Question

If \( f(x) = x^{2}+bx+c \) and \( f(1+k) = f(1-k) \) \( \forall K \in \mathbb{R} \), for two real numbers b and c, then

• A. \( f(1)<f(0)<f(-1) \)
• B. \( f(-1)<f(0)<f(1) \)
• C. \( f(0)<f(-1)<f(1) \)
• D. \( f(0)<f(1)<f(-1) \)

Hint:

For any quadratic function \( f(x) \), the condition \( f(a+k) = f(a-k) \) for all \(k\) immediately tells you that the parabola's axis of symmetry is at \( x=a \). This is a major shortcut.

Answer 1

The Correct Option is A

Given the function is a quadratic, \( f(x) = x^2 + bx + c \).
The condition \( f(1+k) = f(1-k) \) for all real numbers \( k \) implies that the graph of the function, a parabola, is symmetric about the vertical line \( x=1 \).
The axis of symmetry for a parabola \( y = ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \).
In this case, \( a=1 \). So, the axis of symmetry is \( x = -\frac{b}{2} \).
Equating the axis of symmetry to 1, we get:
\( -\frac{b}{2} = 1 \Rightarrow b = -2 \).
So, the function becomes \( f(x) = x^2 - 2x + c \).
Now, we evaluate the function at the points \( x = 1, 0, -1 \).
\( f(1) = (1)^2 - 2(1) + c = 1 - 2 + c = c - 1 \).
\( f(0) = (0)^2 - 2(0) + c = c \).
\( f(-1) = (-1)^2 - 2(-1) + c = 1 + 2 + c = c + 3 \).
Comparing these three values:
\( c - 1<c<c + 3 \).
Therefore, the correct inequality is \( f(1)<f(0)<f(-1) \).