Given that K! = 1 x 2 x 3 x ... X K, which is
the largest among the following numbers?
A. (2!)1/2
B. (3!)1/3
C. (4!)1/4
D.
(3!)
2
Answers
Answered by
0
Answer:
( d ) Hopeing it is the answer
Answered by
0
Answer:
Answer is D.
Step-by-step explanation:
On perfoming individual calculation of the options as in option A) (2!)^(1/2) = (1x2)^(1/2)= (2)^(1/2)= √2 which approximately = 1.414, in case of option B)(3!)^(1/3)=(1x2x3)^(1/3)= (6)^(1/3)= cube root of 6 is approximately = 1.817, in case of option C) (4!)^(1/4)=(1x2x3x4)^(1/4)=(24)^(1/4)= 2.213 and in case of option D) ((3!)/2)m = ((1x2x3)/2)m=(6/2)m= 3m, which is indeed greater than all answers obtained after solving all options.
i.e. 3m > 2.213 > 1.817 > 1.414.
So the answer is option D.
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