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Let `C_(1) and C_(2)` be the graphs of the functions `y=x^(2) and y=2x,` respectively, where `0le x le 1." Let "C_(3)` be the graph of a function y=f(x), where `0lexle1, f(0)=0.` For a point P on `C_(1),` let the lines through P, parallel to the axes, meet `C_(2) and C_(3)` at Q and R, respectively (see figure). If for every position of `P(on C_(1)),` the areas of the shaded regions OPQ and ORP are equal, determine the function f(x). |
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Answer» Let P be on `C_(1),y=x^(2) be (t,t^(2))` `therefore" y co-ordinate of Q is also "t^(2)` `"Now, Q on y =2x where "y=t^(2)` `therefore" "x=t^(2)//2` `therefore" "Q((t^(2))/(2),t^(2))` For point R, x=t and it is on y=f(x) `therefore" "R(t,f(t))` Given that, Area OPQ = Area OPR `rArr" "int_(0)^(t^(2))(sqrt(y)-(y)/(2))dy=int_(0)^(t)(x^(2)-f(x))dx` Differentiating both sides w.r.t. t, we get `(sqrt(t^(2))-(t^(2))/(2))(2t)=t^(2)-f(t)` `rArr" "f(t)=t^(3)-t^(2)` `rArr" "f(x)=x^(3)-x^(2)` |
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