Math JEE Adv. 2025 paper 1

 

To solve the problem, we need to find the real numbers $\alpha$ and $\beta$ such that the given limit equals 2. Here's the step-by-step solution:

Given Limit:

$$\underset{x\to 0}{\lim }\frac{1}{x^3}(\frac{\alpha }{2}{\int }_0^x\frac{1}{1-t^2}dt+\beta x\cos x)=2$$

Step 1: Expand the Integrand

The integrand $\frac{1}{1-t^2}$ can be expanded using the geometric series for $\mathrm{\mid }t\mathrm{\mid }<1$:

$$\frac{1}{1-t^2}=1+t^2+t^4+\cdots$$

Integrate term by term:

$${\int }_0^x\frac{1}{1-t^2}dt={\int }_0^x(1+t^2+t^4+\cdots )dt=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots$$

Step 2: Substitute the Integral into the Limit

Substitute the expanded integral and the Taylor series for $\cos x$ ($\cos x=1-\frac{x^2}{2}+\cdots$) into the limit:

$$\frac{\alpha }{2}(x+\frac{x^3}{3}+\cdots )+\beta x(1-\frac{x^2}{2}+\cdots )$$

Divide by $x^3$:

$$\frac{1}{x^3}(\frac{\alpha }{2}x+\frac{\alpha }{6}{x}^3+\beta x-\frac{\beta }{2}{x}^3+\cdots )$$

Simplify:

$$\frac{\alpha }{2x^2}+\frac{\alpha }{6}+\frac{\beta }{x^2}-\frac{\beta }{2}+\cdots$$

Step 3: Eliminate Lower Order Terms

For the limit to exist as $x\to 0$, the coefficients of $\frac{1}{x^2}$ must cancel out:

$$\frac{\alpha }{2}+\beta =0\Rightarrow \alpha =-2\beta$$

Step 4: Evaluate the Remaining Terms

After cancellation, the remaining terms are:

$$\frac{\alpha }{6}-\frac{\beta }{2}=2$$

Substitute $\alpha =-2\beta$:

$$\frac{-2\beta }{6}-\frac{\beta }{2}=2\Rightarrow -\frac{\beta }{3}-\frac{\beta }{2}=2$$

Combine like terms:

$$-\frac{5\beta }{6}=2\Rightarrow \beta =-\frac{12}{5}$$

Then, $\alpha =-2\beta =\frac{24}{5}$.

Step 5: Compute $\alpha +\beta$

$$\alpha +\beta =\frac{24}{5}-\frac{12}{5}=\frac{12}{5}=2.4$$

Final Answer:

$$\boxed{{\displaystyle 2.4}}$$