MHT CET 2018 Mathematics Question Paper with Answer and Solution

(B) The vertices of the rectangle are $(a, b)$,$(-a, b)$,$(-a, -b)$,and $(a, -b)$.
Since the circle passes through these vertices,the diagonal of the rectangle acts as the diameter of the circle.
Taking the diagonal connecting $(a, b)$ and $(-a, -b)$ as the diameter,the equation of the circle is given by:
$(x - a)(x + a) + (y - b)(y + b) = 0$
$x^2 - a^2 + y^2 - b^2 = 0$
$x^2 + y^2 = a^2 + b^2$

(B) Let $P$ be the statement "Hema gets above $95 \%$ marks" and $Q$ be the statement "Hema gets admission in a good college".
The statement "Getting above $95 \%$ marks is a necessary condition for Hema to get admission in a good college" is equivalent to the implication $Q \implies P$.
The negation of an implication $Q \implies P$ is given by $\sim(Q \implies P) \equiv Q \land \sim P$.
Here,$Q$ is "Hema gets admission in a good college" and $\sim P$ is "Hema does not get above $95 \%$ marks".
Therefore,the negation is: "Hema gets admission in a good college and she does not get above $95 \%$ marks."

(B) Given compound statement is $p \wedge (\sim p \wedge q)$.
Using the associative law,we can rewrite this as $(p \wedge \sim p) \wedge q$.
Since $p \wedge \sim p$ is always false $(F)$,the expression becomes $F \wedge q$.
Since $F \wedge q$ is always false,the statement is a contradiction.

(B) Let $p$ be the statement: "The weather is fine".
Let $q$ be the statement: "My friends will come".
Let $r$ be the statement: "We go for a picnic".
The given statement is $p \rightarrow (q \wedge r)$.
The contrapositive of $p \rightarrow (q \wedge r)$ is $\sim(q \wedge r) \rightarrow \sim p$.
Using De Morgan's Law,$\sim(q \wedge r) \equiv (\sim q \vee \sim r)$.
So,the contrapositive is $(\sim q \vee \sim r) \rightarrow \sim p$.
Translating this back to words: "If my friends do not come or we do not go for a picnic,then the weather will not be fine."

(C) The given equation is $5x^2 + xy - kx - 2y + 2 = 0$. Since this represents a pair of lines and one line is $5x + y - 1 = 0$,we can write the equation as $(5x + y - 1)(ax + by + c) = 0$.
Comparing the coefficients of $y^2$,we see the coefficient is $0$,so the second line must be of the form $ax + c = 0$.
Thus,$(5x + y - 1)(ax + c) = 5ax^2 + axy + (5c - a)x + cy - c = 0$.
Comparing this with $5x^2 + xy - kx - 2y + 2 = 0$:
From the $xy$ term,$a = 1$.
From the constant term,$-c = 2$,so $c = -2$.
From the $y$ term,$c = -2$ (which matches).
From the $x$ term,$-k = 5c - a$.
Substituting the values: $-k = 5(-2) - 1 = -10 - 1 = -11$.
Therefore,$k = 11$.

(C) Let the slopes of the two lines be $m$ and $2m$.
From the equation $ax^2+2hxy+by^2=0$,we have:
Sum of slopes: $m_1+m_2 = m+2m = 3m = -\frac{2h}{b} \Rightarrow m = -\frac{2h}{3b}$.
Product of slopes: $m_1 \times m_2 = m \times 2m = 2m^2 = \frac{a}{b}$.
Substituting the value of $m$ into the product equation:
$2(-\frac{2h}{3b})^2 = \frac{a}{b}$
$2(\frac{4h^2}{9b^2}) = \frac{a}{b}$
$\frac{8h^2}{9b^2} = \frac{a}{b}$
$8h^2 = 9ab$.

(C) The given equation is $x^2 - y^2 + x + 3y - 2 = 0$.
Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=1, h=0, b=-1, g=\frac{1}{2}, f=\frac{3}{2}, c=-2$.
The point of intersection $(x, y)$ of the pair of lines is given by the solution of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial}{\partial x}(x^2 - y^2 + x + 3y - 2) = 2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$.
$\frac{\partial}{\partial y}(x^2 - y^2 + x + 3y - 2) = -2y + 3 = 0 \Rightarrow y = \frac{3}{2}$.
Thus,the point of intersection is $(-\frac{1}{2}, \frac{3}{2})$.

(D) Given the parabola equation $y^2 = -16x$ and parameter $t = \frac{1}{2}$.
Comparing $y^2 = -16x$ with the standard form $y^2 = -4ax$,we get $4a = 16$,which implies $a = 4$.
The parametric coordinates of a point on the parabola $y^2 = -4ax$ are given by $P(t) = (-at^2, 2at)$.
Substituting $a = 4$ and $t = \frac{1}{2}$ into the coordinates:
$x = -a t^2 = -4 \times (\frac{1}{2})^2 = -4 \times \frac{1}{4} = -1$.
$y = 2at = 2 \times 4 \times \frac{1}{2} = 4$.
Therefore,the coordinates of the point are $(-1, 4)$.

