The AI coach, step by step, on three questions

A companion to What works and to the meta-analysis behind it, Techniques that raise test performance after a test. Section numbers below refer to the meta-analysis.

Three questions from AIAT-04 (19 October 2026, EMRS Bhopal Grade 12 JEE, 48 students), one per subject. Each was chosen because many of the class’s wrong answers fell on the same option or two, which suggests an error a lesson can name. On a class of 48 that is a hypothesis, not a finding (section 3.8).

	Question	Attempted	Right	The shared wrong answer
Maths	Q66, unit vectors	32	10	19 chose 6 (answer 9)
Physics	Q16, compound microscope	22	6	15 chose 100 (answer 70)
Chemistry	Q33, shapes of IF_x^n	16	8	split 4 and 4 between two options

For each question, the steps below are what the coach does after the test for a student who got it wrong, in the order the student meets them. They follow the order the evidence supports (section 3): the correct answer and the working first, then the student’s own error, the worked route only for a student weak in the chapter, and a retry on a new question.

Coach step	What happens	Evidence
1. How sure, and your working	A confidence rating, then the first step or a photo of the rough work	3.5, 3.8
2. The correct answer and the working	The full solution, before any explanation	3.5
3. Confirm the wrong idea, then name it	One probe question; if it confirms the belief, the refutation	3.8
4. Where did your working go wrong?	One or two prompts	3.5
5. If weak in the chapter	A recap of the lesson, a worked example, steps missing, practice	3.1 to 3.4
6. Retry, unaided	A new question of the same kind	3.6
7. Judge it later	Recall at day 3, a new question at about day 10	3.10

A student who answers “guessed” in step 1 skips step 3 and goes to step 5. A student who got the question right skips to step 6, with the solution on request, and drops to step 5 on a miss. When most of a class is weak on the idea, the coach flags it to the teacher for a re-teach: the lesson itself (section 3.1) is the teacher’s.

Answers to the blanks are in an answer line under each problem. The reading of why students chose each wrong answer is ours, from the answer pattern; we do not have their working, which is what step 1 asks for.

---

Maths: AIAT-04 Q66 (Vector Algebra)

If a, b and c are unit vectors, then |a − b|² + |b − c|² + |c − a|² does not exceed (A) 4 (B) 9 (C) 8 (D) 6

Answer (B) 9. Tagged medium; 10 of 48 right.

Answer	Students	Average time
(D) 6	19	2 min 58 s
(B) 9, correct	10	1 min 43 s
(C) 8	2	3 min 34 s
(A) 4	1	5 min 19 s
Skipped	16	2 min 0 s on screen

Solution. Each term is |a − b|² = 2 − 2 a·b, so the sum is 6 − 2S, where S = a·b + b·c + c·a. The sum is largest when S is smallest. Since |a + b + c|² = 3 + 2S ≥ 0, S ≥ −3/2, and the sum is at most 9. It reaches 9 when a + b + c = 0: three unit vectors in a plane, 120° apart.

The trap. Setting the dot products to zero (three perpendicular vectors) gives exactly 6. The belief: the dot-product term can only reduce the sum, so the maximum is when it is zero. It fails because dot products can be negative, all three at once. The two students who chose 8 probably made one pair opposite (a = −b, giving 4) and the third perpendicular (2 + 2): closer, but only one dot product negative.

1. How sure, and your working

How sure were you of your answer? Sure / Fairly sure / Guessed

Show me your first step, or send a photo of your rough work.

“Guessed” goes straight to step 5. A first step that sets the dot products to zero, or sketches three perpendicular vectors, is the belief behind 6. One that makes a single pair opposite is the belief behind 8.

2. The correct answer and the working

The coach shows the solution above in full, before any explanation.

3. Confirm the wrong idea, then name it

Probe.

Two unit vectors a and b. Can a·b be negative? If yes, give an example and say what the angle between them is. If no, say why.

What the answer tells you: “no” is the belief behind 6, stated directly. “Yes, when they point in opposite directions” with no mention of other angles is the belief behind 8.

