Search any question & find its solution
Question:
Answered & Verified by Expert
Mapping $f: R \rightarrow R$ which is defined as $f(x)=\cos x, x \in R$ will be
Options:
Solution:
1893 Upvotes
Verified Answer
The correct answer is:
Neither one-one nor onto
Let $x_1, x_2 \in R$, then $f\left(x_1\right)=\cos x_1, f\left(x_2\right)=\cos x_2$, so $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \cos x_1=\cos x_2 \Rightarrow x_1=2 n \pi \pm x_2$
$\Rightarrow x_1 \neq x_2$, so it is not one-one.
Again the value of $f$-image of $x$ lies in between -1 to 1
$\Rightarrow f[R]=\{f(x):-1 \leq f(x) \leq 1)\}$
So other numbers of co-domain (besides -1 and 1 ) is not $f$-image. $f[R] \in R$, so it is also not onto. So this mapping is neither one-one nor onto.
$\Rightarrow \cos x_1=\cos x_2 \Rightarrow x_1=2 n \pi \pm x_2$
$\Rightarrow x_1 \neq x_2$, so it is not one-one.
Again the value of $f$-image of $x$ lies in between -1 to 1
$\Rightarrow f[R]=\{f(x):-1 \leq f(x) \leq 1)\}$
So other numbers of co-domain (besides -1 and 1 ) is not $f$-image. $f[R] \in R$, so it is also not onto. So this mapping is neither one-one nor onto.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.