Question

lf ${\sinh }^{3}x-{\cosh }^{3}x=\frac{k{e}^x-e^{kx}}{1-k}$ then $k=$

• A. $-1$
• B. $0$
• C. $-3$
• D. 2

Answer 1

Answer: (C)

$-3$

We know that,
$e^x=\sinh x+\cosh x$
$\sinh x=\frac{e^x-e^{-x}}{2}$
$\cosh x=\frac{e^x+e^{-x}}{2}$
Let $f\left(x\right)={\sinh }^{3}x-cos{h}^{3}x$
$=(\sinh x-\cosh x)({\sinh }^{2}x+\cos h^{2}x+\sinh x\cosh x)$
$=(-e^{-x})\left(\right(e^x{)}^2-2\sinh x\cosh x+\sinh x\cosh x)$
$=(-e^{-x})(e^{2x}-\sinh x\cosh x)$
$=(-e^{-x})(e^{2x}-\frac{e^{2x}-e^{-2x}}{4})$
$=(-e^{-x})\left(\frac{3e^{2x}+e^{-2x}}{4}\right)$
$f\left(x\right)=\frac{(-3)e^x-e^{(-3)x}}{4}=\frac{(-3)e^x-e^{(-3)x}}{1(-3)}$
$k=-3$