JEE Mains · Maths · STD 12 - 9. differential equations
Let a differentiable function f satisfy the equation \( \int_{0}^{36}f(\frac{tx}{36})dt=4\alpha f(x) \). If \( y=f(x) \) is a standard parabola passing through the points (2, 1) and \( (-4,\beta) \),Then \( \beta^{\alpha} \) is equal to ___ .
- A 16
- B 32
- C 64
- D 128
Answer & Solution
Correct Answer
(C) 64
Step-by-step Solution
Detailed explanation
\(\int_0^{36} f\left(\frac{ tx }{36}\right) dt =4 \alpha f( x )\) Put \(\frac{ tx }{36}= y\) \(\frac{d y}{d t}=\frac{x}{36}\) \(\int_0^{ x } \frac{f( y ) 36 dy }{ x }=4 \alpha f( x )\) \(\int_0^{ x } f( y ) dy =\frac{\alpha f( x ) x }{9}\)…
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