What are the most common mistakes in DSE M1?

Not checking end points for a greatest or least value, dropping the chain-rule factor, forgetting + C, keeping the old limits after a substitution, reversing a conditional probability, using a Poisson mean for the wrong interval, and misreading the normal table.

How can I check my M1 answers quickly?

Differentiate an integral to see whether you get the integrand back, check that probabilities lie between 0 and 1 and that a distribution sums to 1, and sketch the normal curve to see whether the answer should be more or less than 0.5.

Does the M1 normal table give the area to the left of z?

No. It gives A(z), the area between 0 and z. Add 0.5 for the area to the left of a positive z.

Common Mistakes in DSE M1 and How to Avoid Them

Most marks lost in M1 come from a short list of slips, not from hard mathematics. In calculus: end points, the inside derivative, + C, and limits after a substitution. In statistics: the direction of a conditional probability, the Poisson mean for the stated interval, the table area from 0 to z, and σ/√n for a sample mean.

Check my M1 working step by step

f(x) = x e⁻ˣ on 0 ≤ x ≤ 3 from Example 1. The greatest value is at the stationary point x = 1; the least value is at the end point x = 0. Try it yourself →

Ten-Second Checks

Greatest / least valueDid I compare the end points?

Chain ruleDid I multiply by the inside derivative?

Indefinite integralIs + C there?

SubstitutionDid the limits change to u-values?

Conditional probabilityIs it P(A | B) or P(B | A)?

PoissonIs λ for the interval in the question?

Normal tableThe table gives the area from 0 to z

Sample meanStandard deviation is σ/√n

Calculus Slips

Stopping at the stationary point. For a greatest or least value on a ≤ x ≤ b, the answer may be at an end point. For f(x) = x e⁻ˣ on 0 ≤ x ≤ 3 the greatest value is at the stationary point x = 1, but the least value, 0, is at the end point x = 0.
Dropping the inside derivative. (e³ˣ)′ = 3e³ˣ and (ln(3x + 1))′ = 3/(3x + 1).
Multiplying when you should divide. ∫ e⁰·³ᵗ dt = e⁰·³ᵗ/0.3. Differentiate your answer to check.
Leaving out + C, then having no constant for the initial condition to fix.
Keeping x-limits after substituting u.
Asserting over- or under-estimate without d²y/dx². The mark is for the sign of the second derivative.

Statistics Slips

Reversing a conditional probability. "Given that it is defective, find the probability it came from machine B" is P(B | D), which needs Bayes' theorem, not the P(D | B) printed in the question.
Forgetting the complement. P(X ≥ 1) = 1 − P(X = 0) is one term; adding P(1) + P(2) + … is slow and error-prone.
Using the wrong Poisson mean. If the rate is 2.5 per hour and the question asks about 2 hours, λ = 5.
Misreading the normal table. The M1 table gives A(z) = P(0 ≤ Z ≤ z). So P(Z > 1.96) = 0.5 − 0.4750, not 1 − 0.4750.
Using σ for a sample mean. The mean of n observations has standard deviation σ/√n.
Rounding too early. Keep four decimal places in probabilities until the final line.

Worked Examples — Common Mistakes in M1

Example 1: The end-point trap

Find the greatest and least values of f(x) = x e⁻ˣ for 0 ≤ x ≤ 3.

f′(x) = (1 − x)e⁻ˣ = 0 gives x = 1
f(1) = 1/e ≈ 0.368
End points: f(0) = 0, f(3) = 3e⁻³ ≈ 0.149
Compare the three values

Answer: Greatest 1/e at x = 1; least 0 at x = 0

Example 2: Reading the normal table the right way

Given A(1.96) = 0.4750, find P(Z > 1.96) and P(Z < 1.96).

A(z) is the area from 0 to z; each half of the curve has area 0.5
P(Z > 1.96) = 0.5 − 0.4750 = 0.0250
P(Z < 1.96) = 0.5 + 0.4750 = 0.9750

Answer: 0.0250, 0.9750