(C) The word $HULULULU$ contains $8$ letters: $H(1), U(4), L(3)$.
Total number of arrangements $n(S) = \frac{8!}{4!3!} = \frac{40320}{24 \times 6} = \frac{40320}{144} = 280$.
To find the number of arrangements where all three $L$ are together,we treat $(LLL)$ as a single unit.
Now we have $6$ units: $(LLL), H, U, U, U, U$.
The number of arrangements $n(A) = \frac{6!}{4!1!} = \frac{720}{24} = 30$.
The required probability is $P(A) = \frac{n(A)}{n(S)} = \frac{30}{280} = \frac{3}{28}$.

(C) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
The expression is $E = a \cos^2\left(\frac{C}{2}\right) + c \cos^2\left(\frac{A}{2}\right)$.
Using the identity $2 \cos^2\theta = 1 + \cos(2\theta)$,we get:
$E = \frac{a}{2}(1 + \cos C) + \frac{c}{2}(1 + \cos A)$
$E = \frac{1}{2}(a + a \cos C + c + c \cos A)$
By the projection formula,$a \cos C + c \cos A = b$.
$E = \frac{1}{2}(a + c + b)$
Since $a + c = 2b$,we have:
$E = \frac{1}{2}(2b + b) = \frac{3b}{2}$.

(B) The given series is $S_{10} = 9 + 99 + 999 + \ldots$ up to $10$ terms.
This can be written as $S_{10} = (10-1) + (10^2-1) + (10^3-1) + \ldots + (10^{10}-1)$.
$S_{10} = (10 + 10^2 + 10^3 + \ldots + 10^{10}) - (1 + 1 + 1 + \ldots + 1 \text{ (10 times)})$.
The first part is a geometric progression with $a=10$,$r=10$,and $n=10$.
Sum of $GP$ $= a\frac{r^n-1}{r-1} = 10\frac{10^{10}-1}{10-1} = \frac{10}{9}(10^{10}-1)$.
Subtracting the sum of $1$s (which is $10$):
$S_{10} = \frac{10}{9}(10^{10}-1) - 10 = \frac{10(10^{10}-1) - 90}{9} = \frac{10^{11}-10-90}{9} = \frac{10^{11}-100}{9} = \frac{100}{9}(10^9-1)$.

(A) Given $X = \{4^n - 3n - 1 : n \in N\}$ and $Y = \{9(n - 1) : n \in N\}$.
By binomial expansion,$4^n = (1 + 3)^n = 1 + n(3) + \frac{n(n-1)}{2!} (3^2) + \dots + 3^n$.
So,$4^n - 3n - 1 = 1 + 3n + \frac{9n(n-1)}{2} + \dots - 3n - 1 = \frac{9n(n-1)}{2} + \dots = 9 \left[ \frac{n(n-1)}{2} + \dots \right]$.
This shows that every element of $X$ is a multiple of $9$,and since $n(n-1)$ is always even,$\frac{n(n-1)}{2}$ is an integer.
Thus,$X \subseteq Y$.
Alternatively,testing values:
For $n=1, X = \{4^1 - 3(1) - 1\} = \{0\}$.
For $n=2, X = \{4^2 - 3(2) - 1\} = \{16 - 6 - 1\} = \{9\}$.
For $n=3, X = \{4^3 - 3(3) - 1\} = \{64 - 9 - 1\} = \{54\}$.
$X = \{0, 9, 54, \dots\}$.
$Y = \{0, 9, 18, 27, 36, 45, 54, \dots\}$.
Since all elements of $X$ are in $Y$,$X \cap Y = X$.

(A) The line bisects the angle between the coordinate axes in the second quadrant. The angle made by this line with the positive $x$-axis is $135^{\circ}$.
Therefore,the slope of the line is $m = \tan(135^{\circ}) = -1$.
The line passes through the point $(-3, 1)$.
Using the point-slope form,the equation of the line is:
$(y - y_1) = m(x - x_1)$
$(y - 1) = -1(x - (-3))$
$y - 1 = -1(x + 3)$
$y - 1 = -x - 3$
$x + y + 2 = 0$

(B) Given equation: $\sin x + \sin 3x + \sin 5x = 0$
Grouping terms: $(\sin 5x + \sin x) + \sin 3x = 0$
Using $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:
$2 \sin 3x \cos 2x + \sin 3x = 0$
$\sin 3x (2 \cos 2x + 1) = 0$
This implies $\sin 3x = 0$ or $\cos 2x = -\frac{1}{2}$.
Case $1$: $\sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$.
For $x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$,possible values are $x = \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$.
Case $2$: $\cos 2x = -\frac{1}{2} \implies 2x = 2n\pi \pm \frac{2\pi}{3} \implies x = n\pi \pm \frac{\pi}{3}$.
For $x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$,possible values are $x = \frac{2\pi}{3}, \frac{4\pi}{3}$.
Combining both cases,the distinct solutions are $x = \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$.
Thus,the total number of solutions is $3$.

