MHT CET 2009 Mathematics Question Paper with Answer and Solution

(B) The equation of the pair of tangents from a point $(x_{1}, y_{1})$ to the circle $S: x^{2}+y^{2}-r^{2}=0$ is given by $SS_{1}=T^{2}$.
Here,$S = x^{2}+y^{2}-4$,$S_{1} = 3^{2}+2^{2}-4 = 9$,and $T = 3x+2y-4$.
Substituting these,we get $(x^{2}+y^{2}-4)(9) = (3x+2y-4)^{2}$.
Expanding the equation: $9x^{2}+9y^{2}-36 = 9x^{2}+4y^{2}+16+12xy-24x-16y$.
Simplifying: $5y^{2}-12xy+24x+16y-52 = 0$.
For a pair of lines $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,the slopes $m_{1}, m_{2}$ satisfy $bm^{2}+2hm+a=0$ (by dividing by $x^{2}$ and setting $m=y/x$).
Here,$b=5$,$2h=-12$ (so $h=-6$),and $a=24$ (coefficient of $x^{2}$ is $0$,but we treat the quadratic in $y/x$).
Actually,dividing $5y^{2}-12xy+24x+16y-52=0$ by $x^{2}$ gives $5m^{2}-12m+24/x+16m/x-52/x^{2}=0$. This approach is complex.
Using the condition for tangents $y-2 = m(x-3) \Rightarrow mx-y+(2-3m)=0$.
The distance from $(0,0)$ to the line is $r=2$: $\frac{|2-3m|}{\sqrt{m^{2}+1}} = 2$.
Squaring both sides: $(2-3m)^{2} = 4(m^{2}+1)$.
$4-12m+9m^{2} = 4m^{2}+4$.
$5m^{2}-12m = 0$.
$m(5m-12) = 0$.
Thus,the slopes are $m_{1} = 0$ and $m_{2} = \frac{12}{5}$.
Therefore,$|m_{1}-m_{2}| = |0 - \frac{12}{5}| = \frac{12}{5}$.

(C) The equation of the ellipse is $\frac{x^{2}}{32}+\frac{y^{2}}{18}=1$,where $a^{2}=32$ and $b^{2}=18$.
The equation of a tangent with slope $m = -\frac{3}{4}$ is $y = mx \pm \sqrt{a^{2}m^{2}+b^{2}}$.
Substituting the values: $y = -\frac{3}{4}x \pm \sqrt{32 \times (-\frac{3}{4})^{2} + 18}$.
$y = -\frac{3}{4}x \pm \sqrt{32 \times \frac{9}{16} + 18} = -\frac{3}{4}x \pm \sqrt{18 + 18} = -\frac{3}{4}x \pm 6$.
Considering the positive intercept,the equation is $y = -\frac{3}{4}x + 6$,which simplifies to $3x + 4y = 24$.
The intercepts are found by setting $y=0$ and $x=0$:
For $y=0$,$3x=24 \Rightarrow x=8$,so $A = (8, 0)$.
For $x=0$,$4y=24 \Rightarrow y=6$,so $B = (0, 6)$.
The area of $\Delta AOB = \frac{1}{2} \times |OA| \times |OB| = \frac{1}{2} \times 8 \times 6 = 24$ sq unit.

(A) The given equation of the ellipse is $5x^{2} + 9y^{2} = 45$. Dividing by $45$,we get $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$.
Here,$a^{2} = 9$ and $b^{2} = 5$.
The line $4x - 3y + k = 0$ can be written as $y = \frac{4}{3}x + \frac{k}{3}$.
Comparing this with the line $y = mx + c$,we have $m = \frac{4}{3}$ and $c = \frac{k}{3}$.
The condition for the line $y = mx + c$ to touch the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} + b^{2}$.
Substituting the values,we get $(\frac{k}{3})^{2} = 9(\frac{4}{3})^{2} + 5$.
$\frac{k^{2}}{9} = 9(\frac{16}{9}) + 5 = 16 + 5 = 21$.
$k^{2} = 9 \times 21 = 189$.
$k = \pm \sqrt{189} = \pm 3\sqrt{21}$.

(D) Let the polynomial be $f(x) = ax^{2}+bx+c$.
Given $f(0)=8$,we have $a(0)^{2}+b(0)+c=8$,which implies $c=8$.
Thus,the polynomial is $f(x) = ax^{2}+bx+8$.
Given $f(1)=12$,we have $a(1)^{2}+b(1)+8=12$,which simplifies to $a+b=4$ (Equation $i$).
Given $f(2)=18$,we have $a(2)^{2}+b(2)+8=18$,which simplifies to $4a+2b=10$,or $2a+b=5$ (Equation $ii$).
Subtracting Equation $(i)$ from Equation $(ii)$,we get $(2a+b)-(a+b) = 5-4$,which gives $a=1$.
Substituting $a=1$ into Equation $(i)$,we get $1+b=4$,which gives $b=3$.
Therefore,the required polynomial is $f(x) = x^{2}+3x+8$.

