MHT CET 2016 Mathematics Question Paper with Answer and Solution

(D) Let $S_1 \equiv p$ and $S_2 \equiv q$. Then $S_1' \equiv \sim p$ and $S_2' \equiv \sim q$.
The circuit consists of two parallel branches.
The first branch has switches $S_1$ and $S_2'$ in series,which corresponds to $(p \wedge \sim q)$.
The second branch has switches $S_1'$ and $S_2$ in series,which corresponds to $(\sim p \wedge q)$.
Since the branches are in parallel,the total expression is $(p \wedge \sim q) \vee (\sim p \wedge q)$.
This expression is equivalent to the exclusive $OR$ operation,which is $\sim(p \leftrightarrow q)$.

(C) For any real number $x$,the square $x^2$ is always greater than or equal to $0$ $(x^2 \geq 0)$.
Option $A$ is false because for $x=0$,$x^2=0$,which is not positive.
Option $B$ is false because the square of a real number cannot be negative.
Option $C$ is true because there exists a real number $x=0$ such that $x^2=0$,which is not positive.
Option $D$ is false because there exist irrational numbers like $\sqrt{2}$.

(D) Step $1$: Determine the truth value of statement $p$. Every square is a rectangle is a true statement,so $p = T$.
Step $2$: Determine the truth value of statement $q$. Every rhombus is a kite is a true statement,so $q = T$.
Step $3$: Evaluate $p \rightarrow q$. Since $T \rightarrow T$ is $T$,the truth value is $T$.
Step $4$: Evaluate $p \leftrightarrow q$. Since $T \leftrightarrow T$ is $T$,the truth value is $T$.
Therefore,the truth values are $T, T$.

(A) The first quadrant is the region between the positive $x$-axis $(0^\circ)$ and the positive $y$-axis $(90^\circ)$.
Lines that trisect the first quadrant make angles of $30^\circ$ and $60^\circ$ with the positive $x$-axis.
The equations of these lines are $y = \tan(30^\circ)x$ and $y = \tan(60^\circ)x$.
$y = \frac{1}{\sqrt{3}}x \implies x - \sqrt{3}y = 0$
$y = \sqrt{3}x \implies \sqrt{3}x - y = 0$
The joint equation is $(x - \sqrt{3}y)(\sqrt{3}x - y) = 0$.
Expanding this: $\sqrt{3}x^2 - xy - 3xy + \sqrt{3}y^2 = 0$.
$\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$.

(B) The lines $x=5$ (a vertical line) and $y=3$ (a horizontal line) intersect at the point $(5, 3)$.
Since these lines are perpendicular,the angle bisectors are lines passing through $(5, 3)$ that make angles of $45^{\circ}$ and $135^{\circ}$ with the $x$-axis.
The slopes of these bisectors are $m = \tan 45^{\circ} = 1$ and $m = \tan 135^{\circ} = -1$.
The equations of the bisectors are:
$y - 3 = 1(x - 5) \Rightarrow y - x + 2 = 0$
$y - 3 = -1(x - 5) \Rightarrow y + x - 8 = 0$
The joint equation is $(y - x + 2)(y + x - 8) = 0$.
Expanding this: $y^2 + xy - 8y - xy - x^2 + 8x + 2y + 2x - 16 = 0$.
Simplifying,we get $-x^2 + y^2 + 10x - 6y - 16 = 0$,or $x^2 - y^2 - 10x + 6y + 16 = 0$.

(C) general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
For option $A$: $x^2 - x = 0 \implies x(x-1) = 0$,which represents lines $x=0$ and $x=1$.
For option $B$: $xy - x = 0 \implies x(y-1) = 0$,which represents lines $x=0$ and $y=1$.
For option $C$: $y^2 - x + 1 = 0$. This is a parabola,not a pair of lines,as it cannot be factored into two linear factors.
For option $D$: $xy + x + y + 1 = 0 \implies x(y+1) + 1(y+1) = 0 \implies (x+1)(y+1) = 0$,which represents lines $x=-1$ and $y=-1$.
Thus,the equation that does not represent a pair of lines is $y^2 - x + 1 = 0$.

(B) Let $X = (a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$
$= (a^2 + b^2 - 2ab) \cos^2 \frac{C}{2} + (a^2 + b^2 + 2ab) \sin^2 \frac{C}{2}$
$= (a^2 + b^2) (\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}) - 2ab (\cos^2 \frac{C}{2} - \sin^2 \frac{C}{2})$
$= (a^2 + b^2)(1) - 2ab \cos C$
$= a^2 + b^2 - 2ab \cos C$
$= c^2$ (by the Law of Cosines).

(C) Given the equation $\tan^2 x = 1$.
Taking the square root on both sides,we get $\tan x = \pm 1$.
We know that $\tan x = \tan \alpha$ implies $x = n \pi + \alpha$.
For $\tan x = 1$,$x = n \pi + \frac{\pi}{4}$.
For $\tan x = -1$,$x = n \pi - \frac{\pi}{4}$.
Combining these two results,the general solution is $x = n \pi \pm \frac{\pi}{4}$,where $n \in \mathbb{Z}$.

