MHT CET 2008 Mathematics Question Paper with Answer and Solution

(D) Given curve is $y^2=2(x-3)$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 2$,which implies $\frac{dy}{dx} = \frac{1}{y}$.
The slope of the tangent at any point $(x, y)$ is $m_t = \frac{1}{y}$.
The slope of the normal is $m_n = -\frac{1}{m_t} = -y$.
The given line is $y - 2x + 1 = 0$,which can be written as $y = 2x - 1$. The slope of this line is $2$.
Since the normal is parallel to the line,their slopes must be equal,so $-y = 2$,which gives $y = -2$.
Substituting $y = -2$ into the curve equation: $(-2)^2 = 2(x - 3) \Rightarrow 4 = 2(x - 3) \Rightarrow 2 = x - 3 \Rightarrow x = 5$.
Thus,the required point is $(5, -2)$.

(A) The given hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$. Here,$a^{2}=3$ and $b^{2}=2$.
The equation of the tangent parallel to $y-x+5=0$ is of the form $y=x+c$.
Comparing with $y=mx+c$,we have $m=1$.
The condition for the line $y=mx+c$ to be a tangent 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,$c^{2}=3(1)^{2}-2 = 3-2 = 1$.
Thus,$c=\pm 1$.
The equations of the tangents are $y=x+1$ or $y=x-1$,which can be written as $x-y+1=0$ or $x-y-1=0$.
Comparing with the given options,$x-y-1=0$ is the correct choice.

(C) Let the coordinates of the other end of the diameter be $(t, 3-t)$ since it lies on the line $x+y=3$.
Let the centre of the circle be $(h, k)$.
The centre is the midpoint of the diameter with endpoints $(1, 1)$ and $(t, 3-t)$.
So,$h = \frac{1+t}{2}$ and $k = \frac{1+3-t}{2} = \frac{4-t}{2}$.
From these equations,we have $t = 2h-1$ and $t = 4-2k$.
Equating the two expressions for $t$:
$2h-1 = 4-2k$
$2h+2k = 5$.
Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $2x+2y=5$.

(A) Given the circle equation $x^{2}+y^{2}=13$.
Since the abscissa of the points is $x=2$,we substitute this into the circle equation:
$2^{2}+y^{2}=13$
$4+y^{2}=13$
$y^{2}=9$
$y=\pm 3$.
So,the points of contact are $(2, 3)$ and $(2, -3)$.
The equation of the tangent to the circle $x^{2}+y^{2}=r^{2}$ at point $(x_{1}, y_{1})$ is given by $xx_{1}+yy_{1}=r^{2}$.
For point $(2, 3)$: $2x+3y=13$.
For point $(2, -3)$: $2x-3y=13$.
Thus,the equations are $2x+3y=13$ and $2x-3y=13$.

(D) The equation of any normal to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at point $(a \cos \phi, b \sin \phi)$ is given by $ax \sec \phi - by \operatorname{cosec} \phi = a^{2} - b^{2} \quad \dots (i)$
Comparing this with the given line $x \cos \alpha + y \sin \alpha = p$,we have:
$\frac{a \sec \phi}{\cos \alpha} = \frac{-b \operatorname{cosec} \phi}{\sin \alpha} = \frac{a^{2} - b^{2}}{p}$
From this,we get $\sec \phi = \frac{(a^{2} - b^{2}) \cos \alpha}{ap}$ and $\operatorname{cosec} \phi = \frac{-(a^{2} - b^{2}) \sin \alpha}{bp}$.
Using the identity $\cos^{2} \phi + \sin^{2} \phi = 1$,we have $\frac{1}{\sec^{2} \phi} + \frac{1}{\operatorname{cosec}^{2} \phi} = 1$.
Substituting the values,we get $\frac{a^{2} p^{2}}{(a^{2} - b^{2})^{2} \cos^{2} \alpha} + \frac{b^{2} p^{2}}{(a^{2} - b^{2})^{2} \sin^{2} \alpha} = 1$.
Multiplying by $(a^{2} - b^{2})^{2}$,we get $p^{2} (a^{2} \sec^{2} \alpha + b^{2} \operatorname{cosec}^{2} \alpha) = (a^{2} - b^{2})^{2}$.

(C) We need to evaluate $\lim_{x \rightarrow \infty} \sqrt{\frac{x - \sin x}{x + \cos^{2} x}}$.
Divide the numerator and denominator by $x$ inside the square root:
$= \lim_{x \rightarrow \infty} \sqrt{\frac{1 - \frac{\sin x}{x}}{1 + \frac{\cos^{2} x}{x}}}$
Since $\lim_{x \rightarrow \infty} \frac{\sin x}{x} = 0$ and $\lim_{x \rightarrow \infty} \frac{\cos^{2} x}{x} = 0$ (because $\sin x$ and $\cos^{2} x$ are bounded functions):
$= \sqrt{\frac{1 - 0}{1 + 0}}$
$= \sqrt{1} = 1$

(B) Using the trigonometric identity $\cos C - \cos D = -2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$,we have:
$\lim _{x \rightarrow 0} \frac{\cos ax - \cos bx}{x^{2}} = \lim _{x \rightarrow 0} \frac{-2 \sin \left( \frac{ax+bx}{2} \right) \sin \left( \frac{ax-bx}{2} \right)}{x^{2}}$
$= -2 \lim _{x \rightarrow 0} \left( \frac{\sin \left( \frac{a+b}{2} x \right)}{x} \right) \left( \frac{\sin \left( \frac{a-b}{2} x \right)}{x} \right)$
Using the standard limit $\lim _{\theta \rightarrow 0} \frac{\sin k\theta}{\theta} = k$,we get:
$= -2 \left( \frac{a+b}{2} \right) \left( \frac{a-b}{2} \right) = -2 \left( \frac{a^{2} - b^{2}}{4} \right) = \frac{b^{2} - a^{2}}{2}$

(C) The given system of inequations is $x \geq 0$,$y \geq 0$,$y \leq 6$,and $x+y \leq 3$.
$1$. The conditions $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.
$2$. The line $x+y = 3$ passes through $(3, 0)$ and $(0, 3)$. The inequality $x+y \leq 3$ represents the region on or below this line.
$3$. The condition $y \leq 6$ is satisfied by the region defined by $x+y \leq 3$ in the first quadrant,as the maximum value of $y$ in this region is $3$.
$4$. Since the region is enclosed by the axes and the line $x+y=3$,it is a bounded region in the first quadrant.

