Step 1: Let the shortest side = \(x\) cm.

Longest side = \(3x\) cm (3 times shortest side).

Third side = longest side – 2 = \(3x – 2\) cm.

Step 2: Perimeter = sum of all sides = \(x + 3x + (3x – 2) = 7x – 2\).

Step 3: Given perimeter ≥ 61 cm:

\(7x – 2 \geq 61\)

\(7x \geq 63\)

\(x \geq 9\)

Step 4: Since sides must be positive, check: \(x > 0\) and \(3x – 2 > 0\) automatically satisfied for \(x \geq 9\).

Main: \(\left(\frac{2}{x} – \frac{x}{2}\right)^5\)

Using binomial theorem: \((a – b)^5 = a^5 – 5a^4b + 10a^3b^2 – 10a^2b^3 + 5ab^4 – b^5\)

Here \(a = \frac{2}{x}\), \(b = \frac{x}{2}\)

Term 1: \(a^5 = \left(\frac{2}{x}\right)^5 = \frac{32}{x^5}\)

Term 2: \(-5a^4b = -5 \cdot \left(\frac{2}{x}\right)^4 \cdot \frac{x}{2} \)\(= -5 \cdot \frac{16}{x^4} \cdot \frac{x}{2}\)\( = -5 \cdot \frac{16}{x^4} \cdot \frac{x}{2} \)\(= -5 \cdot \frac{8}{x^3} \)\(= -\frac{40}{x^3}\)

Term 3: \(+10a^3b^2\)\( = 10 \cdot \left(\frac{2}{x}\right)^3 \cdot \left(\frac{x}{2}\right)^2\)\( = 10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4}\)\( = 10 \cdot \frac{8}{x^3} \cdot \frac{x^2}{4} \)\(= 10 \cdot \frac{2}{x}\)\( = \frac{20}{x}\)

Term 4: \(-10a^2b^3 \)\(= -10 \cdot \left(\frac{2}{x}\right)^2 \cdot \left(\frac{x}{2}\right)^3 \)\(= -10 \cdot \frac{4}{x^2} \cdot \frac{x^3}{8}\)\( = -10 \cdot \frac{4}{x^2} \cdot \frac{x^3}{8}\)\( = -10 \cdot \frac{x}{2}\)\( = -5x\)

Term 5: \(+5ab^4 \)\(= 5 \cdot \frac{2}{x} \cdot \left(\frac{x}{2}\right)^4 \)\(= 5 \cdot \frac{2}{x} \cdot \frac{x^4}{16}\)\( = 5 \cdot \frac{x^3}{8} \)\(= \frac{5x^3}{8}\)

Term 6: \(-b^5 \)\(= -\left(\frac{x}{2}\right)^5\)\( = -\frac{x^5}{32}\)

Final expansion:

\(\frac{32}{x^5} – \frac{40}{x^3} + \frac{20}{x} – 5x + \frac{5x^3}{8} – \frac{x^5}{32}\)

OR: \((1 – 2x)^5\)

\((1 – 2x)^5 \)\(= 1 – 5(2x) + 10(2x)^2 – 10(2x)^3 + 5(2x)^4 – (2x)^5\)

\(= 1 – 10x + 10 \cdot 4x^2 – 10 \cdot 8x^3 + 5 \cdot 16x^4 – 32x^5\)

\(= 1 – 10x + 40x^2 – 80x^3 + 80x^4 – 32x^5\)

Main: Angle between lines

Line 1: \(\sqrt{3}x + y = 1\) ⇒ \(y = -\sqrt{3}x + 1\) ⇒ slope \(m_1 = -\sqrt{3}\)

Line 2: \(x + \sqrt{3}y = 1\) ⇒ \(\sqrt{3}y = -x + 1\) ⇒ \(y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}\) ⇒ slope \(m_2 = -\frac{1}{\sqrt{3}}\)

Angle \(\theta\) between lines: \(\tan \theta = \left|\frac{m_1 – m_2}{1 + m_1m_2}\right|\)

\(m_1 – m_2 = -\sqrt{3} – \left(-\frac{1}{\sqrt{3}}\right)\)\( = -\sqrt{3} + \frac{1}{\sqrt{3}}\)\( = \frac{-3 + 1}{\sqrt{3}} = -\frac{2}{\sqrt{3}}\)

\(1 + m_1m_2 = 1 + (-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)\)\( = 1 + 1 = 2\)

\(\tan \theta = \left|\frac{-2/\sqrt{3}}{2}\right| = \left|-\frac{1}{\sqrt{3}}\right| = \frac{1}{\sqrt{3}}\)

Therefore, \(\theta = 30^\circ\) or \(\frac{\pi}{6}\) radians.

OR: Equation of parallel line

Given line: \(3x – 4y + 2 = 0\) ⇒ slope = \(\frac{3}{4}\)

Parallel lines have same slope. Equation through (-2,3):

\(y – 3 = \frac{3}{4}(x + 2)\)

\(4(y – 3) = 3(x + 2)\)

\(4y – 12 = 3x + 6\)

\(3x – 4y + 18 = 0\)

Main: Circle equation

Center \((h,k) = (-a,-b)\), radius \(r = \sqrt{a^2 – b^2}\)

Standard form: \((x – h)^2 + (y – k)^2 = r^2\)

\((x + a)^2 + (y + b)^2 = (\sqrt{a^2 – b^2})^2\)

\((x + a)^2 + (y + b)^2 = a^2 – b^2\)

Required equation: \((x + a)^2 + (y + b)^2 = a^2 – b^2\)

OR: Ellipse \(\frac{x^2}{36} + \frac{y^2}{16} = 1\)

Here \(a^2 = 36\) ⇒ \(a = 6\) (major axis along x-axis)

\(b^2 = 16\) ⇒ \(b = 4\)

Eccentricity \(e = \sqrt{1 – \frac{b^2}{a^2}} = \sqrt{1 – \frac{16}{36}} \)\(= \sqrt{\frac{20}{36}} = \frac{\sqrt{20}}{6} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}\)

Foci: \((\pm ae, 0) = (\pm 6 \cdot \frac{\sqrt{5}}{3}, 0) \)\(= (\pm 2\sqrt{5}, 0)\)

Vertices: \((\pm a, 0) = (\pm 6, 0)\)

Length of major axis = \(2a = 12\)

Length of minor axis = \(2b = 8\)

Length of latus rectum = \(\frac{2b^2}{a} = \frac{2 \cdot 16}{6} = \frac{32}{6} = \frac{16}{3}\)

Step 1: Find mid-points (xᵢ) and calculations

Step 2: Calculate mean (x̄)

x̄ = \(\frac{\sum f_i x_i}{\sum f_i} = \frac{1350}{50} = 27\)

Step 3: Calculate variance and standard deviation

Variance (σ²) = \(\frac{\sum f_i (x_i – \bar{x})^2}{\sum f_i} = \frac{6800}{50} = 136\)

Standard deviation (σ) = \(\sqrt{136} = 2\sqrt{34} \approx 11.66\)

Method 1: Using standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\)

\(\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \lim_{x \to 0} \frac{\frac{\sin ax}{ax} \cdot ax}{\frac{\sin bx}{bx} \cdot bx}\)

= \(\lim_{x \to 0} \frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}} \cdot \frac{ax}{bx}\)

= \(\frac{\lim_{ax \to 0} \frac{\sin ax}{ax}}{\lim_{bx \to 0} \frac{\sin bx}{bx}} \cdot \frac{a}{b}\)

= \(\frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b}\)

Method 2: Direct application

\(\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}\) (standard result)