Three unit vectors. Which arrangement makes a·b + b·c + c·a smallest? (i) all three perpendicular to each other (ii) all three in one plane, 120° apart (iii) two opposite, the third perpendicular to both

Correct: (ii), giving −3/2. (i) gives 0 and (iii) gives −1.

If the probe confirms the belief, the refutation.

You chose 6. That is the sum when the three dot products are zero, which happens when the vectors are perpendicular. But the question asks for the largest possible value, and zero is not the smallest the dot products can be: for two unit vectors a·b = cos θ, which is negative for any angle above 90°. Three vectors can’t all be opposite each other, so how negative can the three dot products get together? The identity |a + b + c|² ≥ 0 answers that. Expanding it gives 3 + 2S ≥ 0, so S can be as low as −3/2 and no lower. That happens when the three vectors add to zero, which is the 120° arrangement.

For each wrong option.

Chosen	Feedback
6	“6 is the sum when all three dot products are zero (perpendicular vectors). Dot products can be negative, which makes the sum bigger. Try three vectors 120° apart: what is a·b there?” Then Faded 1.
8	“Good instinct: you made one dot product negative (a = −b). Can you make all three negative at once? Find three unit vectors where every pair is more than 90° apart.” Then the 120° case.
4	“4 is the largest value of a single term |a − b|². The question adds three terms. Start from the case a = −b: what are the other two terms?” Then step 2 of the worked example.

4. Where did your working go wrong?

One or two of these, typed or chosen from a short menu:

• At step 2: Why does the sum get larger when S gets smaller?
• At step 3: Why does |a + b + c|² ≥ 0 tell you anything about the dot products?
• At step 4: Why can’t all three pairs be opposite each other?

5. If weak in the chapter

Lesson recap. |a − b|² = |a|² + |b|² − 2 a·b. A dot product a·b = |a||b| cos θ is negative for any angle above 90°. A square is never negative, so |a + b + c|² ≥ 0 sets how negative the three dot products can be together.

Worked example.

a, b, c are vectors with |a| = 1, |b| = 2, |c| = 3. Find the largest value of |a − b|² + |b − c|² + |c − a|².

Step	Working
1. Expand one term	|a − b|² = |a|² + |b|² − 2 a·b = 1 + 4 − 2 a·b
2. Add all three	Sum = 2(1 + 4 + 9) − 2S = 28 − 2S, where S = a·b + b·c + c·a
3. Bound S	|a + b + c|² = 14 + 2S ≥ 0, so S ≥ −7
4. Largest value, and when	Sum ≤ 28 + 14 = 42, reached when a + b + c = 0, for example a = u, b = 2u, c = −3u

Steps missing. Worth trying, not proven (section 3.3).

Faded 1 (last step blank). |a| = |b| = |c| = 2.

Step	Working
1	|a − b|² = 4 + 4 − 2 a·b
2	Sum = 2(4 + 4 + 4) − 2S = 24 − 2S
3	|a + b + c|² = 12 + 2S ≥ 0, so S ≥ −6
4	Sum ≤ ____ , reached when ____

Answer: 36, when a + b + c = 0 (three vectors of length 2, 120° apart).

Faded 2 (last two steps blank). |a| = |b| = |c| = 3.

Step	Working
1	|a − b|² = 9 + 9 − 2 a·b
2	Sum = 54 − 2S
3	____
4	____

Answer: |a + b + c|² = 27 + 2S ≥ 0, so S ≥ −27/2; sum ≤ 81.

Faded 3 (only the first step given). |a| = 1, |b| = 1, |c| = 2.

Step	Working
1	|a − b|² = 1 + 1 − 2 a·b
2-4	____

Answer: sum = 12 − 2S; |a + b + c|² = 6 + 2S ≥ 0, so S ≥ −3; sum ≤ 18, reached at a = b = u, c = −2u.

A pattern worth letting the student notice after Faded 2: for three vectors of equal length r, the largest value is 9r².

Practice: the same idea among others.

Same tool (“a square is never negative”) in different chapters, in random order with unrelated questions:

1. Minimum of x² − 4x + 7. (3, from (x − 2)² + 3)
2. Minimum of x + 1/x for x > 0. (2, from (√x − 1/√x)² ≥ 0)
3. Largest value of |a − b|² for unit vectors. (4)
4. The day-10 question in step 7. (−25)

6. Retry, unaided

A new question of the same kind, no hints. Students who got Q66 right start here, with the solution on request.

a, b, c are unit vectors. Find the smallest value of |a + b|² + |b + c|² + |c + a|².

