The given matrix \(A\) is upper triangular with eigenvalues \(-1, -1, -1\). Write \(A = -I + N\), where

\(N = \begin{pmatrix} 0 & -1 & 2 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}\)

Then \(N^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\) and \(N^3 = 0\). By the binomial theorem,

\(A^{19} = (-I + N)^{19} = \sum_{k=0}^{2} \binom{19}{k} (-I)^{19-k} N^k\)

\(= (-1)^{19}I + \binom{19}{1}(-1)^{18} N + \binom{19}{2}(-1)^{17} N^2\)

\(= -I + 19N - \binom{19}{2} N^2\), where \(\binom{19}{2} = 171\)

\(A^{19} = -\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} + 19\begin{pmatrix}0&-1&2\\0&0&-1\\0&0&0\end{pmatrix} - 171\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\)

\(= \begin{pmatrix} -1 & -19 & -133 \\ 0 & -1 & -19 \\ 0 & 0 & -1 \end{pmatrix}\)

Sum of all entries = \((-1) + (-19) + (-133) + 0 + (-1) + (-19) + 0 + 0 + (-1) = -174\)

The absolute value is \(174\).

\(\cot \theta + \tan \theta = 2\)

\(\dfrac{\cos \theta}{\sin \theta} + \dfrac{\sin \theta}{\cos \theta} = 2\)

\(\dfrac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} = 2\)

\(\dfrac{1}{\sin \theta \cos \theta} = 2\)

\(\sin \theta \cos \theta = \dfrac{1}{2}\)

This is possible only when \(\tan \theta = 1\)

\(\theta = n\pi + \dfrac{\pi}{4}\)

Length of projection of \(\vec{a}\) on \(\vec{b}\) = \(\left| \dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \right|\)

\(\vec{a} \cdot \vec{b} = (2)(-2) + (3)(1) + (1)(2) = -4 + 3 + 2 = 1\)

\(|\vec{b}| = \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)

Length of projection = \(\left| \dfrac{1}{3} \right| = \dfrac{1}{3}\)

The three cube roots of 27 are \(3\), \(3\omega\), and \(3\omega^2\), where \(\omega\) is the complex cube root of unity with \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\).

Let the matrix be \(A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}\).

Observe that the sum of all rows is \(x + y + z\).

Since \(x + y + z = 3 + 3\omega + 3\omega^2 = 3(1 + \omega + \omega^2) = 0\), the rows are linearly dependent.

Therefore, the determinant is \(0\).

Let the center be \((r, r)\) since the circle touches both axes in the first quadrant, and the radius is \(r\).

The perpendicular distance from center \((r, r)\) to the line \(x - y - 2 = 0\) equals the radius \(r\).

Distance = \(\dfrac{|r - r - 2|}{\sqrt{1^2 + (-1)^2}} = \dfrac{|-2|}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2}\)

So \(r = \sqrt{2}\).

Area of circle = \(\pi r^2 = \pi (\sqrt{2})^2 = 2\pi\)

Let \(X\) be the score. Given \(X \sim N(500, 100^2)\).

We need \(P(450 < X < 500)\).

Convert to standard normal: \(P(450 < X < 500) = P\left(\dfrac{450 - 500}{100} < Z < \dfrac{500 - 500}{100}\right)\)

\(= P(-0.5 < Z < 0)\)

\(= \Phi(0) - \Phi(-0.5)\)  \(= \Phi(0.5) - \Phi(0)\) \(= 0.691 -0.5\)

Therefore, \(P(-0.5 < Z < 0) = 0.191\)

Given \(P(E) = P(E|F) = \dfrac{P(E \cap F)}{P(F)}\)

This implies \(P(E)P(F) = P(E \cap F)\), which means \(E\) and \(F\) are independent.

We know that, if two events \(A\) and \(B\) are independent, then \(A^c\) and \(B\) are also independent. Also \(A^c\) and \(B^c\) are also independent. Now check each option:

(a) If \(E\) and \(F\) are independent, then \(E^c\) and \(F\) are also independent. So (a) is true.

(b) Directly states \(E\) and \(F\) are independent, so it is also true.

(d) \(P(F) = P(F|E) = \dfrac{P(F \cap E)}{P(E)}\) gives \(P(F)P(E) = P(F \cap E)\), same condition, so it is true.

(c) \(P(E^c) P(F^c) \neq P(E^c \cap F^c)\) means \(E^c\) and \(F^c\) are NOT independent.

If \(E\) and \(F\) are independent, then \(E^c\) and \(F^c\) are also independent, so \(P(E^c) P(F^c) = P(E^c \cap F^c)\).

Thus (c) contradicts the independence condition, so it is NOT true.

The given circle is \(x^2 + y^2 - 6x - 10y + 30 = 0\).

Center: \(C = (3, 5)\) and radius \(r = \sqrt{9 + 25 - 30} = 2\)

The image of this circle across the line \(3x + y = 2\) will have the same radius \(2\).

Let the image center be \(C' = (h, k)\), then using the reflection formula:

\(\dfrac{h-3}{3} = \dfrac{k-5}{1} = \dfrac{-2(3 \cdot 3 + 5 - 2)}{3^2 + 1^2} = \dfrac{-2(12)}{10} = \dfrac{-24}{10} = -2.4\)

\(h = 3 + 3(-2.4) = 3 - 7.2 = -4.2\)

\(k = 5 + 1(-2.4) = 5 - 2.4 = 2.6\)

So image center \(C' = (-4.2, 2.6)\).

