Question

The set of values of $$x$$ for which the identity $${\cos ^{ - 1}}x + {\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right) = \frac{\pi }{3}$$         holds good is

A. $$\left[ {0,1} \right]$$
B. $$\left[ {0,\frac{1}{2}} \right]$$
C. $$\left[ {\frac{1}{2} , 1} \right]$$  
D. $$\left\{ { - 1,0,1} \right\}$$
Answer :   $$\left[ {\frac{1}{2} , 1} \right]$$
Solution :
$$\eqalign{ & {\bf{Case 1 :}} \cr & {\text{If }}0 \leqslant x \leqslant \frac{1}{2},\,\,{\text{then }}{\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right) \cr & {\cos ^{ - 1}}\left( {x \times \frac{1}{2} + \sqrt {1 - {x^2}} \frac{{\sqrt 3 }}{2}} \right) \cr & = {\cos ^{ - 1}}x - {\cos ^{ - 1}}\frac{1}{2} \cr} $$
Therefore, the equation is
$$\eqalign{ & {\cos ^{ - 1}}x + {\cos ^{ - 1}}x - {\cos ^{ - 1}}\frac{1}{2} = \frac{\pi }{3} \cr & \Rightarrow x = \frac{1}{2}. \cr & {\bf{Case 2 :}} \cr & {\text{If }}\frac{1}{2} \leqslant x \leqslant 1,{\text{ then}} \cr & {\cos ^{ - 1}}\left( {\frac{x}{2} + \frac{1}{2}\sqrt {3 - 3{x^2}} } \right) = {\cos ^{ - 1}}\frac{1}{2} - {\cos ^{ - 1}}x \cr} $$
Therefore, the equation is
$${\cos ^{ - 1}}x + {\cos ^{ - 1}}\frac{1}{2} - {\cos ^{ - 1}}x = \frac{\pi }{3},$$
which is an identity.
Hence, the identity holds good for $$x \in \left[ {\frac{1}{2},1} \right].$$

Releted MCQ Question on
Trigonometry >> Inverse Trigonometry Function

Releted Question 1

The value of $$\tan \left[ {{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) + {{\tan }^{ - 1}}\left( {\frac{2}{3}} \right)} \right]$$      is

A. $$\frac{6}{{17}}$$
B. $$\frac{7}{{16}}$$
C. $$\frac{16}{{7}}$$
D. none
Releted Question 2

If we consider only the principle values of the inverse trigonometric functions then the value of $$\tan \left( {{{\cos }^{ - 1}}\frac{1}{{5\sqrt 2 }} - {{\sin }^{ - 1}}\frac{4}{{\sqrt {17} }}} \right)$$      is

A. $$\frac{{\sqrt {29} }}{3}$$
B. $$\frac{{29}}{3}$$
C. $$\frac{{\sqrt {3}}}{29}$$
D. $$\frac{{3}}{29}$$
Releted Question 3

The number of real solutions of $${\tan ^{ - 1}}\sqrt {x\left( {x + 1} \right)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$$         is

A. zero
B. one
C. two
D. infinite
Releted Question 4

If $${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - .....} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - .....} \right) = \frac{\pi }{2}$$             for $$0 < \left| x \right| < \sqrt 2 ,$$   then $$x$$ equals

A. $$ \frac{1}{2}$$
B. 1
C. $$ - \frac{1}{2}$$
D. $$- 1$$

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Inverse Trigonometry Function


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