JEE Main Mathematics Question (2026) — Solution

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 = _ 0 ^ 2 |x^2+2x-3|\,dx. Factoring the quadratic expression gives x^2+2x-3 = (x+3)(x-1). For x [0, 1], (x+3)(x-1) 0, so |x^2+2x-3| = 3 - 2x - x^2. For x [1, 2], (x+3)(x-1) 0, so |x^2+2x-3| = x^2+2x-3. Splitting the integral at x = 1: I = _ 0 ^ 1 (3 - 2x - x^2)\,dx + _ 1 ^ 2 (x^2+2x-3)\,dx Evaluating the first integral: _ 0 ^ 1 (3 - 2x - x^2)\,dx = [ 3x - x^2 - x^3 3 ]_0^1 = 3 - 1 - 1 3 = 5 3 Evaluating the second integral: _ 1 ^ 2 (x^2+2x-3)\,dx = [ x^3 3 + x^2 - 3x ]_1^2 = ( 8 3 + 4 - 6 ) - ( 1 3 + 1 - 3 ) = ( 8 3 - 2 ) - ( 1 3 - 2 ) = 7 3 Adding the two parts: I = 5 3 + 7 3 = 12 3 = 4 Answer: 4