Q.

Let f:R→R be a continuous and differentiable function  such that fx2+1x=5 for ∀x∈(0,∞) . Then the value  of f16+y2y24y for y∈(0,∞) is equal to

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a

5

b

25

c

125

d

625

answer is B.

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Detailed Solution

fx2+1x=5   ⇒f16+y2y24y =f4y2+12y2 =52=25
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