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# Sequences And Series Of Real Numbers -4

## 20 Questions MCQ Test IIT JAM Mathematics | Sequences And Series Of Real Numbers -4

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This mock test of Sequences And Series Of Real Numbers -4 for IIT JAM helps you for every IIT JAM entrance exam. This contains 20 Multiple Choice Questions for IIT JAM Sequences And Series Of Real Numbers -4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Sequences And Series Of Real Numbers -4 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Sequences And Series Of Real Numbers -4 exercise for a better result in the exam. You can find other Sequences And Series Of Real Numbers -4 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
QUESTION: 1

### If  then

Solution:

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

QUESTION: 2

Solution:
QUESTION: 3

### A Cauchy sequence is convergent, if it is a

Solution:
QUESTION: 4

If radius of convergence of series is 1, then radius of convergence of series is…

Solution:

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

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

So, both the series have same radius of convergence.

QUESTION: 5

If {xn} and {yn} are two convergent sequence such that xn < yn,  n ∈ N, then

Solution:

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.

QUESTION: 6

Let a = min{x2 + 2x + 3, x ∈ R} and  then the value of

Solution:

QUESTION: 7

The sequence

Solution:

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.

QUESTION: 8

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

Solution:
QUESTION: 9

A convergent sequence is a Cauchy sequence, if it is a

Solution:

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.

QUESTION: 10

Solution:
QUESTION: 11

Solution:
QUESTION: 12

For the given Arithmetic progression find the position of first negative term? 50, 47, 44, 41,............

Solution:

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.

QUESTION: 13

In the given AP series the term at position 11 would be? 5, 8, 11, 14, 17, 20.........50.

Solution:

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

QUESTION: 14

Which amongst the following statements is not true?

Solution:
QUESTION: 15

If sequences  are convergent, then

Solution:
QUESTION: 16

A sequence contains a convergent subsequence, if it is

Solution:
QUESTION: 17

be a sequence converges to 0 and  be a sequence that is bounded, then    is a sequence that

Solution:
QUESTION: 18

Which of the following sequences of functions is uniformly convergent on (0, 1)?

Solution:

QUESTION: 19

The sequence {xn}, where xn = nl/n, converge to

Solution:
QUESTION: 20

Let xn = 22n for all n ∈ N. Then the sequence {xn}

Solution:

xn = 22n for all n ∈ N