Let \(f: R \rightarrow R\) be a twice differentiable function such that the quadratic equation \(\mathrm{f}(\mathrm{x}) \mathrm{m}^{2}-2 \mathrm{f}^{\prime}(\mathrm{x}) \mathrm{m}+\mathrm{f}^{\prime}(\mathrm{x})=0\) in m , has two equal roots for every \(\mathrm{x} \in \mathrm{R}\). If \(\mathrm{f}(0)=1\), \(\mathrm{f}^{\prime}(0)=2\) and \((\alpha, \beta)\) is the largest interval in which the function \(\mathrm{f}\left(\log _{\mathrm{e}} \mathrm{x}-\mathrm{x}\right)\) is increasing, then \(\alpha+\beta\) is equal to

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Application of Derivatives

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

Answer: 1

Equal roots in

m

require the discriminant to vanish:

(f^{'} (x))^{2} = f (x) f^{''} (x)

, i.e.

\frac{f ^{'} ( x )}{f ( x )} = \frac{f ^{''} ( x )}{f ^{'} ( x )}

. Integrating gives

f (x) = C f^{'} (x)

; with

f (0) = 1

and

f^{'} (0) = 2

C = \frac{1}{2}

, so

f^{'} (x) = 2 f (x)

and

f (x) = e^{2 x}

. Then

g (x) = f (ln x - x) = x^{2} e^{- 2 x}

for

x > 0

, and

g^{'} (x) = 2 x e^{- 2 x} (1 - x)

, which is positive precisely on

(0, 1)

. Hence the largest interval of increase is

(0, 1)

, giving

α + β = 1

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Integer answer