(D) In the given circuit,the top branch consists of switches $\sim p$ and $q$ connected in series. The Boolean expression for this branch is $(\sim p \wedge q)$.
The bottom branch consists of switches $p$ and $\sim q$ connected in series. The Boolean expression for this branch is $(p \wedge \sim q)$.
Since these two branches are connected in parallel,the total Boolean polynomial for the circuit is the disjunction of the two expressions:
$(\sim p \wedge q) \vee (p \wedge \sim q)$.

(B) Let $p$ be the statement: "It rains".
Let $q$ be the statement: "$I$ shall go to school".
The given conditional statement is $p \Rightarrow q$.
The negation of a conditional statement $p \Rightarrow q$ is given by $\sim(p \Rightarrow q) \equiv p \wedge \sim q$.
Here,$p \wedge \sim q$ translates to: "It rains and $I$ shall not go to school".
Therefore,the correct option is $B$.

(A) To find the dual of a Boolean expression,we replace $\vee$ with $\wedge$,$\wedge$ with $\vee$,$0$ with $1$,and $1$ with $0$.
Given the expression $\left(x^{\prime} \vee y^{\prime}\right) = x \wedge y$,we replace the operators:
$\vee$ becomes $\wedge$
$\wedge$ becomes $\vee$
Thus,the dual is $\left(x^{\prime} \wedge y^{\prime}\right) = x \vee y$.

(C) The general equation of a pair of straight lines is given by $Ax^{2}+2Hxy+By^{2}+2Gx+2Fy+C=0$. \\ For the pair of lines to be perpendicular,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero. \\ That is,$A+B=0$. \\ Comparing the given equation $12x^{2}+7xy+ay^{2}+13x-y+3=0$ with the general form,we have $A=12$ and $B=a$. \\ Substituting these values into the condition $A+B=0$,we get $12+a=0$. \\ Therefore,$a=-12$.

(B) The given parabola is $y^{2}=16x$. Comparing with $y^{2}=4ax$,we get $4a=16$,so $a=4$.
Let the point on the parabola be $(h, k)$.
Given that the ordinate is twice the abscissa,we have $k=2h$.
Substituting $k=2h$ into the parabola equation: $(2h)^{2}=16h$ $\Rightarrow 4h^{2}=16h$ $\Rightarrow 4h(h-4)=0$.
Thus,$h=0$ or $h=4$.
For $h=0$,$k=0$. For $h=4$,$k=8$.
The point $(0,0)$ is the vertex,where the focal distance is $a=4$.
The point $(4,8)$ is on the parabola,where the focal distance is $h+a = 4+4 = 8$.
Since the question asks for the focal distance of a point (implying a non-vertex point),the answer is $8$.

(A) Given parabola is $y^{2}=8x$,so $4a=8 \Rightarrow a=2$.
Any tangent to the parabola is of the form $y=mx+\frac{a}{m}$,which is $y=mx+\frac{2}{m}$ or $mx-y+\frac{2}{m}=0$.
For this line to be a tangent to the circle $x^{2}+y^{2}=2$ (with center $(0,0)$ and radius $r=\sqrt{2}$),the perpendicular distance from the center to the line must equal the radius.
$\frac{|m(0)-(0)+\frac{2}{m}|}{\sqrt{m^{2}+(-1)^{2}}}=\sqrt{2}$
$\frac{2}{|m|\sqrt{m^{2}+1}}=\sqrt{2}$
Squaring both sides: $\frac{4}{m^{2}(m^{2}+1)}=2 \Rightarrow m^{2}(m^{2}+1)=2$
$m^{4}+m^{2}-2=0$
$(m^{2}+2)(m^{2}-1)=0$
Since $m$ must be real,$m^{2}=1 \Rightarrow m=\pm 1$.
For $m=1$,the tangent is $y=x+\frac{2}{1} \Rightarrow y=x+2$.
For $m=-1$,the tangent is $y=-x+\frac{2}{-1} \Rightarrow y=-x-2$.

(A) Let $P(A)$ be the probability of failing in Physics and $P(B)$ be the probability of failing in Mathematics.
Given $P(A) = \frac{20}{100} = 0.2$ and $P(B) = \frac{10}{100} = 0.1$.
Assuming the events are independent,the probability of failing in at least one subject is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the events are independent,$P(A \cap B) = P(A) \cdot P(B) = 0.2 \times 0.1 = 0.02$.
Therefore,$P(A \cup B) = 0.2 + 0.1 - 0.02 = 0.28$.
Converting to percentage,$0.28 \times 100 = 28 \%$.

(D) Given the curve equation: $y=x^{3}-3x^{2}-9x+5$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = 3x^{2}-6x-9$.
Since the tangent is parallel to the $x$-axis,its slope must be zero:
$\frac{dy}{dx} = 0$.
Substituting the derivative:
$3x^{2}-6x-9 = 0$.
Dividing by $3$:
$x^{2}-2x-3 = 0$.
Factoring the quadratic equation:
$(x-3)(x+1) = 0$.
Thus,the values of $x$ (abscissae) are $x=3$ and $x=-1$.

