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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Determinants

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Determinants

Consider the matrices : A= [ array ll 2 & -5 3 & m array ], B= [...

Consider the matrices : A=A=\left[2−53m\begin{array}{ll}2 & -5 \\ 3 & m\end{array}\right], B=\left[20m\begin{array}{l}20 \\ m\end{array}\right] and X=\left[xy\begin{array}{l}x \\ y\end{array}\right]. Let the set of all mm, for which the system of equations AX=BA X=B has a negative solution (i.e., x<0x < 0 and y<0y < 0 ), be the interval (a,b)(a, b). Then 8∫ab∣A∣dm8 \int_a^b|A| d m is equal to_________

Correct answer

0

Step-by-step solution

A=(2−53m),B=(20m)X=(xy)2x−5y=20…(1)3x+my=m…(2)⇒y=2m−602m+15y<0⇒m∈(−152,30)\begin{aligned} & A=\left(\begin{array}{cc}2 & -5 \\ 3 & m\end{array}\right), B=\left(\begin{array}{c}20 \\ m\end{array}\right) \\ & X=\left(\begin{array}{l}x \\ y\end{array}\right) \\ & 2 x-5 y=20 \ldots(1)\\ & 3 x+m y=m \ldots(2)\\ & \Rightarrow y=\frac{2 m-60}{2 m+15} \\ & y < 0 \Rightarrow m \in\left(\frac{-15}{2}, 30\right)\end{aligned} x=25 m2 m+15x<0⇒m∈(−152,0)⇒m∈(−152,0)∣A∣=2 m+15\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, 8∫−1520(2 m+15)dm=8{ m2+15 m}−1520⇒8{−(2254−2252)}=8×2254=450\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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