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  • Tell me? - Complex numbers and quadratic equations - JEE Main-6

The number of solutions of \frac{\left ( x-b \right )\left ( x-c \right )}{\left (a-b \right )\left (a-c \right )}+\frac{\left ( x-c \right )\left ( x-a \right )}{\left (b-c \right )\left (b-a \right )}+\frac{\left ( x-a \right )\left ( x-b \right )}{\left (c-a \right )\left (c-b \right )}= 1 \left ( Where \; a\neq b\neq c \right ) equals

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    none of these

 

Answers (1)

we can see the above equation has degree 2, so it will be equation, if satisfied for two values of x.

We can check, x = a, x = b and x = c all satisfy the equation, so a quadratic is being satisfied for more than two values of x, so it is an identity hence satisfied for infinite values of x.

\therefore Option (D)

 

Quadratic Equation become an Identity -

Satisfied by more than two values of x

\Leftrightarrow a=b=c=0

 

-

 

 


Option 1)

0

This is incorrect

Option 2)

1

This is incorrect

Option 3)

2

This is incorrect

Option 4)

none of these

This is correct

Posted by

Vakul

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