$l i m x \to \infty [m i n (y^{2} - 4 y + 11) \frac{s i n x}{x}]$ is equal to (where [.] denotes the greatest integer function)

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Solution

The correct option is B 6
$∵ y^{2} - 4 y + 11 = (y - 2)^{2} + 7 ∴ m i n (y^{2} - 4 y + 11) = 7 (a t y = 2) ∴ l i m x \to 0 [m i n (y^{2} - 4 y + 11) \frac{s i n x}{x}] = l i m x \to 0 [\frac{7 s i n x}{x}] N o w, L e t f (x) = [\frac{7 s i n x}{x}] F o r x > 0 S i n x < x ∴ \frac{s i n x}{x} < 1 \Rightarrow \frac{7 s i n x}{x} < 7 \Rightarrow [\frac{7 s i n x}{x}] = 6 ∴ R H L = l i m x \to 0 + f (x) = 6 F o r x < 0 s i n x > x \frac{s i n x}{x} < 1 ∴ \frac{7 s i n x}{x} < 7 \Rightarrow [\frac{7 s i n x}{x}] = 6 H e n c e, l i m x \to 0 [m i n (y^{2} - 4 y + 11) \frac{s i n x}{x}] = 6$

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