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

Options

A. 0
B. 3
C. 2
D. 1

Answer

To solve the problem, we need to determine the value of \( \lambda \) such that the function \( f(x) = \cos^{-1}\left(\frac{\lambda - x^2}{x^2 + 2}\right) \) is surjective from \( \mathbb{R} \) to \( \left[\frac{\pi}{3}, \pi\right) \).

### Step 1: Understanding the Function
The function \( f(x) \) is defined as:
\[ f(x) = \cos^{-1}\left(\frac{\lambda - x^2}{x^2 + 2}\right) \]
For \( f(x) \) to be well-defined, the argument of the \( \cos^{-1} \) function must lie in the interval \([-1, 1]\).

### Step 2: Setting the Range of the Argument
We need to find the values of \( x \) for which:
\[ -1 \leq \frac{\lambda - x^2}{x^2 + 2} \leq 1 \]

### Step 3: Solving the Inequalities
#### Inequality 1:
\[ \frac{\lambda - x^2}{x^2 + 2} \geq -1 \]
Cross-multiplying gives:
\[ \lambda - x^2 \geq - (x^2 + 2) \implies \lambda - x^2 \geq -x^2 - 2 \implies \lambda + 2 \geq 0 \implies \lambda \geq -2 \]

#### Inequality 2:
\[ \frac{\lambda - x^2}{x^2 + 2} \leq 1 \]
Cross-multiplying gives:
\[ \lambda - x^2 \leq x^2 + 2 \implies \lambda \leq 2x^2 + 2 \]

### Step 4: Finding the Range of \( \lambda \)
From the second inequality, we can express \( \lambda \) in terms of \( x^2 \):
\[ \lambda \leq 2x^2 + 2 \]
As \( x \) varies over all real numbers, \( x^2 \) can take any non-negative value. Therefore, the maximum value of \( \lambda \) occurs when \( x^2 \) is maximized.

### Step 5: Considering the Surjectivity Condition
For \( f \) to be surjective onto \( \left[\frac{\pi}{3}, \pi\right) \), we need to ensure that the range of \( f(x) \) covers the entire interval \( \left[\frac{\pi}{3}, \pi\right) \).

The maximum value of \( \frac{\lambda - x^2}{x^2 + 2} \) occurs when \( x^2 \) is minimized. The minimum value of \( x^2 \) is \( 0 \):
\[ \frac{\lambda - 0}{0 + 2} = \frac{\lambda}{2} \]
This must equal \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ \frac{\lambda}{2} = \frac{1}{2} \implies \lambda = 1 \]

### Step 6: Conclusion
Thus, the value of \( \lambda \) that makes \( f(x) \) a surjective function from \( \mathbb{R} \) to \( \left[\frac{\pi}{3}, \pi\right) \) is:
\[ \boxed{1} \]