A) 0
B) abc
C) \[a+b+c\]
D) none of these
Correct Answer: A
Solution :
| Let \[f\left( x \right)=f\left( x \right)=a\frac{\left( x-b \right)\left( x-c \right)}{\left( a-b \right)\left( a-c \right)}\]\[+b\frac{\left( x-c \right)\left( x-a \right)}{\left( b-c \right)\left( b-a \right)}+c\frac{\left( x-a \right)\left( x-b \right)}{\left( c-a \right)\left( c-b \right)}-x\] |
| We have \[f\left( a \right)=0,f\left( b \right)=0\operatorname{and}\,f\left( c \right)=0.\] |
| But, f(x) is quadratic so it could not have more than two zeros. \[\Rightarrow \]\[f(x)\]is identity |
| Hence,\[a\frac{\left( x-b \right)\left( x-c \right)}{\left( a-b \right)\left( a-c \right)}+b\frac{\left( x-a \right)\left( x-c \right)}{\left( b-a \right)\left( b-c \right)}\]\[+c\frac{\left( x-a \right)\left( x-b \right)}{\left( c-a \right)\left( c-b \right)}\equiv 0\,\forall \,x\in R\] |
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