AMC 10 Daily Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 19 Hard

Suppose that f\left( x \right) and g\left( x \right) are both quadratic functions with quadratic coefficient 1. If g\left( 6 \right)=35 and \frac{f\left( -1 \right)}{g\left( -1 \right)}=\frac{f\left( 1 \right)}{g\left( 1 \right)}=\frac{21}{20}, determine f\left( 6 \right).

  • A.

    -35

  • B.

    0

  • C.

    1

  • D.

    35

  • E.

    \frac{147}{4}

Answer:D

If f\left( x \right)={{x}^{2}}+ax+b and g\left( x \right)={{x}^{2}}+cx+d, by the given conditions,

20\left( 1-a+b \right)=21\left( 1-c+d \right)\cdots \cdots①;

20\left( 1+a+b \right)=21\left( 1+c+d \right)\cdots \cdots②.

①+② gives 40+40b=42+42d, so 20b=1+21d;

①-② gives -40a=-42c, 20a=21c.

Also, by g\left( 6 \right)=35, we have 36+6c+d=35.

∴ 36+6\times \frac{20}{21}a+\frac{20b-1}{21}=35 and 6a+b=-1,

∴ f\left( 6 \right)=36+6a+b=35.

Let h\left( x \right)=21g\left( x \right)-20f\left( x \right).

We can readily get h\left( x \right) is also a quadratic function with quadratic coefficient 1.

Also, h\left( -1 \right)=21g\left( -1 \right)-20f\left( -1 \right)=0, h\left( 1 \right)=21g\left( 1 \right)-20f\left( 1 \right)=0,

∴ h(x)=(x+1)(x-1)={{x}^{2}}-1,

∴ h\left( 6 \right)=21g\left( 6 \right)-20f\left( 6 \right)={{6}^{2}}-1=35,

∴ 21\times 35-20f\left( 6 \right)=35 and f\left( 6 \right)=35.

Related Problems
Problem 17 AMC 10 Daily Practice Round 2
Problem 18 AMC 10 Daily Practice Round 2
Problem 20 AMC 10 Daily Practice Round 2
Problem 21 AMC 10 Daily Practice Round 2
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