We know the formula that \(\binom{m}{k} + \binom{m}{k+1} = \binom{m+1}{k+1}\)

Here \(m = 15\) and \(k = 7\), hence the desired sum is \(\binom{16}{8}\)

Thus \(n = 16\) and \(r = 8\).

Suppose number of students reading Hitavad and Hindustan are \(n(A)\) and \(n(B)\), then

\(n(A \cup B) = n(A) + n(B) - n(A \cap B)\)

\(\Rightarrow 50 - 10 = 30 + 35 - n(A \cap B)\)

\(\Rightarrow n(A \cap B) = 25\)

Given that \(A = \{4^x - 3x - 1\}\) and \(B = \{9(x - 1)\}\)

By putting \(x = 1, 2, 3, 4, \ldots\) we see that,

\(A = \{0, 9, 54, \ldots\}\) and \(B = \{0, 9, 18, 27, \ldots\}\)

The set A is a proper subset of B or \(A \subset B\)

Alternately we can expand A as

\((1 + 3)^x - 3x - 1 = 1 + 3x + \binom{x}{2} \cdot 3^2 + \ldots - 3x - 1\) = a multiple of 9.

Here A and B are both multiples of 9, but B contains each multiple of 9 while A does not. Thus A is a subset of B.

A has 3 elements, hence number of subsets = \(2^3 = 8\).

In DELHI all the letters are different. The first letter is fixed, so the remaining 4 letters can be permuted in 4! ways.

Number of such words = 4! = 24.

Out of the 10 friends, 3 are not attending the party, so Naresh has to select 6 friends out of 7 friends.

Number of ways = \(^7C_6 = 7\)

Total number of prizes are 10 and there are 4 recipients, number of ways is equal to the number of solutions of the equation \(x + y + z + w = 10\), where \(x, y, z, w \ge 1\).

Hence the number of solutions = \(^{10-1}C_{4-1} = ^9C_3 = 84\)

→ Read detailed chapter on Beggar's Method.

Suppose the distance AB = BC = CA = \(x\), then total journey = \(3x\)

Total time taken in the journey = \(\dfrac{x}{30} + \dfrac{x}{40} + \dfrac{x}{50} = \dfrac{47x}{600}\).

Hence the average speed = \(\dfrac{\text{Total Journey}}{\text{Total time}} = \dfrac{3x}{\frac{47x}{600}} = \dfrac{1800}{47} = 38.3\)

Probability that B will solve the problem = \(1 - \dfrac{1}{4} = \dfrac{3}{4}\).

Let probability that C will solve the problem = \(p\),

then probability that none of them is able to solve the problem = \(\dfrac{1}{2} \times \dfrac{1}{4} \times (1 - p) = \dfrac{1-p}{8}\).

From the given information,

\(\dfrac{1-p}{8} = \dfrac{1}{64} \Rightarrow p = \dfrac{7}{8}\)

A can select any number from the first 25 numbers with probability \(\dfrac{25}{25}\).

Now B has to select a number different from A, this can be done with probability \(\dfrac{24}{25}\).

Required probability = \(\dfrac{24}{25}\).

Mean of A = \(\dfrac{2 + 3 + 7 + 1 + 3 + 2 + 3}{7} = 3\)

Median of A = 3 and Mode of A = 3

Choice (D) is correct.

The formula for standard deviation is \(\sqrt{\dfrac{\sum fx^2}{n} - (\bar{x})^2}\), where \(\bar{x}\) is the mean of the data and \(\bar{x} = \dfrac{\sum fx}{n}\)

Hence \(\bar{x} = \dfrac{3 \cdot 6 + 6 \cdot 7 + 9 \cdot 8 + \ldots + 4 \cdot 12}{3 + 6 + 9 + \ldots + 4} = 9\)

\(\sum fx^2 = 4012\)

Hence standard deviation = \(\sqrt{\dfrac{4012}{48} - 9^2} = 1.607\)

Given that \(A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}\), then

\(A^2 = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}\)

\(= \begin{bmatrix} \cos^2\alpha - \sin^2\alpha & 2\sin\alpha\cos\alpha \\ -2\sin\alpha\cos\alpha & \cos^2\alpha - \sin^2\alpha \end{bmatrix}\)

\(= \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix}\)

Similarly, it can be proved that

\(A^n = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{bmatrix}\).

