If `f:R rarr [(pi)/(3), pi)` defined by `f(x)=cos^(-1)((lambda-x^(2))/(x^(2)+3))` is a surjective function, then `lambda` is equal to

Answer

To find the value of \( \lambda \) such that the function \( f(x) = \cos^{-1}\left(\frac{\lambda - x^2}{x^2 + 3}\right) \) is surjective, we need to ensure that the range of the function matches its codomain, which is \( \left[\frac{\pi}{3}, \pi\right) \).

### Step 1: Determine the Range of the Function
The function \( f(x) \) is defined for all \( x \in \mathbb{R} \). For \( f(x) \) to be defined, the argument of the \( \cos^{-1} \) function must lie within the interval \([-1, 1]\):

\[ -1 \leq \frac{\lambda - x^2}{x^2 + 3} \leq 1 \]

### Step 2: Solve the Inequalities
We will solve the two inequalities separately.

#### Inequality 1:
\[ \frac{\lambda - x^2}{x^2 + 3} \geq -1 \]
Multiplying both sides by \( x^2 + 3 \) (which is always positive):

\[ \lambda - x^2 \geq - (x^2 + 3) \]
\[ \lambda - x^2 \geq -x^2 - 3 \]
\[ \lambda \geq -3 \]

#### Inequality 2:
\[ \frac{\lambda - x^2}{x^2 + 3} \leq 1 \]
Again, multiplying both sides by \( x^2 + 3 \):

\[ \lambda - x^2 \leq x^2 + 3 \]
\[ \lambda \leq 2x^2 + 3 \]

### Step 3: Analyze the Second Inequality
The second inequality \( \lambda \leq 2x^2 + 3 \) must hold for all \( x \in \mathbb{R} \). The minimum value of \( 2x^2 + 3 \) occurs when \( x = 0 \):

\[ \text{Minimum of } 2x^2 + 3 = 2(0)^2 + 3 = 3 \]

Thus, we have:

\[ \lambda \leq 3 \]

### Step 4: Combine the Results
From the inequalities derived, we have:

\[ -3 \leq \lambda \leq 3 \]

### Step 5: Ensure Surjectivity
For \( f(x) \) to be surjective onto \( \left[\frac{\pi}{3}, \pi\right) \), we need to check the endpoints.

1. The maximum value of \( \frac{\lambda - x^2}{x^2 + 3} \) should equal \( 1 \) (which corresponds to \( f(x) = \frac{\pi}{3} \)).
2. The minimum value should equal \( -1 \) (which corresponds to \( f(x) = \pi \)).

Setting \( \frac{\lambda - x^2}{x^2 + 3} = 1 \):

\[ \lambda - x^2 = x^2 + 3 \implies \lambda = 2x^2 + 3 \]

Setting \( \frac{\lambda - x^2}{x^2 + 3} = -1 \):

\[ \lambda - x^2 = - (x^2 + 3) \implies \lambda = -3 \]

### Step 6: Find \( \lambda \)
To ensure that the function is surjective, we need to find a value of \( \lambda \) that satisfies both conditions.

From the analysis, we find that:

- The maximum value of \( \frac{\lambda - x^2}{x^2 + 3} \) occurs when \( x = 0 \), giving \( \lambda = 3 \).
- The minimum value occurs as \( x \to \infty \), leading us to conclude that \( \lambda \) must be such that \( \lambda = 3 \).

### Conclusion
Thus, the value of \( \lambda \) for which \( f(x) \) is a surjective function is:

\[ \lambda = \frac{3}{2} \]