Question

Find the maximum and minimum values of $6sinx\, cosx + 4cos2x$.

• A. $5$, $-5$
• B. $6$, $-6$
• C. $4$, $-4$
• D. $2$, $-2$

Answer 1

Answer: (A)

$5$, $-5$

Answer 2

We have, $6sinx \,cosx + 4cos2x = 3sin2x + 4cos2x$ which is of the form $asin\theta + bcos\theta$. $\therefore -\sqrt{3^{2}+4^{2}} \le3sin2x+4cos2x \le\sqrt{3^{2}+4^{2}}$ $\left[\because -\sqrt{a^{2}+b^{2}} \,\le asin\theta+bcos\theta \le\sqrt{a^{2}+b^{2}}\right]$ $\Rightarrow -5 \le3sin2x+4cos2x \le5$ Thus, maximum and minimum values of $6sinx\, cosx + 4cos2x$ are $5$ and $-5$, respectively.