(C) Given curves are $r_1 = \sin \theta + \cos \theta$ and $r_2 = 2 \sin \theta$.
To find the intersection,set $r_1 = r_2$:
$\sin \theta + \cos \theta = 2 \sin \theta \Rightarrow \cos \theta = \sin \theta \Rightarrow \tan \theta = 1 \Rightarrow \theta = \frac{\pi}{4}$.
For $r_1 = \sin \theta + \cos \theta$,$\frac{dr_1}{d\theta} = \cos \theta - \sin \theta$. Let $\phi_1$ be the angle between the tangent and the radius vector,then $\tan \phi_1 = \frac{r_1}{dr_1/d\theta} = \frac{\sin \theta + \cos \theta}{\cos \theta - \sin \theta}$.
At $\theta = \frac{\pi}{4}$,$\tan \phi_1 = \frac{1/\sqrt{2} + 1/\sqrt{2}}{1/\sqrt{2} - 1/\sqrt{2}} = \frac{\sqrt{2}}{0} = \infty \Rightarrow \phi_1 = \frac{\pi}{2}$.
For $r_2 = 2 \sin \theta$,$\frac{dr_2}{d\theta} = 2 \cos \theta$. Then $\tan \phi_2 = \frac{r_2}{dr_2/d\theta} = \frac{2 \sin \theta}{2 \cos \theta} = \tan \theta$.
At $\theta = \frac{\pi}{4}$,$\tan \phi_2 = \tan(\frac{\pi}{4}) = 1 \Rightarrow \phi_2 = \frac{\pi}{4}$.
The angle of intersection $\psi = |\phi_1 - \phi_2| = |\frac{\pi}{2} - \frac{\pi}{4}| = \frac{\pi}{4}$.

(B) Let $f(x) = x^{3} - 12x^{2} + 36x + 17$.
Find the derivative: $f'(x) = 3x^{2} - 24x + 36$.
Set $f'(x) = 0$ to find critical points: $3(x^{2} - 8x + 12) = 0 \Rightarrow 3(x - 2)(x - 6) = 0$.
The critical points are $x = 2$ and $x = 6$.
Evaluate the function at the critical points and the interval endpoints $[1, 10]$:
$f(1) = (1)^{3} - 12(1)^{2} + 36(1) + 17 = 1 - 12 + 36 + 17 = 42$.
$f(2) = (2)^{3} - 12(2)^{2} + 36(2) + 17 = 8 - 48 + 72 + 17 = 49$.
$f(6) = (6)^{3} - 12(6)^{2} + 36(6) + 17 = 216 - 432 + 216 + 17 = 17$.
$f(10) = (10)^{3} - 12(10)^{2} + 36(10) + 17 = 1000 - 1200 + 360 + 17 = 177$.
Comparing these values,the maximum value is $177$.

(A) The required area is given by the integral of $|y|$ with respect to $x$ from $x=-1$ to $x=2$.
Since $y=x$,we have $|y|=|x|$.
Required Area $= \int_{-1}^{2} |x| dx$
$= \int_{-1}^{0} |x| dx + \int_{0}^{2} |x| dx$
$= \int_{-1}^{0} (-x) dx + \int_{0}^{2} x dx$
$= \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{2}$
$= (0 - (-1/2)) + (4/2 - 0)$
$= 1/2 + 2 = 5/2$ sq unit.

(C) The curves are $y=x^{2}$ (or $x=\sqrt{y}$) and $x=y^{2}$.
They intersect at $(0,0)$ and $(1,1)$.
When revolving about the $y$-axis,the volume $V$ is given by the formula $V = \pi \int_{a}^{b} (x_{outer}^{2} - x_{inner}^{2}) dy$.
Here,for $y \in [0,1]$,the outer curve is $x = \sqrt{y}$ and the inner curve is $x = y^{2}$.
Thus,$V = \pi \int_{0}^{1} ((\sqrt{y})^{2} - (y^{2})^{2}) dy$
$V = \pi \int_{0}^{1} (y - y^{4}) dy$
$V = \pi \left[ \frac{y^{2}}{2} - \frac{y^{5}}{5} \right]_{0}^{1}$
$V = \pi \left( \frac{1}{2} - \frac{1}{5} \right) = \pi \left( \frac{5-2}{10} \right) = \frac{3}{10} \pi$.

(C) For a function $f(x)$ to be continuous at $x = 0$,the condition $f(0) = \lim_{x \to 0} f(x)$ must be satisfied.
Given $f(0) = k$.
Now,we calculate the limit: $\lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin \frac{1}{x}$.
We know that for all $x \neq 0$,$-1 \leq \sin \frac{1}{x} \leq 1$.
Multiplying by $x$ (where $x > 0$),we get $-x \leq x \sin \frac{1}{x} \leq x$.
By the Squeeze Theorem,as $x \to 0$,both $-x$ and $x$ approach $0$.
Therefore,$\lim_{x \to 0} x \sin \frac{1}{x} = 0$.
Since $f(0) = \lim_{x \to 0} f(x)$,we have $k = 0$.

(C) We use the standard result $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.
Let $f(x) = \log \sin x$.
Then $f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.
Thus,the integral becomes $\int_{\pi / 4}^{\pi / 2} e^x (f(x) + f'(x)) dx = [e^x f(x)]_{\pi / 4}^{\pi / 2}$.
Substituting the limits:
$= [e^x \log \sin x]_{\pi / 4}^{\pi / 2} = e^{\pi / 2} \log \sin(\pi / 2) - e^{\pi / 4} \log \sin(\pi / 4)$.
Since $\sin(\pi / 2) = 1$ and $\log(1) = 0$,the first term is $0$.
$= 0 - e^{\pi / 4} \log(1 / \sqrt{2}) = -e^{\pi / 4} \log(2^{-1/2})$.
$= -e^{\pi / 4} \cdot (-1/2) \log 2 = \frac{1}{2} e^{\pi / 4} \log 2$.

