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
==> $\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, $0 < \alpha < 1$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$f(x)=\left\{\begin{array}{cl}x|x| \sin \left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x=0,\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}1-2 x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text { otherwise }\end{array}\right.$
Let $a, b, c, d \in R$. Define the function $h: R \rightarrow R$ by
$h(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in R$
Match each entry in $List-I$ to the correct entry in $List-II$.
| $List-I$ | $List-II$ |
| ($P$) If $a=0, b=1, c=0$ and $d=0$, then | ($1$) $h$ is one-one |
| ($Q$) If $a=1, b=0, c=0$ and $d=0$, then | ($2$) $h$ is onto. |
| ($R$) If $a=0, b=0, c=1$ and $d=0$, then | ($3$) $h$ is differentiable on $R$. |
| ($S$) If $a=0, b=0, c=0$ and $d=1$, then | ($4$) the range of $h$ is $[0,1]$ |
| ($5$) the range of $h$ is $\{0,1\}$ |
The correct option is