Question

Solution

Correct option is

decreasing on [0, )    Since log function is an increasing function and e < π,

log (e + x) < log (π + x). Thus (e + x) log (e + x) < (e + x)

log (π + x) < (π + x) log (π + x) for all x > 0.

Thus,f’ (x) < 0 for ∀ x > 0 ⇒ f (x) decreases on (0, ∞)

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Q10

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