(A) The given expression is $\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \dots \cos 90^{\circ} \dots \cos 179^{\circ}$.
Since $\cos 90^{\circ} = 0$,the entire product becomes $0$ because any number multiplied by $0$ is $0$.
Therefore,$\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \dots \cos 179^{\circ} = 0$.

(D) Given the equation: $2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$
Using the expansion formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$2 \left(\sin \theta \cos \frac{\pi}{3} + \cos \theta \sin \frac{\pi}{3}\right) = \cos \theta \cos \frac{\pi}{6} + \sin \theta \sin \frac{\pi}{6}$
Substitute the values $\cos \frac{\pi}{3} = \frac{1}{2}$,$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,$\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$,and $\sin \frac{\pi}{6} = \frac{1}{2}$:
$2 \left(\sin \theta \cdot \frac{1}{2} + \cos \theta \cdot \frac{\sqrt{3}}{2}\right) = \cos \theta \cdot \frac{\sqrt{3}}{2} + \sin \theta \cdot \frac{1}{2}$
$\sin \theta + \sqrt{3} \cos \theta = \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta$
Multiply by $2$ to clear the fractions:
$2 \sin \theta + 2 \sqrt{3} \cos \theta = \sqrt{3} \cos \theta + \sin \theta$
Rearrange the terms:
$2 \sin \theta - \sin \theta = \sqrt{3} \cos \theta - 2 \sqrt{3} \cos \theta$
$\sin \theta = -\sqrt{3} \cos \theta$
Divide by $\cos \theta$:
$\tan \theta = -\sqrt{3}$

(B) Given that $A + B + C = \pi$.
Since $A + B = \pi - C$,we take the cotangent on both sides:
$\cot(A + B) = \cot(\pi - C)$.
Using the identity $\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ and $\cot(\pi - C) = -\cot C$,we get:
$\frac{\cot A \cot B - 1}{\cot A + \cot B} = -\cot C$.
Multiplying both sides by $(\cot A + \cot B)$,we obtain:
$\cot A \cot B - 1 = -\cot A \cot C - \cot B \cot C$.
Rearranging the terms,we get:
$\cot A \cot B + \cot B \cot C + \cot A \cot C = 1$.

(A) Given curve: $y^2 = ax^3 + b$.
Since the point $(2,3)$ lies on the curve,we have $3^2 = a(2)^3 + b$,which simplifies to $9 = 8a + b$ (Equation $1$).
Differentiating the curve equation with respect to $x$: $2y \frac{dy}{dx} = 3ax^2$,so $\frac{dy}{dx} = \frac{3ax^2}{2y}$.
At the point $(2,3)$,the slope of the tangent is $\left. \frac{dy}{dx} \right|_{(2,3)} = \frac{3a(2)^2}{2(3)} = \frac{12a}{6} = 2a$.
The given line is $y = 4x - 5$,which has a slope of $4$.
Equating the slopes: $2a = 4$,which gives $a = 2$.
Substituting $a = 2$ into Equation $1$: $9 = 8(2) + b \Rightarrow 9 = 16 + b \Rightarrow b = -7$.
Now,calculate $7a + 2b$: $7(2) + 2(-7) = 14 - 14 = 0$.

(D) Given $f(x) = \frac{x}{x^2+1}$.
To find the interval where $f(x)$ is increasing,we find its derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$.
For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.
Since $(x^2+1)^2$ is always positive for all real $x$,the condition $f'(x) > 0$ implies $1-x^2 > 0$.
This simplifies to $x^2 - 1 < 0$,which is $(x-1)(x+1) < 0$.
Solving this inequality,we get $x \in (-1, 1)$.

(D) Given the function $f(x) = x \log x$.
To find the minimum value,we first find the derivative $f'(x)$.
Using the product rule: $f'(x) = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) = x \cdot \frac{1}{x} + \log x = 1 + \log x$.
For critical points,set $f'(x) = 0$:
$1 + \log x = 0 \implies \log x = -1 \implies x = e^{-1} = \frac{1}{e}$.
Now,find the second derivative $f''(x)$:
$f''(x) = \frac{d}{dx}(1 + \log x) = \frac{1}{x}$.
Evaluate $f''(x)$ at $x = \frac{1}{e}$:
$f''(\frac{1}{e}) = \frac{1}{1/e} = e$.
Since $e > 0$,the function has a local minimum at $x = \frac{1}{e}$.
The minimum value is $f(\frac{1}{e}) = \frac{1}{e} \log(\frac{1}{e}) = \frac{1}{e} \log(e^{-1}) = \frac{1}{e} (-1) = -\frac{1}{e}$.

