JEE Main20268 April 2026Evening ShiftMathematicsContinuity and DifferentiabilityActual
Let f(x) = cases 1 3 , & x /2 b(1- x) ( -2x)^2 , & x > /2 cases . If f is continuous at x= /2 , then the value of ₀^ 3b-6 |x^2+2x-3| ,dx is:
Options
- A5
- B2
- C3
- D4
Correct answer
D. 4
Step-by-step solution
Since f(x) is continuous at x = /2 , the right-hand limit must equal the value of the function at x = /2 . _ x /2^+ f(x) = f( /2) _ x /2^+ b(1- x) ( -2x)^2 = 1 3 Let x = /2 + h . As x /2^+ , h 0^+ . _ h 0^+ b(1- ( /2+h)) ( -2( /2+h))^2 = 1 3 _ h 0^+ b(1- h) (-2h)^2 = 1 3 _ h 0^+ b(1- h) 4h^2 = 1 3 Using the standard limit _ h 0 1- h h^2 = 1 2 , we get: b 4 1 2 = 1 3 b 8 = 1 3 b = 8 3 The upper limit of the integral is 3b - 6 = 3 ( 8 3 ) - 6 = 2 . The integral to evaluate is I = ₀² |x^2+2x-3| ,dx . Factoring the qua