Question

Let the functions \(f\) and \(g\) be defined by \(f(x)=3\sin x,\; -\frac{\pi}{2}\le x\le \frac{\pi}{2}\) and \(g(x)=6-3x^2,\; x\in\mathbb{R}\). Then \(f^{-1}(g(x))=\)

• A. \(\sin^{-1}(2 - x^2)\)
• B. \(3\sin^{-1}(6 - 3x^2)\)
• C. \(3\sin^{-1}(2 - x^2)\)
• D. \(\sin^{-1}(6 - 3x^2)\)
• E. \(\frac{1}{3}\sin^{-1}(6 - 3x^2)\)

Hint:

Always simplify the argument inside the inverse function by factoring out constants. Here, dividing \(6 - 3x^2\) by 3 is the crucial simplification step.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We first need to find the inverse function \(f^{-1}(x)\) and then substitute \(g(x)\) into it.

Step 2: Key Formula or Approach:
1. To find the inverse of \(y = f(x)\), solve for \(x\) in terms of \(y\).
2. Then find \(f^{-1}(g(x))\) by replacing the variable in the inverse function with the expression for \(g(x)\).

Step 3: Detailed Explanation:
Find \(f^{-1}(x)\):
Let \(y = 3\sin x\).
\[ \frac{y}{3} = \sin x \]
\[ x = \sin^{-1}\left(\frac{y}{3}\right) \]
So, \(f^{-1}(x) = \sin^{-1}\left(\frac{x}{3}\right)\).
Now, substitute \(g(x) = 6 - 3x^2\) into the inverse function:
\[ f^{-1}(g(x)) = \sin^{-1}\left(\frac{6 - 3x^2}{3}\right) \]
\[ f^{-1}(g(x)) = \sin^{-1}\left(\frac{3(2 - x^2)}{3}\right) \]
\[ f^{-1}(g(x)) = \sin^{-1}(2 - x^2) \]

Step 4: Final Answer:
The result is \(\sin^{-1}(2 - x^2)\).