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				   a fxn is defined on[-1,1]andarea of sq. is 4.with its vertices at (0,0) and (x,f(x))then f(x) is...
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6 years ago

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										Ans:Area of Square: 4=>The sides of square are of 2 units each.Diagonal Length:$\sqrt{(f(x))^{2}+x^{2}} = 2\sqrt{2}$$(f(x))^{2}+x^{2}} = 8$$f(x) = \pm \sqrt{8-x^{2}}$Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago

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