Question
Consider the functions \( f(x) = x – 3 \) and \( g(x) = x^2 + k^2 \), where \( k \) is a real constant.
(a) Write down an expression for \( (g \circ f)(x) \). [2]
(b) Given that \( (g \circ f)(2) = 10 \), find the possible values of \( k \). [3]
▶️Answer/Explanation
Detail Solution
(a)
Step 1: Identify the composition of functions \( (g \circ f)(x) \).
The composition of functions \( (g \circ f)(x) \) means we will substitute \( f(x) \) into \( g(x) \).
Step 2: Substitute \( f(x) \) into \( g(x) \).
We have \( f(x) = x – 3 \) and \( g(x) = x^2 + k^2 \). Therefore, we substitute \( f(x) \) into \( g \):
\[
(g \circ f)(x) = g(f(x)) = (f(x))^2 + k^2.
\]
$$(g \circ f)(x) = g(f(x)) =(x-3)^2 +k^2 = x^2 – 6x + 9 +k^2$$
$$(g \circ f)(x) = x^2 – 6x + 9 + k^2.$$
Thus, the final expression for \( (g \circ f)(x) \) is:
\[
(g \circ f)(x) = x^2 – 6x + (9 + k^2).
\]
(b)
We have \( f(x) = x – 3 \) and \( g(x) = x^2 + k^2 \).
\[
(g \circ f)(x) = g(f(x)) = (f(x))^2 + k^2.
\]
$$(g \circ f)(x) = g(f(x)) =(x-3)^2 +k^2 = x^2 – 6x + 9 +k^2$$
$$ (g \circ f)(2) = g(f(2))= x^2 – 6x + 9 +k^2$$
$$ (g \circ f)(2) = g(f(2))= 2^2 – 6\times 2 + 9 +k^2$$
$$ (g \circ f)(2) = g(f(2))= 4 – 12+ 9 +k^2$$
$$ (g \circ f)(2) = g(f(2))= 13-12 +k^2$$
$$ (g \circ f)(2) = g(f(2))= 1+k^2$$
Step 1: Set up the equation based on the given condition
We know that \( (g \circ f)(2) = 10 \), so we set up the equation:
$$ 1 + k^2 = 10 $$
Step 2: Solve for \( k^2 \)
Subtract 1 from both sides:
$$ k^2 = 10 – 1 = 9 $$
Step 3: Find the possible values of \( k \)
Taking the square root of both sides gives us:
$$ k = \pm 3 $$
————Markscheme—————–
(a) \( (g \circ f)(x) = (x – 3)^2 + k^2 \)
(b) Substituting \( x = 2 \) into \( (g \circ f)(x) \):
\( (2 – 3)^2 + k^2 = 10 \)
\( 1 + k^2 = 10 \)
\( k^2 = 9 \)
\( k = \pm 3 \)