Let `f(x)=a sin x+c`, where a and c are real numbers and a>0. Then `f(x)lt0, AA x in R` if

Options

A. `c lt -a`
B. `c gt -a`
C. `-a lt c lt a`
D. `c lt a`

Explanation

To solve the problem, we need to determine the conditions under which the function \( f(x) = a \sin x + c \) is less than zero for all \( x \in \mathbb{R} \), given that \( a > 0 \).

### Step-by-Step Solution:

1. **Write the function**:
\[ f(x) = a \sin x + c \]

2. **Set the condition**:
We want to find when \( f(x) < 0 \). Therefore, we can express this as:
\[ a \sin x + c < 0 \]

3. **Rearrange the inequality**:
To isolate \( \sin x \), we can rearrange the inequality:
\[ a \sin x < -c \]

4. **Divide by \( a \)**:
Since \( a > 0 \), we can divide both sides of the inequality by \( a \) without changing the direction of the inequality:
\[ \sin x < -\frac{c}{a} \]

5. **Determine the range of \( \sin x \)**:
The sine function \( \sin x \) has a range of \([-1, 1]\). For the inequality \( \sin x < -\frac{c}{a} \) to hold for all \( x \), the right-hand side must be less than the minimum value of \( \sin x \):
\[ -\frac{c}{a} < -1 \]

6. **Solve for \( c \)**:
To find the condition on \( c \), we can multiply both sides by \(-1\) (which reverses the inequality):
\[ \frac{c}{a} > 1 \]
Multiplying both sides by \( a \) (which is positive) gives:
\[ c > a \]

### Conclusion:
The function \( f(x) = a \sin x + c \) is less than zero for all \( x \in \mathbb{R} \) if:
\[ c > a \]