JEE MAIN - Mathematics (2024 - 4th April Morning Shift - No. 24)
If $$\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(x+5)^{1 / 3}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}=\frac{\mathrm{m} \sqrt{5}}{\mathrm{n}(2 \mathrm{n})^{2 / 3}}$$, where $$\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$, then $$8 \mathrm{~m}+12 \mathrm{n}$$ is equal to _______.
Answer
100
Explanation
$$I=\lim _\limits{x \rightarrow 1} \frac{(5 x+1)^{1 / 3}-(+5)^{1 /}}{(2 x+3)^{1 / 2}-(x+4)^{1 / 2}}$$
From: $$\frac{0}{0}$$, using $$\mathrm{L}-\mathrm{H}$$ rule
$$\begin{aligned} & I=\lim _{x \rightarrow 1} \frac{\frac{1}{3} \times 5(5 x+1)^{-2 / 3}--(+5)}{\frac{1}{2} \times 2(2 x+3)^{-1 / 2}-\frac{1}{2}(x+4)^{-1 / 2}} \\ & =\frac{\left(\left.\frac{5}{3}-\frac{1}{3} \right\rvert\, 6^{-2 / 3}\right.}{\frac{1}{2} 5^{-1 / 2}}=\frac{8}{3} \times \frac{5^{1 / 2}}{6^{2 / 3}}=\frac{m \sqrt{5}}{n(2 n)^{2 / 3}} \\ & \Rightarrow m=8, n=3 \\ & \Rightarrow 8 m+12 n=100 \end{aligned}$$
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