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

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Toolbox:
  • Let $f(x)$ be a function defined on $[a,b]$ such that
  • (i) it is continuous on $[a,b]$
  • (ii) it is differentiable on $(a,b)$
  • Then,there exists $c\in (a,b)$ such that $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
$y=2x^2-5x+3$------(1)
The given points are $A(1,0)$ and $B(2,1)$.
Let $f(x)=y$
$f(x)=2x^2-5x+3,x\in (1,2)$
Being a polynomial $f(x)$ is continuous at $[1,2]$
$f'(x)=4x-5$,which exists in $(2,1)$
Hence it differentiable at $(2,1)$
Hence both the conditions of Lagrange's Mean value theorem is verified.
Step 2:
$f'(c)=4c-5$
$f(2)=2(2)^2-5(2)+3=1$
$f(1)=2(1)^2-5(1)+3=0$
Therefore $4c-5=\large\frac{1-0}{2-1}$
$\Rightarrow 4c-5=1$
$\Rightarrow 4c=6$
$c=\large\frac{6}{4}=\frac{3}{2}$
Since the value of $c$ is the $x$-coordinate of that point on which the tangent is parallel to the chord joining $A(1,0)$ and $B(2,1)$
$x=\large\frac{3}{2}$
Substituting for $x$ in equ(1) we get,
$y=2\big(\large\frac{3}{2}\big)^2-$$5\big(\large\frac{3}{2}\big)$$+3=0$
Hence the required point is $\big(\large\frac{3}{2},$$0\big)$
answered Jul 3, 2013 by sreemathi.v
 

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