Answered

The graph of an inverse trigonometric function passes through the point (1, pi/2). Which of the following could be the equation of the function?

Ay=cos^-1 x
By=cot^-1 x
Cy=sin^-1 x
Dy=tan^-1 x

Verified Answer :

Answer: C) \(y=sin^-1 x\)

Step-by-step explanation:

Since, the graph of an inverse trigonometric function will pass through the point \((1,\frac{\pi}{2})\),

If this point satisfies the function,

For the function \(y=cos^{-1} x\)

If x = 1

\(y=cos^{-1}1=0\)

Thus,  \((1,\frac{\pi}{2})\) is not satisfying function  \(y=cos^{-1} x\),

⇒ The graph of   \(y=cos^{-1} x\) is not passing through the point  \((1,\frac{\pi}{2})\)

For the function \(y=cot^{-1}x\)

If x = 1

\(y=cot^{-1}1=\frac{\pi}{4}\)

Thus,  \((1,\frac{\pi}{2})\) is not satisfying function \(y=cot^{-1}x\),

⇒ The graph of   \(y=cot^{-1}x\) is not passing through the point  \((1,\frac{\pi}{2})\)

For the function \(y=sin^{-1} x\)

If x = 1

\(y=sin^{-1}1=\frac{\pi}{2}\)

Thus,  \((1,\frac{\pi}{2})\) is satisfying function \(y=sin^{-1} x\),

⇒ The graph of   \(y=sin^{-1} x\) is passing through the point  \((1,\frac{\pi}{2})\).

For the function \(y=tan^{-1}x\)

If x = 1

\(y=tan^{-1}1=\frac{\pi}{4}\)

Thus,  \((1,\frac{\pi}{2})\) is not satisfying function  \(y=cos^{-1} x\),

⇒ The graph of   \(y=tan^{-1} x\) is not passing through the point \((1,\frac{\pi}{2})\).

Hence, Option C is correct.