Answer: 6 + 2S ≥ 6 − 3 = 3. The trap here is again 6.

7. Judge it later

Day 3, no notes: For unit vectors, write |a − b|² in terms of a·b. What inequality bounds a·b + b·c + c·a, and what is the bound?

Day 10, new question:

a, b, c are vectors with |a| = 3, |b| = 4, |c| = 5 and a + b + c = 0. Find a·b + b·c + c·a.

Answer: 0 = 9 + 16 + 25 + 2S, so S = −25.

In the exam hall

“Does not exceed” asks for the largest possible value, so any arrangement you can compute is a lower limit on the answer. Try a = −b with c perpendicular to both: 4 + 2 + 2 = 8. That rules out 4 and 6 in under a minute. With two options left, answering is worth +1.5 marks in expectation under +4/−1. Trying 120° gives 9 and settles it. The 16 students who skipped could have got this far.

---

Physics: AIAT-04 Q16 (Ray Optics)

The focal lengths of the eyepiece and the objective of a compound microscope are 5 cm and 1 cm respectively, and the length of the tube is 20 cm. What is the magnifying power of the microscope when the final image is formed at infinity? (Take D = 25 cm) (A) 50 (B) 70 (C) 80 (D) 100

Answer (B) 70. Tagged medium; 6 of 48 right.

Answer	Students	Average time
(D) 100	15	3 min 22 s
(B) 70, correct	6	2 min 39 s
(C) 80	1	6 min 14 s
Skipped	26	1 min 52 s on screen

Solution. For the final image to be at infinity, the objective’s image must sit at the eyepiece’s focus, 5 cm from the eyepiece. So the objective’s image distance is v₀ = 20 − 5 = 15 cm. From the lens formula, 1/u₀ = 1/15 − 1/1 = −14/15, so u₀ = −15/14 cm and the objective magnifies |v₀/u₀| = 14. The eyepiece magnifies D/f_e = 25/5 = 5. Total: 14 × 5 = 70.

The trap. 100 is (20/1) × (25/5): the textbook formula m = (L/f₀)(D/f_e) with the tube length, 20 cm, put in for L. In that formula L is the distance between the objective’s back focal point and the eyepiece’s front focal point, which here is 20 − 1 − 5 = 14 cm. With 14 the formula gives 70. The belief: the L in the formula is the length of the tube.

1. How sure, and your working

How sure were you of your answer? Sure / Fairly sure / Guessed

Show me your first step, or send a photo of your rough work.

“Guessed” goes straight to step 5. A first step that writes 20/1, or uses L = 20, is the belief behind 100. One that first finds where the objective’s image sits is on the right track, and the slip is later.

2. The correct answer and the working

The coach shows the solution above in full, before any explanation.

3. Confirm the wrong idea, then name it

Probe.

In the formula m = (L/f₀)(D/f_e) for a compound microscope with the final image at infinity, what is L? (i) the distance between the two lenses (ii) the distance between the objective’s focal point and the eyepiece’s focal point (iii) the image distance of the objective

Correct: (ii). A student who picks (i) holds the belief behind 100.

Then: where must the objective’s image be for the final image to be at infinity?

Correct: at the eyepiece’s focal point.

If the probe confirms the belief, the refutation.

You chose 100, which comes from putting the tube length, 20 cm, into m = (L/f₀)(D/f_e). In that formula L is not the tube length. It is the gap between the objective’s focal point and the eyepiece’s focal point: 20 − 1 − 5 = 14 cm, which gives 70. You can avoid the formula entirely. For the final image to be at infinity, the objective’s image has to land exactly at the eyepiece’s focus, so it is 5 cm short of the eyepiece: v₀ = 15 cm. For any lens, |m| = v/f − 1, so the objective gives 15/1 − 1 = 14, and the eyepiece gives D/f_e = 5.

For each wrong option.

