Questions
Consider the functions \( f(x)=cos x\) and \(g(x)=sin 2x\), where \(0\leq x\leq \pi \).
The graph of f intersects the graph of g at the point A, the point B \(\left ( \frac{\pi }{2},0 \right )\) and the point C as shown on the following diagram.
(a) Find the x-coordinate of point A and the x-coordinate of point C.
The shaded region R is enclosed by the graph of f and the graph of g between the points B and C.
(b) Find the area of R.
▶️Answer/Explanation
Detailed solution
(a) Find the x-Coordinates of Point A and Point C
The graphs of \( f(x) = \cos x \) and \( g(x) = \sin 2x \) intersect at points A, B, and C. We’re given B at \( \left( \frac{\pi}{2}, 0 \right) \), which we can verify:
\( f\left(\frac{\pi}{2}\right) = \cos \frac{\pi}{2} = 0 \),
\( g\left(\frac{\pi}{2}\right) = \sin \left(2 \cdot \frac{\pi}{2}\right) = \sin \pi = 0 \).
This confirms B is an intersection point. To find A and C, set \( f(x) = g(x) \):
\[
\cos x = \sin 2x
\]
Use the identity \( \sin 2x = 2 \sin x \cos x \):
\[
\cos x = 2 \sin x \cos x
\]
\[
\cos x – 2 \sin x \cos x = 0
\]
\[
\cos x (1 – 2 \sin x) = 0
\]
So:
1. \( \cos x = 0 \), giving \( x = \frac{\pi}{2} \) (which is point B),
2. \( 1 – 2 \sin x = 0 \), so \( 2 \sin x = 1 \), \( \sin x = \frac{1}{2} \).
Solve \( \sin x = \frac{1}{2} \) in \( [0, \pi] \):
\( x = \frac{\pi}{6} \) (since \( \sin \frac{\pi}{6} = \frac{1}{2} \)),
\( x = \pi – \frac{\pi}{6} = \frac{5\pi}{6} \) (since \( \sin (\pi – x) = \sin x \)).
Thus, the intersection points are:
\( x = \frac{\pi}{6} \),
\( x = \frac{\pi}{2} \) (point B),
\( x = \frac{5\pi}{6} \).
From the graph, point A is at the smallest x-value, and point C at the largest within \( [0, \pi] \):
– Point A: \( x = \frac{\pi}{6} \),
– Point C: \( x = \frac{5\pi}{6} \).
Verify coordinates:
At \( x = \frac{\pi}{6} \), \( y = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \), \( g = \sin \left(2 \cdot \frac{\pi}{6}\right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \), matches.
At \( x = \frac{5\pi}{6} \), \( y = \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} \), \( g = \sin \left(2 \cdot \frac{5\pi}{6}\right) = \sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2} \), matches.
(b) Find the Area of Region R
Region R is the area enclosed by \( f(x) \) and \( g(x) \) between points B (\( x = \frac{\pi}{2} \)) and C (\( x = \frac{5\pi}{6} \)), where \( g(x) \geq f(x) \) (as seen in the graph, \( \sin 2x \) is above \( \cos x \)).
The area between two curves from \( a \) to \( b \) where \( g(x) \geq f(x) \) is:
\[
\text{Area} = \int_a^b [g(x) – f(x)] \, dx
\]
Here, \( a = \frac{\pi}{2} \), \( b = \frac{5\pi}{6} \), \( g(x) – f(x) = \sin 2x – \cos x \).
Compute:
\[
\text{Area} = \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (\sin 2x – \cos x) \, dx
\]
Find the antiderivatives:
\( \int \sin 2x \, dx \): Use substitution, let \( u = 2x \), \( du = 2 \, dx \), \( dx = \frac{du}{2} \), so \( \int \sin 2x \, dx = \int \sin u \cdot \frac{du}{2} = \frac{1}{2} (-\cos u) = -\frac{1}{2} \cos 2x \),
\( \int -\cos x \, dx = -\int \cos x \, dx = -\sin x \).
