Differential Equations Question 176
Question: A function $ y=f(x) $ has a second order derivatives $ {{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
[AIEEE 2004]
Options:
A) $ {{(x+1)}^{3}} $
B) $ {{(x-1)}^{3}} $
C) $ {{(x+1)}^{2}} $
D) $ {{(x-1)}^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given $ {{f}’}’(x)=6(x-1) $
$ {f}’(x)=3{{(x-1)}^{2}}+c_1 $
……..(i) But at point (2, 1) the line $ y=3x-5 $ is tangent to the graph $ y=f(x) $ . Hence $ {{. \frac{dy}{dx} |} _{x=2}}=3 $ or $ {f}’(2)=3 $ . Then from (i) $ {f}’(2)=3{{(2-1)}^{2}}+c_1 $
$ 3=3+c_1 $
Therefore $ c_1=0 $ i.e., $ {f}’(x)=3{{(x-1)}^{2}} $
Given $ f(2)=1 $
$ f(x)={{(x-1)}^{3}}+c_2 $
Therefore $ f(2)=1+c_2 $
Therefore $ 1=1+c_2 $
Therefore $ c_2=0 $
Hence $ f(x)={{(x-1)}^{3}} $ .
BETA