Equation of image circle: \((x + 4.2)^2 + (y - 2.6)^2 = 4\)

\(x^2 + 8.4x + 17.64 + y^2 - 5.2y + 6.76 = 4\)

\(x^2 + y^2 + 8.4x - 5.2y + 20.4 = 0\)

So \(\alpha = 8.4\), \(\beta = -5.2\), \(\gamma = 20.4\)

\(\alpha + \beta + \gamma = 8.4 - 5.2 + 20.4 = 23.6\)

Floor value = \(\lfloor 23.6 \rfloor = 23\)

Let \(AB\) be the tower with foot \(B\) and top \(A\).

Let the first point be \(P\) at ground level, and the second point \(Q\) is \(h\) meters above \(P\).

At \(Q\), the angle of depression of the foot \(B\) is \(60^\circ\). Hence \(\angle QBP = 60^\circ\).

Let the horizontal distance be \(x\).

\(\tan 60^\circ = \dfrac{PQ}{PB} = \dfrac{h}{x}\)

\(\Rightarrow x = \dfrac{h}{\tan 60^\circ} = h \cot 60^\circ\)

Total number of ways to arrange 5 captains = \(5! = 120\).

Treat India and Australia as a single block. Then we have 4 items to arrange (the block plus the other 3 captains).

Number of arrangements = \(4! = 24\).

Within the block, India and Australia can be arranged in \(2! = 2\) ways.

So favorable arrangements = \(24 \times 2 = 48\).

Probability = \(\dfrac{48}{120} = \dfrac{2}{5}\)

Write \(5^n = (1 + 4)^n\) and expand using the binomial theorem:

\((1 + 4)^n = 1 + 4n + \binom{n}{2} \times 4^2 + \binom{n}{3} \times 4^3 + \ldots\)

Then \(5^n - 4n - 1 = \binom{n}{2} \times 4^2 + \binom{n}{3} \times 4^3 + \ldots\)

For \(k \geq 2\), each term \(\binom{n}{k} 4^k\) is divisible by \(4^2 = 16\).

Thus \(5^n - 4n - 1\) is divisible by 16 for all \(n \in \mathbb{N}\).

So every element of \(A\) is a multiple of 16.

The set \(B = \{16(n - 1) : n \in \mathbb{N}\}\) contains all non-negative multiples of 16: \(0, 16, 32, 48, \ldots\)

Since \(A\) consists of some specific multiples of 16 (not necessarily all), we have \(A \subseteq B\).

Therefore, \(A \subset B\).

We know that \(|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}\)

Given \(|\vec{a} + \vec{b}| = 1\), so:

\(1^2 = 3^2 + 4^2 + 2\vec{a} \cdot \vec{b}\)

\(1 = 9 + 16 + 2\vec{a} \cdot \vec{b}\)

\(\vec{a} \cdot \vec{b} = -12\)

Now, \(|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}\)

\(= 9 + 16 - 2(-12)\)

\(= 25 + 24 = 49\)

Therefore, \(|\vec{a} - \vec{b}| = \sqrt{49} = 7\)

Let \(I = \int \limits_{0}^{\dfrac{\pi}{2}} \dfrac{1 + 2\cos x}{(2 + \cos x)^2} \, dx\)

Rewrite the integrand by dividing numerator and denominator appropriately.

Let \(t = 2\csc x + \cot x\)

Then \(dt = (-2\csc x \cot x - \csc^2 x) dx = -(2\cot x \csc x + \csc^2 x) dx\)

So \((2\cot x \csc x + \csc^2 x) dx = -dt\)

The numerator \(\csc^2 x + 2\cot x \csc x = (2\cot x \csc x + \csc^2 x)\)

Thus the integrand becomes \(\dfrac{-dt}{t^2}\)

When \(x = 0\), \(t \to \infty\); when \(x = \dfrac{\pi}{2}\), \(t = 2\)

Therefore, \(I = \int_{\infty}^{2} \dfrac{-dt}{t^2} = \int_{2}^{\infty} \dfrac{dt}{t^2} = \left[-\dfrac{1}{t}\right]_{2}^{\infty} = 0 - \left(-\dfrac{1}{2}\right) = \dfrac{1}{2}\)

Since \(\dfrac{1}{2} \in (0, 1)\), the integral lies in the interval \((0, 1)\).

Given \(\vec{a} \times \vec{b} = \vec{c}\)

Take the cross product with \(\vec{c}\) on both sides:

\(\vec{c} \times (\vec{a} \times \vec{b}) = \vec{c} \times \vec{c} = 0\)

Using the vector triple product formula:

\(\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b} = 0\)

Given \(\vec{c} \cdot \vec{a} = 2\) and \(\vec{c} \cdot \vec{b} = 1\), we get:

\(1 \cdot \vec{a} - 2\vec{b} = 0\) or \(\vec{a} = 2\vec{b}\)

Therefore, \(|\vec{a}| = 2|\vec{b}| = 2 \times 1 = 2\)

First line: \(2y = x + 1 \Rightarrow y = \dfrac{1}{2}x + \dfrac{1}{2}\), slope \(m_1 = \dfrac{1}{2}\)

Second line: \(y = 3x + 2\), slope \(m_2 = 3\)

Angle between lines: \(\tan \theta = \left| \dfrac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \dfrac{3 - \dfrac{1}{2}}{1 + \dfrac{1}{2} \cdot 3} \right|\)

\(= \left| \dfrac{\dfrac{5}{2}}{1 + \dfrac{3}{2}} \right| = \left| \dfrac{\dfrac{5}{2}}{\dfrac{5}{2}} \right| = 1\)

So \(\theta = \dfrac{\pi}{4}\) (acute angle)

The obtuse angle = \(\pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}\)

Total outcomes = \(4 \times 4 = 16\)

Event \(A = \{x \geq 2\}\): \(x\) can be 2, 3, or 4 (3 values), \(y\) can be any of 4 values

