IIT JEE Advanced 2025 Moderate Level Question Based on Leibniz Integral Formula and L’Hopital’s Rule

Question:

Let \alpha and \beta be the real numbers such that ,

 {\lim\limits_{x\to 0}{\frac{1}{x^3}(\frac{\alpha}{2}\int_{0}^{x}\frac{1}{1-t^2}dt+\beta x \cos x)}=2}

then the value of \alpha + \beta is__________.

Ans: 2.40 [IIT Adv. 2025 Paper-1, Marks (+4, 0)]

Solution:

First we write the question correctly as below

 {\lim\limits_{x\to 0} {\frac {{\frac{\alpha}{2}\int_{0}^{x}\frac{1}{1-t^2}dt+\beta x \cos x}}{x^3}=2}}

Since the limit is in \frac {0}{0} form therefore, Applying L’Hopital’s Rule, we get

 {\lim\limits_{x\to 0} {\frac {{\frac{\alpha}{2} (\frac{1}{1-x^2})+\beta \cos x-\beta x \sin x}}{3 x^2}=2}}

Now, put x=0 in the limit, if it is in the indeterminate form then only the limit will continue further, so make it in Indeterminate form.

Now on putting x=0 we get,

\frac{\alpha}{2}+\beta =0 therefore \alpha=-2\beta ……..(1)

Now on putting value of   \alpha we get,

 {\lim\limits_{x\to 0} {\frac {\beta {(x^2-1)^{-1}}+\beta \cos x-\beta x \sin x}{3 x^2}=2}}

The limit is again in \frac {0}{0} form therefore, again Applying L’Hopital’s Rule, we get

 {\lim\limits_{x\to 0} {\frac {\beta {(-2x (x^2-1)^{-2}}- \sin x- \sin x-x \cos x)}{6 x}=2}}

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