{d}{dx}((1+x)^{1/2}) + \frac{d}{dx}(x^x)$.
For $x^x$,let $u = x^x$,then $\ln u = x \ln x$.
$\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$,so $\frac{du}{dx} = x^x(1 + \ln x)$.
Thus,$\frac{dy}{dx} = -\frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + x^x(1 + \ln x)$.
At $x = 1$:
$\left[\frac{dy}{dx}\right]_{x=1} = -\frac{1}{2\sqrt{2}\sqrt{2}} + 1^1(1 + \ln 1) = -\frac{1}{4} + 1(1 + 0) = -\frac{1}{4} + 1 = \frac{3}{4}$.

(A) We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$ for $|z| \geq 1$.
Given $y = \operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$.
Using the identity,we can rewrite the first term as $\sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$.
Thus,$y = \sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$.
Using the inverse trigonometric identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$,we get $y = \frac{\pi}{2}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

(B) We are given $y=\tan ^{-1}\left[\frac{1}{1+x(1+x)}\right]+\tan ^{-1}\left[\frac{1}{1+(x+2)(x+1)}\right]$.
Using the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$,we can rewrite the terms:
$y = \tan ^{-1}\left(\frac{(x+1)-x}{1+(x+1)x}\right) + \tan ^{-1}\left(\frac{(x+2)-(x+1)}{1+(x+2)(x+1)}\right)$.
This simplifies to:
$y = (\tan ^{-1}(x+1) - \tan ^{-1} x) + (\tan ^{-1}(x+2) - \tan ^{-1}(x+1))$.
Canceling the common terms,we get:
$y = \tan ^{-1}(x+2) - \tan ^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan ^{-1}(x+2)) - \frac{d}{dx}(\tan ^{-1} x)$.
$\frac{dy}{dx} = \frac{1}{1+(x+2)^2} - \frac{1}{1+x^2}$.
Thus,the correct option is $B$.

(C) Given $y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}$.
Divide the numerator and denominator by $b \cos x$:
$y=\tan ^{-1}\left\{\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b}\tan x}\right\}$.
Let $\frac{a}{b}=\tan \alpha$,then:
$y=\tan ^{-1}\left\{\frac{\tan \alpha-\tan x}{1+\tan \alpha \tan x}\right\}$.
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$y=\tan ^{-1}[\tan(\alpha-x)] = \alpha-x$.
Differentiating with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x}(\alpha-x) = 0-1 = -1$.

(D) Given $y = \tan^{-1} \left( \sqrt{\frac{1+\sin x}{1-\sin x}} \right)$.
Using the identities $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$,we get:
$y = \tan^{-1} \left( \sqrt{\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}} \right) = \tan^{-1} \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right)$.
Dividing numerator and denominator by $\cos \frac{x}{2}$,we get:
$y = \tan^{-1} \left( \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \right) = \tan^{-1} \left( \tan(\frac{\pi}{4} + \frac{x}{2}) \right)$.
Since $0 \leq x < \frac{\pi}{2}$,we have $\frac{\pi}{4} \leq \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$,so $y = \frac{\pi}{4} + \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} + \frac{x}{2}) = 0 + \frac{1}{2} = \frac{1}{2}$.
Thus,at $x = \frac{\pi}{6}$,$\frac{dy}{dx} = \frac{1}{2}$.

