The domain of the function defined by \(f(x)=\sin ^{-1} \sqrt{x-1}\) is

(a)

[1, 2]

(b)

[-1, 1]

(c)

[0, 1]

(d)

[-1, 0]

If \(x = \frac{1}{5}\), the value of cos (cos^-1x+2sin^-1x) is

(a)

\(-\sqrt { \frac { 24 }{ 25 } } \)

(b)

\(\sqrt { \frac { 24 }{ 25 } } \)

(c)

\(\frac{1}{5}\)

(d)

\(-\frac{1}{5}\)

\(\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\) is equal to

(a)

\(\frac { 1 }{ 2 } \ { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)

(b)

\(\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)

(c)

\(\frac { 1 }{ 2 } {tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)

(d)

\({ tan}^{ -1 }\left( \frac { 1}{ 2 } \right) \)

\(\sin ^{-1}\left(\tan \frac{\pi}{4}\right)-\sin ^{-1}\left(\sqrt{\frac{3}{x}}\right)=\frac{\pi}{6}\). Then x is a root of the equation

(a)

x²−x−6 = 0

(b)

x²−x−12 = 0

(c)

x²+x−12 = 0

(d)

x²+x−6 = 0

sin (tan^-1x), |x| < 1 is equal to

(a)

\(\frac{x}{\sqrt{1-x^2}}\)

(b)

\(\frac{1}{\sqrt{1-x^2}}\)

(c)

\(\frac{1}{\sqrt{1+x^2}}\)

(d)

\(\frac{x}{\sqrt{1+x^2}}\)