Question

The set of values of parameter $a$ so that the equation $({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3=a{\pi }^3$ has a solution.

• A. $[\frac{-1}{32},\frac{7}{8}]$
• B. $[\frac{1}{32},,\frac{9}{8}]$
• C. $[0,\frac{7}{8}]$
• D. $[\frac{1}{32},\frac{7}{8}]$

Answer 1

The correct option is D $[\frac{1}{32},\frac{7}{8}]$

Let $y=({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3=({\sin }^{-1}x+{\cos }^{-1}x)\left[\right({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2-{\sin }^{-1}x.{\cos }^{-1}x]$
$\Rightarrow y=\frac{\pi }{2}\left[\right({\sin }^{1-}x+{\cos }^{-1}x{)}^2-3{\sin }^{-1}x.{\cos }^{-1}x]$
$\Rightarrow y=\frac{\pi }{2}[\frac{{\pi }^2}{4}-3{\sin }^{-1}x(\frac{\pi }{2}-{\sin }^{-1}x)]$
$\Rightarrow y=\frac{\pi }{2}[\frac{{\pi }^2}{16}+3{(\frac{\pi }{4}-{\sin }^{-1}x)}^2]$
Now we know $-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$
Thus Range of $y$ is $[\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8}]$
Thus given equation to have solution $a\in [\frac{1}{32},\frac{7}{8}]$