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Question 104 of 148

Q.The value of 'c' of Lagranges Mean value theorem for f(x)=xf(x) = \sqrt{x}, when a=1a = 1 and b=4b = 4 is :

(a) 12\dfrac{1}{2}
(b) 94\dfrac{9}{4}
(c) 14\dfrac{1}{4}
(d) 32\dfrac{3}{2}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2018MCQ· 1mImportance★★★★★
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Applying Lagrange's Mean Value Theorem to f(x)=xf(x)=\sqrt x on [1,4][1,4] and solving f′(c)f'(c) equal to the average rate of change gives c=94c=\dfrac94.

  1. LMVT states: for ff continuous on [a,b][a,b] and differentiable on (a,b)(a,b), there exists c∈(a,b)c\in(a,b) with f′(c)=f(b)−f(a)b−af'(c) = \dfrac{f(b)-f(a)}{b-a}.
  2. Here f(x)=xf(x)=\sqrt x, a=1a=1, b=4b=4. Compute f(1)=1=1f(1)=\sqrt1=1 and f(4)=4=2f(4)=\sqrt4=2.
  3. The average rate of change is f(4)−f(1)4−1=2−13=13\dfrac{f(4)-f(1)}{4-1} = \dfrac{2-1}{3} = \dfrac13. …

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