The area bounded by the lines y = 2, x = 1, x = a and the curve y = f(x), which cuts the last two lines above the first line for all a≥1, is equal to 23((2a)3/2−3a+3−2√2) Then f(x) =
A
2√2x;x≥1
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B
√2x;x≥1
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C
2√x;x≥1
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D
None of these
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Solution
The correct option is A2√2x;x≥1 Required Area =∫a1[f(x)−2]dx=23[(2a3/2)−3a−3−2√2] Differentiate w.r.t. 'a', ⇒f(a)−2=23[32√2a×2−3] ⇒f(a)=2√2a;q≥1 or f(x)=2√2x;x≥1