IIT-JEE Advanced 2022 Maths Paper 1 Problem 2

Problem

Let α be a positive real number. Let f: R → R and g: (α, ∞) → R be the functions defined by f(x) = sin(πx/12) and g(x) = 2loge(√x – √α) / loge(ex – e√α). Then find the value of \( \displaystyle \lim_{x \to \alpha^+}\)f(g(x)).

Solution

Since f(x) is continuous everywhere, we simply need to evaluate f(β), where β = \( \displaystyle \lim_{x \to \alpha^+}\)g(x).

Now, if we look at g(x), it takes the -∞/-∞ form when x tends to α+. So, we’ll evaluate it using l’ôpital’s rule.

After differentiating both the numerator and denominator, we’ll get g(x) = \( \displaystyle \lim_{x \to \alpha^+}\) [2/(√x – √α) × 1/2√x] / [1/(e√x – e√α) x e√x × 1/2√x]

Let’s clean this up. We’ll get 2(e√x – e√α)/(√x – √α) × 1/e√x.

The first term in the product can be converted into a standard limit.

This is what we need to find \( \displaystyle \lim_{x \to \alpha^+}\) 2(e√x – √α – 1)/(√x – √α) × eα/e√x.

The first term is a known limit, which evaluates to 1. The second term evaluates to 1 as well. So, we’re left with 2 as the overall limit, i.e. the value of β.

Finally, the required limit equals sin(2π/12) or sin(π/6), which equals 1/2.

Comments

Nothing much to say here. The problem was simple enough – pretty laborious to type, but not as laborious to solve.

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