Let f be a twice differentiable function such that `f''(x)gt 0 AA x in R`. Let h(x) is defined by `h(x)=f(sin^(2)x)+f(cos^(2)x)` where `|x|lt (pi)/(2)`.
The number of critical points of h(x) are

Options

A. 1
B. 2
C. 3
D. more than 3

Explanation

To find the number of critical points of the function \( h(x) = f(\sin^2 x) + f(\cos^2 x) \), where \( f \) is a twice differentiable function with \( f''(x) > 0 \) for all \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Differentiate \( h(x) \)

To find the critical points, we first need to compute the derivative \( h'(x) \).

\[ h'(x) = \frac{d}{dx}[f(\sin^2 x)] + \frac{d}{dx}[f(\cos^2 x)] \]

Using the chain rule, we have:

\[ h'(x) = f'(\sin^2 x) \cdot \frac{d}{dx}(\sin^2 x) + f'(\cos^2 x) \cdot \frac{d}{dx}(\cos^2 x) \]

Calculating the derivatives of \( \sin^2 x \) and \( \cos^2 x \):

\[ \frac{d}{dx}(\sin^2 x) = 2\sin x \cos x = \sin(2x) \]
\[ \frac{d}{dx}(\cos^2 x) = 2\cos x (-\sin x) = -\sin(2x) \]

Thus, we can write:

\[ h'(x) = f'(\sin^2 x) \cdot \sin(2x) - f'(\cos^2 x) \cdot \sin(2x) \]

Factoring out \( \sin(2x) \):

\[ h'(x) = \sin(2x) \left( f'(\sin^2 x) - f'(\cos^2 x) \right) \]

### Step 2: Set \( h'(x) = 0 \)

To find the critical points, we need to solve:

\[ \sin(2x) \left( f'(\sin^2 x) - f'(\cos^2 x) \right) = 0 \]

This gives us two cases:

1. \( \sin(2x) = 0 \)
2. \( f'(\sin^2 x) - f'(\cos^2 x) = 0 \)

### Step 3: Solve \( \sin(2x) = 0 \)

The equation \( \sin(2x) = 0 \) implies:

\[ 2x = n\pi \quad \text{for } n \in \mathbb{Z} \]
\[ x = \frac{n\pi}{2} \]

Considering the domain \( |x| < \frac{\pi}{2} \), the valid solutions are:

- \( x = 0 \)

### Step 4: Solve \( f'(\sin^2 x) = f'(\cos^2 x) \)

Since \( f''(x) > 0 \) implies that \( f'(x) \) is an increasing function, the equality \( f'(\sin^2 x) = f'(\cos^2 x) \) can only hold if:

\[ \sin^2 x = \cos^2 x \]

This leads to:

\[ \tan^2 x = 1 \quad \Rightarrow \quad \tan x = \pm 1 \]

Thus, the solutions are:

\[ x = \frac{\pi}{4}, -\frac{\pi}{4} \]

### Step 5: Compile the critical points

The critical points we found are:

1. \( x = 0 \)
2. \( x = \frac{\pi}{4} \)
3. \( x = -\frac{\pi}{4} \)

### Conclusion

Thus, the total number of critical points of \( h(x) \) is:

\[ \boxed{3} \]