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Q.

The point on the graph of the curve y=(x−3)2, where the tangent is parallel to the line joining of (3,0), (4, 1) is

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a

72,14

b

(1, 4)

c

(2,1)

d

12,254

answer is A.

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Detailed Solution

dydx=1−04−3⇒2(x−3)=1⇒x−3=12⇒x=72⇒y=14
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