Roots are \(m\) and \(n\), thus \(m + n = \dfrac{2b}{a}\), \(mn = \dfrac{c}{a}\)

Now \(\dfrac{b}{an^2 + c} + \dfrac{b}{am^2 + c} = \dfrac{b}{a}\left(\dfrac{1}{n^2 + c/a} + \dfrac{1}{m^2 + c/a}\right)\)

\(= \dfrac{m + n}{2}\left(\dfrac{1}{n^2 + mn} + \dfrac{1}{m^2 + mn}\right)\)

\(= \dfrac{m + n}{2}\left(\dfrac{m + n}{mn(m + n)}\right) = \dfrac{m + n}{2mn} = \dfrac{b}{c}\).

To possess a non-zero solution, \(\begin{vmatrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{vmatrix} = 0\)

\(\Rightarrow 4(2) - k(k - 2) + 2(2k - 8) = 0\)

\(\Rightarrow k^2 - 6k + 8 = 0\)

Hence \(k = 2, 4\). There are two possible values.

Only third choice is wrong, the correct statement should be \(\det(cA) = c^n\det(A)\)

where \(n\) is the order of the determinant.

The equation of the ellipse is \(\dfrac{x^2}{16} + \dfrac{y^2}{1} = 1\) and the slope of the tangent = \(\tan 60° = \sqrt{3}\).

If the equation of the tangent is \(y = mx + c\), then \(c^2 = a^2m^2 + b^2\)

\(\Rightarrow c^2 = 16 \times 3 + 1 = 49\) Or \(c = \pm 7\)

Equation of the tangent is \(y = \sqrt{3}x \pm 7\)

Or \(\sqrt{3}x - y \pm 7 = 0\)

The centre of the ellipse is \((0, 1)\) and its major axis lies on y-axis.

The circle passes through one of the vertices \((0, -2)\) of the ellipse.

Hence, there will be two intersection points.

First term \(a = 3\), let common difference = \(d\)

Sum of first \(n\) terms = \(\dfrac{n}{2}[2a + (n-1)d]\)

Given: \(S_8 = 2S_5\)

\(\dfrac{8}{2}[2(3) + 7d] = 2 \times \dfrac{5}{2}[2(3) + 4d]\)

\(4[6 + 7d] = 5[6 + 4d]\;\Rightarrow d = \dfrac{3}{4}\)

\(\dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab} = \dfrac{a^3 + b^3 + c^3}{abc}\)

\(= \dfrac{a^3 + b^3 + c^3 - 3abc + 3abc}{abc}\)

\(= \dfrac{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}{abc} + 3= 3\)

We know that \(-1 \le \sin\left(\dfrac{1}{x}\right) \le 1\), so

\(e^{-1} \le e^{\sin\left(\dfrac{1}{x}\right)} \le e^1\)

\(\Rightarrow \displaystyle \lim_{x \to 0} x^2 e^{\sin\left(\dfrac{1}{x}\right)} = \lim_{x \to 0} x^2\lambda\), where \(\lambda\) is a finite number that lies between \(e^{-1}\) and \(e\).

Therefore \(\displaystyle \lim_{x \to 0} x^2 e^{\sin\left(\dfrac{1}{x}\right)} = 0\).

The graph of the function shows that the point at \(x = 0\) is the point of minima.

For \(x \le 0\): \(f(x) = x^2\) (parabola opening upward)

For \(x > 0\): \(f(x) = 2\sin x\) (oscillating but starts from 0)

At \(x = 0\), both parts meet and the function has a minimum value of 0.

The value of the function \(g(x)\) when \(x \ne 0\) is \(\dfrac{x^2 - 2x}{2x} = \dfrac{x}{2} - 1\).

When \(x = 0 + h\) or \(x = 0 - h\), then \(g(x) = -1\).

\(\lim_{x \to 0} g(x) = \lim_{x \to 0} \left(\dfrac{x}{2} - 1\right) = -1\)

As the function is continuous, so \(k = -\dfrac{1}{2}\).

The graph is given by \(y = 2x^3 e^x\)

\(\dfrac{dy}{dx} = 6x^2 e^x + 2x^3 e^x = 2x^2 e^x(3 + x)\)

For increasing, \(\dfrac{dy}{dx} > 0\)

\(\Rightarrow 2x^2 e^x(3 + x) > 0\)

Since \(x^2 \ge 0\) and \(e^x > 0\) always, we need \(3 + x > 0\) or \(x > -3\)

\(\int \sec^2 x \operatorname{cosec}^4 x \, dx = \int \dfrac{1}{\cos^2 x \sin^4 x} \, dx\)

\(= \int \dfrac{1}{\cot^2 x \sin^6 x} dx = \int \dfrac{\operatorname{cosec}^6 x}{\cot^2 x} \, dx\)

