MCQ
The maximum value of $3\cos \theta - 4\sin \theta $ is
- A$3$
- B$4$
- ✓$5$
- DNone of these
✓
Answer
Correct option: C.
$5$
c
(c) Let $3 = r\cos \alpha ,4 = r\sin \alpha ,$so $r = 5$
(c) Let $3 = r\cos \alpha ,4 = r\sin \alpha ,$so $r = 5$
$f(\theta ) = r.(\cos \alpha \cos \theta + \sin \alpha \sin \theta ) = 5.\cos (\theta - \alpha )$
$\therefore $ The maximum value of $f(\theta ) = 5.1 = 5.$
{Since the maximum value of $\cos (\theta - \alpha ) = 1$}.
Aliter : As we know that, the maximum value of $a\sin \theta + b\cos \theta $ is $ + \sqrt {{a^2} + {b^2}} $
and the minimum value is $ - \sqrt {{a^2} + {b^2}} $.
Therefore, the maximum value is $(3\cos \theta + 4\sin \theta ) = + \sqrt {{3^2} + {{( - 4)}^2}} = 5$
and the minimum value is $-5.$
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