Free Objective Test 01 Practice Test - 11th and 12th 

Question 1

Let P(n) denote the statement that n2 + n is odd. It

is seem that P(n)  ⇒ P(n + 1), Pn is true for all

A.

n > 1

B.

n

C.

n > 2

D.

None of these

SOLUTION

Solution : D

P(n) = n2 + n. It is always odd (statement) but

square of any odd number is always odd and 

also, sum of odd number is always even. So

for no any 'n' for which this statement is true.

Question 2

For natural number n, (n!)2 > nn, if

A.

n > 3

B.

n > 4

C.

n≥ 4

D.

n≥ 3

SOLUTION

Solution : D

Check through option, condition (n!)2 > nn is

true when n ≥ 3.

Question 3

For every positive integral value of n, 3n > n3 when

A.

n > 2

B.

n≥3

C.

n≥ 4

D.

n < 4

SOLUTION

Solution : C

Check through option, the condition 3n > n3 is

true when n ≥ 4.

Question 4

For every positive integer n, 2n < n! when

A.

n < 4

B.

n≥ 4

C.

n < 3

D.

None of these

SOLUTION

Solution : B

Check through option, the condition 2n < n! is

true when n ≥ 4.

Question 5

For positive integer n, 10n−2 > 81n, if

A.

n > 5

B.

n≥ 5

C.

n < 5

D.

n > 6

SOLUTION

Solution : B

Check through option, the condition

10n−2 > 81n is satisfied if n ≥ 5.

Question 6

If n is a natural number then (n+12)n ≥ n ! is true

when

A.

n > 1

B.

n≥ 1

C.

n > 2 

D.

n≥2

SOLUTION

Solution : B

Check through option, the condition

 (n+12)n ≥ n ! is true for n  ≥ 1.

Question 7

For every natural number n

A.

n>2n

B.

n<2n

C.

n≥2n

D.

Can't be determined.

SOLUTION

Solution : B

Let n = 1 then option (a) and (d) is eliminated.

Equality can't be attained for any value of n so,

option (b) satisfied.

Question 8

For every natural number n, n(n2−1) is divisible by

 

A.

4

B.

6

C.

10

D.

None of these

SOLUTION

Solution : B

n(n2−1) = (n - 1)(n)(n + 1)

It is product of three consecutive natural

numbers, so according to Langrange's theorem

it is divisible by 3 ! i.e., 6.

Question 9

If n ∈ N, then 11n+2 + 122n+1 is divisible by

 

A.

113

B.

123

C.

133

D.

None of these

SOLUTION

Solution : C

Putting n = 1 in 11n+2+122n+1

We get, 111+2+122×1+1 = 113+123 = 3059, which

is divisible by 133.

 

Question 10

If n ∈ N, then 72n + 23n−3.3n−1 is always divisible by

 

A.

25

B.

35

C.

45

D.

None of these

SOLUTION

Solution : A

Putting n = 1 in 72n+23n−3.3n−1

=50, divisible by 25

Question 11

If n ∈ N, then x2n−1+y2n−1 is divisible by 

A.

x + y

B.

x - y

C.

x2 + y2

D.

x2+xy

SOLUTION

Solution : A

x2n−1+y2n−1 is always contain equal odd power.

So it is always divisible by x + y.

Question 12

For all positive integral values of n, 32n - 2n + 1 is 

divisible by

A.

2

B.

4

C.

8

D.

12

SOLUTION

Solution : A

Putting n = 2 in 32n - 2n + 1 then,

32×2 - 2×2+1 = 81 - 4 + 1 = 78, which is divisible

by 2.

Question 13

For each n ∈ N, the correct statement is

A.

2n < n

B.

n2 > 2n

C.

n4 < 10n

D.

23n > 7n + 1

SOLUTION

Solution : C

Let n = 1, then option (a), (b) and (d)

eliminated. Only option (c) satisfied.

Question 14

For natural number n, 2n(n-1) ! < nn, if

A.

n < 2

B.

n > 2

C.

n≥ 2

D.

Never

SOLUTION

Solution : B

Check through option, the condition

2n(n-1)!<nn is satisfied for n > 2.

Question 15

To find:
12+22+32...................+n2

A.

=n32

B.

>n32

C.

>n33

D.

=n33

SOLUTION

Solution : C

Let n=1
n32=12;    n33=13
Let n=2,⇒12+22=5
n32=4;    n33=83
Let n=3,⇒12+22+32=14
n32=272;    n33=9
Let n=4,⇒12+22+32+42=30
n32=32;    n33=643
12+22+32+⋯+n2>n33
P(n):12+22+32+⋯+n2>n33
P(1) is true
Let P(k) be true.
⇒12+22+32+⋯+k2>k33
12+22+32+⋯+k2+(k+1)2>k33+k2+2k+1
=k3+3k2+6k+33
=k3+3k2+3k+1+3k+23
=(k+1)33+k+23
12+22+32+⋯+k2+(k+1)2>(k+1)33
P(k+1) is true.
P(n) is true ∀ n∈N

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