Question

The maximum value of $\frac{3}{5sinx-12\cos x+19}$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

Let $f\left(x\right)=\frac{3}{5sinx-12cosx+19}$
We know that $-13\le 5sinx-12cosx\le 13$
Hence, the maximum value of $f\left(x\right)$ will be when $5sinx-12cosx=-13$
Hence, $f_{max}=\frac{3}{-13+19}=\frac{1}{2}$
Hence, option B is correct.