Question:

<p><strong>Step 1: Find condition on a</strong></p><p>Combine the fractions: $$\lim_{x \to 1}\left(\frac{1}{x-1} - \frac{1}{x^a - 2x + 1}\right) = \lim_{x \to 1}\frac{(x^a - 2x + 1) - (x-1)}{(x-1)(x^a - 2x + 1)}$$</p><p>$$= \lim_{x \to 1}\frac{x^a - 3x + 2}{(x-1)(x^a - 2x + 1)}$$</p><p><strong>Step 2: Apply limit existence condition</strong></p><p>For the limit to exist and be non-zero, at x=1: numerator = 0 and denominator ≠ 0.</p><p>Numerator: $1 - 3 + 2 = 0$ ✓ (satisfied for any positive integer a)</p><p>Denominator at x=1: $(1-1)(x^a - 2 + 1) = 0$ always.</p><p>So we need the factor (x-1) to cancel. For this, $x^a - 2x + 1$ must be divisible by (x-1).</p><p>Check: $1^a - 2(1) + 1 = 1 - 2 + 1 = 0$ ✓ for all a.</p><p><strong>Step 3: Determine a using multiplicity</strong></p><p>For finite non-zero limit, (x-1) must appear exactly once in numerator and denominator (after one cancellation).</p><p>Numerator: $x^a - 3x + 2 = (x-1)(x^{a-1} + x^{a-2} + \cdots + x - 2)$ at x=1 gives derivative: $ax^{a-1} - 3|_{x=1} = a - 3$</p><p>For simple zero: $a - 3 = 0 \Rightarrow a = 3$</p><p><strong>Step 4: Calculate b using L'Hôpital's Rule with a=3</strong></p><p>$$\lim_{x \to 1}\frac{x^3 - 3x + 2}{(x-1)(x^3 - 2x + 1)} = \lim_{x \to 1}\frac{3x^2 - 3}{(x^3 - 2x + 1) + (x-1)(3x^2 - 2)}$$</p><p>At x=1: numerator derivative = $6 - 3 = 3$; denominator = $(1-2+1) + 0 = 0$</p><p>Apply L'Hôpital again: $\lim_{x \to 1}\frac{6x}{3x^2 - 2 + 3x^2 - 2 + (x-1)(6x)} = \frac{6}{6-2+6-2} = \frac{6}{8} = \frac{3}{4}$</p><p>So $b = \frac{3}{4}$</p><p><strong>Step 5: Verify option A</strong></p><p>$\tan^{-1}(\tan 3)$ where $3 \approx 171.89°$ is in the interval $(\pi/2, \pi)$</p><p>Since $3 > \pi$ and $3 < 2\pi$: $\tan^{-1}(\tan 3) = 3 - \pi$ ✓</p><p><strong>∴ Answer:</strong> A</p>