Question

Minimum value of $3\sin \theta +4\cos \theta$ is

• A. $1$
• B. $3$
• C. $-5$
• D. $5$

Answer 1

Answer: (C)

$-5$

Let,

$f\left(\theta \right)=3\sin \theta +4\cos \theta$.

Now, $f^{\prime }\left(\theta \right)=3\cos \theta -4\sin \theta$........(1)

Again $f^{\prime \prime }\left(\theta \right)=-3\sin \theta -4\cos \theta$........(2)

Now for maximum of minimum value of $f\left(\theta \right)$ we must have,

$f^{\prime }\left(\theta \right)=0$

or, $3\cos \theta -4\sin \theta =0$

or, $\frac{\sin \theta }{3}=\frac{\cos \theta }{4}=\pm \frac{\sqrt{{\sin }^2\theta +{\cos }^2\theta }}{\sqrt{3^2+4^2}}$

or, $\sin \theta =\pm \frac{3}{5}$, $\cos \theta =\pm \frac{4}{5}$

Now it is evident from (2) that for $\sin \theta =-\frac{3}{5}$, $\cos \theta =-\frac{4}{5}$ we have $f^{\prime \prime }\left(\theta \right)>0$.

In this case we will get minimum value of $f\left(\theta \right)$.

Now minimum value of $f\left(\theta \right)=-\frac{3^2+4^2}{5}=-5$.