Let a differentiable function \(f\) satisfy the equation \(\int_{0}^{36} f\left(\frac{\mathrm{tx}}{36}\right) \mathrm{dt}=4 \alpha f(\mathrm{x})\). If \(\mathrm{y}=f(\mathrm{x})\) is a standard parabola passing through the points \((2,1)\) and \((-4, \beta)\), Then \(\beta^{\alpha}\) is equal to ______.

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Definite Integrals

Let a differentiable function

f

satisfy the equation

\int_{0}^{36} f (\frac{tx}{36}) dt = 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 ______.

Answer: 64

The substitution

u = \frac{t x}{36}

turns

\int_{0}^{36} f (\frac{t x}{36}) d t

into

\frac{36}{x} \int_{0}^{x} f (u) d u

, so the equation becomes

\int_{0}^{x} f (u) d u = \frac{α x}{9} f (x)

. Differentiating,

f (x) = \frac{α}{9} (f (x) + x f^{'} (x))

, hence

(9 - α) f (x) = α x f^{'} (x)

and

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

. Integrating gives

f (x) = c x^{9/ α - 1}

. For

y = f (x)

to be a standard parabola the exponent must be

2

, so

\frac{9}{α} - 1 = 2

, giving

α = 3

and

f (x) = c x^{2}

. The point

(2, 1)

gives

c = \frac{1}{4}

, so

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

and

β = f (- 4) = 4

. Therefore

β^{α} = 4^{3} = 64

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Integer answer