Differentiability
Consider h(x) = |x - 1| + |x + 1| on the real line; at how many points does h fail to be differentiable?
Select the correct option:
Solution
2
Differentiability of an absolute-value combination can break only where an inside expression changes sign, since each |.| contributes a corner at its own zero. The zeros are at x = 1 for |x-1| and at x = -1 for |x+1|, giving two candidate corner points. Between and outside these points h is a sum of linear pieces and is smooth. At x = -1 the slope changes from -2 to 0, and at x = 1 it changes from 0 to +2, so both are genuine non-differentiable corners, as the flat-bottomed graph shows. Hence there are exactly 2 such points. Option 0 ignores the corners created at the kinks of the moduli. Option 1 counts only one zero, forgetting the second modulus. Infinitely many is wrong because a finite sum of absolute values has finitely many kinks. The governing JEE pattern is locating corner points of piecewise-linear modulus sums. Plausibility check: for |x| \ge 1 the function grows linearly with slope \pm 2, while on [-1,1] it is constant at 2, and a constant-to-slope transition is precisely a corner, confirming two failure points.
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About This Question
- Subject
- mathematics
- Chapter
- limit, continuity and differentiability
- Topic
- differentiability
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
2
Differentiability of an absolute-value combination can break only where an inside expression changes sign, since each |.| contributes a corner at its own zero. The zeros are at x = 1 for |x-1| and at x = -1 for |x+1|, giving two candidate corner points. Between and outside these points h is a sum of linear pieces and is smooth. At x = -1 the slope changes from -2 to 0, and at x = 1 it changes from 0 to +2, so both are genuine non-differentiable corners, as the flat-bottomed graph shows. Hence there are exactly 2 such points. Option 0 ignores the corners created at the kinks of the moduli. Option 1 counts only one zero, forgetting the second modulus. Infinitely many is wrong because a finite sum of absolute values has finitely many kinks. The governing JEE pattern is locating corner points of piecewise-linear modulus sums. Plausibility check: for |x| \ge 1 the function grows linearly with slope \pm 2, while on [-1,1] it is constant at 2, and a constant-to-slope transition is precisely a corner, confirming two failure points.
This easy difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of differentiability. It appeared in the 2025 exam.
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