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
n > 1
n
n > 2
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
n > 3
n > 4
n≥ 4
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
n > 2
n≥3
n≥ 4
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
n < 4
n≥ 4
n < 3
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
n > 5
n≥ 5
n < 5
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
n > 1
n≥ 1
n > 2
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
n>2n
n<2n
n≥2n
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
4
6
10
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
113
123
133
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
25
35
45
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
x + y
x - y
x2 + y2
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
2
4
8
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
2n < n
n2 > 2n
n4 < 10n
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
n < 2
n > 2
n≥ 2
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
=n32
>n32
>n33
=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