Question

lf $f\left(x\right)=coshx+$ sinhx and $f\left(x\right)f\left(y\right)=f\left(k\right)$ then $k=$

• A. $xy$
• B. $\frac{x}{y}$
• C. $x-y$
• D. $x+y$

Answer 1

Answer: (D)

$x+y$

Expansion of $coshx=1+\frac{x^2}{2!}+\frac{x^4}{4!}+......$
Expansion of $sinhx=x+\frac{x^3}{3!}+\frac{x^5}{5!}+......$
$f\left(x\right)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+........+\frac{x^n}{n!}=e^x$ [$\because$ known expansion of $e^x$]
$\therefore f\left(y\right)=1+y+\frac{y^2}{2!}+.........+\frac{y^n}{n!}=e^y$
$\therefore f\left(x\right)f\left(y\right)=f\left(k\right)$
$e^{x+y}=f\left(k\right)$
$\therefore k=x+y$