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Question

Find a point on the curve y = x2 + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2).

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Solution

​Let:
fx=x2+x

The tangent to the curve is parallel to the chord joining the points 0, 0 and 1, 2.

Assume that the chord joins the points a, fa and b, fb.

a=0, b=1

The polynomial function is everywhere continuous and differentiable.

So, fx=x2+x is continuous on 0, 1 and differentiable on 0, 1.

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists c0, 1 such that f'c=f1-f01-0.

Now,
fx=x2+xf'x=2x+1, f1=2, f0=0

f'x=f1-f01-02x+1=2-01-02x=1x=12

Thus, c=120,1 such that ​f'c=f1-f01-0.

Clearly,
fc=122+12=34.

Thus, c, fc, i.e.​ 12, 34, is a point on the given curve where the tangent is parallel to the chord joining the points (4, 0) and (5, 1).

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