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Sequences And Series Of Real Numbers -4 - IIT JAM MCQ


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20 Questions MCQ Test - Sequences And Series Of Real Numbers -4

Sequences And Series Of Real Numbers -4 for IIT JAM 2024 is part of IIT JAM preparation. The Sequences And Series Of Real Numbers -4 questions and answers have been prepared according to the IIT JAM exam syllabus.The Sequences And Series Of Real Numbers -4 MCQs are made for IIT JAM 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Sequences And Series Of Real Numbers -4 below.
Solutions of Sequences And Series Of Real Numbers -4 questions in English are available as part of our course for IIT JAM & Sequences And Series Of Real Numbers -4 solutions in Hindi for IIT JAM course. Download more important topics, notes, lectures and mock test series for IIT JAM Exam by signing up for free. Attempt Sequences And Series Of Real Numbers -4 | 20 questions in 60 minutes | Mock test for IIT JAM preparation | Free important questions MCQ to study for IIT JAM Exam | Download free PDF with solutions
Sequences And Series Of Real Numbers -4 - Question 1
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If  then

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 1

Correct Answer :- D

Explanation : For |a|<1 

b = 1/a, |b|>1 

lim (n→∞) an = lim (n→∞) (1/b)n

= lim(n→∞) 1/bn

= 1/(±∞)

= 0

 

Sequences And Series Of Real Numbers -4 - Question 2
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Which amongst the following statements is true?

Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 2

Option c will be correct answer of this 

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Sequences And Series Of Real Numbers -4 - Question 3
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A Cauchy sequence is convergent, if it is a

Sequences And Series Of Real Numbers -4 - Question 4
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If radius of convergence of series is 1, then radius of convergence of series is…
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 4

By the differentiation’s theorem, we have
...(i)

Differentiating term by term, we obtain
...(ii)

So, both the series have same radius of convergence.

Sequences And Series Of Real Numbers -4 - Question 5
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If {xn} and {yn} are two convergent sequence such that xn < yn,  n ∈ N, then
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 5

The correct option is Option A.

Let lim xn = x and lim yn = y and zn = yn - xn

      n->∞                n->∞

Then (zn) is a convergent sequence such that zn > 0 ∇ n ∊N

and lim zn = y - x. 

       n->∞

Now (zn) is a convergent sequence of real numbers and zn > 0 ∇ n ∊N

So, lim zn ≥ 0

      n->∞

So, y - x ≥ 0

     => y ≥ x

     => lim yn ≥ lim xn

          n->∞      n->∞

 

Hence, proved.

Sequences And Series Of Real Numbers -4 - Question 6
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Let a = min{x2 + 2x + 3, x ∈ R} and  then the value of 
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 6

Sequences And Series Of Real Numbers -4 - Question 7
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The sequence 
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 7

Correct Answer :- b

Explanation : Suppose otherwise, that there exists a number L implies R and a positive integer N such that

| f(n) - L | < e {for all } e > 0 {for all } n > N. 

Since N is a positive integer, we know 4N > N and 4N+2 > N. 

But,

f(4N) = cos (2N pi) = 1 and f(4N+2) = cos((2N+1)pi) = -1

Taking e = 1/2 

= |1 - L| < 1/2 and |-1-L| < 1/2 

=> |1+L| < 1/2

But, these imply

|1-L| + |1+L| < 1.

By the triangle inequality we then have

|1 - L + 1 + L | < 1

=> 2 < 1

a contradiction. Hence, there is no such limit L.

Therefore, the sequence converges to zero.

Sequences And Series Of Real Numbers -4 - Question 8
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If for any e > 0, there exists a positive integer m such that I an - am | < e whenever n ≥ m . The sequence an > is called
Sequences And Series Of Real Numbers -4 - Question 9
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A convergent sequence is a Cauchy sequence, if it is a
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 9

ANSWER :- b

Solution :- If {an}∞n=1 is a cauchy sequence of real numbers and if there is a sub-sequence of this sequence, {anj}∞j=1 which converges to a real number L, then I need to show that the sequence {an}∞n=1 converges to the real number L.

Sequences And Series Of Real Numbers -4 - Question 10
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Sequences And Series Of Real Numbers -4 - Question 11
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Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 11


Sequences And Series Of Real Numbers -4 - Question 12
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For the given Arithmetic progression find the position of first negative term? 50, 47, 44, 41,............ 
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 12

Let nth term=0, the next term would be first negative term.
0 = 50 + (n - 1) – 3, n = 17.66.. therefore at n = 18 the first negative term would occur.

Sequences And Series Of Real Numbers -4 - Question 13
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In the given AP series the term at position 11 would be? 5, 8, 11, 14, 17, 20.........50. 
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 13

nth term = a + (n – 1)d, nth term = 5 + (11 - 1)3 = 35.

*Multiple options can be correct
Sequences And Series Of Real Numbers -4 - Question 14
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Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 14




 

Sequences And Series Of Real Numbers -4 - Question 15
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If sequences  are convergent, then
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 15

- Convergent sequences have limits as n→∞.
- Let sequences an and bn converge to L and M respectively.

- The sequence (an​+bn​) converges to L + M.

- The sequence (an​bn​) converges to L x M.


- Thus, an + bn is always convergent, confirming option A.

Sequences And Series Of Real Numbers -4 - Question 16
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A sequence contains a convergent subsequence, if it is
Sequences And Series Of Real Numbers -4 - Question 17
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 be a sequence converges to 0 and  be a sequence that is bounded, then    is a sequence that
Sequences And Series Of Real Numbers -4 - Question 18
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Which of the following sequences of functions is uniformly convergent on (0, 1)?
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 18


 

Sequences And Series Of Real Numbers -4 - Question 19
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The sequence {xn}, where xn = nl/n, converge to
Sequences And Series Of Real Numbers -4 - Question 20
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Let xn = 22n for all n ∈ N. Then the sequence {xn}
Detailed Solution for Sequences And Series Of Real Numbers -4 - Question 20

xn = 22n for all n ∈ N

 

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