(C) Given the equation $\sin 2x + \cos 2x = 0$.
Dividing by $\cos 2x$ (assuming $\cos 2x \neq 0$),we get $\tan 2x = -1$.
Since $\tan \theta = -1$ implies $\theta = n\pi - \frac{\pi}{4}$,we have $2x = n\pi - \frac{\pi}{4}$.
Thus,$x = \frac{n\pi}{2} - \frac{\pi}{8} = \frac{(4n - 1)\pi}{8}$.
For $\pi < x < 2\pi$:
If $n = 3$,$x = \frac{(12 - 1)\pi}{8} = \frac{11\pi}{8}$.
If $n = 4$,$x = \frac{(16 - 1)\pi}{8} = \frac{15\pi}{8}$.
Both values lie in the interval $(\pi, 2\pi)$.

(A) Given the curve equation $6y = x^3 + 2$.
Differentiating both sides with respect to $t$,we get $6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}$.
This simplifies to $2 \frac{dy}{dt} = x^2 \frac{dx}{dt}$.
We are given that the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate,so $\frac{dy}{dt} = 8 \frac{dx}{dt}$.
Substituting this into the derivative equation: $2(8 \frac{dx}{dt}) = x^2 \frac{dx}{dt}$.
Assuming $\frac{dx}{dt} \neq 0$,we have $16 = x^2$,which gives $x = 4$ or $x = -4$.
If $x = 4$,then $6y = (4)^3 + 2 = 64 + 2 = 66$,so $y = 11$. The point is $(4, 11)$.
If $x = -4$,then $6y = (-4)^3 + 2 = -64 + 2 = -62$,so $y = -31/3$.
Comparing with the given options,the correct point is $(4, 11)$.

(A) Using the linear approximation formula: $f(x+h) \approx f(x) + h f'(x)$.
Here,let $x=1$ and $h=0.1$.
First,calculate $f(1) = (1)^3 + 5(1)^2 - 7(1) + 9 = 1 + 5 - 7 + 9 = 8$.
Next,find the derivative $f'(x) = 3x^2 + 10x - 7$.
Calculate $f'(1) = 3(1)^2 + 10(1) - 7 = 3 + 10 - 7 = 6$.
Now,substitute these values into the formula:
$f(1.1) \approx f(1) + 0.1 \times f'(1)$
$f(1.1) \approx 8 + 0.1 \times 6$
$f(1.1) \approx 8 + 0.6 = 8.6$.

(D) Given function is $f(x)=e^x(\sin x-\cos x)$.
First,we check the conditions of Rolle's theorem:
$1$. $f(x)$ is continuous on $\left[\frac{\pi}{4}, \frac{5 \pi}{4}\right]$ and differentiable on $\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$.
$2$. $f\left(\frac{\pi}{4}\right) = e^{\pi/4}(\sin(\pi/4) - \cos(\pi/4)) = e^{\pi/4}(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$.
$3$. $f\left(\frac{5\pi}{4}\right) = e^{5\pi/4}(\sin(5\pi/4) - \cos(5\pi/4)) = e^{5\pi/4}(-\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}})) = 0$.
Since $f\left(\frac{\pi}{4}\right) = f\left(\frac{5\pi}{4}\right) = 0$,Rolle's theorem is applicable.
Now,find $f'(x)$:
$f'(x) = e^x(\sin x - \cos x) + e^x(\cos x + \sin x) = 2e^x \sin x$.
Set $f'(c) = 0$ for $c \in \left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$:
$2e^c \sin c = 0$.
Since $e^c \neq 0$,we have $\sin c = 0$.
In the interval $\left(\frac{\pi}{4}, \frac{5 \pi}{4}\right)$,$\sin c = 0$ at $c = \pi$.

(B) To find the area bounded by the curve $y=2x-x^2$ and the $x$-axis,we first find the intersection points with the $x$-axis by setting $y=0$.
$2x-x^2=0 \Rightarrow x(2-x)=0$,which gives $x=0$ and $x=2$.
The area $A$ is given by the integral:
$A = \int_0^2 (2x-x^2) dx$
$A = [x^2 - \frac{x^3}{3}]_0^2$
$A = (2^2 - \frac{2^3}{3}) - (0^2 - \frac{0^3}{3})$
$A = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3} \text{ sq units}$.

(C) The feasible region is determined by the constraints $x_1+x_2 \leq 10$,$-2x_1+3x_2 \leq 15$,$x_1 \leq 6$,and $x_1, x_2 \geq 0$. The corner points of the feasible region are $O(0,0)$,$E(6,0)$,$F(6,4)$,$G(3,7)$,and $D(0,5)$.
We evaluate the objective function $z=x_1+x_2$ at these corner points:
$z(O) = 0+0 = 0$
$z(E) = 6+0 = 6$
$z(F) = 6+4 = 10$
$z(G) = 3+7 = 10$
$z(D) = 0+5 = 5$
The maximum value of $z$ is $10$,which occurs at both corner points $F(6,4)$ and $G(3,7)$.
Since the objective function is maximized at two distinct corner points,it will have the same maximum value at every point on the line segment joining these two points.