(D) To find $\Delta^{5} u_{x}$,we construct the forward difference table:

$x, u_{x}$	$\Delta u_{x}, \Delta^{2} u_{x}, \Delta^{3} u_{x}, \Delta^{4} u_{x}, \Delta^{5} u_{x}$
$0, 3$	$9, 60, -10, -259, 755$
$1, 12$	$69, 50, -269, 496$
$2, 81$	$119, -219, 227$
$3, 200$	$-100, 8$
$4, 100$	$-92$
$5, 8$	-

The forward difference $\Delta^{5} u_{0}$ is calculated as follows:
$\Delta u_{0} = u_{1} - u_{0} = 12 - 3 = 9$
$\Delta u_{1} = u_{2} - u_{1} = 81 - 12 = 69$
$\Delta u_{2} = u_{3} - u_{2} = 200 - 81 = 119$
$\Delta u_{3} = u_{4} - u_{3} = 100 - 200 = -100$
$\Delta u_{4} = u_{5} - u_{4} = 8 - 100 = -92$
Continuing this process for higher-order differences,we obtain the final value $\Delta^{5} u_{0} = 755$.

(B) We are given the condition $x_{1} = 1, x_{2} = 0, x_{3} = 0$.
We evaluate the given options:
For option $B$: $x_{1} \cdot x_{2}' = 1 \cdot (0)' = 1 \cdot 1 = 1$.
Thus,the function $f(x_{1}, x_{2}, x_{3}) = x_{1} \cdot x_{2}'$ satisfies the condition.

(C) Given $D_{30} = \{1, 2, 3, 5, 6, 10, 15, 30\}$.
We need to calculate $f(2, 5, 15) = (2+5) \cdot (5^{\prime} + 15)$.
First,calculate $2+5 = \operatorname{LCM}(2, 5) = 10$.
Next,calculate $5^{\prime} = \frac{30}{5} = 6$.
Then,$5^{\prime} + 15 = 6 + 15 = \operatorname{LCM}(6, 15) = 30$.
Finally,$f(2, 5, 15) = 10 \cdot 30 = \operatorname{GCD}(10, 30) = 10$.

(A) Given the foci are $(\pm ae, 0) = (\pm 3, 0)$,we have $ae = 3$,so $a^{2}e^{2} = 9$.
For a hyperbola,$b^{2} = a^{2}e^{2} - a^{2}$,thus $a^{2} + b^{2} = a^{2}e^{2} = 9$ (Equation $i$).
The equation of the tangent $y = mx + c$ is $y = -2x + 4$,where $m = -2$ and $c = 4$.
The condition for tangency to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} - b^{2}$.
Substituting the values: $4^{2} = a^{2}(-2)^{2} - b^{2} \Rightarrow 4a^{2} - b^{2} = 16$ (Equation $ii$).
Adding Equation $(i)$ and $(ii)$: $(a^{2} + b^{2}) + (4a^{2} - b^{2}) = 9 + 16$ $\Rightarrow 5a^{2} = 25$ $\Rightarrow a^{2} = 5$.
Substituting $a^{2} = 5$ into Equation $(i)$: $5 + b^{2} = 9 \Rightarrow b^{2} = 4$.
The equation of the hyperbola is $\frac{x^{2}}{5} - \frac{y^{2}}{4} = 1$,which simplifies to $4x^{2} - 5y^{2} = 20$.

(C) Given the limit: $\lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4$
We can rewrite the expression as: $\lim _{x \rightarrow 0} \left( \frac{e^{k x}-1}{x} \cdot \frac{\sin k x}{x} \right) = 4$
Using the standard limits $\lim _{x \rightarrow 0} \frac{e^{k x}-1}{x} = k$ and $\lim _{x \rightarrow 0} \frac{\sin k x}{x} = k$:
$k \cdot k = 4$
$k^{2} = 4$
$k = \pm 2$

(A) Given,$\lim_{x \rightarrow \infty} f(x) = 1$
$\Rightarrow \lim_{x}$ ${\rightarrow \infty} \frac{ax + b}{x + 1} = 1$
$\Rightarrow \lim_{x}$ ${\rightarrow \infty} \frac{a + \frac{b}{x}}{1 + \frac{1}{x}} = 1$
$\Rightarrow a = 1$
Given,$\lim_{x \rightarrow 0} f(x) = 2$
$\Rightarrow \frac{a(0) + b}{0 + 1} = 2$
$\Rightarrow b = 2$
Thus,$f(x) = \frac{x + 2}{x + 1}$
Therefore,$f(-2) = \frac{-2 + 2}{-2 + 1} = \frac{0}{-1} = 0$

(B) Let $L = \lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$.
As $x \rightarrow 1$,$\log _{e} x \rightarrow 0$,so the expression takes the form $1^{\infty}$.
We use the standard limit formula: $\lim _{u \rightarrow 0} (1 + u)^{1/u} = e$.
Let $u = \log _{e} x$. As $x \rightarrow 1$,$u \rightarrow 0$.
Substituting this into the limit,we get $\lim _{u \rightarrow 0} (1 + u)^{1/u} = e$.
Therefore,the correct option is $B$.

