Question
Question: Find the interval for which \(f\left( x \right)=\sin x\) is one-one \(\left[ 0,\pi \right]\)...
Find the interval for which f(x)=sinx is one-one [0,π]
Solution
To solve this question we need to know the concept of differentiation. A function is said to be one-one when that function maps distinct elements of its domain to distinct elements of its codomain. This type of function is also called an injective function. Differentiation of sinx is cosx
The function cosx is positive only in the first and fourth quadrant.
Complete step by step solution:
The question asks us to find the interval in which the function sinx is one-one when angle x is in the interval[0,π].
The first part is to differentiate the function so as to find the interval in which the function comes to be one-one. On differentiating the function f(x)=sinx with respect to x, we get:
f′(x)=dxd(f(x))
Here f(x) is sinx, on differentiating we get:
⇒dxd(sinx)
The differentiation of the function sinx is cosx , so putting the value we get:
⇒cosx
Now for f(x) to be one-one, f(x) has to be strictly increasing or increasing, which means or mathematically would be written as f′(x)≥0 or f′(x)>0 respectively.
We have got f′(x)=cosx , for function to be one-one value of cosx should be greater than equal to zero, which would be written as:
cosx≥0
So the value of cosx will be positive when the value of xin the interval[0,2π].
∴The interval for which f(x)=sinx is one-one [0,π] is [0,2π].
Note: We need to know the formula of differentiation to solve the question. There are many applications of derivatives and finding the function to be one-one is one of them. We should know where the trigonometric function will give the positive or negative value. Always keep in mind that in a one-one function the answer never repeats for any value in a certain interval.