Given line: \(x = 3 - 3y\). Substitute the value of \(x\) into the equation of the circle:

\((3 - 3y)^2 + y^2 + 10(3 - 3y) - 12y + 51 = 0\)

\(9 - 18y + 9y^2 + y^2 + 30 - 30y - 12y + 51 = 0\)

\(\Rightarrow y^2 - 6y + 9 = 0\)

\((y - 3)^2 = 0 \Rightarrow y = 3\) and \(x=-6\)

Therefore, point of intersection is \((-6, 3)\)

Using \(\sin C+\sin D\) identity: \(\sin x + \sin 5x = 2\sin 3x \cos 2x\)

So equation becomes: \(2\sin 3x \cos 2x = \sin 3x\)

\(\sin 3x (2\cos 2x - 1) = 0\)

Case 1: \(\sin 3x = 0 \Rightarrow 3x = n\pi \Rightarrow x = \dfrac{n\pi}{3}\)

In \([0,\pi]\): \(x = 0, \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi\): 4 solutions

Case 2: \(\cos 2x = \dfrac{1}{2} \Rightarrow 2x = 2n\pi \pm \dfrac{\pi}{3}\)

\(x = n\pi \pm \dfrac{\pi}{6}\)

In \([0,\pi]\): \(x = \dfrac{\pi}{6}\) (for \(n=0\)), \(x = \dfrac{5\pi}{6}\) (for \(n=1\)): 2 solutions

As none of the solutions is common, so there are 6 solutions.

For an acute triangle, the minimum value of \(\sec A + \sec B + \sec C\) occurs when the triangle is equilateral.

When \(A = B = C = 60°\):

Therefore, \(\sec A + \sec B + \sec C = 2 + 2 + 2 = 6\)

Thus, the least value is 6.

For set P: \(\sin \theta - \cos \theta = \sqrt{2}\cos \theta\)

\(\sin \theta = \cos \theta + \sqrt{2}\cos \theta = \cos \theta(1 + \sqrt{2})\)

\(\tan \theta = 1 + \sqrt{2}\)

For set Q: \(\sin \theta + \cos \theta = \sqrt{2}\sin \theta\)

\(\cos \theta = \sqrt{2}\sin \theta - \sin \theta = \sin \theta(\sqrt{2} - 1)\)

\(\cot \theta = \sqrt{2} - 1\)

\(\tan \theta = \dfrac{1}{\sqrt{2} - 1} = \dfrac{1(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \sqrt{2} + 1\)

Both sets satisfy \(\tan \theta = \sqrt{2} + 1\)

Therefore \(P = Q\)

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

Since \(x + y + z = \pi\), we have \(\tan x +\tan y+ \tan z = \tan x \tan y \tan z\)

\(\Rightarrow 2k +3k +5k=(2k)(3k)(5k) \) or \(k^2=\dfrac{1}{3}\)

Then \(\tan^2 x + \tan^2 y + \tan^2 z = 4k^2 + 9k^2 + 25k^2 = 38k^2\)

\(= 38 \times \dfrac{1}{3} = \dfrac{38}{3}\)

The centre and the radius of the first circle are: \(C_1(a,0)\),\(r_1 = \sqrt{a^2 - c^2}\)

For the second circle, centre and the radius are: \(C_2(0,b)\), radius \(r_2 = \sqrt{b^2 - c^2}\)

Distance between centres: \(d = \sqrt{a^2 + b^2}\)

To touch externally, \(d = r_1 + r_2\)

\(\sqrt{a^2 + b^2} = \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2}\)

Squaring both sides:

\(a^2 + b^2 = a^2 - c^2 + b^2 - c^2 + 2\sqrt{(a^2 - c^2)(b^2 - c^2)}\)

\(c^2 = \sqrt{(a^2 - c^2)(b^2 - c^2)}\)

Squaring again: \(c^4 = (a^2 - c^2)(b^2 - c^2)\)

\(c^4 = a^2 b^2 - a^2 c^2 - b^2 c^2 + c^4\)

\(0 = a^2 b^2 - a^2 c^2 - b^2 c^2\)

Dividing by \(a^2 b^2 c^2\): \(\dfrac{1}{c^2} = \dfrac{1}{b^2} + \dfrac{1}{a^2}\)

The given lines are:

\(\ell_1: (1+p)x - py + p(1+p) = 0\), let this line cuts \(x\) axis at \(A\), then \(A = (-p,0)\)

Slope of this line is \(\dfrac{1+p}{p}\)

