11th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-1

  • question_answer 15) Let a, b be the roots of the equations \[(x-a)(x-b)+c,c\ne 0,\] Then the roots of the equations - \[\left( \mathbf{x}-\mathbf{a} \right)\left( \mathbf{x}-\mathbf{b} \right)+\mathbf{c}=\mathbf{0}\]are:-

    A) \[\left( a,c \right)\]                       

    B) \[\left( b,c \right)\]          

    C) \[\left( a+c,b+c \right)\]              

    D) \[a,b\]

    Correct Answer: D

    Solution :

    [d]   \[\because \left( x-a \right)\left( x-b \right)=c\] \[\Rightarrow {{x}^{2}}-\left( a+b \right)x+a.b-c=0\]              ?..(i) \[\because \alpha \And \beta \]be the root of the equ.(i), we have \[\alpha +\beta =\frac{-b}{a}=a+b\] \[\alpha -\beta =\frac{c}{a}=ab-c\] Now, \[\because \left( x-\alpha  \right)\left( x-\beta  \right)+c=0\] \[{{x}^{2}}-\left( \alpha +\beta  \right)x+\alpha .\beta +c=0\] \[{{x}^{2}}-\left( a+b \right)x+ab-c+c=0\] \[{{x}^{2}}-\left( a+b \right)x+a.b=0\] \[\left( x-a \right)\left( x-b \right)=0\] If \[\left( x-a \right)=0\]                       \[~\therefore x=a\] If \[\left( x-b \right)=0\]                       \[~\therefore x=b\] So, \[~x=a,b\]. Hence option [d] is correct.

adversite



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