Question

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be the function given by $f(x) = \cos(\tan^{-1} x)$. Which one of the following statements is TRUE?

• A. $f$ is decreasing for $x > 0$
• B. $f$ is decreasing for $x < 0$
• C. $f$ is decreasing on $\mathbb{R}$
• D. $f$ is decreasing on the interval $(-1, 1)$

Hint:

An alternate graphical way is to recognize that $f(x) = \frac{1}{\sqrt{1+x^2}}$ is an even function symmetric about the y-axis, with its maximum at $x=0$.
As $x$ moves away from 0 in the positive direction, the denominator increases, which means $f(x)$ must decrease for $x > 0$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We can simplify $f(x)$ using trigonometry. Let $\theta = \tan^{-1} x \implies \tan \theta = x$.
Step 2: Simplifying f(x):
If $\tan \theta = x/1$, then the hypotenuse is $\sqrt{1 + x^2}$. Therefore, $\cos \theta = \frac{1}{\sqrt{1 + x^2}}$. So, $f(x) = (1 + x^2)^{-1/2}$.
Step 3: Differentiating to check Monotonicity:
$f'(x) = -\frac{1}{2}(1 + x^2)^{-3/2} \cdot (2x) = \frac{-x}{(1 + x^2)^{3/2}}$.
For $x > 0$: $f'(x) < 0$ (Function is decreasing).
For $x < 0$: $f'(x) > 0$ (Function is increasing).

Step 4: Final Answer:
The function f is decreasing for x > 0.