Question

# $$\cos ^{ -1 }{ \left\{ \cfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\cfrac { { x }^{ 2 } }{ 4 } } \right\} } =\cos ^{ -1 }{ \cfrac { x }{ 2 } -\cos ^{ -1 }{ x } }$$ hold for

A
x[1,1]
B
xR
C
x[0,1]
D
x[1,0]

Solution

## The correct option is C $$x\in \left[ 0,1 \right]$$Given , $$\cos ^{ -1 }{ \left\{ \cfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\cfrac { { x }^{ 2 } }{ 4 } } \right\} } =\cos ^{ -1 }{ \cfrac { x }{ 2 } -\cos ^{ -1 }{ x } }$$Since, $$\cfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\cfrac { { x }^{ 2 } }{ 4 } }$$ is positive and defined for all $$x\in \left[ -1,1 \right]$$Now, $$\cos^{-1} {\frac{x}{2}}$$ is defined for $$-1\le \frac{x}{2} \le 1$$$$\Rightarrow -2 \le x \le 2$$Also, $$\cos^{-1} x$$ is defined for $$-1\le x \le 1$$Also, LHS$$\ge 0$$$$\Rightarrow \cos ^{ -1 }{ \cfrac { x }{ 2 } -\cos ^{ -1 }{ x } } \ge 0\Rightarrow \cos ^{ -1 }{ \cfrac { x }{ 2 } \ge \cos ^{ -1 }{ x } }$$$$\Rightarrow x\in \left[ 0,1 \right]$$So, LHS is defined for $$x\in [0,1]$$Mathematics

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