(B) Let $I = \int_{0}^{\pi} x \sin^{3} x \, dx$ ...$(i)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get:
$I = \int_{0}^{\pi} (\pi - x) \sin^{3}(\pi - x) \, dx = \int_{0}^{\pi} (\pi - x) \sin^{3} x \, dx$ ...(ii)
Adding $(i)$ and (ii):
$2I = \int_{0}^{\pi} (x + \pi - x) \sin^{3} x \, dx = \pi \int_{0}^{\pi} \sin^{3} x \, dx$
Using $\sin 3x = 3 \sin x - 4 \sin^{3} x$,we have $\sin^{3} x = \frac{3 \sin x - \sin 3x}{4}$.
$2I = \frac{\pi}{4} \int_{0}^{\pi} (3 \sin x - \sin 3x) \, dx$
$2I = \frac{\pi}{4} \left[ -3 \cos x + \frac{\cos 3x}{3} \right]_{0}^{\pi}$
$2I = \frac{\pi}{4} \left[ (-3(-1) + \frac{-1}{3}) - (-3(1) + \frac{1}{3}) \right]$
$2I = \frac{\pi}{4} \left[ (3 - \frac{1}{3}) - (-3 + \frac{1}{3}) \right] = \frac{\pi}{4} \left[ \frac{8}{3} + \frac{8}{3} \right] = \frac{\pi}{4} \times \frac{16}{3} = \frac{4 \pi}{3}$
$I = \frac{2 \pi}{3}$

(A) Given the integral $\int_{1}^{2} \frac{dx}{x}$ with $n=4$ sub-intervals.
The step size $h = \frac{b-a}{n} = \frac{2-1}{4} = 0.25$.
The values of $x_i$ are $x_0=1, x_1=1.25, x_2=1.5, x_3=1.75, x_4=2$.
The corresponding values of $y_i = \frac{1}{x_i}$ are:
$y_0 = \frac{1}{1} = 1$
$y_1 = \frac{1}{1.25} = 0.8$
$y_2 = \frac{1}{1.5} = 0.6667$
$y_3 = \frac{1}{1.75} = 0.5714$
$y_4 = \frac{1}{2} = 0.5$
Using Simpson's $\frac{1}{3}$ rule: $\int_{a}^{b} y dx \approx \frac{h}{3} [y_0 + y_n + 4(y_1 + y_3 + \dots) + 2(y_2 + y_4 + \dots)]$.
$\int_{1}^{2} \frac{dx}{x} \approx \frac{0.25}{3} [y_0 + y_4 + 4(y_1 + y_3) + 2(y_2)]$.
$\int_{1}^{2} \frac{dx}{x} \approx \frac{0.25}{3} [1 + 0.5 + 4(0.8 + 0.5714) + 2(0.6667)]$.
$\int_{1}^{2} \frac{dx}{x} \approx \frac{0.25}{3} [1.5 + 4(1.3714) + 1.3334]$.
$\int_{1}^{2} \frac{dx}{x} \approx \frac{0.25}{3} [1.5 + 5.4856 + 1.3334] = \frac{0.25}{3} [8.319] = 0.69325 \approx 0.6932$.

(C) Given $f(x) = \frac{1}{1+x}$,interval $[0, 1]$,and number of sub-intervals $n = 4$.
Width of each sub-interval $h = \frac{b-a}{n} = \frac{1-0}{4} = 0.25$.

$i$	$x_i$	$y_i = \frac{1}{1+x_i}$
$0$	$0$	$1$
$1$	$0.25$	$0.8$
$2$	$0.5$	$0.6667$
$3$	$0.75$	$0.5714$
$4$	$1$	$0.5$

Using the Trapezoidal rule formula:
$\int_{0}^{1} f(x) dx \approx \frac{h}{2} [y_0 + 2(y_1 + y_2 + y_3) + y_4]$
$\int_{0}^{1} f(x) dx \approx \frac{0.25}{2} [1 + 2(0.8 + 0.6667 + 0.5714) + 0.5]$
$\int_{0}^{1} f(x) dx \approx 0.125 [1 + 2(2.0381) + 0.5]$
$\int_{0}^{1} f(x) dx \approx 0.125 [1 + 4.0762 + 0.5]$
$\int_{0}^{1} f(x) dx \approx 0.125 [5.5762] = 0.697025 \approx 0.6977$ (rounding to given options).

(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$.
To find the degree,we must eliminate the radical sign.
Rewrite the equation as $\sqrt{\frac{dy}{dx}} = 4 \frac{dy}{dx} + 7x$.
Squaring both sides,we get:
$(\frac{dy}{dx}) = (4 \frac{dy}{dx} + 7x)^2$.
$(\frac{dy}{dx}) = 16(\frac{dy}{dx})^2 + 49x^2 + 56x \frac{dy}{dx}$.
Rearranging the terms,we get $16(\frac{dy}{dx})^2 + (56x - 1)\frac{dy}{dx} + 49x^2 = 0$.
The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$.
The highest power of the highest order derivative is $2$,so the degree is $2$.
Thus,the order is $1$ and the degree is $2$.

(A) The given differential equation is $(3xy + y^2)dx + (x^2 + xy)dy = 0$.
Rearranging the terms,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: $v + x\frac{dv}{dx} = -\frac{3x^2v + x^2v^2}{x^2 + x^2v} = -\frac{3v + v^2}{1 + v}$.
$x\frac{dv}{dx} = -\frac{3v + v^2}{1 + v} - v = -\frac{3v + v^2 + v + v^2}{1 + v} = -\frac{2v^2 + 4v}{1 + v} = -\frac{2v(v + 2)}{v + 1}$.
Separating the variables: $\frac{v + 1}{v(v + 2)} dv = -\frac{2}{x} dx$.
Using partial fractions: $\frac{v + 1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2} \Rightarrow v + 1 = A(v + 2) + Bv$.
For $v = 0$,$1 = 2A \Rightarrow A = 1/2$. For $v = -2$,$-1 = -2B \Rightarrow B = 1/2$.
So,$\int (\frac{1}{2v} + \frac{1}{2(v + 2)}) dv = -\int \frac{2}{x} dx$.
$\frac{1}{2} \ln|v| + \frac{1}{2} \ln|v + 2| = -2 \ln|x| + C'$.
$\ln|v(v + 2)| = -4 \ln|x| + 2C' \Rightarrow \ln|v(v + 2)x^4| = C''$.
$v(v + 2)x^4 = c^2$.
Substituting $v = y/x$: $\frac{y}{x}(\frac{y}{x} + 2)x^4 = c^2 \Rightarrow y(y + 2x)x^2 = c^2 \Rightarrow x^2(y^2 + 2xy) = c^2$.

