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MathematicsNumerical+3Question 2 of 54

JEE Advanced 2022 · Paper 1, Mathematics question 2

Let α\alpha be a positive real number. Let f:R→Rf : \mathbb{R} \to \mathbb{R} and g:(α,∞)→Rg : (\alpha, \infty) \to \mathbb{R} be the functions defined by
f(x)=sin⁡(πx12)andg(x)=2log⁡e(x−α)log⁡e(ex−eα).f(x) = \sin\left(\frac{\pi x}{12}\right) \quad\text{and}\quad g(x) = \frac{2\log_e\left(\sqrt x - \sqrt\alpha\right)}{\log_e\left(e^{\sqrt x} - e^{\sqrt\alpha}\right)}.
Then the value of lim⁡x→α+f(g(x))\displaystyle\lim_{x\to\alpha^+} f(g(x)) is _____.

Answer and worked solution

Official answer: 0.49

Write h=x−α→0+h=\sqrt x-\sqrt\alpha\to0^+.

Denominator.
log⁡e(ex−eα)=α+log⁡e(eh−1)=α+log⁡eh+log⁡eeh−1h.\log_e\left(e^{\sqrt x}-e^{\sqrt\alpha}\right)=\sqrt\alpha+\log_e\left(e^{h}-1\right)=\sqrt\alpha+\log_e h+\log_e\frac{e^h-1}{h}.
The last term tends to 0.

Limit of g.
g(x)=2log⁡ehα+log⁡eh+o(1)→2(since log⁡eh→−∞).g(x)=\frac{2\log_e h}{\sqrt\alpha+\log_e h+o(1)}\to2\quad(\text{since }\log_e h\to-\infty).

ff is continuous, so the limit is f(2)=sin⁡π6=0.5f(2)=\sin\frac\pi6=0.5.

Topics:LimitsLogarithms
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