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Differentiability

Mediummathematics

Define f(x) = x^2 for x \le 1 and f(x) = ax + b for x > 1; find a, b making f differentiable everywhere, then a+b.

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About This Question

Subject
mathematics
Chapter
limit, continuity and differentiability
Topic
differentiability
Difficulty
Medium
Year
2025
Tags
advanced-calculus-drilldifferentiabilitypiecewise-functionsmooth-junctionslope-matching

Solution

Correct Answer:

Differentiability of a piecewise function at the join x = 1 demands two matching conditions: the values must agree for continuity, and the one-sided derivatives must agree for a common tangent. Continuity at x = 1 gives 1^2 = a(1) + b, so a + b = 1. Differentiability requires the left derivative 2x|_{x=1} = 2 to equal the right derivative a, so a = 2. Substituting back, 2 + b = 1, hence b = -1, and the pair (a,b) = (2,-1) makes f smooth across the junction. Therefore a + b = 1, exactly the continuity equation. Option 0 satisfies neither matching condition consistently. Option 2 uses only the derivative match a = 2 and mislabels it as the sum. Option 3 adds the two derivatives instead of using the continuity sum. The governing JEE pattern is the dual value-and-slope matching for differentiable piecewise functions. Plausibility check: with a = 2, b = -1 the right piece is 2x - 1, whose value at x = 1 is 1 and whose slope is 2, matching the parabola's value 1 and slope 2 precisely, so the graph joins smoothly and a + b = 1 holds.

This medium 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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