Given curve: \(x^2 y^2 - 2x = 4(1 - y)\). Differentiating both sides with respect to \(x\):

\(2xy^2 + 2x^2 y\dfrac{dy}{dx} - 2 = -4\dfrac{dy}{dx}\)

\(\dfrac{dy}{dx}(2x^2 y + 4) = 2 - 2xy^2\)

\(\dfrac{dy}{dx} = \dfrac{2 - 2xy^2}{2x^2 y + 4}\)

At point \((2, -2)\):

\(\dfrac{dy}{dx} = \dfrac{2 - 2(2)(-2)^2}{2(2)^2(-2) + 4} = \dfrac{2 - 16}{-16 + 4} = \dfrac{-14}{-12} = \dfrac{7}{6}\)

Equation of tangent at \((2, -2)\):

\(y + 2 = \dfrac{7}{6}(x - 2) \Rightarrow 6y + 12 = 7x - 14 \Rightarrow 7x - 6y = 26\)

Checking each option: Only choice (A) \((-2, -7)\) does not satisfy the equation.

Verification: \(7(-2) - 6(-7) = -14 + 42 = 28 \ne 26\)

Simplify the expression inside the square root:

\(1 + 2\cot x(\text{cosec} x + \cot x) = 1 + 2\cdot\dfrac{\cos x}{\sin x}\left(\dfrac{1}{\sin x} + \dfrac{\cos x}{\sin x}\right)\)

\(= 1 + \dfrac{2\cos x}{\sin x}\cdot\dfrac{1 + \cos x}{\sin x} = 1 + \dfrac{2\cos x(1 + \cos x)}{\sin^2 x}\)

\(= 1 + \dfrac{2\cos x(1 + \cos x)}{1 - \cos^2 x} = 1 + \dfrac{2\cos x(1 + \cos x)}{(1 - \cos x)(1 + \cos x)}\)

\(= 1 + \dfrac{2\cos x}{1 - \cos x} = \dfrac{1 - \cos x + 2\cos x}{1 - \cos x} = \dfrac{1 + \cos x}{1 - \cos x}\)

Using identities: \(1 + \cos x = 2\cos^2\dfrac{x}{2}\) and \(1 - \cos x = 2\sin^2\dfrac{x}{2}\)

Thus the expression becomes \(\dfrac{2\cos^2\dfrac{x}{2}}{2\sin^2\dfrac{x}{2}} = \cot^2\dfrac{x}{2}\)

Therefore \(\sqrt{\cot^2\dfrac{x}{2}} = \left|\cot\dfrac{x}{2}\right| = \cot\dfrac{x}{2}\) (since \(0 < x < \dfrac{\pi}{2}\))

The integral becomes \(\int \cot\dfrac{x}{2}\, dx = 2\int \cot t\, dt\) where \(t = \dfrac{x}{2}\)

\(= 2\log|\sin t| + C = 2\log\left|\sin\dfrac{x}{2}\right| + C\)

The word QUEEN has letters: E, E, N, Q, U. In alphabetical order: E, E, N, Q, U.

Words starting with E: Fix E at first place. Remaining letters Q, U, E, N (4 distinct letters) can be arranged in \(4! = 24\) ways.

Words starting with N: Fix N at first place. Remaining letters Q, U, E, E (with E repeating twice) can be arranged in \(\dfrac{4!}{2!} = 12\) ways.

Words starting with QE: Fix QE at first two places, the remaining U, E and N can be arranged in \(3! = 6\) ways.

Words starting with QN: Fix QN at the starting two places, the remaining letters U, E, E can be arranged in \(\dfrac{3!}{2!} = 3\) ways.

Next word is QUEEN.

Number of words before QUEEN is: \(24 + 12 + 6 + 3 = 45\)

Hence QUEEN is the 46th word in dictionary order.

Given equation: \(y\, dx = (x + 3y^2)\, dy\)

\(\Rightarrow \dfrac{dx}{dy} = \dfrac{x + 3y^2}{y} = \dfrac{x}{y} + 3y\)

This is a linear differential equation in \(x\): \(\dfrac{dx}{dy} - \dfrac{1}{y}x = 3y\)

Integrating factor \(IF = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \dfrac{1}{y}\)

Multiplying both sides by IF:

\(\dfrac{1}{y}\dfrac{dx}{dy} - \dfrac{1}{y^2}x = 3\)

Integrating: \(\dfrac{x}{y} = \int 3\, dy = 3y + C\)

So \(x = 3y^2 + Cy\)

Using point \((1, 1)\): \(1 = 3(1)^2 + C(1) \Rightarrow C = -2\)

Thus the curve is: \(x = 3y^2 - 2y\)

Checking choice (C): \(x = 3\left(\dfrac{1}{3}\right)^2 - 2\left(\dfrac{1}{3}\right) = \dfrac{1}{3} - \dfrac{2}{3} = -\dfrac{1}{3}\) ✓

Rationalize both numerator and denominator:

\(\displaystyle \lim_{x \to 3} \dfrac{(\sqrt{3x} - 3)(\sqrt{3x} + 3)(\sqrt{2x - 4} + \sqrt{2})}{(\sqrt{3x}+ 3)(\sqrt{2x - 4} - \sqrt{2})(\sqrt{2x - 4} + \sqrt{2})}\)

\(= \displaystyle \lim_{x \to 3} \dfrac{(3x - 9)(\sqrt{2x - 4} + \sqrt{2})}{(\sqrt{3x}+ 3)(2x - 4 - 2)}\)

\(= \displaystyle \lim_{x \to 3}\dfrac{3(x - 3)(\sqrt{2x - 4} + \sqrt{2})}{(\sqrt{3x}+ 3) \cdot 2(x - 3)}\)

\(= \displaystyle \lim_{x \to 3} \dfrac{3(\sqrt{2x - 4} + \sqrt{2})}{2(\sqrt{3x}+ 3)}\)

\(= \dfrac{3(\sqrt{2} + \sqrt{2})}{2(3 + 3)} = \dfrac{3 \cdot 2\sqrt{2}}{2 \cdot 6} = \dfrac{6\sqrt{2}}{12} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}\)

Let the two unknown observations be \(x\) and \(y\).