\(\ell_2: (1+q)x - qy + q(1+q) = 0\), let this line cuts \(x\) axis at \(B\), then \(B = (-q,0)\)

Slope of this line is \(\dfrac{1+q}{q}\)

\(\ell_3: y = 0\) (x-axis)

Equation of the line that is perpendicular to the line \(\ell_2\) and passing through \(A\) is:

\(y-0=\dfrac{-q}{1+q}(x+p)\) or \(y(1+q)+qx+qp=0\)

Equation of the line that is perpendicular to the line \(\ell_1\) and passing through \(B\) is:

\(y-0=\dfrac{-p}{1+p}(x+q)\) or \(y(1+p)+px+qp=0\)

Subtracting the second equation from the first, we get,

\(y(q-p)+x(q-p)=0\) or \(y=-x\), this is an equation of a straight line.

Circle: \(x^2 + y^2 - 8x = 0 \Rightarrow (x-4)^2 + y^2 = 16\)

Centre \((4,0)\), radius 4

Hyperbola: \(\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1\)

Let tangent be \(y = mx + c\) with \(m > 0\)

For circle: \(\dfrac{|4m + c|}{\sqrt{m^2 + 1}} = 4 \Rightarrow (4m + c)^2 = 16(m^2 + 1)\) ... (1)

For hyperbola: \(c^2 = 9m^2 - 4\) ... (2)

From (2): \(c = \sqrt{9m^2 - 4}\) (positive case)

Substituting in (1) and solving: \(495m^4 + 104m^2 - 400 = 0\)

Solving: \(m^2 = \dfrac{4}{5} \Rightarrow m = \dfrac{2}{\sqrt{5}}\)

Then \(c = \sqrt{9 \times \dfrac{4}{5} - 4} = \sqrt{\dfrac{16}{5}} = \dfrac{4}{\sqrt{5}}\)

Tangent: \(y = \dfrac{2}{\sqrt{5}}x + \dfrac{4}{\sqrt{5}}\)

Multiply by \(\sqrt{5}\): \(\sqrt{5}y = 2x + 4\)

Therefore: \(2x - \sqrt{5}y + 4 = 0\)

Curve \(y^2 = x\) is a rightward opening parabola

For \(x \ge 0\): \(y = |x| = x\)

Intersection: \(x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1\), so the intersection points are \((0,0)\) and \((1,1)\)

Area in first quadrant = \(\int \limits_0^1 (\sqrt{x} - x)\, dx\)

\(= \left[\dfrac{2x^{3/2}}{3} - \dfrac{x^2}{2}\right]_0^1 = \dfrac{2}{3} - \dfrac{1}{2} = \dfrac{1}{6}\)

Therefore, area = \(\dfrac{1}{6}\) sq. unit

Given line: \(x - 2y = 1 \Rightarrow y = \dfrac{x}{2} - \dfrac{1}{2}\)

Slope \(m = \dfrac{1}{2}\)

Perpendicular slope = \(-2\) (since product of slopes = -1)

Line through \((1,1)\) with slope -2:

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

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

\(y = -2x + 3\)

Given \(A = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}\).

First, calculate \(A^2 = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\).

So \(A^n = 0\) for all \(n \geq 2\).

Then \(f(A) = I + A + A^2 + \ldots + A^{16} = I + A\).

\(I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}\).

There are 9 boxes. 3 boxes are small. 5 specific balls cannot fit into these 3 small boxes. So these 5 balls must go into the remaining 6 boxes.

First, arrange the 5 restricted balls into the 6 boxes: number of ways = \(P(6,5) = 6 \times 5 \times 4 \times 3 \times 2 = 720\).

After placing these 5 balls, we have 4 balls left (the unrestricted ones) and 4 boxes left (since 5 boxes are already filled). These can be arranged in \(4! = 24\) ways.

Total ways = \(720 \times 24 = 17280\).

A function is inverse of itself if \(f(f(x)) = x\) for all \(x\) in domain. We can check the options and find the correct one

For option (a):\(f(f(x)) = f\left(\dfrac{1-x}{1+x}\right) = \dfrac{1 - \dfrac{1-x}{1+x}}{1 + \dfrac{1-x}{1+x}}\)

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

So it is self-inverse. Rest of the functions are not self-inverse. So (a) is the correct choice.

We need to select and assign positions.

Choose the President: 10 ways.

Then Vice-President from remaining 9: 9 ways.

Secretary from remaining 8: 8 ways.

