MHT CET 2017 Mathematics Question Paper with Answer and Solution

(A) To find the dual of a compound statement,we replace $\wedge$ with $\vee$,$\vee$ with $\wedge$,$t$ (tautology) with $c$ (contradiction),and $c$ (contradiction) with $t$ (tautology).
Given the statement: $\sim p \wedge (q \vee c)$.
Replacing $\wedge$ with $\vee$,$\vee$ with $\wedge$,and $c$ with $t$:
The dual is $\sim p \vee (q \wedge t)$.

(C) tautology is a statement pattern that is always true for all possible truth values of its components.
Let us evaluate each option:
$(A) \ p \vee (q \rightarrow p) \equiv p \vee (\sim q \vee p) \equiv (p \vee p) \vee \sim q \equiv p \vee \sim q$. This is not a tautology as it is false when $p$ is $F$ and $q$ is $T$.
$(B) \ \sim q \rightarrow \sim p \equiv q \vee \sim p$. This is not a tautology as it is false when $q$ is $F$ and $p$ is $T$.
$(D) \ p \wedge \sim p \equiv F$. This is a contradiction.
$(C) \ (q \rightarrow p) \vee (\sim p \leftrightarrow q)$. Let us construct the truth table:

$p, q$	$(q \rightarrow p) \vee (\sim p \leftrightarrow q)$
$T, T$	$T \vee (F \leftrightarrow T) = T \vee F = T$
$T, F$	$T \vee (F \leftrightarrow F) = T \vee T = T$
$F, T$	$F \vee (T \leftrightarrow T) = F \vee T = T$
$F, F$	$T \vee (T \leftrightarrow F) = T \vee F = T$

Since the result is always $T$,option $(C)$ is a tautology.

(B) We evaluate the given options to find the one equivalent to $(\sim p \wedge q)$.
Consider option $B$: $(p \vee q) \wedge \sim p$.
By the Distributive Law,this is equivalent to $(p \wedge \sim p) \vee (q \wedge \sim p)$.
Since $(p \wedge \sim p) = F$ (Complementary Law),we have $F \vee (q \wedge \sim p)$.
By the Identity Law,$F \vee (q \wedge \sim p) = (q \wedge \sim p)$.
By the Commutative Law,$(q \wedge \sim p) = (\sim p \wedge q)$.
Thus,the statement pattern $(\sim p \wedge q)$ is logically equivalent to $(p \vee q) \wedge \sim p$.

(A) The given equation is $px^2 - qy^2 = 0$.
Comparing this with the general homogeneous equation of the second degree $ax^2 + 2hxy + by^2 = 0$,we get $a = p$,$h = 0$,and $b = -q$.
The lines represented by the equation are real and distinct if $h^2 - ab > 0$.
Substituting the values,we get $0^2 - (p)(-q) > 0$.
This simplifies to $pq > 0$.

(C) The given equation is $K x^2 + 5 x y + y^2 = 0$.
Comparing this with the general equation $a x^2 + 2 h x y + b y^2 = 0$,we have $a = K$,$2h = 5$,and $b = 1$.
Let the slopes of the lines be $m_1$ and $m_2$.
Then,$m_1 + m_2 = -\frac{2h}{b} = -5$ and $m_1 m_2 = \frac{a}{b} = K$.
It is given that the difference between the slopes is $1$,i.e.,$|m_1 - m_2| = 1$.
Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4 m_1 m_2$,we substitute the values:
$1^2 = (-5)^2 - 4(K)$.
$1 = 25 - 4K$.
$4K = 24$.
$K = 6$.

(D) Let $O(0,0), A(1,2), B(3,4)$.
$1$. Median from $O$ to $AB$:
The midpoint $D$ of $AB$ is $(\frac{1+3}{2}, \frac{2+4}{2}) = (2,3)$.
The equation of the median $OD$ passing through $(0,0)$ and $(2,3)$ is $y = \frac{3}{2}x$,which simplifies to $3x - 2y = 0$.
$2$. Altitude from $O$ to $AB$:
The slope of $AB$ is $m_{AB} = \frac{4-2}{3-1} = \frac{2}{2} = 1$.
The slope of the altitude $OP$ is $m_{OP} = -\frac{1}{m_{AB}} = -1$.
The equation of the altitude $OP$ passing through $(0,0)$ with slope $-1$ is $y = -x$,which simplifies to $x + y = 0$.
$3$. Joint equation:
The joint equation of the altitude and median is $(3x - 2y)(x + y) = 0$.
Expanding this,we get $3x^2 + 3xy - 2xy - 2y^2 = 0$,which is $3x^2 + xy - 2y^2 = 0$.

(C) Given $\sin^2 A + \sin^2 B = \sin^2 C$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
Substituting these into the given equation: $(\frac{a}{2R})^2 + (\frac{b}{2R})^2 = (\frac{c}{2R})^2 \Rightarrow a^2 + b^2 = c^2$.
This implies that $\triangle ABC$ is a right-angled triangle with the hypotenuse $c = l(AB) = 10$.
The area of the triangle is $Area = \frac{1}{2}ab$.
Since $c = 10$,we have $a = 10 \sin A$ and $b = 10 \cos A$.
$Area = \frac{1}{2}(10 \sin A)(10 \cos A) = 50 \sin A \cos A = 25 \sin(2A)$.
The maximum value of $\sin(2A)$ is $1$ (when $2A = 90^{\circ}$ or $A = 45^{\circ}$).
Therefore,the maximum area is $25 \times 1 = 25$.