Mean = 5 gives \(\dfrac{1+2+6+x+y}{5} = 5 \Rightarrow x + y = 16\)

Variance = 124 gives \(\dfrac{\sum x_i^2}{5} - (5)^2 = 124 \Rightarrow \dfrac{1^2+2^2+6^2+x^2+y^2}{5} = 149\)

\(\dfrac{1+4+36+x^2+y^2}{5} = 149 \Rightarrow 41 + x^2 + y^2 = 745 \Rightarrow x^2 + y^2 = 704\)

Now \((x+y)^2 = x^2 + y^2 + 2xy \Rightarrow 256 = 704 + 2xy \Rightarrow 2xy = -448 \Rightarrow xy = -224\)

Thus \(x\) and \(y\) are roots of \(t^2 - 16t - 224 = 0\)

Solving: \(t = \dfrac{16 \pm \sqrt{256 + 896}}{2} = \dfrac{16 \pm \sqrt{1152}}{2} = \dfrac{16 \pm 24\sqrt{2}}{2} = 8 \pm 12\sqrt{2}\)

Both values are greater than 5.

Mean deviation from mean = \(\dfrac{|1-5| + |2-5| + |6-5| + |x-5| + |y-5|}{5}\)

\(= \dfrac{4 + 3 + 1 + (x-5) + (y-5)}{5} = \dfrac{8 + (x+y-10)}{5} = \dfrac{8 + (16-10)}{5} = \dfrac{14}{5} = 2.8\)

Let total number of experts be \(n\).

Number who voted for A = \(\dfrac{n}{2}\), for B = \(\dfrac{2n}{3}\)

Both = 10, neither = 6

Using the principle of inclusion-exclusion:

\(n = n(A) + n(B) - n(A \cap B) + n(\text{neither})\)

\(n = \dfrac{n}{2} + \dfrac{2n}{3} - 10 + 6\)

\(n = \dfrac{3n + 4n}{6} - 4 \Rightarrow n = \dfrac{7n}{6} - 4\)

Multiplying by 6: \(6n = 7n - 24 \Rightarrow n = 24\)

Given: \(\alpha (2\vec{a} - \vec{b}) + \beta (\vec{a} + 2\vec{b}) = 8\vec{b} - \vec{a}\)

Expanding the left-hand side:

\((2\alpha + \beta)\vec{a} + (-\alpha + 2\beta)\vec{b} = -\vec{a} + 8\vec{b}\)

Comparing coefficients of \(\vec{a}\) and \(\vec{b}\) on both sides:

For \(\vec{a}\): \(2\alpha + \beta = -1\) ... (1)

For \(\vec{b}\): \(-\alpha + 2\beta = 8\) ... (2)

From equation (2): \(\alpha = 2\beta - 8\)

Substituting in equation (1): \(2(2\beta - 8) + \beta = -1\)

\(4\beta - 16 + \beta = -1 \Rightarrow 5\beta = 15 \Rightarrow \beta = 3\)

Therefore \(\alpha = 2(3) - 8 = -2\)

Force vector \(\vec{F} = 78 \cdot \dfrac{(2,2,1)}{\sqrt{2^2+2^2+1^2}} = 78 \cdot \dfrac{(2,2,1)}{3} = (52, 52, 26)\)

Point of application \(P = (2,3,5)\)

Moment about origin: \(\vec{M}_O = \vec{r} \times \vec{F}\) where \(\vec{r} = (2,3,5)\)

\(\vec{M}_O = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ 52 & 52 & 26 \end{vmatrix}\)

\(= \hat{i}(78 - 260) - \hat{j}(52 - 260) + \hat{k}(104 - 156)\)

\(= (-182, 208, -52)\)

Direction vector from origin to \(Q(12,3,4)\): \(\vec{d} = (12,3,4)\)

Unit vector: \(\hat{d} = \dfrac{(12,3,4)}{13}\)

Moment about the line = \(|\vec{M}_O \cdot \hat{d}|\)

\(= \left|\dfrac{(-182)(12) + (208)(3) + (-52)(4)}{13}\right| = \left|\dfrac{-2184 + 624 - 208}{13}\right| = \left|\dfrac{-1768}{13}\right| = 136\)

For all real \(x\), \(5^x > 0\) and \(5^{-x} > 0\).

By AM-GM inequality: \(5^x + 5^{-x} \ge 2\sqrt{5^x \cdot 5^{-x}} = 2\sqrt{1} = 2\)

So the right-hand side \(\ge 2\).

The left-hand side is \(\sin(e^x)\), which lies between \(-1\) and \(1\) for all real \(x\). Therefore the equation has no real solutions.

First circle: \(x^2 + y^2 = 16\) has centre \(C_1(0,0)\) and radius \(r_1 = 4\).

Second circle: \(x^2 + y^2 - 2y = 0 \Rightarrow x^2 + (y-1)^2 = 1\) has centre \(C_2(0,1)\) and radius \(r_2 = 1\).