\(= \int \dfrac{(1 + \cot^2 x)^2 \operatorname{cosec}^2 x}{\cot^2 x} dx\)

Now put \(\cot x = t\), then \(-\operatorname{cosec}^2 x \, dx = dt\)

\(\Rightarrow -\int \dfrac{(1 + t^2)^2}{t^2} \, dt = -\dfrac{t^3}{3} - 2t + \dfrac{1}{t}\)

\(= -\dfrac{1}{3}\cot^3 x - 2\cot x + \tan x\)

Hence \(k = 1\)

\(\int e^x \left(\dfrac{1 + \sin x \cos x}{\cos^2 x}\right) dx = \int e^x (\sec^2 x + \tan x) \, dx\)

Using the result \(\int e^x[f(x) + f'(x)] dx = e^x f(x) + c\)

Here \(f(x) = \tan x\) and \(f'(x) = \sec^2 x\)

The answer is: \(e^x \tan x + c\).

\(I_n = \int \limits_0^a (x^2 - a^2)^n dx\)

Integrating by parts,

\(= \left[x(x^2 - a^2)^n\right]_0^a - \int \limits_0^a n(x^2 - a^2)^{n-1} \cdot 2x^2 dx\)

\(\Rightarrow I_n = 0 - 2n\int \limits_0^a x^2(x^2 - a^2)^{n-1} dx\)

\(\Rightarrow I_n = -2n\int \limits_0^a (x^2 - a^2 + a^2)(x^2 - a^2)^{n-1} dx\)

\(= -2n\int \limits_0^a (x^2 - a^2)^n dx + 2na^2\int \limits_0^a (x^2 - a^2)^{n-1} dx\)

\(\Rightarrow I_n = -2nI_n + 2na^2 I_{n-1}\)

Hence \(I_n = \dfrac{2na^2}{2n+1} I_{n-1}\)

\(\int \limits_{-2}^{2} (ax^5 + bx^3 + c) dx\)

\(= \left(\dfrac{ax^6}{6} + \dfrac{bx^4}{4} + cx\right)_{-2}^{2}\)

\(= 4c\)

Therefore, the answer depends upon the value of \(c\).

The line \(y = 3 - x\) intersects the x-axis at \(x = 3\) (when \(y = 0\)).

The parabola \(y = x^2 - 9\) intersects \(y = 0\) at \(x = \pm 3\).

Given constraints: \(x \ge -1\) and \(y \ge 0\).

The bounded region forms a triangle with vertices at appropriate points.

The area = \(\dfrac{4 \times 4}{2} = 8\).

\((\vec{a} + \vec{b} + \vec{c}) \cdot [(\vec{a} + \vec{b}) \times (\vec{a} + \vec{c})]\) can be expanded as:

\(= (\vec{a} + \vec{b} + \vec{c}) \cdot [\vec{a} \times \vec{a} + \vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{b} \times \vec{c}]\)

\(= (\vec{a} + \vec{b} + \vec{c}) \cdot [\vec{a} \times \vec{c} + \vec{b} \times \vec{a} + \vec{b} \times \vec{c}]\)

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

\(= [\vec{a} \, \vec{b} \, \vec{c}] + [\vec{b} \, \vec{a} \, \vec{c}] + [\vec{c} \, \vec{b} \, \vec{a}]\)

\(= [\vec{a} \, \vec{b} \, \vec{c}] - [\vec{a} \, \vec{b} \, \vec{c}] - [\vec{a} \, \vec{b} \, \vec{c}] = -[\vec{a} \, \vec{b} \, \vec{c}]\)

Resultant force of two vectors \(F_1, F_2\) is \(\sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos\theta}\)

When \(\theta = 90°\), then resultant is \(\sqrt{F_1^2 + F_2^2}\).

Thus \(\theta = 90°\).

Given that \(\vec{a} + \vec{b} + \vec{c}\) is collinear with \(\vec{d}\), so \(\vec{a} + \vec{b} + \vec{c} = \lambda\vec{d}\) …(1)

Also \(\vec{b} + \vec{c} + \vec{d}\) is collinear with \(\vec{a}\), so \(\vec{b} + \vec{c} + \vec{d} = \mu\vec{a}\) …(2)

From equation (1) – equation (2),

\(\vec{a} - \vec{d} = \lambda\vec{d} - \mu\vec{a}\)

By comparison, \(\lambda = -1, \mu = -1\)

Putting the value of \(\lambda\) in equation (1), we have,

\(\vec{a} + \vec{b} + \vec{c} = -\vec{d}\)

\(\Rightarrow \vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}\)

Three forces are:

\(f_1 = 5\left(\dfrac{6\hat{i} + 2\hat{j} + 3\hat{k}}{\sqrt{6^2 + 2^2 + 3^2}}\right) = \dfrac{5}{7}(6\hat{i} + 2\hat{j} + 3\hat{k})\)

\(f_2 = 3\left(\dfrac{3\hat{i} - 2\hat{j} + 6\hat{k}}{\sqrt{3^2 + (-2)^2 + 6^2}}\right) = \dfrac{3}{7}(3\hat{i} - 2\hat{j} + 6\hat{k})\)

\(f_3 = 1\left(\dfrac{2\hat{i} - 3\hat{j} - 6\hat{k}}{\sqrt{2^2 + (-3)^2 + (-6)^2}}\right) = \dfrac{1}{7}(2\hat{i} - 3\hat{j} - 6\hat{k})\)

Net force \(f = \dfrac{1}{7}(41\hat{i} + \hat{j} + 27\hat{k})\)

Displacement = \(\vec{d} = (3\hat{i} + 4\hat{k})\)

Work done = \(f \cdot d = \dfrac{123 + 108}{7} = 33\)

This is a direct formula. Position vector of the point \(p\) is \(\dfrac{n\vec{a} + m\vec{b}}{m + n}\).

Given that \(\vec{a} + 2\vec{b}\) is collinear with \(\vec{c}\)

\(\Rightarrow \vec{a} + 2\vec{b} = \lambda\vec{c}\) …(1)

Also \(\vec{b} + 3\vec{c}\) is collinear with \(\vec{a}\)

\(\Rightarrow \vec{b} + 3\vec{c} = \mu\vec{a}\) …(2)

Equating the value of \(2\vec{b}\) from the above two equations, we have

\(\lambda\vec{c} - \vec{a} = 2\mu\vec{a} - 6\vec{c}\)

By comparison, we have \(\lambda = -6\).

Putting this value in the first equation, we have

\(\vec{a} + 2\vec{b} = -6\vec{c}\)

Or \(|\vec{a} + 2\vec{b} + 6\vec{c}| = 0\)

Suppose the vertices are A, B and C, then

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

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

\(\vec{CA} = -2\hat{i} - \hat{j}\)

Here \(\vec{BC} \cdot \vec{CA} = -2 + 2 = 0\)

Hence the triangle is right angled.

Volume = \(\begin{vmatrix} 2 & 3 & 4 \\ 1 & \alpha & 2 \\ 1 & 2 & \alpha \end{vmatrix} = 15\)

\(\Rightarrow 2(\alpha^2 - 4) - 3(\alpha - 2) + 4(2 - \alpha) = 15\)

\(\Rightarrow 2\alpha^2 - 7\alpha - 9 = 0\)

\(\Rightarrow (\alpha + 1)(2\alpha - 9) = 0\)

Hence \(\alpha = -1, \dfrac{9}{2}\)

Given that \(\sin^2 x - \sin x - 2 = 0\)

\(\Rightarrow (\sin x - 2)(\sin x + 1) = 0\)

\(\Rightarrow \sin x = -1\) (since \(\sin x = 2\) is not possible)

Hence \(x = \dfrac{3\pi}{2}\)

Let \(\tan x = 2k, \tan y = 3k, \tan z = 5k\)

We know that when \(x + y + z = \pi\),

\(\tan x + \tan y + \tan z = \tan x \tan y \tan z\)

\(\Rightarrow 2k + 3k + 5k = (2k)(3k)(5k)\)

\(\Rightarrow 10k = 30k^3\) or \(k^2 = \dfrac{1}{3}\)

Hence \(\tan^2 x + \tan^2 y + \tan^2 z = 4k^2 + 9k^2 + 25k^2\)

\(= 38k^2 = \dfrac{38}{3}\)

Multiply and divide by \(\sin 72°\), we get

\(\dfrac{\sin 12° \sin 48° \sin 72° \sin 54°}{\sin 72°}\)

\(= \dfrac{\sin 36° \sin 54°}{4\sin 72°}\) (using \(\sin\theta \sin(60° - \theta) \sin(60° + \theta) = \dfrac{\sin 3\theta}{4}\))

\(= \dfrac{2\sin 36° \sin 54°}{8\sin 72°} = \dfrac{2\sin 36° \cos 36°}{8\sin 72°} = \dfrac{1}{8}\)

Eliminating \(y\) from the given relations,

\(\cos x = \tan y = \dfrac{1}{\cot y} = \dfrac{1}{\tan z} = \cot z = \tan x\)

