Let $f, g:(0, \infty) \rightarrow \mathbb{R}$ be two functions defined by $f(x)=\int_{-x}^{x}\left(|t|-t^{2}\right) e^{-t^{2}} dt$ and $g(x)=\int_{0}^{x^{2}} t^{1/2} e^{-t} dt$. Then the value of $9\left(\mathrm{f}\left(\sqrt{\log_{\mathrm{e}} 9}\right)+\mathrm{g}\left(\sqrt{\log_{\mathrm{e}} 9}\right)\right)$ is equal to

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Definite Integrals

Let

f, g : (0, \infty) \to R

be two functions defined by

f (x) = \int_{- x}^{x} (∣ t ∣ - t^{2}) e^{- t^{2}} d t

and

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

. Then the value of

9 (f (lo g_{e} 9) + g (lo g_{e} 9))

is equal to

Answer: (C)

Work with the sum

f (x) + g (x)

, because its awkward pieces cancel. In

f

the term

∣ t ∣ e^{- t^{2}}

is even, so

\int_{- x}^{x} ∣ t ∣ e^{- t^{2}} d t = 2 \int_{0}^{x} t e^{- t^{2}} d t = 1 - e^{- x^{2}}

, while the even term

t^{2} e^{- t^{2}}

contributes

- 2 \int_{0}^{x} t^{2} e^{- t^{2}} d t

. In

g

, the substitution

t = u^{2}

turns

\int_{0}^{x^{2}} t e^{- t} d t

into

2 \int_{0}^{x} u^{2} e^{- u^{2}} d u

. These two

t^{2} e^{- t^{2}}

integrals are equal and opposite, so they cancel and leave

f (x) + g (x) = 1 - e^{- x^{2}}

. At

x = lo g_{e} 9

one has

x^{2} = lo g_{e} 9

and

e^{- x^{2}} = \frac{1}{9}

, so

f + g = 1 - \frac{1}{9} = \frac{8}{9}

, and multiplying by

9

as the problem requires gives

8

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Single correct option