Circumference of the circular wire = \(2\pi r = 2\pi \times 7.5 = 15\pi\) cm

This wire is bent to form an arc on the hoop of radius 120 cm.

Arc length = \(15\pi\) cm

Using the formula: Arc length = \(r\theta\) where \(\theta\) is in radians.

\(15\pi = 120\theta\)

\(\theta = \dfrac{15\pi}{120} = \dfrac{\pi}{8}\) radians

Converting to degrees: \(\theta = \dfrac{\pi}{8} \times \dfrac{180}{\pi} = \dfrac{180}{8} = 22.5^{\circ}\)

To form a parallelogram, we need to select 2 lines from one set of parallel lines and 2 lines from another set of parallel lines.

We have 4 parallel lines in one direction.

Number of ways to select 2 lines from 4 = \(^4C_2 = 6\)

We have 3 parallel lines in another direction.

Number of ways to select 2 lines from 3 = \(^3C_2 = 3\)

Total number of parallelograms = \(6 \times 3 = 18\)

Side of square = 16, so radius of each quadrant = 16

Area of the square = \(16^2 = 256\)

Area of one quadrant = \(\dfrac{1}{4}\pi (16)^2 = 64\pi\)

Area of two quadrants = \(2 \times 64\pi = 128\pi\)

The common area between the two quadrants (lens-shaped region):

Common area = Area of 2 quadrants - Area of square = \(128\pi - 256\)

Shaded area \(=256 -(128\pi-256)\)

\(=128(4-\pi)\)

For a fair die, \(X\) can be \(1, 2, 3, 4, 5\) or \(6\), each with probability \(\dfrac{1}{6}\).

Expected value \(E(X) = \sum x_i P(x_i)\)

\(E(X) = 1 \cdot \dfrac{1}{6} + 2 \cdot \dfrac{1}{6} + 3 \cdot \dfrac{1}{6} + 4 \cdot \dfrac{1}{6} + 5 \cdot \dfrac{1}{6} + 6 \cdot \dfrac{1}{6}\)

\(E(X) = \dfrac{1 + 2 + 3 + 4 + 5 + 6}{6} = \dfrac{21}{6} = \dfrac{7}{2}\)

Word: INDEPENDENCE (12 letters)

Letters: I(1), N(3), D(2), E(4), P(1), C(1)

Part 1: Arrangements beginning with I and ending with P

Fix I at start and P at end. Arrange remaining 10 letters: N(3), D(2), E(4), C(1)

Number of arrangements = \(\dfrac{10!}{3! \cdot 2! \cdot 4!}\)

Calculating: \(\dfrac{3628800}{6 \cdot 2 \cdot 24} = \dfrac{3628800}{288} = 12600\)

Part 2: Vowels never together

Vowels: I, E, E, E, E (5 vowels); Consonants: N, D, P, N, D, N, C (7 consonants). The required answer = total number of words – those when all the vowels are together. Total number of words \(=\dfrac{10!}{4!\times 3!\times 2!}=1663200 \)

Number of words when all the vowels are together. Assume all the vowels as one group, there will be 8 objects to be arranged in a line, number of such words is

\(\dfrac{5!}{4!}\times \dfrac{8!}{3!\times 2!}=16800\)

Hence the number of words = \(1663200-16800=1646400\)

Diameter of coin = 2 cm, Angular diameter of moon = \(31'\)

Converting \(31'\) to radians: \(31' = \dfrac{31}{60}\) degrees = \(\dfrac{31}{60} \times \dfrac{\pi}{180} = \dfrac{31\pi}{10800}\) radians

For small angles, \(\theta \approx \dfrac{d}{r}\) where \(d\) is diameter and \(r\) is distance.

