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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
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 the chord $AB$. Also, find that point.
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Answer

We have, $y = 2x^2 - 5x + 3 $which is continuous in [1, 2] as it is a polynomial function.
Also, y’ = 4x - 5, which exists in (1, 2).
By mean value theorem, $\exists\text{ c}\in(1,2)$ at which drawn tangent is parallel to the chord AB,
where A and B are (1, 0) and (2, 1), respectively.
$\therefore\ \text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow\ 4\text{c}-5=\frac{(8-10+3)-(2-5+3)}{1}$
$\Rightarrow\ 4\text{c}-5=1$
$\Rightarrow\ \text{c}=\frac{6}{4}=\frac{3}{2}\in(1,2)$
For $\text{x}=\frac{3}{2},\ \text{y}=2\Big(\frac{3}{2}\Big)^2-5\Big(\frac{3}{2}\Big)+3$
$=2\times\frac{9}{4}-\frac{15}{2}+3=\frac{9-15+6}{2}=0$
Hence, $\Big(\frac{3}{2},0\Big)$ is the point on the curve $y = 2x^2 - 5x + 3 $between the points
A(1, 0) and B(2, 1) where tangent is parallel to the chord AB.

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