(C) Given differential equation is $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$.
Let $v = x-y$. Then $\frac{dv}{dx} = 1 - \frac{dy}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.
Substituting this into the equation: $1 - \frac{dv}{dx} = \frac{v+3}{2v+5}$.
Rearranging gives $\frac{dv}{dx} = 1 - \frac{v+3}{2v+5} = \frac{2v+5-v-3}{2v+5} = \frac{v+2}{2v+5}$.
Separating variables: $\int \frac{2v+5}{v+2} dv = \int dx$.
Rewriting the integrand: $\int \left( \frac{2(v+2) + 1}{v+2} \right) dv = \int dx$.
This simplifies to $\int (2 + \frac{1}{v+2}) dv = \int dx$.
Integrating both sides: $2v + \log|v+2| = x + c$.
Substituting $v = x-y$ back: $2(x-y) + \log|x-y+2| = x + c$.

(A) Given equation is $x^{3}+y^{3}-3 a x y=0$.
On differentiating both sides with respect to $x$,we get:
$\frac{d}{d x}(x^{3}) + \frac{d}{d x}(y^{3}) - 3 a \frac{d}{d x}(x y) = 0$
$3 x^{2} + 3 y^{2} \frac{d y}{d x} - 3 a \left( x \frac{d y}{d x} + y \right) = 0$
Dividing by $3$,we get:
$x^{2} + y^{2} \frac{d y}{d x} - a x \frac{d y}{d x} - a y = 0$
Grouping the terms containing $\frac{d y}{d x}$:
$\frac{d y}{d x} (y^{2} - a x) = a y - x^{2}$
Therefore,$\frac{d y}{d x} = \frac{a y - x^{2}}{y^{2} - a x}$.

(C) Given,$v = \frac{ds}{dt} = 6t - \frac{t^2}{6}$.
Integrating both sides with respect to $t$,we get:
$s = \int (6t - \frac{t^2}{6}) dt = 3t^2 - \frac{t^3}{18} + C$.
Given that $s = 0$ at $t = 0$,we substitute these values to find the constant $C$:
$0 = 3(0)^2 - \frac{(0)^3}{18} + C \implies C = 0$.
Thus,the displacement function is $s(t) = 3t^2 - \frac{t^3}{18}$.
To find the distance traveled in $3 \ s$,we calculate $s(3)$:
$s(3) = 3(3)^2 - \frac{(3)^3}{18} = 3(9) - \frac{27}{18} = 27 - \frac{3}{2} = \frac{54 - 3}{2} = \frac{51}{2}$.

(B) Given,$x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$.
On differentiating $x$ and $y$ with respect to $t$,we get:
$\frac{dx}{dt} = \frac{(1+t^{2})(-2t) - (1-t^{2})(2t)}{(1+t^{2})^{2}} = \frac{-2t - 2t^{3} - 2t + 2t^{3}}{(1+t^{2})^{2}} = \frac{-4t}{(1+t^{2})^{2}}$.
$\frac{dy}{dt} = \frac{(1+t^{2})(2a) - (2at)(2t)}{(1+t^{2})^{2}} = \frac{2a + 2at^{2} - 4at^{2}}{(1+t^{2})^{2}} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}}$.
Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a(1-t^{2})}{(1+t^{2})^{2}} \times \frac{(1+t^{2})^{2}}{-4t}$.
$\frac{dy}{dx} = \frac{2a(1-t^{2})}{-4t} = \frac{a(1-t^{2})}{-2t} = \frac{a(t^{2}-1)}{2t}$.

(A) Given,$y=x^{n} \log x+x(\log x)^{n}$.
Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ to both terms:
$\frac{dy}{dx} = \frac{d}{dx}(x^n \log x) + \frac{d}{dx}(x(\log x)^n)$
$= (x^n \cdot \frac{1}{x} + \log x \cdot nx^{n-1}) + (x \cdot n(\log x)^{n-1} \cdot \frac{1}{x} + (\log x)^n \cdot 1)$
$= (x^{n-1} + nx^{n-1} \log x) + (n(\log x)^{n-1} + (\log x)^n)$
$= x^{n-1}(1 + n \log x) + (\log x)^{n-1}(n + \log x)$.

(A) Given,$y = \log_{10} x + \log_{x} 10 + \log_{x} x + \log_{10} 10$.
Using the change of base formula $\log_{a} b = \frac{\log_{e} b}{\log_{e} a}$,we can rewrite the expression as:
$y = \frac{\log_{e} x}{\log_{e} 10} + \frac{\log_{e} 10}{\log_{e} x} + 1 + 1$.
$y = \frac{1}{\log_{e} 10} \cdot \log_{e} x + \log_{e} 10 \cdot (\log_{e} x)^{-1} + 2$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{1}{\log_{e} 10} \cdot \frac{d}{dx}(\log_{e} x) + \log_{e} 10 \cdot \frac{d}{dx}((\log_{e} x)^{-1}) + 0$.
$\frac{dy}{dx} = \frac{1}{\log_{e} 10} \cdot \frac{1}{x} + \log_{e} 10 \cdot (-1)(\log_{e} x)^{-2} \cdot \frac{1}{x}$.
$\frac{dy}{dx} = \frac{1}{x \log_{e} 10} - \frac{\log_{e} 10}{x(\log_{e} x)^{2}}$.

(A) We know that $\Delta \log f(x) = \log f(x+h) - \log f(x) = \log \left[\frac{f(x+h)}{f(x)}\right]$.
Since $f(x+h) = (1 + \Delta) f(x)$,we have $\Delta \log f(x) = \log \left[\frac{(1 + \Delta) f(x)}{f(x)}\right] = \log \left[1 + \frac{\Delta f(x)}{f(x)}\right]$.
Next,for the second term,$\Delta^{2}(3^{x}) = (E - 1)^{2} 3^{x} = (E^{2} - 2E + 1) 3^{x}$.
$= E^{2}(3^{x}) - 2E(3^{x}) + 3^{x} = 3^{x+2} - 2 \cdot 3^{x+1} + 3^{x}$.
$= 3^{x}(9 - 6 + 1) = 4 \cdot 3^{x}$.
Combining both terms,the result is $\log \left[1 + \frac{\Delta f(x)}{f(x)}\right] + 4 \cdot 3^{x}$.