\(P(A) = \dfrac{12}{16} = \dfrac{3}{4}\) ✓

Event \(B = \{y > x\}\): Favorable outcomes are (1,2), (1,3), (1,4), (2,3), (2,4), (3,4) = 6 outcomes

\(P(B) = \dfrac{6}{16} = \dfrac{3}{8}\) ✓

Event \(A \cap B = \{x \geq 2 \text{ and } y > x\}\): (2,3), (2,4), (3,4) = 3 outcomes

\(P(A \cap B) = \dfrac{3}{16} \neq \dfrac{1}{4}\) ✗ (This is NOT true)

Check independence: \(P(A)P(B) = \dfrac{3}{4} \times \dfrac{3}{8} = \dfrac{9}{32} \neq \dfrac{3}{16}\)

So \(A\) and \(B\) are not independent ✓

\(8^{x-1} = 4^{-x}\)

\(2^{3(x-1)} = 2^{-2x}\)

Equating exponents: \(3x - 3 = -2x\)

\(x = \dfrac{3}{5}\)

Then \(x + 1 = \dfrac{3}{5} + 1 = \dfrac{8}{5}\) and \(1 - x = 1 - \dfrac{3}{5} = \dfrac{2}{5}\)

Now evaluate the expression:

\(\dfrac{1}{\log_{x+1} 4 - \log_{x+1} 5} + \dfrac{1}{\log_{1-x} 4 - \log_{1-x} 5}\)

\(= \dfrac{1}{\log_{8/5} 4 - \log_{8/5} 5} + \dfrac{1}{\log_{2/5} 4 - \log_{2/5} 5}\)

\(= \dfrac{1}{\log_{8/5} \dfrac{4}{5}} + \dfrac{1}{\log_{2/5} \dfrac{4}{5}}\)

\(= \log_{4/5} \dfrac{8}{5} + \log_{4/5} \dfrac{2}{5}\)

\(= \log_{4/5} \left(\dfrac{8}{5} \times \dfrac{2}{5}\right)\)

\(= \log_{4/5} \dfrac{16}{25}\)

\(= 2 \log_{4/5} \dfrac{4}{5}=2\)

Given \(x^{(8 \log_5 x - 24)} = 5^{-4}\)

Let \(t = \log_5 x\), then \(x = 5^t\)

LHS = \((5^t)^{(8t - 24)} = 5^{t(8t - 24)}\)

So \(5^{t(8t - 24)} = 5^{-4}\)

Equating exponents: \(t(8t - 24) = -4\)

\(8t^2 - 24t + 4 = 0\)

Divide by 4: \(2t^2 - 6t + 1 = 0\)

Sum of roots: \(t_1 + t_2 = \dfrac{6}{2} = 3\)

Since \(x = 5^t\), the product of all possible values of \(x\) is:

\(x_1 \times x_2 = 5^{t_1} \times 5^{t_2} = 5^{t_1 + t_2} = 5^3 = 125\)

Let tower height = \(h\). If \(x_1, x_2, x_3\) are distances of A, B, C from the foot, then:

\(x_1 = h\cot \alpha,\; x_2 = h\cot 2\alpha,\; x_3 = h\cot 3\alpha\)

\(AB = x_1 - x_2 = h(\cot \alpha - \cot 2\alpha)\)

\(BC = x_2 - x_3 = h(\cot 2\alpha - \cot 3\alpha)\)

\(\dfrac{AB}{BC} = \dfrac{\cot \alpha - \cot 2\alpha}{\cot 2\alpha - \cot 3\alpha}\)

Using \(\cot A - \cot B = \dfrac{\sin(B-A)}{\sin A \sin B}\):

Numerator = \(\dfrac{\sin \alpha}{\sin \alpha \sin 2\alpha}\), Denominator = \(\dfrac{\sin \alpha}{\sin 2\alpha \sin 3\alpha}\)

Thus \(\dfrac{AB}{BC} = \dfrac{\sin 3\alpha}{\sin \alpha}\)

Now \(\dfrac{\sin 3\alpha}{\sin \alpha} = 3 - 4\sin^2 \alpha = 4\cos^2 \alpha - 1 = 1 + 2(2\cos^2 \alpha - 1) = 1 + 2\cos 2\alpha\)

Number of female students = 40, average height = 5.15 feet

Total height of female students = \(40 \times 5.15 = 206\) feet

Number of male students = 20, average height = 5.66 feet

Total height of male students = \(20 \times 5.66 = 113.2\) feet

Total students = \(40 + 20 = 60\)

Total height of all students = \(206 + 113.2 = 319.2\) feet

Average height of all students = \(\dfrac{319.2}{60} = 5.32\) feet

Factor \(C\):

\(C = BA + 2B + 2A + 4I = B(A + 2I) + 2(A + 2I) = (B + 2I)(A + 2I)\)

From \(A^2 + 5A + 5I = 0\):

\(A^2 + 5A + 6I = I\)

\((A + 2I)(A + 3I) = I\)

So \((A + 2I)^{-1} = A + 3I\)

From \(B^2 + 3B + I = 0\):

\(B^2 + 3B + 2I = I\)

\((B + 2I)(B + I) = I\)

So \((B + 2I)^{-1} = B + I\)

Thus:

\(C^{-1} = [(B + 2I)(A + 2I)]^{-1} = (A + 2I)^{-1}(B + 2I)^{-1}\)

\(= (A + 3I)(B + I)\)

\(= AB + A + 3B + 3I\)

Let the first term be \(a\) and common difference be \(d\).

