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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)=sinxf\left( x \right)=\sin x is one-one [0,π]\left[ 0,\pi \right]

Explanation

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\sin x is cosx\cos x
The function cosx\cos x 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\sin x is one-one when angle xx is in the interval[0,π]\left[ 0,\pi \right].
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)=sinxf\left( x \right)=\sin x with respect to x, we get:
f(x)=d(f(x))dxf'\left( x \right)=\dfrac{d\left( f\left( x \right) \right)}{dx}
Here f(x)f\left( x \right) is sinx\sin x, on differentiating we get:
d(sinx)dx\Rightarrow \dfrac{d\left( \sin x \right)}{dx}
The differentiation of the function sinx\sin x is cosx\cos x , so putting the value we get:
cosx\Rightarrow \cos x
Now for f(x)f\left( x \right) to be one-one, f(x)f\left( x \right) has to be strictly increasing or increasing, which means or mathematically would be written as f(x)0f'\left( x \right)\ge 0 or f(x)>0f'\left( x \right)>0 respectively.
We have got f(x)=cosxf'\left( x \right)=\cos x , for function to be one-one value of cosxcos x should be greater than equal to zero, which would be written as:
cosx0\cos x\ge 0
So the value of cosx\cos x will be positive when the value of xxin the interval[0,π2]\left[ 0,\dfrac{\pi }{2} \right].
\therefore The interval for which f(x)=sinxf\left( x \right)=\sin x is one-one [0,π]\left[ 0,\pi \right] is [0,π2]\left[ 0,\dfrac{\pi }{2} \right].

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.