Continuity and Differentiability — Class 12 Maths Solution

We have, $y = 2{x^2} - 5x + 3,$ which is continuous in [1,2] as it is a polynomial function.

Also, ${y^\prime } = 4x - 5,$ which exists in (1,2).

By mean value theorem, $\exists c \in (1,2)$ at which drawn tangent is parallel to the chord AB, where $A$ and $B$ are (1,0) and (2,1) , respectively.

therefore,${f^\prime }(c) = \frac{{f(2) - f(1)}}{{2 - 1}}$

$\Rightarrow$ $4c - 5 = \frac{{(8 - 10 + 3) - (2 - 5 + 3)}}{1}$

$\Rightarrow$ $4c - 5 = 1$

therefore,$c = \frac{6}{4} = \frac{3}{2} \in (1,2)$

For $x = \frac{3}{2},$

$y = 2{\left( {\frac{3}{2}} \right)^2} - 5\left( {\frac{3}{2}} \right) + 3$

$= 2 \times \frac{9}{4} - \frac{{15}}{2} + 3 = \frac{{9 - 15 + 6}}{2} = 0$

Hence, $\left( {\frac{3}{2},0} \right)$ is the point on the curve $y = 2{x^2} - 5x + 3$ between the points $A(1,0)$ and $B(2,1),$ where tangent is parallel to the chord AB.

Long Answer Questions(L.A.)