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

Answer

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

### Step 1: Understand the Range of the Function
The function \(f(x)\) is defined using the inverse cosine function. For \(f(x)\) to be defined, the argument of the cosine inverse must lie within the range \([-1, 1]\):
\[ -1 \leq \frac{\lambda - x^2}{x^2 + 2} \leq 1 \]

### Step 2: Set Up the Inequalities
We can break this down into two inequalities:
1. \(\frac{\lambda - x^2}{x^2 + 2} \geq -1\)
2. \(\frac{\lambda - x^2}{x^2 + 2} \leq 1\)

#### For the first inequality:
\[ \lambda - x^2 \geq - (x^2 + 2) \]
This simplifies to:
\[ \lambda \geq -2 \]

#### For the second inequality:
\[ \lambda - x^2 \leq x^2 + 2 \]
This simplifies to:
\[ \lambda \leq 2x^2 + 2 \]

### Step 3: Analyze the Range of \(x\)
Since \(x\) can take any real value, we need to ensure that \(\lambda\) can accommodate all possible values of \(x^2\). The term \(2x^2 + 2\) can take any value greater than or equal to \(2\) as \(x^2\) ranges from \(0\) to \(\infty\).

### Step 4: Determine the Conditions for Surjectivity
For \(f(x)\) to be surjective onto \(\left[\frac{\pi}{3}, \pi\right)\), the range of the function must match this interval. We need to find the values of \(y\) such that:
\[ \frac{\lambda - x^2}{x^2 + 2} = \cos(y) \]
where \(y\) ranges from \(\frac{\pi}{3}\) to \(\pi\).

### Step 5: Find the Corresponding Values of \(\cos(y)\)
- At \(y = \frac{\pi}{3}\), \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).
- At \(y = \pi\), \(\cos(\pi) = -1\).

Thus, we have:
\[ \text{Range of } \cos(y) = [-1, \frac{1}{2}] \]

### Step 6: Equate the Ranges
From the inequalities, we have:
\[ -1 \leq \frac{\lambda - x^2}{x^2 + 2} \leq \frac{1}{2} \]
This gives us the conditions:
1. \(\lambda - x^2 \geq - (x^2 + 2) \Rightarrow \lambda \geq -2\)
2. \(\lambda - x^2 \leq x^2 + 2 \Rightarrow \lambda \leq 2\)

### Step 7: Find the Value of \(\lambda\)
To ensure that the maximum value of \(\frac{\lambda - x^2}{x^2 + 2}\) reaches \(\frac{1}{2}\):
\[ \lambda - 2 \leq 2 \Rightarrow \lambda \leq 2 \]
To ensure that the minimum value reaches \(-1\):
\[ \lambda + 2 \geq 0 \Rightarrow \lambda \geq -2 \]

### Conclusion
The only value of \(\lambda\) that satisfies both conditions and allows the function to be surjective is:
\[ \lambda = 1 \]

### Final Answer
Thus, the value of \(\lambda\) is \(1\).