2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    If \[\alpha ,\beta \]and \[\gamma \]are three consecutive terms of a non-constant G.P. such that the equations \[\alpha {{x}^{2}}+2\beta x+\gamma =0\]and \[{{x}^{2}}+x1=0\] have a common root, then \[\alpha (\beta +\gamma )\]is equal to: [JEE Main 12-4-2019 Afternoon]

    A) \[\beta \gamma \]        

    B) \[0\]

    C) \[\alpha \gamma \]                              

    D) \[\alpha \beta \]

    Correct Answer: A

    Solution :

    \[\alpha {{x}^{2}}+2\beta x+\gamma =0\]             Let\[\beta =\alpha t,\gamma =\alpha {{t}^{2}}\]             \[\therefore \]\[\alpha {{x}^{2}}+2\alpha tx+\alpha {{t}^{2}}=0\]             \[\Rightarrow {{x}^{2}}+2tx+{{t}^{2}}=0\]             \[\Rightarrow {{(x+t)}^{2}}=0\]             \[\Rightarrow x=-t\] it must be root of equation \[{{x}^{2}}+x1=0\] \[\therefore {{t}^{2}}-t-1=0\]                                                  (1) Now \[\alpha (\beta +\gamma )={{\alpha }^{2}}(t+{{t}^{2}})\] Option \[\beta \gamma =\alpha t.\alpha {{t}^{2}}={{\alpha }^{2}}{{t}^{3}}={{a}^{2}}({{t}^{2}}+t)\] (from equation 1)                    


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