Distance between centres: \(d = \sqrt{(0-0)^2 + (0-1)^2} = 1\)

Now \(r_1 + r_2 = 5\) and \(|r_1 - r_2| = 3\)

Since \(d = 1 < 3 = |r_1 - r_2|\), the smaller circle lies completely inside the larger circle without touching.

In this case, there are no common tangents.

Check continuity at \(x = 0\):

Left-hand limit: \(\displaystyle \lim_{x\to 0^-} f(x) = 0\) (since \(f(x)=0\) for \(x\leq 0\))

Right-hand limit: \(\displaystyle \lim_{x\to 0^+} x\sin\left(\dfrac{1}{x}\right) = 0\) (since \(\left|\sin\left(\dfrac{1}{x}\right)\right|\leq 1\) and \(x\to 0\))

Also \(f(0) = 0\). Therefore \(f\) is continuous at \(x=0\).

Check differentiability at \(x = 0\):

Left-hand derivative: \(\displaystyle \lim_{h\to 0^-} \dfrac{f(0+h)-f(0)}{h} = \displaystyle \lim_{h\to 0^-} \dfrac{0-0}{h} = 0\)

Right-hand derivative: \(\displaystyle \lim_{h\to 0^+} \dfrac{h\sin(1/h)-0}{h} = \displaystyle \lim_{h\to 0^+} \sin(1/h)\)

This limit does not exist as it oscillates between -1 and 1.

Hence \(f\) is not differentiable at \(x=0\).

Starting point: \(z_0 = 1 + 2i\)

Moving horizontally away from origin by 5 units means increasing the real part by 5:

\(z = (1+5) + 2i = 6 + 2i\)

Then vertically away from origin by 3 units means increasing the imaginary part by 3:

\(z_1 = 6 + (2+3)i = 6 + 5i\)

From \(z_1\), moving \(\sqrt{2}\) units in the direction of \(\hat{i} + \hat{j}\) (unit vector \(\dfrac{1}{\sqrt{2}}(\hat{i}+\hat{j})\)) means adding:

\(\sqrt{2} \times \dfrac{1}{\sqrt{2}}(1+i) = 1 + i\)

So new point: \(z = (6+1) + (5+1)i = 7 + 6i\)

Now moving through an angle \(\dfrac{\pi}{2}\) anti-clockwise on a circle with centre at origin means multiplying by \(e^{i\pi/2} = i\):

\(z_2 = i \times (7 + 6i) = 7i + 6i^2 = 7i - 6 = -6 + 7i\)

Area of triangle \(\Delta = \dfrac{1}{2}bc\sin A\). Also given \(\Delta = a^2 - (b-c)^2\).

Using cosine rule: \(a^2 = b^2 + c^2 - 2bc\cos A\)

Then \(a^2 - (b-c)^2 = (b^2 + c^2 - 2bc\cos A) - (b^2 + c^2 - 2bc) = 2bc(1 - \cos A)\)

Thus \(\dfrac{1}{2}bc\sin A = 2bc(1 - \cos A)\)

Assuming \(bc \neq 0\): \(\dfrac{1}{2}\sin A = 2(1 - \cos A)\)

\(\sin A = 4(1 - \cos A)\)

Using \(\sin A = 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}\) and \(1 - \cos A = 2\sin^2\dfrac{A}{2}\):

\(2\sin\dfrac{A}{2}\cos\dfrac{A}{2} = 4 \times 2\sin^2\dfrac{A}{2} = 8\sin^2\dfrac{A}{2}\)

If \(\sin\dfrac{A}{2} \neq 0\), divide by \(2\sin\dfrac{A}{2}\): \(\cos\dfrac{A}{2} = 4\sin\dfrac{A}{2}\)

Thus \(\tan\dfrac{A}{2} = \dfrac{1}{4}\)

Then \(\tan A = \dfrac{2\tan\dfrac{A}{2}}{1 - \tan^2\dfrac{A}{2}} = \dfrac{2 \times \dfrac{1}{4}}{1 - \dfrac{1}{16}} = \dfrac{\dfrac{1}{2}}{\dfrac{15}{16}} = \dfrac{1}{2} \times \dfrac{16}{15} = \dfrac{8}{15}\)

Total number of ways to choose two distinct numbers from 1 to 30 is \(\binom{30}{2} = 435\).

We need \(a^2 - b^2 = (a-b)(a+b)\) divisible by 3. This happens if either both the numbers \(a\) and \(b\) are divisible by 3 or none of them is divisible by 3. Because when none of them is a multiple of 3, either their sum or difference will be a multiple of 3.

As there are 10 numbers which are divisible by 3 and there are 20 numbers which are not divisible by 3. Hence total number of favourable ways = \(\binom{10}{2} + \binom{20}{2} = 235\)

Probability = \(\dfrac{235}{435} = \dfrac{47}{87}\)

Given: \(\sin^2 x\tan x + \cos^2 x\cot x - 2\sin 2x = 1 + \tan x + \cot x\)

Rewriting: \(\dfrac{\sin^3 x}{\cos x} + \dfrac{\cos^3 x}{\sin x} - 2\sin 2x = 1 + \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x}\)

\(\dfrac{\sin^4 x + \cos^4 x}{\sin x\cos x} - 2\sin 2x = 1 + \dfrac{1}{\sin x\cos x}\)

Now \(\sin^4 x + \cos^4 x = 1 - 2\sin^2 x\cos^2 x\)

So LHS = \(\dfrac{1 - 2\sin^2 x\cos^2 x}{\sin x\cos x} - 2\sin 2x = \dfrac{1}{\sin x\cos x} - 2\sin x\cos x - 4\sin x\cos x\)