\(\dfrac{31\pi}{10800} = \dfrac{2}{r}\) (diameter in cm)

or \(r = \dfrac{2 \times 10800}{31\pi} = \dfrac{21600}{31 \times 3.14159} \approx 221.7\) cm

Hence \(r = 2.217\) m

\(\sqrt{1 + \cos 2x} = \sqrt{2\cos^2 x} = \sqrt{2}|\cos x|\)

The equation becomes: \(\sqrt{2}|\cos x| = \sqrt{2}\cos^{-1}(\cos x)\)

Hence \(|\cos x| = \cos^{-1}(\cos x)\)

For \(x \in [0, \pi]\), \(\cos^{-1}(\cos x) = x\)

For \(x \in \left[-\dfrac{\pi}{2}, 0\right)\), \(\cos^{-1}(\cos x) = -x\)

Plotting both the graphs  \(y=\cos x\) and \(y=\cos^{-1}(\cos x)\) we see that there are only two solutions.

This is a first-order linear differential equation in the form \(\dfrac{dy}{dx} + Py = Q\).

Here, \(P = 1\) and \(Q = 1\).

Integrating factor: I.F. = \(e^{\int P dx} = e^{\int 1 dx} = e^x\)

Multiplying the equation by I.F.:

\(e^x \dfrac{dy}{dx} + e^x y = e^x\)

\(\dfrac{d}{dx}(ye^x) = e^x\)

Integrating both sides: \(ye^x = \int e^x dx = e^x + C\)

\(y = 1 + Ce^{-x}\)

Applying initial condition \(y(0) = 2\):

\(2 = 1 + Ce^0 = 1 + C\), so \(C = 1\)

Solution: \(y = 1 + e^{-x}\)

When \(x = 1\): \(y(1) = 1 + e^{-1}\)

Projections on \(x, y, z\) axes are \(12, 4, 3\) respectively.

Length of the line segment: \(d = \sqrt{12^2 + 4^2 + 3^2} =  \sqrt{169} = 13\)

Direction cosines are given by \(\ell = \dfrac{a}{d}, m = \dfrac{b}{d}, n = \dfrac{c}{d}\) where \(a, b, c\) are the projections.

\(\ell = \dfrac{12}{13}\), \(m = \dfrac{4}{13}\), \(n = \dfrac{3}{13}\)

First box must contain exactly 1 ball.

Number of ways to choose 1 ball from 20 = \(^{20}C_1 = 20\)

Remaining 19 balls can be distributed into remaining 4 boxes.

Each of the 19 balls can go into any of the 4 boxes, number of ways = \(4^{19}\)

Total number of ways = \(20 \times 4^{19}\)

There are 5 days: Tuesday to Saturday. Each person can visit on any of these 5 days.

Total possible outcomes = \(5 \times 5 = 25\)

For consecutive days, the pairs are: (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat)

That's 4 pairs. But either person can visit first, so favorable outcomes are:

Ruchana visits before Bhakati : 4 ways

Bhakati visits before Ruchana: 4 ways

Total favorable outcomes = \(4 + 4 = 8\)

Probability = \(\dfrac{8}{25}\)

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

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

Required ratio = \(\dfrac{^{2n}C_{n}}{^{2n-1}C_{n}}\)

\(= \dfrac{(2n!)(n!)(n-1)!}{(2n-1)!(n!)(n!)}=2\)

Therefore, the ratio is \(2:1\)

Area of equilateral triangle = \(\dfrac{\sqrt{3}}{4}a^2 = 49\sqrt{3}\)

Hence  \(a^2 = 196\), or \(a = 14\) cm

Radius of each circle = \(\dfrac{a}{2} = 7\) cm

Each angle of equilateral triangle = \(60^{\circ}\)

Area of sector at each vertex = \(\dfrac{60}{360} \times \pi r^2 = \dfrac{49\pi}{6}\)

Total area of three sectors inside triangle = \(3 \times \dfrac{49\pi}{6}=\dfrac{49\pi}{2}\) cm²

Area of triangle = \(49 \sqrt{3}\) cm²

Area not covered by circles = \(\dfrac{49\pi}{2}-49\sqrt{3}=7.94\) cm².

Check each point in the equation \(y^2 = 4x\), only \(4,4\) satisfies the equation of the parabola.