(C) Since $f(x)$ is continuous at $x = 0$,we must have $\lim_{x \to 0} f(x) = f(0)$.
Therefore,$\lim_{x \to 0} \frac{\text{log}(1 + 2x) \cdot \sin x^{\circ}}{x^2} = k$.
We know that $\sin x^{\circ} = \sin(\frac{\pi x}{180})$.
So,$\lim_{x \to 0} \frac{\text{log}(1 + 2x)}{x} \cdot \frac{\sin(\frac{\pi x}{180})}{x} = k$.
Multiplying and dividing by $2$ and $\frac{\pi}{180}$ respectively:
$\lim_{x \to 0} \left( 2 \cdot \frac{\text{log}(1 + 2x)}{2x} \right) \cdot \left( \frac{\pi}{180} \cdot \frac{\sin(\frac{\pi x}{180})}{\frac{\pi x}{180}} \right) = k$.
Using the standard limits $\lim_{u \to 0} \frac{\text{log}(1+u)}{u} = 1$ and $\lim_{v \to 0} \frac{\sin v}{v} = 1$:
$2 \cdot 1 \cdot \frac{\pi}{180} \cdot 1 = k$.
Thus,$k = \frac{\pi}{90}$.

(A) Given the function $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$.
Since $f(x)$ is continuous at $x = 0$,the condition $\lim_{x \to 0} f(x) = f(0)$ must hold.
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 $x$,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 $f(0) = k$,we have $k = 0$.

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{(\sec x)^{1/n}}{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}} dx \dots (i)$
Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$I = \int_0^{\frac{\pi}{2}} \frac{(\sec(\frac{\pi}{2}-x))^{1/n}}{(\sec(\frac{\pi}{2}-x))^{1/n} + (\operatorname{cosec}(\frac{\pi}{2}-x))^{1/n}} dx$
$I = \int_0^{\frac{\pi}{2}} \frac{(\operatorname{cosec} x)^{1/n}}{(\operatorname{cosec} x)^{1/n} + (\sec x)^{1/n}} dx \dots (ii)$
Adding $(i)$ and $(ii)$:
$2I = \int_0^{\frac{\pi}{2}} \frac{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}}{(\sec x)^{1/n} + (\operatorname{cosec} x)^{1/n}} dx$
$2I = \int_0^{\frac{\pi}{2}} 1 dx$
$2I = [x]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$
$I = \frac{\pi}{4}$

(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$.
Consider the function $f(x) = \log \left(\frac{2-\sin x}{2+\sin x}\right)$.
Now,check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \log \left(\frac{2-\sin(-x)}{2+\sin(-x)}\right) = \log \left(\frac{2+\sin x}{2-\sin x}\right)$.
Since $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we have:
$f(-x) = -\log \left(\frac{2-\sin x}{2+\sin x}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) d x = 0$.
Therefore,$I = 0$.

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

(B) Since the circle touches the $y$-axis at the origin,its center must lie on the $x$-axis. Let the center be $(a, 0)$ and the radius be $a$.
The equation of the circle is $(x-a)^2 + (y-0)^2 = a^2$.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0$ ... $(i)$.
Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $2a = 2x + 2y \frac{dy}{dx}$ ... $(ii)$.
Substituting the value of $2a$ from $(ii)$ into $(i)$,we get $x^2 + y^2 - x(2x + 2y \frac{dy}{dx}) = 0$.
This simplifies to $x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.
Thus,the differential equation is $y^2 - x^2 - 2xy \frac{dy}{dx} = 0$,or $x^2 - y^2 + 2xy \frac{dy}{dx} = 0$.

(D) The integrating factor $(I.F.)$ of the linear differential equation $\frac{dy}{dx} + Py = Q$ is given by $e^{\int P dx}$.
Given that the integrating factor is $\sin x$,we have:
$e^{\int P dx} = \sin x$
Taking the natural logarithm on both sides:
$\int P dx = \ln(\sin x)$
Differentiating both sides with respect to $x$:
$P = \frac{d}{dx}[\ln(\sin x)]$
Using the chain rule:
$P = \frac{1}{\sin x} \cdot \frac{d}{dx}(\sin x)$
$P = \frac{1}{\sin x} \cdot \cos x$
$P = \cot x$
Therefore,the correct option is $D$.

(A) Given the differential equation: $y(1+\log x) \frac{dx}{dy} - x \log x = 0$
Rearranging the terms to separate the variables:
$y(1+\log x) \frac{dx}{dy} = x \log x$
$\frac{1+\log x}{x \log x} dx = \frac{dy}{y}$
Integrating both sides:
$\int \frac{1+\log x}{x \log x} dx = \int \frac{dy}{y}$
Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1+u}{u} du = \int (\frac{1}{u} + 1) du = \log u + u$.
Substituting back $u = \log x$:
$\log(\log x) + \log x = \log y + C$
Alternatively,using substitution $v = x \log x$,$dv = (1 + \log x) dx$:
$\int \frac{dv}{v} = \int \frac{dy}{y}$
$\log(x \log x) = \log y + \log C$
$\log(x \log x) = \log(Cy)$
$x \log x = Cy$
Given $x = e$ and $y = e^2$:
$e \log e = C(e^2)$
$e(1) = Ce^2 \Rightarrow C = \frac{1}{e}$
Substituting $C$ back into the equation:
$x \log x = \frac{y}{e}$
$y = ex \log x$

(C) Given,$y = e^{m \sin^{-1} x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{d}{dx}(m \sin^{-1} x) = y \cdot \frac{m}{\sqrt{1 - x^2}}$.
Squaring both sides,we get:
$(\frac{dy}{dx})^2 = \frac{m^2 y^2}{1 - x^2}$.
Multiplying both sides by $(1 - x^2)$,we obtain:
$(1 - x^2) (\frac{dy}{dx})^2 = m^2 y^2$.
Comparing this with the given equation $(1 - x^2) (\frac{dy}{dx})^2 = A y^2$,we find that $A = m^2$.

