Question

If ${\int }_0^1{e}^{x^2}(x-\alpha )dx=0$, then

• A. $1<\alpha <2$
• B. $\alpha <2$
• C. $0<\alpha <1$
• D. $\alpha =0$

Answer 1

Answer: (C)

$0<\alpha <1$

$\because {\int }_0^1{e}^{x^2}(x-\alpha )dx=0,$

$\therefore e^{x^2}(x-\alpha )$ must be $+ve$ and $-ve$ both for $x\in (0,1)$

i.e. $e^x(x-\alpha )=0$ for one $x\in (0,1)$

$\therefore \alpha \in (0,1)$