Q96 / 150 · MHT CET 2024 (15 May Shift 2) · Mathematics

$\lim _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}}=$

  • Correct answer
    A
    $-\frac{4}{3}$
  • B
    $\frac{4}{3}$
  • C
    $\frac{2}{3}$
  • D
    $-\frac{4}{9}$

Solution

$\begin{aligned} & \lim _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{3-x}-3^{\frac{x}{2}}} \\ & =\lim _{x \rightarrow 2} \frac{\left(3^x\right)^2+3^3-12\left(3^x\right)}{3^3-\left(3^{\frac{x}{2}}\right)^3} \\ & =\lim _{x \rightarrow 2} \frac{\left(3^x-9\right)\left(3^x-3\right)}{\left(3-3^{\frac{x}{2}}\right)\left(9+3.3^{\frac{x}{2}}+3^x\right)} \\ & =-\lim _{x \rightarrow 2} \frac{\left(3^{\frac{x}{2}}+3\right)\left(3^{\frac{x}{2}}-3\right)\left(3^x-3\right)}{\left(3^{\frac{x}{2}}-3\right)\left(3^x+3.3^{\frac{x}{2}}+9\right)} \\ & =-\frac{(3+3)(9-3)}{9+9+9}\end{aligned}$ $=\frac{-4}{3}$

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

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