(D) Given $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$.
Using the trigonometric identities $1+\cos x = 2 \cos^2 \frac{x}{2}$ and $1-\cos x = 2 \sin^2 \frac{x}{2}$,we get:
$y=\tan ^{-1} \sqrt{\frac{2 \cos ^2 \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}}$
$y=\tan ^{-1} \sqrt{\cot^2 \frac{x}{2}} = \tan ^{-1} \left(\cot \frac{x}{2}\right)$
Since $\cot \theta = \tan \left(\frac{\pi}{2} - \theta\right)$,we have:
$y=\tan ^{-1} \left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right] = \frac{\pi}{2}-\frac{x}{2}$
Differentiating with respect to $x$:
$\frac{d y}{d x} = \frac{d}{d x} \left(\frac{\pi}{2} - \frac{x}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2}$

(D) Given $y=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(ex^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$.
Using properties of logarithms,$\log(e/x^2) = \log e - \log x^2 = 1 - 2\log x$ and $\log(ex^2) = \log e + \log x^2 = 1 + 2\log x$.
Let $u = 2\log x$. Then the first term is $\tan^{-1}\left(\frac{1-u}{1+u}\right) = \tan^{-1}(1) - \tan^{-1}(u) = \frac{\pi}{4} - \tan^{-1}(2\log x)$.
For the second term,$\tan^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right] = \tan^{-1}(3) + \tan^{-1}(2\log x)$.
Adding these,$y = \frac{\pi}{4} - \tan^{-1}(2\log x) + \tan^{-1}(3) + \tan^{-1}(2\log x) = \frac{\pi}{4} + \tan^{-1}(3)$.
Since $y$ is a constant,$\frac{dy}{dx} = 0$ and $\frac{d^2 y}{dx^2} = 0$.

(B) Let $f(x) = \cot^{-1}(\sqrt{\cos 2x})$.
Using the chain rule,the derivative is $f'(x) = \frac{-1}{1 + (\sqrt{\cos 2x})^2} \times \frac{d}{dx}(\sqrt{\cos 2x})$.
$f'(x) = \frac{-1}{1 + \cos 2x} \times \frac{1}{2\sqrt{\cos 2x}} \times (-\sin 2x \times 2)$.
$f'(x) = \frac{\sin 2x}{(1 + \cos 2x)\sqrt{\cos 2x}}$.
At $x = \frac{\pi}{6}$,$2x = \frac{\pi}{3}$.
$f'\left(\frac{\pi}{6}\right) = \frac{\sin(\pi/3)}{(1 + \cos(\pi/3))\sqrt{\cos(\pi/3)}}$.
$f'\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}/2}{(1 + 1/2)\sqrt{1/2}} = \frac{\sqrt{3}/2}{(3/2)(1/\sqrt{2})} = \frac{\sqrt{3}}{2} \times \frac{2}{3} \times \sqrt{2} = \frac{\sqrt{6}}{3} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}} = \left(\frac{2}{3}\right)^{1/2}$.

(D) Given $y = 2 \sin x + 3 \cos x$.
First,find the first derivative: $\frac{dy}{dx} = 2 \cos x - 3 \sin x$.
Next,find the second derivative: $\frac{d^2y}{dx^2} = -2 \sin x - 3 \cos x$.
We can rewrite this as $\frac{d^2y}{dx^2} = -(2 \sin x + 3 \cos x) = -y$.
Rearranging the equation,we get $y + \frac{d^2y}{dx^2} = 0$.
Comparing this with the given equation $y + A \frac{d^2y}{dx^2} = B$,we find $A = 1$ and $B = 0$.

(C) Given $y^2 = ax^2 + bx + c$.
Differentiating with respect to $x$,we get:
$2y \frac{dy}{dx} = 2ax + b$
$\frac{dy}{dx} = \frac{2ax + b}{2y}$
Differentiating again with respect to $x$ using the quotient rule:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{2ax + b}{2y} \right) = \frac{2a(2y) - (2ax + b)(2 \frac{dy}{dx})}{4y^2}$
Substituting $\frac{dy}{dx} = \frac{2ax + b}{2y}$:
$\frac{d^2y}{dx^2} = \frac{4ay - (2ax + b) \frac{2ax + b}{y}}{4y^2} = \frac{4ay^2 - (2ax + b)^2}{4y^3}$
Now,multiply by $y^3$:
$y^3 \frac{d^2y}{dx^2} = \frac{4ay^2 - (2ax + b)^2}{4} = \frac{4a(ax^2 + bx + c) - (4a^2x^2 + 4abx + b^2)}{4}$
$y^3 \frac{d^2y}{dx^2} = \frac{4a^2x^2 + 4abx + 4ac - 4a^2x^2 - 4abx - b^2}{4} = \frac{4ac - b^2}{4}$
Since $a, b, c$ are constants,$\frac{4ac - b^2}{4}$ is a constant.

