Cumulative Assessment

Cumulative I - Foundations (Phases 0–4)

25 interleaved questions spanning multiple phases. Cumulative review is one of the most powerful retention tools - it forces retrieval of older material right when it is starting to fade.

Q1 Medium

The present value of a continuous stream c(t) discounted at rate r over [0,T] is:

∫₀ᵀ c(t)dt

∫₀ᵀ c(t)e^{−rt}dt

∫₀ᵀ c(t)e^{rt}dt

c(T)e^{−rT}

Q2 Easy

You have been staring at a proof for 15 minutes with no progress. The evidence-based next move is to:

Push harder in focused mode until it yields

Reread the chapter from the start

Deliberately disengage and let diffuse mode work, then return

Look up the full solution immediately

Q3 Medium

For \(A=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), the eigenvalues are:

2 and 2

1 and 3

0 and 4

-1 and 3

Q4 Medium

In an SM-2 style scheduler, a FAILED recall causes the interval to:

Grow by the ease factor

Stay the same

Reset to ~1 day

Double

Q5 Medium

For proving every integer n≥2 has a prime factorization, the natural technique is:

weak induction

strong induction

a single counterexample

direct computation only

Q6 Medium

Which sequence converges by the Monotone Convergence Theorem?

a_n=(−1)^n

a_n=n

a_n=1−1/n

a_n=sin n

Q7 Medium

The most valuable field in an error log entry is usually:

The date

The cause of the error (concept vs method vs algebra vs misread)

The problem number

How long it took

Q8 Medium

Which statement is most naturally attacked by contrapositive?

√2 is irrational

if n² is odd then n is odd

there are infinitely many primes

the empty set is a subset of every set

Q9 Medium

The system \(Ax=b\) has a solution if and only if:

\(b=0\)

\(A\) is square

\(b\in\operatorname{col}(A)\)

\(\det A\neq0\)

Q10 Easy

The linear (first-order Taylor) approximation of f about a is:

f(a)

f(a)+f′(a)(x−a)

f′(a)(x−a)

f(a)+f″(a)(x−a)²/2

Q11 Medium

A relation that is reflexive, symmetric, and transitive necessarily produces:

a total order

a partition of the set into disjoint classes

a bijection

a Cartesian product

Q12 Easy

The Lagrange condition for minimizing \(f\) on \(g=c\) states that at the optimum:

\(\nabla f=0\)

\(\nabla f=\lambda\nabla g\)

\(f=g\)

\(\nabla g=0\)

Q13 Medium

For a convex function, a point with \(\nabla f=0\) is:

Only a local minimum

A global minimum

Possibly a saddle

Never an extremum

Q14 Easy

If \(\det A=0\) then:

\(A\) is invertible

\(A\) has 0 as an eigenvalue

\(A=0\)

\(A\) is symmetric

Q15 Medium

For a bond, ΔP/P ≈ −DΔy + ½CΔy². The convexity term CΔy² is:

the linear risk

the second-order Taylor curvature correction

always negligible

the same as duration

Q16 Medium

The value of \(\int_{-\infty}^\infty e^{-x^2}\,dx\) is:

\(\pi\)

\(\sqrt{\pi}\)

\(2\pi\)

\(1\)

Q17 Medium

Differentiating x²+y²=25 implicitly gives 2x+2yy′=0, so y′ equals:

−x/y

x/y

−y/x

−2x

Q18 Medium

Which statement about the function f: ℝ→ℝ, f(x)=x² is correct?

It is injective

It is surjective onto ℝ

It is neither injective nor surjective onto ℝ

It is a bijection

Q19 Easy

The main mechanical benefit of notation fluency is that it:

Makes proofs shorter

Frees working-memory slots for reasoning instead of decoding

Eliminates the need for definitions

Guarantees correct answers

Q20 Medium

A bidirectional flashcard set for a formula includes, at minimum:

Only symbol→meaning

Only meaning→symbol

Both concept→equation and equation→concept

A copy of the textbook page

Q21 Medium

A 13 ft ladder slides down a wall; the base moves out. Which relation should you differentiate w.r.t. t for a related-rates problem?

x+y=13

x²+y²=13²

xy=13

x²−y²=13

Q22 Medium

Proof by contradiction of a statement S proceeds by:

assuming S and deriving S again

assuming ¬S and deriving a logical impossibility

assuming the hypothesis and reaching the conclusion

checking finitely many cases

Q23 Medium

After Gram–Schmidt, the residual \(x-\operatorname{proj}_U x\) lies in:

\(U\)

\(U^\perp\)

the zero subspace only

the whole space with no structure

Q24 Medium

A square matrix has a nontrivial null space. Then it is:

Invertible

Not invertible (singular)

Symmetric

Orthogonal

Q25 Medium

‘A function is integrable only if it is bounded (on a closed interval).’ This tells us:

boundedness is sufficient for integrability

boundedness is necessary for integrability

integrability is sufficient for boundedness

boundedness and integrability are equivalent

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