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Question 120 of 148

Q.Find the value in the interval (12,2)\left(\dfrac{1}{2}, 2\right) satisfied by the Rolle's theorem for the function f(x)=x+1x,x∈[12,2]f(x)=x+\dfrac{1}{x}, x\in\left[\dfrac{1}{2}, 2\right].

Puducherry TnboardTamil Nadu HSC (DGE) Board 2020Subjective· 2mImportance★★★★★
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Verifies Rolle's theorem's hypotheses, then solves f′(x)=0f'(x)=0 and picks the root inside the given open interval.

  1. f(x)=x+1xf(x)=x+\dfrac1x is a rational function, so it is continuous on [12,2]\left[\tfrac12,2\right] and differentiable on (12,2)\left(\tfrac12,2\right) (it has no discontinuity in this interval, since x≠0x\neq0 there).
  2. Check the equal end-value condition: f(12)=12+2=52f\left(\tfrac12\right)=\tfrac12+2=\tfrac52 and f(2)=2+12=52f(2)=2+\tfrac12=\tfrac52, so f(12)=f(2)f\left(\tfrac12\right)=f(2).
  3. All hypotheses of Rolle's theorem hold, so there exists c∈(12,2)c\in\left(\tfrac12,2\right) with f′(c)=0f'(c)=0. …

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