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Matrices

Question
CBSEENMA12036217

If a2 + b2 + c2 = -2 and straight f left parenthesis straight x right parenthesis space equals space open vertical bar table row cell 1 plus straight a squared straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row cell left parenthesis 1 plus straight a squared right parenthesis straight x end cell cell 1 plus straight b squared straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row cell left parenthesis 1 plus straight a squared right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell 1 plus straight c squared straight x end cell end table close vertical bar then f(x) is a polynomial of degree

  • 1

  • 0

  • 2

  • 3

Solution

C.

2

straight f left parenthesis straight x right parenthesis space equals space open vertical bar table row cell 1 space plus left parenthesis straight a squared plus straight b squared plus straight c squared plus 2 right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis straight x end cell row cell 1 plus left parenthesis straight a squared plus straight b squared space plus straight c squared space plus 2 right parenthesis straight x end cell cell 1 plus straight b squared straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis straight x end cell row cell 1 plus space left parenthesis straight a squared plus straight b squared space plus straight c squared plus 2 right parenthesis straight x end cell cell left parenthesis 1 plus straight b squared right parenthesis straight x end cell cell 1 plus straight c squared straight x end cell end table close vertical bar
Applying space straight C subscript 1 space rightwards arrow with space on top space straight C subscript 1 space plus straight C subscript 2 plus space straight C subscript 3
space equals space open vertical bar table row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row 1 cell 1 plus straight b squared space straight x end cell cell left parenthesis 1 plus straight c squared right parenthesis space straight x end cell row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell 1 plus straight c squared space straight x end cell end table close vertical bar space therefore space straight a squared space plus straight b squared space plus straight c squared space plus 2 space equals space 0
straight f left parenthesis straight x right parenthesis space equals space space open vertical bar table row 0 cell straight x minus 1 end cell 0 row 0 cell 1 minus straight x end cell cell straight x minus 1 end cell row 1 cell left parenthesis 1 plus straight b squared right parenthesis space straight x end cell cell 1 plus straight c squared space straight x end cell end table close vertical bar space space semicolon space Applying space straight R subscript 1 space rightwards arrow space straight R subscript 1 minus straight R subscript 2
straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight x minus 1 right parenthesis squared
Hence space degree space equals space 2

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