Chapter 23: Real Analysis and Series of Functions (Set-2)
On the real line, which inequality shows x is to the left of y?
A x ≥ y
B x ≤ y
C x > y
D x ≠ y
On the real line, ordering matches position. If x ≤ y, then x lies left of y or at the same point. This is the basic meaning of the order relation.
Which set contains its endpoints?
A (2, 5)
B (2, 5]
C [2, 5]
D [2, 5)
A closed interval includes both endpoints. So [2,5] contains 2 and 5 as well as all points between them, making it a standard example of a closed set.
Which interval excludes 3 but includes 7?
A [3, 7]
B (3, 7]
C [3, 7)
D (3, 7)
The bracket “]” means the endpoint is included, while “(” means excluded. So (3,7] contains 7 but does not contain 3, by definition.
Which statement best describes an open set (basic idea)?
A Contains all boundary
B Has maximum element
C Always finite set
D Every point has neighborhood
A set is open if each point inside it has some small interval around it still entirely inside the set. This “neighborhood” idea is the simplest introduction to openness.
A closed set (basic idea) is one that
A Contains no points
B Is always bounded
C Contains all limits
D Is always open
Informally, a closed set contains its limit points: if a sequence from the set converges, its limit stays in the set. This is a core basic view of closed sets.
Which number is an upper bound of S = {x : x ≤ 4}?
A 4
B 3
C 0
D −2
Every element of S is ≤ 4, so 4 is an upper bound. Numbers larger than 4 are also upper bounds, but 4 is the smallest such bound.
For S = (0, 1), which is a lower bound?
A 1
B 0
C 2
D 1/2
All elements of (0,1) are greater than 0, so 0 is a lower bound. It is not in the set, but it still bounds the set from below.
For S = (0, 1), the infimum is
A 1
B 1/2
C 0
D −1
The greatest lower bound of (0,1) is 0. Any number bigger than 0 fails to be a lower bound, and 0 is the best possible lower bound.
For S = (0, 1), the supremum is
A 0
B 2
C 1/2
D 1
The least upper bound of (0,1) is 1. Even though 1 is not included, it is the smallest number that is ≥ every element of the set.
Which set is bounded below but not bounded above?
A (−∞, 2]
B [0, ∞)
C (−∞, ∞)
D (−5, 5)
[0,∞) has a lower bound 0, since all elements are ≥ 0. But it goes to infinity, so there is no finite upper bound.
If A and B are bounded above, then A ∪ B is
A Bounded above
B Not bounded
C Always empty
D Only finite
If A has an upper bound M and B has an upper bound N, then max(M,N) bounds both sets. Therefore their union still has a finite upper bound.
If A and B are bounded, then A + B is
A Always unbounded
B Only lower bounded
C Bounded
D Only upper bounded
If A is bounded in [m1, M1] and B in [m2, M2], then sums a+b lie in [m1+m2, M1+M2]. So the sum set remains bounded.
Which is a bounded set?
A (−∞, 0)
B {x: |x|<3}
C (2, ∞)
D All integers
|x|<3 means −3
The set {1/n : n∈ℕ} has supremum
A 0
B 2
C Does not exist
D 1
Values are 1, 1/2, 1/3, … so the largest value is 1. Since 1 is in the set, it is both maximum and supremum.
The set {1/n : n∈ℕ} has infimum
A 1
B −1
C 0
D 1/2
The terms approach 0 but never reach it, so 0 is the greatest lower bound. Any positive number fails to bound all terms from below.
Which sequence converges to 0?
A xₙ = n
B xₙ = 1/n
C xₙ = (−1)ⁿ
D xₙ = n²
1/n becomes arbitrarily small as n grows, so it tends to 0. The others either grow without bound or oscillate without settling to one value.
Which sequence is bounded but not convergent?
A xₙ = (−1)ⁿ
B xₙ = 1/n
C xₙ = 5
D xₙ = 1/n²
(−1)ⁿ stays between −1 and 1, so it is bounded. But it keeps jumping between −1 and 1, so it does not approach a single limit.
A sequence is monotone increasing if
A xₙ₊₁ ≤ xₙ
B xₙ = 0 always
C xₙ₊₁ ≥ xₙ
D xₙ alternates
Monotone increasing means terms never decrease: each next term is at least the previous one. This property plus boundedness guarantees convergence in ℝ.