(A) Using the distributive law,we have:
$(p \vee q) \wedge (p \vee \sim q) = p \vee (q \wedge \sim q)$
By the complement law,$q \wedge \sim q = F$ (where $F$ denotes a contradiction or false statement).
So,the expression becomes $p \vee F$.
Since $F$ is the identity element for the disjunction operator $\vee$,$p \vee F = p$.
Thus,the simplified form is $p$.

(A) The given circuit consists of two parallel blocks connected in series,which are then connected in parallel with a wire (which acts as an open circuit or is not present in the simplified logic).
Looking at the circuit:
$1$. The first block has switches $p$ and $q$ in parallel,represented by $(p \vee q)$.
$2$. The second block has switches $p$ and $r$ in parallel,represented by $(p \vee r)$.
$3$. These two blocks are connected in series,so the expression is $(p \vee q) \wedge (p \vee r)$.
Using the distributive law of Boolean algebra:
$(p \vee q) \wedge (p \vee r) = p \vee (q \wedge r)$.

(C) We know that the implication $A \rightarrow B$ is equivalent to $\sim A \vee B$.
Therefore,$\sim p \rightarrow q \equiv \sim(\sim p) \vee q \equiv p \vee q$.
Now,applying the negation: $\sim(\sim p \rightarrow q) \equiv \sim(p \vee q)$.
By De Morgan's Law,$\sim(p \vee q) \equiv \sim p \wedge \sim q$.

(A) The angle $\varphi$ between the lines represented by $a x^{2}+2 h x y+b y^{2}=0$ is given by $\tan \varphi = \left| \frac{2 \sqrt{h^{2}-a b}}{a+b} \right|$.
For the equation $x^{2}+2 x y \sec \theta+y^{2}=0$,we have $a=1$,$b=1$,and $h=\sec \theta$.
Substituting these values into the formula:
$\tan \varphi = \left| \frac{2 \sqrt{\sec^{2} \theta - (1)(1)}}{1+1} \right|$
$\tan \varphi = \left| \frac{2 \sqrt{\tan^{2} \theta}}{2} \right|$
$\tan \varphi = \tan \theta$.
Therefore,the angle between the lines is $\theta$.

(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing $hxy + 10x + 6y + 4 = 0$ with the general form,we get $a = 0, b = 0, 2h' = h \Rightarrow h' = h/2, g = 5, f = 3, c = 4$.
For the equation to represent a pair of lines,the determinant must be zero:
$\Delta = \begin{vmatrix} a & h' & g \\ h' & b & f \\ g & f & c \end{vmatrix} = 0$
$\begin{vmatrix} 0 & h/2 & 5 \\ h/2 & 0 & 3 \\ 5 & 3 & 4 \end{vmatrix} = 0$
$0(0 - 9) - \frac{h}{2}(2h/2 - 15) + 5(3h/2 - 0) = 0$
$-\frac{h}{2}(h - 15) + \frac{15h}{2} = 0$
$-\frac{h^2}{2} + \frac{15h}{2} + \frac{15h}{2} = 0$
$-\frac{h^2}{2} + 15h = 0$
$-h^2 + 30h = 0 \Rightarrow h(30 - h) = 0$
Therefore,$h = 0$ or $h = 30$.

(A) The given equation is $x^{2} - 4xy + 3y^{2} = 0$.
Factoring the equation: $x^{2} - 3xy - xy + 3y^{2} = 0$ $\Rightarrow x(x - 3y) - y(x - 3y) = 0$ $\Rightarrow (x - y)(x - 3y) = 0$.
The lines are $x - y = 0$ and $x - 3y = 0$.
Lines parallel to these passing through $(3, -2)$ are of the form $(x - y + k_{1}) = 0$ and $(x - 3y + k_{2}) = 0$.
For $(x - y + k_{1}) = 0$ passing through $(3, -2)$: $3 - (-2) + k_{1} = 0$ $\Rightarrow 5 + k_{1} = 0$ $\Rightarrow k_{1} = -5$.
For $(x - 3y + k_{2}) = 0$ passing through $(3, -2)$: $3 - 3(-2) + k_{2} = 0$ $\Rightarrow 3 + 6 + k_{2} = 0$ $\Rightarrow k_{2} = -9$.
The joint equation is $(x - y - 5)(x - 3y - 9) = 0$.
Expanding: $x(x - 3y - 9) - y(x - 3y - 9) - 5(x - 3y - 9) = 0$.
$x^{2} - 3xy - 9x - xy + 3y^{2} + 9y - 5x + 15y + 45 = 0$.
$x^{2} + 3y^{2} - 4xy - 14x + 24y + 45 = 0$.

(B) The given equation of the parabola is $y^{2}=12x$.
Comparing this with $y^{2}=4ax$,we get $4a=12$,which implies $a=3$.
For the point $P(x, y)$,the ordinate is given as $y=6$.
Since the point $P$ lies on the parabola,we substitute $y=6$ into the equation:
$(6)^{2}=12x$ $\Rightarrow 36=12x$ $\Rightarrow x=3$.
The focal distance of a point $P(x, y)$ on the parabola $y^{2}=4ax$ is given by the formula $x+a$.
Substituting the values $x=3$ and $a=3$,we get:
Focal distance $= 3+3=6$.

