JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 7)
The sum of the absolute minimum and the absolute maximum values of the
function f(x) = |3x $$-$$ x2 + 2| $$-$$ x in the interval [$$-$$1, 2] is :
function f(x) = |3x $$-$$ x2 + 2| $$-$$ x in the interval [$$-$$1, 2] is :
$${{\sqrt {17} + 3} \over 2}$$
$${{\sqrt {17} + 5} \over 2}$$
5
$${{9 - \sqrt {17} } \over 2}$$
Explanation
$f(x)=\left|x^2-3 x-2\right|-x \forall x \in[-1,2]$
$\Rightarrow f(x)=\left\{\begin{array}{l}x^2-4 x-2 \text { if }-1 \leq x<\frac{3-\sqrt{17}}{2} \\ -x^2+2 x+2 \text { if } \frac{3-\sqrt{17}}{2} \leq x \leq 2\end{array}\right.$
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$f(x)_{\max }=3$
$f(x)_{\min }=f\left(\frac{3-\sqrt{17}}{2}\right)$ $=\frac{\sqrt{17}-3}{2}$
$\Rightarrow f(x)=\left\{\begin{array}{l}x^2-4 x-2 \text { if }-1 \leq x<\frac{3-\sqrt{17}}{2} \\ -x^2+2 x+2 \text { if } \frac{3-\sqrt{17}}{2} \leq x \leq 2\end{array}\right.$
$f(x)_{\max }=3$
$f(x)_{\min }=f\left(\frac{3-\sqrt{17}}{2}\right)$ $=\frac{\sqrt{17}-3}{2}$
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