Verified Solution • 2025–26 Session moderate

QJEEM23Ma31c6: JEE Main 2023 Mathematics - Class 12 JEEM23Ma31c6 Solution

lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3\lim\limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^6+(\sqrt{3 x+1}-\sqrt{3 x-1})^6}{\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6} x^3

Multiple Choice Options

  • A is equal to 9
  • B is equal to 272\frac{27}{2}
  • C does not exist
  • D is equal to 27

Step-by-Step Verified Solution

lim⁡x→∞(3x+1+3x−1)6+(3x+1−3x−1)6(x+x2−1)6+(x−x2−1)6x3\lim \limits_{x \rightarrow \infty} \frac{(\sqrt{3 x+1}+\sqrt{3 x-1})^{6}+(\sqrt{3 x+1}-\sqrt{3 x-1})^{6}}{\left(x+\sqrt{x^{2}-1}\right)^{6}+\left(x-\sqrt{x^{2}-1}\right)^{6}} x^{3}

amp;=lim⁡x→∞x3×{x3{(3+1x+3−1x)6+(3+1x−3−1x)6}x6{(1+1−1x2)6+(1−1−1x2)6}}amp;=(23)6+026+0=33=27\begin{aligned}& = \lim \limits_{x \rightarrow \infty} x^{3} \times\left\{\frac{x^{3}\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^{6}+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^{6}\right\}}{x^{6}\left\{\left(1+\sqrt{1-\frac{1}{x^{2}}}\right)^{6}+\left(1-\sqrt{1-\frac{1}{x^{2}}}\right)^{6}\right\}}\right\} \\\\& =\frac{(2 \sqrt{3})^{6}+0}{2^{6}+0}=3^{3}=27\end{aligned}

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