Suppose `f(x)=e^(ax)+e^(bx)," where " a ne b,` and that f''(x) -2f'(x)-15f(x)=0 for all x. Then the product ab is

Options

A. 25
B. 9
C. -15
D. -9

Answer

C

Explanation

We have
`f(x)=e^(ax)+b^(bx)`
`Rightarrow f'(x)=a e^(ax)+b e^(bx) and f''(x)=a^(2) e^(ax)+b^(2) e^(bx)`
`therefore f''(x)-2f'(x)-15f(x)=0"for all x"`
`Rightarrow (a^(2)-2a-15)e^(ax)+(b^(2)-2b-15)e^(bx)=0"for all x"`
`Rightarrow a^(2)-2a-15=0 and b^(2)-2b-15=0`
`a=5,-3 and b=5,-3`
`Rightarrow a=5,b=-3 or a=-3, b=5 Rightarrow ab=15 `