Piecewise Continuity Fixing a Definite Integral – JEE Main 2026 PYQ

Quick Summary

The function is defined as 1/3 for x≤π/2, and b(1−sinx)/(π−2x)² for x>π/2. Continuity at π/2 requires the limit of the second branch to equal 1/3. Substituting x=π/2+h and using 1−sinx≈h²/2, (π−2x)²=4h² exactly, the limit reduces to b/8, so b = 8/3. This makes the integral’s upper limit 3b−6 = 2 exactly. Then ∫₀² |x²+2x−3| dx splits at the root x=1 (since x²+2x−3=(x+3)(x−1) is negative on [0,1) and positive on (1,2]), giving a total of 4.

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Why This Answer

This question is really two independent skills chained together: first a continuity condition that determines an unknown constant b, then a definite integral whose LIMITS depend on that constant. Missing either half sinks the whole question — get b wrong and the integral’s upper limit changes entirely.

The limit itself, (1−sinx)/(π−2x)² as x→(π/2)⁺, is a classic disguised 0/0 form: substituting x=π/2+h turns 1−sinx into 1−cos(h) (via sin(π/2+h)=cos h), which is a standard small-angle expansion ≈h²/2, while (π−2x)²=(−2h)²=4h² exactly — no approximation needed on that side. The ratio of the two collapses cleanly to 1/8, independent of any further limiting process.

Core Concept

Continuity at a branch point fixes unknown constants. For a piecewise function to be continuous where the pieces meet, the limit of the “other side” formula must equal the defined value at that point — this is routinely used to solve for one or more unknown parameters in the formula.

Absolute value inside a definite integral. |g(x)| requires locating the sign changes of g(x) within the interval of integration (via its roots) and splitting the integral there, flipping the sign of g(x) on whichever sub-interval it is negative.

Step-by-Step Solution

1. Set up the continuity condition. lim_x→(π/2)+ f(x) must equal f(π/2)=1/3.

2. Evaluate the limit with x=π/2+h, h→0+:

   1 – sin(π/2+h) = 1 – cos(h) ≈ h²/2 (for small h)
   (π – 2(π/2+h))² = (-2h)² = 4h²
   limit = b(h²/2)/(4h²) = b/8

3. Solve for b: b/8 = 1/3 ⇒ b = 8/3.

4. Compute the integral’s upper limit: 3b−6 = 3(8/3)−6 = 8−6 = 2.

5. Factor the integrand: x²+2x−3 = (x+3)(x−1). On [0,2], this is negative for x∈[0,1) and positive for x∈(1,2].

6. Split and integrate:

   ∫₀¹ -(x²+2x-3)dx = -[x³/3+x²-3x]₀¹ = -[(1/3+1-3)] = 5/3
   ∫₁² (x²+2x-3)dx = [x³/3+x²-3x]₁² = (8/3+4-6)-(1/3+1-3) = 7/3
   Total = 5/3 + 7/3 = 4

Final Answer

The value of the integral is exactly 4. In the exam, when a definite integral’s limit depends on a constant defined earlier in the question, always solve for that constant completely and simplify the limit numerically before setting up the integral — an error there corrupts everything downstream.

Key Facts to Remember

• sin(π/2+h) = cos(h), useful for limits near x=π/2.
• 1−cos(h) ≈ h²/2 for small h (standard small-angle expansion).
• Always factor the expression inside |·| to find its sign changes before splitting a definite integral.
• x²+2x−3 = (x+3)(x−1), negative between the roots, positive outside.

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Prerequisites

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

• Continuity condition lim(x→c) f(x) = f(c) for piecewise functions.
• Small-angle approximations, especially 1−cos(h) ≈ h²/2.
• Integrating |g(x)| by factoring and splitting at sign changes.

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

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