Question

The maximum value of $3 \sin x + 4 \cos x + 5$ is:

• A. 5
• B. 10
• C. 12
• D. 15

Hint:

$(3, 4, 5)$ is a standard Pythagorean triple, so $\sqrt{3^2+4^2}$ is instantly $5$. Add $5$ to get $10$.

Answer 1

The Correct Option is B

Step 1: Range of $a\sin x + b\cos x$.
Any expression $a\sin x + b\cos x$ stays between $-\sqrt{a^2+b^2}$ and $+\sqrt{a^2+b^2}$. So its largest value is $\sqrt{a^2+b^2}$.

Step 2: Read off $a$ and $b$.
Here $a = 3$ and $b = 4$.

Step 3: Find the amplitude.
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Step 4: Handle the extra constant.
The $+5$ at the end simply shifts everything up by 5.

Step 5: Add the parts.
The biggest value is amplitude plus shift: \[ 5 + 5 = 10 \]

Step 6: Conclusion.
\[ \boxed{ \text{Maximum} = 10 } \]