Joint-Secretary from remaining 7: 7 ways.

Now we have 6 members left, from which we need to choose 2 Executive Committee members. Since these two positions are identical, number of ways = \(^6\text{C}_2 = 15\).

Total ways = \(10 \times 9 \times 8 \times 7 \times 15 = 75600\).

Here are the three sets representing the Venn diagrams for Mathematics, Business and Literature. Number of students who like all three subjects is 5.

It is clear from the diagram that the number of students who like exactly one subjects is \(20+25+20=65\)

Numbers containing only digits \(1\) and \(2\).

Numbers less than \(2 \times 10^8 = 200,000,000\) include:

One-digit numbers which contain only the digits 1 and 2 \(2^1\)

Two-digit numbers which contain only the digits 1 and 2 \(2^2\) and so on up to 8 digit numbers. It also include 9-digit numbers which start with 1 only.

Nine-digit numbers \(=1\times 2^8\)

The total numbers are \(2^1+2^2+2^3+...+2^8+2^8=766\)

We know that \(\displaystyle \lim_{x\to a} \left(f(x)\right)^{g(x)}\), where \(f(a)\to 1\) and \(g(a) \to \infty\) is:

\(\displaystyle \lim_{x\to a} e^{g(x)[f(x)-1]}\)

Hence the value of \(\displaystyle \lim_{x\to \infty} \left(1 + \dfrac{a}{x} + \dfrac{b}{x^{2}}\right)^{2x}\) is \(\displaystyle \lim_{x\to \infty}e^{2x[\frac{a}{x}+\frac{b}{x^2}]}\)

\(\displaystyle \lim_{x\to \infty} e^{2a+\frac{2b}{x}}=e^{2a}\)

Given that the value of the limit is \(e^2\), hence \(a=1\) and \(b\) can take any value.

Note that \(f(-x) = \sin^{5}(-x) + \sin^{3}(-x) = -\sin^{5}x - \sin^{3}x = -f(x)\).

So \(f(x) + f(-x) = 0\).

Hence the integrand \([f(x) + f(-x)][g(x) + g(-x)]\) is identically zero.

Therefore the integral is zero.

We know \(\dfrac{dx}{dy} = \dfrac{1}{\dfrac{dy}{dx}}\).

Then \(\dfrac{d^{2}x}{dy^{2}} = \dfrac{d}{dy}\left(\dfrac{dx}{dy}\right) = \dfrac{d}{dx}\left(\dfrac{dx}{dy}\right) \cdot \dfrac{dx}{dy}\)

\(= \dfrac{d}{dx}\left(\dfrac{1}{y'}\right) \cdot \dfrac{1}{y'}\) where \(y' = \dfrac{dy}{dx}\).

\(\dfrac{d}{dx}\left(\dfrac{1}{y'}\right) = -\dfrac{y''}{(y')^{2}}\). So

\(\dfrac{d^{2}x}{dy^{2}} = -\dfrac{y''}{(y')^{2}} \cdot \dfrac{1}{y'} = -\dfrac{y''}{(y')^{3}}\).

Thus \(\dfrac{d^{2}x}{dy^{2}} = -\left(\dfrac{d^{2}y}{dx^{2}}\right) \left(\dfrac{dy}{dx}\right)^{-3}\).

Let \(u = \log_{10}x = \dfrac{\ln x}{\ln 10}\) and \(v = \log_{x}10 = \dfrac{\ln 10}{\ln x}\).

Then \(\dfrac{du}{dx} = \dfrac{1}{x \ln 10}\) and \(\dfrac{dv}{dx} = -\dfrac{\ln 10}{x (\ln x)^{2}}\).

We need \(\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{\frac{1}{x \ln 10}}{-\frac{\ln 10}{x (\ln x)^{2}}}\)

\(= -\dfrac{1}{x \ln 10} \cdot \dfrac{x (\ln x)^{2}}{\ln 10} = -\dfrac{(\ln x)^{2}}{(\ln 10)^{2}}\).

Since \(\ln x = \log_{e} x\), hence choice (a) is correct.

