MCQ
If $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ then
- A$1 < \alpha < 2$
- B$\alpha < 0$
- ✓$0 < \alpha < 1$
- DNone of these
✓
Answer
Correct option: C.
$0 < \alpha < 1$
c
(c) $\int_0^1 {{e^{{x^2}}}} (x - \alpha )\,dx = 0$
(c) $\int_0^1 {{e^{{x^2}}}} (x - \alpha )\,dx = 0$
==> $\frac{1}{2}\int_0^1 {2x.{e^{{x^2}}}dx = \alpha \int_0^1 {{e^{{x^2}}}dx} } $
==> $\frac{1}{2}|{e^{{x^2}}}|_0^1 = \alpha \int_0^1 {{e^{{x^2}}}dx} $
==> $\frac{1}{2}(e - 1) = \alpha \,\int_0^1 {{e^{{x^2}}}dx} $
==> $\alpha = \frac{{\frac{1}{2}(e - 1)}}{{\int_0^1 {{e^{{x^2}}}dx} }} > 0$ and $\alpha < 1$.
So, $0 < \alpha < 1$.
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