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What is the value of ab if  \mathrm{ \lim _{x \rightarrow 0} \frac{\sqrt[3]{a x+b}-2}{x}=\frac{7}{12}}

Option: 1

56


Option: 2

12


Option: 3

36


Option: 4

25


Answers (1)

best_answer

                                        \mathrm{ y=\frac{\sqrt[3]{a x+b}-2}{x} }

using Taylor expansion around \mathrm{x=0} or binomial theorem

                             \mathrm{ \sqrt[3]{a x+b}=\sqrt[3]{b}+\frac{a x}{3 b^{2 / 3}}-\frac{a^2 x^2}{9 b^{5 / 3}}+O\left(x^3\right) }
making

         \mathrm{ y=\frac{\sqrt[3]{b}-2+\frac{a x}{3 b^{2 / 3}}-\frac{a^2 x^2}{9 b^{5 / 3}}+O\left(x^3\right)}{x}=\frac{\sqrt[3]{b}-2}{x}+\frac{a}{3 b^{2 / 3}}-\frac{a^2 x}{9 b^{5 / 3}}+O\left(x^2\right) }

So, \mathrm{\sqrt[3]{b}-2 \Longrightarrow b=8 \, \, and\, \, \frac{a}{3 b^{2 / 3}}=\frac{7}{12} \Longrightarrow a=7.}

                           \mathrm{ a b=7 \times 8=56 }

Posted by

Irshad Anwar

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