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Problem Solving Approach in Calculus-Part-I
Dhrubajyoti Mandal
Email: dhurbajyoti@nitsikkim
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in

Abstract
This is a collection of a few problems of College Level Calculus
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I request you to go through the note once
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1
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Which of the following is true ?
a) f is one-one in the interval [-1,1]
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c) f is NOT one-one in the interval [-4,0]
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Answer: First notice that f is a continuous function
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Now let us think when this kind of
situation happens
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Then there exist a neighbourhood of x = a say (a − δ, a + δ) where on both sides of x = a the value of the function
decreases (or increases) continuously if x = a is a local minima (or maxima)
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So f must not be one-one in
(a − δ, a + δ), in fact f must not be one-one in any interval containing (a − δ, a + δ)
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So first let us find out the local maxima and minima of f
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Now, x = 0 and x = 3 are interior points of
[−1, 1] and [2, 4] respectively
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Therefore options (a) and (b) are incorrect
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So [−4, 0] contains no local maxima or minima as its interior
point
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Lastly, [0, 4] contains the local minima x = 3 as its interior point
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Therefore the option (d) is the correct answer
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Question Suppose {an } be a sequence of positive real numbers
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n→∞

an+1
an
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Now
lim

n→∞

an+1
= l < 1 =⇒ ln < 1, ∀n
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Therefore
n→∞

lim an = 0
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Question: Let S is the sum of a convergent series
convergent ? If convergent then find the sum in terms of S,
Answer: As

∞
P

∞
P

an
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Then is

n=1
a1 , a2
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Again

n=1
N
X

tn = 3

n=1

Therefore

tn

n=1

an converges therefore the sequence N-th partial sum

n=1

∞
P

N
X

an − (2a1 + a2 )
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So

∞
X

tn = 3S − 2a1 − a2

n=1

4
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Which of the following statements are true?
a) There exist a continuous function f : [a b] → (a b) such that f is one-one
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c) There exist a continuous function f : (a b) → [a b] such that f is one-one
d) There exist a continuous function f : (a b) → [a b] such that f is onto

Answer: If [a b] and [c d] are two different closed and bounded intervals then there always exist a map f : [a b] →
[c d] defined by
f (x) = c1 x + c2
where c1 , c2 are functions of a, b, c, d and can be determined by imposing conditions like f (a) = c, f (b) = d
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Then it can be easily seen that this map is one-one (Can you see it!)
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Similarly
it can be shown that, given two different open and bounded intervals we always have a one-one and onto
map between them
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For option (a) take a closed interval [c d] ⊂ (a b)
and we are done
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The option (b) is incorrect as we
know that image of a compact set under a continuous function is compact
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Now define a map f : (a b) → [a b]
as follows:




a
x ∈ (a c)



f (x) = g(x) x ∈ [c d]




b
x ∈ (d b)
Then f : (a b) → [a b] is an onto continuous map and hence the option (d) is correct
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Question: Let f : [0 2] → R be a continuous function such that f (0) = f (2) > f (1)
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This again leads to the fact that
φ(x) = f (x) − f (x + 1) has a root in (0 1)
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Moreover φ : [0 1] → R is continuous
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Hence we
are done (Thanks to Bolzano’s theorem !)
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Question: Let f be a real valued function of real variable, such that |f n (0)| ≤ k for all n ∈ N where k > 0
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n

1

a) | f n!(0) | n → 0 as n → ∞
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c) f ( x) exists for all x ∈ R and for all n ∈ N
∞
P
f n (0)
d)
(n−1)! is absolutely convergent
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Therefore from (1) we get | f n!(0) | n → 0 as n → ∞ by Sandwich theorem
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For option (c) try to construct function which is not differentiable at x = 1 but all order derivative exist and
bounded at x = 0
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Can you see that f is not differentiable at x = 1 and
|f n (0)| ≤ 1
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At last,

∞
X
n=1

as

∞
P
n=1

1
(n−1)!

|

∞
X
f n (0)
1
|≤k
<∞
(n − 1)!
(n − 1)!
n=1

is convergent by d’Alembert’s ratio test
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7
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Which of the
following options may be correct
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b) The range of f is a finite set consisting of more than one element
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d) The range of f is a singleton set
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e
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Also note that D is path-connected (if we take any two points in D then there always exist
a path between them which lies entirely in D) and hence it is connected
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Now as f is continuous and D is compact therefore the range of f i
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f (D) must be a compact subset of R
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Hence the option (c) cannot be true
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e
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Hence option (c) is excluded because
any finite subset of R which contain more than one element is not connected
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Then f (D) is a singleton set
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i
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f (D) must be a closed and bounded interval
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8
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The radius
∞
∞
P
P
of convergence of the power series
an xn is 4, then the power series
bn x n
n=0

n=0

a) Converges for all x with |x| < 2
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c) does not converge for any x with |x| > 2
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Answer : If x = 0, then

∞
P

bn xn = 0
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Now let us find a power series

n=0

radius of convergence 4
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Hence radius of convergence of

n=0

is 4, i
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an =

∞
P

∞
P
n=0

( x4 )n

is a good choice
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Therefore if we take bn = n+1
4n , then this bn satisfies all the given conditions
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Now what if x = 4 and x = 3
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Therefore the options (b) and (c) are false
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So
option (a) correct (Nice! Still want to verify option (a)
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Hint: n+1
4n <
∞
∞
P
P
n
n
the radius of convergence of
bn x and
cn x )
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Question The set A = { 1+x
: −1 < x < 1}, as a subset of R
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b) compact and not connected
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d) not compact and not connected
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(−1, 1) is a connected set in R
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Again as x → 1−, y → 12
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So A is not closed and hence

A is not compact in R (can you see that A is unbounded also)
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