1. **Expand \(\frac{1}{(x-2)^7}\) to the fourth term using the binomial theorem:** The binomial theorem for negative powers states: $$ (1 + u)^n = \sum_{k=0}^\infty \binom{n}{k} u^k $$ where \( n = -7 \) and \( u = -\frac{x}{2} \) if we rewrite \( (x-2)^{-7} = 2^{-7} (1 - \frac{x}{2})^{-7} \). 1. Write the expression as: $$ \frac{1}{(x-2)^7} = 2^{-7} \left(1 - \frac{x}{2}\right)^{-7} $$ 2. Use the binomial expansion for \( (1 + u)^n \) with \( n = -7 \) and \( u = -\frac{x}{2} \): $$ \binom{-7}{k} = (-1)^k \binom{7 + k - 1}{k} = (-1)^k \binom{6 + k}{k} $$ 3. The first four terms (\(k=0\) to \(k=3\)) are: $$ T_k = 2^{-7} \binom{-7}{k} \left(-\frac{x}{2}\right)^k $$ Calculate each term: - \(k=0\): $$ T_0 = 2^{-7} \times 1 \times 1 = \frac{1}{128} $$ - \(k=1\): $$ \binom{-7}{1} = -7, \quad T_1 = 2^{-7} \times (-7) \times \left(-\frac{x}{2}\right) = \frac{7x}{256} $$ - \(k=2\): $$ \binom{-7}{2} = \frac{-7 \times (-8)}{2} = 28, \quad T_2 = 2^{-7} \times 28 \times \left(-\frac{x}{2}\right)^2 = 2^{-7} \times 28 \times \frac{x^2}{4} = \frac{7x^2}{256} $$ - \(k=3\): $$ \binom{-7}{3} = \frac{-7 \times (-8) \times (-9)}{3!} = -84, \quad T_3 = 2^{-7} \times (-84) \times \left(-\frac{x}{2}\right)^3 = 2^{-7} \times (-84) \times \left(-\frac{x^3}{8}\right) = \frac{21x^3}{512} $$ **Final expansion to the fourth term:** $$ \frac{1}{(x-2)^7} = \frac{1}{128} + \frac{7x}{256} + \frac{7x^2}{256} + \frac{21x^3}{512} + \cdots $$ --- 2.1 **Differentiate \( y = x^4 - 3x + 6 \) using first principles:** 1. The definition of derivative from first principles is: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ 2. Calculate \( f(x+h) \): $$ (x+h)^4 - 3(x+h) + 6 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 - 3x - 3h + 6 $$ 3. Substitute into the difference quotient: $$ \frac{f(x+h) - f(x)}{h} = \frac{(x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 - 3x - 3h + 6) - (x^4 - 3x + 6)}{h} $$ 4. Simplify numerator: $$ 4x^3h + 6x^2h^2 + 4xh^3 + h^4 - 3h $$ 5. Divide by \( h \): $$ 4x^3 + 6x^2h + 4xh^2 + h^3 - 3 $$ 6. Take the limit as \( h \to 0 \): $$ f'(x) = 4x^3 - 3 $$ --- 2.2 **Evaluate the limit:** $$ \lim_{x \to 0} \frac{\sin^2 x}{1 - \cos x} $$ 1. Use the identity: $$ 1 - \cos x = 2 \sin^2 \frac{x}{2} $$ 2. Rewrite the limit: $$ \lim_{x \to 0} \frac{\sin^2 x}{2 \sin^2 \frac{x}{2}} $$ 3. Use \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \), so: $$ \sin^2 x = 4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2} $$ 4. Substitute: $$ \lim_{x \to 0} \frac{4 \sin^2 \frac{x}{2} \cos^2 \frac{x}{2}}{2 \sin^2 \frac{x}{2}} = \lim_{x \to 0} 2 \cos^2 \frac{x}{2} $$ 5. As \( x \to 0 \), \( \cos \frac{x}{2} \to 1 \), so: $$ 2 \times 1^2 = 2 $$ --- 2.3 **Find new measurements when area doubles with equal increase in length and width:** 1. Original dimensions: Length \( L = 10 \) m, Width \( W = 5 \) m 2. Original area: $$ A = L \times W = 10 \times 5 = 50 $$ 3. Let increase be \( x \) meters, new dimensions: $$ L' = 10 + x, \quad W' = 5 + x $$ 4. New area is double original: $$ (10 + x)(5 + x) = 2 \times 50 = 100 $$ 5. Expand: $$ 50 + 10x + 5x + x^2 = 100 $$ $$ x^2 + 15x + 50 = 100 $$ 6. Simplify: $$ x^2 + 15x - 50 = 0 $$ 7. Solve quadratic: $$ x = \frac{-15 \pm \sqrt{15^2 - 4 \times 1 \times (-50)}}{2} = \frac{-15 \pm \sqrt{225 + 200}}{2} = \frac{-15 \pm \sqrt{425}}{2} $$ 8. Approximate \( \sqrt{425} \approx 20.62 \), so: $$ x = \frac{-15 + 20.62}{2} = 2.81 \quad \text{(positive root)} $$ 9. New dimensions: $$ L' = 10 + 2.81 = 12.81 \text{ m}, \quad W' = 5 + 2.81 = 7.81 \text{ m} $$ --- 2.4 **Differentiate:** $$ y = \frac{3}{2-x} + \frac{\pi}{e^{3x}} - \frac{1}{\csc 3x} + \frac{1}{\cot x} - \frac{1}{\sec x} + \frac{\sin 2x}{\cos x} + \frac{3}{2} \ln x - \frac{1}{3p} $$ 1. Differentiate each term: - \( \frac{3}{2-x} = 3(2-x)^{-1} \), derivative: $$ 3 \times (-1)(2-x)^{-2} \times (-1) = \frac{3}{(2-x)^2} $$ - \( \frac{\pi}{e^{3x}} = \pi e^{-3x} \), derivative: $$ \pi \times (-3) e^{-3x} = -3\pi e^{-3x} $$ - \( -\frac{1}{\csc 3x} = -\sin 3x \), derivative: $$ -3 \cos 3x $$ - \( \frac{1}{\cot x} = \tan x \), derivative: $$ \sec^2 x $$ - \( -\frac{1}{\sec x} = -\cos x \), derivative: $$ \sin x $$ - \( \frac{\sin 2x}{\cos x} = \sin 2x \sec x \), use product rule: $$ \frac{d}{dx}(\sin 2x) \sec x + \sin 2x \frac{d}{dx}(\sec x) = 2 \cos 2x \sec x + \sin 2x \sec x \tan x $$ - \( \frac{3}{2} \ln x \), derivative: $$ \frac{3}{2x} $$ - \( -\frac{1}{3p} \) is constant w.r.t \( x \), derivative: $$ 0 $$ 2. Combine all: $$ y' = \frac{3}{(2-x)^2} - 3\pi e^{-3x} - 3 \cos 3x + \sec^2 x + \sin x + 2 \cos 2x \sec x + \sin 2x \sec x \tan x + \frac{3}{2x} $$