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The point $P$ on the curve $y = x^{2} + 1$ such that the tangent at $P$ , $y = 0, x = 0$ and x=1 form a trapezium of the greatest area is

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$(\frac{1}{2}, \frac{5}{4})$

$(\frac{- 1}{2}, 1)$

$(1, 2)$

$(- 1, 2)$

answer is A.

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Detailed Solution

$P (t, t^{2} + 1)$

Tangent at $P$ is $y = 2 t x + 1 - t^{2}$
Area of trapezium = $\frac{1}{2} | 2 + 2 t - 2 t^{2} | = | 1 + t - t^{2} |$
Max area if $t = \frac{1}{2}$
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