Chosen	Feedback
100	“You used the tube length as L. In m = (L/f₀)(D/f_e), L is the gap between the two focal points: 20 − 1 − 5 = 14. Better: find where the objective’s image must be (5 cm short of the eyepiece) and use the lens formula.” Then step 5.
80	“80 is 16 × 5. The eyepiece part (5) is right; the objective gives 15/1 − 1 = 14. Check steps 2 and 3 of the worked example.”
50	“50 is 10 × 5. The eyepiece part is right. Recheck the objective: v₀ = 15 cm and f₀ = 1 cm.”

4. Where did your working go wrong?

• At step 1: Why must the intermediate image sit exactly at the eyepiece’s focus?
• At step 3: Where does |m| = v/f − 1 come from? (From 1/v − 1/u = 1/f and m = v/u.)
• After Faded 1: Why is the trap answer always bigger than the right one?

5. If weak in the chapter

Lesson recap. With the final image at infinity, the objective’s image sits at the eyepiece’s focus. The magnifying power is the objective’s |v₀/u₀| times the eyepiece’s D/f_e, and for the objective |m₀| = v₀/f₀ − 1.

Worked example.

f₀ = 2 cm, f_e = 5 cm, lenses 25 cm apart, final image at infinity, D = 25 cm. Find the magnifying power.

Step	Working
1. Where is the intermediate image?	At the eyepiece’s focus: 5 cm from the eyepiece
2. Objective’s image distance	v₀ = 25 − 5 = 20 cm
3. Objective magnification	|m₀| = v₀/f₀ − 1 = 20/2 − 1 = 9
4. Total	m = 9 × D/f_e = 9 × 5 = 45

Check with the formula: L = 25 − 2 − 5 = 18, and (18/2)(25/5) = 45.

Steps missing. Worth trying, not proven (section 3.3).

Faded 1 (last step blank). f₀ = 1 cm, f_e = 5 cm, lenses 16 cm apart.

Step	Working
1	Intermediate image 5 cm from the eyepiece
2	v₀ = 16 − 5 = 11 cm
3	|m₀| = 11/1 − 1 = 10
4	m = ____

Answer: 10 × 5 = 50. (The trap answer would be 16 × 5 = 80.)

Faded 2 (last two steps blank). f₀ = 2 cm, f_e = 5 cm, lenses 17 cm apart.

Step	Working
1	Intermediate image 5 cm from the eyepiece
2	v₀ = 17 − 5 = 12 cm
3	____
4	____

Answer: |m₀| = 12/2 − 1 = 5; m = 5 × 5 = 25.

Faded 3 (only the first step given). f₀ = 1 cm, f_e = 2.5 cm, lenses 12.5 cm apart.

Step	Working
1	Intermediate image 2.5 cm from the eyepiece
2-4	____

Answer: v₀ = 10 cm; |m₀| = 9; m = 9 × 10 = 90.

Practice: the same idea among others.

Magnification questions from across optics, so the student has to decide which formula applies:

1. Astronomical telescope in normal adjustment, f₀ = 100 cm, f_e = 5 cm. Magnifying power and tube length? (20; 105 cm)
2. Simple magnifier, f = 5 cm, image at the near point. (1 + 25/5 = 6)
3. The same magnifier, image at infinity. (25/5 = 5)
4. Faded 2 above, without the steps. (25)

6. Retry, unaided

A new question of the same kind, no hints. Students who got Q16 right start here, with the solution on request.

Same microscope (f₀ = 1 cm, f_e = 5 cm, lenses 20 cm apart), but the final image is formed at D = 25 cm. Find the magnifying power.

Answer: the eyepiece now forms its image at −25 cm, so its object is at 25/6 cm and magnifies 1 + D/f_e = 6. v₀ = 20 − 25/6 = 95/6 cm, so |m₀| = 95/6 − 1 = 89/6. Total = (89/6) × 6 = 89.

7. Judge it later

Day 3, no notes: For a compound microscope with the final image at infinity, where is the objective’s image? In m = (L/f₀)(D/f_e), what is L?

Day 10, new question:

A compound microscope has f₀ = 2 cm and f_e = 4 cm, with the lenses 24 cm apart. Find the magnifying power with the final image at infinity (D = 25 cm).

