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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $A$ and $B$ are square matrices such that $B = -A^{-1} BA,$ then $(A + B)^2 =$
Let $E$ be an event of a sample space $S$ of an experiment then $P(S \mid E)=$
Objective function of a LPP is:
The area bounded by the parabola $y^2 = 4x$ and the line $2x - 3y + 4 = 0$, in square unit, is
Find the value of x, if $\begin{bmatrix}2&5\\3&\text{x}\end{bmatrix}=\begin{bmatrix}\text{x}&-1\\5&3\end{bmatrix}$ is:
Solve:$\sin { \left( { \tan }^{ -1 }\text{x} \right) } ,\left| \text{x} \right| <1$ is equal to:
Let $p$ and $q$ be the position vectors of $P$ and $Q$ respectively with respect to $O$ and $|p|\, = p,\,\,|q|\,\, = q.$ The points $R$ and $S$ divide $PQ$ internally and externally in the ratio $2 : 3$ respectively. If $\overrightarrow {OR} $ and $\overrightarrow {OS} $ are perpendicular, then
The edge of a cube is of length $‘a’$ then the shortest distance between the diagonal of a cube and an edge skew to it is
Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be three unit vectors, such that $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=1$ and $\vec{\text{a}}$ is perpendicular to $\vec{\text{b}}.$ If $\vec{\text{c}}$ makes angle $\alpha$ and $\beta$ with $\vec{\text{a}}$ and $\vec{\text{b}}$ respectively, then $\cos\alpha+\cos\beta=$
Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .