
Find the interval for which $f\left( x \right)=\sin x$ is one-one $\left[ 0,\pi \right]$
Answer
519.9k+ views
Hint: 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 $\sin x$ is $\cos x$
The function $\cos x$ 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 $\sin x$ is one-one when angle $x$ is in the interval$\left[ 0,\pi \right]$.
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 $f\left( x \right)=\sin x$ with respect to x, we get:
$f'\left( x \right)=\dfrac{d\left( f\left( x \right) \right)}{dx}$
Here $f\left( x \right)$ is $\sin x$, on differentiating we get:
$\Rightarrow \dfrac{d\left( \sin x \right)}{dx}$
The differentiation of the function $\sin x$ is $\cos x$ , so putting the value we get:
$\Rightarrow \cos x$
Now for $f\left( x \right)$ to be one-one, $f\left( x \right)$ has to be strictly increasing or increasing, which means or mathematically would be written as $f'\left( x \right)\ge 0$ or $f'\left( x \right)>0$ respectively.
We have got $f'\left( x \right)=\cos x$ , for function to be one-one value of $cos x$ should be greater than equal to zero, which would be written as:
$\cos x\ge 0$
So the value of $\cos x$ will be positive when the value of $x$in the interval$\left[ 0,\dfrac{\pi }{2} \right]$.
$\therefore $The interval for which $f\left( x \right)=\sin x$ is one-one $\left[ 0,\pi \right]$ is $\left[ 0,\dfrac{\pi }{2} \right]$.
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.
The function $\cos x$ 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 $\sin x$ is one-one when angle $x$ is in the interval$\left[ 0,\pi \right]$.
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 $f\left( x \right)=\sin x$ with respect to x, we get:
$f'\left( x \right)=\dfrac{d\left( f\left( x \right) \right)}{dx}$
Here $f\left( x \right)$ is $\sin x$, on differentiating we get:
$\Rightarrow \dfrac{d\left( \sin x \right)}{dx}$
The differentiation of the function $\sin x$ is $\cos x$ , so putting the value we get:
$\Rightarrow \cos x$
Now for $f\left( x \right)$ to be one-one, $f\left( x \right)$ has to be strictly increasing or increasing, which means or mathematically would be written as $f'\left( x \right)\ge 0$ or $f'\left( x \right)>0$ respectively.
We have got $f'\left( x \right)=\cos x$ , for function to be one-one value of $cos x$ should be greater than equal to zero, which would be written as:
$\cos x\ge 0$
So the value of $\cos x$ will be positive when the value of $x$in the interval$\left[ 0,\dfrac{\pi }{2} \right]$.
$\therefore $The interval for which $f\left( x \right)=\sin x$ is one-one $\left[ 0,\pi \right]$ is $\left[ 0,\dfrac{\pi }{2} \right]$.
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.
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