JEE Main 2024 — Determinants Question with Solution
JEE Main 2024 (09 Apr Shift 2)
Question
Consider the matrices : and . Let the set of all , for which the system of equations has a negative solution (i.e., and ), be the interval . Then is equal to_________
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Show full solutionCorrect answer: 450
Correct answer
450
Step-by-step explanation
$\begin{aligned}
& \mathrm{x}=\frac{25 \mathrm{~m}}{2 \mathrm{~m}+15} \\
& \mathrm{x} < 0 \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) \\
& \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 0\right) \\
& |\mathrm{A}|=2 \mathrm{~m}+15
\end{aligned}$
Now, $\begin{aligned} & 8 \int_{\frac{-15}{2}}^0(2 \mathrm{~m}+15) \mathrm{dm}=8\left\{\mathrm{~m}^2+15 \mathrm{~m}\right\}_{\frac{-15}{2}}^0 \\ & \Rightarrow 8\left\{-\left(\frac{225}{4}-\frac{225}{2}\right)\right\} \\ & =8 \times \frac{225}{4}=450 \end{aligned}$
Now, $\begin{aligned} & 8 \int_{\frac{-15}{2}}^0(2 \mathrm{~m}+15) \mathrm{dm}=8\left\{\mathrm{~m}^2+15 \mathrm{~m}\right\}_{\frac{-15}{2}}^0 \\ & \Rightarrow 8\left\{-\left(\frac{225}{4}-\frac{225}{2}\right)\right\} \\ & =8 \times \frac{225}{4}=450 \end{aligned}$
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This is a previous-year question from JEE Main 2024, covering the Determinants chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.