(A) Given,$\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$.
Applying the definition of logarithm,we get $\frac{x^2-y^2}{x^2+y^2} = 10^2 = 100$.
Cross-multiplying,we have $x^2 - y^2 = 100(x^2 + y^2)$.
$x^2 - y^2 = 100x^2 + 100y^2$.
Rearranging the terms,we get $x^2 - 100x^2 = 100y^2 + y^2$,which simplifies to $-99x^2 = 101y^2$.
Now,differentiate both sides with respect to $x$:
$\frac{d}{dx}(-99x^2) = \frac{d}{dx}(101y^2)$.
$-99(2x) = 101(2y) \frac{dy}{dx}$.
$-198x = 202y \frac{dy}{dx}$.
Therefore,$\frac{dy}{dx} = -\frac{198x}{202y} = -\frac{99x}{101y}$.

(B) Let $y_1 = \log (\sec \theta + \tan \theta)$.
Then,$\frac{dy_1}{d\theta} = \frac{1}{\sec \theta + \tan \theta} \cdot (\sec \theta \tan \theta + \sec^2 \theta)$.
$\frac{dy_1}{d\theta} = \frac{\sec \theta (\tan \theta + \sec \theta)}{\sec \theta + \tan \theta} = \sec \theta$.
Let $y_2 = \sec \theta$.
Then,$\frac{dy_2}{d\theta} = \sec \theta \tan \theta$.
We need to find $\frac{dy_1}{dy_2} = \frac{dy_1/d\theta}{dy_2/d\theta} = \frac{\sec \theta}{\sec \theta \tan \theta} = \frac{1}{\tan \theta} = \cot \theta$.
At $\theta = \frac{\pi}{4}$,$\frac{dy_1}{dy_2} = \cot \frac{\pi}{4} = 1$.

(D) Let $u = \tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ and $v = \sin ^{-1}\left(3 x-4 x^3\right)$.
Substitute $x = \sin \theta$,where $\theta = \sin ^{-1} x$.
Then $u = \tan ^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\right) = \tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \tan ^{-1}(\tan \theta) = \theta$.
And $v = \sin ^{-1}(3 \sin \theta - 4 \sin ^3 \theta) = \sin ^{-1}(\sin 3 \theta) = 3 \theta$.
We need to find $\frac{du}{dv}$.
Since $u = \theta$ and $v = 3 \theta$,we have $\frac{du}{d\theta} = 1$ and $\frac{dv}{d\theta} = 3$.
Therefore,$\frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{1}{3}$.

(A) Given the integral equation: $\int \frac{f(x)}{\log (\sin x)} d x = \log [\log \sin x] + c$.
To find $f(x)$,we differentiate both sides with respect to $x$:
$\frac{d}{d x} \left( \int \frac{f(x)}{\log (\sin x)} d x \right) = \frac{d}{d x} (\log [\log \sin x] + c)$.
Using the Fundamental Theorem of Calculus on the left side,we get:
$\frac{f(x)}{\log (\sin x)} = \frac{d}{d x} (\log [\log \sin x])$.
Applying the chain rule on the right side:
$\frac{d}{d x} (\log [\log \sin x]) = \frac{1}{\log \sin x} \cdot \frac{d}{d x} (\log \sin x)$.
Since $\frac{d}{d x} (\log \sin x) = \frac{1}{\sin x} \cdot \cos x = \cot x$,we have:
$\frac{f(x)}{\log \sin x} = \frac{1}{\log \sin x} \cdot \cot x$.
Comparing both sides,we obtain $f(x) = \cot x$.

(C) Let $I = \int \frac{(x^2+2) a^{(x+\tan^{-1} x)}}{x^2+1} dx$.
We can rewrite the integrand as:
$I = \int \left( \frac{x^2+1+1}{x^2+1} \right) a^{(x+\tan^{-1} x)} dx = \int \left( 1 + \frac{1}{x^2+1} \right) a^{(x+\tan^{-1} x)} dx$.
Let $u = x + \tan^{-1} x$.
Then,$du = (1 + \frac{1}{1+x^2}) dx$.
Substituting these into the integral:
$I = \int a^u du$.
Using the standard integral formula $\int a^u du = \frac{a^u}{\ln a} + C$,we get:
$I = \frac{a^u}{\ln a} + C$.
Substituting back $u = x + \tan^{-1} x$:
$I = \frac{a^{x+\tan^{-1} x}}{\ln a} + C$.