\(= \dfrac{1}{\sin x\cos x} - 6\sin x\cos x\)

RHS = \(1 + \dfrac{1}{\sin x\cos x}\)

Equating: \(-6\sin x\cos x = 1\)

\(\sin x\cos x = -\dfrac{1}{6}\)

\(\dfrac{1}{2}\sin 2x = -\dfrac{1}{6} \Rightarrow \sin 2x = -\dfrac{1}{3}\)

In \((0,\pi)\), \(2x \in (0,2\pi)\). For \(\sin 2x = -\dfrac{1}{3}\), the solutions are \(2x = \pi + \arcsin\dfrac{1}{3}\) or \(2x = 2\pi - \arcsin\dfrac{1}{3}\)

This gives \(x = \dfrac{5\pi}{12}, \dfrac{11\pi}{12}\)

Given \(|\vec{a}| = 2\), \(\vec{a}\cdot\vec{b} = 1\), angle = \(\dfrac{\pi}{3}\)

From \(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\dfrac{\pi}{3}\): \(1 = 2|\vec{b}| \cdot \dfrac{1}{2} = |\vec{b}|\)

Thus \(|\vec{b}| = 1\)

\(|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\dfrac{\pi}{3} = 2 \cdot 1 \cdot \dfrac{\sqrt{3}}{2} = \sqrt{3}\)

Let \(\vec{r} + 2\vec{a} - 10\vec{b} = \lambda(\vec{a}\times\vec{b})\)

Taking dot product with \(\vec{a}\times\vec{b}\):

\((\vec{r} + 2\vec{a} - 10\vec{b})\cdot(\vec{a}\times\vec{b}) = \lambda |\vec{a}\times\vec{b}|^2\)

Given LHS = 6:

\(6 = \lambda (\sqrt{3})^2 = 3\lambda\)

Therefore \(\lambda = 2\)

Total players = \(n + 2\). Each pair plays 2 games.

Games among men = \(2 \times \binom{n}{2} = 2 \times \dfrac{n(n-1)}{2} = n(n-1)\)

Games between men and women: Each man plays 2 games with each woman.

Number of such games = \(2 \times (n \times 2) = 4n\)

Given: Games among men = Games (men vs women) + 66

\(n(n-1) = 4n + 66\)

\(n^2 - n = 4n + 66\)

\(n^2 - 5n - 66 = 0\)

\((n - 11)(n + 6) = 0\)

Since \(n > 0\): \(n = 11\)

Total players = \(11 + 2 = 13\)

Total elements in S can be counted using the A's or B's.

From A's:

Each \(A_i\) has 5 elements, total element occurrences = \(30 \times 5 = 150\)

Each element belongs to exactly 10 A's, so number of distinct elements = \(\dfrac{150}{10} = 15\)

From B's:

Each \(B_j\) has 3 elements, total element occurrences = \(n \times 3 = 3n\)

Each element belongs to exactly 9 B's, so number of distinct elements = \(\dfrac{3n}{9} = \dfrac{n}{3}\)

Since both represent the same set S:

\(15 = \dfrac{n}{3}\)

Therefore \(n = 45\)

For continuity at \(x = 0\), we need \(\displaystyle \lim_{x\to 0^-} f(x) = f(0) = \displaystyle \lim_{x\to 0^+} f(x)\)

\(f(0) = \cos[0] = \cos 0 = 1\)

Right-hand limit: \(\displaystyle \lim_{x\to 0^+} \cos[x] = \cos 0 = 1\) (since for \(0 \leq x < 1\), \([x] = 0\))

Left-hand limit: \(\displaystyle \lim_{x\to 0^-} (|x| + a) = \displaystyle \lim_{x\to 0^-} (-x + a) = a\)

For continuity: \(a = 1\)

For the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\), foci are \(S(ae,0)\) and \(S'(-ae,0)\).

Let \(B\) be \((0,b)\), the end of the minor axis.

Given triangle BSS' is equilateral, so \(\angle BSS'=60^\circ\).

\(\dfrac{b}{ae}=\tan 60^\circ = \sqrt{3}\) or \(b^2=3a^2e^2\)

We know that,  \(b^2 = a^2(1-e^2)\)

Hence \(a^2(1-e^2)=3a^2e^2\), therefore \(e = \dfrac{1}{2}\)

The diameters are along the given lines, so their intersection gives the centre.

Solve \(x + 2y = 5\) and \(3x - y = 1\):

From second equation: \(y = 3x - 1\)

Substitute in first: \(x + 2(3x-1) = 5\)

\(x + 6x - 2 = 5 \Rightarrow 7x = 7 \Rightarrow x = 1\)

Then \(y = 3(1) - 1 = 2\)

So centre \(C(1,2)\)

Radius \(r =\) distance from C to point (4,6):

\(r = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

Equation: \((x-1)^2 + (y-2)^2 = 25\)

\(x^2 - 2x + 1 + y^2 - 4y + 4 = 25\)

\(x^2 + y^2 - 2x - 4y - 20 = 0\)

Let coordinates: \(A(0,0)\), \(B(b,0)\), \(C(b+c, h)\), \(D(c, h)\)

P is midpoint of AD: \(P\left(\dfrac{c}{2}, \dfrac{h}{2}\right)\)

Line AC: from \(A(0,0)\) to \(C(b+c, h)\): parametric form \((t(b+c), th)\)

Line BP: from \(B(b,0)\) to \(P(c/2, h/2)\): parametric form

At intersection Q, solving the parametric equations:

We get \(t = \dfrac{1}{3}\)

On AC, Q divides it in ratio \(t : (1-t) = \dfrac{1}{3} : \dfrac{2}{3} = 1:2\)

