Verified Solution • 2025–26 Session moderate

QJEEM22Mad591: JEE Main 2022 Mathematics - Class 12 JEEM22Mad591 Solution: The sum of the absolute minimum and the absolute maximum values …

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 :

Multiple Choice Options

  • A 17+32{{\sqrt {17} + 3} \over 2}
  • B 17+52{{\sqrt {17} + 5} \over 2}
  • C 5
  • D 9−172{{9 - \sqrt {17} } \over 2}

Step-by-Step Verified Solution

f(x)=∣x2−3x−2∣−x∀x∈[−1,2]f(x)=\left|x^2-3 x-2\right|-x \forall x \in[-1,2]

⇒f(x)={x2−4x−2 if −1≤xlt;3−172−x2+2x+2 if 3−172≤x≤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.

JEE Main 2022 (Online) 26th June Morning Shift Mathematics - Application of Derivatives Question 79 English Explanation
f(x)max⁡=3f(x)_{\max }=3

f(x)min⁡=f(3−172)f(x)_{\min }=f\left(\frac{3-\sqrt{17}}{2}\right) =17−32=\frac{\sqrt{17}-3}{2}

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