(B) Let $I = \int \left( \frac{4 e^x - 25}{2 e^x - 5} \right) dx$.
We can rewrite the numerator as:
$4 e^x - 25 = 2(2 e^x - 5) - 15$.
So,$I = \int \left( \frac{2(2 e^x - 5) - 15}{2 e^x - 5} \right) dx$.
$I = \int \left( 2 - \frac{15}{2 e^x - 5} \right) dx$.
Wait,let us re-evaluate the expression:
$I = \int \left( \frac{2(2 e^x - 5) - 15}{2 e^x - 5} \right) dx = \int 2 dx - 15 \int \frac{1}{2 e^x - 5} dx$.
Alternatively,using the form $\frac{4 e^x - 25}{2 e^x - 5} = \frac{2(2 e^x - 5) - 15}{2 e^x - 5} = 2 - \frac{15}{2 e^x - 5}$.
Let us check the provided solution steps:
$I = \int \left( \frac{10 e^x - 25 - 6 e^x}{2 e^x - 5} \right) dx = \int \left( \frac{5(2 e^x - 5) - 6 e^x}{2 e^x - 5} \right) dx = \int 5 dx - 6 \int \frac{e^x}{2 e^x - 5} dx$.
Let $u = 2 e^x - 5$,then $du = 2 e^x dx$,so $e^x dx = \frac{du}{2}$.
$I = 5x - 6 \int \frac{1}{u} \cdot \frac{du}{2} = 5x - 3 \int \frac{1}{u} du = 5x - 3 \log |2 e^x - 5| + C$.
Comparing with $Ax + B \log |2 e^x - 5| + C$,we get $A = 5$ and $B = -3$.

(D) Let $I = \int \frac{dx}{\sqrt{8+2x-x^2}}$.
First,complete the square for the quadratic expression inside the square root:
$8+2x-x^2 = -(x^2-2x-8) = -(x^2-2x+1-9) = -( (x-1)^2 - 9 ) = 9 - (x-1)^2$.
Now,substitute this back into the integral:
$I = \int \frac{dx}{\sqrt{9-(x-1)^2}} = \int \frac{dx}{\sqrt{3^2-(x-1)^2}}$.
Using the standard integral formula $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\left(\frac{x}{a}\right)+c$,we get:
$I = \sin^{-1}\left(\frac{x-1}{3}\right)+c$.

(B) Given the equation: $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$.
Using the formula $2 \tan^{-1} \theta = \tan^{-1} \left( \frac{2 \theta}{1 - \theta^2} \right)$,we get:
$\tan^{-1} \left( \frac{2 \cos x}{1 - \cos^2 x} \right) = \tan^{-1} (2 \csc x)$.
Since $1 - \cos^2 x = \sin^2 x$,the equation becomes:
$\frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x}$.
Assuming $\sin x \neq 0$,we can simplify:
$\frac{\cos x}{\sin x} = 1 \Rightarrow \cot x = 1$.
Thus,$x = \frac{\pi}{4}$.
Now,calculate $\sin x + \cos x$:
$\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) $1$. The shaded region is bounded by the lines $x=0$ ($Y$-axis),$y=0$ ($X$-axis),$x=5$,$y=3$,and the line passing through $(0,0)$ and $(3,3)$.
$2$. The line passing through $(0,0)$ and $(3,3)$ has the equation $y=x$,which can be written as $x-y=0$. Since the shaded region lies below this line,the constraint is $x-y \geq 0$.
$3$. The vertical line $x=5$ bounds the region on the right,so $x \leq 5$.
$4$. The horizontal line $y=3$ bounds the region on the top,so $y \leq 3$.
$5$. Since the region is in the first quadrant,$x \geq 0$ and $y \geq 0$.
$6$. Combining these,the constraints are $x, y \geq 0, x-y \geq 0, x \leq 5, y \leq 3$.

(B) The sum of the product of elements of any row (or column) with their corresponding cofactors is equal to the determinant of the matrix,but the sum of the product of elements of any row (or column) with the cofactors of any other row (or column) is always $0$.
Here,we are calculating $a_{11} A_{21} + a_{12} A_{22} + a_{13} A_{23}$,which is the sum of the products of the elements of the first row with the cofactors of the second row.
By the property of determinants,this sum must be $0$.
Verification:
$a_{11} = 1, a_{12} = 1, a_{13} = 0$
$A_{21} = (-1)^{2+1} \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = -1(1-0) = -1$
$A_{22} = (-1)^{2+2} \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1(1-0) = 1$
$A_{23} = (-1)^{2+3} \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = -1(2-1) = -1$
$a_{11} A_{21} + a_{12} A_{22} + a_{13} A_{23} = 1(-1) + 1(1) + 0(-1) = -1 + 1 + 0 = 0$.

