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Bunuel
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Given that x is even, the quickest approach is plug in 2 for x.

Only D can produce an odd value.
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are there any way I needn't try one by one?
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JoeAa
are there any way I needn't try one by one?

Given that the question asks to find an option that could be odd, we need to evaluate each option until you reach an option that could produce an odd result. Unless, of course, you right away stumble to the correct option.
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Even X Odd = Even ; So, it can't produce odd, unless I found out 2^2X3/4 -- in that case only it produces odd. I think if one is unable to get that -- they could mark the wrong one. Is there any other efficient way as it took me 3+ mins to do it.
shameekv1989
a) A can be rewritten in the form \(\frac{x(x+1)(x+2)}{4}\) i.e. 3 consecutive integers -> Since x is even then the (x+2) will be divisible by 4 (whenever it is possible; won't be when x is -2 or 0 but the expression won't be odd in such cases) and the expression will be even.

b) \(x^3−3x^2+2x\) = \(x*(x^2-3x+2)\) = \(x*(x-1)*(x-2)\) -> Again multiplication of 3 consecutive integers -> Always even

c) \(x^2-4x+6\) -> Since x is even this term will always be even as even*even = even and even + even = even

d) \(\frac{x^3+x^2}{4}\) = \(\frac{x^2*(x+1)}{4}\) -> \(x^2\) will always be even and will always be divisible by 4 since x is a multiple of 2. x+1 will always be odd since x is even. Hence this could be ODD ---> Answer

e) \(\frac{(x-2)(x^2+2x)}{4}\) = \(\frac{(x-2)x(x+2)}{4}\) -> again since x is even x+2 will be divisible by 4 as in case a)
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