Question

The function$f:X\to Y$defined by $f\left(x\right)=\sin x$ is one-one but not onto if$X$and$Y$ are respectively equal to.

• A. $\mathrm{ℝ}$ and $\mathrm{ℝ}$
• B. $[0,\mathrm{\pi }]$and$[0,1]$
• C. $\left[0,\frac{\mathrm{\pi }}{2}\right]$and$\left[-1,1\right]$
• D. $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$and $[-1,1]$

Answer 1

The correct option is C

$\left[0,\frac{\mathrm{\pi }}{2}\right]$and$\left[-1,1\right]$

Explanation for the correct option:

Option C: $\left[0,\frac{\mathrm{\pi }}{2}\right]$and$\left[-1,1\right]$

$f:X\to Y$

$X$: domain and $Y$: Co-domain

Domain =$X=$ $\left[0,\frac{\mathrm{\pi }}{2}\right]$ and co-domain=$Y=$$\left[-1,1\right]$

By graph we can clearly observe function is one -one but range for given domian is $\left[0,1\right]$.

For a unique value of $x$ there exist a unique value of $y$, but for $y$ in the interval $[-1,0)$ does not have any pre-image.

So, the function is one-one but not onto.

Explanation for the in-correct option:

Option B $[0,\mathrm{\pi }]$and$[0,1]$

In the given domain $\sin \left(\frac{\mathrm{\pi }}{6}\right)=\sin \left(\frac{5\mathrm{\pi }}{6}\right)=\frac{1}{2}$,so function is not one one

and range of function in given domain is $[0,1]$

So function is not one -one but onto.

Option A :$\mathrm{ℝ}$ and $\mathrm{ℝ}$

$Sinx$ has range only $\left[-1,1\right]$.So, function is onto

and $Sinx$is periodic function,so one -one is not possible.

Therefore function is neither one-one nor onto.

Option D : $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$and $[-1,1]$

$\sin x$is one-one function in the given domain and range of function also is $[-1,1]$as shown in figure.

Therefore function is one-one and onto in the given domain and co-domain.

Hence, the correct option is $\mathrm{Option}\left(\mathrm{C}\right)$.