g10-13-01 · Additional mathematics
Two denominators, one identity
How can substituting a forbidden value still find a coefficient?
1 MIN · PREDICT ON PAPER
Decompose (mx+n)/[(x−a)(x−b)]; predict what remains in the numerator identity at x=a.
Discuss or write first. No typed answer is required to continue.
Polynomial part=0; A=1, B=-1; x≠0,-1
2 MIN · EXPLAIN AND REVISE
Which condition does your claim need?
Compare at least two cases. Write a reason, then check the explanation together.
Reveal explanation and conditions
After clearing denominators, mx+n=A(x−b)+B(x−a) is a polynomial identity. Substitution there is valid even at a or b. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.
Conditions: a≠b; the original fractions require x≠a,b.
Mathematical justification: Putting x=a gives A=(ma+n)/(a−b); x=b gives B=(mb+n)/(b−a). These coefficients also satisfy A+B=m and −Ab−Ba=n.
1 MIN · EXIT CHECK
Decompose 1/[x(x+1)].
Check after answering
1/x−1/(x+1), retaining x≠0,−1.
Teacher guidance and extension
Before: Rational domains and coefficient identities
Ask: Point explicitly to the expression in which the substitution is performed.
Watch for: Do not substitute a forbidden value into the original rational expression.
Challenge: Recover A and B by coefficient comparison instead of special-value substitution.
Suggested timing: predict 1 min, compare 3 min, explain 2 min, exit check 1 min. Observations can suggest a claim; they do not prove it for every input.
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