(A) We use the property of matrix inversion: $(XY)^{-1} = Y^{-1} X^{-1}$.
Applying this to the given expression:
$(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1}$
Since $(M^{-1})^{-1} = M$,we have:
$(B^{-1} A^{-1})^{-1} = A B$
Now,calculate the product $AB$:
$AB = \begin{bmatrix} 2 & 3 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
$AB = \begin{bmatrix} (2)(0) + (3)(1) & (2)(-1) + (3)(0) \\ (-3)(0) + (2)(1) & (-3)(-1) + (2)(0) \end{bmatrix}$
$AB = \begin{bmatrix} 0 + 3 & -2 + 0 \\ 0 + 2 & 3 + 0 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 2 & 3 \end{bmatrix}$

(C) Given $AX = I$,then $X = A^{-1}$.
First,calculate the determinant of $A$:
$|A| = (1 \times 3) - (2 \times 4) = 3 - 8 = -5$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj } A = \begin{bmatrix} 3 & -2 \\ -4 & 1 \end{bmatrix}$.
Now,calculate $X = A^{-1} = \frac{1}{|A|} \text{adj } A$:
$X = \frac{1}{-5} \begin{bmatrix} 3 & -2 \\ -4 & 1 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} -3 & 2 \\ 4 & -1 \end{bmatrix}$.

(A) This is a binomial distribution problem where $n = 10$ and the probability of success $p = \frac{1}{2}$ (since it is a True/False test). The probability of failure is $q = 1 - p = \frac{1}{2}$.
We need to find $P(X \geq 7) = P(X=7) + P(X=8) + P(X=9) + P(X=10)$.
The formula for binomial probability is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.
$P(X=7) = {}^{10}C_{7} (\frac{1}{2})^{7} (\frac{1}{2})^{3} = \frac{120}{1024}$.
$P(X=8) = {}^{10}C_{8} (\frac{1}{2})^{8} (\frac{1}{2})^{2} = \frac{45}{1024}$.
$P(X=9) = {}^{10}C_{9} (\frac{1}{2})^{9} (\frac{1}{2})^{1} = \frac{10}{1024}$.
$P(X=10) = {}^{10}C_{10} (\frac{1}{2})^{10} (\frac{1}{2})^{0} = \frac{1}{1024}$.
Summing these probabilities: $P(X \geq 7) = \frac{120 + 45 + 10 + 1}{1024} = \frac{176}{1024} = \frac{11}{64}$.

(B) Given that $X$ follows a binomial distribution $B(n, p)$ with $n=5$ and $p=\frac{1}{3}$.
Since $q = 1 - p$,we have $q = 1 - \frac{1}{3} = \frac{2}{3}$.
The probability mass function is given by $P(X=k) = { }^n C_k p^k q^{n-k}$.
We need to find $P(2 < X < 4)$,which is equivalent to $P(X=3)$.
Substituting the values into the formula:
$P(X=3) = { }^5 C_3 \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^{5-3}$
$P(X=3) = \frac{5 \times 4}{2 \times 1} \times \left(\frac{1}{27}\right) \times \left(\frac{2}{3}\right)^2$
$P(X=3) = 10 \times \frac{1}{27} \times \frac{4}{9}$
$P(X=3) = \frac{40}{243}$

(C) The cumulative distribution function $F(x)$ is defined as $F(x) = P(X \leq x)$.
For $x = 0$,we have $F(0) = P(X \leq 0)$.
From the given probability distribution table:
$P(X \leq 0) = P(X = -2) + P(X = -1) + P(X = 0)$
$P(X \leq 0) = 0.2 + 0.5 + 0.15 = 0.85$.
Alternatively,we know that the sum of all probabilities is $1$,so $P(X \leq 0) + P(X > 0) = 1$.
Therefore,$F(0) = P(X \leq 0) = 1 - P(X > 0)$.
Thus,the correct option is $C$.

(B) The probability density function is given by $f(x) = \frac{1}{5}$ for $0 \leq x \leq 5$ and $0$ otherwise.
We need to find the probability that the waiting time $X$ is not more than $4$ minutes,which is $P(X \leq 4)$.
This is calculated by integrating the probability density function from the lower bound to $4$:
$P(X \leq 4) = \int_{0}^{4} f(x) \, dx$
$P(X \leq 4) = \int_{0}^{4} \frac{1}{5} \, dx$
$P(X \leq 4) = \left[ \frac{x}{5} \right]_{0}^{4}$
$P(X \leq 4) = \frac{4}{5} - \frac{0}{5} = \frac{4}{5} = 0.8$
Therefore,the probability is $0.8$.

(B) Given that $X$ follows a binomial distribution $B(n, p)$.
For a binomial distribution,the mean $E(X) = np = 5$ and the variance $\operatorname{Var}(X) = npq = 2.5$.
Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{2.5}{5} = 0.5 = \frac{1}{2}$.
Since $p + q = 1$,we have $p = 1 - 0.5 = 0.5 = \frac{1}{2}$.
Substituting $p = \frac{1}{2}$ into $np = 5$,we get $n \times \frac{1}{2} = 5$,which implies $n = 10$.
We need to find $P(X < 1)$,which is equivalent to $P(X = 0)$.
The probability mass function is given by $P(X = k) = {}^{n}C_{k} p^{k} q^{n-k}$.
For $k = 0$,$P(X = 0) = {}^{10}C_{0} \left(\frac{1}{2}\right)^{0} \left(\frac{1}{2}\right)^{10-0} = 1 \times 1 \times \left(\frac{1}{2}\right)^{10} = \left(\frac{1}{2}\right)^{10}$.

