Let a differentiable function f satisfy the equation ∫_ 0 ^ 36 f((t x)/(36)) d t=4 α f(x) . If y=f(x) is a standard parabola passing through the points (2,1) and (-4, β) , then β^ α is equal to \_\_\_\_ .

Correct answer: 0 From ∫_0^ 36 f((tx)/(36)) dt = 4α f(x) , substitute u = (tx)/(36) to get (36)/(x) ∫_0^x f(u) du = 4α f(x) . Differentiating: f(x) = (α)/(9)[f(x) + xf'(x)] , which simplifies to (9-α)f(x) = α x f'(x) . This gives (f'(x))/(f(x)) = (9-α)/(α x) , integrating to f(x) = Kx^ (9-α)/(α) . For f to be a standard parabola: (9-α)/(α) = 2 α = 3 . Thus f(x) = Kx^2 . Using point (2,1) : K = (1)/(4) , so f(x) = (x^2)/(4) . At (-4, β) : β = (16)/(4) = 4 . Therefore β^α = 4^3 = 64

JEE Main 202624 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Definite Integration

Let a differentiable function f satisfy the equation ∫_ 0 ^ 36 f((t...

Let a differentiable function

f

satisfy the equation

\int_{0}^{36} f\left(\frac{t x}{36}\right) d t=4 \alpha f(x)

. If

y=f(x)

is a standard parabola passing through the points

(2,1)

and

(-4, \beta)

, then

\beta^{\alpha}

is equal to

\_\_\_\_

Correct answer

Step-by-step solution

From

\int_0^{36} f\left(\frac{tx}{36}\right) dt = 4\alpha f(x)

, substitute

u = \frac{tx}{36}

to get

\frac{36}{x} \int_0^x f(u) du = 4\alpha f(x)

.
Differentiating:

f(x) = \frac{\alpha}{9}[f(x) + xf'(x)]

, which simplifies to

(9-\alpha)f(x) = \alpha x f'(x)

.
This gives

\frac{f'(x)}{f(x)} = \frac{9-\alpha}{\alpha x}

, integrating to

f(x) = Kx^{\frac{9-\alpha}{\alpha}}

.
For f to be a standard parabola:

\frac{9-\alpha}{\alpha} = 2 \Rightarrow \alpha = 3

.
Thus

f(x) = Kx^2

. Using point

(2,1)

K = \frac{1}{4}

, so

f(x) = \frac{x^2}{4}

.
At

(-4, \beta)

\beta = \frac{16}{4} = 4

.
Therefore

\beta^\alpha = 4^3 = 64

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