(D) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are given by:
$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$.
The direction ratios of the median $AD$ are given by the difference of the coordinates of $D$ and $A$:
$AD = \left( \frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5 \right) = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} \right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be proportional to $(1, 1, 1)$ or $(\pm 1, \pm 1, \pm 1)$.
Thus,we have:
$\frac{\lambda - 5}{2} = 1 \implies \lambda - 5 = 2 \implies \lambda = 7$.
$\frac{\mu - 8}{2} = 1 \implies \mu - 8 = 2 \implies \mu = 10$.
Therefore,the values are $\lambda = 7$ and $\mu = 10$.

(D) The points of trisection divide the line segment $AB$ in the ratio $1:2$ and $2:1$ internally.
Let the points be $P$ and $Q$ such that $P$ divides $AB$ in ratio $1:2$ and $Q$ divides $AB$ in ratio $2:1$.
The $z$-coordinate of a point dividing the segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in ratio $m:n$ is given by $\frac{mz_2 + nz_1}{m+n}$.
For $z_1$ (ratio $1:2$):
$z_1 = \frac{1(6) + 2(4)}{1 + 2} = \frac{6 + 8}{3} = \frac{14}{3}$.
For $z_2$ (ratio $2:1$):
$z_2 = \frac{2(6) + 1(4)}{2 + 1} = \frac{12 + 4}{3} = \frac{16}{3}$.
Therefore,$z_1 + z_2 = \frac{14}{3} + \frac{16}{3} = \frac{30}{3} = 10$.

(D) The given equation is $\tan 2\theta = 1$.
Since $\tan 2\theta$ is positive,$2\theta$ lies in the first or third quadrant.
For the principal solutions,we consider $0 \le \theta < 2\pi$,which implies $0 \le 2\theta < 4\pi$.
Thus,$2\theta = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$.
Dividing by $2$,we get $\theta = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$.
Hence,there are $4$ principal solutions.

(C) Given the curve $y = \sqrt{x - 1}$.
Finding the derivative,we get $\frac{dy}{dx} = \frac{1}{2\sqrt{x - 1}}$. Let this be the slope of the tangent $m_1 = \frac{1}{2\sqrt{x - 1}}$.
The equation of the line is $2x + y - 5 = 0$,which can be written as $y = -2x + 5$. The slope of this line is $m_2 = -2$.
Since the tangent is perpendicular to the line,the product of their slopes must be $-1$,i.e.,$m_1 \times m_2 = -1$.
Substituting the values,we have $\left(\frac{1}{2\sqrt{x - 1}}\right) \times (-2) = -1$.
This simplifies to $\frac{-1}{\sqrt{x - 1}} = -1$,which implies $\sqrt{x - 1} = 1$.
Squaring both sides,we get $x - 1 = 1$,so $x = 2$.
Substituting $x = 2$ into the curve equation,$y = \sqrt{2 - 1} = \sqrt{1} = 1$.
Thus,the required point is $(2, 1)$.

(A) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$. Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.
Given $\frac{dV}{dt} = 4 \pi \ cm^3/sec$,we have $4 \pi r^2 \frac{dr}{dt} = 4 \pi$,which implies $\frac{dr}{dt} = \frac{1}{r^2}$.
When the volume $V = 288 \pi \ cm^3$,we have $\frac{4}{3} \pi r^3 = 288 \pi$,so $r^3 = 216$,which means $r = 6 \ cm$.
Now,the surface area $A = 4 \pi r^2$. Differentiating with respect to $t$,we get $\frac{dA}{dt} = 8 \pi r \frac{dr}{dt}$.
Substituting $r = 6$ and $\frac{dr}{dt} = \frac{1}{r^2}$,we get $\frac{dA}{dt} = 8 \pi r \times \frac{1}{r^2} = \frac{8 \pi}{r}$.
For $r = 6$,$\frac{dA}{dt} = \frac{8 \pi}{6} = \frac{4}{3} \pi \ cm^2/sec$.

(C) Let $Z = ax + by$ be the objective function.
According to the Fundamental Theorem of Linear Programming,if an optimal solution exists for a linear programming problem,it must occur at one of the corner points (vertices) of the feasible region.
Even if the optimal value is attained at more than one point,it is guaranteed to be attained at at least one of the corner points of the convex polygon formed by the constraints.
Therefore,the objective function attains its optimum value at at least one of the corner points.

(B) Given function is $f(x) = \frac{\log x}{x}$.
To find the maximum value,we find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{x \cdot \frac{d}{dx}(\log x) - \log x \cdot \frac{d}{dx}(x)}{x^2} = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}$.
For critical points,set $f'(x) = 0$:
$\frac{1 - \log x}{x^2} = 0 \Rightarrow 1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e$.
To confirm this is a maximum,we check the second derivative or the sign change of $f'(x)$. Since $f'(x) > 0$ for $x < e$ and $f'(x) < 0$ for $x > e$,$x = e$ is a point of local maximum.
The maximum value is $f(e) = \frac{\log e}{e} = \frac{1}{e}$.

(D) The region is bounded by $y=3x+1$ (upper line),$y=2x+1$ (lower line),and $x=4$. The lines intersect at $x=0$,where $y=1$. Thus,the vertices of the triangle are $(0, 1)$,$(4, 9)$,and $(4, 13)$.
The area $A$ can be calculated using integration:
$A = \int_{0}^{4} [(3x+1) - (2x+1)] \, dx$
$A = \int_{0}^{4} x \, dx$
$A = \left[ \frac{x^2}{2} \right]_{0}^{4}$
$A = \frac{16}{2} - 0 = 8 \text{ sq. units}$.
Alternatively,using the formula for the area of a triangle with base on the line $x=4$:
Base length $= 13 - 9 = 4$.
Height $= 4 - 0 = 4$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \text{ sq. units}$.

(A) The objective function is $Z = 4 x_1 + 5 x_2$.
The constraints are:
$1) 2 x_1 + x_2 \geq 7$
$2) 2 x_1 + 3 x_2 \leq 15$
$3) x_2 \leq 3$
$4) x_1, x_2 \geq 0$
To find the feasible region,we determine the intersection points of the boundary lines:
For $2 x_1 + x_2 = 7$,the intercepts are $(3.5, 0)$ and $(0, 7)$.
For $2 x_1 + 3 x_2 = 15$,the intercepts are $(7.5, 0)$ and $(0, 5)$.
The line $x_2 = 3$ is a horizontal line.
Solving for intersection points of the feasible region:
- Intersection of $2 x_1 + x_2 = 7$ and $x_2 = 3$: $2 x_1 + 3 = 7 \implies 2 x_1 = 4 \implies x_1 = 2$. Point: $(2, 3)$.
- Intersection of $2 x_1 + 3 x_2 = 15$ and $x_2 = 3$: $2 x_1 + 9 = 15 \implies 2 x_1 = 6 \implies x_1 = 3$. Point: $(3, 3)$.
- The $x_1$-intercepts are $(3.5, 0)$ and $(7.5, 0)$.
The corner points of the feasible region are $(3.5, 0), (7.5, 0), (3, 3), (2, 3)$.
Evaluating $Z = 4 x_1 + 5 x_2$ at these points:

Corner Point	$Z = 4 x_1 + 5 x_2$
$(3.5, 0)$	$4(3.5) + 5(0) = 14$
$(7.5, 0)$	$4(7.5) + 5(0) = 30$
$(3, 3)$	$4(3) + 5(3) = 12 + 15 = 27$
$(2, 3)$	$4(2) + 5(3) = 8 + 15 = 23$

The minimum value of $Z$ is $14$,which occurs at the point $(3.5, 0)$. Since the $x_2$-coordinate is $0$,this point lies on the $x_1$-axis.

(A) Continuity at $x = 0$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} x = 0$
$\lim_{x \to 0^+} f(x) = 0$
$f(0) = 0$
Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$,the function is continuous at $x = 0$.
Differentiability at $x = 0$:
Left-hand derivative $(LHD)$ = $\lim_{h \to 0^+} \frac{f(0-h) - f(0)}{-h} = \lim_{h \to 0^+} \frac{-h - 0}{-h} = 1$
Right-hand derivative $(RHD)$ = $\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{0 - 0}{h} = 0$
Since $LHD \neq RHD$,the function is not differentiable at $x = 0$.
Conclusion: The function is continuous but not differentiable at $x = 0$.

(B) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$K = \lim_{x \to 0} \log(\sec^2 x)^{\cot^2 x}$
Using the property $\log(a^b) = b \log a$,we get:
$K = \lim_{x \to 0} \cot^2 x \cdot \log(\sec^2 x)$
Since $\sec^2 x = 1 + \tan^2 x$,we have:
$K = \lim_{x \to 0} \cot^2 x \cdot \log(1 + \tan^2 x)$
$K = \lim_{x \to 0} \frac{\log(1 + \tan^2 x)}{\tan^2 x}$
Using the standard limit $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$,where $u = \tan^2 x$ as $x \to 0$:
$K = 1$.

(C) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
Given $f(0) = K$,so $K = \lim_{x \to 0} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}$.
Using the formula $\tan(\frac{\pi}{4} + x) = \frac{1 + \tan x}{1 - \tan x}$,we get:
$K = \lim_{x \to 0} (\frac{1 + \tan x}{1 - \tan x})^{\frac{1}{x}}$.
This is of the form $1^{\infty}$,so we use the formula $\lim_{x \to 0} (1 + g(x))^{h(x)} = e^{\lim_{x \to 0} g(x)h(x)}$.
$K = \lim_{x \to 0} [1 + (\frac{1 + \tan x}{1 - \tan x} - 1)]^{\frac{1}{x}} = \lim_{x \to 0} [1 + \frac{2 \tan x}{1 - \tan x}]^{\frac{1}{x}}$.
$K = e^{\lim_{x \to 0} (\frac{2 \tan x}{1 - \tan x} \cdot \frac{1}{x})}$.
Since $\lim_{x \to 0} \frac{\tan x}{x} = 1$,we have:
$K = e^{\lim_{x \to 0} (\frac{2}{1 - \tan x} \cdot 1)} = e^{\frac{2}{1 - 0}} = e^{2}$.

(D) We are given that $\int_0^{\frac{\pi}{2}} \log \cos x \, dx = \frac{\pi}{2} \log \left(\frac{1}{2}\right)$.
We need to evaluate $I = \int_0^{\frac{\pi}{2}} \log \sec x \, dx$.
Since $\sec x = \frac{1}{\cos x}$,we have $\log \sec x = \log \left(\frac{1}{\cos x}\right) = \log 1 - \log \cos x = -\log \cos x$.
Therefore,$I = \int_0^{\frac{\pi}{2}} -\log \cos x \, dx$.
$I = -\int_0^{\frac{\pi}{2}} \log \cos x \, dx$.
Substituting the given value,$I = -\left[\frac{\pi}{2} \log \left(\frac{1}{2}\right)\right]$.
Since $\log \left(\frac{1}{2}\right) = \log (2^{-1}) = -\log 2$,we get $I = -\left[\frac{\pi}{2} (-\log 2)\right] = \frac{\pi}{2} \log 2$.