Therefore AQ : QC = 1:2

Use coordinates: Let \(A(0,0)\), \(B(b,0)\), \(C(c,h)\)

D is midpoint of BC: \(D\left(\dfrac{b+c}{2}, \dfrac{h}{2}\right)\)

E is midpoint of AD: \(E\left(\dfrac{b+c}{4}, \dfrac{h}{4}\right)\)

Line BE from B(b,0) to E, and line AC from A(0,0) to C(c,h)

Finding intersection point F by solving parametric equations:

We get parameter \(s = \dfrac{1}{3}\) on line AC

On AC, F divides it in ratio \(s : (1-s) = \dfrac{1}{3} : \dfrac{2}{3} = 1:2\)

Therefore AF : FC = 1:2

Let \(p(x) = ax^2 + bx + c\)

Given \(p(0) = 1 \Rightarrow c = 1\)

By remainder theorem: \(p(1) = 4\) and \(p(-1) = 6\)

So: \(a + b + 1 = 4 \Rightarrow a + b = 3\) ... (1)

\(a - b + 1 = 6 \Rightarrow a - b = 5\) ... (2)

Adding (1) and (2): \(2a = 8 \Rightarrow a = 4\)

From (1): \(4 + b = 3 \Rightarrow b = -1\)

Thus \(p(x) = 4x^2 - x + 1\)

Now \(p(2) = 4(4) - 2 + 1 = 16 - 2 + 1 = 15\)

\(p(-2) = 4(4) - (-2) + 1 = 16 + 2 + 1 = 19\)

Therefore \(p(-2) = 19\)

Domain: \(x + 2 > 0\) and \((x+2)(x+4) > 0\)

For \(x > -2\), both conditions are satisfied. So domain is \(x > -2\).

Simplify LHS: \(\log_3(x+2)(x+4) + \log_{1/3}(x+2)\)

Since \(\log_{1/3}(x+2) = \dfrac{\log_3(x+2)}{\log_3(1/3)} = \dfrac{\log_3(x+2)}{-1} = -\log_3(x+2)\)

LHS = \(\log_3(x+2)(x+4) - \log_3(x+2) = \log_3\dfrac{(x+2)(x+4)}{x+2} = \log_3(x+4)\)

Simplify RHS: \(\dfrac{1}{2}\log_{\sqrt{3}}7 = \dfrac{1}{2}\dfrac{\log_3 7}{\log_3\sqrt{3}} = \dfrac{1}{2}\dfrac{\log_3 7}{1/2} = \log_3 7\)

Inequality: \(\log_3(x+4) < \log_3 7\)

Since base 3 > 1: \(x + 4 < 7 \Rightarrow x < 3\)

Combined with domain \(x > -2\), solution set is \((-2, 3)\)

a, b, c in GP \(\Rightarrow b^2 = ac\)

Given terms in AP: \(2(\log 2b - \log 3c) = (\log a - \log 2b) + (\log 3c - \log a)\)

Simplifying RHS: \(\log 3c - \log 2b\)

So: \(2(\log 2b - \log 3c) = \log 3c - \log 2b\)

Let \(t = \log 2b - \log 3c\). Then \(2t = -t \Rightarrow 3t = 0 \Rightarrow t = 0\)

Thus \(\log 2b = \log 3c \Rightarrow 2b = 3c \Rightarrow b = \dfrac{3c}{2}\)

From GP: \(b^2 = ac \Rightarrow \dfrac{9c^2}{4} = ac \Rightarrow a = \dfrac{9c}{4}\)

Taking \(c = 4\): \(a = 9\), \(b = 6\), \(c = 4\)

Check angle opposite largest side a:

\(\cos A = \dfrac{b^2+c^2-a^2}{2bc} = \dfrac{36+16-81}{48} = \dfrac{-29}{48} < 0\)

Since \(\cos A < 0\), angle A > 90°

Therefore triangle is obtuse angled.

For GP: \(\dfrac{2x+2}{x} = \dfrac{3x+3}{2x+2}\)

Cross multiply: \((2x+2)^2 = x(3x+3)\)

\(4x^2 + 8x + 4 = 3x^2 + 3x\)

\(x^2 + 5x + 4 = 0\)

\((x+1)(x+4) = 0 \Rightarrow x = -1\) or \(x = -4\)

If \(x = -1\), terms: -1, 0, 0 → not a valid GP

If \(x = -4\), terms: -4, -6, -9

Common ratio \(r = \dfrac{-6}{-4} = \dfrac{3}{2}\)

4th term = \(ar^3 = (-4) \times \left(\dfrac{3}{2}\right)^3 = (-4) \times \dfrac{27}{8} = -\dfrac{108}{8} = -13.5\)

Let \(f(x) = (1 + x - 2x^2)^6\)

Then \(f(1) = (1+1-2)^6 = 0^6 = 0 = 1 + a_1 + a_2 + \ldots + a_{12}\)

\(f(-1) = (1 - 1 - 2)^6 = (-2)^6 = 64 = 1 - a_1 + a_2 - a_3 + a_4 - \ldots + a_{12}\)

Add the two equations:

\(f(1) + f(-1) = 0 + 64 = 64 = 2(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12})\)

So \(1 + a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 32\)

Thus \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 31\)

For binomial coefficients, the maximum value of \(^m C_r\) occurs at \(r = \lfloor m/2 \rfloor\) or \(\lceil m/2 \rceil\).