(B) The equation of the tangent to the parabola $y^{2} = 4ax$ at point $(x_{1}, y_{1})$ is given by $yy_{1} = 2a(x + x_{1})$.
Here,the parabola is $y^{2} = 16x$,so $4a = 16$,which implies $a = 4$.
The point is $(x_{1}, y_{1}) = (3, 6)$.
Substituting these values into the formula:
$y(6) = 2(4)(x + 3)$
$6y = 8(x + 3)$
$6y = 8x + 24$
Dividing by $2$:
$3y = 4x + 12$
Rearranging the terms:
$3y - 4x - 12 = 0$.

(A) The probability that exactly one of the events $A$ or $B$ occurs is given by $P(A \Delta B) = P(A \cup B) - P(A \cap B)$.
Given $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.2$.
Therefore,the probability of exactly one event occurring is $0.6 - 0.2 = 0.4$.

(B) The locus of the point of intersection of perpendicular tangents to a circle is known as its director circle.
For a circle given by the equation $x^{2}+y^{2}=r^{2}$,the equation of the director circle is $x^{2}+y^{2}=2r^{2}$.
Here,the given circle is $x^{2}+y^{2}=16$,so $r^{2}=16$.
Substituting this into the director circle equation,we get $x^{2}+y^{2}=2(16) = 32$.
Thus,the required locus is $x^{2}+y^{2}=32$.

(B) Given curve is $y=4 x e^{x}$.
First,find the derivative $\frac{dy}{dx}$ using the product rule:
$\frac{dy}{dx} = 4e^{x} + 4x e^{x} = 4e^{x}(1+x)$.
Now,evaluate the slope at the point $\left(-1, -\frac{4}{e}\right)$:
$\left(\frac{dy}{dx}\right)_{(-1, -4/e)} = 4e^{-1}(1 + (-1)) = 4e^{-1}(0) = 0$.
Since the slope is $0$,the tangent is a horizontal line.
The equation of the tangent line is given by $y - y_1 = m(x - x_1)$:
$y - (-\frac{4}{e}) = 0(x - (-1))$.
$y + \frac{4}{e} = 0$.
Therefore,$y = -\frac{4}{e}$.

(B) Let $f(x) = x^{1/3}$. We need to find the value of $f(28)$.
Let $x = 27$ and $\Delta x = 1$,so that $x + \Delta x = 28$.
We know that $f(x) = x^{1/3}$,so $f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}$.
The formula for approximation is $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.
Here,$f(27) = (27)^{1/3} = 3$.
$f'(27) = \frac{1}{3(27)^{2/3}} = \frac{1}{3(9)} = \frac{1}{27}$.
Therefore,$f(28) \approx 3 + \frac{1}{27} \times 1$.
$f(28) \approx 3 + 0.037037... \approx 3.037$.

(A) The given curves are $x^{2}=y$ and $y=4x$.
To find the intersection points,substitute $y=x^{2}$ into $y=4x$:
$x^{2}=4x \implies x^{2}-4x=0 \implies x(x-4)=0$.
So,$x=0$ and $x=4$.
When $x=0, y=0$ and when $x=4, y=16$.
The intersection points are $(0,0)$ and $(4,16)$.
The required area $A$ is given by the integral:
$A = \int_{0}^{4} (4x - x^{2}) dx$.
Integrating the terms:
$A = \left[ \frac{4x^{2}}{2} - \frac{x^{3}}{3} \right]_{0}^{4} = \left[ 2x^{2} - \frac{x^{3}}{3} \right]_{0}^{4}$.
Evaluating at the limits:
$A = \left( 2(4)^{2} - \frac{(4)^{3}}{3} \right) - (0) = \left( 32 - \frac{64}{3} \right) = \frac{96-64}{3} = \frac{32}{3} \text{ sq unit}$.

(A) Let $I = \int_{0}^{\pi / 2} \frac{\sin x - \cos x}{1 - \sin x \cos x} d x$ $(i)$
Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $x$ with $(\frac{\pi}{2} - x)$:
$I = \int_{0}^{\pi / 2} \frac{\sin(\frac{\pi}{2} - x) - \cos(\frac{\pi}{2} - x)}{1 - \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x)} dx$
Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we get:
$I = \int_{0}^{\pi / 2} \frac{\cos x - \sin x}{1 - \cos x \sin x} dx$
$I = - \int_{0}^{\pi / 2} \frac{\sin x - \cos x}{1 - \sin x \cos x} dx$
$I = -I$ (ii)
Adding equation $(i)$ and (ii):
$I + I = 0$
$2I = 0$
$I = 0$

(C) We evaluate each integral one by one:
$(a)$ $\int_{0}^{1} e^{x} dx = [e^{x}]_{0}^{1} = e^{1} - e^{0} = e - 1$. This is false.
$(b)$ $\int_{0}^{1} 2^{x} dx = [\frac{2^{x}}{\log_{e} 2}]_{0}^{1} = \frac{1}{\log 2} \cdot (2^{1} - 2^{0}) = \frac{1}{\log 2}$. This is false.
$(c)$ $\int_{0}^{1} \sqrt{x} dx = \int_{0}^{1} x^{1/2} dx = [\frac{x^{3/2}}{3/2}]_{0}^{1} = \frac{2}{3} [1^{3/2} - 0^{3/2}] = \frac{2}{3}$. This is true.
$(d)$ $\int_{0}^{1} x dx = [\frac{x^{2}}{2}]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}$. This is false.

