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Question

The tangent to the curve y=xex2 is passing through the point (1,e) and also passes through the point :

A
(2,3e)
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B
(3,6e)
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C
(43,2e)
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D
(53,2e)
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Solution

The correct option is C (43,2e)
Given curve, y=xex2
And its slope of tangent =dydx
dydx=x2xex2+ex2
=2x2ex2+ex2
=ex2(1+2x2)
At point (1,e), dydx=3e
For given point (1,e) equation of the tangent is
(ye)=3e(x1)
Hence, the point (43,2e) lies on the above line.

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