AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 20 Hard

Suppose that f\left( x \right) and g\left( x \right) are both quadratic functions with quadratic coefficient 2. If g\left( 10 \right)=25 and \frac{f\left( -2 \right)}{g\left( -2 \right)}=\frac{f\left( 2 \right)}{g\left( 2 \right)}=\frac{16}{15}, determine f\left( 10 \right).

  • A.

    -25

  • B.

    0

  • C.

    1

  • D.

    25

  • E.

    \frac{208}{15}

Answer:E

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

15\left( 8-2a+b \right)=16\left( 8-2c+d \right)\cdots \cdots①;

15\left( 8+2a+b \right)=16\left( 8+2c+d \right)\cdots \cdots②.

①+② gives 240+30b=256+32d, so d=\frac{15}{16} b-\frac12;

②-① gives 60a=64c, c=\frac{15}{16} a.

Also, by g\left( 10 \right)=25, we have 200+10c+d=25.

∴ 200+10\times \frac{15}{16}a+\frac{15}{16}b-\frac12=25 and 10a+b=-\frac{2792}{15},

∴ f\left( 10 \right)=200+10a+b=\frac{208}{15}.

The answer is \text{E}

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