(B) The region is bounded by the parabola $x^2=4y$,the lines $y=1$ and $y=4$,and the y-axis in the first quadrant.
From $x^2=4y$,we have $x = \sqrt{4y} = 2\sqrt{y}$ (since $x > 0$ in the first quadrant).
The required area $A$ is given by the integral with respect to $y$:
$A = \int_{1}^{4} x \, dy = \int_{1}^{4} 2\sqrt{y} \, dy$
$A = 2 \int_{1}^{4} y^{1/2} \, dy$
$A = 2 \left[ \frac{y^{3/2}}{3/2} \right]_{1}^{4}$
$A = 2 \times \frac{2}{3} \left[ y^{3/2} \right]_{1}^{4}$
$A = \frac{4}{3} (4^{3/2} - 1^{3/2})$
$A = \frac{4}{3} (8 - 1)$
$A = \frac{4}{3} (7) = \frac{28}{3}$ square units.

(A) To find the maximum value of $Z = 2x + y$,we identify the corner points of the feasible region defined by the constraints $3x + 5y \leq 26$,$5x + 3y \leq 30$,$x \geq 0$,and $y \geq 0$.
$1$. Find the intersection of the lines $3x + 5y = 26$ and $5x + 3y = 30$:
Multiply the first equation by $5$ and the second by $3$:
$15x + 25y = 130$
$15x + 9y = 90$
Subtracting the equations: $16y = 40 \implies y = \frac{40}{16} = 2.5$.
Substitute $y = 2.5$ into $3x + 5(2.5) = 26 \implies 3x + 12.5 = 26 \implies 3x = 13.5 \implies x = 4.5$.
So,point $B$ is $(4.5, 2.5)$.
$2$. The corner points of the feasible region are $(0, 0)$,$(6, 0)$,$(4.5, 2.5)$,and $(0, 5.2)$.
$3$. Evaluate $Z = 2x + y$ at each corner point:
At $(0, 0): Z = 2(0) + 0 = 0$
At $(6, 0): Z = 2(6) + 0 = 12$
At $(4.5, 2.5): Z = 2(4.5) + 2.5 = 9 + 2.5 = 11.5$
At $(0, 5.2): Z = 2(0) + 5.2 = 5.2$
The maximum value is $12$ at the point $(6, 0)$.

(D) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{e^{x^2} - \cos x}{x^2}$
To evaluate this limit,we add and subtract $1$ in the numerator:
$f(0) = \lim_{x \to 0} \frac{(e^{x^2} - 1) + (1 - \cos x)}{x^2}$
$f(0) = \lim_{x \to 0} \frac{e^{x^2} - 1}{x^2} + \lim_{x \to 0} \frac{1 - \cos x}{x^2}$
Using standard limits $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$:
$f(0) = 1 + \frac{1}{2} = \frac{3}{2}$.

(C) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit and the function value at $x = 0$.
$\lim_{x \to 0^+} (x^2 + \alpha) = \lim_{x \to 0^-} (2\sqrt{x^2 + 1} + \beta)$
$0^2 + \alpha = 2\sqrt{0^2 + 1} + \beta$
$\alpha = 2 + \beta \implies \alpha - \beta = 2 . . . (1)$
Given $f(\frac{1}{2}) = 2$. Since $\frac{1}{2} \ge 0$,we use the first part of the function:
$f(\frac{1}{2}) = (\frac{1}{2})^2 + \alpha = 2$
$\frac{1}{4} + \alpha = 2$
$\alpha = 2 - \frac{1}{4} = \frac{7}{4}$
Substitute $\alpha = \frac{7}{4}$ into equation $(1)$:
$\frac{7}{4} - \beta = 2$
$\beta = \frac{7}{4} - 2 = -\frac{1}{4}$
Now,calculate $\alpha^2 + \beta^2$:
$\alpha^2 + \beta^2 = (\frac{7}{4})^2 + (-\frac{1}{4})^2$
$\alpha^2 + \beta^2 = \frac{49}{16} + \frac{1}{16} = \frac{50}{16} = \frac{25}{8}$

(B) We use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x$ and $dv = \sec^2 x \, dx$.
Then $du = dx$ and $v = \tan x$.
Applying the formula:
$\int_0^{\frac{\pi}{4}} x \sec^2 x \, dx = [x \tan x]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} \tan x \, dx$.
Evaluating the first term:
$[x \tan x]_0^{\frac{\pi}{4}} = (\frac{\pi}{4} \tan \frac{\pi}{4}) - (0 \cdot \tan 0) = \frac{\pi}{4} \cdot 1 - 0 = \frac{\pi}{4}$.
Evaluating the integral of $\tan x$:
$\int \tan x \, dx = \ln |\sec x|$.
So,$\int_0^{\frac{\pi}{4}} \tan x \, dx = [\ln |\sec x|]_0^{\frac{\pi}{4}} = \ln |\sec \frac{\pi}{4}| - \ln |\sec 0| = \ln \sqrt{2} - \ln 1 = \ln \sqrt{2} - 0 = \ln \sqrt{2}$.
Combining the results:
$\frac{\pi}{4} - \ln \sqrt{2}$.

(C) Given the integral $\int_{0}^{k} \frac{dx}{2 + 18x^2} = \frac{\pi}{24}$.
Factor out $2$ from the denominator: $\frac{1}{2} \int_{0}^{k} \frac{dx}{1 + 9x^2} = \frac{\pi}{24}$.
Multiply both sides by $2$: $\int_{0}^{k} \frac{dx}{1 + (3x)^2} = \frac{\pi}{12}$.
Using the formula $\int \frac{dx}{1 + a^2x^2} = \frac{1}{a} \tan^{-1}(ax) + C$,we get:
$\left[ \frac{1}{3} \tan^{-1}(3x) \right]_{0}^{k} = \frac{\pi}{12}$.
$\frac{1}{3} \tan^{-1}(3k) - \frac{1}{3} \tan^{-1}(0) = \frac{\pi}{12}$.
Since $\tan^{-1}(0) = 0$,we have $\frac{1}{3} \tan^{-1}(3k) = \frac{\pi}{12}$.
$\tan^{-1}(3k) = \frac{\pi}{4}$.
$3k = \tan(\frac{\pi}{4}) = 1$.
Therefore,$k = \frac{1}{3}$.

(D) The general equation of a parabola with latus rectum $4a$ and axis parallel to the $x$-axis is given by $(y-k)^2 = 4a(x-h)$.
Here,$a$ is a fixed parameter (given as the latus rectum length),while $h$ and $k$ are arbitrary constants representing the coordinates of the vertex $(h, k)$.
Since there are $2$ arbitrary constants,the order of the differential equation is equal to the number of arbitrary constants.
Therefore,the order of the differential equation is $2$.

(B) Given $y = (\tan^{-1} x)^2$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{1}{1+x^2}$.
Multiplying both sides by $(1+x^2)$,we have:
$(1+x^2) \frac{dy}{dx} = 2 \tan^{-1} x$.
Differentiating again with respect to $x$ using the product rule on the left side:
$(1+x^2) \frac{d^2 y}{dx^2} + \frac{dy}{dx} \cdot (2x) = 2 \cdot \frac{1}{1+x^2}$.
Multiplying the entire equation by $(1+x^2)$,we get:
$(1+x^2)^2 \frac{d^2 y}{dx^2} + 2x(1+x^2) \frac{dy}{dx} = 2$.
Thus,the value is $2$.

(B) Let $x+y = v$.
Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$.
Substituting this into the given differential equation: $\frac{dv}{dx} - 1 = \cos v$.
Rearranging the terms,we get $\frac{dv}{dx} = 1 + \cos v$.
Using the trigonometric identity $1 + \cos v = 2 \cos^2 \left(\frac{v}{2}\right)$,we have $\frac{dv}{dx} = 2 \cos^2 \left(\frac{v}{2}\right)$.
Separating the variables: $\frac{dv}{2 \cos^2 \left(\frac{v}{2}\right)} = dx$,which simplifies to $\frac{1}{2} \sec^2 \left(\frac{v}{2}\right) dv = dx$.
Integrating both sides: $\int \frac{1}{2} \sec^2 \left(\frac{v}{2}\right) dv = \int dx$.
This gives $\tan \left(\frac{v}{2}\right) = x + c$.
Substituting $v = x+y$ back,we get $\tan \left(\frac{x+y}{2}\right) = x + c$.

(A) Given $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$.
Differentiating $x$ with respect to $\theta$ using the product rule:
$\frac{dx}{d\theta} = e^\theta(\cos \theta + \sin \theta) + e^\theta(\sin \theta - \cos \theta) = e^\theta(\cos \theta + \sin \theta + \sin \theta - \cos \theta) = 2e^\theta \sin \theta$.
Differentiating $y$ with respect to $\theta$ using the product rule:
$\frac{dy}{d\theta} = e^\theta(\cos \theta - \sin \theta) + e^\theta(\sin \theta + \cos \theta) = e^\theta(\cos \theta - \sin \theta + \sin \theta + \cos \theta) = 2e^\theta \cos \theta$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2e^\theta \cos \theta}{2e^\theta \sin \theta} = \cot \theta$.
Evaluating at $\theta = \frac{\pi}{4}$:
$\left. \frac{dy}{dx} \right|_{\theta=\frac{\pi}{4}} = \cot \left( \frac{\pi}{4} \right) = 1$.

(D) Given,$\log _{10}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2$
$\Rightarrow \frac{x^3-y^3}{x^3+y^3} = 10^2 = 100$
$\Rightarrow x^3 - y^3 = 100x^3 + 100y^3$
$\Rightarrow -99x^3 = 101y^3$
$\Rightarrow y^3 = -\frac{99}{101}x^3$
Differentiating both sides with respect to $x$:
$3y^2 \frac{dy}{dx} = -\frac{99}{101} \cdot 3x^2$
$\frac{dy}{dx} = -\frac{99}{101} \cdot \frac{x^2}{y^2}$
Since $y^3 = -\frac{99}{101}x^3$,we have $-\frac{99}{101} = \frac{y^3}{x^3}$
Substituting this into the derivative:
$\frac{dy}{dx} = \frac{y^3}{x^3} \cdot \frac{x^2}{y^2} = \frac{y}{x}$

(C) Given the function $f(x) = \frac{x^2-4}{x-2}$.
Since the domain is $R-\{2\}$,we can simplify the expression for $x \neq 2$:
$f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$.
As $x$ can take any real value except $2$,the value of $x+2$ can take any real value except $2+2 = 4$.
Therefore,the range of the function is $R-\{4\}$.

