Question

The function f(x) = tanx - x

• A. is a decreasing function on [0, $\frac{\pi}{2}$)
• B. is an increasing function on [0, $\frac{\pi}{2}$)
• C. is a constant function
• D. is neither increasing nor decreasing function on [0, $\frac{\pi}{2}$)

Hint:

When checking for increasing/decreasing behavior, finding the first derivative is the standard method. Remembering basic trigonometric identities is crucial for simplifying the derivative and analyzing its sign.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
To determine if a function is increasing or decreasing on an interval, we examine the sign of its first derivative on that interval. If f'(x)>0, the function is increasing. If f'(x)<0, it is decreasing.
Step 2: Key Formula or Approach:
1. Find the derivative of f(x), which is f'(x).
2. Analyze the sign of f'(x) in the given interval [0, $\frac{\pi}{2}$).
Step 3: Detailed Explanation:
The given function is f(x) = tanx - x.
First, find the derivative f'(x):
\[ f'(x) = \frac{d}{dx}(\tan x - x) = \sec^2 x - 1 \] We know the trigonometric identity $1 + \tan^2 x = \sec^2 x$.
Substituting this into the derivative expression:
\[ f'(x) = (1 + \tan^2 x) - 1 = \tan^2 x \] Now, we need to analyze the sign of f'(x) = $\tan^2 x$ on the interval [0, $\frac{\pi}{2}$).
For any value of x in this interval, tanx is defined. Since $\tan^2 x$ is a square, its value is always greater than or equal to zero.
\[ f'(x) = \tan^2 x \ge 0 \] The derivative is zero only at x = 0 (since tan(0) = 0). For all other values of x in (0, $\frac{\pi}{2}$), tanx is positive, so $\tan^2 x$ is strictly positive.
Since f'(x) $\ge$ 0 on the interval [0, $\frac{\pi}{2}$) and is not identically zero over any subinterval, the function f(x) is increasing on this interval.
Step 4: Final Answer:
The function f(x) = tanx - x is an increasing function on [0, $\frac{\pi{2}$)}.