For \(x \ge 0\): \(f(x) = x + x = 2x\)

For \(x < 0\): \(f(x) = x + (-x) = 0\)

At \(x = 0\):

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

Right-hand limit: \(\displaystyle \lim_{x\to 0^+} f(x) = 0\)

\(f(0) = 0 + |0| = 0\)

Since left-hand limit = right-hand limit = \(f(0)\), the function is continuous at \(x = 0\), therefore, \(f\) is continuous for all \(x \in (-\infty, \infty)\)

\(|a - b|^2 = |a|^2 + |b|^2 - 2a\cdot b = 2 - 2a\cdot b\) (since \(|a| = |b| = 1\))

Similarly: \(|b - c|^2 = 2 - 2b\cdot c\) and \(|c - a|^2 = 2 - 2c\cdot a\)

Sum = \(6 - 2(a\cdot b + b\cdot c + c\cdot a)\)...(1)

Let \(S = a\cdot b + b\cdot c + c\cdot a\)

We know \((a + b + c)^2 = |a|^2 + |b|^2 + |c|^2 + 2(a\cdot b + b\cdot c + c\cdot a) = 3 + 2S\)

Since \((a + b + c)^2 \ge 0\): \(3 + 2S \ge 0 \Rightarrow S \ge -\dfrac{3}{2}\)

Putting the value of \(S\) in equation (1), maximum value of the sum = \(6 - 2S = 6 - 2(-\dfrac{3}{2}) = 6 + 3 = 9\)

For angle bisector: \(\vec{a}\) is proportional to \(\dfrac{\vec{b}}{|\vec{b}|} + \dfrac{\vec{c}}{|\vec{c}|}\)

\(|\vec{b}| = \sqrt{2}\), \(|\vec{c}| = \sqrt{2}\)

\(\dfrac{\vec{b}}{\sqrt{2}} + \dfrac{\vec{c}}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}[(\hat{i} + \hat{j}) + (\hat{j} + \hat{k})] = \dfrac{1}{\sqrt{2}}(\hat{i} + 2\hat{j} + \hat{k})\)

So \(\vec{a}\) should be proportional to \(\hat{i} + 2\hat{j} + \hat{k}\)

Therefore \(\alpha = 1, \beta = 1\)

Net force: \(\vec{F} = (4\hat{i} - 3\hat{j} + 7\hat{k}) + (-2\hat{i} + 2\hat{j} - 8\hat{k}) = 2\hat{i} - \hat{j} - \hat{k}\)

Displacement vector: \(\vec{d} = (2-5)\hat{i} + (5-7)\hat{j} + (-6-1)\hat{k} = -3\hat{i} - 2\hat{j} - 7\hat{k}\)

Work done = \(\vec{F} \cdot \vec{d} = (2)(-3) + (-1)(-2) + (-1)(-7)\)

\(= -6 + 2 + 7 = 3\)

Initial position: \((1, 2, 3)\) and the final height of the bird is 13 m, so the displacement of the bird along \(z\) axis is 10 m.

Velocity of the bird \(=10\vec{i} + 6\vec{j} + \vec{k}\) km/hr, that means the velocity in \(z\) direction is \(v_z = 1\) km/hr \(=\dfrac{5}{18}\) m/s

Time taken to cover a distance of 10 m in the \(z\) direction \(=\dfrac{10}{\frac{5}{18}}=36\) seconds.

We have \(P(A\cup B) = P(A) + P(B) - P(A\cap B)\)

\(\dfrac{5}{6} = P(A) + \dfrac{1}{2} - \dfrac{1}{3}\)

\(\dfrac{5}{6} = P(A) + \dfrac{3}{6} - \dfrac{2}{6} = P(A) + \dfrac{1}{6}\)

\(P(A) = \dfrac{5}{6} - \dfrac{1}{6} = \dfrac{4}{6} = \dfrac{2}{3}\)

Check independence: \(P(A)P(B) = \dfrac{2}{3} \times \dfrac{1}{2} = \dfrac{1}{3}\)

Since \(P(A \cap B) = \dfrac{1}{3} = P(A)P(B)\), events \(A\) and \(B\) are independent.

By AM-GM inequality:

\(\dfrac{\alpha_1 + \alpha_2 + \ldots + 2\alpha_n}{n} \ge (\alpha_1\alpha_2\ldots 2\alpha_n)^{1/n} = (2C)^{1/n}\)

Therefore: \(\alpha_1 + \alpha_2 + \ldots + 2\alpha_n \ge n(2C)^{1/n}\)

For the cubic \(x^3 - 3x^2 + 3x + 7 = 0\):

Sum of roots: \(a + b + c = 3\)

Sum of pairwise products: \(ab + bc + ca = 3\)

Product of roots: \(abc = -7\)

Now expand \((a+1)(b+1)(c+1)\):

\(= abc + ab + bc + ca + a + b + c + 1\)

