99 Percentile Qs Bank for JEE MainMathematicsApplication of Derivatives

Define f(x)= 1 2 [| x|+ x], 0 < x 2 . Then, f is

Options

Aincreasing in ( 2 , 3 2 )
Bdecreasing in (0, 2 ) and increasing in ( 2 , )
Cincreasing in (0, 2 ) and decreasing in ( 2 , )
Dincreasing in (0, 4 ) and decreasing in ( 4 , )

Correct answer

C. increasing in (0, 2 ) and decreasing in ( 2 , )

Step-by-step solution

Given, f(x)= 1 2 [| x|+ x], 0 < x 2 Case I, when, 0 < x aligned f(x) & = 1 2 [ x+ x]= x f^ (x) & = x x & >0, for 0 < x < 2 (increasing) x & < 0, for 2 < x < (decreasing) aligned Case II When < x 2 f(x)= 1 2 [- x+ x]=0

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