2 marksMathematicsLimits, continuity and differentiabilityDifferentiability and its link to continuityModerateabout 2 min

IIT JAM 2022 Question 34: Limits, continuity and differentiability

Let $f$ be defined by $f(x) = |x| + \left|\cos\left(\dfrac{\pi}{2} - x\right)\right|$ for $x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ . Then

Select all that apply.

Answer and worked solution

Answer: (A) and (C) and (D)

Since $\cos\left(\tfrac{\pi}{2} - x\right) = \sin x$ , we have $f(x) = |x| + |\sin x|$ , which is continuous on the whole interval, so (A) and (D) hold and $\lim_{x\to0} f(x) = f(0) = 0$ .

Near $0$ , for $x > 0$ , $f(x) = x + \sin x$ with derivative $1 + \cos x \to 2$ ; for $x < 0$ , $f(x) = -x - \sin x$ with derivative $-1 - \cos x \to -2$ . The one-sided derivatives at $0$ differ, so $f$ is not differentiable at $0$ (statement (B) is false) but is differentiable everywhere else (statement (C) is true). The correct statements are (A), (C) and (D).

The trap: (B)

f

is differentiable at

x = 0

The two absolute values create a corner at $x=0$ ; the left and right derivatives ( $-2$ and $+2$ ) do not match, so differentiability at $0$ fails even though the function is continuous there.

Plain-text version

Let f be defined by f(x) = |x| + |cos (π/2 - x)| for x ∈ (-π/2, π/2). Then (A) f is continuous on (-π/2, 0) ∪ (0, π/2). (B) f is differentiable at x = 0. (C) f is differentiable everywhere except x = 0. (D) lim_(x→ 0) f(x) = 0.

Also written as: Let f be defined by f(x) = |x| + |cos (pi/2 - x)| for x in (-pi/2, pi/2). Then

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What this question tests

This question is on Differentiability and its link to continuity, part of Limits, continuity and differentiability in Mathematics.

When a derivative exists and why differentiability implies continuity but not conversely.

TestStop has 12 past-paper questions on this subtopic; the nearest in year are below.

IIT JAM 2021 · Q35Let f be defined on (-π/2, π/2) by f(x) = (cos (π/2 + |x|))/(sin (π/2 - |x|)). Then,

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