\(= (-7) + 3 + 3 + 1=0\)

We know that \(1+2x+3x^2+4x^3+......+\infty=\dfrac{1}{(1-x)^2} \), therefore \(1 - 2x + 3x^2 - 4x^3 + \ldots = (1 + x)^{-2}\) for \(|x| < 1\)

So the expression becomes:

\([(1+x)^{-2}]^{-n} = (1+x)^{2n}\)

Coefficient of \(x^n\) in \((1+x)^{2n}\) is:

\(^{2n}\text{C}_n = \dfrac{(2n)!}{n! \cdot n!} = \dfrac{(2n)!}{(n!)^2}\)

From first equation: \(\alpha + \beta = p\), \(\alpha\beta = r\)

From second equation: \(\dfrac{\alpha}{2} + 2\beta = q\) and \(\dfrac{\alpha}{2} \times 2\beta = \alpha\beta = r\)

From \(\dfrac{\alpha}{2} + 2\beta = q\), multiply by 2: \(\alpha + 4\beta = 2q\)

We have \(\alpha + \beta = p\)

Subtracting: \((\alpha + 4\beta) - (\alpha + \beta) = 2q - p\)

\(3\beta = 2q - p \Rightarrow \beta = \dfrac{2q - p}{3}\)

Then \(\alpha = p - \beta = p - \dfrac{2q - p}{3} = \dfrac{3p - 2q + p}{3} = \dfrac{4p - 2q}{3} = \dfrac{2(2p - q)}{3}\)

Now \(r = \alpha\beta = \dfrac{2(2p - q)}{3} \times \dfrac{2q - p}{3} = \dfrac{2(2p - q)(2q - p)}{9}\)

Let \(\theta = \text{cosec}^{-1}\dfrac{5}{3}\), then \(\text{cosec}\theta = \dfrac{5}{3} \Rightarrow \sin\theta = \dfrac{3}{5}\)

Since \(\theta\) is in principal range: \(\cos\theta = \dfrac{4}{5}\)

So \(\cot\theta = \dfrac{\cos\theta}{\sin\theta} = \dfrac{4/5}{3/5} = \dfrac{4}{3}\) or \(\tan\theta = \dfrac{3}{4}\)

Let \(\phi = \tan^{-1}\dfrac{2}{3}\). Then \(\tan\phi = \dfrac{2}{3}\)

We need \(\cot(\theta + \phi) = \dfrac{1}{\tan(\theta + \phi)}\)

\(\tan(\theta + \phi) = \dfrac{\tan\theta + \tan\phi}{1 - \tan\theta \tan\phi} = \dfrac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}}\)

\(= \dfrac{\frac{9 + 8}{12}}{1 - \frac{1}{2}} = \dfrac{17}{6}\)

Therefore \(\cot(\theta + \phi) = \dfrac{6}{17}\)

Rewrite the equation as

\(\sin(\theta + \alpha - \alpha) = 3\sin(\theta + \alpha + \alpha)\)

\(\sin(\theta + \alpha)\cos\alpha - \cos(\theta + \alpha)\sin\alpha = 3[\sin(\theta + \alpha)\cos\alpha + \cos(\theta + \alpha)\sin\alpha]\)

\(\Rightarrow \sin(\theta + \alpha)\cos\alpha - \cos(\theta + \alpha)\sin\alpha\)

\(= 3\sin(\theta + \alpha)\cos\alpha + 3\cos(\theta + \alpha)\sin\alpha\)

\(\Rightarrow \sin(\theta + \alpha)\cos\alpha + 2\cos(\theta + \alpha)\sin\alpha = 0\)

Divide both sides by \(\cos(\theta + \alpha)\cos\alpha\)

\(\dfrac{\sin(\theta + \alpha)\cos\alpha}{\cos(\theta + \alpha)\cos\alpha} + \dfrac{2\cos(\theta + \alpha)\sin\alpha}{\cos(\theta + \alpha)\cos\alpha} = 0\)

\(\Rightarrow \tan(\theta + \alpha) + 2\tan\alpha = 0\)

Let \(AB = c\), \(AC = b\), \(BC = a\)

Since \(\angle A = 90°\): \(a^2 = b^2 + c^2\) (Pythagorean theorem)

D is midpoint of AC, so \(AD = \dfrac{b}{2}\) and \(DC = \dfrac{b}{2}\)

In right triangle ABD:

\(BD^2 = AB^2 + AD^2 = c^2 + \left(\dfrac{b}{2}\right)^2 = c^2 + \dfrac{b^2}{4}\)