(B) We have $I = \int x \sin x \sec ^{3} x \, dx = \int x \tan x \sec ^{2} x \, dx$.
Let $u = x$ and $dv = \tan x \sec ^{2} x \, dx$.
Then $du = dx$ and $v = \int \tan x \sec ^{2} x \, dx = \frac{\tan ^{2} x}{2}$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$:
$I = x \cdot \frac{\tan ^{2} x}{2} - \int \frac{\tan ^{2} x}{2} \, dx$
$I = \frac{x \tan ^{2} x}{2} - \frac{1}{2} \int (\sec ^{2} x - 1) \, dx$
$I = \frac{x(\sec ^{2} x - 1)}{2} - \frac{1}{2} (\tan x - x) + c$
$I = \frac{x \sec ^{2} x}{2} - \frac{x}{2} - \frac{\tan x}{2} + \frac{x}{2} + c$
$I = \frac{1}{2} [x \sec ^{2} x - \tan x] + c$.

(C) Let $I = \int \frac{x^{e-1}+e^{x-1}}{x^{e}+e^{x}} d x$.
Substitute $t = x^{e}+e^{x}$.
Differentiating both sides with respect to $x$,we get $dt = (e x^{e-1} + e^{x-1} \cdot \ln(e)) dx$. Since $\ln(e) = 1$,this simplifies to $dt = e(x^{e-1} + e^{x-1}) dx$.
Therefore,$(x^{e-1} + e^{x-1}) dx = \frac{1}{e} dt$.
Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{1}{e} dt = \frac{1}{e} \int \frac{1}{t} dt$.
Integrating,we get $I = \frac{1}{e} \log|t| + c$.
Substituting back $t = x^{e}+e^{x}$,we get $I = \frac{1}{e} \log(x^{e}+e^{x}) + c$.

(A) The constraints are $5x_{1} + 10x_{2} \geq 0$,$x_{1} + x_{2} \leq 1$,$x_{2} \leq 4$,and $x_{1}, x_{2} \geq 0$.
Since $x_{1}, x_{2} \geq 0$,the constraint $5x_{1} + 10x_{2} \geq 0$ is always satisfied in the first quadrant.
The feasible region is defined by the intersection of $x_{1} + x_{2} \leq 1$ and $x_{1}, x_{2} \geq 0$.
This region is a triangle with vertices at $(0, 0)$,$(1, 0)$,and $(0, 1)$.
Since the feasible region is a closed and bounded polygon,the $LPP$ has a bounded solution.

(B) The merchant earns a profit of $Rs \ 40$ per quintal on rice and $Rs \ 25$ per quintal on wheat.
Given that he stores $x$ quintals of rice and $y$ quintals of wheat.
The total profit $Z$ is given by the sum of the profit from rice and wheat.
Therefore,the objective function is $Z = 40x + 25y$.

(D) Given the matrix equation: $\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]$
Multiplying the matrices,we get the system of linear equations:
$x + z = 1$ $(i)$
$-x + y = 1$ $(ii)$
$-y + z = 2$ $(iii)$
Adding $(ii)$ and $(iii)$,we get:
$(-x + y) + (-y + z) = 1 + 2$
$-x + z = 3$ $(iv)$
Now,adding $(i)$ and $(iv)$:
$(x + z) + (-x + z) = 1 + 3$
$2z = 4 \Rightarrow z = 2$
Substituting $z = 2$ into $(i)$:
$x + 2 = 1 \Rightarrow x = -1$
Substituting $x = -1$ into $(ii)$:
$-(-1) + y = 1 \Rightarrow 1 + y = 1 \Rightarrow y = 0$
Thus,the solution is $(x, y, z) = (-1, 0, 2)$.

(A) Let $A = \left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]$.
First,we calculate the determinant $|A|$:
$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.
Since $|A| \neq 0$,the inverse exists.
Next,we find the matrix of cofactors $C_{ij}$:
$C_{11} = +(2-3) = -1, C_{12} = -(1-9) = 8, C_{13} = +(1-6) = -5$.
$C_{21} = -(1-2) = 1, C_{22} = +(0-6) = -6, C_{23} = -(0-3) = 3$.
$C_{31} = +(3-4) = -1, C_{32} = -(0-2) = 2, C_{33} = +(0-1) = -1$.
Thus,$\operatorname{adj}(A) = \left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]^T = \left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]$.
Finally,$A^{-1} = \frac{1}{|A|} \operatorname{adj}(A) = \frac{1}{-2} \left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right] = \left[\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right]$.

(C) Let the sample space for two children be $S = \{BB, BG, GB, GG\}$,where $B$ denotes a boy and $G$ denotes a girl. Each outcome is equally likely with probability $1/4$.
Let $A$ be the event that at least one child is a boy. Then $A = \{BB, BG, GB\}$,so $P(A) = 3/4$.
Let $B$ be the event that both children are boys. Then $B = \{BB\}$.
We want to find the conditional probability $P(B|A)$,which is the probability that both are boys given that at least one is a boy.
$P(B|A) = \frac{P(B \cap A)}{P(A)}$.
Since $B \subset A$,$B \cap A = B = \{BB\}$,so $P(B \cap A) = 1/4$.
Therefore,$P(B|A) = \frac{1/4}{3/4} = 1/3$.