(D) $I = \int \frac{1}{\sin x \cdot \cos^2 x} \, dx$
$I = \int \frac{\sin^2 x + \cos^2 x}{\sin x \cdot \cos^2 x} \, dx$
$I = \int \frac{\sin^2 x}{\sin x \cdot \cos^2 x} \, dx + \int \frac{\cos^2 x}{\sin x \cdot \cos^2 x} \, dx$
$I = \int \frac{\sin x}{\cos^2 x} \, dx + \int \frac{1}{\sin x} \, dx$
$I = \int \tan x \sec x \, dx + \int \operatorname{cosec} x \, dx$
$I = \sec x + \ln |\operatorname{cosec} x - \cot x| + c$

(A) Given integral: $I = \int e^x \left[ \frac{2 + \sin 2x}{1 + \cos 2x} \right] dx$
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$:
$I = \int e^x \left[ \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} \right] dx$
$I = \int e^x \left[ \frac{2(1 + \sin x \cos x)}{2 \cos^2 x} \right] dx$
$I = \int e^x \left[ \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right] dx$
$I = \int e^x (\sec^2 x + \tan x) dx$
Since $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$,where $f(x) = \tan x$ and $f'(x) = \sec^2 x$:
$I = e^x \tan x + C$

(D) We know that $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + C$.
Given integral is $I = \int \frac{dx}{\sqrt{4^2 - (3x)^2}}$.
Using the substitution $u = 3x$,we have $du = 3dx$,so $dx = \frac{du}{3}$.
$I = \int \frac{du/3}{\sqrt{4^2 - u^2}} = \frac{1}{3} \int \frac{du}{\sqrt{4^2 - u^2}}$.
$I = \frac{1}{3} \sin^{-1}(\frac{u}{4}) + C = \frac{1}{3} \sin^{-1}(\frac{3x}{4}) + C$.
Comparing this with $A \sin^{-1}(Bx) + C$,we get $A = \frac{1}{3}$ and $B = \frac{3}{4}$.
Therefore,$A + B = \frac{1}{3} + \frac{3}{4} = \frac{4 + 9}{12} = \frac{13}{12}$.

(C) Given the equation: $\tan ^{-1} 2x + \tan ^{-1} 3x = \frac{\pi}{4}$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan ^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} \right) = \frac{\pi}{4}$
$\tan ^{-1} \left( \frac{5x}{1 - 6x^2} \right) = \frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{5x}{1 - 6x^2} = \tan \left( \frac{\pi}{4} \right) = 1$
$5x = 1 - 6x^2$
$6x^2 + 5x - 1 = 0$
Factoring the quadratic equation:
$6x^2 + 6x - x - 1 = 0$
$6x(x + 1) - 1(x + 1) = 0$
$(6x - 1)(x + 1) = 0$
This gives $x = \frac{1}{6}$ or $x = -1$.
Since $\tan ^{-1} 2x + \tan ^{-1} 3x = \frac{\pi}{4} > 0$,$x$ must be positive. Therefore,$x = -1$ is rejected.
Thus,$x = \frac{1}{6}$.

(C) We know that the sum of the product of elements of any row (or column) with their corresponding cofactors is equal to the determinant of the matrix,i.e.,$a_{i1} A_{i1} + a_{i2} A_{i2} + a_{i3} A_{i3} = |A|$.
Here,$a_{31} A_{31} + a_{32} A_{32} + a_{33} A_{33} = |A|$.
Now,we calculate the determinant of $A$ by expanding along the first row:
$|A| = 1(1 \times 7 - 5 \times 4) - 2(1 \times 7 - 5 \times 2) + 3(1 \times 4 - 1 \times 2)$
$|A| = 1(7 - 20) - 2(7 - 10) + 3(4 - 2)$
$|A| = 1(-13) - 2(-3) + 3(2)$
$|A| = -13 + 6 + 6$
$|A| = -1$
Therefore,the value is $-1$.

