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Question: How do I use the Mean Value Theorem so \(2x-1-\sin \left( x \right)=0\) has exactly one real root?...

How do I use the Mean Value Theorem so 2x−1−sin⁡(x)=02x-1-\sin \left( x \right)=0 has exactly one real root?

Explanation

Solution

In this question we have been asked to show that for the expression 2x−1−sin⁡(x)=02x-1-\sin \left( x \right)=0 there exists only one root. We will first use the intermediate value theorem to show that there exists a root of the given expression by considering a range of two numbers. We will then try to find two roots using Rolle’s theorem and prove by contradiction that only one root exists.

Complete step-by-step solution:
We have the expression given to us as:
⇒2x−1−sin⁡(x)=0\Rightarrow 2x-1-\sin \left( x \right)=0
We will consider the function to be f(x)f\left( x \right) therefore, it can be written as:
⇒f(x)=2x−1−sin⁡(x)=0\Rightarrow f\left( x \right)=2x-1-\sin \left( x \right)=0
Now consider a range [0,1][0,1] for the function.
We will substitute the values in the expression.
On substituting the value 00 in the expression, we get:
⇒f(0)=2(0)−1−sin⁡(0)\Rightarrow f\left( 0 \right)=2\left( 0 \right)-1-\sin \left( 0 \right)
On simplifying, we get:
⇒f(0)=−1<0\Rightarrow f\left( 0 \right)=-1<0
On substituting the value 11 in the expression, we get:
⇒f(1)=2(1)−1−sin⁡(1)\Rightarrow f\left( 1 \right)=2\left( 1 \right)-1-\sin \left( 1 \right)
On simplifying, we get:
⇒f(1)=1−sin⁡(1)\Rightarrow f\left( 1 \right)=1-\sin \left( 1 \right)
Since sin⁡(1)\sin \left( 1 \right) will have a value lesser than 11, because sin⁡(π2)=1\sin \left( \dfrac{\pi }{2} \right)=1, we get:
⇒f(1)=1−sin⁡(1)>0\Rightarrow f\left( 1 \right)=1-\sin \left( 1 \right)>0
Therefore, by the intermediate value theorem there exists a number cc, such that f(c)=0f\left( c \right)=0.
Now we need to prove that there is only one root. We will prove it by contradiction therefore, we will consider two roots.
Consider aa and bb such that f(a)=0f\left( a \right)=0 and f(b)=0f\left( b \right)=0.
By Rolle’s theorem there exists a cc such that f′(c)=0f'\left( c \right)=0
On differentiating f(x)f\left( x \right), we get:
⇒f′(x)=2−cos⁡x\Rightarrow f'\left( x \right)=2-\cos x
Now from the derivative we can see that f′(x)f'\left( x \right) will always be positive since cos⁡x\cos x has a range of −1-1 to 11.
Therefore, we can write:
⇒f′(x)>0\Rightarrow f'\left( x \right)>0, for all xx.
So, by contradiction there are not two real roots aa and bb, but since we proved by the intermediate value theorem there exists a root, there exists only one root, hence proved.

Note: It is to be remembered when doing any proof when there is something to be proven true, every true case has to be shown which can be done by using algebra or limits. But in the case of proving something false, only one case has to be shown, which is the main reason why proof by contradiction is so commonly used.