Question

Using mean value theorem,prove that there is a point on the curve y=2x2−5x+3 between the points A(1,0) and B(2,1)where tangent is parallel to chord AB. Also find the point.

Answer 1

sorry but I didn't understood the question.

Answer 2

The point is (1.5, 0)

Proof:

The given curve is

$y=2x^2-5x+3$

or, $f(x)=2x^2-5x+3$

The slope of the curve is given by

$\frac{dy}{dx}$

or, $\frac{dy}{dx}=4x-5$

or, $f'(x)=4x-5$

Now, by mean value theorem, there exists a point between (1, 0) and (2, 1) where

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Thus

$f'(c)=\frac{f(2)-f(1)}{2-1}$

$\implies f'(c)=\frac{(2\times 2^2-5\times 2+3)-(2\times 1^2-5\times 1+3)}{(2-1)}$

$\implies f'(c)=(8-10+3)-(2-5+3)$

$\implies f'(c)=1$

That means at point x = c, the slope must be 1

Thus

$f'(x)=1$

or, $4x-5=1$

$\implies x=\frac{3}{2}$

$\implies x=1.5$

Clearly x = 1.5 lies between x = 1 and x = 2

At x = 1.5

$y=2\times (1.5)^2-5\times 1.5+3$

$\implies y=0$

Thus the point is (1.5, 0) which lies between (1, 0) and (2, 1)[/tex]

Hope this answer is helpful.

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