Question
यदि $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ तो
==> $\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} $
==> $\alpha = \frac{{\frac{1}{2}(e - 1)}}{{\int_0^1 {{e^{{x^2}}}dx} }} > 0$ व $\alpha < 1$.
$\therefore$ $0 < \alpha < 1$.
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तथा $\mathrm{g}(\mathrm{x})=\left\{\begin{array}{cc}\min \{\mathrm{f}(\mathrm{t})\}, & 0 < \mathrm{t} \leq \mathrm{x} \text { और } 0 < \mathrm{x} \leq 1 \\ \frac{3}{2}+\mathrm{x}, & 1<\mathrm{x} < 2\end{array}\right.$
द्वारा परिभाषित फलन $\mathrm{g}(\mathrm{x})$ का विचार कीजिए। तो,