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If  a_{1}< a_{2}< a_{3}< a_{4}< a_{5}< a_{6}  then number of real roots of equation \left ( x-a_{1} \right )\left ( x-a_{3} \right )\left ( x-a_{5} \right )+\left ( x-a_{2} \right )\left ( x-a_{4} \right )\left ( x-a_{6} \right )= 0 equals

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    3

 

Answers (1)

Let f\left ( x \right )=\left ( x-a_{1} \right )\left ( x-a_{3} \right )\left ( x-a_{5} \right )+\left ( x-a_{2} \right )\left ( x-a_{4} \right )\left ( x-a_{6} \right )

There are sign changes in f\left ( a_{1} \right )\: and\: f\left ( a_{2} \right )

There are sign changes in f\left ( a_{3} \right )\: and\: f\left ( a_{4} \right )

There are sign changes in f\left ( a_{5} \right )\: and\: f\left ( a_{6} \right )

\therefore Being a cubic exactly one root in \left ( a_{1},a_{2} \right ), one in \left ( a_{3},a_{4} \right ) and one in \left ( a_{5},a_{6} \right ) So all those roots are real

So, Option (D)

 

Number of roots of polynomial equation -

For a polynomial equation  P\left ( x \right )= 0 if  P\left ( a \right )  and  P\left ( b \right )  are of opposite sign then odd number of roots lie between a and b, if they are of same sign then either no root or even number of roots lie between them.

-

 

 


Option 1)

0

This is incorrect

Option 2)

1

This is incorrect

Option 3)

2

This is incorrect

Option 4)

3

This is correct

Posted by

Vakul

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