JEE Mathematics 2024 Question with Solution

Given:

• $f(x) = \int_{-x}^{x} (|t| - t^2)e^{-t^2} \, dt$
• $g(x) = \int_{0}^{x^2} t^{1/2}e^{-t} \, dt$
• $x = \sqrt{\ln 9}$

Find:

• The value of $f(\sqrt{\ln 9}) + g(\sqrt{\ln 9})$

From the solution, the working concludes that

$$f\left( \sqrt{\log_e 9} \right) + g\left( \sqrt{\log_e 9} \right) = 8$$

and the correct option is C.

Also,

$$x^2 = \ln 9$$

so the upper limit in $g(x)$ becomes $\ln 9$.

The solution uses symmetry for $f(x)$ and states the required sum directly. Therefore, the final value is $8$.

Hence, the correct option is C.

Given:

• $f(x) = \int_{-x}^{x} (|t| - t^2)e^{-t^2} \, dt$
• $g(x) = \int_{0}^{x^2} t^{1/2}e^{-t} \, dt$
• $x = \sqrt{\log_e 9}$

Find:

• $f(x) + g(x)$ at this value of $x$

the solution rewrites

$$f(x) = \int_{-x}^{0} (-t - t^2)e^{-t^2} \, dt + \int_{0}^{x} (t - t^2)e^{-t^2} \, dt$$

and then invokes symmetry to simplify the expression.

It also uses

$$x^2 = \log_e 9$$

and hence

$$e^{x^2} = 9$$

For the second integral,

$$g(x) = \int_{0}^{x^2} t^{1/2}e^{-t} \, dt$$

so at the given value,

$$g\left(\sqrt{\log_e 9}\right) = \int_{0}^{\log_e 9} t^{1/2}e^{-t} \, dt$$

The extracted solution does not provide full intermediate evaluation steps for the individual integrals, but it explicitly concludes that

$$f\left( \sqrt{\log_e 9} \right) + g\left( \sqrt{\log_e 9} \right) = 8$$

Therefore, the required value is $8$, so the correct option is C.