MCQ
Mapping $f:R \to R$ which is defined as $f(x) = \cos x,\;x \in R$ will be
- ✓Neither one-one nor onto
- BOne-one
- COnto
- DOne-one onto
✓
Answer
Correct option: A.
Neither one-one nor onto
a
(a) Let ${x_1},\,{x_2} \in R,$ then $f({x_1}) = \cos {x_1}$, $f({x_2}) = \cos {x_2}$,
(a) Let ${x_1},\,{x_2} \in R,$ then $f({x_1}) = \cos {x_1}$, $f({x_2}) = \cos {x_2}$,
so $f({x_1}) = f({x_2})$
$⇒$ $\cos {x_1} = \cos {x_2}$ $⇒$ ${x_1} = 2n\pi \pm {x_2}$
$⇒$ ${x_1} \ne {x_2}$, so it is not one-one.
Again the value of $f$-image of $x$ lies in between $-1$ to $1$
$⇒$ $f[R] = \left\{ {f(x): - 1 \le f(x) \le 1)} \right\}$
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.
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