JEE Main 12 Jan 2019 Evening Question 17
Question: If the function f given by $ f(x)=x^{3}-3(a-2)x^{2}+3ax+7, $ for some $ a\in R $ is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, $ \frac{f(x)-14}{{{(x-1)}^{2}}}=0(x\ne 1) $ is [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 7
B) $ -7 $
C) 5
D) 6
Show Answer
Answer:
Correct Answer: A
Solution:
- Here $ f(x)=x^{3}-3(a-2)x^{2}+3ax+7 $
$ \therefore $ $ f’(x)=3x^{2}-6(a-2)x+3a $
Since $ f’(x)\ge 0\forall x\in (0,1] $ and $ f’(x)\le 0\forall x\in [1,5) $
$ \therefore $ $ f’(x)=0 $ at $ x=1 $
$ \Rightarrow $ $ 3-6(a-2)+3a=0\Rightarrow -3a+15=0 $
$ \Rightarrow $ $ a=5 $ So, $ f(x)=x^{3}-9x^{2}+15x+7 $
Now, $ \frac{f(x)-14}{{{(x-1)}^{2}}}=0\Rightarrow \frac{x^{3}-9x^{2}+15x-7}{{{(x-1)}^{2}}}=0 $
$ \Rightarrow $ $ \frac{{{(x-1)}^{2}}(x-7)}{{{(x-1)}^{2}}}=0 $
$ \Rightarrow $ $ x=7 $
