Question

The maximum value of 3cos $\theta$ - 4 sin$\theta$ is

• A. 3
• B. 4
• C. 5
• D. None of these

Answer 1

The correct option is C

5

Let 3 = $rcos\alpha$, 4 = $rsin\alpha$, so r = 5

$f\left(\theta \right)$ = $r.(cos\alpha cos\theta +sin\alpha sin\theta )$ = $5.cos(\theta -\alpha )$

∴ The maximum value of $f\left(\theta \right)$ = 5.1 = 5.

{Since the maximum value of $cos(\theta -\alpha )$ = 1}.

Aliter: As we know that, the maximum value of $asin\theta +bcos\theta$ is +$\sqrt{a^2+b^2}$ and the minimum value

is -$\sqrt{a^2+b^2}$. Therefore, the maximum value is $(3cos\theta +4sin\theta )$ = + $\sqrt{3^2+(-4{)}^2}$ = 5 and the minimum value is -5.