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}}
The limit is still in \frac {0}{0} form therefore, again Applying L’Hopital’s Rule, we get
{\lim\limits_{x\to 0} {\frac {\beta {(-2 (x^2-1)^{-2}}+4x.2x {(x^2-1)^{-3}}-2\cos x+x \sin x-\cos x)}{6 }=2}}
Now, the limit is in non of the indeterminate forms, therefore putting x=0 we get,
\frac {\beta {(-2+0-2+0-1)}}{6}=2
\Rightarrow \beta = -\frac {12}{5}
\therefore \alpha +\beta = -2\beta + \beta =-\beta = \frac {12}{5} = 2.40
This is the final answer, given by IIT JEE Advanced.
Alternate Method for solving the above question
The given question can also be solved using Expansion Series
{\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 {(1-x^2)^{-1}}+\beta \cos x-\beta x \sin x}{3 x^2}=2}}
Now, applying expansion series formulas
using Binomial Theorem for Negative and Fractional Index, sine expansion series formula and cosine expansion series formula
\lim\limits_{x\to 0} {\frac{\beta [-(1+(-1)(-x^{2})+\frac{(-1)(-1-1)}{2!}(-x^{2})^{2}+...+(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-...)-x(x-\frac{x^{3}}{3!}+..)]}{3x^{2}}}=2
Now comparing coefficient of x^{2} in numerator and denominator, we get the desired value of limit
\therefore \frac{\beta(-1-\frac{-1}{2!}-1)}{3}=\frac{-5\beta}{6}=2
\Rightarrow \beta = -\frac {12}{5}
\therefore \alpha +\beta = -2\beta + \beta =-\beta = \frac {12}{5} = 2.40
This is the final answer, given by IIT JEE Advanced.
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