JEE Main & Advanced JEE Main Paper Phase-I Held on 07-1-2020 Morning

  • question_answer
    If \[g(x)={{x}^{2}}+x-1\] and \[(gof)(x)=4{{x}^{2}}-10x+5,\]then \[f\left( \frac{5}{4} \right)\] is equal to: [JEE MAIN Held on 07-01-2020 Morning]

    A) \[-\frac{1}{2}\] 

    B) \[\frac{3}{2}\]

    C) \[\frac{1}{2}\]              

    D) \[-\frac{3}{2}\]

    Correct Answer: A

    Solution :

    [a]        \[g\left( f\left( \frac{5}{4} \right) \right)=4{{\left( \frac{5}{4} \right)}^{2}}-10\left( \frac{5}{4} \right)+5=\frac{-5}{4}\] Now,  \[g\left( f\left( \frac{5}{4} \right) \right)={{f}^{2}}\left( \frac{5}{4} \right)+f\left( \frac{5}{4} \right)-1\] Let \[f\left( \frac{5}{4} \right)=t\] \[\Rightarrow {{t}^{2}}+t-1=\frac{-5}{4}\] \[{{t}^{2}}+t+\frac{1}{4}=0\] \[\therefore {{\left( t+\frac{1}{2} \right)}^{2}}=0\] i.e., \[f\left( \frac{5}{4} \right)=\frac{-1}{2}\]      


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