If , then a + b + m is equal to: [2026]

Question

If $\lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m$ , then a + b + m is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

We have, $\lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m$

$\Rightarrow \lim_{x \to 2} \frac{(\sqrt{x - 1} - 1) \sin (x^{3} - 5 x^{2} + a x + b)}{[(x - 1) - 1] \log_{e} (1 + (x - 2))} = m$

$\Rightarrow 2 \lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{{(x - 2)}^{2}} = m$ ... (i)

Let $f (x) = x^{3} - 5 x^{2} + a x + b$

$\Rightarrow f' (x) = 3 x^{2} - 10 x + a$

For limit to be finite, $f (2) = 0, f' (2) = 0$

For $f (2) = 0 \Rightarrow 8 - 20 + 2 a + b, f' (2) = 0$

$\Rightarrow 2 a + b = 12$

For $f' (2) = 0 \Rightarrow 3 (4) - 10 (2) + a = 0$

$\Rightarrow a = 8 ∴ b = - 4$

$∴$ From (i),

$2 \lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + 8 x - 4)}{{(x - 2)}^{2}} = m$

$\Rightarrow 2 \lim_{x \to 2} \frac{\sin [{(x 2)}^{2} (x - 1)]}{{(x - 2)}^{2} (x - 1)} (x - 1) = m$

$\Rightarrow 2 [1] [2 - 1] = m \Rightarrow m = 2$

$∴ a + b + m = 8 - 4 + 2 = 6$

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