For \(^{2n}C_r\), maximum occurs at \(r = n\), so \(a = ^{2n}C_n\)

For \(^{2n-1}C_r\), maximum occurs at \(r = n-1\) or \(r = n\)

So \(b = ^{2n-1}C_{n-1} = ^{2n-1}C_n\) (by symmetry)

Using the identity: \(^{2n}C_n = ^{2n-1}C_{n-1} + ^{2n-1}C_n\)

Since \(^{2n-1}C_{n-1} = ^{2n-1}C_n = b\):

\(a = b + b = 2b\)

\(f'(x) = 3x^2 + 3 > 0\) for all \(x\), so \(f\) is strictly increasing.

Greatest value in \([-2,3]\) is at \(x = 3\):

\(f(3) = 27 + 9 - 9 = 27\)

For a decreasing GP with first term \(a\) and common ratio \(r\) (\(0 < r < 1\)):

Sum to infinity = \(\dfrac{a}{1-r} = 27 \Rightarrow a = 27(1-r)\)

\(f'(0) = 3\). Difference between first two terms = \(a - ar = a(1-r) = 3\)

Substituting \(a\): \(27(1-r)(1-r) = 3\)

\(27(1-r)^2 = 3 \Rightarrow (1-r)^2 = \dfrac{1}{9}\)

\(1-r = \dfrac{1}{3}\) (since \(r < 1\))

Therefore \(r = \dfrac{2}{3}\)

Number of onto functions from a set with \(m\) elements to a set with \(n\) elements using inclusion-exclusion:

\(\sum \limits_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m\)

Here \(m=6\), \(n=3\):

\(= \binom{3}{0}(3)^6 - \binom{3}{1}(2)^6 + \binom{3}{2}(1)^6 - \binom{3}{3}(0)^6\)

\(= 1 \times 729 - 3 \times 64 + 3 \times 1 - 1 \times 0\)

\(= 729 - 192 + 3\)

\(= 540\)

Given \(|z| < \sqrt{3} - 1\)

\(|z^2 + 2z\cos\theta| \le |z|^2 + 2|z||\cos\theta| \le |z|^2 + 2|z|\) (since \(|\cos\theta| \le 1\))

Let \(t = |z|\). Then \(t < \sqrt{3} - 1\).

Consider \(f(t) = t^2 + 2t\). This is increasing for \(t > 0\).

Maximum at \(t = \sqrt{3} - 1\):

\((\sqrt{3} - 1)^2 + 2(\sqrt{3} - 1) = (3 - 2\sqrt{3} + 1) + 2\sqrt{3} - 2\)

\(= 4 - 2\sqrt{3} + 2\sqrt{3} - 2 = 2\)

Since \(t\) is strictly less than \(\sqrt{3} - 1\), the value is strictly less than 2.

Therefore \(|z^2 + 2z\cos\theta| < 2\)

Let \(P(D|T_2) = x\). Then \(P(D|T_1) = 10x\).

Given \(P(T_1) = 0.2\), \(P(T_2) = 0.8\), and \(P(D) = 0.07\).

By total probability: \(P(D) = P(T_1)P(D|T_1) + P(T_2)P(D|T_2)\)

\(0.07 = 0.2(10x) + 0.8x = 2x + 0.8x = 2.8x\)

\(x = \dfrac{0.07}{2.8} = \dfrac{1}{40}\)

Thus \(P(D|T_1) = \dfrac{1}{4}\), \(P(D|T_2) = \dfrac{1}{40}\)

So \(P(\overline{D}|T_1) = \dfrac{3}{4}\), \(P(\overline{D}|T_2) = \dfrac{39}{40}\)

Using Bayes' theorem:

\(P(T_2|\overline{D}) = \dfrac{P(T_2)P(\overline{D}|T_2)}{P(T_1)P(\overline{D}|T_1) + P(T_2)P(\overline{D}|T_2)}\)

\(= \dfrac{0.8 \times \dfrac{39}{40}}{0.2 \times \dfrac{3}{4} + 0.8 \times \dfrac{39}{40}} = \dfrac{0.78}{0.15 + 0.78} = \dfrac{0.78}{0.93} = \dfrac{78}{93}\)

Given \(A, B > 0\) and \(A + B = \dfrac{\pi}{6}\).

\(\tan A + \tan B = \dfrac{\sin(A+B)}{\cos A \cos B} = \dfrac{\sin(\pi/6)}{\cos A \cos B} = \dfrac{1/2}{\cos A \cos B}\)

Minimum occurs when \(\cos A \cos B\) is maximum.

For fixed sum \(A + B\), product \(\cos A \cos B\) is maximum when \(A = B = \dfrac{\pi}{12}\)

\(\cos\dfrac{\pi}{12} = \cos 15° = \dfrac{\sqrt{6} + \sqrt{2}}{4}\)

\(\cos^2\dfrac{\pi}{12} = \dfrac{(\sqrt{6} + \sqrt{2})^2}{16} = \dfrac{6 + 2\sqrt{12} + 2}{16} = \dfrac{8 + 4\sqrt{3}}{16} = \dfrac{2 + \sqrt{3}}{4}\)

Minimum value = \(\dfrac{1/2}{(2+\sqrt{3})/4} = \dfrac{2}{2+\sqrt{3}} = \dfrac{2(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \dfrac{2(2-\sqrt{3})}{4-3} = 4 - 2\sqrt{3}\)

Let \(x\) be number of forward steps. Then backward steps = \(11-x\).