(B) Using integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \tan^{-1} x$ and $dv = x \, dx$. Then $du = \frac{1}{1+x^2} \, dx$ and $v = \frac{x^2}{2}$.
$\int_0^1 x \tan^{-1} x \, dx = \left[ \frac{x^2}{2} \tan^{-1} x \right]_0^1 - \int_0^1 \frac{x^2}{2(1+x^2)} \, dx$
$= \left( \frac{1}{2} \cdot \frac{\pi}{4} - 0 \right) - \frac{1}{2} \int_0^1 \frac{x^2+1-1}{1+x^2} \, dx$
$= \frac{\pi}{8} - \frac{1}{2} \int_0^1 \left( 1 - \frac{1}{1+x^2} \right) \, dx$
$= \frac{\pi}{8} - \frac{1}{2} \left[ x - \tan^{-1} x \right]_0^1$
$= \frac{\pi}{8} - \frac{1}{2} \left( (1 - \frac{\pi}{4}) - (0 - 0) \right)$
$= \frac{\pi}{8} - \frac{1}{2} + \frac{\pi}{8} = \frac{\pi}{4} - \frac{1}{2}$.

(A) The integral is defined as $\int_0^3 [x] \, dx$.
Since $[x]$ changes its value at every integer,we split the integral into intervals:
$\int_0^3 [x] \, dx = \int_0^1 [x] \, dx + \int_1^2 [x] \, dx + \int_2^3 [x] \, dx$
In the interval $[0, 1)$,$[x] = 0$.
In the interval $[1, 2)$,$[x] = 1$.
In the interval $[2, 3)$,$[x] = 2$.
Substituting these values:
$= \int_0^1 0 \, dx + \int_1^2 1 \, dx + \int_2^3 2 \, dx$
$= 0 + [x]_1^2 + [2x]_2^3$
$= (2 - 1) + (6 - 4)$
$= 1 + 2 = 3$.

(A) The general equation of a parabola with its axis along the $y$-axis is given by $(x - 0)^2 = 4a(y - k)$,where $a$ and $k$ are arbitrary constants.
This simplifies to $x^2 = 4ay - 4ak$.
Differentiating both sides with respect to $x$,we get:
$2x = 4a \frac{dy}{dx}$
$\Rightarrow x = 2a \frac{dy}{dx}$
$\Rightarrow \frac{1}{2a} = \frac{1}{x} \frac{dy}{dx}$
Differentiating again with respect to $x$ to eliminate the arbitrary constant $a$:
$\frac{d}{dx} \left( \frac{1}{x} \cdot \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{1}{2a} \right)$
Using the product rule on the left side:
$\frac{1}{x} \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \left( - \frac{1}{x^2} \right) = 0$
Multiplying the entire equation by $x^2$:
$x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$

(B) Given the differential equation: $\frac{dy}{dx} = \tan \left(\frac{y}{x}\right) + \frac{y}{x}$.
Let $v = \frac{y}{x}$,so $y = vx$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into the original equation:
$v + x \frac{dv}{dx} = \tan v + v$.
Subtracting $v$ from both sides:
$x \frac{dv}{dx} = \tan v$.
Separating the variables:
$\frac{dv}{\tan v} = \frac{dx}{x}$,which is $\cot v \, dv = \frac{dx}{x}$.
Integrating both sides:
$\int \cot v \, dv = \int \frac{1}{x} \, dx$.
$\log |\sin v| = \log |x| + \log |c|$.
$\log |\sin v| = \log |cx|$.
Taking the exponential of both sides:
$\sin v = cx$.
Substituting $v = \frac{y}{x}$ back:
$\sin \left(\frac{y}{x}\right) = cx$.

(B) Given differential equation is $x dy + 2y dx = 0$.
Dividing by $xy$,we get $\frac{dy}{y} + \frac{2 dx}{x} = 0$.
Integrating both sides,we get $\int \frac{dy}{y} + 2 \int \frac{dx}{x} = \int 0$.
This gives $\ln|y| + 2 \ln|x| = C_1$.
Using the property of logarithms,$\ln|y| + \ln|x^2| = C_1$,which implies $\ln|yx^2| = C_1$.
Thus,$yx^2 = e^{C_1} = C$.
Given $x = 2$ and $y = 1$,we substitute these values: $(1)(2)^2 = C$,so $C = 4$.
Therefore,the particular solution is $x^2 y = 4$.

(C) Given equations are $x = a (t - 1/t)$ and $y = a (t + 1/t)$.
Squaring and subtracting the equations:
$y^2 - x^2 = a^2 [(t + 1/t)^2 - (t - 1/t)^2]$
Using the identity $(a+b)^2 - (a-b)^2 = 4ab$,we get:
$y^2 - x^2 = a^2 [4 \cdot t \cdot (1/t)] = 4a^2$.
Differentiating both sides with respect to $x$:
$d/dx (y^2 - x^2) = d/dx (4a^2)$
$2y (dy/dx) - 2x = 0$
$2y (dy/dx) = 2x$
Therefore,$dy/dx = x/y$.