(B) Let the direction ratios of the required line be $a, b, c$.
Since the line is perpendicular to the lines with direction ratios $(-1, 2, 2)$ and $(0, 2, 1)$,we have the following equations:
$-a + 2b + 2c = 0$ $(i)$
$0a + 2b + c = 0$ (ii)
From (ii),we get $c = -2b$.
Substituting $c = -2b$ into $(i)$:
$-a + 2b + 2(-2b) = 0$
$-a + 2b - 4b = 0$
$-a - 2b = 0 \implies a = -2b$.
Now,we have $a = -2b$ and $c = -2b$.
If we let $b = 1$,then $a = -2$ and $c = -2$.
The direction ratios are $(-2, 1, -2)$,which is proportional to $(2, -1, 2)$.
Alternatively,using the cross product of the two direction vectors $\vec{n_1} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n_2} = 0\hat{i} + 2\hat{j} + 1\hat{k}$:
$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 0 & 2 & 1 \end{vmatrix} = \hat{i}(2 - 4) - \hat{j}(-1 - 0) + \hat{k}(-2 - 0) = -2\hat{i} + 1\hat{j} - 2\hat{k}$.
The direction ratios are proportional to $(-2, 1, -2)$,which is equivalent to $(2, -1, 2)$.

(A) The given equation of the line is $\frac{x+2}{2} = \frac{2y-5}{3}, z = -1$.
First,rewrite the line in standard symmetric form:
$\frac{x+2}{2} = \frac{y - 5/2}{3/2} = \frac{z+1}{0}$.
The direction ratios of the line are $(a, b, c) = (2, 3/2, 0)$.
To simplify,multiply by $2$: $(4, 3, 0)$.
The magnitude of the direction vector is $\sqrt{4^2 + 3^2 + 0^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
The direction cosines $(l, m, n)$ are given by $\left( \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \right)$.
Thus,the direction cosines are $\left( \frac{4}{5}, \frac{3}{5}, 0 \right)$.

(A) The coordinates of point $Q$ are $(a, b, c)$.
The foot of the perpendicular from $Q(a, b, c)$ to the $yz$-plane $(x=0)$ is $A(0, b, c)$.
The foot of the perpendicular from $Q(a, b, c)$ to the $zx$-plane $(y=0)$ is $B(a, 0, c)$.
The origin is $O(0, 0, 0)$.
The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant:
$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right|=0$
Substituting $O(0, 0, 0)$,$A(0, b, c)$,and $B(a, 0, c)$:
$\left|\begin{array}{ccc} x & y & z \\ 0 & b & c \\ a & 0 & c \end{array}\right|=0$
Expanding the determinant along the first row:
$x(bc - 0) - y(0 - ac) + z(0 - ab) = 0$
$bcx + acy - abz = 0$
Dividing the entire equation by $abc$ $(abc \neq 0)$:
$\frac{bcx}{abc} + \frac{acy}{abc} - \frac{abz}{abc} = 0$
$\frac{x}{a} + \frac{y}{b} - \frac{z}{c} = 0$

(A) The angle $\theta$ between a line with direction vector $\bar{b}$ and a plane with normal vector $\bar{n}$ is given by $\sin \theta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.
Given the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$,the direction vector is $\bar{b} = \hat{i}+\hat{j}+\hat{k}$.
Given the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$,the normal vector is $\bar{n} = 2\hat{i}-\hat{j}+\hat{k}$.
Calculate the magnitudes: $|\bar{b}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{4+1+1} = \sqrt{6}$.
Calculate the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.
Substitute into the formula: $\sin \theta = \frac{|2|}{\sqrt{3} \cdot \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$.
Therefore,$\theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)$.

(B) We know the principal values of the inverse trigonometric functions:
$\tan ^{-1}(\sqrt{3}) = \frac{\pi}{3}$
$\sec ^{-1}(-2) = \pi - \sec ^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\operatorname{cosec}^{-1}(-\sqrt{2}) = -\operatorname{cosec}^{-1}(\sqrt{2}) = -\frac{\pi}{4}$
$\cos ^{-1}\left(-\frac{1}{2}\right) = \pi - \cos ^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
Substituting these values into the expression:
$\frac{\frac{\pi}{3} - \frac{2\pi}{3}}{-\frac{\pi}{4} + \frac{2\pi}{3}} = \frac{-\frac{\pi}{3}}{\frac{-3\pi + 8\pi}{12}} = \frac{-\frac{\pi}{3}}{\frac{5\pi}{12}} = -\frac{\pi}{3} \times \frac{12}{5\pi} = -\frac{4}{5}$

(B) We know that in any triangle,the centroid $(G)$,orthocenter $(H)$,and circumcenter $(P)$ are collinear,forming the Euler line.
$G$ divides the line segment $PH$ internally in the ratio $1:2$.
Using the section formula for the position vector $\vec{g}$ of the centroid $G$:
$\vec{g} = \frac{1 \cdot \vec{h} + 2 \cdot \vec{p}}{1 + 2}$
$\vec{g} = \frac{\vec{h} + 2\vec{p}}{3}$
Multiplying both sides by $3$:
$3\vec{g} = \vec{h} + 2\vec{p}$
Rearranging the terms to the form $x\vec{p} + y\vec{h} + z\vec{g} = 0$:
$2\vec{p} + 1\vec{h} - 3\vec{g} = 0$
Comparing this with $x\vec{p} + y\vec{h} + z\vec{g} = 0$,we get $x = 2$,$y = 1$,and $z = -3$.
Thus,$(x, y, z) = (2, 1, -3)$.

