Question ID: #606
Let $f: \mathbb{R} \to \mathbb{R}$ be a thrice differentiable odd function satisfying $f'(x) \ge 0$, $f”(x) = f(x)$, $f(0)=0$, $f'(0)=3$. Then $9f(\log_{e}3)$ is equal to
Solution:
Given $f”(x) = f(x)$. Multiply both sides by $2f'(x)$:
$$ 2f'(x)f”(x) = 2f(x)f'(x) $$
Integrate both sides:
$$ (f'(x))^2 = (f(x))^2 + C $$
Using $f(0)=0$ and $f'(0)=3$:
$$ 9 = 0 + C \Rightarrow C = 9 $$
$$ f'(x) = \sqrt{(f(x))^2 + 9} \quad (\text{Since } f'(x) \ge 0) $$
Solve the differential equation $\frac{dy}{dx} = \sqrt{y^2+9}$:
$$ \int \frac{dy}{\sqrt{y^2+9}} = \int dx $$
$$ \ln|y + \sqrt{y^2+9}| = x + K $$
Using $y(0)=0$:
$$ \ln(3) = K $$
$$ y + \sqrt{y^2+9} = e^{x+\ln 3} = 3e^x $$
At $x = \ln 3$:
$$ y + \sqrt{y^2+9} = 3e^{\ln 3} = 9 $$
$$ \sqrt{y^2+9} = 9 – y $$
Squaring both sides:
$$ y^2 + 9 = 81 – 18y + y^2 $$
$$ 18y = 72 \Rightarrow y = 4 $$
Value required:
$$ 9f(\ln 3) = 9(4) = 36 $$
Ans. (36)
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