A bounded monotone increasing sequence converges to
A Its infimum
B Its supremum
C Always 0
D Always 1
For an increasing sequence that is bounded above, the limit exists and equals the least upper bound of its values. This is the standard monotone convergence result.
Bolzano–Weierstrass tells that a bounded sequence has
A A maximum
B A derivative
C A uniform limit
D A convergent subsequence
Even if a bounded sequence does not converge, ℝ guarantees it contains a subsequence that does converge. This is a key compactness-style property of real numbers.
If a real sequence converges, it must be
A Unbounded
B Strictly increasing
C Bounded
D Alternating
Every convergent sequence stays close to its limit after some point, so it cannot go to infinity. Hence convergence implies boundedness, though boundedness alone does not imply convergence.
A Cauchy sequence in ℝ is
A Always divergent
B Always convergent
C Always monotone
D Always periodic
Completeness of ℝ ensures every Cauchy sequence has a real limit. This is a central reason ℝ is “complete” and ℚ is not.
Which statement is TRUE in ℚ?
A Some Cauchy fails converge
B Every Cauchy converges
C Every bounded has sup in ℚ
D Every set has max
In rational numbers, a Cauchy sequence may approach an irrational limit (like √2), so it does not converge within ℚ. This shows ℚ is not complete.
If fₙ → f uniformly, then for large n
A Error depends on x
B Only one point works
C Error bounded for all x
D Only endpoints work
Uniform convergence means: given ε>0, you can choose N so that for all n≥N and all x in the domain, |fₙ(x)−f(x)|<ε. One N works globally.
Uniform convergence implies pointwise because
A Uses same ε only
B Works for each x
C Requires differentiability
D Requires bounded domain
If the difference is small for all x once n is large, then it is certainly small for any fixed x. So uniform convergence automatically gives pointwise convergence.
A simple example of pointwise but not uniform convergence is
A fₙ(x)=x
B fₙ(x)=0
C fₙ(x)=1
D fₙ(x)=xⁿ on [0,1]
On [0,1], xⁿ → 0 for x<1 but equals 1 at x=1, so limit function is 0 on [0,1) and 1 at 1. Convergence is not uniform.
For fₙ(x)=xⁿ on [0,1], what is fₙ(1) as n→∞?
A Goes to 0
B Does not exist
C Goes to 1
D Becomes negative
At x=1, we have fₙ(1)=1ⁿ=1 for every n, so the limit is 1. The non-uniform behavior comes from points near 1, not from x=1 itself.
Weierstrass M-test concludes uniform convergence if
A ∑ Mₙ converges
B ∑ fₙ diverges
C fₙ are discontinuous
D Domain is infinite
If |fₙ(x)| ≤ Mₙ for all x and the numeric series ∑ Mₙ converges, then ∑ fₙ(x) converges uniformly. It gives an easy domination-based proof.
If ∑ fₙ converges uniformly and each fₙ is continuous, then sum is
A Discontinuous
B Continuous
C Always constant
D Undefined
Uniform convergence preserves continuity: partial sums are continuous, and uniform limit of continuous functions remains continuous. This is one main reason uniform convergence is important.
Uniform convergence is most directly used to justify
A Increasing functions only
B Only integer bounds
C Limit swaps safely
D Only finite sums
Uniform convergence provides strong control of errors, which allows passing limits through certain operations like continuity and integration (with standard conditions). Pointwise convergence alone can fail.
For ∑ aₙ(x−a)ⁿ, the center is
A x
B a
C n
D aₙ
A power series is centered at a because powers involve (x−a). The behavior depends on distance from a. Many formulas and radius of convergence are measured from this center point.
If a power series has radius R, it converges for
A |x−a|
B |x−a|>R
C x=a only
D |x−a|=0 only
Inside the radius, a power series converges absolutely. Outside it diverges. Boundary points |x−a|=R are special and must be tested separately.
Ratio test for power series mainly tests
A Term signs
B Only endpoints
C Growth of coefficients
D Only integrals
Ratio test examines |aₙ₊₁/aₙ| to estimate how terms behave. This helps find R by converting convergence into a condition on |x−a| relative to a limit.
Root test uses limsup of
A |aₙ₊₁−aₙ|
B √[n]{|term|}
C |aₙ|+|aₙ₊₁|
D n·aₙ
Root test considers L = limsup √[n]{|aₙ(x−a)ⁿ|}. If L<1, the series converges absolutely; if L>1, it diverges. It’s useful when ratio is messy.
Inside convergence radius, a power series is
A Not continuous
B Always divergent
C Only piecewise
D Continuous function
Within |x−a|
Term-by-term integration produces coefficients
A Multiply by n
B Always same
C Divide by n+1
D Replace by zero
Integrating ∑ aₙ(x−a)ⁿ gives ∑ aₙ(x−a)ⁿ⁺¹/(n+1) plus a constant. This is valid inside the radius and keeps the same convergence radius.
Term-by-term differentiation changes coefficients by
A Divide by n
B Multiply by n
C Add 1 always
D Remove x−a
Differentiating ∑ aₙ(x−a)ⁿ gives ∑ n aₙ(x−a)ⁿ⁻¹. This is valid for |x−a|
A power series can be treated as a series of
A Functions
B Matrices
C Integers only
D Sets only
For each n, the term aₙ(x−a)ⁿ is a function of x. So a power series is a series of functions, and ideas like uniform convergence can apply on intervals inside the radius.
If aₙ does not tend to 0, then ∑ aₙ
A Must converge
B Converges absolutely
C Must diverge
D Depends only on sign
A necessary condition for a series to converge is aₙ → 0. If terms fail to approach 0, partial sums cannot settle to a finite limit, so the series diverges.
Geometric series ∑ rⁿ converges when
A |r|≥1
B |r|<1
C r=1 only
D r=−1 only
The geometric series has sum 1/(1−r) for |r|<1. If |r|≥1, partial sums do not settle. This is a basic benchmark for comparisons.
Comparison test needs nonnegative terms because it compares
A Partial sums signs
B Derivatives only
C Integrals only
D Sizes of terms
The basic comparison test uses 0 ≤ aₙ ≤ bₙ to compare term sizes. This ensures partial sums behave in a controlled way and allows concluding convergence from a known convergent series.
If 0 ≤ aₙ ≤ bₙ and ∑ bₙ converges, then ∑ aₙ
A Diverges
B Converges
C Oscillates
D Undefined
Since aₙ is dominated by bₙ term-by-term and bₙ has finite total sum, the smaller series cannot “outgrow” it. Hence ∑ aₙ must also converge.
Alternating series test requires bₙ to be
A Decreasing to 0
B Increasing
C Constant
D Unbounded
For ∑ (−1)ⁿbₙ with bₙ≥0, if bₙ decreases and bₙ→0, then the series converges. The alternating signs create cancellation that controls partial sums.
Absolute convergence implies
A Conditional convergence only
B Divergence
C Convergence
D No sum exists
If ∑ |aₙ| converges, then ∑ aₙ converges automatically. Absolute convergence is a stronger property than ordinary convergence and gives more stability under rearrangement.
In ℝ, a set is compact (basic) if it is
A Open and unbounded
B Closed and bounded
C Dense and open
D Countable only
In real numbers, Heine–Borel says a set is compact exactly when it is closed and bounded. Compactness links to subsequences and guarantees many “max/min exists” results.
Which interval is compact in ℝ?
A (0,1)
B (0,1]
C [0,1)
D [0,1]
[0,1] is closed and bounded, so it is compact in ℝ. Open intervals like (0,1) are not compact because they do not include their boundary points.
A continuous function on [a,b] must attain
A Only supremum
B Only infimum
C Maximum and minimum
D Only zero value
On a compact set like [a,b], continuous functions achieve both maximum and minimum values. This is a key application of compactness and is used widely in analysis.
Maclaurin series expansion uses derivatives at
A x=1
B x=0
C x=−1
D x=π
Maclaurin is the Taylor series centered at 0. Coefficients come from derivatives at 0, giving a power series representation near 0 whenever the function is sufficiently analytic.
Taylor series remainder idea measures
A Error after truncation
B Exact sum always
C Only radius value
D Only endpoint test
The remainder term estimates how far a finite Taylor polynomial is from the true function. It helps judge accuracy of approximation and shows when a series truly represents the function.
Power series for sin x and cos x converge for
A |x|<1 only
B x in [0,1]
C All real x
D x in integers
The Maclaurin series for sin x and cos x have infinite radius of convergence, so they converge for every real x. This makes them reliable for approximations across ℝ.