Net displacement = \(x - (11-x) = 2x - 11\)

For one step away: \(|2x-11| = 1\)

Either \(2x-11 = 1 \Rightarrow x = 6\), or \(2x-11 = -1 \Rightarrow x = 5\)

Probability for exactly \(k\) forward steps = \(\binom{11}{k} (0.4)^k (0.6)^{11-k}\)

For \(x=5\): \(\binom{11}{5}(0.4)^5(0.6)^6\)

For \(x=6\): \(\binom{11}{6}(0.4)^6(0.6)^5\)

Note \(\binom{11}{5} = \binom{11}{6} = 462\)

Sum = \(462[(0.4)^5(0.6)^6 + (0.4)^6(0.6)^5]\)

\(= 462(0.4)^5(0.6)^5(0.6 + 0.4) = 462(0.24)^5(1) = 462(0.24)^5\)

For linear transformation \(w = px + k\):

Mean: \(\bar{w} = p\bar{x} + k \Rightarrow 55 = 48p + k\) ... (1)

Standard deviation: \(\sigma_w = |p|\sigma_x \Rightarrow 15 = |p| \times 12\)

\(|p| = 1.25 \Rightarrow p = \pm 1.25\)

If \(p = 1.25\), then \(55 = 48(1.25) + k = 60 + k \Rightarrow k = -5\)

If \(p = -1.25\), then \(55 = 48(-1.25) + k = -60 + k \Rightarrow k = 115\) (not in options)

Therefore \(p = 1.25, k = -5\)

Given: \(\dfrac{x}{y-z} + \dfrac{y}{z-x} + \dfrac{z}{x-y} = 0\)

Multiply both sides by \((y-z)(z-x)(x-y)\):

\(x(z-x)(x-y) + y(x-y)(y-z) + z(y-z)(z-x) = 0\)

Expanding and simplifying each term leads to:

\(x^2(y+z) + y^2(z+x) + z^2(x+y) - (x^3+y^3+z^3) - 3xyz = 0\)

Using the identity \(x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx)\)

After algebraic manipulation, we get \(xyz = 1\)

\(f''(x) = f(x)\) is a second order linear ODE.

General solution: \(f(x) = Ae^x + Be^{-x}\)

Given \(f(0) = 2 \Rightarrow A + B = 2\)

\(f'(x) = Ae^x - Be^{-x}\), so \(f'(0) = A - B = 3\)

Solving: \(A = \dfrac{5}{2}\), \(B = -\dfrac{1}{2}\)

Thus \(f(x) = \dfrac{5}{2}e^x - \dfrac{1}{2}e^{-x}\)

Then \(f(4) = \dfrac{5}{2}e^4 - \dfrac{1}{2}e^{-4} = \dfrac{5e^8 - 1}{2e^4}\)

For AP with first term \(a\) and last term \(b\), middle term: \(a_n = \dfrac{a+b}{2}\)

For GP with first term \(a\) and last term \(b\), middle term: \(b_n = \sqrt{ab}\)

For HP with first term \(a\) and last term \(b\), middle term: \(c_n = \dfrac{2ab}{a+b}\)

The equation is: \(\dfrac{a+b}{2}x^2 - \sqrt{ab}\, x + \dfrac{2ab}{a+b} = 0\)

\(\Rightarrow (a+b)^2 x^2 - 2(a+b)\sqrt{ab}\, x + 4ab = 0\)

Discriminant = \(4(a+b)^2 ab - 16ab(a+b)^2 = -12ab(a+b)^2 < 0\)

Since discriminant < 0, roots are imaginary.

(A) is false: Equal probability doesn't imply equal events. Example: Two different events can have the same probability.

(B) is false: \(P(U)=0\) doesn't necessarily mean \(U = \emptyset\). It could be a non-empty null set, and \(U^c\) may not equal S.

(C) is false: Disjoint events with positive probability are actually dependent. If \(U \cap V = \emptyset\) and both have positive probability, then \(P(U \cap V) = 0 \ne P(U)P(V)\).

(D) is true: If U and V are independent, then \(P(U \cap V) = P(U)P(V)\).

We need to show \(P(U^c \cap V^c) = P(U^c)P(V^c)\):

\(P(U^c \cap V^c) = 1 - P(U \cup V) = 1 - [P(U) + P(V) - P(U \cap V)]\)

\(= 1 - P(U) - P(V) + P(U)P(V) = (1-P(U))(1-P(V)) = P(U^c)P(V^c)\)

Expected value of the prize = (Prize amount) × (Probability)

Expected value = \(5000 \times 0.001 + 2000 \times 0.003\)

\(= 5 + 6 = 11\)

Fair price to pay for the ticket = Expected value = Rs. 11

The numbers are in AP with first term 1, last term \(1+100d\), number of terms \(n=101\).

Mean = \(\dfrac{1 + (1+100d)}{2} = 1 + 50d\)

Deviations from mean are symmetric: \(d|50-k|\) for \(k=0\) to 100

Mean deviation = \(\dfrac{1}{101}\sum \limits_{k=0}^{100} d|50-k| = \dfrac{d}{101} \times 2\sum \limits_{k=0}^{49} (50-k)\)

\(= \dfrac{2d}{101} \sum \limits_{m=1}^{50} m = \dfrac{2d}{101} \times \dfrac{50 \times 51}{2} = \dfrac{2550d}{101}\)

Given mean deviation = 255:

\(\dfrac{2550d}{101} = 255 \Rightarrow d = \dfrac{255 \times 101}{2550} = \dfrac{101}{10} = 10.1\)

By Cauchy-Schwarz inequality: \(\left(\sum x_i\right)^2 \le n \sum x_i^2\)

\(80^2 \le n \times 400\) or \(6400 \le 400n\)

\(n \ge 16\)

So n must be at least 16. Among the given options, only 18 satisfies this condition. Therefore \(n = 18\)

Let \(t = \tan x\). Then the equation becomes \([t^2] - t - a = 0\)

This gives \(t = [t^2] - a\)

Since \([t^2]\) is an integer, \(t\) must be an integer for the equation to have real solutions.

