Aussie Maths & Science Teachers: Save your time with SmarterEd
It is given that \(h(x)=\sin^2 x-\sin x\) is a composite function of \(f(x)\) and \(g(x)\) where \(f(x)=x^2-x\) and \(g(x)=\sin x\) for all real values of \(x\).
For what values of \(k\) does the equation \(\sin^2 x-\sin x=k\) have real solutions?
\(D\)
\(f(x)=x^2-x,\ \ g(x)=\sin x\ \ \Rightarrow\ \ f(g(x))=\sin^2 x-\sin x\)
\(h(x)=f(g(x))\)
\(\text{Since}\ \ g(x)=\sin x\ \ \Rightarrow\ \ -1 \leq g(x) \leq 1\)
\(\text{Consider}\ \ f(u)=u^2-u\ \ \text{for}\ \ -1 \le u \le 1:\)
\(\text{Vertex at}\ \ u=\dfrac{1}{2}:\ \ f\left(\dfrac{1}{2}\right)=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\)
\(\text{Endpoints:}\ \ f(-1)=2, \ \ f(1)=0\)
\(\therefore\ \text{Real solutions exist for}\ \ -\dfrac{1}{4} \le k \le 2\)
\(\Rightarrow D\)