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STD 12 - 6. Application of derivatives — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 6. Application of derivatives4 Marks
MCQ
If $x =1$ is a critical point of the function $f(x)=\left(3 x^{2}+a x-2-a\right) e^{x},$ then 
  • ✓
    $x =1$ is a local minima and $x =-\frac{2}{3}$ is a local maxima of $f$.
  • B
    $x =1$ is a local maxima and $x =-\frac{2}{3}$ is a local minima of $f$
  • C
    $x =1$ and $x =-\frac{2}{3}$ are local minima of $f$
  • D
    $x =1$ and $x =-\frac{2}{3}$ are local maxima of $f$
✓

Answer

Correct option: A.
$x =1$ is a local minima and $x =-\frac{2}{3}$ is a local maxima of $f$.
a
$f(x)=\left(3 x^{2}+a x-2-a\right) e^{x}$

$f^{\prime}(x)=\left(3 x^{2}+a x-2-a\right) e^{x}+e^{x}(6 x+a)$

$=e^{x}\left(3 x^{2}+x(6+a)-2\right)$

$f^{\prime}(x)=0$ at $x=1$

$\Rightarrow 3+(6+a)-2=0$

$a=-7$

$f^{\prime}(x)=e^{x}\left(3 x^{2}-x-2\right)$

$=e^{x}(x-1)(3 x+2)$

$x =1$ is point of local minima

$x =\frac{-2}{3}$ is point of local maxima

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