2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    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]

    A) 7                                             

    B) \[-7\]   

    C) 5                                 

    D)   6

    Correct Answer: A

    Solution :

    Here \[f(x)={{x}^{3}}-3(a-2){{x}^{2}}+3ax+7\] \[\therefore \]\[f'(x)=3{{x}^{2}}-6(a-2)x+3a\] 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,\[f(x)={{x}^{3}}-9{{x}^{2}}+15x+7\] Now,\[\frac{f(x)-14}{{{(x-1)}^{2}}}=0\Rightarrow \frac{{{x}^{3}}-9{{x}^{2}}+15x-7}{{{(x-1)}^{2}}}=0\] \[\Rightarrow \]\[\frac{{{(x-1)}^{2}}(x-7)}{{{(x-1)}^{2}}}=0\]\[\Rightarrow \]\[x=7\]


You need to login to perform this action.
You will be redirected in 3 sec spinner