Skip to main content

Let f: R → R be a twice differentiable function such that the quadratic equation f(x) m^2-2 f^prime(x) m+f^prime prime(x)=0 in m, has two equal roots…

JEE Main 2026 — Mathematics Calculus

2026integerhard

Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R be a twice differentiable function such that the quadratic equation f(x)m2−2f′(x)m+f′′(x)=0f(x) \mathrm{m}^{2}-2 f^{\prime}(x) \mathrm{m}+f^{\prime \prime}(x)=0f(x)m2−2f′(x)m+f′′(x)=0 in m, has two equal roots for every x∈Rx \in \mathbf{R}x∈R. If f(0)=1,f′(0)=2f(0)=1, f^{\prime}(0)=2f(0)=1,f′(0)=2, and (α,β)(\alpha, \beta)(α,β) is the largest interval in which the function f(log⁡ex−x)f\left(\log _{\mathrm{e}} x-x\right)f(loge​x−x) is increasing, then α+β\alpha+\betaα+β is equal to ____\_\_\_\_____.

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Calculus in other years

More JEE Main Calculus questions

Other Mathematics topics for JEE Main

Keep practicing Calculus

Work through every JEE Main Calculus PYQ, year by year.

Browse all Mathematics topics