(A) Given differential equation is $\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1-\left(\frac{d y}{d x}\right)^{4}}$.
To find the order and degree,we first eliminate the radical sign by cubing both sides:
$\left(\frac{d^{2} y}{d x^{2}}\right)^{3} = 1 - \left(\frac{d y}{d x}\right)^{4}$.
The order of a differential equation is the order of the highest derivative present. Here,the highest derivative is $\frac{d^{2} y}{d x^{2}}$,so the order is $2$.
The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Here,the power of $\frac{d^{2} y}{d x^{2}}$ is $3$.
Therefore,the order is $2$ and the degree is $3$.

(B) Given the family of lines: $y = mx + \frac{4}{m}$ $(i)$
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = m$
Substitute $m = \frac{dy}{dx}$ into equation $(i)$:
$y = x\left(\frac{dy}{dx}\right) + \frac{4}{\frac{dy}{dx}}$
Multiply the entire equation by $\frac{dy}{dx}$ to clear the denominator:
$y\left(\frac{dy}{dx}\right) = x\left(\frac{dy}{dx}\right)^{2} + 4$
Rearranging the terms to form the differential equation:
$x\left(\frac{dy}{dx}\right)^{2} - y\frac{dy}{dx} + 4 = 0$
Thus,the required differential equation is $x\left(\frac{dy}{dx}\right)^{2} - y\frac{dy}{dx} + 4 = 0$.

(B) We know that for any real number $z$,$\tan^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$.
Given the equation $xy = \tan^{-1}(xy) + \cot^{-1}(xy)$,we can substitute $z = xy$ to get $xy = \frac{\pi}{2}$.
Now,differentiate both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(xy) = \frac{d}{dx}(\frac{\pi}{2})$
$x \frac{dy}{dx} + y(1) = 0$
$x \frac{dy}{dx} = -y$
$\frac{dy}{dx} = -\frac{y}{x}$

(C) Given the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.
Let $x+y = t$.
Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting this into the original equation:
$\frac{dt}{dx} - 1 = \frac{t+1}{t-1}$
$\frac{dt}{dx} = \frac{t+1}{t-1} + 1 = \frac{t+1+t-1}{t-1} = \frac{2t}{t-1}$.
Separating the variables,we get $\left(\frac{t-1}{2t}\right) dt = dx$.
This simplifies to $\left(\frac{1}{2} - \frac{1}{2t}\right) dt = dx$.
Integrating both sides:
$\int \left(\frac{1}{2} - \frac{1}{2t}\right) dt = \int dx$
$\frac{1}{2}t - \frac{1}{2} \log |t| = x + C_1$.
Multiplying by $2$:
$t - \log |t| = 2x + 2C_1$.
Substituting $t = x+y$ back:
$(x+y) - \log |x+y| = 2x + C$ (where $C = 2C_1$).
$y - x = \log |x+y| + C$,which can be rearranged as $y = x + \log |x+y| + C$.

(A) Given the equation of motion: $s = 2t^{3} - 9t^{2} + 12t$.
To find the velocity $v$,we differentiate $s$ with respect to $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(2t^{3} - 9t^{2} + 12t) = 6t^{2} - 18t + 12$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(6t^{2} - 18t + 12) = 12t - 18$.
For the acceleration to be zero,we set $a = 0$:
$12t - 18 = 0$
$12t = 18$
$t = \frac{18}{12} = \frac{3}{2} \ s = 1.5 \ s$.
Thus,the acceleration of the particle will be zero after $1.5 \ s$.

(B) We know that $\Delta = E - 1$,so $\Delta^{2} = (E - 1)^{2} = E^{2} - 2E + 1$.
Given expression is $\left(\frac{\Delta^{2}}{E}\right) x^{3} = \left(\frac{E^{2} - 2E + 1}{E}\right) x^{3}$.
This simplifies to $(E - 2 + E^{-1}) x^{3}$.
Applying the operators: $E(x^{3}) = (x+1)^{3}$,$-2(x^{3}) = -2x^{3}$,and $E^{-1}(x^{3}) = (x-1)^{3}$.
Summing these: $(x+1)^{3} - 2x^{3} + (x-1)^{3}$.
Expanding the terms: $(x^{3} + 3x^{2} + 3x + 1) - 2x^{3} + (x^{3} - 3x^{2} + 3x - 1)$.
Combining like terms: $(x^{3} - 2x^{3} + x^{3}) + (3x^{2} - 3x^{2}) + (3x + 3x) + (1 - 1) = 6x$.

