Let f: R arrow 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 (α, β) is the largest interval in

Correct answer: 0 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 (α, β) = (0, 1) α + β = 1

JEE Main 202621 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Application of Derivatives

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Application of Derivatives

Let f: R arrow R be a twice differentiable function such that the...

Let

f: \mathbf{R} \rightarrow \mathbf{R} be a twice differentiable function such that the quadratic equation

f(x) \mathrm{m}^{2}-2 f^{\prime}(x) \mathrm{m}+f^{\prime \prime}(x)=0

in m, has two equal roots for every

x \in \mathbf{R}

. If

f(0)=1, f^{\prime}(0)=2

, and

(\alpha, \beta)

is the largest interval in which the function

\left(\log _{\mathrm{e}} x-x\right)

is increasing, then

\alpha+\beta

is equal to

\_\_\_\_

Correct answer

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

\frac{f''(x)}{f'(x)} = \frac{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(\ln x - x)

g'(x) = 2e^{2(\ln x - x)} \cdot \left(\frac{1}{x} - 1\right)

g'(x) > 0

when

\frac{1-x}{x} > 0

, i.e.,

0 < x < 1

(\alpha, \beta) = (0, 1)

\alpha + \beta = 1

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