`acosx+bsinx` lies between

Options

A. a and b
B. `-(a^(2)+b^(2))and(a^(2)+b^(2))`
C. `-sqrt(a^(2)+b^(2))andsqrt(a^(2)+b^(2))`
D. `-sqrt(a+b)andsqrt(a+b)`

Answer

To find the range of the expression \( A \cos x + B \sin x \), we can follow these steps:

### Step 1: Write the Expression
We start with the expression:
\[ y = A \cos x + B \sin x \]

### Step 2: Multiply and Divide by \(\sqrt{A^2 + B^2}\)
To simplify the expression, we multiply and divide by \(\sqrt{A^2 + B^2}\):
\[ y = \sqrt{A^2 + B^2} \left( \frac{A}{\sqrt{A^2 + B^2}} \cos x + \frac{B}{\sqrt{A^2 + B^2}} \sin x \right) \]

### Step 3: Define \(\cos \theta\) and \(\sin \theta\)
Let:
\[ \cos \theta = \frac{B}{\sqrt{A^2 + B^2}}, \quad \sin \theta = \frac{A}{\sqrt{A^2 + B^2}} \]
This is based on the right triangle formed where the height is \(A\) and the base is \(B\).

### Step 4: Substitute \(\cos \theta\) and \(\sin \theta\)
Substituting these values into the expression gives:
\[ y = \sqrt{A^2 + B^2} \left( \sin \theta \cos x + \cos \theta \sin x \right) \]

### Step 5: Use the Sine Addition Formula
Using the sine addition formula, we can rewrite the expression:
\[ y = \sqrt{A^2 + B^2} \sin(x + \theta) \]

### Step 6: Determine the Range of \(\sin(x + \theta)\)
The sine function oscillates between -1 and 1:
\[ -1 \leq \sin(x + \theta) \leq 1 \]

### Step 7: Multiply by \(\sqrt{A^2 + B^2}\)
Multiplying the entire inequality by \(\sqrt{A^2 + B^2}\) gives:
\[ -\sqrt{A^2 + B^2} \leq y \leq \sqrt{A^2 + B^2} \]

### Conclusion
Thus, the range of the expression \( A \cos x + B \sin x \) is:
\[ \boxed{[-\sqrt{A^2 + B^2}, \sqrt{A^2 + B^2}]} \]