(C) The function is defined as $f(x) = [x]$.
For any integer $n$,the floor function $[x]$ is discontinuous at $x = n$ because the left-hand limit is $n-1$ and the right-hand limit is $n$.
Given the domain $x \in (-1, 2)$,the integers within this interval are $0$ and $1$.
Therefore,the function $f(x) = [x]$ is discontinuous at $x = 0$ and $x = 1$.

(C) The function is defined as $f(x) = \frac{1}{\sqrt{x+|x|}}$.
For the function to be defined,the expression inside the square root must be strictly greater than zero:
$x + |x| > 0$.
Case $1$: If $x > 0$,then $|x| = x$. Substituting this,we get $x + x = 2x > 0$,which implies $x > 0$.
Case $2$: If $x < 0$,then $|x| = -x$. Substituting this,we get $x + (-x) = 0$. Since $0$ is not greater than $0$,this case yields no solution.
Case $3$: If $x = 0$,then $x + |x| = 0 + 0 = 0$. Since the denominator cannot be zero,$x = 0$ is not in the domain.
Therefore,the domain of the function is $(0, \infty)$.

(B) The function $f(x) = \sqrt{x-1} + \sqrt{6-x}$ is defined if and only if the expressions under the square roots are non-negative.
For $\sqrt{x-1}$ to be defined,$x - 1 \geq 0$,which implies $x \geq 1$.
For $\sqrt{6-x}$ to be defined,$6 - x \geq 0$,which implies $x \leq 6$.
Combining these two conditions,we get $1 \leq x \leq 6$.
Therefore,the domain of $f(x)$ is $[1, 6]$.

(B) For the function $f(x) = \log _{10}(x^2-5x+6)$ to be defined,the argument of the logarithm must be strictly positive.
$x^2-5x+6 > 0$
Factorizing the quadratic expression:
$(x-2)(x-3) > 0$
Using the wavy curve method or sign scheme,the expression is positive when $x$ lies outside the roots $2$ and $3$.
Thus,$x < 2$ or $x > 3$.
Therefore,the domain is $x \in (-\infty, 2) \cup (3, \infty)$.

(C) Given the set $A = \{10, 11, 12, 14, 26\}$.
We find the prime factorization for each element in $A$:
$10 = 2 \times 5$,so the highest prime factor is $5$.
$11 = 11$,so the highest prime factor is $11$.
$12 = 2^2 \times 3$,so the highest prime factor is $3$.
$14 = 2 \times 7$,so the highest prime factor is $7$.
$26 = 2 \times 13$,so the highest prime factor is $13$.
Thus,the range of $f$ is the set of all these highest prime factors: $\{3, 5, 7, 11, 13\}$.
Therefore,the correct option is $C$.

(A) Given the function $f(x) = 3 + 2^x + 4^x$.
Since $2^x > 0$ for all $x \in \mathbb{R}$,it follows that $4^x = (2^x)^2 > 0$.
Let $t = 2^x$,where $t \in (0, \infty)$.
Then the function becomes $g(t) = 3 + t + t^2$.
Since $t > 0$,the minimum value of $t + t^2$ approaches $0$ as $t \to 0^+$.
As $t \to \infty$,$t + t^2 \to \infty$.
Therefore,the range of $t + t^2$ for $t > 0$ is $(0, \infty)$.
Adding $3$ to this range,the range of $f(x)$ is $(3 + 0, 3 + \infty) = (3, \infty)$.