AMC 10 Daily Practice - Polynomial Functions

Complete problem set with solutions and individual problem pages

Problem 1 Easy

Given that f\left( x \right)={{x}^{5}}+a{{x}^{3}}+bx-8 (a,b are constants) and f\left( -2 \right)=10, then f\left( 2 \right)=            .

  • A.

    -26

  • B.

    -18

  • C.

    -10

  • D.

    10

Answer:A

Let g\left( x \right)={{x}^{5}}+a{{x}^{3}}+bx, then g\left( x \right) is odd and g\left( 2 \right)=-g\left( -2 \right).

\therefore g\left( -2 \right)+g\left( 2 \right)=0.

\therefore f\left( -2 \right)+8+f\left( 2 \right)+8=0.

\because f\left( -2 \right)=10.

\therefore f\left( 2 \right)=-26.

Related Problems
Problem 2 AMC 10 Daily Practice - Polynomial Functions
Problem 3 AMC 10 Daily Practice - Polynomial Functions
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