Given \(\sum_{\ell=0}^{18} t_{3\ell+1} = 1197\):

\(t_1 + t_4 + t_7 + \ldots + t_{55} = 1197\) (19 terms with common difference 3d)

Sum = \(\dfrac{19}{2}[2a + 18(3d)] = \dfrac{19}{2}[2a + 54d] = 1197\)

\(2a + 54d = 126\)

\(a + 27d = 63\) ...(1)

Given \(t_7 + 3t_{22} = 174\):

\((a + 6d) + 3(a + 21d) = 174\)

\(a + 6d + 3a + 63d = 174\)

\(4a + 69d = 174\) ...(2)

From equations (1) and (2): \(d = 2\) and \(a = 9\)

Now \(\sum_{\ell=1}^9 t_\ell^2 = \sum_{\ell=1}^9 [9 + (\ell-1)2]^2 = \sum_{\ell=1}^9 (2\ell + 7)^2\)

\(= \sum_{\ell=1}^9 (4\ell^2 + 28\ell + 49)\)

\(= 4\sum_{\ell=1}^9 \ell^2 + 28\sum_{\ell=1}^9 \ell + 49 \times 9\)

\(\sum_{\ell=1}^9 \ell^2 = \dfrac{9 \times 10 \times 19}{6} = 285\)

\(\sum_{\ell=1}^9 \ell = \dfrac{9 \times 10}{2} = 45\)

Sum = \(4(285) + 28(45) + 441 = 1140 + 1260 + 441 = 2841\)

So \(947b = 2841 \Rightarrow b = \dfrac{2841}{947} = 3\)

Given \(x = t^2 + 3t - 8\) and \(y = 2t^2 - 2t - 5\)

For point \((2, -1)\), find \(t\):

\(t^2 + 3t - 8 = 2\)

\(t^2 + 3t - 10 = 0\)

\((t + 5)(t - 2) = 0\)

\(t = -5\) or \(t = 2\)

Check \(y\) value:

For \(t = 2\): \(y = 2(4) - 2(2) - 5 = 8 - 4 - 5 = -1\) ✓

For \(t = -5\): \(y = 2(25) - 2(-5) - 5 = 50 + 10 - 5 = 55\) ✗

So \(t = 2\)

\(\dfrac{dx}{dt} = 2t + 3 = 2(2) + 3 = 7\)

\(\dfrac{dy}{dt} = 4t - 2 = 4(2) - 2 = 6\)

Slope of tangent = \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{6}{7}\)

Slope of normal = \(-\dfrac{1}{\text{slope of tangent}} = -\dfrac{7}{6}\)

Given \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\), \(\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}\), \(\vec{c} = \hat{i} - 2\hat{j} + \hat{k}\)

\(2\vec{a} = 2\hat{i} + 2\hat{j} + 2\hat{k}\)

\(3\vec{c} = 3\hat{i} - 6\hat{j} + 3\hat{k}\)

\(2\vec{a} - \vec{b} + 3\vec{c} = (2\hat{i} + 2\hat{j} + 2\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} - 6\hat{j} + 3\hat{k})\)

\(= (2 - 2 + 3)\hat{i} + (2 + 1 - 6)\hat{j} + (2 - 3 + 3)\hat{k}\)

\(= 3\hat{i} - 3\hat{j} + 2\hat{k}\)

Magnitude = \(\sqrt{3^2 + (-3)^2 + 2^2} = \sqrt{9 + 9 + 4} = \sqrt{22}\)

Since the magnitude is already \(\sqrt{22}\), the required vector is \(3\hat{i} - 3\hat{j} + 2\hat{k}\).

Given \(MF_2 \perp F_1F_2\), so \(M\) is vertically above \(F_2\).

Foci: \(F_1(ae, 0)\), \(F_2(-ae, 0)\). Let \(M = (-ae, k)\).

From the hyperbola equation: \(\dfrac{(-ae)^2}{a^2} - \dfrac{k^2}{b^2} = 1\)

Since \(b^2 = a^2(e^2 - 1)\):

\(k^2 = b^2(e^2 - 1) = a^2(e^2 - 1)^2\) or \(k = a(e^2 - 1)\)

Triangles \(F_1OH\) and \(F_1F_2M\) are similar, so

\(\dfrac{OH}{F_1H}=\dfrac{F_2M}{F_1F_2}\)

\(\dfrac{\lambda ae}{\sqrt{a^2e^2-\lambda^2a^2e^2}}=\dfrac{a(e^2-1)}{2ae}\)

Solving the quadratic in \(e\), we get

\(e = \dfrac{\lambda + 1}{\sqrt{1 - \lambda^2}}\)

For \(\lambda = \dfrac{2}{5}\): \(e = \dfrac{7/5}{\sqrt{21}/5} = \sqrt{\dfrac{7}{3}}\)

For \(\lambda = \dfrac{3}{5}\): \(e = \dfrac{8/5}{4/5} = 2\)

Thus \(e \in \left(\sqrt{\dfrac{7}{3}}, 2\right)\)

Variance is the square of the standard deviation.

If observations are measured in units \(u\), then:

- Standard deviation is in units \(u\)

- Variance is in units \(u^2\) (squared units)

Therefore, statement (A) is NOT correct because variance is expressed in squared units, not the same units as the observations.

All other statements are correct properties of statistical measures.

For Poisson distribution, mean per week = 3

Mean for two weeks = \(3 \times 2 = 6\)

Let \(X\) be the number of accidents in two weeks. Then \(X \sim \text{Poisson}(6)\)

The probability mass function: \(P(X = k) = \dfrac{e^{-\lambda} \lambda^k}{k!}\)

\(P(X = 3) = \dfrac{e^{-6} \cdot 6^3}{3!} = \dfrac{e^{-6} \cdot 216}{6} = 36e^{-6}\)

Given \(P(X = 3) = ke^{-6}\)

Thus \(36e^{-6} = ke^{-6}\)

\(k = 36\)

Given \(y = \dfrac{x}{1 + x \tan x}\)

Differentiating using the quotient rule:

\(\dfrac{dy}{dx} = \dfrac{(1 + x \tan x)(1) - x(\tan x + x \sec^2 x)}{(1 + x \tan x)^2}\)

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

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

For critical points, set \(\dfrac{dy}{dx} = 0\):

\(1 - x^2 \sec^2 x = 0\)

\(x^2 \sec^2 x = 1\)

\(x^2 = \cos^2 x\)

\(x = \pm \cos x\)

Using the second derivative test (or analyzing the sign change), we find:

At \(x = \cos x\), the curve attains a maximum.

