Q101 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

$\lim _{x \rightarrow 3 / 2} \frac{\left(4 x^2-6 x\right)\left(4 x^2+6 x+9\right)}{\sqrt[3]{2 x}-\sqrt[3]{3}}=$

  • Correct answer
    A
    $\sqrt[3]{3^{17}}$
  • B
    $\sqrt[3]{3^{16}}$
  • C
    $\sqrt[3]{3^{15}}$
  • D
    $\sqrt[3]{3^{14}}$

Solution

$\begin{aligned} & \text { } \lim _{x \rightarrow \frac{3}{2}} \frac{\left(4 x^2-6 x\right)\left(4 x^2+6 x+9\right)}{(2 x)^{1 / 3}-3^{1 / 3}} \\ & =\lim _{x \rightarrow \frac{3}{2}} \frac{2 x(2 x-3)\left(4 x^2+6 x+9\right)\left[(2 x)^{2 / 3}+3^{2 / 3}+(6 x)^{1 / 3}\right]}{(2 x-3)} \\ & =\lim _{x \rightarrow \frac{3}{2}} 2 x\left(4 x^2+6 x+9\right)\left[(2 x)^{2 / 3}+3^{2 / 3}+(6 x)^{1 / 3}\right] \\ & =3 \times(9+9+9)\left(3^{2 / 3}+3^{2 / 3}+3^{2 / 3}\right)=3^5 \times 3^{2 / 3}=3^{\frac{17}{3}} \end{aligned}$

Hence, the correct answer is option: 'A'.

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