(B) Given the expression $(A^2 - 5A)A^{-1}$.
Distributing $A^{-1}$ inside the parentheses,we get:
$(A^2 \cdot A^{-1}) - (5A \cdot A^{-1})$
Since $A^2 \cdot A^{-1} = A$ and $A \cdot A^{-1} = I$ (where $I$ is the identity matrix),the expression simplifies to:
$A - 5I$
Now,substitute the matrix $A$ and the identity matrix $I$:
$\begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$= \begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$= \begin{bmatrix} 1-5 & 2-0 & 3-0 \\ -1-0 & 1-5 & 2-0 \\ 1-0 & 2-0 & 4-5 \end{bmatrix}$
$= \begin{bmatrix} -4 & 2 & 3 \\ -1 & -4 & 2 \\ 1 & 2 & -1 \end{bmatrix}$

(C) The possible outcomes on a die are ${1, 2, 3, 4, 5, 6}$.
The perfect squares among these are ${1, 4}$.
Thus,the probability of getting a perfect square in a single throw is $p = \frac{2}{6} = \frac{1}{3}$.
The probability of not getting a perfect square in a single throw is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
For $n = 4$ independent throws,the probability of not getting a perfect square in any of the four throws is $q^4 = (\frac{2}{3})^4 = \frac{16}{81}$.
The probability of getting a perfect square in at least one throw is $1 - P(\text{no perfect square}) = 1 - \frac{16}{81} = \frac{65}{81}$.

(D) When a coin is tossed three times,the total number of outcomes is $2^3 = 8$. The sample space is $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$.
Let $n(H)$ be the number of heads and $n(T)$ be the number of tails. $X = |n(H) - n(T)|$.
For $HHH: n(H)=3, n(T)=0, X=|3-0|=3$.
For $HHT: n(H)=2, n(T)=1, X=|2-1|=1$.
For $HTH: n(H)=2, n(T)=1, X=|2-1|=1$.
For $HTT: n(H)=1, n(T)=2, X=|1-2|=1$.
For $THH: n(H)=2, n(T)=1, X=|2-1|=1$.
For $THT: n(H)=1, n(T)=2, X=|1-2|=1$.
For $TTH: n(H)=1, n(T)=2, X=|1-2|=1$.
For $TTT: n(H)=0, n(T)=3, X=|0-3|=3$.
The outcomes where $X=1$ are $\{HHT, HTH, HTT, THH, THT, TTH\}$.
There are $6$ such outcomes.
Therefore,$P(X=1) = \frac{6}{8} = \frac{3}{4}$.

(D) Given $X \sim B(n, p)$ with $n = 10$ and $p = 0.4$.
Since $q = 1 - p$,we have $q = 1 - 0.4 = 0.6$.
The mean of a binomial distribution is $E(X) = np = 10 \times 0.4 = 4$.
The variance of a binomial distribution is $V(X) = npq = 10 \times 0.4 \times 0.6 = 2.4$.
We know the relationship $V(X) = E(X^2) - (E(X))^2$.
Substituting the values,we get $2.4 = E(X^2) - (4)^2$.
$2.4 = E(X^2) - 16$.
Therefore,$E(X^2) = 16 + 2.4 = 18.4$.

(B) The possible outcomes of rolling a die are $\{1, 2, 3, 4, 5, 6\}$. We determine the number of positive divisors for each outcome as follows:

Outcome	Positive Divisors	Number of Divisors $(X)$
$1$	$1$	$1$
$2$	$1, 2$	$2$
$3$	$1, 3$	$2$
$4$	$1, 2, 4$	$3$
$5$	$1, 5$	$2$
$6$	$1, 2, 3, 6$	$4$

The set of values taken by the random variable $X$ is the range. From the table,the values are $\{1, 2, 3, 4\}$. Thus,the range of $X$ is $\{1, 2, 3, 4\}$.

(C) Let the direction angles be $\alpha, \beta, \gamma$ with $X, Y, Z$ axes respectively.
Given $\alpha = 120^{\circ}$ and $\gamma = 60^{\circ}$.
We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values: $\cos^2(120^{\circ}) + \cos^2 \beta + \cos^2(60^{\circ}) = 1$.
$(-\frac{1}{2})^2 + \cos^2 \beta + (\frac{1}{2})^2 = 1$.
$\frac{1}{4} + \cos^2 \beta + \frac{1}{4} = 1$.
$\cos^2 \beta = 1 - \frac{1}{2} = \frac{1}{2}$.
$\cos \beta = \pm \frac{1}{\sqrt{2}}$.
Thus,$\beta = 45^{\circ}$ or $\beta = 135^{\circ}$.
Since $135^{\circ}$ is one of the options,the correct answer is $135^{\circ}$.

(D) The direction ratios (drs) of line segment $PQ$ are given by $(3-4, y-5, 4-x) = (-1, y-5, 4-x)$.
The direction ratios (drs) of line segment $QR$ are given by $(5-3, 8-y, 0-4) = (2, 8-y, -4)$.
Since points $P, Q,$ and $R$ are collinear,the direction ratios must be proportional:
$\frac{-1}{2} = \frac{y-5}{8-y} = \frac{4-x}{-4}$.
From $\frac{-1}{2} = \frac{y-5}{8-y}$:
$-1(8-y) = 2(y-5)$
$-8+y = 2y-10$
$y = 2$.
From $\frac{-1}{2} = \frac{4-x}{-4}$:
$4 = 2(4-x)$
$4 = 8-2x$
$2x = 4$
$x = 2$.
Therefore,$x+y = 2+2 = 4$.

