Question

Let $f\left(x\right)=e^{ax}+e^{bx},$ where $a\ne b$ and $f^{\prime \prime }\left(x\right)-2f^{\prime }\left(x\right)-15f\left(x\right)=0$ for all $x\in R$. Then the product $ab$ is equal to

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

Answer 1

The correct option is C $-15$

$f\left(x\right)=e^{ax}+e^{bx}$
On differentiating with respect to $x$,
$f^{\prime }\left(x\right)=a{e}^{ax}+b{e}^{bx}$
Again differentiating with respect to $x$,
$f^{\prime \prime }\left(x\right)=a^2\cdot e^{ax}+b^2\cdot e^{bx}$

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