Question

Suppose, $f\left(x\right)=e^{ax}+e^{bx}$, where $a\ne b$ and $f^{\prime \prime }\left(x\right)-2f^{\prime }\left(x\right)-15f\left(x\right)=0$ for all $xϵR$. Then, find ab.

• A. $-15$
• B. $15$
• C. $10$
• D. $-10$

Answer 1

Answer: (A)

$-15$

$f\left(x\right)=e^{ax}+e^{bx}$
$\Rightarrow f^{\prime }\left(x\right)=a{e}^{ax}+b{e}^{bx},f^{\prime \prime }\left(x\right)=a^2{e}^{ax}+b^2{e}^{bx}$
$\therefore f^{\prime \prime }\left(x\right)-2f^{\prime }\left(x\right)-15f\left(x\right)=0$
$\Rightarrow \{a^2{e}^{ax}+b^2{e}^{bx}\}-2\{a{e}^{ax}+b{e}^{bx}\}-15\{e^{ax}+e^{bx}\}=0$, 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 $b^2-2b-15=0$
$\Rightarrow (a-5)(a+3)=0$ and $(b-5)(b+3)=0$
$\Rightarrow a=5$ or -3 and $b=5$ or -3
But $a\ne b$, hence, $a=5,b=-3$
or $a=-3,b=5\Rightarrow ab=-15$