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.
is neither increasing nor decreasing function on [0, $\frac{\pi}{2}$)
Show Solution
The Correct Option isB
Solution and Explanation
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, $\frac{\pi}{2}$).