JEE PYQ: Limits Question 15
Question 15 - 2020 (05 Sep Shift 1)
If $\alpha$ is the positive root of the equation $p(x) = x^2 - x - 2 = 0$, then $\lim_{x \to \alpha^+} \frac{\sqrt{1 - \cos(p(x))}}{x + \alpha - 4}$ is equal to:
(1) $\frac{3}{2}$
(2) $\frac{3}{\sqrt{2}}$
(3) $\frac{1}{\sqrt{2}}$
(4) $\frac{1}{2}$
Show Answer
Answer: (1)
Solution
$\alpha = 2$. $p(x) = (x-2)(x+1)$. $\lim_{x\to 2^+}\frac{\sqrt{2}\sin\left(\frac{x^2-x-2}{2}\right)}{x-2} \times \frac{(x^2-x-2)}{2(x-2)} \times \frac{2}{x^2-x-2} = \frac{1}{\sqrt{2}} \times 1 \times 3 = \frac{3}{\sqrt{2}}$.
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