If `f:R rarr [pi/6,pi/2], f(x)=sin^(-1)((x^(2)-a)/(x^(2)+1))` is an onto function, the set of values `a` is

Options

A. `{-1/2}`
B. `[-1/2,-1)`
C. `(-1, infty)`
D. None of these

Answer

To determine the set of values for \( a \) such that the function \( f(x) = \sin^{-1}\left(\frac{x^2 - a}{x^2 + 1}\right) \) is onto from \( \mathbb{R} \) to \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \), we will follow these steps:

### Step 1: Understand the Range of \( f(x) \)
Since \( f(x) \) is defined to be onto, we need to ensure that the output of \( f(x) \) covers the entire interval \( \left[\frac{\pi}{6}, \frac{\pi}{2}\right] \).

### Step 2: Set Up the Inequalities
We know that:
\[ \frac{\pi}{6} \leq f(x) \leq \frac{\pi}{2} \]
This translates to:
\[ \frac{\pi}{6} \leq \sin^{-1}\left(\frac{x^2 - a}{x^2 + 1}\right) \leq \frac{\pi}{2} \]

### Step 3: Apply the Sine Function
Taking the sine of all parts of the inequality, we get:
\[ \sin\left(\frac{\pi}{6}\right) \leq \frac{x^2 - a}{x^2 + 1} \leq \sin\left(\frac{\pi}{2}\right) \]
This simplifies to:
\[ \frac{1}{2} \leq \frac{x^2 - a}{x^2 + 1} \leq 1 \]

### Step 4: Solve the First Inequality
Starting with the left inequality:
\[ \frac{1}{2} \leq \frac{x^2 - a}{x^2 + 1} \]
Multiplying through by \( 2(x^2 + 1) \) (which is positive for all \( x \)):
\[ 1(x^2 + 1) \leq 2(x^2 - a) \]
This simplifies to:
\[ x^2 + 1 \leq 2x^2 - 2a \]
Rearranging gives:
\[ 0 \leq x^2 - 2a - 1 \]
This implies:
\[ x^2 \geq 2a + 1 \]

### Step 5: Solve the Second Inequality
Now, we solve the right inequality:
\[ \frac{x^2 - a}{x^2 + 1} \leq 1 \]
Again, multiplying through by \( x^2 + 1 \):
\[ x^2 - a \leq x^2 + 1 \]
This simplifies to:
\[ -a \leq 1 \quad \Rightarrow \quad a \geq -1 \]

### Step 6: Combine the Results
From the first inequality, we have \( x^2 \geq 2a + 1 \) which must hold for all \( x \in \mathbb{R} \). The minimum value of \( x^2 \) is \( 0 \), hence:
\[ 0 \geq 2a + 1 \quad \Rightarrow \quad 2a \leq -1 \quad \Rightarrow \quad a \leq -\frac{1}{2} \]

### Step 7: Final Result
Combining both results, we find:
\[ -1 \leq a \leq -\frac{1}{2} \]
Thus, the set of values for \( a \) is:
\[ a \in \left[-1, -\frac{1}{2}\right] \]