At \(x = -\cos x\), the curve attains a minimum.

The word BANGLORE has 8 letters: B, A, N, G, L, O, R, E.

The string ANGLE consists of 5 letters: A, N, G, L, E.

Treat ANGLE as a single block.

Remaining letters: B, O, R

Total objects to arrange: [ANGLE block], B, O, R = 4 objects

Number of arrangements = \(4! = 24\)

Note: Within the block ANGLE, the letters must appear in that exact order (not rearranged), so no additional permutations within the block.

Let the point on the ground be \(P\). Let the lower airplane be at height \(h\) and the upper airplane at height \(4000\) m. Let the horizontal distance from \(P\) to the airplanes be \(x\).

For the lower airplane (angle of elevation \(30^\circ\)):

\(\tan 30^\circ = \dfrac{h}{x}\)

\(x = h\sqrt{3}\)

For the upper airplane (angle of elevation \(60^\circ\)):

\(\tan 60^\circ = \dfrac{4000}{x}\)

\(x = \dfrac{4000}{\sqrt{3}}\)

Equating the two expressions for \(x\):

\(h\sqrt{3} = \dfrac{4000}{\sqrt{3}}\)

\(h = \dfrac{4000}{3}\)

Vertical distance between airplanes = \(4000 - h = 4000 - \dfrac{4000}{3} = \dfrac{12000 - 4000}{3} = \dfrac{8000}{3}\) m

For a number to be a multiple of 6, it must be divisible by both 2 and 3.

The last digit must be even: 2, 4, or 6

The sum of all three digits must be divisible by 3.

Case 1: Last digit = 2

Sum of remaining two digits must make total divisible by 3.

Possible pairs: (1,3), (3,4), (6,1), (6,4) → 4 pairs × 2 orders = 8 numbers

Case 2: Last digit = 4

Possible pairs: (2,3), (3,5), (2,6), (5,6) → 4 pairs × 2 orders = 8 numbers

Case 3: Last digit = 6

Possible pairs: (1,2), (1,5), (2,4), (4,5) → 4 pairs × 2 orders = 8 numbers

Total = 8 + 8 + 8 = 24

Let \(t = x + 1\). As \(x \to \infty\), \(t \to \infty\).

The given limit becomes: \(\lim_{t \to \infty} -t \left( e^{\dfrac{1}{t}} - 1 \right)\)

Using the Taylor expansion: \(e^{\dfrac{1}{t}} = 1 + \dfrac{1}{t} + \dfrac{1}{2t^2} + \dfrac{1}{6t^3} + \cdots\)

Then \(e^{\dfrac{1}{t}} - 1 = \dfrac{1}{t} + \dfrac{1}{2t^2} + \dfrac{1}{6t^3} + \cdots\)

Multiplying by \(-t\):

\(-t \left( e^{\dfrac{1}{t}} - 1 \right) = -t \left( \dfrac{1}{t} + \dfrac{1}{2t^2} + \dfrac{1}{6t^3} + \cdots \right)\)

\(= -1 - \dfrac{1}{2t} - \dfrac{1}{6t^2} - \cdots\)

As \(t \to \infty\), all terms except \(-1\) tend to \(0\).

Therefore, the limit is \(-1\).

Even integers between 99 and 999 are from 100 to 998 inclusive.

Total even numbers = \(\dfrac{998 - 100}{2} + 1 = 450\)

Let \(A\) = even numbers divisible by 3 (i.e., divisible by 6):

First = 102, Last = 996

Count = \(\dfrac{996 - 102}{6} + 1 = 150\)

Let \(B\) = even numbers divisible by 5 (i.e., divisible by 10):

First = 100, Last = 990

Count = \(\dfrac{990 - 100}{10} + 1 = 90\)

Numbers divisible by both 3 and 5 (i.e., divisible by 30):

First = 120, Last = 990

Count = \(\dfrac{990 - 120}{30} + 1 = 30\)

By inclusion-exclusion principle:

Numbers divisible by 3 or 5 = \(150 + 90 - 30 = 210\)

Therefore, numbers NOT divisible by 3 or 5 = \(450 - 210 = 240\)

Given \(\vec{a}\) and \(\vec{c}\) are unit vectors, so \(|\vec{a}| = 1\) and \(|\vec{c}| = 1\).

The angle between \(\vec{a}\) and \(\vec{c}\) is \(\cos^{-1}\left(\dfrac{1}{4}\right)\), therefore:

\(\vec{a} \cdot \vec{c} = \cos\left(\cos^{-1}\left(\dfrac{1}{4}\right)\right) = \dfrac{1}{4}\)

We have \(\vec{b} = 2\vec{c} + \lambda \vec{a}\) and \(|\vec{b}| = 4\)

Taking the dot product of \(\vec{b}\) with itself:

\(|\vec{b}|^2 = (2\vec{c} + \lambda \vec{a}) \cdot (2\vec{c} + \lambda \vec{a})\)

\(16 = 4|\vec{c}|^2 + \lambda^2 |\vec{a}|^2 + 4\lambda (\vec{a} \cdot \vec{c})\)

\(16 = 4(1) + \lambda^2(1) + 4\lambda\left(\dfrac{1}{4}\right)\)

\(16 = 4 + \lambda^2 + \lambda\)

\(\lambda^2 + \lambda - 12 = 0\)

\((\lambda + 4)(\lambda - 3) = 0\)

\(\lambda = 3\) or \(\lambda = -4\)

Since \(\lambda > 0\), we get \(\lambda = 3\)

Since \(h(x) = \text{sgn}(g(x))\) is defined for all real \(x\) where \(g(x)\) is defined, the domain of \(h\) is exactly the same as the domain of \(g\).