(B) Given,$E = \{x \text{ is a prime number}\} = \{2, 3, 5, 7\}$.
$P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.
Given,$F = \{x < 4\} = \{1, 2, 3\}$.
$P(F) = P(1) + P(2) + P(3) = 0.15 + 0.23 + 0.12 = 0.50$.
The intersection $E \cap F = \{x \text{ is prime and } x < 4\} = \{2, 3\}$.
$P(E \cap F) = P(2) + P(3) = 0.23 + 0.12 = 0.35$.
Using the formula $P(E \cup F) = P(E) + P(F) - P(E \cap F)$:
$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

(C) The position vectors are $\vec{A} = 2\hat{i}+\hat{j}-\hat{k}$,$\vec{B} = 3\hat{i}-2\hat{j}+\hat{k}$,and $\vec{C} = \hat{i}+4\hat{j}-3\hat{k}$.
$\overrightarrow{AB} = \vec{B} - \vec{A} = (3-2)\hat{i} + (-2-1)\hat{j} + (1-(-1))\hat{k} = \hat{i} - 3\hat{j} + 2\hat{k}$.
$|\overrightarrow{AB}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1+9+4} = \sqrt{14}$.
$\overrightarrow{BC} = \vec{C} - \vec{B} = (1-3)\hat{i} + (4-(-2))\hat{j} + (-3-1)\hat{k} = -2\hat{i} + 6\hat{j} - 4\hat{k}$.
$|\overrightarrow{BC}| = \sqrt{(-2)^2 + 6^2 + (-4)^2} = \sqrt{4+36+16} = \sqrt{56} = 2\sqrt{14}$.
$\overrightarrow{AC} = \vec{C} - \vec{A} = (1-2)\hat{i} + (4-1)\hat{j} + (-3-(-1))\hat{k} = -\hat{i} + 3\hat{j} - 2\hat{k}$.
$|\overrightarrow{AC}| = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1+9+4} = \sqrt{14}$.
Since $\overrightarrow{AB} + \overrightarrow{AC} = (\hat{i} - 3\hat{j} + 2\hat{k}) + (-\hat{i} + 3\hat{j} - 2\hat{k}) = 0$,we have $\overrightarrow{AB} = -\overrightarrow{AC}$,which implies that the vectors are parallel and the points $A, B, C$ lie on the same line.
Thus,the points are collinear.

(C) Let the direction ratios of the required line be $a, b, c$. Since the line lies in both planes,it is perpendicular to the normals of both planes. Thus,$3a + 2b + c = 0$ and $a + b - 2c = 0$.
Using the cross product method,we have:
$\frac{a}{(2)(-2) - (1)(1)} = \frac{b}{(1)(1) - (3)(-2)} = \frac{c}{(3)(1) - (2)(1)}$
$\frac{a}{-4-1} = \frac{b}{1+6} = \frac{c}{3-2}$
$\frac{a}{-5} = \frac{b}{7} = \frac{c}{1}$.
To find a point on the line,we set $z = 0$ in the given plane equations:
$3x + 2y = 5$ and $x + y = 3$.
Solving these,we multiply the second by $2$: $2x + 2y = 6$.
Subtracting this from the first: $(3x - 2x) = 5 - 6 \Rightarrow x = -1$.
Substituting $x = -1$ into $x + y = 3$,we get $-1 + y = 3 \Rightarrow y = 4$.
So,the point is $(-1, 4, 0)$.
The symmetric equation is $\frac{x - (-1)}{-5} = \frac{y - 4}{7} = \frac{z - 0}{1}$,which is $\frac{x+1}{-5} = \frac{y-4}{7} = \frac{z-0}{1}$.

(C) The equation of a plane passing through the point $(-1, 3, -2)$ is $a(x+1) + b(y-3) + c(z+2) = 0$.
Since the plane contains the line with direction ratios $(-3, 2, 1)$,we have $-3a + 2b + c = 0$.
The plane also passes through the point $(0, 7, -7)$,so substituting these coordinates into the plane equation gives $a(0+1) + b(7-3) + c(-7+2) = 0$,which simplifies to $a + 4b - 5c = 0$.
Solving the system of equations $-3a + 2b + c = 0$ and $a + 4b - 5c = 0$ using cross-multiplication:
$\frac{a}{2(-5) - 1(4)} = \frac{b}{1(1) - (-3)(-5)} = \frac{c}{-3(4) - 2(1)}$
$\frac{a}{-10-4} = \frac{b}{1-15} = \frac{c}{-12-2}$
$\frac{a}{-14} = \frac{b}{-14} = \frac{c}{-14} \implies a=1, b=1, c=1$.
Substituting these values into the plane equation: $1(x+1) + 1(y-3) + 1(z+2) = 0 \implies x+y+z=0$.

(D) Let the two lines be $L_1$ and $L_2$. The line $L_1$ passes through $A = 6 \overrightarrow{a}-4 \overrightarrow{b}+4 \overrightarrow{c}$ and $B = -4 \overrightarrow{c}$. The direction vector of $L_1$ is $\overrightarrow{d_1} = B - A = -6 \overrightarrow{a}+4 \overrightarrow{b}-8 \overrightarrow{c}$. The equation of $L_1$ is $\overrightarrow{r} = 6 \overrightarrow{a}-4 \overrightarrow{b}+4 \overrightarrow{c} + m(-6 \overrightarrow{a}+4 \overrightarrow{b}-8 \overrightarrow{c})$ ... $(i)$.
The line $L_2$ passes through $C = -\overrightarrow{a}-2 \overrightarrow{b}-3 \overrightarrow{c}$ and $D = \overrightarrow{a}+2 \overrightarrow{b}-5 \overrightarrow{c}$. The direction vector of $L_2$ is $\overrightarrow{d_2} = D - C = 2 \overrightarrow{a}+4 \overrightarrow{b}-2 \overrightarrow{c}$. The equation of $L_2$ is $\overrightarrow{r} = -\overrightarrow{a}-2 \overrightarrow{b}-3 \overrightarrow{c} + n(2 \overrightarrow{a}+4 \overrightarrow{b}-2 \overrightarrow{c})$ ... $(ii)$.
Equating the two expressions for $\overrightarrow{r}$:
$6 \overrightarrow{a}-4 \overrightarrow{b}+4 \overrightarrow{c} + m(-6 \overrightarrow{a}+4 \overrightarrow{b}-8 \overrightarrow{c}) = -\overrightarrow{a}-2 \overrightarrow{b}-3 \overrightarrow{c} + n(2 \overrightarrow{a}+4 \overrightarrow{b}-2 \overrightarrow{c})$
Comparing coefficients of $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$:
$6 - 6m = -1 + 2n \implies 6m + 2n = 7$ ... $(iii)$
$-4 + 4m = -2 + 4n \implies 4m - 4n = 2 \implies 2m - 2n = 1$ ... $(iv)$
$4 - 8m = -3 - 2n \implies 8m - 2n = 7$ ... $(v)$
Adding $(iii)$ and $(iv)$: $8m = 8 \implies m = 1$. Substituting $m=1$ in $(iv)$: $2(1) - 2n = 1 \implies 2n = 1 \implies n = 1/2$. Checking in $(v)$: $8(1) - 2(1/2) = 8 - 1 = 7$. The values satisfy the system.
Substituting $m=1$ in $(i)$: $\overrightarrow{r} = 6 \overrightarrow{a}-4 \overrightarrow{b}+4 \overrightarrow{c} - 6 \overrightarrow{a}+4 \overrightarrow{b}-8 \overrightarrow{c} = -4 \overrightarrow{c}$.