(A) Given $x=f(t)$ and $y=g(t)$.
First,find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dx}{dt} = f^{\prime}(t)$ and $\frac{dy}{dt} = g^{\prime}(t)$.
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g^{\prime}(t)}{f^{\prime}(t)}$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{g^{\prime}(t)}{f^{\prime}(t)} \right) = \frac{d}{dt} \left( \frac{g^{\prime}(t)}{f^{\prime}(t)} \right) \cdot \frac{dt}{dx}$.
Using the quotient rule for $\frac{d}{dt} \left( \frac{g^{\prime}(t)}{f^{\prime}(t)} \right)$:
$= \frac{f^{\prime}(t) \cdot g^{\prime \prime}(t) - g^{\prime}(t) \cdot f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^2} \cdot \frac{1}{f^{\prime}(t)}$.
$= \frac{f^{\prime}(t) \cdot g^{\prime \prime}(t) - g^{\prime}(t) \cdot f^{\prime \prime}(t)}{\left[f^{\prime}(t)\right]^3}$.

(A) Given that $g(x)$ is the inverse of $f(x)$,we have $f(g(x)) = x$.
Differentiating both sides with respect to $x$ using the chain rule,we get $f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.
We are given $f^{\prime}(x) = \frac{1}{1+x^4}$.
Substituting $g(x)$ for $x$ in the expression for $f^{\prime}(x)$,we get $f^{\prime}(g(x)) = \frac{1}{1+[g(x)]^4}$.
Substituting this into the differentiated equation: $\frac{1}{1+[g(x)]^4} \cdot g^{\prime}(x) = 1$.
Therefore,$g^{\prime}(x) = 1+[g(x)]^4$.

(B) $I = \int \frac{\sec^{8} x}{\text{cosec} x} dx$
Since $\sec x = \frac{1}{\cos x}$ and $\text{cosec} x = \frac{1}{\sin x}$,we have:
$I = \int \frac{\sin x}{\cos^{8} x} dx$
Let $u = \cos x$,then $du = -\sin x dx$,which implies $\sin x dx = -du$.
Substituting these into the integral:
$I = \int -\frac{du}{u^{8}} = -\int u^{-8} du$
Using the power rule $\int u^{n} du = \frac{u^{n+1}}{n+1}$:
$I = -\left( \frac{u^{-7}}{-7} \right) + c = \frac{1}{7u^{7}} + c$
Substituting $u = \cos x$ back:
$I = \frac{1}{7 \cos^{7} x} + c = \frac{\sec^{7} x}{7} + c$

(D) To evaluate $\int \sqrt{\frac{x - 5}{x - 7}} dx$,we multiply the numerator and denominator by $\sqrt{x - 5}$:
$\int \frac{x - 5}{\sqrt{(x - 5)(x - 7)}} dx = \int \frac{x - 5}{\sqrt{x^2 - 12x + 35}} dx$
Next,we express the numerator in terms of the derivative of the quadratic expression $x^2 - 12x + 35$,which is $2x - 12$:
$= \frac{1}{2} \int \frac{2x - 10}{\sqrt{x^2 - 12x + 35}} dx = \frac{1}{2} \int \frac{2x - 12 + 2}{\sqrt{x^2 - 12x + 35}} dx$
$= \frac{1}{2} \int \frac{2x - 12}{\sqrt{x^2 - 12x + 35}} dx + \int \frac{1}{\sqrt{x^2 - 12x + 35}} dx$
$= \sqrt{x^2 - 12x + 35} + \int \frac{1}{\sqrt{(x - 6)^2 - 1}} dx$
Using the standard integral $\int \frac{1}{\sqrt{t^2 - a^2}} dt = \log |t + \sqrt{t^2 - a^2}|$,we get:
$= \sqrt{x^2 - 12x + 35} + \log |x - 6 + \sqrt{(x - 6)^2 - 1}| + C$
$= 1 \cdot \sqrt{x^2 - 12x + 35} + \log |x - 6 + \sqrt{x^2 - 12x + 35}| + C$
Comparing this with the given form,we find $A = 1$.

(A) We use the method of partial fractions to decompose the integrand: $\frac{1}{(x^{2} + 4)(x^{2} + 9)} = \frac{1}{5} \left( \frac{1}{x^{2} + 4} - \frac{1}{x^{2} + 9} \right)$.
Integrating both sides with respect to $x$:
$\int \frac{1}{(x^{2} + 4)(x^{2} + 9)} dx = \frac{1}{5} \int \frac{1}{x^{2} + 2^{2}} dx - \frac{1}{5} \int \frac{1}{x^{2} + 3^{2}} dx$.
Using the standard formula $\int \frac{1}{x^{2} + a^{2}} dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + C$:
$= \frac{1}{5} \left( \frac{1}{2} \tan^{-1} \frac{x}{2} \right) - \frac{1}{5} \left( \frac{1}{3} \tan^{-1} \frac{x}{3} \right) + C$.
$= \frac{1}{10} \tan^{-1} \frac{x}{2} - \frac{1}{15} \tan^{-1} \frac{x}{3} + C$.
Comparing this with $A \tan^{-1} \frac{x}{2} + B \tan^{-1} \frac{x}{3} + C$,we get $A = \frac{1}{10}$ and $B = -\frac{1}{15}$.
Therefore,$A - B = \frac{1}{10} - (-\frac{1}{15}) = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$.