Thus statement (C) is FALSE.

Statement (A) is TRUE: Domain of \(h(x)\) = Domain of \(g(x)\)

Statement (B) is TRUE: The continuity of \(h\) fails precisely at points where \(g(x) = 0\) (where the signum function has a jump discontinuity), while \(g\) may be continuous at those points.

Statement (D) is TRUE: The signum function has a discontinuity at zero, so \(h\) is discontinuous at points where \(g(x) = 0\).

Using the sum-to-product identity: \(\sin C + \sin D = 2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)\)

\(\sin x + \sin(x+1) = 2\sin\left(x + \dfrac{1}{2}\right)\cos\left(\dfrac{1}{2}\right)\)

The maximum value of \(\sin\left(x + \dfrac{1}{2}\right)\) is \(1\).

Therefore, the maximum value = \(2 \times 1 \times \cos\dfrac{1}{2} = 2\cos\dfrac{1}{2}\)

Hence \(k = 2\)

Let \(F(x) = \int \limits_{\sin^2 x}^{2\sin x} e^{t^2} \, dt\)

Using Leibniz rule for differentiation under the integral sign:

\(\dfrac{dF}{dx} = e^{(2\sin x)^2} \cdot \dfrac{d}{dx}(2\sin x) - e^{(\sin^2 x)^2} \cdot \dfrac{d}{dx}(\sin^2 x)\)

At \(x = \pi\): \(\sin \pi = 0\)

First term:

\(\dfrac{d}{dx}(2\sin x) = 2\cos x\)

At \(x = \pi\): \(2\cos \pi = 2(-1) = -2\)

So first term = \(e^{0} \times (-2) = -2\)

Second term:

\(\dfrac{d}{dx}(\sin^2 x) = 2\sin x \cos x\)

At \(x = \pi\): \(2\sin \pi \cos \pi = 2(0)(-1) = 0\)

So second term = \(e^{0} \times 0 = 0\)

Therefore, \(\dfrac{dF}{dx}\bigg|_{x=\pi} = -2 - 0 = -2\)

From \(x + y = 1\) and \(x^2 + y^2 = 2\):

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

\(2 = 1 - 2xy\)

\(xy = -\dfrac{1}{2}\)

\(x\) and \(y\) are roots of \(t^2 - t - \dfrac{1}{2} = 0\), giving two ordered pairs, so \(N = 2\)

Now calculate \(x^5 + y^5\):

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

\(x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1 - 3\left(-\dfrac{1}{2}\right)(1) = 1 + \dfrac{3}{2} = \dfrac{5}{2}\)

\(x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = 4 - 2\left(\dfrac{1}{4}\right) = \dfrac{7}{2}\)

\(x^5 + y^5 = (x^2 + y^2)(x^3 + y^3) - x^2y^2(x + y)\)

\(= 2 \cdot \dfrac{5}{2} - \dfrac{1}{4}(1) = 5 - \dfrac{1}{4} = \dfrac{19}{4}\)

Thus \(A = \dfrac{19}{4}\) and \(AN = \dfrac{19}{4} \times 2 = \dfrac{19}{2}\)

Let \(O(0, 0)\) be the vertex of the parabola and one vertex of the equilateral triangle.

Let the side length be \(a\). The other two vertices \(A\) and \(B\) are symmetric about the \(x\)-axis. For an equilateral triangle with one vertex at the origin and side length \(a\), if the opposite side is perpendicular to the axis of symmetry, the other vertices are at distance \(a\) from the origin and at distance \(\dfrac{a\sqrt{3}}{2}\) along the axis.

Let \(A = \left(\dfrac{a\sqrt{3}}{2}, \dfrac{a}{2}\right)\) (in the first quadrant)

Since \(A\) lies on \(y^2 = x\):

\(\left(\dfrac{a}{2}\right)^2 = \dfrac{a\sqrt{3}}{2}\)

\(a = 2\sqrt{3}\)

The \(x\)-coordinate of the midpoint of \(AB\) is \(\dfrac{a\sqrt{3}}{2} = \dfrac{2\sqrt{3} \cdot \sqrt{3}}{2} = 3\)

In an equilateral triangle, the centroid divides the median in ratio \(2:1\) from the vertex.

Distance from \(O\) to midpoint of \(AB\) = 3

Centroid \(G\) is at \(\dfrac{2}{3} \times 3 = 2\) from \(O\) along the \(x\)-axis.

Thus \(G = (2, 0)\)

For a three-digit number, the hundreds place cannot be 0.

Available digits: 0, 1, 2, 3, 5

Hundreds place: Can be filled by 1, 2, 3, or 5 → 4 choices

Tens place: Can be filled by any of 0, 1, 2, 3, 5 → 5 choices

Units place: Can be filled by any of 0, 1, 2, 3, 5 → 5 choices

Since repetition is allowed:

Total number of three-digit numbers = \(4 \times 5 \times 5 = 100\)

Let \( t = \sin^2 \alpha \), where \( 0 \leq t \leq 1 \).

Then \( \cos^2 \alpha = 1 - t \), so \( \cos^4 \alpha = (1 - t)^2 \).

Thus \( B = t + (1 - t)^2 \).

