If `f'(x^2-4x+3)gt0 " for all " x in (2,3) `then f(sinx) is increasing on

Options

A. `underset (n in Z)cup (" 2n " pi, (4n+1) (pi)/2)`
B. `underset (n in Z)cup ((4n-1)(pi)/2, " 2n " pi)`
C. R
D. none of these

Explanation

To solve the problem, we need to determine the intervals on which the function \( f(\sin x) \) is increasing, given that \( f'(x^2 - 4x + 3) > 0 \) for all \( x \) in the interval \( (2, 3) \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
We have the condition \( f'(x^2 - 4x + 3) > 0 \) for \( x \in (2, 3) \). This means that the derivative of the function \( f \) is positive for the values of \( x^2 - 4x + 3 \) when \( x \) is in the interval \( (2, 3) \).

2. **Finding the Roots of the Quadratic**:
We need to find the roots of the quadratic equation \( x^2 - 4x + 3 = 0 \). Factoring gives:
\[ (x - 3)(x - 1) = 0 \]
Thus, the roots are \( x = 1 \) and \( x = 3 \).

3. **Analyzing the Interval**:
The quadratic \( x^2 - 4x + 3 \) opens upwards (since the coefficient of \( x^2 \) is positive). We can analyze the sign of the quadratic in the intervals determined by the roots:
- For \( x < 1 \): \( x^2 - 4x + 3 > 0 \)
- For \( 1 < x < 3 \): \( x^2 - 4x + 3 < 0 \)
- For \( x > 3 \): \( x^2 - 4x + 3 > 0 \)

Since we are interested in the interval \( (2, 3) \), we find that \( x^2 - 4x + 3 < 0 \) in this interval.

4. **Finding the Values of \( x^2 - 4x + 3 \)**:
We evaluate \( x^2 - 4x + 3 \) at the endpoints of the interval:
- At \( x = 2 \):
\[ 2^2 - 4 \cdot 2 + 3 = 4 - 8 + 3 = -1 \]
- At \( x = 3 \):
\[ 3^2 - 4 \cdot 3 + 3 = 9 - 12 + 3 = 0 \]

Thus, \( x^2 - 4x + 3 \) is negative in the interval \( (2, 3) \).

5. **Finding the Derivative of \( g(x) = f(\sin x) \)**:
We denote \( g(x) = f(\sin x) \). To find when \( g(x) \) is increasing, we compute its derivative:
\[ g'(x) = f'(\sin x) \cdot \cos x \]
For \( g(x) \) to be increasing, we need \( g'(x) > 0 \).

6. **Conditions for \( g'(x) > 0 \)**:
Since \( f'(\sin x) > 0 \) when \( \sin x \) is in the interval where \( x^2 - 4x + 3 < 0 \), we need to find when \( \sin x \) lies in the interval \( (-1, 0) \) (since \( f' \) is positive for values less than 0).

7. **Finding Intervals for \( \sin x \)**:
The sine function is positive in the intervals \( (0, \pi) \) and negative in \( (\pi, 2\pi) \). Therefore, we need to find where \( \sin x < 0 \):
\[ \sin x < 0 \quad \text{for } x \in (\pi, 2\pi) \]

8. **Final Result**:
Therefore, \( f(\sin x) \) is increasing on the intervals where \( x \) satisfies:
\[ x \in (2n\pi + \pi, 2n\pi + 2\pi) \quad \text{for } n \in \mathbb{Z} \]