NIMCET 2011 Question Paper with solutions

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Question : 74 of 120

Marks: +1, -0

Suppose

X=2^{100}, Y=3^{100}

and

Z=4^{100}

. Exactly one of the following is true. Which is it?

$X+Y=Z$
$X+Y < Z$
$X+Y > Z$
$X Y=Z$

Solution:

Suppose,

2^{2}+3^{2}<4^{2}( ∵ 4+9<16)

⇒ \;\; 2^{3}+3^{3}<4^{3}( ∵ 8+27<64)

⇒ \;\; 2^{4}+3^{4}<4^{4}( ∵ 16+81<256)

So, solving the pattern be conclude that for higher value

n>1

2^{n}+3^{n}<4^{n}

Then,

X+Y < Z

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