Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

If $m$ and $M$ are respectively minimum and maximum values of $f(x)= |2-|x||, -3≤x≤3$, then:

Options:

m = 0 and M = 2

m = 1 and M = 2

m = 0 and M = 4

m = 0 and M = 1

Correct Answer:

m = 0 and M = 2

Explanation:

$f(x)= |2-|x||,\,\, -3≤x≤3$

$⇒0≤|x|≤3$

$⇒-3≤-|x|≤0$

$⇒-1≤2-|x|≤2$

so $0≤|2-|x||≤2$

m = 0 and M = 2