For the same f, g(x)=xf(x) has minimum −1 on [−1,1]. Find all a.

01 / Standard#Your turn

Official paper · jm01-2024 · II.4(b) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Both absolute-value arguments are negative on this interval.

Hint 2

Check whether the quadratic vertex lies inside the interval.

Worked solution

Reduce to a quadratic and locate its vertex.
$g(x)=2x^2+(a-10)x,\quad x_0=(10-a)/4$
For 6≤a≤14 the minimum is the vertex value.
$-\frac{(a-10)^2}{8}=-1\implies a=10\pm2\sqrt2$
Outside this range the minimum is at an endpoint; each resulting candidate violates its case.
$a<6:\ a-8=-1\implies a=7;\quad a>14:\ 12-a=-1\implies a=13$

a=10±2√2.

Checks and common pitfalls: The unconstrained vertex formula requires its vertex to be in [−1,1].

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Reduce to a quadratic and locate its vertex.
$g(x)=2x^2+(a-10)x,\quad x_0=(10-a)/4$
For 6≤a≤14 the minimum is the vertex value.
$-\frac{(a-10)^2}{8}=-1\implies a=10\pm2\sqrt2$
Outside this range the minimum is at an endpoint; each resulting candidate violates its case.
$a<6:\ a-8=-1\implies a=7;\quad a>14:\ 12-a=-1\implies a=13$

Think first. Reveal a hint when the class is ready.

2024 JM01