If `f(x)=cosx+sinx` and `g(x)=x^(2)-1`, then `g(f(x))` is injective in the interval

Options

A. `[0, (pi)/(2)]`
B. `[-(pi)/(4), (pi)/(4)]`
C. `[-(pi)/(2),(pi)/(2)]`
D. `[0, pi]`

Answer

To determine the interval in which the function \( g(f(x)) \) is injective, we will follow these steps:

### Step 1: Define the functions
We have:
- \( f(x) = \cos x + \sin x \)
- \( g(x) = x^2 - 1 \)

### Step 2: Find \( g(f(x)) \)
We need to compute \( g(f(x)) \):
\[ g(f(x)) = g(\cos x + \sin x) = (\cos x + \sin x)^2 - 1 \]

### Step 3: Expand \( g(f(x)) \)
Now, let's expand \( (\cos x + \sin x)^2 \):
\[ (\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2\sin x \cos x \]
Using the identity \( \cos^2 x + \sin^2 x = 1 \):
\[ (\cos x + \sin x)^2 = 1 + 2\sin x \cos x \]
Thus,
\[ g(f(x)) = 1 + 2\sin x \cos x - 1 = 2\sin x \cos x \]
Using the double angle identity for sine:
\[ g(f(x)) = \sin(2x) \]

### Step 4: Determine the intervals for injectivity
The function \( \sin(2x) \) is periodic with a period of \( \pi \). To find the intervals where it is injective, we need to consider the intervals where it is either strictly increasing or strictly decreasing.

The function \( \sin(2x) \) is strictly increasing in the intervals:
- \( \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \)

### Step 5: Conclusion
Thus, the function \( g(f(x)) = \sin(2x) \) is injective in the interval:
\[ \boxed{\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)} \]