(A) Given,$x = 2 \cos \theta - \cos 2 \theta$ and $y = 2 \sin \theta - \sin 2 \theta$.
Differentiating with respect to $\theta$:
$\frac{dx}{d\theta} = -2 \sin \theta + 2 \sin 2 \theta = 2(\sin 2 \theta - \sin \theta)$.
$\frac{dy}{d\theta} = 2 \cos \theta - 2 \cos 2 \theta = 2(\cos \theta - \cos 2 \theta)$.
Now,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2(\cos \theta - \cos 2 \theta)}{2(\sin 2 \theta - \sin \theta)} = \frac{\cos \theta - \cos 2 \theta}{\sin 2 \theta - \sin \theta}$.
Using trigonometric identities $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$\frac{dy}{dx} = \frac{-2 \sin \frac{3 \theta}{2} \sin \frac{-\theta}{2}}{2 \cos \frac{3 \theta}{2} \sin \frac{\theta}{2}} = \frac{2 \sin \frac{3 \theta}{2} \sin \frac{\theta}{2}}{2 \cos \frac{3 \theta}{2} \sin \frac{\theta}{2}} = \tan \frac{3 \theta}{2}$.

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

(A) Let $u = \cos^{3} x$ and $v = \sin^{3} x$.
Applying the chain rule,we find the derivatives with respect to $x$:
$\frac{du}{dx} = 3 \cos^{2} x (-\sin x) = -3 \cos^{2} x \sin x$
$\frac{dv}{dx} = 3 \sin^{2} x (\cos x) = 3 \sin^{2} x \cos x$
Now,the derivative of $u$ with respect to $v$ is given by:
$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-3 \cos^{2} x \sin x}{3 \sin^{2} x \cos x}$
Simplifying the expression:
$\frac{du}{dv} = -\frac{\cos x}{\sin x} = -\cot x$

(A) To evaluate the integral $\int x \log x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \log x$ and $dv = x \, dx$.
Then,$du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$.
Applying the formula:
$\int x \log x \, dx = (\log x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$
$= \frac{x^2}{2} \log x - \frac{1}{2} \int x \, dx$
$= \frac{x^2}{2} \log x - \frac{1}{2} \cdot \frac{x^2}{2} + c$
$= \frac{x^2}{2} \log x - \frac{x^2}{4} + c$
$= \frac{x^2}{4} (2 \log x - 1) + c$.

(C) We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$.
The given integral is $\int e^{x} \left(\frac{x-1}{x^{2}}\right) dx$.
This can be rewritten as $\int e^{x} \left(\frac{1}{x} - \frac{1}{x^{2}}\right) dx$.
Here,let $f(x) = \frac{1}{x}$,then $f'(x) = -\frac{1}{x^{2}}$.
Applying the formula,we get $\int e^{x} \left(\frac{1}{x} - \frac{1}{x^{2}}\right) dx = e^{x} \left(\frac{1}{x}\right) + c = \frac{e^{x}}{x} + c$.

(B) Let $I = \int_{5}^{10} \frac{1}{(x-1)(x-2)} dx$.
Using partial fractions,we write $\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.
Solving for constants,we get $A = -1$ and $B = 1$.
Thus,$I = \int_{5}^{10} \left( \frac{-1}{x-1} + \frac{1}{x-2} \right) dx$.
Integrating,we get $I = [-\log|x-1| + \log|x-2|]_{5}^{10}$.
$I = [\log|\frac{x-2}{x-1}|]_{5}^{10}$.
Substituting the limits,$I = \log|\frac{10-2}{10-1}| - \log|\frac{5-2}{5-1}|$.
$I = \log(\frac{8}{9}) - \log(\frac{3}{4})$.
$I = \log(\frac{8/9}{3/4}) = \log(\frac{8}{9} \times \frac{4}{3}) = \log(\frac{32}{27})$.

(C) Let $I = \int [\sin (\log x) + \cos (\log x)] \, dx$.
We know that $\frac{d}{dx} [x \sin (\log x)] = \sin (\log x) \cdot \frac{d}{dx}(x) + x \cdot \cos (\log x) \cdot \frac{d}{dx}(\log x)$.
$= \sin (\log x) \cdot 1 + x \cdot \cos (\log x) \cdot \frac{1}{x}$.
$= \sin (\log x) + \cos (\log x)$.
Therefore,$\int [\sin (\log x) + \cos (\log x)] \, dx = \int \frac{d}{dx} [x \sin (\log x)] \, dx$.
$= x \sin (\log x) + c$.

(C) The feasible region is determined by the constraints $2x + 3y \leq 18$,$2x + y \leq 10$,$x \geq 0$,and $y \geq 0$. The vertices of the feasible region are $O(0, 0)$,$A(5, 0)$,$B(3, 4)$,and $C(0, 6)$.
We evaluate the objective function $z = 9x + 13y$ at each vertex:
$1$. At $O(0, 0)$: $z = 9(0) + 13(0) = 0$
$2$. At $A(5, 0)$: $z = 9(5) + 13(0) = 45$
$3$. At $B(3, 4)$: $z = 9(3) + 13(4) = 27 + 52 = 79$
$4$. At $C(0, 6)$: $z = 9(0) + 13(6) = 78$
Comparing these values,the maximum value of $z$ is $79$.

