CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image