(A) Let the point of intersection divide $AB$ in the ratio $\lambda : 1$ and $CD$ in the ratio $\mu : 1$.
The position vector of a point on $AB$ is $\vec{r} = \frac{\lambda(-4 \vec{c}) + 1(6 \vec{a}-4 \vec{b}+4 \vec{c})}{\lambda+1} = \frac{6 \vec{a}-4 \vec{b} + (4-4 \lambda) \vec{c}}{\lambda+1}$.
The position vector of a point on $CD$ is $\vec{r} = \frac{\mu(\vec{a}+2 \vec{b}-5 \vec{c}) + 1(-\vec{a}-2 \vec{b}-3 \vec{c})}{\mu+1} = \frac{(\mu-1) \vec{a} + (2 \mu-2) \vec{b} + (-5 \mu-3) \vec{c}}{\mu+1}$.
Equating the two expressions for $\vec{r}$:
$\frac{6 \vec{a}-4 \vec{b} + (4-4 \lambda) \vec{c}}{\lambda+1} = \frac{(\mu-1) \vec{a} + (2 \mu-2) \vec{b} + (-5 \mu-3) \vec{c}}{\mu+1}$.
Comparing coefficients of $\vec{a}$ and $\vec{b}$:
$\frac{6}{\lambda+1} = \frac{\mu-1}{\mu+1}$ and $\frac{-4}{\lambda+1} = \frac{2 \mu-2}{\mu+1}$.
Dividing the two equations: $\frac{6}{-4} = \frac{\mu-1}{2(\mu-1)} \Rightarrow -\frac{3}{2} = \frac{1}{2}$,which implies $\mu=1$.
Substituting $\mu=1$ into the first equation: $\frac{6}{\lambda+1} = 0$,which is impossible unless $\lambda \to \infty$.
However,checking the point $B$ (where $\lambda=0$): $\vec{r} = 6 \vec{a}-4 \vec{b}+4 \vec{c}$.
Checking the point $C$ (where $\mu=0$): $\vec{r} = -\vec{a}-2 \vec{b}-3 \vec{c}$.
Re-evaluating the intersection: The point $B$ corresponds to $\lambda=0$ on $AB$. If we set $\mu=1$ in the $CD$ line equation,we get $\vec{r} = \frac{0 \vec{a} + 0 \vec{b} - 8 \vec{c}}{2} = -4 \vec{c}$,which is exactly point $B$. Thus,the intersection is $B$.

(C) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.
Since $M$ is the midpoint of diagonal $AC$,its position vector is $\vec{m} = \frac{\vec{a} + \vec{c}}{2}$.
Since $N$ is the midpoint of diagonal $BD$,its position vector is $\vec{n} = \frac{\vec{b} + \vec{d}}{2}$.
Now,consider the expression $\overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{CB} + \overrightarrow{CD}$:
$= (\vec{b} - \vec{a}) + (\vec{d} - \vec{a}) + (\vec{b} - \vec{c}) + (\vec{d} - \vec{c})$
$= 2\vec{b} + 2\vec{d} - 2\vec{a} - 2\vec{c}$
$= 2[(\vec{b} + \vec{d}) - (\vec{a} + \vec{c})]$
$= 4 \left[ \frac{\vec{b} + \vec{d}}{2} - \frac{\vec{a} + \vec{c}}{2} \right]$
$= 4(\vec{n} - \vec{m})$
$= 4 \overrightarrow{MN}$

(C) Given $\vec{c} = m \vec{a} + n \vec{b}$.
Substituting the vectors,we get:
$3 \hat{i} + 0 \hat{j} - \hat{k} = m(\hat{i} + \hat{j} - 2 \hat{k}) + n(2 \hat{i} - \hat{j} + \hat{k})$
$3 \hat{i} + 0 \hat{j} - \hat{k} = (m + 2n) \hat{i} + (m - n) \hat{j} + (-2m + n) \hat{k}$
Comparing the coefficients of $\hat{i}, \hat{j}, \text{ and } \hat{k}$ on both sides:
$m + 2n = 3 \quad (i)$
$m - n = 0 \quad (ii)$
$-2m + n = -1 \quad (iii)$
From equation $(ii)$,we have $m = n$.
Substituting $m = n$ into equation $(i)$:
$m + 2m = 3 \Rightarrow 3m = 3 \Rightarrow m = 1$.
Since $m = n$,we have $n = 1$.
Thus,$m + n = 1 + 1 = 2$.

(A) The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is given by the determinant of the components of the vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$.
$\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & \lambda & 1 \\ 1 & -1 & 4 \end{vmatrix} = 10$
Expanding the determinant along the first row:
$1(4\lambda - (-1)) - 1(8 - 1) + 1(-2 - \lambda) = 10$
$1(4\lambda + 1) - 1(7) + 1(-2 - \lambda) = 10$
$4\lambda + 1 - 7 - 2 - \lambda = 10$
$3\lambda - 8 = 10$
$3\lambda = 18$
$\lambda = 6$