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MathsJEE Main 24 Jan 2026, Shift 1·Q21·Numerical·Tough

Differential Equations · Variable separable differential equations

Let a differentiable function ff satisfy the equation ∫036f(tx36)dt=4αf(x).\int_0^{36} f\left(\dfrac{tx}{36}\right)dt = 4\alpha f(x). If y=f(x)y = f(x) is a standard parabola passing through the points (2,1)(2, 1) and (−4,β)(-4, \beta), Then βα\beta^{\alpha} is equal to .

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