For all real numbers x and y, the custom binary operator \Delta is defined by x \Delta y =… — GMAT

For all real numbers $x$ and $y$ , the custom binary operator $\Delta$ is defined by $x \Delta y = x^2 - 5xy + 4y^2$ . The function $f$ is defined by $f(x) = x \Delta 1$ . What is the positive integer value of $k$ such that $f(f(k)) = 0$ ?

Cevap: 5

Cevap

The positive integer value of

k

5

Substituting

y = 1

into

x \Delta y = x^2 - 5xy + 4y^2

gives

f(x) = x^2 - 5x + 4

. For

f(f(k)) = 0

, the outer function evaluation requires

f(k)

to be a root of

f(x) = 0

. Solving

x^2 - 5x + 4 = 0

yields roots

1

and

4

. Setting

f(k) = 4

yields

k^2 - 5k = 0

, which has roots

k = 0

and

k = 5

. Since

k

must be a positive integer,

k = 5

. Setting

f(k) = 1

yields irrational values, so

5

is the unique solution.

Adım Adım Çözüm

Substitute

y = 1

into the operator definition to find

f(x)

f(x) = x^2 - 5x + 4

f(x) = x \Delta 1 = x^2 - 5x(1) + 4(1)^2

Set

f(u) = 0

for

u = f(k)

and solve for

u

u = 1

u = 4

Factoring

u^2 - 5u + 4 = 0

yields

(u - 1)(u - 4) = 0

Solve

f(k) = 4

for

k

k = 0

k = 5

k^2 - 5k + 4 = 4 \implies k^2 - 5k = 0 \implies k(k - 5) = 0

Solve

f(k) = 1

for

k

and check for integer solutions.

k = \frac{5 \pm \sqrt{13}}{2}

(irrational roots)

k^2 - 5k + 4 = 1 \implies k^2 - 5k + 3 = 0

Select the positive integer solution.

k = 5

The value

k = 0

is not positive, and the roots from

f(k) = 1

are not integers.

Anahtar Kavram

Custom operator evaluation, composite/nested functions, and quadratic root analysis.

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru