question_answer

Set of all real values of x satisfying the in equation\[\frac{{{\log }_{2}}({{x}^{2}}-5x+4)}{{{\log }_{2}}({{x}^{2}}+1)}>1\] is

A)  \[\left( -\infty ,\frac{3}{5} \right)-\{0\}\]             

B)  \[(-\infty ,1)-\{0\}\]

C)  \[\left( \frac{3}{5},\infty  \right)\]                          

D)  \[\left( 4,\infty  \right)\]

Correct Answer: A

Solution :

 Idea Students are advised to stick with definition of the function, to solve this question. \[\therefore \]Here \[f(x)={{\log }_{a}}x\] is defined when a > 0 and x > 0. Here, it is given that\[\frac{{{\log }_{2}}({{x}^{2}}-5x+4)}{{{\log }_{2}}({{x}^{2}}+1)}>1\] \[{{x}^{2}}-5x+4>0\Rightarrow (x-4)(x-1)>0\] \[\Rightarrow \]\[x\in (=\infty ,1)\cup (4,\infty )\]                                           ?(i) \[{{x}^{2}}+1>0\]which is true\[\forall x\in R\]                   ?(ii) \[{{\log }_{2}}({{x}^{2}}-5x+4)>{{\log }_{2}}({{x}^{2}}+1)\] \[{{x}^{2}}-5x+4>{{x}^{2}}+1\] \[-5x+4>{{x}^{2}}+1\] \[-5x+3>0\] \[x<\frac{3}{5}\]                                                                               ..(iii) From Eqs. (i), (ii) and (iii), we get\[x\in \left( -\infty ,\frac{3}{4} \right)-\{0\}\] TEST Edge Generally, in JEE Main, domain and range related question are asked from this concept. Students are advised to understand basic concept of the function given and also acquainted yourself with the concept of wavy curve method, i.e., Let \[(x-1){{(x-2)}^{2}}(x-3)\ge 0\]Then x= 1, 2, 3 Solutions \[(-,\infty ]\cup [3,\infty )\cup \{2\}\]