Antiderivative of \( \sin 2x – \cos x \):
\[
-\frac{1}{2} \cos 2x – \sin x
\]
Evaluate from \( \frac{\pi}{2} \) to \( \frac{5\pi}{6} \):
\[
\left[ -\frac{1}{2} \cos 2x – \sin x \right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}}
\]
At \( x = \frac{5\pi}{6} \):
\( \cos \left(2 \cdot \frac{5\pi}{6}\right) = \cos \frac{5\pi}{3} = \cos \frac{\pi}{3} = \frac{1}{2} \),
\( \sin \frac{5\pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2} \),
So: \( -\frac{1}{2} \cdot \frac{1}{2} – \frac{1}{2} = -\frac{1}{4} – \frac{1}{2} = -\frac{3}{4} \).
At \( x = \frac{\pi}{2} \):
\( \cos \left(2 \cdot \frac{\pi}{2}\right) = \cos \pi = -1 \),
\( \sin \frac{\pi}{2} = 1 \),
So: \( -\frac{1}{2} \cdot (-1) – 1 = \frac{1}{2} – 1 = -\frac{1}{2} \).
Area:
\[
\left(-\frac{3}{4}\right) – \left(-\frac{1}{2}\right) = -\frac{3}{4} + \frac{1}{2} = -\frac{3}{4} + \frac{2}{4} = -\frac{1}{4}
\]
The area is positive, so take the absolute value:
\[
\left| -\frac{1}{4} \right| = \frac{1}{4}
\]
……………………….Markscheme…………………………
Ans:
(a) recognising \(cos x=2sin x cos x\)
\(cos x\neq 0\) so \(sin x=\frac{1}{2}\) OR one correct value (accept degrees)
x-coordinates \(\frac{\pi }{6}\) and \(\frac{5\pi }{6}\)
(b) METHOD 1
attempt to integrate \(\pm (cos x-sin 2x)\)
\(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}(cos x – sin 2x)dx\) OR \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}(cos x – 2 sin x cos x)dx\)
\(=\left [ sin x +\frac{1}{2}cos 2x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\) OR \(=\left [ sin x – sin^{2}x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\)
attempt to substitute their limits into their integral and subtract
\(=\left ( sin\left ( \frac{5\pi }{6} \right )+\frac{1}{2}cos\left ( \frac{5\pi }{3} \right ) \right )-\left ( sin\left ( \frac{\pi }{2} \right )+\frac{1}{2}cos(\pi ) \right )\) OR \(=\left ( sin\left ( \frac{5\pi }{6} \right )-sin^{2}\left ( \frac{5\pi }{6} \right ) \right )-\left ( sin\left ( \frac{\pi }{2} \right )-sin^{2}\left ( \frac{\pi }{2} \right )\right)\)
\(=\left ( \frac{1}{2}+\frac{1}{4} \right )-\left ( 1-\frac{1}{2} \right )\) OR \(=\left ( \frac{1}{2}-\frac{1}{4} \right )-\left ( 1-1 \right )\)
area = \(\frac{1}{4}\)
METHOD 2
\(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}cos xdx=\left [ sin x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\) and \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}sin2xdx=\left [ -\frac{1}{2}cos2x \right ]_{\frac{\pi }{2}}^{\frac{5\pi }{6}}\)
attempt to substitute their limits into their integral and subtract (for both integrals)
\(sin\left ( \frac{5\pi }{6} \right )-sin\left ( \frac{\pi }{2} \right )\) and \(-\frac{1}{2}cos\left ( \frac{5\pi }{3} \right )+\frac{1}{2}cos\left ( \pi \right )\)
attempt to subtract the two integrals in either order (seen anywhere)
\(\left ( sin\left ( \frac{5\pi }{6} \right )-sin\left ( \frac{\pi }{2} \right ) \right )-\left ( -\frac{1}{2}cos\left ( \frac{5\pi }{3} \right )+\frac{1}{2}cos(\pi ) \right )\) OR \(\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}cos xdx-\int_{\frac{\pi }{2}}^{\frac{5\pi }{6}}sin2xdx\)
\(=\left ( \frac{1}{2}-1 \right )-\left ( \frac{1}{4}-\frac{1}{2} \right )\left ( =-\frac{1}{4} \right )\)
area = \(\frac{1}{4}\)