(A) We are given the integral $\int \frac{1}{\sqrt{9-16 x^2}} d x$.
Rewrite the denominator as $\sqrt{3^2-(4 x)^2}$.
Using the standard formula $\int \frac{1}{\sqrt{a^2-u^2}} d u = \sin^{-1}(\frac{u}{a}) + c$,where $u = 4x$ and $du = 4 dx$,we get:
$\int \frac{1}{\sqrt{3^2-(4 x)^2}} d x = \frac{1}{4} \int \frac{1}{\sqrt{3^2-(4 x)^2}} d(4x) = \frac{1}{4} \sin^{-1}(\frac{4x}{3}) + c$.
Comparing this with $\alpha \sin^{-1}(\beta x) + c$,we identify $\alpha = \frac{1}{4}$ and $\beta = \frac{4}{3}$.
Now,calculate $\alpha + \frac{1}{\beta} = \frac{1}{4} + \frac{1}{4/3} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$.

(A) We are given the expression $\cos ^{-1}\left(\cot \left(\frac{\pi}{2}\right)\right)+\cos ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$.
First,evaluate the inner trigonometric functions:
$\cot \left(\frac{\pi}{2}\right) = 0$.
$\sin \left(\frac{2 \pi}{3}\right) = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.
Now,substitute these values back into the expression:
$\cos ^{-1}(0) + \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
We know that $\cos ^{-1}(0) = \frac{\pi}{2}$ and $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.
Adding these values gives:
$\frac{\pi}{2} + \frac{\pi}{6} = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.

(C) We are given that $A(\operatorname{adj} A) = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$.
This can be written as $A(\operatorname{adj} A) = 10 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 10I$,where $I$ is the identity matrix of order $2 \times 2$.
We know the fundamental property of the adjoint of a matrix: $A(\operatorname{adj} A) = |A|I$.
Comparing the two expressions,we get $|A|I = 10I$.
Therefore,$|A| = 10$.

(B) Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$.
First,calculate the determinant $|A| = 1(3(-1) - 0(2)) - 0 + 0 = -3$.
Next,find the matrix of cofactors $C_{ij}$:
$C_{11} = +((-1)(3) - 0) = -3$,$C_{12} = -(3(-1) - 0) = 3$,$C_{13} = +(3(2) - 3(5)) = -9$.
$C_{21} = -(0(-1) - 0(2)) = 0$,$C_{22} = +(1(-1) - 0(5)) = -1$,$C_{23} = -(1(2) - 0(5)) = -2$.
$C_{31} = +(0(0) - 3(0)) = 0$,$C_{32} = -(1(0) - 0(3)) = 0$,$C_{33} = +(1(3) - 0(3)) = 3$.
Thus,$\text{Adj}(A) = \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$.
Finally,$A^{-1} = \frac{1}{|A|} \text{Adj}(A) = \frac{1}{-3} \begin{bmatrix} -3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3 \end{bmatrix}$.

(D) Let $A = \begin{bmatrix} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{bmatrix}$.
The inverse of a matrix $A$ does not exist if and only if the determinant of the matrix is zero,i.e.,$|A| = 0$.
Calculating the determinant $|A|$:
$|A| = \alpha(3 \times 3 - 1 \times 2) - 14(2 \times 3 - 1 \times 6) + (-1)(2 \times 2 - 3 \times 6)$
$|A| = \alpha(9 - 2) - 14(6 - 6) - 1(4 - 18)$
$|A| = \alpha(7) - 14(0) - 1(-14)$
$|A| = 7\alpha + 14$
Setting $|A| = 0$ for the inverse to not exist:
$7\alpha + 14 = 0$
$7\alpha = -14$
$\alpha = -2$
Thus,the value of $\alpha$ is $-2$.

(C) Let $X$ be the random variable representing the number of heads in $3$ tosses of a fair coin. The possible values of $X$ are $0, 1, 2, 3$.
The probability distribution of $X$ is given by the binomial distribution $B(n=3, p=0.5)$:
$P(X=0) = \binom{3}{0} (0.5)^3 = \frac{1}{8}$
$P(X=1) = \binom{3}{1} (0.5)^3 = \frac{3}{8}$
$P(X=2) = \binom{3}{2} (0.5)^3 = \frac{3}{8}$
$P(X=3) = \binom{3}{3} (0.5)^3 = \frac{1}{8}$
Let $Y$ be the gain,where $Y = 2X$. The expected gain $E[Y]$ is given by $E[2X] = 2E[X]$.
For a binomial distribution,$E[X] = np = 3 \times 0.5 = 1.5$.
Therefore,$E[Y] = 2 \times 1.5 = 3$.
Alternatively,using the table:

$X$	$0, 1, 2, 3$
$Y = 2X$	$0, 2, 4, 6$
$P(Y)$	$\frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \frac{1}{8}$

$E[Y] = 0(\frac{1}{8}) + 2(\frac{3}{8}) + 4(\frac{3}{8}) + 6(\frac{1}{8}) = \frac{0+6+12+6}{8} = \frac{24}{8} = 3$.

(B) Let $p$ be the probability that a person develops immunity,so $p = 0.8$.
Since the events of developing immunity among $8$ different people are independent,the probability that all $8$ people develop immunity is given by the product of their individual probabilities.
Therefore,the required probability is $p \times p \times p \times p \times p \times p \times p \times p = (0.8)^8$.

(A) The probability $P(3 < X \leq 5)$ is given by $F(5) - F(3)$.
From the given table:
$F(5) = 0.85$
$F(3) = 0.48$
Therefore,$P(3 < X \leq 5) = F(5) - F(3) = 0.85 - 0.48 = 0.37$.

