Question:
If $p(x)=x 2-2 \sqrt{2} x+1$, then $p(2 \sqrt{2})$ is equal to
(a) 0
(b) 1
(c) $4 \sqrt{2}$
(d) $8 \sqrt{2}+1$
Solution:
(b) Given, $p(x)=x^{2}-2 \sqrt{2} x+1 \ldots$ (i)
On putting $x=2 \sqrt{2}$ in Eq. (i), we get
$P(2 \sqrt{2})=(2 \sqrt{2})^{2}-(2 \sqrt{2})(2 \sqrt{2})+1=8-8+1=1$
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