JEE Main
Mathematics
Application of Derivatives
2026
JEE Main 2026 (21 January Shift 1)
JEE Main Mathematics Question (2026) — Solution
Question
Let f: R R be a twice differentiable function such that the quadratic equation f(x) m ^ 2 -2 f^ (x) m +f^ (x)=0 in m, has two equal roots for every x R . If f(0)=1, f^ (0)=2, and ( , ) is the largest interval in which the function f ( _ e x-x ) is increasing, then + is equal to \_\_\_\_.
Step-by-step solution
For equal roots in f(x)m^2 - 2f'(x)m + f''(x) = 0, discriminant = 0: [f'(x)]^2 = f(x)f''(x) This gives f''(x) f'(x) = f'(x) f(x) Integrating: f'(x) = kf(x), so f(x) = Ae^ kx Using f(0) = 1 and f'(0) = 2: A = 1, k = 2 f(x) = e^ 2x For g(x) = f( x - x): g'(x) = 2e^ 2( x - x) ( 1 x - 1 ) g'(x) > 0 when 1-x x > 0, i.e., 0 ( , ) = (0, 1) + = 1
Practice more on Quantrex App →
Related: Mathematics — Application of Derivatives · All PYQ Banks