(C) Let the vertices of the triangle be $O(0), A(\overrightarrow{a}+\overrightarrow{b}),$ and $B(\overrightarrow{a}-\overrightarrow{b})$.
The area of the triangle is given by $\frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}|$.
Substituting the vectors,we get $\text{Area} = \frac{1}{2} |(\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b})|$.
Expanding the cross product: $(\overrightarrow{a}+\overrightarrow{b}) \times (\overrightarrow{a}-\overrightarrow{b}) = \overrightarrow{a} \times \overrightarrow{a} - \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{a} - \overrightarrow{b} \times \overrightarrow{b}$.
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$ and $\overrightarrow{b} \times \overrightarrow{b} = 0$,and $\overrightarrow{b} \times \overrightarrow{a} = -(\overrightarrow{a} \times \overrightarrow{b})$,we have:
$= 0 - (\overrightarrow{a} \times \overrightarrow{b}) - (\overrightarrow{a} \times \overrightarrow{b}) - 0 = -2(\overrightarrow{a} \times \overrightarrow{b})$.
Thus,$\text{Area} = \frac{1}{2} |-2(\overrightarrow{a} \times \overrightarrow{b})| = |\overrightarrow{a} \times \overrightarrow{b}|$.
Since $\overrightarrow{a} \perp \overrightarrow{b}$,$|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin(90^{\circ}) = 2 \times 3 \times 1 = 6$.

(B) Let the diagonals of the parallelogram be $\overrightarrow{d_1} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\overrightarrow{d_2} = -\hat{i} + 2\hat{j}$.
The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $\text{Area} = \frac{1}{2} |\overrightarrow{d_1} \times \overrightarrow{d_2}|$.
First,calculate the cross product $\overrightarrow{d_1} \times \overrightarrow{d_2}$:
$\overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix}$
$= \hat{i}((-3)(0) - (2)(2)) - \hat{j}((1)(0) - (2)(-1)) + \hat{k}((1)(2) - (-3)(-1))$
$= \hat{i}(0 - 4) - \hat{j}(0 + 2) + \hat{k}(2 - 3)$
$= -4\hat{i} - 2\hat{j} - \hat{k}$.
Now,calculate the magnitude of the cross product:
$|\overrightarrow{d_1} \times \overrightarrow{d_2}| = \sqrt{(-4)^2 + (-2)^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$.
Therefore,the area of the parallelogram is $\frac{1}{2} \times \sqrt{21} = \frac{\sqrt{21}}{2}$.

(D) Using the scalar triple product property $(\overrightarrow{A} \times \overrightarrow{B}) \cdot \overrightarrow{C} = \overrightarrow{A} \cdot (\overrightarrow{B} \times \overrightarrow{C})$:
$(\overrightarrow{a} \times \hat{j}) \cdot (2 \hat{j} - 3 \hat{k}) = \overrightarrow{a} \cdot \{\hat{j} \times (2 \hat{j} - 3 \hat{k})\}$
$= \overrightarrow{a} \cdot \{2(\hat{j} \times \hat{j}) - 3(\hat{j} \times \hat{k})\}$
Since $\hat{j} \times \hat{j} = 0$ and $\hat{j} \times \hat{k} = \hat{i}$:
$= \overrightarrow{a} \cdot \{2(0) - 3(\hat{i})\}$
$= \overrightarrow{a} \cdot (-3 \hat{i})$
$= -3(\overrightarrow{a} \cdot \hat{i})$
Given $\overrightarrow{a} \cdot \hat{i} = 4$,we substitute this value:
$= -3(4) = -12$

(D) We use the vector triple product formula: $\overrightarrow{u} \times (\overrightarrow{v} \times \overrightarrow{w}) = (\overrightarrow{u} \cdot \overrightarrow{w})\overrightarrow{v} - (\overrightarrow{u} \cdot \overrightarrow{v})\overrightarrow{w}$.
First,evaluate the inner part: $\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b}) = (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{a} - (\overrightarrow{a} \cdot \overrightarrow{a})\overrightarrow{b}$.
Now,substitute this back into the original expression:
$\overrightarrow{a} \times [\overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{b})] = \overrightarrow{a} \times [(\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{a} - (\overrightarrow{a} \cdot \overrightarrow{a})\overrightarrow{b}]$
$= (\overrightarrow{a} \cdot \overrightarrow{b})(\overrightarrow{a} \times \overrightarrow{a}) - (\overrightarrow{a} \cdot \overrightarrow{a})(\overrightarrow{a} \times \overrightarrow{b})$
Since $\overrightarrow{a} \times \overrightarrow{a} = 0$,the first term becomes zero:
$= 0 - (\overrightarrow{a} \cdot \overrightarrow{a})(\overrightarrow{a} \times \overrightarrow{b})$
$= (\overrightarrow{a} \cdot \overrightarrow{a})(\overrightarrow{b} \times \overrightarrow{a})$.