2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    If \[f\left( \frac{x-4}{x+2} \right)=2x+1,(x\in R=\{1,-2\})\], \[\int_{{}}^{{}}{f(x)dx}\] is equal to (where C is a constant of integration)                                                              [JEE Online 15-04-2018]

    A) \[12{{\log }_{e}}|1-x|-3x+c\]

    B) \[-12{{\log }_{e}}|1-x|-3x+c\]

    C) \[-12{{\log }_{e}}|1-x|+3x+c\]

    D) \[12{{\log }_{e}}|1-x|+3x+c\]

    Correct Answer: A

    Solution :

                Let \[\frac{x-4}{x+2}=y\Rightarrow x-4=yx+2y\Rightarrow x(1-y)=2y+4\Rightarrow x=\frac{2y+4}{1-y}\] This gives us\[f(y)=2(\frac{2y+4}{1-y})+1\] So, we have \[f(x)=2(\frac{2x+4}{1-x})+1=\frac{3x+9}{1-x}=-3(\frac{x-1+4}{x-1})=-3-\frac{12}{x-1}\] Thus \[\int_{{}}^{{}}{f(x)dx=12{{\log }_{e}}|1-x|-3x+c}\] So, the correct answer is option A.


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