Answer: v₀ = 24 − 4 = 20 cm, so |m₀| = 20/2 − 1 = 9; the eyepiece gives 25/4; total 56.25. The trap, the tube length as L, gives (24/2)(25/4) = 75.

In the exam hall

The objective’s image is always closer than the far end of the tube, and |m₀| = v₀/f₀ − 1 is always smaller than tube length/f₀. So the answer must be below (20/1) × 5 = 100, which rules out the option 15 students chose. The full calculation is then three lines.

---

Chemistry: AIAT-04 Q33 (Chemical Bonding)

If IF_x^n species are planar and non-polar, which of the following is correct? (x is the number of F atoms and n is the charge on the species) (A) x = 2 and n = +1 (B) x = 3 and n = 0 (C) x = 2 and n = −1 (D) x = 5 and n = 0

Answer (C), IF₂⁻. Tagged hard; 8 of 48 right.

Answer	Students	Average time
(C) IF₂⁻, correct	8	2 min 39 s
(A) IF₂⁺	4	5 min 15 s
(B) IF₃	4	1 min 59 s
Skipped	32	1 min 38 s on screen

Solution. Count the electrons on iodine: 7 valence electrons, minus the charge.

Species	Electrons on I	Bonds	Lone pairs	Electron pairs	Shape	Planar?	Non-polar?
IF₂⁺	6	2	2	4	bent	yes	no
IF₃	7	3	2	5	T-shaped	yes	no
IF₂⁻	8	2	3	5	linear	yes	yes
IF₅	7	5	1	6	square pyramidal	no	no

Only IF₂⁻ is both planar and non-polar.

The trap. The 8 wrong answers split evenly, so the data does not show one shared belief; the likely reason below is a candidate from back-solving the options, to be checked in the lesson. Both wrong answers are what you get by reading the shape off the number of fluorines and ignoring lone pairs. “Two F atoms, so linear” gives IF₂⁺ (which is bent, because the + charge leaves two lone pairs). “Three F atoms, so trigonal planar like BF₃” gives IF₃ (T-shaped, because of two lone pairs). The belief: the shape follows from the number of bonded atoms. Two-thirds of the class skipped the question entirely.

1. How sure, and your working

How sure were you of your answer? Sure / Fairly sure / Guessed

Show me your first step, or send a photo of your rough work.

“Guessed” goes straight to step 5. A first step that draws the shape from the number of F atoms, with no lone pairs counted, is the likely belief behind both wrong answers.

2. The correct answer and the working

The coach shows the solution above in full, before any explanation.

3. Confirm the wrong idea, then name it

Probe.

IF₂⁺ and IF₂⁻ both have two fluorine atoms. Do they have the same shape? Give a one-line reason.

A student who says “same, both linear” holds the belief behind option (A).

If the probe confirms the belief, the refutation.

You chose IF₂⁺ (or IF₃). The shape of a molecule is not set by how many atoms are attached. It is set by how many electron pairs surround the central atom, bonding pairs and lone pairs together. The charge changes the lone pairs: IF₂⁺ has two, so its four electron pairs make a bent shape like water. IF₂⁻ has three, so its five pairs make a trigonal bipyramid, the three lone pairs spread round the middle, and the two F atoms end up in a straight line. A straight line of identical atoms is non-polar. The same goes for IF₃: two lone pairs make it T-shaped, not trigonal planar.

For each wrong option.

Chosen	Feedback
IF₂⁺	“Two F atoms does not make a molecule linear. With a +1 charge iodine has 6 electrons: 2 bonds and 2 lone pairs, 4 pairs in all, so the shape is bent, like water. Bent is polar. Now try IF₂⁻.” Then step 5.
IF₃	“IF₃ is not trigonal planar like BF₃: iodine still has 2 lone pairs, so there are 5 electron pairs and the shape is T-shaped. Planar, but polar.” Then Faded 1 (BrF₃ has the same shape).
IF₅	“IF₅ has 6 electron pairs, one of them a lone pair: square pyramidal. That is neither planar nor non-polar.”

