Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 15 / c

Put $q(x)=\sqrt{{\pi }^2/16+x}$. The original PDF confirms that the $+x$ is inside the radical. For the composite cosine,At $x=0$, $q=\pi /4$ and $\sin q=\cos q=\sqrt{2}/2$. Substituting into $f(0)+f^{\prime }(0)x+f^{\prime \prime }(0)x^2/2$ givesAll three displayed coefficients are nonzero.