JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let $f : R \to R$ 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 $x \in R$ . If $f (0) = 1, f^{'} (0) = 2$ and $(α, β)$ is the largest interval in which the function $f (lo g_{e} x - x)$ is increasing, then $α + β$ is equal to

Your answer

Solution

Equal roots: discriminant zero, so $(f^{'} (x))^{2} = f (x) f^{''} (x)$ . This is equivalent to $\frac{d}{d x} (\frac{f ^{'} ( x )}{f ( x )}) = 0$ , so $\frac{f ^{'} ( x )}{f ( x )} =$ constant $= k$ .

From $f (0) = 1, f^{'} (0) = 2$ : $k = 2$ . Hence $f (x) = e^{2 x}$ .

Let $g (x) = f (ln x - x) = e^{2 (l n x - x)}$ for $x > 0$ . Then $g^{'} (x) = 2 g (x) (\frac{1}{x} - 1)$ , which is $\geq 0$ when $\frac{1 - x}{x} \geq 0$ , i.e. $x \in (0, 1]$ .

So $α = 0$ , $β = 1$ , and $α + β = 1$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →