MCQ
જો $f\left( x \right) = x{e^{x\left( {1 - x} \right)}},\,x \in R$ , તો $f(x)$ એ . . .
- A$[-1 /2, 1]$ પર ઘટતું વિધેય છે .
- B$R$ પર ઘટતું વિધેય છે .
- C$[-1 /2, 1]$ પર વધતું વિધેય છે.
- D$R$ પર વધતું વિધેય છે.
$f^{\prime}(x)=e^{x(1-x)}\left[1+x-2 x^{2}\right]$
$=-e^{x(1-x)}\left[2 x^{2}-x-1\right]$
$=-2 e^{x(1-x} \cdot\left[\left(x+\frac{1}{2}\right)(x-1]\right)$
$f^{\prime}(x)=-2 e^{x(1-x)}-A$
where $A=\left(x+\frac{1}{2}\right)(x-1)$
Now, exponential function is always +ve and $f^{\prime}(x)$ will be opposite to the sign of $A$
which is -ve in $\left[-\frac{1}{2}, 1\right]$
Hence, $f^{\prime}(x)$ is +ve in $\left[-\frac{1}{2}, 1\right]$
$\therefore f(x)$ is increasing on $\left[-\frac{1}{2}, 1\right]$
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