For all real numbers x and y, the custom operation \odot is defined by x \odot y = x^2 y -… — GRE General Test

For all real numbers $x$ and $y$ , the custom operation $\odot$ is defined by $x \odot y = x^2 y - y^2 x$ . The function $f$ is defined by $f(t) = t \odot 3$ . If $t$ is a positive real number such that $f(f(t)) = 0$ and $f(t) \neq 0$ , what is the value of $t$ ?

A
$3$
B
$\frac{3 + \sqrt{5}}{2}$
$\frac{3 + \sqrt{13}}{2}$ Answer
D
$\frac{3 + \sqrt{13}}{4}$
E
$\frac{9 + \sqrt{13}}{2}$

Answer

The correct value of

t

\frac{3 + \sqrt{13}}{2}

Applying the custom binary operator gives

f(t) = 3t^2 - 9t

. Substituting

u = f(t)

into

f(u) = 0

yields

3u(u - 3) = 0

, so

u = 0

u = 3

. Because

f(t) \neq 0

, it must be that

f(t) = 3

. Setting

3t^2 - 9t = 3

leads to

t^2 - 3t - 1 = 0

. Applying the quadratic formula yields the positive value

\frac{3 + \sqrt{13}}{2}

Step-by-Step Solution

Express

f(t)

using the custom symbol definition.

f(t) = t \odot 3 = t^2(3) - (3)^2 t = 3t^2 - 9t

Apply the rule

x \odot y = x^2 y - y^2 x

with

x = t

and

y = 3

Analyze the nested function condition

f(f(t)) = 0

Let

u = f(t)

. Then

f(u) = 3u^2 - 9u = 3u(u - 3) = 0

, which yields

u = 0

u = 3

Evaluate the outer function

f

at the argument

u = f(t)

Apply the problem constraints to determine the exact value of

f(t)

Since

f(t) \neq 0

u = f(t) = 3

. Thus,

3t^2 - 9t = 3

Eliminate

f(t) = 0

based on the explicit condition given in the problem.

Solve the quadratic equation for

t > 0

Dividing

3t^2 - 9t - 3 = 0

by 3 gives

t^2 - 3t - 1 = 0

. Using the quadratic formula,

t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}

. Since

t > 0

t = \frac{3 + \sqrt{13}}{2}

Find the positive real root of the simplified quadratic equation.

Key Concept

Nested Function Evaluation and Custom Symbol Operations

Estimated Time:2m 30s

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