A figure is bounded by the curve $y = x^{2} + 1,$ the axes of co-ordinates and the line x=1. Determine the co-ordinates of a point P at which a tangent be drawn to the curve so as to cut off a trapezium of greatest area from the figure.

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

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

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(- \frac{1}{2}, \frac{5}{2}) .

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(\frac{1}{2}, - \frac{5}{4}) .

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Solution

The correct option is B $(\frac{1}{2}, \frac{5}{4}) .$
The given equation $x^{2} = y - 1$ represents a parabola with vertex at $(0, 1)$
Let the co-ordinates of any point p on the curve $y = 1 + x^{2} b e (h, 1 + h^{2}) .$
Now $\frac{d y}{d x} = 2 x = 2 h$ Tangent at P is $y - (1 + h^{2}) = 2 h (x - h)$ or $y - 2 x h = 1 - h^{2}$
Substituting $x = 0$ , we get $y = (1 - h^{2}) O L$
Substituting $x = 1$ , we get $y = 1 + 2 h - h^{2}$
$∴ M i s (1, 1 + 2 h - h^{2}), N (1, 0)$
A=area of trapezium $= \frac{1}{2} (O L + M N) . O N$ $= \frac{1}{2} [1 - h^{2} + 12 h - h^{2}] .1$
$A = (1 + h - h^{2})$
$∴ \frac{d A}{d h} = 1 - 2 h = 0$
$∴ h = \frac{1}{2}$ and $\frac{d^{2} A}{d h^{2}} = - 2 = - i v e$ Hence max.
$∴$ Point P is $(\frac{1}{2} . \frac{5}{4}) .$
Ans: B

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