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Question
Let f(x) = `sqrt(1 + x^2)`, then ______.
Options
f(xy) = f(x) . f(y)
f(xy) ≥ f(x) . f(y)
f(xy) ≤ f(x) . f(y)
None of these
Solution
Let f(x) = `sqrt(1 + x^2)`, then f(xy) ≤ f(x) . f(y).
Explanation:
Given that: f(x) = `sqrt(1 + x^2)`
⇒ f(xy) = `sqrt(1 + x^2y^2)`
And f(x) . f(y) = `sqrt(1 + x^2) * sqrt(1 + y^2)`
= `sqrt(1 + x^2 + y^2 + x^2y^2)`
∵ `sqrt(1 + x^2y^2) ≤ sqrt(1 + x^2 + y^2 + x^2y^2)`
⇒ f(xy) ≤ f(x) . f(y)
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