The maximum value of `f(x)=a sin x +b cos x` is

Options

A. `-sqrt(a^(2)+b^(2))`
B. `sqrt((a^(2)+b^(2))`
C. `(-1)/(sqrt(a^(2)+b^(2)))`
D. `(1)/(sqrt(a^(2)+b^(2)))`

Answer

To find the maximum value of the function \( f(x) = a \sin x + b \cos x \), we can follow these steps:

### Step 1: Understanding the Function
We start with the function:
\[ f(x) = a \sin x + b \cos x \]
We want to find the maximum value of this function.

### Step 2: Finding the Derivative
We differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = a \cos x - b \sin x \]

### Step 3: Setting the Derivative to Zero
To find the critical points, we set the derivative equal to zero:
\[ a \cos x - b \sin x = 0 \]
This can be rearranged to:
\[ a \cos x = b \sin x \]
Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)):
\[ \tan x = \frac{a}{b} \]

### Step 4: Finding the Values of \( \sin x \) and \( \cos x \)
From the equation \( \tan x = \frac{a}{b} \), we can represent \( \sin x \) and \( \cos x \) in terms of \( a \) and \( b \):
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
Let \( \sin x = \frac{a}{\sqrt{a^2 + b^2}} \) and \( \cos x = \frac{b}{\sqrt{a^2 + b^2}} \).

### Step 5: Substituting Back to Find Maximum Value
Now we substitute \( \sin x \) and \( \cos x \) back into the function:
\[ f(x) = a \left(\frac{a}{\sqrt{a^2 + b^2}}\right) + b \left(\frac{b}{\sqrt{a^2 + b^2}}\right) \]
This simplifies to:
\[ f(x) = \frac{a^2 + b^2}{\sqrt{a^2 + b^2}} = \sqrt{a^2 + b^2} \]

### Conclusion
Thus, the maximum value of \( f(x) = a \sin x + b \cos x \) is:
\[ \boxed{\sqrt{a^2 + b^2}} \]

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