Simplifying, \( B = t + 1 - 2t + t^2 = t^2 - t + 1 \).

Completing the square, \( B = \left(t - \dfrac{1}{2}\right)^2 + \dfrac{3}{4} \).

At \( t = \dfrac{1}{2} \), the minimum value is \( B = \dfrac{3}{4} \).

At \( t = 0 \), \( B = 1 \).

Therefore, \( \dfrac{3}{4} \leq B \leq 1 \).

For continuity at \( x = 0 \), we require \( \lim_{x \to 0} f(x) = f(0) = 0 \).

Consider \( \lim_{x \to 0} x^\alpha \sin \left( \dfrac{1}{x^\beta} \right) \).

The sine term oscillates between \( -1 \) and \( 1 \).

Hence \( |x^\alpha \sin(\cdot)| \leq |x^\alpha| \).

The limit \( x^\alpha \to 0 \) as \( x \to 0 \) if and only if \( \alpha > 0 \), irrespective of the value of \( \beta \).

Therefore \( \lim_{x \to 0} f(x) = 0 \) for all \( \alpha > 0 \) and any \( \beta \in \mathbb{R} \).

Thus \( f \) is continuous at \( x = 0 \) for all \( \alpha > 0 \) and \( \beta \in \mathbb{R} \).

Options (A) and (D) are false since \( \beta \) does not affect continuity and \( \alpha > 0 \) is necessary.

Option (B) is false because differentiability requires stronger conditions on \( \alpha \) and \( \beta \).

Rewrite the expression as

\((\cos 10^\circ \cos 50^\circ \cos 70^\circ)(\cos 10^\circ \cos 20^\circ \cos 40^\circ)\)

\(= \dfrac{\cos (3\times10^\circ)}{4} \left( \dfrac{\sin 80^\circ}{8 \sin 10^\circ}\right)\)

\(= \dfrac{\sqrt{3}\sin 80^\circ}{64 \sin 10^\circ}\)

Given that \(\dfrac{\sqrt{3}\sin 80^\circ}{64 \sin 10^\circ}=\dfrac{\sqrt{3}}{16} \cos 10^\circ + a\)

\(\Rightarrow a = \dfrac{\sqrt{3}}{64} \left[\dfrac{\sin 80^\circ}{\sin 10^\circ}-4\cos 10^\circ \right]\)

\(\Rightarrow a = \dfrac{\sqrt{3}}{64} \left[\dfrac{\sin 80^\circ-4\cos 10^\circ \sin 10^\circ }{\sin 10^\circ}\right]\)

\(\Rightarrow a = \dfrac{\sqrt{3}}{64} \left[\dfrac{\sin 80^\circ- \sin 20^\circ- \sin 20^\circ }{\sin 10^\circ}\right]\)

\(\Rightarrow a = \dfrac{\sqrt{3}}{64} \left[\dfrac{2\cos 50^\circ \sin 30^\circ- \sin 20^\circ }{\sin 10^\circ}\right]\)

\(\Rightarrow a = \dfrac{\sqrt{3}}{64} \left[\dfrac{\cos 50^\circ - \sin 20^\circ }{\sin 10^\circ}\right]\)

\(\Rightarrow a = \dfrac{\sqrt{3}}{64} \left[\dfrac{\sin 40^\circ - \sin 20^\circ }{\sin 10^\circ}\right]\)

\(\Rightarrow a = \dfrac{\sqrt{3}}{64} \left[\dfrac{2\cos 30^\circ \sin 10^\circ }{\sin 10^\circ}\right]\)

\(\Rightarrow a=\dfrac{3}{64}\)

Thus \( 3a^{-1} = 3 \times \dfrac{64}{3} = 64 \).

We require \(x \in A\) and \(y \in A\) such that \(y = 2x - 7\).

Since \(y \in A\), we must have \(1 \leq 2x - 7 \leq 20\).

From \(2x - 7 \geq 1\), we get \(2x \geq 8\) or \(x \geq 4\).

From \(2x - 7 \leq 20\), we get \(2x \leq 27\) or \(x \leq 13\).

Thus \(x\) can take integer values from 4 to 13.

The number of such integers is \(13 - 4 + 1 = 10\).

For each such \(x\), the corresponding \(y = 2x - 7\) lies in \(A\).

Hence, the number of elements in \(R\) is \(10\).

Let height of multistoried building = \(h\) m and horizontal distance = \(x\) m.

Angle of depression of top (at height 8 m) = 30°:

\(\tan 30^\circ = \dfrac{h - 8}{x}\)

\(\dfrac{1}{\sqrt{3}} = \dfrac{h - 8}{x}\) ...(1)

Angle of depression of bottom = 45°:

\(\tan 45^\circ = \dfrac{h}{x}\)

\(1 = \dfrac{h}{x}\), so \(h = x\) ...(2)

Substituting (2) into (1):

\(\dfrac{1}{\sqrt{3}} = \dfrac{x - 8}{x}\)

\(x = \sqrt{3}(x - 8)\)

\(x = \sqrt{3}x - 8\sqrt{3}\)

\(x - \sqrt{3}x = -8\sqrt{3}\)

\(x(1 - \sqrt{3}) = -8\sqrt{3}\)

\(x = \dfrac{-8\sqrt{3}}{1 - \sqrt{3}} = \dfrac{8\sqrt{3}}{\sqrt{3} - 1}\)

Rationalizing: \(x = \dfrac{8\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \dfrac{8\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \dfrac{8\sqrt{3}(\sqrt{3} + 1)}{2}\)

\(= 4\sqrt{3}(\sqrt{3} + 1) = 4(3 + \sqrt{3})\)

