Inverse Trigonometric Functions Question 212
Question: The greatest and the least value of $ {{({{\sin }^{-1}}x)}^{3}}+{{({{\cos }^{-1}}x)}^{3}} $ are
Options:
A) $ -\frac{\pi }{2},\frac{\pi }{2} $
B) $ -\frac{{{\pi }^{3}}}{8},\frac{{{\pi }^{3}}}{8} $
C) $ \frac{7{{\pi }^{3}}}{8},\frac{{{\pi }^{3}}}{32} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ {{({{\sin }^{-1}}x)}^{3}}+{{({{\cos }^{-1}}x)}^{3}} $
$ ={{({{\sin }^{-1}}x+{{\cos }^{-1}}x)}^{3}} $
$ -3{{\sin }^{-1}}x{{\cos }^{-1}}x({{\sin }^{-1}}x+{{\cos }^{-1}}x) $
= $ \frac{{{\pi }^{3}}}{8}-3({{\sin }^{-1}}x{{\cos }^{-1}}x)\frac{\pi }{2} $
= $ \frac{{{\pi }^{3}}}{8}-\frac{3\pi }{2}{{\sin }^{-1}}x( \frac{\pi }{2}-{{\sin }^{-1}}x ) $
= $ \frac{{{\pi }^{3}}}{8}-\frac{3{{\pi }^{2}}}{4}{{\sin }^{-1}}x+\frac{3\pi }{2}{{({{\sin }^{-1}}x)}^{2}} $
= $ \frac{{{\pi }^{3}}}{8}+\frac{3\pi }{2}[ {{({{\sin }^{-1}}x)}^{2}}-\frac{\pi }{2}{{\sin }^{-1}}x ] $
$ =\frac{{{\pi }^{3}}}{8}+\frac{3\pi }{2}[ {{( {{\sin }^{-1}}x-\frac{\pi }{4} )}^{2}} ]-\frac{3{{\pi }^{3}}}{32} $
$ =\frac{{{\pi }^{3}}}{32}+\frac{3\pi }{2}{{( {{\sin }^{-1}}x-\frac{\pi }{4} )}^{2}} $
The least value is $ \frac{{{\pi }^{3}}}{32} $
and since $ {{( {{\sin }^{-1}}x-\frac{\pi }{4} )}^{2}}\le {{( \frac{3\pi }{4} )}^{2}} $
The greatest value is $ \frac{{{\pi }^{3}}}{32}+\frac{9{{\pi }^{2}}}{16}\times \frac{3\pi }{2}=\frac{7{{\pi }^{3}}}{8} $ .
BETA
