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  • For the function f(x)= x+1/x the value of c for mean value theorem is A. 1

For the function \mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}, \mathrm{x} \in[1,3]  the value of c for mean value theorem is
A. 1

B. \sqrt3
C. 2
D. None of these
 

Answers (1)

B)

Mean Value Theorem states that, Let f : [a, b] → R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that

\\ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \\\text { We have, } f(x)=x+\frac{1}{x}

Since, f(x) is a polynomial function it is continuous on [1,3] and differentiable on (1,3).

Now, as per Mean value Theorem, there exists at least one c ∈ (1,3), such that 

\\ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \Rightarrow 1-\frac{1}{c^{2}}=\frac{\left(3+\frac{1}{3}\right)-(1+1)}{3-1}\left[\because f^{\prime}(x)=1+\frac{1}{x^{2}}\right] \\ \Rightarrow \quad \frac{c^{2}-1}{c^{2}}=\frac{\left(\frac{10}{3}\right)-(2)}{2} \\ \Rightarrow \quad \frac{c^{2}-1}{c^{2}}=\frac{\left(\frac{10}{3}\right)-(2)}{2}=\frac{\frac{10-6}{3}}{2}=\frac{2}{3} \\ \Rightarrow 3\left(c^{2}-1\right)=2 c^{2} \\ \Rightarrow 3 c^{2}-2 c^{2}=3 \\ \Rightarrow c^{2}=3 \\ \Rightarrow c=\pm \sqrt{3} \\ \Rightarrow c=\sqrt{3} \in(1,3)

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