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The area of a trapezium is defined by function f and given by f(x) = (10 + x)√(100 − x2), then the area when it is maximised is :(a) 75 cm2(b) 7√3 cm2(c) 75√3 cm2(d) 5 cm2 |
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Answer» Option : (c) \(f'(x)=\frac{-2x^2-10x+100}{\sqrt{100-x^2}}\) \(f'(x)=0\Rightarrow x=-10\,or\,5,\) If \(x=-10\) then area = \(f(x)=0\) which is not maximum Therefore, \( x=5\) Now, \(f''(x)=\frac{2x^3-300x-1000}{(100-x^2)^{\frac{3}{2}}}\) \(\Rightarrow f''(5)=\frac{-30}{\sqrt{75}}<0\) \(\Rightarrow\) Maximum area of trapezium is \(75\sqrt3cm^2\) when x = 5 |
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