If a and b are natural numbers such that a2 - b2 is a prime number, then
a2 - b2 = 1
a2 - b2 = 2
a2 - b2 = a - b
a2 - b2 = a + b
The statement P (n ): ( 1 x 1! ) + (2 x 2! ) + (3 x 3! ) + .... + ( n x n !) = ( n + 1 )! - 1' is
True for all values of n > 1
Not true for any value of n
True for all values of n ∈ N
None of these
P (n) = P (n + 1 ) for all natural numbers n, then P (n) is true ?
For all n
For all n > 1
For all n > m
Nothing can be said
The number 101 x 102 x 103 x 104 x ..... x 107 is divisible by .
4000
4050
5040
5050
If x 3 > ( x2 + x + 2 ), then
x < 2
x ≥ 2
x > 2
x ≤ 2
The solution of the inequality 2x2 + x - 15 ≥ 0 is
The expression 3 2n + 2 - 8n - 9 is divisible by 64 for all
n ∈ N
n ∈ N, n < 2
n ∈ N n ≥ 2
n ∈ N, n > 2
If n > 1 and x ≠ 0. then expression ( 1 + x)n - nx -1 is divisible by
x2
x3
x5
x7
The solution of the inequality is.
( 2/3, 8 )
( -2, 8/3 )
All possible two - factor products are from the digits 1,2,3,4, ...., 200. The number of factors out of the total obtained, which are multiples of 5, is
8040
7180
6150
4040
