I was recently solving a A level Mathematics 9709 past paper when I found this problem:
The function $f$ is such that $f(x)=a+b\cos x$ for $0\leq x \leq2\pi$. It is given that $f\big(\frac{1}{3}\pi \big) = 5$and $f(\pi)=11$.
(i) Find the values of the constants $a$ and $b$.
(ii) Find the set of values of $k$ for which the equation $f(x)=k$ has no soutions.
I was able to solve part (i) making two simultaneous equations getting $a=7$ and $b=-4$. By this way, we get the function $f(x)=7-4 \cos{x}$. The second part is confusing, I know of making discriminant less than 0, I worked as follows: $$ \begin{align*} & 7-4 \cos{x}=k\\ & (7-4cos{x})^2=k^2\\ & 16 \cos^2{x}-56 \cos{x}+\big(49-k^2\big)=0 \end{align*} $$
Here, $\Delta < 0 \implies k < 0 \forall \cos{x}$. My confusions here are:
First observe that you have an error in your calculation. In fact, $\Delta=(-56)^2-4\cdot16\cdot (49-k^2)=64k^2$, which is always nonnegative (and therefore could place no restrictions on solutions).
I will explain how your argument with the discriminant was flawed. Observe that the proof does not anywhere involve the action of the $\cos$ function. So it is equivalently making an argument about a real variable $u=\cos x$ that takes values in $(-\infty,+\infty)$. Following this, observe that the argument begins from $7-4u=k$, where $k$ is a constant, which is a linear equation! So there cannot possibly be any values of $k$ that prohibit it from having a solution, and everything thereafter was irrelevant. And it is therefore not surprising that the discriminant was $\Delta\ge 0$ since the solution for the linear function guarantees a solution for the quadratic.
So the argument was a non-starter. As a more fruitful approach, use what you know about the range of $y=\cos x$ and your knowledge of linear transformations of graphs, particularly $y=f(x)\mapsto y=A\times f(x)$ and $\mapsto y=f(x)+B$, to work out what your graph would look like.