Thus \(h = x = 4(3 + \sqrt{3})\) m

Given: \(\alpha + \beta = -a\) and \(\alpha\beta = b\)

New roots: \(p = \dfrac{1}{\alpha^3 + \alpha} = \dfrac{1}{\alpha(\alpha^2 + 1)}\) and \(q = \dfrac{1}{\beta(\beta^2 + 1)}\)

Product of roots:

\(P = pq = \dfrac{1}{\alpha\beta(\alpha^2 + 1)(\beta^2 + 1)} = \dfrac{1}{b[(\alpha\beta)^2 + \alpha^2 + \beta^2 + 1]}\)

Since \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = a^2 - 2b\):

\(P = \dfrac{1}{b[b^2 + (a^2 - 2b) + 1]} = \dfrac{1}{b(b^2 + a^2 - 2b + 1)}\)

Sum of roots:

\(S = p + q = \dfrac{\alpha^3 + \beta^3 + \alpha + \beta}{b(b^2 + a^2 - 2b + 1)}\)

\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = (-a)^3 - 3b(-a) = -a^3 + 3ab\)

Numerator = \((-a^3 + 3ab) + (-a) = -a^3 - a + 3ab\)

\(S = \dfrac{-a^3 - a + 3ab}{b(b^2 + a^2 - 2b + 1)}\)

The quadratic equation is \(x^2 - Sx + P = 0\).

Multiplying by \(b(b^2 + a^2 - 2b + 1)\):

\(b(b^2 + a^2 - 2b + 1)x^2 - (a^3 + a - 3ab)x + 1 = 0\)

The curves \(y = \sin x\) and \(y = \cos x\) intersect when \(\sin x = \cos x\), which gives \(x = \dfrac{\pi}{4}\).

For \(0 \leq x \leq \dfrac{\pi}{4}\): \(\cos x \geq \sin x\)

For \(\dfrac{\pi}{4} \leq x \leq \dfrac{\pi}{2}\): \(\sin x \geq \cos x\)

Area = \(\int \limits_0^{\pi/4} (\cos x - \sin x) \, dx + \int \limits_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx\)

First integral:

\(\int \limits_0^{\pi/4} (\cos x - \sin x) \, dx = [\sin x + \cos x]_0^{\pi/4}\)

\(= \left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right) - (0 + 1) = \sqrt{2} - 1\)

Second integral:

\(\int \limits_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = [-\cos x - \sin x]_{\pi/4}^{\pi/2}\)

\(= (0 - 1) - \left(-\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\right) = -1 + \sqrt{2} = \sqrt{2} - 1\)

Total area = \((\sqrt{2} - 1) + (\sqrt{2} - 1) = 2(\sqrt{2} - 1)\)

Volume of parallelepiped = \(|\vec{a} \cdot (\vec{b} \times \vec{c})| = 5\)

First, compute \(\vec{b} \times \vec{c}\):

\(\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & 1 & \lambda \end{vmatrix}\)

\(= \hat{i}(2\lambda - (-1)) - \hat{j}(\lambda - (-3)) + \hat{k}(1 - 6)\)

\(= (2\lambda + 1)\hat{i} - (\lambda + 3)\hat{j} - 5\hat{k}\)

Now compute \(\vec{a} \cdot (\vec{b} \times \vec{c})\):

\(\vec{a} \cdot (\vec{b} \times \vec{c}) = 2(2\lambda + 1) + (-3)(-(\lambda + 3)) + 4(-5)\)

\(= 4\lambda + 2 + 3\lambda + 9 - 20\)

\(= 7\lambda - 9\)

Given \(|7\lambda - 9| = 5\):

Case 1: \(7\lambda - 9 = 5\) → \(7\lambda = 14\) → \(\lambda = 2\)

Case 2: \(7\lambda - 9 = -5\) → \(7\lambda = 4\) → \(\lambda = \dfrac{4}{7}\)

Since \(\dfrac{4}{7}\) is not among the options, we take \(\lambda = 2\).

From \(\dfrac{x}{4} + \dfrac{y}{2} = 1\), \(A = (4, 0)\) and \(B = (0, 2)\) and the midpoint  of the line \(AB,\; M =(2, 1)\).

Given that \(AP = BP\), so the point P lies on the perpendicular bisector of the line \(AB\). Slope of \(AB = \dfrac{2-0}{0-4} = -\dfrac{1}{2}\)

Perpendicular bisector through M has slope 2, its equation is: \(y - 1 = 2(x - 2)\) or \(y = 2x - 3\)

The point P is the intersection of this line with \(x+y=1\). The point P is: \(x = \dfrac{4}{3}\), \(y = -\dfrac{1}{3}\)

If \(M'\) is the mirror image of the point \(M\), then \(PM'=PM\).

The distance \(PM'=PM=\sqrt{\left(\dfrac{4}{3} - 2\right)^2 + \left(-\dfrac{1}{3} - -1\right)^2}\)

\(= \sqrt{\dfrac{16}{9} + \dfrac{4}{9}} = \sqrt{\dfrac{20}{9}} = \dfrac{2\sqrt{5}}{3}\)

Let \(B\) be the event of choosing the blue coin and \(R\) be the event of choosing the red coin.

Since the coin is chosen randomly: \(P(B) = P(R) = \dfrac{1}{2}\)

Let \(H\) be the event of getting heads. Then:

\(P(H|B) = 0.99\) and \(P(H|R) = 0.01\)

By the total probability theorem:

\(P(H) = P(B) \cdot P(H|B) + P(R) \cdot P(H|R)\)

\(P(H) = \dfrac{1}{2} \times 0.99 + \dfrac{1}{2} \times 0.01\)

\(P(H) = \dfrac{1}{2}(0.99 + 0.01) = \dfrac{1}{2} \times 1 = 0.5\)