Question

Which one of the following is the correct graphical representation for functions, \(\sin(x)\) and \(\sin^2(x)\) for \(0\leq x\leq \pi\)?

• A. Graph A
• B. Graph B
• C. Graph C
• D. Graph D

Hint:

If \(0<a<1\), then \(a^2<a\). Since \(\sin(x)\) lies between \(0\) and \(1\) on \((0,\pi)\), we get \(\sin^2(x)<\sin(x)\).

Answer 1

The Correct Option is C

Step 1: Recall the range of \(\sin(x)\).
For
\[ 0\leq x\leq \pi, \] the function \(\sin(x)\) varies from \(0\) to \(1\) and then back to \(0\). Its maximum value occurs at
\[ x=\frac{\pi}{2} \] where
\[ \sin\left(\frac{\pi}{2}\right)=1. \]

Step 2: Analyze the function \(\sin^2(x)\).
Since
\[ \sin^2(x)=(\sin x)^2, \] all values remain non-negative. Also,
\[ 0\leq \sin^2(x)\leq 1. \]

Step 3: Compare \(\sin(x)\) and \(\sin^2(x)\).
For all values in the interval \(0<x<\pi\),
\[ 0<\sin(x)<1. \] Squaring a number between \(0\) and \(1\) decreases its value. Hence,
\[ \sin^2(x)<\sin(x) \] for all
\[ 0<x<\pi, \] except at
\[ x=\frac{\pi}{2}, \] where both are equal to \(1\).

Step 4: Identify the intersection points.
The graphs meet at
\[ x=0,\quad x=\frac{\pi}{2},\quad x=\pi. \] At \(x=0\) and \(x=\pi\), both functions are zero. At \(x=\frac{\pi}{2}\), both equal \(1\).

Step 5: Examine the graphical behavior.
The graph of \(\sin(x)\) must lie above the graph of \(\sin^2(x)\) throughout the interval except at the common points.

Step 6: Match with the given options.
Among the four graphs, only option (C) correctly shows
\[ \sin^2(x)<\sin(x) \] for most of the interval while touching at the correct points.

Step 7: Final conclusion.
Therefore, the correct graphical representation is option (C).
\[ \boxed{\text{Graph C}} \]
Hence, the correct answer is option (C).