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Question

The tangent to the parabola y=x2 has been drawn so that the abscissa x0 of the point of tangency belong to the interval [1,2]. The x0 for which the triangle bounded by the tangent, the axis of ordinates and the straight line y=x20 has the greatest area is

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Solution

y=x2dydx=2x
Equation of the tangent at (x0,x20) is yx20=2x0(xx0).
It meets yaxis in R(0,x20),Q is(0,x20)
Z= area of the triangle PQR
=122x20x0=x30,1x02
dZdx0=3x20 in 1x02
Z is an increasing function in [1,2].
Hence, Z, i.e., the area of ΔPQR is greatest at x0=2.

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