JEE Advanced 2022 Paper 1, Mathematics Section 1 Q2: L’Hospital’s Rule

JEE Advanced2022Paper 1Mathematics Section 1

Let $α$ be a positive real number. Let $f : R \to R$ and $g : (α, \infty) \to R$ be the functions defined by

$f (x) = sin (\frac{π x}{12}) and g (x) = \frac{2 lo g _{e} ( x - α )}{lo g _{e} ( e ^{x} - e ^{α} )} .$

Then the value of $x \to α^{+} lim f (g (x))$ is ______.

Your answer

Solution

Set $u = x$ and $a = α$ . As $x \to α^{+}$ , $u \to a^{+}$ , so

$g (x) = \frac{2 lo g _{e} ( u - a )}{lo g _{e} ( e ^{u} - e ^{a} )} .$

Applying L'Hospital's rule on the indeterminate form $\frac{- \infty}{- \infty}$ , differentiate numerator and denominator with respect to $u$ :

$u \to a^{+} lim \frac{2/ ( u - a )}{e ^{u} / ( e ^{u} - e ^{a} )} = u \to a^{+} lim \frac{2 ( e ^{u} - e ^{a} )}{( u - a ) e ^{u}} .$

Using $u \to a lim \frac{e ^{u} - e ^{a}}{u - a} = e^{a}$ , the inner limit equals $\frac{2 e ^{a}}{e ^{a}} = 2$ .

Hence $x \to α^{+} lim f (g (x)) = f (2) = sin \frac{π}{6} = \frac{1}{2} = 0.50$ .

Concept behind this question

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