If `I=int(dx)/(x^(2)-2x+5)=(1)/(2)tan^(-1)(f(x))+C` (where, C is the constant of integration) and `f(2)=(1)/(2)`, then the maximum value of `y=f(sinx)AA x in R` is

Options

A. 4
B. 2
C. 0
D. `-1`

Answer

To solve the problem step by step, we will analyze the given integral and the function \( f(x) \) to find the maximum value of \( y = f(\sin x) \).

### Step 1: Analyze the Integral
We start with the integral:
\[ I = \int \frac{dx}{x^2 - 2x + 5} \]
We can rewrite the denominator:
\[ x^2 - 2x + 5 = (x^2 - 2x + 1) + 4 = (x - 1)^2 + 4 \]
Thus, we have:
\[ I = \int \frac{dx}{(x - 1)^2 + 2^2} \]

### Step 2: Use Substitution
Let \( t = x - 1 \), then \( dx = dt \). The integral becomes:
\[ I = \int \frac{dt}{t^2 + 2^2} \]
This is a standard integral:
\[ \int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \]
where \( a = 2 \). Therefore, we have:
\[ I = \frac{1}{2} \tan^{-1} \left( \frac{t}{2} \right) + C \]

### Step 3: Substitute Back
Substituting back \( t = x - 1 \):
\[ I = \frac{1}{2} \tan^{-1} \left( \frac{x - 1}{2} \right) + C \]

### Step 4: Define the Function \( f(x) \)
From the integral, we can define:
\[ f(x) = \frac{x - 1}{2} \]

### Step 5: Given Condition
We are given that \( f(2) = \frac{1}{2} \). Let's verify:
\[ f(2) = \frac{2 - 1}{2} = \frac{1}{2} \]
This condition holds true.

### Step 6: Find Maximum Value of \( y = f(\sin x) \)
Now, we need to find the maximum value of:
\[ y = f(\sin x) = \frac{\sin x - 1}{2} \]
The sine function \( \sin x \) varies between -1 and 1. Therefore, we analyze:
\[ \sin x - 1 \text{ varies from } -2 \text{ (when } \sin x = -1) \text{ to } 0 \text{ (when } \sin x = 1) \]
Dividing by 2 gives:
\[ \frac{\sin x - 1}{2} \text{ varies from } -1 \text{ to } 0 \]

### Step 7: Conclusion
Thus, the maximum value of \( y = f(\sin x) \) is:
\[ \boxed{0} \]