Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.


Let f be a function from R to R, such that f(x) = cos (x + 2). Is f invertible? Justify your answer.



Let and y be two elements in the domain (R), such that


$\Rightarrow \cos (x+2)=\cos (y+2)$

$\Rightarrow x+2=y+2$ or $x+2=2 \pi-(y+2)$

$\Rightarrow x=y$ or $x+2=2 \pi-y-2$

$\Rightarrow x=y$ or $x=2 \pi-y-4$

So, we cannot say that $x=y$

For example,

$\cos \frac{\pi}{2}=\cos \frac{3 \pi}{2}=0$

So, $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$ have the same image 0 .

$\Rightarrow f$ is not one-one.

$\Rightarrow f$ is not a bijection.

Thus, $f$ is not invertible.

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