WBJEE · Maths · Basic of Mathematics
Let \(a>b>0\) and \(f(n)=a^{1 / n}-b^{1 / n}\),\(J(n)=(a-b)^{1 / n}\) for all \(n \geq 2,\) Then
- A \(f(n) < J(n)\)
- B \(f(n)>J(n)\)
- C \(f(n)=J(n)\)
- D \(f(n)+J(n)=0\)
Answer & Solution
Correct Answer
(A) \(f(n) < J(n)\)
Step-by-step Solution
Detailed explanation
Let \(a=4, b=1\) and \(n=2\) \(\begin{array}{l} \therefore \quad I(n)=a^{1 / n}-b^{1 / n} \\ =4^{1 / 2}-1^{1 / 2}=2-1=1 \\ \text { and } J(n)=(a-b)^{1 / n} \\ =(4-1)^{1 / 2}=\sqrt{3} \\ \therefore \quad J(n)>I(n) \end{array}\) Similarly, we can prove for other values of…
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