\(\Rightarrow \cos x = \tan x\)

Or \(\sin x = \cos^2 x = 1 - \sin^2 x\)

\(\Rightarrow \sin^2 x + \sin x - 1 = 0\)

\(\Rightarrow \sin x = \dfrac{-1 + \sqrt{5}}{2}\)

\(\tan\left(45° + \dfrac{\theta}{2}\right) = \dfrac{1 + \tan\dfrac{\theta}{2}}{1 - \tan\dfrac{\theta}{2}}\)

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

\(= \dfrac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta\)

The value of \(\sin 10° \sin 50° \sin 70°\)

\(= \dfrac{\sin(3 \times 10°)}{4} = \dfrac{\sin 30°}{4} = \dfrac{1}{8}\)

(using \(\sin\theta \sin(60° - \theta) \sin(60° + \theta) = \dfrac{\sin 3\theta}{4}\))

The given expression is:

\(\dfrac{\dfrac{\sin x}{\cos x}}{1 - \dfrac{\cos x}{\sin x}} + \dfrac{\dfrac{\cos x}{\sin x}}{1 - \dfrac{\sin x}{\cos x}}\)

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

\(= \dfrac{\sin^3 x - \cos^3 x}{\sin x \cos x(\sin x - \cos x)}\)

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

\(= \dfrac{1 + \sin x \cos x}{\sin x \cos x} = 1 + \sec x \operatorname{cosec} x\)

Let the tower is PQ, and its height is \(h\), then

\(AQ = h\sqrt{3}\), \(BQ = h\)

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

\(\Rightarrow \dfrac{AB}{BC} = \dfrac{h\sqrt{3} - h}{h - \dfrac{h}{\sqrt{3}}} = \dfrac{3 - \sqrt{3}}{\sqrt{3} - 1} = \sqrt{3}\)

The graphs will intersect at the points \((1, 1)\) and \((0, 0)\)

Shaded area = \(\int_0^1 (\sqrt{x} - x) \, dx = \left[\dfrac{x^{3/2}}{3/2} - \dfrac{x^2}{2}\right]_0^1\)

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

\(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \left(\dfrac{5}{2} - x\right) = \dfrac{1}{2}\)

\(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \left(x - \dfrac{3}{2}\right) = \dfrac{1}{2}\)

\(f(2) = 1\)

Since \(\lim_{x \to 2} f(x) = \dfrac{1}{2} \ne f(2) = 1\)

The function is discontinuous at \(x = 2\).

The given expression can be written as

\(2\tan^{-1}\left[\operatorname{cosec}\left(\operatorname{cosec}^{-1}\left(\dfrac{\sqrt{1+x^2}}{x}\right)\right) - \tan\left(\tan^{-1}\left(\dfrac{1}{x}\right)\right)\right]\)

\(= 2\tan^{-1}\left[\dfrac{\sqrt{1+x^2}}{x} - \dfrac{1}{x}\right] = 2\tan^{-1}\left[\dfrac{\sqrt{1+x^2} - 1}{x}\right]\)

Put \(x = \tan\theta\),

\(= 2\tan^{-1}\left[\dfrac{\sec\theta - 1}{\tan\theta}\right] = 2\tan^{-1}\left(\dfrac{1 - \cos\theta}{\sin\theta}\right)\)

\(= 2\tan^{-1}\left(\dfrac{2\sin^2\dfrac{\theta}{2}}{2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}}\right) = 2\tan^{-1}\left(\tan\dfrac{\theta}{2}\right)\)

\(= \theta = \tan^{-1} x\)

Given that \(\dfrac{3}{5}\sin x + \dfrac{4}{5}\cos x = 1\)

Since \(\sin^2 x + \cos^2 x = 1\), so \(\sin x = \dfrac{3}{5}, \cos x = \dfrac{4}{5}\)

\(\Rightarrow \dfrac{1 - \tan^2\dfrac{x}{2}}{1 + \tan^2\dfrac{x}{2}} = \dfrac{4}{5}\)

or \(\tan\dfrac{x}{2} = \dfrac{1}{3}\)

\(\Rightarrow 6\tan\dfrac{x}{2} - 9\tan^2\dfrac{x}{2} = \dfrac{6}{3} - \dfrac{9}{9} = 1\).

Given that \(A \subseteq B\) and \(B \subseteq C\), hence \(A \subseteq B \subseteq C\)

Therefore \(A \cup B \cup C = C\).

So, cardinality will be same as that of \(C\).