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: Concept Review:
A function's monotonicity (increasing or decreasing) on an interval is determined by the sign of its first derivative on that interval. A positive first derivative (f'(x) > 0) indicates an increasing function, while a negative first derivative (f'(x) < 0) indicates a decreasing function.
Step 2: Methodology:
1. Compute the derivative of the given function f(x) to obtain f'(x).
2. Evaluate the sign of f'(x) over the specified interval [0, $\frac{\pi}{2}$).
Step 3: Derivation and Analysis:
Given function: f(x) = tanx - x.
Calculate the first derivative, f'(x):
\[ f'(x) = \frac{d}{dx}(\tan x - x) = \sec^2 x - 1 \] Employing the trigonometric identity $\sec^2 x = 1 + \tan^2 x$:
\[ f'(x) = (1 + \tan^2 x) - 1 = \tan^2 x \] Analyze the sign of f'(x) = $\tan^2 x$ on the interval [0, $\frac{\pi}{2}$).
For all x within this interval where tanx is defined, $\tan^2 x$ will always be non-negative.
\[ f'(x) = \tan^2 x \ge 0 \] At x = 0, f'(x) = 0. For all other x in (0, $\frac{\pi}{2}$), tanx is positive, making $\tan^2 x$ strictly positive.
Since f'(x) $\ge$ 0 on [0, $\frac{\pi}{2}$) and is not zero over any subinterval, f(x) is determined to be increasing on this interval.
Step 4: Conclusion:
Conclusion: f(x) = tanx - x is an increasing function on [0, $\frac{\pi}{2}$).