Then \(BC^2 - BD^2 = a^2 - \left(c^2 + \dfrac{b^2}{4}\right)\)

\(= (b^2 + c^2) - c^2 - \dfrac{b^2}{4} = b^2 - \dfrac{b^2}{4} = \dfrac{3b^2}{4}\)

Since \(AD^2 = \left(\dfrac{b}{2}\right)^2 = \dfrac{b^2}{4}\):

\(BC^2 - BD^2 = 3 \times \dfrac{b^2}{4} = 3AD^2\)

Let the curve be \(y = f(x)\). At point \((x,y)\), the rectangle formed has area \(xy\).

Area under curve: \(\int \limits_0^x y\, dx\)

Remaining area: \(xy - \int \limits_0^x y\, dx\)

Case 1: \(\int \limits_0^x y\, dx = \dfrac{2}{3}xy\)

Differentiating with respect to x: \(y = \dfrac{2}{3}(y + xy')\)

\(3y = 2y + 2xy' \Rightarrow y = 2xy'\)

\(\dfrac{dy}{y} = \dfrac{dx}{2x} \Rightarrow \ln y = \dfrac{1}{2}\ln x + c\)

\(y^2 = kx\) (a parabola)

Case 2: \(\int \limits_0^x y\, dx = \dfrac{1}{3}xy\)

This also leads to \(y = kx^2\) (a parabola), therefore, the family represents parabolas.

Let the line through \((4,2)\) have intercept form: \(\dfrac{x}{a} + \dfrac{y}{b} = 1\)

It meets axes at \(P(a,0)\) and \(Q(0,b)\)

Since (4,2) lies on the line: \(\dfrac{4}{a} + \dfrac{2}{b} = 1\)

Triangle OPQ is right-angled at O (origin)

For a right triangle, circumcenter is the midpoint of hypotenuse PQ

Circumcenter = \(\left(\dfrac{a}{2}, \dfrac{b}{2}\right)\)

Let \((h,k) = \left(\dfrac{a}{2}, \dfrac{b}{2}\right) \Rightarrow a = 2h, b = 2k\)

Substituting in line condition: \(\dfrac{4}{2h} + \dfrac{2}{2k} = 1\)

\(\dfrac{2}{h} + \dfrac{1}{k} = 1\), therefore, locus is \(\dfrac{2}{x} + \dfrac{1}{y} = 1\)

Let \(x=\dfrac{1}{\sqrt{2}} \left[ \sqrt{1-\dfrac{1}{\sqrt{2}}\sqrt{1-\dfrac{1}{\sqrt{2}}\sqrt{1-....\infty}} }\right]\), then

\(x=\dfrac{1}{\sqrt{2}}\sqrt{(1-x)} \)

\(\Rightarrow x^2=\dfrac{1-x}{2}\) or \(2x^2+x-1=0 \)

\(\Rightarrow x=\dfrac{1}{2}\) as \(x\) is positive.

So the number is \(6+\log_{\frac{1}{4}}\dfrac{1}{2}=6+\log_{\frac{1}{4}}\left[\dfrac{1}{4}\right]^\frac{1}{2}\)

\(=6+\dfrac{1}{2}=\dfrac{13}{2}\)

\(f(-x) = \log \left(-x + \sqrt{x^2 + 1}\right)\)

Multiply numerator and denominator by conjugate \((x + \sqrt{x^2+1})\):

\(f(-x) = \log \left(\dfrac{(-x + \sqrt{x^2+1})(x + \sqrt{x^2+1})}{x + \sqrt{x^2+1}}\right)\)

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

\(= \log \left(\dfrac{1}{x + \sqrt{x^2+1}}\right)\)

\(= -\log(x + \sqrt{x^2+1}) = -f(x)\)

Therefore, \(f\) is an odd function

Let \(x\) and \(y\) be arrival times of A and B in minutes after 6 P.M.

So \(0 \le x, y \le 60\)

They meet if \(|x - y| \le 20\)

Sample space: square \([0,60] \times [0,60]\) with area \(= 3600\)

Meeting region: strip between lines \(y = x - 20\) and \(y = x + 20\)

Non-meeting region: two corner triangles, each with area \(\dfrac{1}{2} \times 40 \times 40 = 800\)

Total non-meeting area \(= 1600\)

Meeting area \(= 3600 - 1600 = 2000\)

Probability \(=\dfrac{2000}{3600} = \dfrac{5}{9}\)

Note: There is a direct shortcut for this type of question, probability that they will meet \(=1-\dfrac{(60-m)^2}{m^2}\), where \(m\) is the minutes for which each of them waits.

We use the identity: \(a^3 + b^3 + c^3 - 3abc = \dfrac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]\)

Numbers 1 to 99 have 33 numbers in each category \(3k, 3k+1, 3k+2\)

The expression is divisible by 3 when:

Case 1: All three numbers are from the same category

Number of ways: \(3 \times ^{33}C_3\)

Case 2: One number from each category

Number of ways: \(^{33}C_1 \times ^{33}C_1 \times ^{33}C_1 = (^{33}C_1)^3\)

Total favorable outcomes = \(3 \cdot ^{33}C_3 + (^{33}C_1)^3\)

Total ways = \(^{99}C_3\)

Probability = \(\dfrac{3 \cdot ^{33}C_3 + (^{33}C_1)^3}{^{99}C_3}\)

Total number of ways to choose numbers: \(25 \times 25 = 625\)

Number of ways they match (both choose same number): 25

Probability they win = \(\dfrac{25}{625} = \dfrac{1}{25}\)

Probability they do not win = \(1 - \dfrac{1}{25} = \dfrac{24}{25}\)

We know \(\sin 18° = \dfrac{\sqrt{5} - 1}{4}\)

So \(\sin^2 18° = \left(\dfrac{\sqrt{5} - 1}{4}\right)^2 = \dfrac{6 - 2\sqrt{5}}{16} = \dfrac{3 - \sqrt{5}}{8}\)

Also \(\cos 36° = \dfrac{\sqrt{5} + 1}{4}\)

So \(\cos^2 36° = \left(\dfrac{\sqrt{5} + 1}{4}\right)^2 = \dfrac{6 + 2\sqrt{5}}{16} = \dfrac{3 + \sqrt{5}}{8}\)

Sum of roots = \(\dfrac{3 - \sqrt{5}}{8} + \dfrac{3 + \sqrt{5}}{8} = \dfrac{6}{8} = \dfrac{3}{4}\)

Product of roots = \(\dfrac{3 - \sqrt{5}}{8} \times \dfrac{3 + \sqrt{5}}{8} = \dfrac{9 - 5}{64} = \dfrac{4}{64} = \dfrac{1}{16}\)

Quadratic equation: \(x^2 - (\text{sum})x + \text{product} = 0\)

\(x^2 - \dfrac{3}{4}x + \dfrac{1}{16} = 0\) or \(16x^2 - 12x + 1 = 0\)

Let first term \(= a\), common ratio \(= r\) (with \(|r| < 1\) for convergence)

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

Sum of first two terms = \(a + ar = a(1+r)\)

Given: \(\dfrac{a}{1-r} = 2a(1+r)\)

\(\Rightarrow 1 - r^2 = \dfrac{1}{2}\) or \(r = \pm \dfrac{1}{\sqrt{2}}\)

Assuming the sequence is in Harmonic Progression (HP), the reciprocals are in AP.

Let \(b_k = \dfrac{1}{a_k}\). Then \(b_m = \dfrac{1}{n}\) and \(b_n = \dfrac{1}{m}\)

Since b's are in AP: \(b_m - b_n = (m-n)d\)

\(\dfrac{1}{n} - \dfrac{1}{m} = (m-n)d\)

\(\dfrac{m-n}{mn} = (m-n)d\) or \(d = \dfrac{1}{mn}\)

Then \(b_{m+n} = b_m + (m+n-m)d = \dfrac{1}{n} + n \cdot \dfrac{1}{mn}\)

\(= \dfrac{1}{n} + \dfrac{1}{m} = \dfrac{m+n}{mn}\)

Therefore \(a_{m+n} = \dfrac{1}{b_{m+n}} = \dfrac{mn}{m+n}\)

For a skew symmetric matrix: \(A^T = -A\)

Taking transpose of inverse:

\((A^{-1})^T = (A^T)^{-1} = (-A)^{-1} = -A^{-1}\)

This shows \((A^{-1})^T = -A^{-1}\)

Therefore, \(A^{-1}\) is also skew symmetric

Mean of squares of first \(n\) natural numbers:

\(\dfrac{1^2 + 2^2 + \ldots + n^2}{n} = \dfrac{n(n+1)(2n+1)}{6n} = \dfrac{(n+1)(2n+1)}{6}\)

