First slide

Mean value theorems

Question

$A point on the curve y = (x - 3)^{2}, where the tangent is parallel to the chord joining the points (3, 0) and (4, 1)$

Moderate

A

$(\frac{7}{2}, \frac{2}{3})$
B

$(\frac{7}{2}, \frac{3}{2})$
C

$(\frac{7}{2}, \frac{3}{5})$
D

$(\frac{7}{2}, \frac{1}{4})$

Solution

$we have, y = (x - 3)^{2}, which is polynomial function. So it is continuous and$
differentiable. Thus conditions of mean value theorem are satisfied. Hence, there exists at least one $c \in (3, 4)$ such that,
$\begin{aligned} f^{'} (c) = \frac{f (4) - f (3)}{4 - 3} \\ \Rightarrow 2 (c - 3) = \frac{1 - 0}{1} \\ \Rightarrow c - 3 = \frac{1}{2} \\ \Rightarrow c = \frac{7}{2} \in (3, 4) \end{aligned}$
$\begin{array}{l} \Rightarrow x = \frac{7}{2}, where tangent is parallel to the chord joining points (3, 0) and (4, 1) \\ For x = \frac{7}{2}, y = {(\frac{7}{2} - 3)}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4} \end{array}$
$So, (\frac{7}{2}, \frac{1}{4}) is the point on the curve, where tangent drawn is parallel to the chord joining the$ $points (3, 0) and (4, 1)$

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