Find a point on the curvey y = (x - 3)2, where the tangent is parallel to the chord joining the points (3,0) and (4,1).

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asked Mar 28, 2018 in Class XII Maths by rahul152 (-2,842 points)

Find a point on the curvey y = (x - 3)2, where the tangent is parallel to the chord joining the points (3,0) and (4,1). 

 

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answered Mar 28, 2018 by nikita74 (-1,017 points)

We have, y = (x-3)2, which is polynomial function .
So it is continuous and differentiable .
Thus condition of mean value theorem are satisfied.
Hence, there exists atleast one c ∈ (3,4) such that , 

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