Question

The Roll's theorem is applicable in the interval – 1 ≤ x ≤ 1 for the function

(a) f(x) = x 

(b) f(x) = x² 

(c) f(x) = 2x² + 3 

(d) f(x) = |x|

Answer 1

The correct option (b) f(x) = x²   

Explanation:

Let f(x) = |x|

Here f(– 1) = |– 1| = 1 and f(1) = 1 ⇒ f(– 1) = f(1)

f(x) is continuous in given interval, but not differentiable at x = 0

∴ Roll's theorem not applied.

f(x) = x² is continuous and differentiable in the interval [– 1, 1]

∴ Roll's theorem is applicable.