It is given that \(y=a\cos3x+b\), where \(a>0\), has a minimum value of \(-3\) and a maximum value of \(5\).
Without solving for \(b\), explain why \(a=4\).[1]
Hence, or otherwise, find the value of \(b\).[1]
State the period of \(y\) in radians.[1]
Sketch the graph of \(y=3\sin\left(\dfrac x2\right)-1\), for \(0\leq x\leq3\pi\).[3]
Sam claimed that the equation \(6\pi\sin\left(\dfrac x2\right)-3x=0\) has only one solution for \(0\leq x\leq3\pi\). Do you agree with him? Draw a suitable line on the same axes in part (b)(i) and justify your answer.[3]
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Answer:(a)(i) amplitude \(4\); (ii) \(b=1\); (iii) \(\dfrac{2\pi}3\) radians. (b)(ii) Disagree: two solutions, found using \(y=\dfrac{3x}{2\pi}-1\).