Question

find p(0) and p(2) for p(y)=y^2 -y+1​

Answer 1

Given:-

• p(y) = y² - y + 1

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To find:-

• p(0) and p(2)

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Solution:-

★Case I:-

$\large{\tt{\longrightarrow{p(0) = 0^2 - 0 + 1}}}$

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$\large{\tt{\longrightarrow{p(0) = 0 - 0 + 1}}}$

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$\large{\tt{\longrightarrow{p(0) = 0 - 1}}}$

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$\boxed{\large{\tt{\longrightarrow{\red{p(0) = -1}}}}}$

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★Case II:-

$\large{\tt{\longrightarrow{p(2) = 2^2 - 2 +1}}}$

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$\large{\tt{\longrightarrow{p(2) = 4 - 2 + 1}}}$

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$\large{\tt{\longrightarrow{p(2) = 4 - 3}}}$

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$\boxed{\large{\tt{\longrightarrow{\red{p(2) = 1}}}}}$

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Hence,

• Value of p(0) = -1
• Value of p(2) = 1

Answer 2

Given Equation

• $\sf{p(y) = y^2 - y + 1}$

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To find

• $\sf{Value\: of\: p(0)}$
• $\sf{Value\: of\: p(2)}$

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Solution

• In this question, we have given an equation i.e., p(y) = y² - y + 1.

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★ We will solve this question in two cases.

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$\large{\boxed{\boxed{\sf{Case\: I}}}}$

$\tt\longmapsto{p(y) = y^2 - y + 1}$

$\tt\longmapsto{p(0) = (0)^2 - (0) + 1}$

$\tt\longmapsto{p(0) = 0 - 0 + 1}$

$\tt\longmapsto{p(0) = 0 - 1}$

$\bf\longmapsto{p(0) = -1}$

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$\large{\boxed{\boxed{\sf{Case\: II}}}}$

$\tt\longmapsto{p(y) = y^2 - y + 1}$

$\tt\longmapsto{p(2) = (2)^2 - (2) + 1}$

$\tt\longmapsto{p(2) = 4 - 2 + 1}$

$\tt\longmapsto{p(2) = 4 - 3}$

$\bf\longmapsto{p(2) = 1}$

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Hence, the values of

• $\sf{p(0) = -1}$
• $\sf{p(2) = 1}$

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