If `f(x)=cot^-1 x ; R^+ -> (0,pi/2) and g(x)=2x-x^2 ; R-> R` . Then the range of the function `f(g(x)) ` where verdefined is

Options

A. `(0,(pi)/2)`
B. `(0,(pi)/4)`
C. `[(pi)/4,(pi)/2)`
D. `{(pi)/4}`

Explanation

To find the range of the function \( f(g(x)) \) where \( f(x) = \cot^{-1}(x) \) and \( g(x) = 2x - x^2 \), we will follow these steps:

### Step 1: Determine the range of \( g(x) \)

The function \( g(x) = 2x - x^2 \) is a quadratic function. We can rewrite it in standard form:

\[ g(x) = -x^2 + 2x \]

This is a downward-opening parabola. To find the maximum value, we can use the vertex formula \( x = -\frac{b}{2a} \):

Here, \( a = -1 \) and \( b = 2 \):

\[ x = -\frac{2}{2 \cdot -1} = 1 \]

Now, we can find \( g(1) \):

\[ g(1) = 2(1) - (1)^2 = 2 - 1 = 1 \]

Next, we find the x-intercepts by setting \( g(x) = 0 \):

\[ 2x - x^2 = 0 \implies x(2 - x) = 0 \]

This gives us \( x = 0 \) and \( x = 2 \). Therefore, the function \( g(x) \) reaches its maximum value of 1 at \( x = 1 \) and has x-intercepts at \( x = 0 \) and \( x = 2 \). The range of \( g(x) \) is:

\[ [0, 1] \]

### Step 2: Find the range of \( f(g(x)) \)

Now, we need to find the range of \( f(g(x)) = \cot^{-1}(g(x)) \). Since \( g(x) \) takes values from 0 to 1, we will evaluate \( f(g(x)) \) at these endpoints.

1. **At \( g(x) = 0 \)**:
\[ f(0) = \cot^{-1}(0) = \frac{\pi}{2} \]

2. **At \( g(x) = 1 \)**:
\[ f(1) = \cot^{-1}(1) = \frac{\pi}{4} \]

Since \( f(x) = \cot^{-1}(x) \) is a decreasing function, the range of \( f(g(x)) \) as \( g(x) \) varies from 0 to 1 will be from \( f(1) \) to \( f(0) \):

\[ \text{Range of } f(g(x)) = \left[ \frac{\pi}{4}, \frac{\pi}{2} \right] \]

### Final Answer:

The range of the function \( f(g(x)) \) is:

\[ \left[ \frac{\pi}{4}, \frac{\pi}{2} \right] \]