Question

The maximum value of a cosx+b sinx is

[MNR 1991; MP PET 1999; UPSEAT 2000]

• A. a+b
• B. a-b
• C.
• D.

Answer 1

The correct option is D

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

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

Now, f(x) = $rsin\theta cosx+rcos\theta sinx$

= $\sqrt{a^2+b^2}$ {$sin(\theta +x)$}

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

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

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