(A) Given $a > b > 0$ and $n \geq 2$. Let $a = 4, b = 1$, and $n = 2$. Then $f(2) = 4^{1/2} - 1^{1/2} = 2 - 1 = 1$. And $J(2) = (4 - 1)^{1/2} = \sqrt{

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(A) Given $a > b > 0$ and $n \geq 2$. Let $a = 4, b = 1$, and $n = 2$. Then $f(2) = 4^{1/2} - 1^{1/2} = 2 - 1 = 1$. And $J(2) = (4 - 1)^{1/2} = \sqrt{3} \approx 1.732$. Since $1 In general, for $a > b > 0$ and $n \geq 2$, by the property of powers, $(a - b)^{1/n} > a^{1/n} - b^{1/n}$ holds true. Thus, $f(n) < J(n)$.

Class 11 Mathematics Binomial Theorem Mix Examples-Binomial Theorem

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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:

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A
$f(n) < J(n)$
B
$f(n) > J(n)$
C
$f(n) = J(n)$
D
$f(n) + J(n) = 0$

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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:

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