MCQ
The sum of the absolute minimum and the absolute maximum values of the function $f(x)=\left|3 x-x^{2}+2\right|-x$ in the interval $[-1,2]$ is
  • $\frac{\sqrt{17}+3}{2}$
  • B
    $\frac{\sqrt{17}+5}{2}$
  • C
    $5$
  • D
    $\frac{9-\sqrt{17}}{2}$

Answer

Correct option: A.
$\frac{\sqrt{17}+3}{2}$
a
$f(x)=\left\{\begin{array}{ll}x^{2}-4 x-2, & \forall x \in\left(-1, \frac{3-\sqrt{17}}{2}\right) \\ -x^{2}+2 x+2, & \forall x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\end{array}\right.$

$f^{\prime}(x)$ when $x \in\left(-1, \frac{3-\sqrt{17}}{2}\right)$

$f^{\prime}(x)=2 x-4=0 \Rightarrow x=2$

$f^{\prime}(x)=2(x-2) \quad \Rightarrow f^{\prime}(x)$ is always $\downarrow$

$f(2)=2$

$f(-1)=3$

$f\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}$

$f^{\prime}(x)$ when $x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)$

$f^{\prime}(x)=-2 x+2$

$f^{\prime}(x)=-2(x-1)$

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

$f(1)=3$

absolute minimum value $=\frac{\sqrt{17}-3}{2}$

absolute maximum value $=3$

Sum $=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}$

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