Question

The maximum value of ${\sin }^{3}x+{\cos }^{3}x$ is-

• A. $1$
• B. $2$
• C. $3/2$
• D. none of these

Answer 1

The correct option is A $1$

$f\left(x\right)={sin}^{3}x+{cos}^{3}x$
$f^{{}^{\prime }}\left(x\right)=3{sin}^{2}xcosx-3{cos}^{2}xsinx=0$
$3sinxcosx(sinx-cosx)=0$
It can also be written as
$\left(3/2\right)sin2x(sinx-cosx)=0$
$sin2x=0$ or $sinx-cosx=0$
$x=\pi /4$
Check for second derivative
$3(2sinx{cos}^{2}x+{sin}^{2}x(-sinx\left)\right)-3({cos}^{2}xcosx+2cosx(-sinx\left)sinx\right)$
$(2sinx{cos}^{2}x-{sin}^{3}x-{cos}^{3}x+2{sin}^{2}xcosx)$
Put$sinx=1/\sqrt{2}$, $cosx=1/\sqrt{2}$
$=\sqrt{2}-1/\sqrt{2}>0$
Therefore $sin2x=0$will give maximum
$2x=n\pi$
$x=n\pi /2$
Take value of n=1
$x=\pi /2$
$f\left(x\right)={sin}^3\left(\pi /2\right)+{cos}^3\left(\pi /2\right)=1$