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Question
Find a point on the curve y = (x – 3)2, where the tangent is parallel to the chord joining the points (3, 0) and (4, 1)
Solution
We have, y = (x – 3)2, which is polynomial function.
So it is continuous and differentiable.
Thus conditions of mean value theorem are satisfied.
Hence, there exists atleast one c ∈ (3, 4) such that,
f'(c) = `("f"(4) - "f"(3))/(4 - 3)`
⇒ 2(c – 3) = `(1 - 0)/1`
⇒ c – 3 = `1/2`
⇒ c = `7/2 ∈ (3, 4)`
⇒ x = `7/2`, where tangent is parallel to the chord joining points (3, 0) and (4, 1).
For x = `7/2`, y = `(7/2 - 3)^2`
= `(1/2)^2`
= `1/4`
So, `(7/2, 1/4)` is the point on the curve, where tangent drawn is parallel to the chord joining the points (3, 0) and (4, 1).
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