(B) Given,$A=\left[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]$.
Since $AB=BA=I$,$B$ is the inverse of $A$,i.e.,$B=A^{-1}$.
The determinant of $A$ is $|A| = \cos^2 \theta - (-\sin^2 \theta) = \cos^2 \theta + \sin^2 \theta = 1$.
The inverse of a $2 \times 2$ matrix $\left[\begin{array}{rr}a & b \\ c & d\end{array}\right]$ is $\frac{1}{ad-bc} \left[\begin{array}{rr}d & -b \\ -c & a\end{array}\right]$.
Applying this to $A$,we get $B = A^{-1} = \frac{1}{1} \left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$.
Thus,$B = \left[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$.

(A) The expression $a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$ represents the expansion of the determinant of matrix $A$ along the first row.
Therefore,$a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} = |A|$.
$|A| = 3 \begin{vmatrix} 2 & 1 \\ 2 & 6 \end{vmatrix} - 2 \begin{vmatrix} 1 & 1 \\ 3 & 6 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix}$.
$|A| = 3(12 - 2) - 2(6 - 3) + 4(2 - 6)$.
$|A| = 3(10) - 2(3) + 4(-4)$.
$|A| = 30 - 6 - 16 = 8$.

(B) Total number of ways to form a committee of $5$ members from $8$ boys and $3$ girls is given by $C(11, 5) = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$.
If $2$ particular girls must be included,we need to select the remaining $3$ members from the remaining $9$ people ($8$ boys and $1$ girl).
The number of ways to select the remaining $3$ members is $C(9, 3) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Therefore,the required probability is $\frac{84}{462} = \frac{2}{11}$.

(C) We are given $P(A)=0.5, P(B)=0.4$,and $P(A \cap B)=0.3$.
We need to find $P(A^{\prime} \mid B^{\prime})$.
Using the definition of conditional probability,$P(A^{\prime} \mid B^{\prime}) = \frac{P(A^{\prime} \cap B^{\prime})}{P(B^{\prime})}$.
By De Morgan's Law,$A^{\prime} \cap B^{\prime} = (A \cup B)^{\prime}$,so $P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 - P(A \cup B)$.
First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.4 - 0.3 = 0.6$.
Then,$P(A^{\prime} \cap B^{\prime}) = 1 - 0.6 = 0.4$.
Also,$P(B^{\prime}) = 1 - P(B) = 1 - 0.4 = 0.6$.
Therefore,$P(A^{\prime} \mid B^{\prime}) = \frac{0.4}{0.6} = \frac{2}{3}$.

(A) Let point $P(x, y, z)$ divide the line segment joining the points $A(5, -3, -2)$ and $B(1, 2, -2)$ in the ratio $m:1$.
Since the point $P$ lies on the $XOZ$-plane,its $y$-coordinate must be $0$.
Using the section formula,the $y$-coordinate of $P$ is given by $\frac{m(2) + 1(-3)}{m + 1} = 0$.
$\Rightarrow 2m - 3 = 0 \Rightarrow m = \frac{3}{2}$.
Now,we find the $x$ and $z$ coordinates using $m = \frac{3}{2}$:
$x = \frac{m(1) + 1(5)}{m + 1} = \frac{\frac{3}{2}(1) + 5}{\frac{3}{2} + 1} = \frac{\frac{3+10}{2}}{\frac{5}{2}} = \frac{13}{5}$.
$z = \frac{m(-2) + 1(-2)}{m + 1} = \frac{\frac{3}{2}(-2) - 2}{\frac{3}{2} + 1} = \frac{-3 - 2}{\frac{5}{2}} = \frac{-5}{\frac{5}{2}} = -2$.
Therefore,the required point is $\left(\frac{13}{5}, 0, -2\right)$.

(B) Let the direction cosines of the line $\overrightarrow{OR}$ be $l, m, n$.
Since the line makes angles $\theta_{1}, \theta_{2}, \theta_{3}$ with the planes $XOY, YOZ, ZOX$ respectively,the angles with the normals to these planes (which are the $Z, X, Y$ axes) are $\frac{\pi}{2}-\theta_{1}, \frac{\pi}{2}-\theta_{2}, \frac{\pi}{2}-\theta_{3}$.
Thus,$|l| = \cos(\frac{\pi}{2}-\theta_{2}) = \sin \theta_{2}$,$|m| = \cos(\frac{\pi}{2}-\theta_{3}) = \sin \theta_{3}$,and $|n| = \cos(\frac{\pi}{2}-\theta_{1}) = \sin \theta_{1}$.
We know that for direction cosines,$l^{2}+m^{2}+n^{2}=1$.
Substituting the values,we get $\sin^{2} \theta_{2} + \sin^{2} \theta_{3} + \sin^{2} \theta_{1} = 1$.
Using the identity $\sin^{2} \theta = 1 - \cos^{2} \theta$,we have $(1-\cos^{2} \theta_{2}) + (1-\cos^{2} \theta_{3}) + (1-\cos^{2} \theta_{1}) = 1$.
$3 - (\cos^{2} \theta_{1} + \cos^{2} \theta_{2} + \cos^{2} \theta_{3}) = 1$.
Therefore,$\cos^{2} \theta_{1} + \cos^{2} \theta_{2} + \cos^{2} \theta_{3} = 3 - 1 = 2$.