(D) $x$ : number of defective pens.
Two pens are taken from the box.
Therefore,$x$ can take values $0, 1, 2$.
$P(x=0) = \frac{{}^4C_2}{{}^6C_2} = \frac{6}{15}$
$P(x=1) = \frac{{}^2C_1 \times {}^4C_1}{{}^6C_2} = \frac{8}{15}$
$P(x=2) = \frac{{}^2C_2}{{}^6C_2} = \frac{1}{15}$

$x_i$	$P(x_i)$	$x_i P(x_i)$	$x_i^2 P(x_i)$
$0$	$\frac{6}{15}$	$0$	$0$
$1$	$\frac{8}{15}$	$\frac{8}{15}$	$\frac{8}{15}$
$2$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{4}{15}$

$E(x) = \sum x_i P(x_i) = 0 + \frac{8}{15} + \frac{2}{15} = \frac{10}{15} = \frac{2}{3}$
$E(x^2) = \sum x_i^2 P(x_i) = 0 + \frac{8}{15} + \frac{4}{15} = \frac{12}{15} = \frac{4}{5}$
$\text{Variance}(x) = E(x^2) - [E(x)]^2 = \frac{4}{5} - (\frac{2}{3})^2 = \frac{4}{5} - \frac{4}{9} = \frac{36-20}{45} = \frac{16}{45}$
$\text{Standard Deviation} = \sqrt{\text{Variance}(x)} = \sqrt{\frac{16}{45}} = \frac{4}{3 \sqrt{5}}$

(B) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np = 18$ and the variance is given by $Var(X) = npq = 12$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{12}{18}$.
This simplifies to $q = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p = \frac{1}{3}$ into the mean equation $np = 18$,we get $n \times \frac{1}{3} = 18$,which implies $n = 54$.
The random variable $X$ can take values from $0, 1, 2, \dots, n$.
Therefore,the possible values of $X$ are $0, 1, 2, \dots, 54$.
The total number of such values is $n + 1 = 54 + 1 = 55$.

(A) Given equation: $x^2 + 5|x| - 6 = 0$.
Since $x^2 = |x|^2$,we can write: $|x|^2 + 5|x| - 6 = 0$.
Factoring the quadratic: $(|x| + 6)(|x| - 1) = 0$.
This gives $|x| = 1$ or $|x| = -6$.
Since $|x|$ cannot be negative,we have $|x| = 1$,which implies $x = 1$ or $x = -1$.
Let $\alpha = 1$ and $\beta = -1$.
Then,$|\tan^{-1} \alpha - \tan^{-1} \beta| = |\tan^{-1}(1) - \tan^{-1}(-1)|$.
$= |\frac{\pi}{4} - (-\frac{\pi}{4})| = |\frac{\pi}{4} + \frac{\pi}{4}| = |\frac{\pi}{2}| = \frac{\pi}{2}$.

(B) Let the general point on the first line $L_1$ be $P(r)$.
$\frac{x - 1}{2} = \frac{y + 1}{2} = \frac{z - 1}{4} = r$
$x = 2r + 1, y = 2r - 1, z = 4r + 1$
So,$P = (2r + 1, 2r - 1, 4r + 1)$.
Since the lines intersect,this point must satisfy the equation of the second line $L_2$: $\frac{x - 3}{1} = \frac{y - 6}{2} = \frac{z}{1} = k$.
Substituting the coordinates of $P$ into $L_2$:
$\frac{2r + 1 - 3}{1} = \frac{2r - 1 - 6}{2} = \frac{4r + 1}{1}$
$\frac{2r - 2}{1} = \frac{2r - 7}{2} = 4r + 1$
Taking the first and third parts: $2r - 2 = 4r + 1 \Rightarrow -3 = 2r \Rightarrow r = -\frac{3}{2}$.
Now,substitute $r = -\frac{3}{2}$ into the coordinates of $P$:
$x = 2(-\frac{3}{2}) + 1 = -3 + 1 = -2$
$y = 2(-\frac{3}{2}) - 1 = -3 - 1 = -4$
$z = 4(-\frac{3}{2}) + 1 = -6 + 1 = -5$
Thus,the intersection point is $(-2, -4, -5)$.

(A) The equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Given that the line passes through $(-3, 2, -5)$,we have $x_1 = -3, y_1 = 2, z_1 = -5$.
Since the line is equally inclined to the positive coordinate axes,the direction cosines $(l, m, n)$ are equal,i.e.,$l = m = n$.
Thus,the direction ratios can be taken as $a = 1, b = 1, c = 1$.
Substituting these values into the equation,we get $\frac{x - (-3)}{1} = \frac{y - 2}{1} = \frac{z - (-5)}{1}$.
Therefore,the equation of the line is $\frac{x+3}{1} = \frac{y-2}{1} = \frac{z+5}{1}$.

(A) The equation of the plane is $\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$,which can be written in Cartesian form as $3x + 2y + 6z - 13 = 0$.
Given the point $(x_1, y_1, z_1) = (2, 3, \lambda)$.
The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$ is given by $d = \left| \frac{ax_1 + by_1 + cz_1 + d}{\sqrt{a^2 + b^2 + c^2}} \right|$.
Substituting the values,we get $5 = \left| \frac{3(2) + 2(3) + 6(\lambda) - 13}{\sqrt{3^2 + 2^2 + 6^2}} \right|$.
$5 = \left| \frac{6 + 6 + 6\lambda - 13}{\sqrt{9 + 4 + 36}} \right|$.
$5 = \left| \frac{6\lambda - 1}{\sqrt{49}} \right|$.
$5 = \left| \frac{6\lambda - 1}{7} \right|$.
$35 = |6\lambda - 1|$.
This gives two cases: $6\lambda - 1 = 35$ or $6\lambda - 1 = -35$.
Case $1$: $6\lambda = 36 \implies \lambda = 6$.
Case $2$: $6\lambda = -34 \implies \lambda = -\frac{34}{6} = -\frac{17}{3}$.
Thus,$\lambda = 6, -\frac{17}{3}$.

