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Directions for the next 3 questions: For three distinct real positive numbers x, y and z, let
f(x, y, z) = min (max(x, y), max (y, z), max (z, x))
g(x, y, z) = max (min(x, y), min (y, z), min (z, x))
h(x, y, z) = max (max(x, y), max(y, z), max (z, x))
j(x, y, z) = min (min (x, y), min(y, z), min (z, x))
m(x, y, z) = max (x, y, z)
n(x, y, z) = min (x, y, z)
From the given functions we can make out that function h and m give max value , function n and j give min value , function f and g give middle value.So in option B , j cancels out n and h cancels out m . So the denominator becomes 0 and value is indeterminable.
Suppose x>y>z
f(x,y,z) = y
g(x,y,z) = y
h(x,y,z) = x
j(x,y,z) = z
m(x,y,z) = x
n(x,y,z) = z
The denominator of the second option becomes 0, hence making it indeterminate.
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