Verified Solution • 2025–26 Session moderate QJEEM24Ma3829: JEE Main 2024 Mathematics - Class 12 JEEM24Ma3829 Solution: If lim x → 1 ( 5 x + 1 ) 1 / 3 − ( x + 5 ) 1 / 3 ( 2 x + 3 ) 1… If limx→1(5x+1)1/3−(x+5)1/3(2x+3)1/2−(x+4)1/2=m5n(2n)2/3, where gcd(m,n)=1\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1gcd(m,n)=1 , then 8 m+12n8 \mathrm{~m}+12 \mathrm{n}8 m+12n is equal to _______. Step-by-Step Verified Solution I=limx→1(5x+1)1/3−(+5)1/(2x+3)1/2−(x+4)1/2 From: 00\frac{0}{0}00 , using L−H\mathrm{L}-\mathrm{H}L−H rule amp;I=limx→113×5(5x+1)−2/3−−(+5)12×2(2x+3)−1/2−12(x+4)−1/2amp;=(53−13∣ 6−2/3125−1/2=83×51/262/3=m5n(2n)2/3amp;⇒m=8,n=3amp;⇒8m+12n=100\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}amp;I=x→1lim21×2(2x+3)−1/2−21(x+4)−1/231×5(5x+1)−2/3−−(+5)amp;=215−1/2(35−316−2/3=38×62/351/2=n(2n)2/3m5amp;⇒m=8,n=3amp;⇒8m+12n=100 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring