Applications Of Derivatives Question 235
Question: A function $ y=f( x ) $ has a second order derivative $ f’’( x )=6( x-1 ) $ . If its graph passes through the point (2, 1) and at that point the tangent to the graph is $ y=3x-5, $ then the function is
Options:
A) $ {{(x-1)}^{2}} $
B) $ {{(x-1)}^{3}} $
C) $ {{(x+1)}^{3}} $
D) $ {{(x+1)}^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ f’’(x)=6(x-1) $
$ \therefore f’(x)=3{{(x-1)}^{2}}+c $ …(i)
At the point (2, 1), the tangent to the graph is $ y=3x-5 $ The slope of the tangent is 3,
$ \therefore f’(2)=3{{(2-1)}^{2}}+c=3. $
$ \Rightarrow 3+c=3\Rightarrow c=0 $ Therefore, from (i), we get $ f’(x)=3{{(x-1)}^{2}} $ …(ii)
$ \Rightarrow f(x)={{(x-1)}^{3}}+k $ Since the graph passes through (2, 1), $ 1={{(2-1)}^{2}}+k $
$ \Rightarrow k=0 $
Hence, the equation of the function is $ f(x)={{(x-1)}^{3}} $
BETA
