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 first line for all `a gt= 1,` is equal to `2/3[(2a)^(3//2)-3a+3-2sqrt2],` then `f(x)=`A. `2sqrt(2x), x ge 1`B. `sqrt(2x), x ge 1`C. ` 2 sqrtx , x ge 1`D. None

The area bounded by the lines `y = 2, x =1, x = a` and the curve `y =

1.	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 first line for all `a gt= 1,` is equal to `2/3[(2a)^(3//2)-3a+3-2sqrt2],` then `f(x)=`A. `2sqrt(2x), x ge 1`B. `sqrt(2x), x ge 1`C. ` 2 sqrtx , x ge 1`D. None
Answer» Correct Answer - A

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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 first line for all `a gt= 1,` is equal to `2/3[(2a)^(3//2)-3a+3-2sqrt2],` then `f(x)=`A. `2sqrt(2x), x ge 1`B. `sqrt(2x), x ge 1`C. ` 2 sqrtx , x ge 1`D. None

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