(C) Let the direction vectors of the two given lines be $\vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} - 2\hat{j} + 2\hat{k}$.
Since the required line is perpendicular to both lines,its direction vector $\vec{b}$ is given by the cross product $\vec{b_1} \times \vec{b_2}$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -2 & 2 \end{vmatrix} = \hat{i}(-4 + 2) - \hat{j}(4 - 1) + \hat{k}(-4 + 2) = -2\hat{i} - 3\hat{j} - 2\hat{k}$.
We can take the direction ratios as $(2, 3, 2)$ by multiplying by $-1$.
The line passes through the point $(3, -1, 2)$.
Thus,the equation of the line is $\frac{x - 3}{2} = \frac{y - (-1)}{3} = \frac{z - 2}{2}$,which simplifies to $\frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 2}{2}$.

(A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ intersect if and only if the determinant of the matrix formed by the differences of their points and their direction ratios is zero:
$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$
Given the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z-0}{1}$,we have $(x_1, y_1, z_1) = (1, -1, 1)$ and $(x_2, y_2, z_2) = (3, k, 0)$.
The direction ratios are $(a_1, b_1, c_1) = (2, 3, 4)$ and $(a_2, b_2, c_2) = (1, 2, 1)$.
Substituting these into the determinant condition:
$\left|\begin{array}{ccc} 3-1 & k-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$
$\Rightarrow \left|\begin{array}{ccc} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right| = 0$
Expanding along the first row:
$2(3(1) - 4(2)) - (k+1)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$
$2(3-8) - (k+1)(2-4) - 1(4-3) = 0$
$2(-5) - (k+1)(-2) - 1(1) = 0$
$-10 + 2k + 2 - 1 = 0$
$2k - 9 = 0$
$2k = 9$
$k = \frac{9}{2}$

(C) Since the given planes pass through a straight line,the determinant of the coefficients of the planes must be zero.
$\begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1( (-1)(-1) - (a)(a) ) - (-c)( (c)(-1) - (a)(b) ) + (-b)( (c)(a) - (-1)(b) ) = 0$
$1(1 - a^2) + c(-c - ab) - b(ac + b) = 0$
$1 - a^2 - c^2 - abc - abc - b^2 = 0$
$1 - a^2 - b^2 - c^2 - 2abc = 0$
Therefore,$a^2 + b^2 + c^2 = 1 - 2abc$.

(D) The normal vectors to the planes are $\vec{n}_1 = p \hat{i} - \hat{j} + 2 \hat{k}$ and $\vec{n}_2 = 2 \hat{i} - p \hat{j} - \hat{k}$.
Given that the angle between the planes is $\theta = \frac{\pi}{3}$,the cosine of the angle between their normals is given by $\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$.
$\vec{n}_1 \cdot \vec{n}_2 = (p)(2) + (-1)(-p) + (2)(-1) = 2p + p - 2 = 3p - 2$.
$|\vec{n}_1| = \sqrt{p^2 + (-1)^2 + 2^2} = \sqrt{p^2 + 5}$.
$|\vec{n}_2| = \sqrt{2^2 + (-p)^2 + (-1)^2} = \sqrt{p^2 + 5}$.
Thus,$\cos \left( \frac{\pi}{3} \right) = \frac{|3p - 2|}{\sqrt{p^2 + 5} \sqrt{p^2 + 5}}$.
$\frac{1}{2} = \frac{|3p - 2|}{p^2 + 5}$.
Case $1$: $3p - 2 = \frac{1}{2}(p^2 + 5) \implies 6p - 4 = p^2 + 5 \implies p^2 - 6p + 9 = 0 \implies (p - 3)^2 = 0 \implies p = 3$.
Case $2$: $-(3p - 2) = \frac{1}{2}(p^2 + 5) \implies -6p + 4 = p^2 + 5 \implies p^2 + 6p + 1 = 0 \implies p = \frac{-6 \pm \sqrt{36 - 4}}{2} = -3 \pm 2\sqrt{2}$.
Since $3$ is the only option provided in the choices,the value of $p$ is $3$.

(C) Given the position vectors of points $L$ and $M$ are $\vec{l} = 2\vec{a} - \vec{b}$ and $\vec{m} = \vec{a} + 2\vec{b}$ respectively.
Point $N$ divides the line segment $LM$ externally in the ratio $m:n = 2:1$.
The formula for the position vector of a point dividing a line segment externally is $\vec{n} = \frac{m\vec{m} - n\vec{l}}{m - n}$.
Substituting the values,we get:
$\vec{n} = \frac{2(\vec{a} + 2\vec{b}) - 1(2\vec{a} - \vec{b})}{2 - 1}$
$\vec{n} = \frac{2\vec{a} + 4\vec{b} - 2\vec{a} + \vec{b}}{1}$
$\vec{n} = 5\vec{b}$.