Multiple choice

Let $f(x)=4{x}^{2}-4ax+{a}^{2}-2a+2$ be a quadratic polynomial in $x$, $a$ be any real number. If x-coordinate of vertex of parabola $y=f(x)$ is less than $0$ and $f(x)$ has minimum value $3$ for $x\in[0,2]$, then value of $a$ is

  1. $1+\sqrt{2}$
  2. $1-\sqrt{2}$
  3. $1-\sqrt{3}$
  4. $1+\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vertex x = -b/2a = 4a/8 = a/2. Given a/2 < 0, so a < 0. Minimum value of quadratic on [0, 2] is at vertex if vertex is in range, or at endpoints. Since a < 0, vertex is at negative x, so minimum on [0, 2] is at x=0. f(0) = a^2 - 2a + 2 = 3. a^2 - 2a - 1 = 0. a = (2 +/- sqrt(4+4))/2 = 1 +/- sqrt(2). Since a < 0, a = 1 - sqrt(2).

AI explanation

The vertex of the parabola is at x = 4a / 8 = a/2, so x < 0 implies a < 0. If the minimum on [0, 2] is 3, it must occur at x = 0 because the parabola opens upward and decreases for all x < 0. Setting f(0) = a^2 - 2a + 2 = 3 gives a^2 - 2a - 1 = 0, which yields a = 1 plus or minus sqrt(2). Since a must be negative, the value of a is 1 - sqrt(2).

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