Given that, \(\dfrac{(n+1)(2n+1)}{6} = 11\)

\((n+1)(2n+1) = 66\)

\(2n^2 + 3n - 65 = 0\)

Factoring: \((2n+13)(n-5) = 0\) or \(n = 5\)

For \(x \ge 0\): \(f(x) = \dfrac{x}{1+x}\), which is differentiable for all \(x \ge 0\)

For \(x < 0\): \(f(x) = \dfrac{x}{1-x}\), which is differentiable for all \(x < 0\)

At \(x = 0\), check differentiability:

Left derivative: \(\displaystyle \lim_{h\to 0^-} \dfrac{f(0+h)-f(0)}{h} = \displaystyle \lim_{h\to 0^-} \dfrac{\frac{h}{1-h}}{h} = \displaystyle \lim_{h\to 0^-} \dfrac{1}{1-h} = 1\)

Right derivative: \(\displaystyle \lim_{h\to 0^+} \dfrac{f(0+h)-f(0)}{h} = \displaystyle \lim_{h\to 0^+} \dfrac{\frac{h}{1+h}}{h} = \displaystyle \lim_{h\to 0^+} \dfrac{1}{1+h} = 1\)

Since left derivative = right derivative, \(f\) is differentiable at 0

Therefore, \(f\) is differentiable everywhere on \((-\infty,\infty)\)

Let \(I = \int \limits_{0}^{\pi} x f(\sin x)\, dx\)

Using property \(\int \limits_{0}^{a} f(x)\, dx = \int \limits_{0}^{a} f(a-x)\, dx\):

\(I = \int \limits_{0}^{\pi} (\pi - x) f(\sin(\pi - x))\, dx\)

Since \(\sin(\pi - x) = \sin x\):

\(I = \int \limits_{0}^{\pi} (\pi - x) f(\sin x)\, dx\)

Adding both expressions for \(I\):

\(2I = \int \limits_{0}^{\pi} [x + (\pi - x)] f(\sin x)\, dx = \pi \int \limits_{0}^{\pi} f(\sin x)\, dx\)

Therefore: \(I = \dfrac{\pi}{2} \int \limits_{0}^{\pi} f(\sin x)\, dx\)

For \(x < 0\): \(f(x) = x + 2\)

For \(x \ge 0\): \(f(x) = |x-2| = \begin{cases} 2-x & \text{if } 0 \le x < 2\\ x-2 & \text{if } x \ge 2 \end{cases}\)

\(\int \limits_{-2}^{3} f(x)\, dx = \int \limits_{-2}^{0} (x+2)\, dx + \int \limits_{0}^{2} (2-x)\, dx + \int \limits_{2}^{3} (x-2)\, dx\)

First integral: \(\left[\dfrac{x^2}{2} + 2x\right]_{-2}^{0} = 0 - (2 - 4) = 2\)

Second integral: \(\left[2x - \dfrac{x^2}{2}\right]_{0}^{2} = (4 - 2) - 0 = 2\)

Third integral: \(\left[\dfrac{x^2}{2} - 2x\right]_{2}^{3} = \left(\dfrac{9}{2} - 6\right) - (2 - 4) = -\dfrac{3}{2} + 2 = \dfrac{1}{2}\)

Total = \(2 + 2 + 0.5 = 4.5\)

The equation \(6x^2 - xy - 2y^2 = 0\) represents a pair of lines through origin

Dividing by \(x^2\) (assuming \(x \ne 0\)): \(6 - m - 2m^2 = 0\) where \(m = \dfrac{y}{x}\)

\(2m^2 + m - 6 = 0\)

For roots \(m_1, m_2\): \(m_1 + m_2 = -\dfrac{1}{2}\) and \(m_1 m_2 = -3\)

Difference of slopes: \(|m_1 - m_2| = \sqrt{(m_1 - m_2)^2} = \sqrt{(m_1+m_2)^2 - 4m_1 m_2}\)

\(= \sqrt{\dfrac{1}{4}+12} = \dfrac{7}{2}\)

Area of circle: \(A = \pi r^2\)

Differentiating with respect to time:

\(\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}\)

Given: \(\dfrac{dr}{dt} = -\dfrac{2}{\pi}\) m/sec and \(r = 10\) m

Substituting:

\(\dfrac{dA}{dt} = 2\pi \times 10 \times \left(-\dfrac{2}{\pi}\right)\)

\(= 20\pi \times \left(-\dfrac{2}{\pi}\right) = 20 \times (-2) = -40\) m²/sec