Find the minimum and maximum value of 3sin x + 4cos x + 5 — JEE Mathematics

The expression $a\sin x + b\cos x$ always lies within the range $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.
Here, $a = 3$ and $b = 4$.

$$\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$$

So, $-5 \le 3\sin x + 4\cos x \le 5$.

Now add 5 across the inequality:

$$-5 + 5 \le 3\sin x + 4\cos x + 5 \le 5 + 5$$

$$0 \le 3\sin x + 4\cos x + 5 \le 10$$

Minimum value = 0, Maximum value = 10.

Answer: Min = 0, Max = 10