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  • <img alt="\text { Let } a, n \in N \text { such that } a \geq n^3 \text {. Then } \sqrt[3]{a+1}-\sqrt[3]{a} \text { is always }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Ctext%20%7B%2

\text { Let } a, n \in N \text { such that } a \geq n^3 \text {. Then } \sqrt[3]{a+1}-\sqrt[3]{a} \text { is always }

Option: 1

\text { less than } \frac{1}{3 n^2}


Option: 2

\text { less than } \frac{1}{2 n^3}


Option: 3

\text { more than } \frac{1}{n^3}


Option: 4

\text { more than } \frac{1}{4 n^2}


Answers (1)

best_answer

 Let \mathrm{f(x)=x^{1 / 3} }
\mathrm{\Rightarrow f^{\prime}(x)=\frac{1}{3 x^{2 / 3}} }
Applying Lagrange's Mean Value Theorem in [a, a+1], we get atleast one \mathrm{c \in(a, a+1) }
\mathrm{\begin{aligned} & f^{\prime}(c)=\frac{f(a+1)-f(a)}{a+1-a} \\ & \Rightarrow \sqrt[3]{a+1}-\sqrt[3]{a}=\frac{1}{3 c^{2 / 3}}<\frac{1}{3 a^{2 / 3}}<\frac{1}{3 n^2} \\ & \Rightarrow \sqrt[3]{a+1}-\sqrt[3]{a}<\frac{1}{3 n^2} \forall a \geq n^3 \end{aligned} }
 

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Rakesh

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