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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