^You might be aware of the fact that the exponential functions grow much faster than polynomial functions.
Let us assume for now that for all values of n, e < d and e < c.
Now lets illustrate by counterexample that our assumption based on our intuition is wrong!
Consider some large value of n.
Let n = 2^32
c) (2^32)^1.75 = 7.2057594e+16
d) (2^32)(log 2^32)^9 = 3.0675921e+18 //base 10
e) 1.0000001^(2^32) = 3.373265e+186
Hence, our intuition based assumption that for all values of n, e < d and e < c is false.
Let's visualise these functions:
//Observe colors infront of the functions to find the corresponding curve in the graph
// Eg: Red <=> n^1.75
Above graph justifies our intuition that 1.0000001^n is approx 1^n i.e. is 1.
Lets zoom out a bit.
Still the curve of 1.0000001^n is hovering around 1.
Let's zoom out more.
Still nothing.
Voilla!
Observe how 1.0000001^n has stopped hovering around 1 and grown so quickly as n has become large.
As we keep on increasing n, at some point e) will overtake c) and d) as we have seen in our
calculation performed above.
PS: 1) Don't confuse between 1.0000001^32 and 1.0000001^(2^32) .
^ ^
| |
= 1 = 3.373265e+186
2) From above calculation as well as graphs, it appears that c < d, therefore option(c)
looks like more apt. answer.