NIMCET 2019 Question Paper with solutions

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Question : 8 of 120

Marks: +1, -0

If x, y, z are distinct real numbers and

|\table\x,x^2,2+x^3;y,y^2,2+y^3;z,z^2,2+z^3|

= 0, then xyz =

1
-1
2
-2

Solution:

|\table\x,x^2,2+x^3;y,y^2,2+y^3;z,z^2,2+z^3|

= 0

⇒|\table x,x^2,2;y,y^2,2;z,z^2,2|+|\table x,x^2,x^3;y,y^2,y^3;z,z^2,z^3|=0

⇒2|\table x,x^2,1;y,y^2,1;z,z^2,1|+xyz|\table 1,x,x^2;1,y,y^2;1,z,z^2|=0

⇒(xyz+2)|\table x,x^2,1;y,y^2,1;z,z^2,1|=0

⇒xyz+2=0

⇒xyz=-2

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