If 0.01 < a < 0.02, and 100 < b < 200, which of the following can be

The given expression is as follows : \(\frac{a^{3}*b + b^{3}*a}{a^{3}*b^{3}}\)
It can be further simplified as \(\frac{ab(a^{2} + b^{2})}{ab(a^{2}*b^{2})}\) = \(\frac{a^{2} + b^{2}}{a^{2}*b^{2}}\)

Now coming to values that a and b can take :
a will be of form \(x * 10^{-2}\) and b will be of form \(y * 10^2\)

Substituting these values,
\(\frac{a^{2} + b^{2}}{a^{2}*b^{2}}\) = \(\frac{(x^2 * 10^{-4}) + (y^2 * 10^{4})}{(x^2 * 10^{-4}) * (y^2 * 10^{4})}\) = \(\frac{(x^2 * 10^{-4}) + (y^2 * 10^{4})}{(x^2 * 10^{-4}) * (y^2 * 10^{4})}\)

This equation can be approximated to (since x and y are extremely small)
\(\frac{(10^{-4}) + (10^{4})}{(10^{-4}) * (10^{4})}\) = \(\frac{(10^{-4}) + (10^{4})}{(10^{-4 + 4})}\) (because \(a^m*a^n = a^{m+n}\))
= \(\frac{(10^{-4}) + (10^{4})}{(10^{0})}\) = \(10^4 = 10000\) which is the largest value possible for the expression

The value closest to this will be Option E.

ekniv
Equation can be written as 1/a^2 + 1/b^2


I HAVE THE WAY WHICH YOU ARE LOOKING FOR-

a belongs (0.01,0.02)
1/a belongs ( 50,100)
1/a^2 belongs ( 2500, 100 00 )

b belongs (100,200)
1/b^2 belongs (0.25x 10^-4 , 10^-4 )

Lets find min value of given equation.
2500 + 0.25x10^-4 = 2500.0 something
>2500 which is E

equations can be resolved to
=> 1/a^2 + 1/b^2

1/b^2 is negligible
1/a^2 can be even >5000 if a=0.15

THerefore IMO E

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