IIT-JEE 2011 · Paper 1 · Mathematics Q51

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Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2$ , $y=0$ , and $x=0$ into two parts $R_1$ $(0\le x\le b)$ and $R_2$ $(b\le x\le1)$ such that $R_1-R_2=\frac14$ . Then $b$ equals

\dfrac34

\dfrac12

\dfrac13

\dfrac14

Answer and worked solution

Official answer: B

Areas. $R_2=\displaystyle\int_b^1(1-x)^2dx=\frac{(1-b)^3}3$ and $R_1=\frac13-R_2$ .

Condition. $R_1-R_2=\frac{1-2(1-b)^3}3=\frac14$ gives $(1-b)^3=\frac18$ , $b=\frac12$ .

(B)

Topics:Area under a curve

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IIT-JEE 2011 · Paper 1, Mathematics question 51