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NCERT Exemplar · Q66

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

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The tangent's slope equals the chord's slope exactly where the Mean Value Theorem promises a point — solving 2(x−3)=12(x-3)=1 gives the point (72,14)\left(\frac72,\frac14\right).

Setting Up

A tangent parallel to a chord means their slopes are equal. The chord's slope is the average rate of change of yy between the two given points; the tangent's slope at any xx is the derivative there. Finding where they match is exactly the geometric content of the Mean Value Theorem — for the smooth curve y=(x−3)2y=(x-3)^2 on [3,4][3,4], MVT guarantees at least one such interior point.

Step 1 — Slope of the chord

The chord joins (3,0)(3,0) and (4,1)(4,1), both on the curve (check: (3−3)2=0(3-3)^2=0, (4−3)2=1(4-3)^2=1):

mchord=1−04−3=1.m_{\text{chord}}=\frac{1-0}{4-3}=1.

Step 2 — Slope of the tangent

Differentiating y=(x−3)2y=(x-3)^2:

dydx=2(x−3).\frac{dy}{dx}=2(x-3). …

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