Let f(x) = x + 1/2. Then, the number of real values of x for which the three unequal terms f(x), f(2x), f(4x) are in HP is
1
0
3
2
The remainder obtained when 1! + 2! + 3! + ... + 11! is divided by 12 is
9
8
7
6
A.
9
Let S = 1! + 2! + 3! + 4! + ... + 11!
Here, we see that from 4! to 11!, we always get a 12 factor, so it is always divisible by 12
Now, 1! + 2! + 3! = 1 + 2 + 6 = 9
Hence, when S is divided by 12, the remainder is 9.
For every real number x,
let f(x) = Then, the equation f(x) = 0 has
no real solution
exactly one real solution
exactly two real solutions
infinite number of real solutions