The function f:X→ Y defined by f(x)= sin x is one-one but not onto, if X and Y are respectively equal to

Question

The function f:X→ Y defined by f(x)= sin x is one-one but not onto, if X and Y are respectively equal to (A) R and R (B)  [0,π ] and [0,1] (C)  [ 0,(π /2) ] and [-1,1]  (D)  [ (-π /2),(π /2) ] and [-1,1]

Tardigrade Admin · Accepted Answer

Since, f:X→ Y, then f(x)=sin text x Now, take option (c). Domain =[ 0,(π /2) ], Range =[-1,1] For every value of x, we get unique value of y. But the value of y in [-1,0) does not have any pre-image. ∴ Function is one-one but not onto.

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

R and R

$[0, π]$ and [0,1]

$[0, \frac{π}{2}]$ and $[- 1, 1]$

$[\frac{- π}{2}, \frac{π}{2}]$ and $[- 1, 1]$

Since, $f : X \to Y,$ then $f (x) = s in x$ Now, take option (c). Domain $= [0, \frac{π}{2}],$ Range $= [- 1, 1]$ For every value of $x,$ we get unique value of y. But the value of y in $[- 1, 0)$ does not have any pre-image. $∴$ Function is one-one but not onto.

Q. The function f:X→Y defined by f(x)=sinx is one-one but not onto, if X and Y are respectively equal to

R and R

[0,π] and [0,1]

[0,2π​] and [−1,1]

[2−π​,2π​] and [−1,1]

Since, f:X→Y, then f(x)=sinx Now, take option (c). Domain =[0,2π​], Range =[−1,1] For every value of x, we get unique value of y. But the value of y in [−1,0) does not have any pre-image. ∴ Function is one-one but not onto.

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

$[0, π]$ and [0,1]

$[0, \frac{π}{2}]$ and $[- 1, 1]$

$[\frac{- π}{2}, \frac{π}{2}]$ and $[- 1, 1]$

Since, $f : X \to Y,$ then $f (x) = s in x$ Now, take option (c). Domain $= [0, \frac{π}{2}],$ Range $= [- 1, 1]$ For every value of $x,$ we get unique value of y. But the value of y in $[- 1, 0)$ does not have any pre-image. $∴$ Function is one-one but not onto.