Which one of the following functions is one-to-one?

Question

Which one of the following functions is one-to-one? (A)  f(x)= sin x,x∈ [-π ,π ]  (B)  f(x)= sin x,x∈ [ -(3π /2),-(π /4) ]  (C)  f(x)= cos x,x∈ [ -(π /2),(π /2) ]  (D)  f(x)= cos x,x∈ [π ,2π )

Tardigrade Admin · Accepted Answer

In the given options (a), (b), (c), (e) the curves are decreasing and increasing in the given intervals, so it is not one-to-one function. But in option (d), the curve is only increasing in the given intervals, so it is one-to-one function.

Q. Which one of the following functions is one-to-one?

$f (x) = sin x, x \in [- π, π]$

$f (x) = sin x, x \in [- \frac{3 π}{2}, - \frac{π}{4}]$

$f (x) = cos x, x \in [- \frac{π}{2}, \frac{π}{2}]$

$f (x) = cos x, x \in [π, 2 π)$

$f (x) = cos x, x \in [- \frac{π}{4}, \frac{π}{4}]$

In the given options (a), (b), (c), (e) the curves are decreasing and increasing in the given intervals, so it is not one-to-one function. But in option (d), the curve is only increasing in the given intervals, so it is one-to-one function.

Q. Which one of the following functions is one-to-one?

f(x)=sinx,x∈[−π,π]

f(x)=sinx,x∈[−23π​,−4π​]

f(x)=cosx,x∈[−2π​,2π​]

f(x)=cosx,x∈[π,2π)

f(x)=cosx,x∈[−4π​,4π​]

In the given options (a), (b), (c), (e) the curves are decreasing and increasing in the given intervals, so it is not one-to-one function. But in option (d), the curve is only increasing in the given intervals, so it is one-to-one function.

$f (x) = sin x, x \in [- π, π]$

$f (x) = sin x, x \in [- \frac{3 π}{2}, - \frac{π}{4}]$

$f (x) = cos x, x \in [- \frac{π}{2}, \frac{π}{2}]$

$f (x) = cos x, x \in [π, 2 π)$

$f (x) = cos x, x \in [- \frac{π}{4}, \frac{π}{4}]$