Point: \(P(1, 2, 3)\)

Line passes through \(A(6, 7, 7)\) with direction ratios \((3, 2, -2)\)

Vector \(\vec{AP} = (1-6, 2-7, 3-7) = (-5, -5, -4)\)

Direction vector of line: \(\vec{b} = (3, 2, -2)\)

Perpendicular distance = \(\dfrac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}\)

Cross product: \(\vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & -5 & -4 \\ 3 & 2 & -2 \end{vmatrix}\)

\(= 18\hat{i} - 22\hat{j} + 5\hat{k}\)

\(|\vec{AP} \times \vec{b}| = \sqrt{18^2 + 22^2 + 5^2} = \sqrt{324 + 484 + 25} = \sqrt{833}\)

\(|\vec{b}| = \sqrt{9 + 4 + 4} = \sqrt{17}\)

Distance = \(\dfrac{\sqrt{833}}{\sqrt{17}} = \sqrt{\dfrac{833}{17}} = \sqrt{49} = 7\)

Arc length formula: \(s = r\theta\) where \(\theta\) is in radians.

\(\theta = 30 \times \dfrac{\pi}{180} = \dfrac{\pi}{6}\) radians

Given arc length \(s = 17.6\) cm

Hence \(17.6 = r \times \dfrac{\pi}{6}\)

\(r = \dfrac{17.6 \times 6}{\pi} = \dfrac{105.6}{3.14159} \approx 33.6\) cm

The domain of \(\cos^{-1}(t)\) is \([-1, 1]\).

For \(\cos^{-1}(2x - 3)\) to be defined:

\(-1 \leq 2x - 3 \leq 1\)

Hence \(1 \leq x \leq 2\)

Domain = \([1, 2]\)

Let the radii be \(r_1\) and \(r_2\).

Sum of areas: \(\pi r_1^2 + \pi r_2^2 = 130\pi\)

So \(r_1^2 + r_2^2 = 130\) ... (1)

Circles touch externally: \(r_1 + r_2 = 14\) ... (2)

From (2): \((r_1 + r_2)^2 = 196\)

\(r_1^2 + 2r_1r_2 + r_2^2 = 196\)

Using (1): \(130 + 2r_1r_2 = 196\)

or \(r_1r_2 = 33\) ... (3)

From (2) and (3), \(r_1\) and \(r_2\) are roots of:

\(t^2 - 14t + 33 = 0\)

Factoring: \((t - 11)(t - 3) = 0\)

So \(r_1 = 11\) and \(r_2 = 3\)

Let initial population = \(P_0\), Growth rate = 10% per year

Population after \(t\) years: \(P = P_0(1.1)^t\)

For population to triple: \(P = 3P_0\)

\(3P_0 = P_0(1.1)^t\) or \(3 = (1.1)^t\)

Taking logarithm: \(\log 3 = t \log 1.1\)

\(t = \dfrac{\log 3}{\log 1.1}\)

Using the identity: \(\tan^{-1} \left( \dfrac{1-x}{1+x} \right) = \dfrac{\pi}{4} - \tan^{-1} x\) for \(x > 0\)

The equation becomes:

\(\dfrac{\pi}{4} - \tan^{-1} x = \dfrac{1}{2} \tan^{-1} x\)

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

\(\dfrac{\pi}{4} = \dfrac{3}{2} \tan^{-1} x\)

\(\tan^{-1} x = \dfrac{\pi}{6}\)

\(x = \tan \dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}\)

We need to find \(P(X > 2) = \int \limits_2^6 f(x) dx\)

\(P(X > 2) = \int \limits_2^6 \dfrac{x}{18} dx = \dfrac{1}{18} \int \limits_2^6 x \, dx\)

\(= \dfrac{1}{18} \left[ \dfrac{x^2}{2} \right]_2^6 = \dfrac{1}{36} [x^2]_2^6\)

\(= \dfrac{1}{36} (36 - 4) = \dfrac{32}{36} = \dfrac{8}{9}\)

Let incomes of A, B, C be \(7x, 9x, 12x\)

Let expenditures be \(8y, 9y, 15y\)

A saves \(\dfrac{1}{4}\) of his income:

Hence  \(7x-8y=\dfrac{7x}{4}\)

