Applications Of Derivatives Question 408
Question: The function $ f(x)={{(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
Options:
A) $ ( \frac{7}{2},\frac{1}{2} ) $
B) $ ( \frac{7}{2},\frac{1}{4} ) $
C) (1, 4)
D) (4, 1)
Show Answer
Answer:
Correct Answer: B
Solution:
Let the point be $ (x_1,,y_1). $
Therefore $ y_1={{(x_1-3)}^{2}} $ …………..(i)
Now slope of the tangent at $ (x_1,,y_1) $ is $ 2(x_1-3), $ but it is equal to 1.
Therefore, $ 2(x_1-3)=1\Rightarrow x_1=\frac{7}{2} $
$ \therefore y_1={{( \frac{7}{2}-3 )}^{2}}=\frac{1}{4} $ .
Hence the point is $ ( \frac{7}{2},\frac{1}{4} ) $ .
BETA
