Verified Solution • 2025–26 Session moderate QJEEM24Ma84f0: JEE Main 2024 Mathematics - Class 12 JEEM24Ma84f0 Solution: The values of , for which , lie in the interval The values of α\alphaα , for which ∣1amp;32amp;α+321amp;13amp;α+132α+3amp;3α+1amp;0∣=0\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0112α+3amp;23amp;31amp;3α+1amp;α+23amp;α+31amp;0=0 , lie in the interval Multiple Choice Options A (−2,1)(-2,1)(−2,1) B (−32,32)\left(-\frac{3}{2}, \frac{3}{2}\right)(−23,23) C (−3,0)(-3,0)(−3,0) D (0,3)(0,3)(0,3) Step-by-Step Verified Solution ∣1amp;32amp;α+321amp;13amp;α+132α+3amp;3α+1amp;0∣=0\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\1 & \frac{1}{3} & \alpha+\frac{1}{3} \\2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0112α+3amp;23amp;31amp;3α+1amp;α+23amp;α+31amp;0=0 amp;⇒(2α+3){7α6}−(3α+1){−76}=0amp;⇒(2α+3)⋅7α6+(3α+1)⋅76=0amp;⇒2α2+3α+3α+1=0amp;⇒2α2+6α+1=0amp;⇒α=−3+72,−3−72\begin{aligned}& \Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0 \\& \Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0 \\& \Rightarrow 2 \alpha^2+3 \alpha+3 \alpha+1=0 \\& \Rightarrow 2 \alpha^2+6 \alpha+1=0 \\& \Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2}\end{aligned}amp;⇒(2α+3){67α}−(3α+1){6−7}=0amp;⇒(2α+3)⋅67α+(3α+1)⋅67=0amp;⇒2α2+3α+3α+1=0amp;⇒2α2+6α+1=0amp;⇒α=2−3+7,2−3−7 Hence option (C) is correct. 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