Question
The range of $$y = \left( {{{\cot }^{ - 1}}x} \right)\left( {{{\cot }^{ - 1}}\left( { - x} \right)} \right){\text{ is}}$$
A.
$$\left( {0,\frac{{{\pi ^2}}}{4}} \right]$$
B.
$$\left( {0,\pi } \right)$$
C.
$$\left( {0, 2\pi } \right]$$
D.
$$\left( {0,1 } \right]$$
Answer :
$$\left( {0,\frac{{{\pi ^2}}}{4}} \right]$$
Solution :
$$\eqalign{
& y = \left( {{{\cot }^{ - 1}}x} \right)\left( {{{\cot }^{ - 1}}\left( { - x} \right)} \right) \cr
& = {\cot ^{ - 1}}\left( x \right)\left( {\pi - {{\cot }^{ - 1}}\left( x \right)} \right) \cr
& {\text{Now, }}{\cot ^{ - 1}}\left( x \right){\text{ and }}\left( {\pi - {{\cot }^{ - 1}}\left( x \right)} \right) > 0 \cr
& {\text{Using A}}{\text{.M}}{\text{.}} \geqslant {\text{G}}{\text{.M}}{\text{., we get}} \cr
& \frac{{{{\cot }^{ - 1}}x + \left( {\pi - {{\cot }^{ - 1}}x} \right)}}{2} \geqslant \sqrt {\left( {{{\cot }^{ - 1}}x} \right)\left( {\pi - {{\cot }^{ - 1}}x} \right)} \cr
& \Rightarrow 0 < {\cot ^{ - 1}}\left( x \right)\left( {\pi - {{\cot }^{ - 1}}\left( x \right)} \right) \leqslant \left( {\frac{{{{\cot }^{ - 1}}x + \left( {\pi - {{\cot }^{ - 1}}x} \right)}}{2}} \right) = \frac{{{\pi ^2}}}{4} \cr
& \Rightarrow 0 < y \leqslant \frac{{{\pi ^2}}}{4} \cr} $$