Question

The maximum value of $a\cos x + b\sin x$is

• A. $a + b$
• B. $a - b$
• C. $|a| + |b|$
• D. $(a^{2} + b^{2})^{1/2}$

Answer 1

Answer: (D)

$(a^{2} + b^{2})^{1/2}$

Answer 2

Let $f(x) = a\cos x + b\sin x$

Suppose that$a = r\sin\theta$and $b = r\cos\theta$ i.e.,$r = \sqrt{a^{2} + b^{2}}$

Now, $f(x) = r\sin\theta\cos x + r\cos\theta\sin x$ $= \sqrt{a^{2} + b^{2}}\{\sin(\theta + x)\}$

But $- 1 \leq \sin(\theta + x) \leq 1 \Rightarrow - r \leq r\sin(\theta + x) \leq r$

$\Rightarrow - \sqrt{a^{2} + b^{2}} \leq \sqrt{a^{2} + b^{2}}\sin(\theta + x) \leq \sqrt{a^{2} + b^{2}}$

Thus, maximum value of $f(x)$ is $\sqrt{a^{2} + b^{2}}$.