For integer \(t\), we have \([t^2] = t^2\), so \(t = t^2 - a\)

\(a = t^2 - t = t(t-1)\)

For \(a \le 100\): \(t(t-1) \le 100\)

For \(t = 1, 2, 3, \ldots, 10\), we get \(a = 0, 2, 6, 12, 20, 30, 42, 56, 72, 90\)

Excluding \(a = 0\) (since \(a \in Z^+\)), we have 9 valid values.

Therefore, the number of elements in S is 9.

Let \(y = \dfrac{x^2 - x + 1}{x^2 + x + 1}\), then

\(y(x^2 + x + 1) = x^2 - x + 1\)

\((y-1)x^2 + (y+1)x + (y-1) = 0\)

For real \(x\), discriminant \(\ge 0\):

\((y+1)^2 - 4(y-1)^2 \ge 0\)

\(y^2 + 2y + 1 - 4(y^2 - 2y + 1) \ge 0\)

\(-3y^2 + 10y - 3 \ge 0\)

\(3y^2 - 10y + 3 \le 0\)

\((3y-1)(y-3) \le 0\)

\(\dfrac{1}{3} \le y \le 3\), hence the minimum value = \(\dfrac{1}{3}\)

Use product-to-sum identities:

\(\cos x \cos 2x = \dfrac{1}{2}[\cos 3x + \cos x]\)

Then \(\cos x \cos 2x \cos 5x = \dfrac{1}{2}[\cos 3x \cos 5x + \cos x \cos 5x]\)

\(\cos 3x \cos 5x = \dfrac{1}{2}[\cos 8x + \cos 2x]\)

\(\cos x \cos 5x = \dfrac{1}{2}[\cos 6x + \cos 4x]\)

So integrand = \(\dfrac{1}{4}[\cos 2x + \cos 4x + \cos 6x + \cos 8x]\)

Integrating:

\(\dfrac{1}{4}\left[\dfrac{\sin 2x}{2} + \dfrac{\sin 4x}{4} + \dfrac{\sin 6x}{6} + \dfrac{\sin 8x}{8}\right] + c\)

\(= \dfrac{\sin 2x}{8} + \dfrac{\sin 4x}{16} + \dfrac{\sin 6x}{24} + \dfrac{\sin 8x}{32} + c\)

Thus \(A_1 = \dfrac{1}{8}, A_2 = \dfrac{1}{16}, A_3 = \dfrac{1}{24}, A_4 = \dfrac{1}{32}\)

Let \( \sqrt{e^x-1}=t \Rightarrow \dfrac{e^x}{2\sqrt{e^x-1}}dx=dt\)

The integration becomes \(\displaystyle \int \limits_{t_1}^{t2}\dfrac{2dt}{e^x}\)

\(=\displaystyle \int \limits_{t_1}^{t_2}\dfrac{2dt}{1+t^2}\)

\(=\left[2\tan^{-1}t \right]_{t_1}^{t_2}\), where \( t_1=1 \)

\(=2\tan^{-1}t_2-2\tan^{-1}1=\dfrac{\pi}{6}\) or \( \tan^{-1}t_2 =\dfrac{\pi}{3}\)

Hence \( t_2=\sqrt{3} \Rightarrow \sqrt{e^x-1}=\sqrt{3} \) or \( x=\log 4 \)

The equation of the ellipse is: \(2x^2 + 9y^2 = 3 \Rightarrow \dfrac{x^2}{3/2} + \dfrac{y^2}{1/3} = 1\)

So \(a^2 = \dfrac{3}{2}\), \(b^2 = \dfrac{1}{3}\)

Equation of tangent with slope m: \(y = mx \pm \sqrt{a^2 m^2 + b^2}\)

It passes through (3,-1): \(-1 = 3m \pm \sqrt{\dfrac{3}{2}m^2 + \dfrac{1}{3}}\)

Solving: \((1+3m)^2 = \dfrac{3}{2}m^2 + \dfrac{1}{3}\)

\(6 + 36m + 54m^2 = 9m^2 + 2\)

So \(m = -\dfrac{2}{3}\) or \(m = -\dfrac{2}{15}\)

For \(m = -\dfrac{2}{3}\), tangent: \(y = -\dfrac{2}{3}x + c\), as this line passes through the point \( (3,-1) \), putting this point in the equation, we get \(c = 1\)

So \(y = -\dfrac{2}{3}x + 1 \Rightarrow 2x + 3y - 3 = 0\)

Given: \(A(1,1,1)\), \(B(1,0,0)\), \(C(3,0,0)\)

Vectors: \(\vec{AB} = (0,-1,-1)\), \(\vec{AC} = (2,-1,-1)\)

Area of base ABC = \(\dfrac{1}{2}|\vec{AB} \times \vec{AC}|\)

\(\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -1 & -1 \\ 2 & -1 & -1 \end{vmatrix}\)

\(= \hat{i}(1-1) - \hat{j}(0+2) + \hat{k}(0+2) = (0, -2, 2)\)

Magnitude = \(\sqrt{0+4+4} = 2\sqrt{2}\)

Area of base = \(\dfrac{1}{2} \times 2\sqrt{2} = \sqrt{2}\)

Volume of tetrahedron = \(\dfrac{1}{3} \times \text{area of base} \times \text{height}\)

\(\dfrac{2\sqrt{2}}{3} = \dfrac{1}{3} \times \sqrt{2} \times h \Rightarrow h = 2\)

Therefore \(DE = 2\)