Or  \(y= \dfrac{21x}{32}\)

Putting the value of \(y\) in terms of \(x\), we get the ratio of their savings as \( 56:99:69\)

The equation is \((y^2 + 3xy)dx + (x^2 + xy)dy = 0\)

This is a homogeneous equation. Let \(y = vx\), so \(dy = vdx + xdv\)

Substituting:

\((v^2x^2 + 3vx^2)dx + (x^2 + vx^2)(vdx + xdv) = 0\)

\(x^2(v^2 + 3v)dx + x^2(1 + v)(vdx + xdv) = 0\)

Dividing by \(x^2\):

\((v^2 + 3v)dx + (1 + v)(vdx + xdv) = 0\)

\((v^2 + 3v + v + v^2)dx + (1 + v)xdv = 0\)

\((2v^2 + 4v)dx + (1 + v)xdv = 0\)

\(\dfrac{dx}{x} = -\dfrac{(1 + v)dv}{2v(v + 2)}\)

Integrating and applying initial condition \(x = 1, y = 1\) (so \(v = 1\)):

After integration and simplification: \(y = \dfrac{x + 1}{x^2}\)

This is a binomial distribution problem with \(n = 10\) and \(p = 0.5\).

Probability of exactly 6 correct = \(\binom{10}{6} (0.5)^6 (0.5)^4\)

\(P(X = 6) = \binom{10}{6} (0.5)^{10}\)

\(\binom{10}{6} = \dfrac{10!}{6! \cdot 4!} = 210\)

\(P(X = 6) = 210 \times \dfrac{1}{1024} = \dfrac{210}{1024} \approx 0.205\)

Side of cube: \(a^3 = 64\), so \(a = 4\) cm

When joined face to face, the cuboid has dimensions:

Length = \(8\) cm, Width = \(4\) cm, Height = \(4\) cm

Surface area = \(2(lb + bh + hl)\)

\(= 2(8 \times 4 + 4 \times 4 + 4 \times 8)\)

\(= 2(32 + 16 + 32) =  160\text{ cm}^2\)

Relative speed of trains = \(60 + 90 = 150\) km/h, initital distnace between the trains is \(300\) km, hence 

Time until trains meet = \(\dfrac{300}{150}=2\) hours

Distance traveled by bird in this time = \(120 \times 2=240\)

Given family: \(y = ax^2\) so \(\dfrac{dy}{dx} = 2ax\)

From the original equation: \(a = \dfrac{y}{x^2}\)

Substituting:

\(\dfrac{dy}{dx} = 2 \cdot \dfrac{y}{x^2} \cdot x = \dfrac{2y}{x}\)

Therefore: \(y' = \dfrac{2y}{x}\)

For binomial distribution: Variance = \(np(1-p)\)

As the number of trianls is \(n = 9\) and variance = 2

\(9p(1-p) = 2\)

\(9p - 9p^2 = 2\)

\(9p^2 - 9p + 2 = 0\) or \(p = \dfrac{1}{3} \text{ or } \dfrac{2}{3}\), as probability of success is greater, so the answer is  \(\dfrac{2}{3}\).

Cosine rule: \(c^2 = a^2 + b^2 - 2ab\cos C\)

\(\Rightarrow \cos C = \dfrac{a^2 + b^2 - c^2}{2ab}\)

\(\cos C = \dfrac{9^2 + 7^2 - 12^2}{2 \times 9 \times 7}\)

\(= \dfrac{81 + 49 - 144}{126} = \dfrac{-14}{126} = -\dfrac{1}{9}\)

\(C = \cos^{-1}\left(-\dfrac{1}{9}\right) \approx 96.37^\circ\)

Area of sector = \(\dfrac{\theta}{360} \times \pi r^2\) where \(\theta\) is in degrees

Given: \(r = 6\) cm, \(\theta = 60^\circ\)

\(\text{Area} = \dfrac{60}{360} \times \pi \times 6^2\)

\(= \dfrac{1}{6} \times \pi \times 36 = 6\pi \text{ cm}^2\)