Q80 / 180 · KCET 2019 · Mathematics

If \( P(n): 2^{n} < n ! \) then the smallest positive integer for which \( P(n) \) is true, is

  • A
    \( 03 \)
  • B
    \( 05 \)
  • C
    \( 02 \)
  • Correct answer
    D
    \( 04 \)

Solution

$\$ P(\mathrm{n}): 2^{\wedge}\{\mathrm{n}\}$ Since $P(1): 2 < 1$ is false. $P(2): 2^{2} < 2 !$ is false $P(3): 2^{3} < 3 !$ is false $P(4): 2^{4} < 4 !$ is true. $P(5): 2^{5} < 5 !$ is true. So, the smallest positive integer is $4 .$

Hence, the correct answer is option: 'D'.