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  • Explain solution for RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xvi) math

Explain solution for  RD Sharma class 12 chapter Mean value theorem exercise 14.1 question 3 sub question (xvi) math

Answers (1)

Answer:

               c = 2.5 \in (1,4)

Hint:

               f(1) = f(4) , so there exists atc \in (1,4)

Given:

               f(x)=x^{2}-5 x+4 on[1,4]

Explanation:

f(x)=x^{2}-5 x+4 on[1,4]

Since given function f(x) is a polynomial, it is continuous and differentiable everywhere i.e. on R.

Let us find the values at extreme            

f(1)=1^{2}-5(1)+4 \\\\ \therefore \quad f(1)=1-5+4=-4+4 \\\\ \therefore \quad f(1)=0 \\\\ f(4)=4^{2}-5(4)+4 \\\\ f(4)=16-20+4=-4+4 \\\\ \therefore \quad f(4)=0

We havef(1)=f(4), so there exists atc \in (1,4) , such thatf' (c) = 0

Let’s find the derivative off (x)

\begin{array}{l} \Rightarrow \quad f^{\prime}(x)=\frac{d\left(x^{2}-5 x+4\right)}{d x} \\\\ \Rightarrow \quad f^{\prime}(x)=2 x-5 \ \ \ \ \ \ \ {\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]} \end{array}

We havef' (c) = 0

\\ \Rightarrow 2c- 5 = 0 \\\\ \Rightarrow 2c = 5 \\\\ \Rightarrow c = \frac{5}{2} \\\\ \Rightarrow c = 2.5 \in (1,4)

Hence Theorem is verified.

Posted by

infoexpert22

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