Question

Let f:R$\to$ R be defined by f(x)=$\cos$ (5x+2). Is f invertible?
if yes enter 1 else enter 0

Answer 1

For a function to be invertible it is necessary that it is bijective i.e. one-one and into. The function given here is neither surjective nor injective as shown below. Since $-1\cos (5x+2)\le 1$, the range of f ={y:y is real $,-1\le y\le 1$. which is a proper subset of the co-domain R. Hence f is into so that it is not surjective. f is many-one since $\cos$ (5x+2)has the same value for many values of x. Thus $f(x+\frac{2}{5}n\pi )=\cos \{5(x+\frac{2}{5}n\pi )+2\}=\cos 2n\pi +5x+2=\cos (5x+2)=f\left(x\right)$. for all $n=0,\pm 1,\pm 2,\pm 3,$....Since f is not bijective, it is not invertible