(A) The equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
Since the normal makes equal acute angles with the coordinate axes,its direction cosines are equal,i.e.,$l = m = n$. Since $l^2 + m^2 + n^2 = 1$,we have $3l^2 = 1$,so $l = m = n = \frac{1}{\sqrt{3}}$.
Thus,the normal vector can be taken as $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.
The given point is $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$.
Substituting these into the equation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:
$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = (-\hat{i} + \hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$
$= (-1)(1) + (1)(1) + (2)(1) = -1 + 1 + 2 = 2$.
Therefore,the equation is $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2$.

(C) The angle $\theta$ between two planes $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$ is given by $\cos \theta = \left| \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1| |\vec{n}_2|} \right|$.
Here,$\vec{n}_1 = m\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{n}_2 = 2\hat{i} - m\hat{j} + \hat{k}$.
The angle $\theta = \frac{\pi}{3}$,so $\cos \theta = \cos \frac{\pi}{3} = \frac{1}{2}$.
$\vec{n}_1 \cdot \vec{n}_2 = (m)(2) + (-1)(-m) + (2)(1) = 2m + m + 2 = 3m + 2$.
$|\vec{n}_1| = \sqrt{m^2 + (-1)^2 + 2^2} = \sqrt{m^2 + 5}$.
$|\vec{n}_2| = \sqrt{2^2 + (-m)^2 + 1^2} = \sqrt{m^2 + 5}$.
Substituting these into the formula: $\frac{1}{2} = \left| \frac{3m + 2}{m^2 + 5} \right|$.
$m^2 + 5 = 2|3m + 2|$.
Case $1$: $m^2 + 5 = 6m + 4 \Rightarrow m^2 - 6m + 1 = 0$. Roots are $m = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2}$.
Case $2$: $m^2 + 5 = -(6m + 4) \Rightarrow m^2 + 6m + 9 = 0 \Rightarrow (m + 3)^2 = 0 \Rightarrow m = -3$.
Given the options,$m=3$ is not a solution for the corrected equation. Re-evaluating the dot product: $\vec{n}_1 \cdot \vec{n}_2 = 2m + m + 2 = 3m + 2$. If the question intended $\vec{n}_2 = 2\hat{i} - m\hat{j} - \hat{k}$,then $\vec{n}_1 \cdot \vec{n}_2 = 2m + m - 2 = 3m - 2$. Then $|3m - 2| = \frac{1}{2}(m^2 + 5) \Rightarrow m^2 - 6m + 9 = 0 \Rightarrow m = 3$.

(C) For a vector $\vec{r}$ to be equally inclined to the coordinate axes, the direction cosines must satisfy $|l| = |m| = |n|$.
Since $l^2 + m^2 + n^2 = 1$, we substitute $|l| = |m| = |n|$ to get $l^2 + l^2 + l^2 = 1$.
This simplifies to $3l^2 = 1$, which gives $l^2 = \frac{1}{3}$.
Thus, $l = \pm \frac{1}{\sqrt{3}}$, $m = \pm \frac{1}{\sqrt{3}}$, and $n = \pm \frac{1}{\sqrt{3}}$.
Each of the direction cosines $l, m, n$ has $2$ possible values $(\pm \frac{1}{\sqrt{3}})$.
Therefore, the total number of such vectors is $2 \times 2 \times 2 = 8$.

(C) Let the position vectors of points $P, Q, R, S, M, N$ be $\vec{p}, \vec{q}, \vec{r}, \vec{s}, \vec{m}, \vec{n}$ respectively.
Since $M$ is the midpoint of $PQ$,we have $\vec{m} = \frac{\vec{p} + \vec{q}}{2}$,which implies $\vec{p} + \vec{q} = 2\vec{m}$.
Since $N$ is the midpoint of $RS$,we have $\vec{n} = \frac{\vec{r} + \vec{s}}{2}$,which implies $\vec{r} + \vec{s} = 2\vec{n}$.
Now,consider the expression $\overrightarrow{PS} + \overrightarrow{QR} = (\vec{s} - \vec{p}) + (\vec{r} - \vec{q})$.
Rearranging the terms,we get $(\vec{s} + \vec{r}) - (\vec{p} + \vec{q})$.
Substituting the values derived from the midpoints,we get $2\vec{n} - 2\vec{m}$.
This simplifies to $2(\vec{n} - \vec{m}) = 2\overrightarrow{MN}$.

(C) If the points $O(0,0,0)$,$P(2,3,4)$,$Q(1,2,3)$,and $R(x, y, z)$ are co-planar,then the scalar triple product of the vectors $\vec{OR}$,$\vec{OP}$,and $\vec{OQ}$ must be zero.
$\implies [\vec{OR} \quad \vec{OP} \quad \vec{OQ}] = 0$
Using the determinant form:
$\left| \begin{array}{ccc} x & y & z \\ 2 & 3 & 4 \\ 1 & 2 & 3 \end{array} \right| = 0$
Expanding along the first row:
$x(3 \times 3 - 4 \times 2) - y(2 \times 3 - 4 \times 1) + z(2 \times 2 - 3 \times 1) = 0$
$x(9 - 8) - y(6 - 4) + z(4 - 3) = 0$
$x(1) - y(2) + z(1) = 0$
$x - 2y + z = 0$
Thus,the correct equation is $x - 2y + z = 0$.