4. Where did your working go wrong?

• At step 1: Why does a negative charge add an electron to the central atom?
• At step 4: In a trigonal bipyramid, why do lone pairs go to the equatorial positions?
• At step 5: Why is a planar molecule not automatically non-polar? (IF₂⁺ and IF₃ are planar and polar.)

5. If weak in the chapter

Lesson recap. Shape follows the number of electron pairs on the central atom, bonding and lone pairs together. In IF_x^n, iodine has 7 − n electrons; x of them go into bonds and the rest form (7 − n − x)/2 lone pairs. In a trigonal bipyramid the lone pairs take the equatorial positions.

Worked example.

Is XeF₄ planar and non-polar?

Step	Working
1. Electrons on the central atom	Xe has 8; charge 0, so 8
2. Bonds and lone pairs	4 bonds use 4; (8 − 4)/2 = 2 lone pairs
3. Electron pairs and arrangement	4 + 2 = 6: octahedral
4. Shape	Two lone pairs go opposite each other: square planar
5. Planar? Polar?	Planar, and the four bond dipoles cancel: non-polar

Steps missing. Worth trying, not proven (section 3.3).

Faded 1 (last step blank). BrF₃.

Step	Working
1	Br has 7; charge 0, so 7
2	3 bonds; (7 − 3)/2 = 2 lone pairs
3	3 + 2 = 5: trigonal bipyramidal
4	Two lone pairs go equatorial: T-shaped
5	Planar? ____ Polar? ____

Answer: planar, polar.

Faded 2 (last two steps blank). ICl₄⁻.

Step	Working
1	I has 7; charge −1, so 8
2	4 bonds; (8 − 4)/2 = 2 lone pairs
3	4 + 2 = 6: octahedral
4	____
5	____

Answer: square planar; planar and non-polar.

Faded 3 (only the first step given). SF₄.

Step	Working
1	S has 6; charge 0, so 6
2-5	____

Answer: 4 bonds, 1 lone pair, 5 pairs; see-saw; not planar, polar.

Practice: the same idea among others.

Shape and polarity questions where the trick differs each time:

1. Is CO₂ polar? Is SO₂? (CO₂ linear, non-polar; SO₂ has a lone pair, bent, polar)
2. Which has the smaller bond angle, NH₃ or H₂O? (H₂O: two lone pairs)
3. Hybridisation of Xe in XeF₄. (sp³d²)
4. Faded 2 above, without the steps. (square planar, non-polar)

6. Retry, unaided

A new question of the same kind, no hints. Students who got Q33 right start here, with the solution on request.

Which one of IF₄⁺, IF₄⁻ and IF₆⁺ is planar and non-polar?

Answer: IF₄⁻ (8 electrons, 4 bonds, 2 lone pairs, square planar). IF₄⁺ has 1 lone pair and is see-saw; IF₆⁺ has none and is octahedral, non-polar but not planar.

7. Judge it later

Day 3, no notes: How do you find the number of lone pairs on the central atom of IF_x^n? What shape do 2 bonds and 3 lone pairs give?

Day 10, new question:

Which one of ICl₂⁻, ICl₂⁺, ClF₃ and BrF₅ is planar and non-polar?

Answer: ICl₂⁻ (8 electrons on I, 2 bonds, 3 lone pairs, linear). ICl₂⁺ is bent and polar, ClF₃ is T-shaped and polar, BrF₅ is square pyramidal and not planar.

In the exam hall

Check one property at a time. “Non-polar” needs a symmetric shape, which needs the lone pairs to sit symmetrically. IF₅ is out at once (square pyramidal is not planar). For the rest, count electrons on iodine (6, 7, 8 for A to C), which gives 2, 2 and 3 lone pairs. Only two bonds with three lone pairs puts the F atoms in a straight line. That takes under a minute. With one option eliminated, answering is already worth +0.67 marks in expectation; 32 students left it blank.

---

What to look for afterwards

The test is not the score on the lesson, which the research says will look worse than it is (Deslauriers et al. 2019). It is whether, two or three weeks later, a student gets a new question built on the same idea right, unaided, and whether the class stops choosing the trap answer on the next question of that kind. For these three: 6 on vector sums, the full tube length in microscope magnification, and the number of bonded atoms in shapes.