Differentiability of e^(sin|x|−|x|) – JEE Main 2026 PYQ

Quick Summary

f(x) = e^(sin|x| − |x|) splits into two smooth branches at x=0: for x≥0, the exponent is sinx−x; for x<0, it is x−sinx (since |x|=−x there and sin|x|=−sinx). Both branches have the SAME one-sided derivative at 0 (namely 0), so despite the modulus signs, f is differentiable everywhere on ℝ — Statement I is true. For x<0, the exponent’s derivative is 1−cos(x), which is strictly positive except at multiples of 2π; on (−π,−π/2) it never touches 0, so f is strictly increasing there — Statement II is also true. Both true → option (1).

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Correct Answer

Why This Answer

The instinct with a modulus inside a function is to expect a “corner” (non-differentiable point) wherever the modulus’s argument changes sign — here at x=0. That instinct is usually right for |x| itself, but sin|x|−|x| is special: writing g(x)=sin|x|−|x|, the right-side branch is sinx−x with derivative cosx−1, which equals exactly 0 at x=0; the left-side branch is x−sinx with derivative 1−cosx, which is ALSO exactly 0 at x=0. The one-sided derivatives coincide, so g — and hence f=e^g, since exp is smooth — is differentiable at the join point too, not just elsewhere. This is the kind of function where the modulus creates a symmetric “valley” shape smooth enough to have a well-defined tangent even at the vertex.

For Statement II, the relevant branch is the x<0 one: g(x)=x−sin(x), g′(x)=1−cos(x). Since cos(x)<1 for any x that is not an integer multiple of 2π, and the interval (−π,−π/2) contains no such multiple, g′(x)>0 strictly throughout — g is strictly increasing there, and since exp is an increasing function, f=e^g inherits the same strict increase.

Core Concept

Differentiability of |x|-based functions. A function built from |x| is not automatically non-differentiable at the sign-change point — check the one-sided derivatives explicitly. If sin|x| is present, remember sin|x|=sin(x) for x≥0 but sin|x|=−sin(x) for x<0 (since sin is odd), which is what creates the perfectly matching branches here.

Monotonicity through composition. If h(x)=e^{g(x)}, then h′(x)=e^{g(x)}·g′(x), and since e^{g(x)}>0 always, the sign of h′ matches the sign of g′ exactly. So “is f increasing” reduces to “is g increasing,” a purely algebraic/trigonometric question about 1−cos(x) or cos(x)−1.

Step-by-Step Solution

1. Split the exponent by sign of x. For x≥0: sin|x|−|x| = sinx−x. For x<0: sin|x|−|x| = sin(−x)−(−x) = −sinx+x = x−sinx.

2. Check one-sided derivatives of g at x=0.

   Right: g(x)=sinx−x, g'(x)=cosx−1, g'(0+) = cos0−1 = 0
   Left: g(x)=x−sinx, g'(x)=1−cosx, g'(0-) = 1−cos0 = 0

   Both equal 0, so g is differentiable at 0 with g′(0)=0. Combined with g being smooth away from 0 (each branch is a difference of smooth functions), g is differentiable on all of ℝ.

3. Lift to f = e^g. Since exp is differentiable everywhere and g is differentiable everywhere, the composition f is differentiable everywhere too. Statement I confirmed true.

4. Examine monotonicity on (−π, −π/2). Here x<0, so g(x)=x−sinx, g′(x)=1−cos(x). Since x∈(−π,−π/2) never equals a multiple of 2π, cos(x)≠1, so g′(x)>0 strictly.

5. Conclude f is increasing there. f′(x) = e^{g(x)}·g′(x) > 0 on (−π,−π/2), since e^{g(x)}>0 always and g′(x)>0 there. Statement II confirmed true.

Final Answer

f is differentiable on all of ℝ and strictly increasing on (−π,−π/2). The lesson: with |x| inside a function, always check the one-sided derivatives explicitly rather than assuming a corner — some modulus combinations, like this one, are perfectly smooth.

Key Facts to Remember

• sin|x| = sin(x) for x≥0 and −sin(x) for x<0 (sin is an odd function).
• Differentiability at a modulus join point requires checking BOTH one-sided derivatives match — don’t assume a corner exists.
• For h=e^g, sign of h′ always matches sign of g′, since e^g>0 everywhere.
• 1−cos(x) ≥ 0 always, with equality only at integer multiples of 2π.

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Prerequisites

Before practising this question type, make sure the following are in place:

• One-sided derivatives and the definition of differentiability at a point.
• Splitting |x|-based functions into branches by sign of x, and the oddness of sin(x).
• Chain rule for e^{g(x)} and how its sign tracks g′(x).

Revise these from the Continuity and Differentiability chapter, then attempt the related questions below.

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