Question

The maximum value of $1+4\sin \theta +3\cos \theta$ is:

• A. $-3$
• B. $-4$
• C. $5$
• D. $6$

Answer 1

The correct option is D $6$

Let $y=1+4\sin \theta +3\cos \theta$

Since max value of $a\sin \theta +b\cos \theta =\sqrt{a^2+b^2}$

max value of $y=1+$ max value of $(4\sin \theta +3\cos \theta )$

$=1+\sqrt{4^2+3^2}=1+5=6$

and minimum of $y=1+$ minimum of $(4\sin \theta +3\cos \theta )$

$=1-\sqrt{3^2+4^2}=1-5=-4$