Question

The function $f\left(x\right)=(x-3{)}^2$ satisfies all the conditions of mean value theorem in [3, 4]. A point on $y=(x-3{)}^2,$ where the tangent is parallel to the chord joining (3, 0) and (4, 1) is

• A. $(\frac{7}{2},\frac{1}{2})$
• B. $(\frac{7}{2},\frac{1}{4})$
• C. (1, 4)
• D. (4, 1)

Answer 1

The correct option is B $(\frac{7}{2},\frac{1}{4})$

Slope of the chord joining (3, 0) and (4, 1) = $\frac{1-0}{4-3}$ = 1
Slope of the tangent at any point on the given curve -
f'(x) = 2.(x-3)
So, 2(x-3) = 1
or x = $\frac{7}{2}$
Therefore the point will be = (7/2, 1/4)