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

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.

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.

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

A.

n > 2

B.

n3

C.

n 4

D.

n < 4

#### SOLUTION

Solution : C

Check through option, the condition 3n > n3 is

true when n ≥ 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.

For positive integer n, 10n2 > 81n, if

A.

n > 5

B.

n 5

C.

n < 5

D.

n > 6

#### SOLUTION

Solution : B

Check through option, the condition

10n2 > 81n is satisfied if n ≥ 5.

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

when

A.

n > 1

B.

n 1

C.

n > 2

D.

n2

#### SOLUTION

Solution : B

Check through option, the condition

(n+12)n ≥ n ! is true for n  ≥ 1.

For every natural number n

A.

n>2n

B.

n<2n

C.

n2n

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.

For every natural number n, n(n21) is divisible by

A.

4

B.

6

C.

10

D.

None of these

#### SOLUTION

Solution : B

n(n21) = (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.

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.

If n ∈ N, then 72n + 23n3.3n1 is always divisible by

A.

25

B.

35

C.

45

D.

None of these

#### SOLUTION

Solution : A

Putting n = 1 in 72n+23n3.3n1

=50, divisible by 25

If n ∈ N, then x2n1+y2n1 is divisible by

A.

x + y

B.

x - y

C.

x2 + y2

D.

x2+xy

#### SOLUTION

Solution : A

x2n1+y2n1 is always contain equal odd power.

So it is always divisible by x + y.

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.

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.

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.

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  nN