Let \(\alpha\) and \(\beta\) be the real numbers such that \[ \lim _{x \rightarrow 0} \frac{1}{x^{3}}\left(\frac{\alpha}{2} \int_{0}^{x} \frac{1}{1-t^{2}} d t+\beta x \cos x\right)=2 \] Then the value of \(\alpha+\beta\) is ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Limits XII

Let

α

and

β

be the real numbers such that

x \to 0 lim \frac{1}{x ^{3}} (\frac{α}{2} \int_{0}^{x} \frac{1}{1 - t ^{2}} d t + β x cos x) = 2

Then the value of

α + β

is ____.

Answer: 2.4

Apply L'Hôpital's rule:

x \to 0 lim \frac{\frac{α}{2} \cdot \frac{1}{1 - x ^{2}} + β cos x - β x sin x}{3 x ^{2}}

Expand using Taylor series:

\frac{1}{1 - x ^{2}} = 1 + x^{2} + x^{4} + \dots

cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots

x sin x = x^{2} - \frac{x ^{4}}{6} + \dots

Substituting:

\frac{α}{2} (1 + x^{2} + \dots) + β (1 - \frac{x ^{2}}{2} + \dots) - β (x^{2} - \frac{x ^{4}}{6} + \dots)

= (\frac{α}{2} + β) + x^{2} (\frac{α}{2} - \frac{β}{2} - β) + \dots

= (\frac{α}{2} + β) + x^{2} (\frac{α - 3 β}{2}) + \dots

For the limit to equal 2:

\frac{α}{2} + β = 0 and \frac{α - 3 β}{6} = 2

From the first equation:

α = - 2 β

Substituting into the second:

\frac{- 2 β - 3 β}{6} = 2 ⟹ β = - \frac{12}{5}

Therefore:

α = \frac{24}{5}

α + β = \frac{24}{5} - \frac{12}{5} = \frac{12}{5} = 2.40

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer