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Let \( f(x)=a x^{2}+b x+c \). If \( x_{1}, x_{2}, x_{3} \) are three distinct real numbers such that \( f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)=0 \), then \( f(x) \equiv 0 \),
If \( x_{1}, x_{2}, x_{3} \) are three distinct real numbers and
\[
\begin{array}{l}
g(x)=\frac{\left(x-x_{2}\right)\left(x-x_{3}\right)}{\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right)}+\frac{\left(x-x_{3}\right)\left(x-x_{1}\right)}{\left(x_{2}-x_{3}\right)\left(x_{2}-x_{1}\right)} \\
+\frac{\left(x-x_{1}\right)\left(x-x_{2}\right)}{\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)}
\end{array}
\]
then \( g(x) \) is identically equal to
(a) 0
(b) 1
(c) \( \left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right) \)
(d) none of these.
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