Continuity and Differentiability — Class 12 Maths Solution

We have, $y = {(x - 3)^2},$ which is continuous in ${x_1} = 3$ and ${x_2} = 4$ i.e., [3,4].

Also, ${y^\prime } = 2(x - 3) \cdot 1 = 2(x - 3)$ which exists in (3,4).

Hence, by mean value theorem there exists a point on the curve at which tangent drawn is parallel to the chord joining the points (3,0) and (4,1) .

Thus, $f'(c) = \frac{{f(4) - f(3)}}{{4 - 3}}$
$\Rightarrow$ $2(c - 3) = \frac{{{{(4 - 3)}^2} - {{(3 - 3)}^2}}}{{4 - 3}}$

$\Rightarrow$ $2c - 6 = \frac{{1 - 0}}{1} \Rightarrow c = \frac{7}{2}$

For $x = \frac{7}{2},$ $y = {\left( {\frac{7}{2} - 3} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}$

So, $\left( {\frac{7}{2},\frac{1}{4}} \right)$ is the point on the curve at which tangent drawn is parallel to the chord joining the points (3,0) and (4,1).