Question

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Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{b) \ 1}$

$\bold\orange{Given:}$

$\bold{=>p(x)=x^{2}-2\sqrt2x+1}$

$\bold\pink{To \ find:}$

$\bold{=>p(2\sqrt2)}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{=>p(x)=x^{2}-2\sqrt2x+1}$

$\bold{=>p(2\sqrt2)=(2\sqrt2)^{2}-2\sqrt2×2\sqrt2+1}$

$\bold{=>p(2\sqrt2)=8-8+1}$

$\bold{=>p(2\sqrt2)=1}$

$\bold\purple{\tt{\therefore{p(2\sqrt2)=1}}}$

Answer 2

$\underline{ \underline \red{{ \bf{ANSWER}}}}$

Given

• p(x) = x² - 2√2x + 1

• p(2√2) = ?

If p(x) = x² - 2√2x + 1

Then p(2√2) = ?

Substituting x = 2√2 in p(x)

$\small{\to \rm{p(2 \sqrt{2} ) = {(2 \sqrt{2} )}^{2} - (2 \sqrt{2} )(2 \sqrt{2} ) + 1}} \\ \small \to \rm{p(2 \sqrt{2} ) = \cancel{ {(2 \sqrt{2} )}^{2} } - \cancel{ {(2 \sqrt{2} )}^{2}} + 1} \\ \\ \small \to \bf{p(2 \sqrt{2} ) = 1}$

Hence, p(2√2) = 1

$\therefore$ Option B is your answer