(A) Let the equation of the plane be $A(x + 2) + B(y - 2) + C(z - 2) = 0$.
Since it passes through $(2, -2, -2)$,we have $A(2 + 2) + B(-2 - 2) + C(-2 - 2) = 0$,which simplifies to $4A - 4B - 4C = 0$,or $A - B - C = 0$ ... $(i)$.
The plane is perpendicular to $9x - 13y - 3z = 0$,so the normal vector of the required plane $(A, B, C)$ is perpendicular to the normal vector of the given plane $(9, -13, -3)$.
Thus,$9A - 13B - 3C = 0$ ... $(ii)$.
Solving $(i)$ and $(ii)$ using cross product:
$\frac{A}{(-1)(-3) - (-1)(-13)} = \frac{B}{(-1)(9) - (1)(-3)} = \frac{C}{(1)(-13) - (-1)(9)}$
$\frac{A}{3 - 13} = \frac{B}{-9 + 3} = \frac{C}{-13 + 9}$
$\frac{A}{-10} = \frac{B}{-6} = \frac{C}{-4}$
Simplifying the ratios,we get $A:B:C = 5:3:2$.
Substituting these into the plane equation: $5(x + 2) + 3(y - 2) + 2(z - 2) = 0$.
$5x + 10 + 3y - 6 + 2z - 4 = 0$.
$5x + 3y + 2z = 0$.

(A) Given the equation: $2 \overrightarrow{a} + 3 \overrightarrow{b} - 5 \overrightarrow{c} = \overrightarrow{0}$
Rearranging the terms,we get: $5 \overrightarrow{c} = 2 \overrightarrow{a} + 3 \overrightarrow{b}$
Dividing by $5$,we have: $\overrightarrow{c} = \frac{2 \overrightarrow{a} + 3 \overrightarrow{b}}{5}$
Since $2 + 3 = 5$,we can write this as: $\overrightarrow{c} = \frac{2 \overrightarrow{a} + 3 \overrightarrow{b}}{2 + 3}$
This is the section formula for internal division,which states that if a point $C$ divides the line segment $AB$ in the ratio $m:n$,its position vector is given by $\overrightarrow{c} = \frac{n \overrightarrow{a} + m \overrightarrow{b}}{m + n}$.
Comparing $\overrightarrow{c} = \frac{2 \overrightarrow{a} + 3 \overrightarrow{b}}{2 + 3}$ with the formula $\overrightarrow{c} = \frac{n \overrightarrow{a} + m \overrightarrow{b}}{m + n}$,we identify $n = 2$ and $m = 3$.
Therefore,the point $C$ divides the segment $AB$ in the ratio $m:n = 3:2$ internally.

(B) Given $\overrightarrow{p} = x \overrightarrow{a} + y \overrightarrow{b} + z \overrightarrow{c}$.
Substituting the vectors,we get:
$3 \hat{i} + 2 \hat{j} + 4 \hat{k} = x(\hat{i} + \hat{j}) + y(\hat{j} + \hat{k}) + z(\hat{i} + \hat{k})$
$3 \hat{i} + 2 \hat{j} + 4 \hat{k} = (x + z) \hat{i} + (x + y) \hat{j} + (y + z) \hat{k}$
Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides:
$x + z = 3$ $(i)$
$x + y = 2$ $(ii)$
$y + z = 4$ $(iii)$
Adding $(i), (ii),$ and $(iii)$:
$2(x + y + z) = 3 + 2 + 4 = 9 \Rightarrow x + y + z = 4.5$
Subtracting $(iii)$ from the sum: $x = 4.5 - 4 = 0.5 = \frac{1}{2}$
Subtracting $(i)$ from the sum: $y = 4.5 - 3 = 1.5 = \frac{3}{2}$
Subtracting $(ii)$ from the sum: $z = 4.5 - 2 = 2.5 = \frac{5}{2}$
Thus,$x = \frac{1}{2}, y = \frac{3}{2}, z = \frac{5}{2}$.

(B) The volume of a parallelepiped with coterminous edges represented by vectors $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\overrightarrow{OC}$ is given by the scalar triple product $|[\overrightarrow{OA} \overrightarrow{OB} \overrightarrow{OC}]|$.
Given the vertices $O(0,0,0)$, $A(2,-2,1)$, $B(5,-4,4)$, and $C(1,-2,4)$, the vectors are:
$\overrightarrow{OA} = 2\hat{i} - 2\hat{j} + \hat{k}$
$\overrightarrow{OB} = 5\hat{i} - 4\hat{j} + 4\hat{k}$
$\overrightarrow{OC} = \hat{i} - 2\hat{j} + 4\hat{k}$
The volume is the absolute value of the determinant:
$V = \left| \begin{vmatrix} 2 & -2 & 1 \\ 5 & -4 & 4 \\ 1 & -2 & 4 \end{vmatrix} \right|$
Expanding along the first row:
$V = |2((-4)(4) - (4)(-2)) - (-2)((5)(4) - (4)(1)) + 1((5)(-2) - (-4)(1))|$
$V = |2(-16 + 8) + 2(20 - 4) + 1(-10 + 4)|$
$V = |2(-8) + 2(16) + 1(-6)|$
$V = |-16 + 32 - 6|$
$V = |10| = 10 \text{ cubic units}$.