Question

Let \( f: \mathbf{R} \rightarrow \mathbf{R} \) be a thrice differentiable odd function satisfying \( f'(x) \ge 0 \), \( f''(x)=f(x) \), \( f(0)=0 \), \( f'(0)=3 \). Then \( 9f(\ln 3) \) equals:

• A. \(24\)
• B. \(30\)
• C. \(36\)
• D. \(42\)

Hint:

For \(f''=f\), always convert to hyperbolic form; odd/even properties immediately eliminate half the constants.

Answer 1

The Correct Option is C

Concept: The differential equation \(f''(x)=f(x)\) is a standard second-order linear homogeneous differential equation with constant coefficients. Its general solution is: \[ f(x)=Ae^x+Be^{-x} \] Also, the function is given to be odd, so: \[ f(-x)=-f(x) \] This condition strongly restricts the form of the solution.

Step 1: Solve the differential equation. The auxiliary equation is: \[ m^2-1=0 \Rightarrow m=\pm 1 \] So, \[ f(x)=Ae^x+Be^{-x} \]

Step 2: Use odd function property. \[ f(-x)=Ae^{-x}+Be^{x} \] Odd condition: \[ Ae^{-x}+Be^{x}=-(Ae^x+Be^{-x}) \] Comparing coefficients of \(e^x\) and \(e^{-x}\): \[ A=-B \] So, \[ f(x)=A(e^x-e^{-x}) \] \[ f(x)=2A\sinh x \]

Step 3: Use initial condition \(f'(0)=3\). \[ f(x)=2A\sinh x \] \[ f'(x)=2A\cosh x \] At \(x=0\): \[ f'(0)=2A\cdot 1=2A=3 \] \[ A=\frac{3}{2} \] Thus, \[ f(x)=3\sinh x \]

Step 4: Evaluate \(f(\ln 3)\). \[ f(\ln 3)=3\sinh(\ln 3) \] Using identity: \[ \sinh t=\frac{e^t-e^{-t}}{2} \] So, \[ \sinh(\ln 3)=\frac{3-\frac{1}{3}}{2} =\frac{\frac{8}{3}}{2} =\frac{4}{3} \] \[ f(\ln 3)=3 \cdot \frac{4}{3}=4 \]

Step 5: Final required value. \[ 9f(\ln 3)=9\times 4=36 \] \[ \boxed{36} \]