Let $f(x)=e^{a x}+e^{b x}$, where $a \neq b$. If $f''(x)-2 f'(x)-15 f(x)=0$ for all x, then the product ab is equal to - ______.
Answer & explanation
We have,
$f(x)=e^{a x}+e^{b x}$
$\Rightarrow f'(x)=a e^{a x}+b e^{b x}$ and $f''(x)=a^2 e^{a x}+b^2 e^{b x}$
∴ $f''(x)-2 f'(x)-15 f(x)=0$ for all x
$\Rightarrow \left(a^2-2 a-15\right) e^{a x}+\left(b^2-2 b-15\right) e^{b x}=0$ for all x
$\Rightarrow a^2-2 a-15=0$ and $b^2-2 b-15=0$
$\Rightarrow a=5,-3$ and $b=5,-3$
$\Rightarrow a=5, b=-3$ or $a=-3, b=5 \Rightarrow a b=-15$