Question

The range of the function $f(x) = \frac{1}{40\cos x + 9\sin x + 50}$ is $(x \in R)$

Answer 1

$\left[\frac{1}{91},\frac{1}{9}\right]$

Answer 2

We rewrite the denominator as:

$$D(x)=40\cos x+9\sin x+50$$

The term $40\cos x+9\sin x$ has a maximum and minimum value determined by:

$$\sqrt{40^2+9^2}=\sqrt{1600+81}=\sqrt{1681}=41.$$

Thus, its range is $[-41,41]$.

Adding 50, we get:

$$D(x)\in[50-41,50+41]=[9,91].$$

Since $f(x)=\frac{1}{D(x)}$ and $1/y$ is a decreasing function for $y>0$, the range of $f(x)$ is:

$$\left[\frac{1}{91},\frac{1}{9}\right].$$

Explanation (minimal):

1. Maximum of $40\cos x+9\sin x =41$ and minimum $-41$.
2. Denom range becomes $[9,91]$.
3. Taking reciprocal, range of $f(x)=\frac{1}{D(x)}$ is $\left[\frac{1}{91},\frac{1}{9}\right]$.