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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
Consider the matrices $:\mathrm{A}=\left[\begin{array}{cc}2 & -5 \\ 3 & \mathrm{~m}\end{array}\right], \mathrm{B}=\left[\begin{array}{l}20 \\ \mathrm{~m}\end{array}\right]$ and $X=\left[\begin{array}{l}x \\ y\end{array}\right]$. Let the set of all $m$, for which the system of equations $\mathrm{AX}=\mathrm{B}$ has a negative solution $($i.e., $\mathrm{x}<0$ and $\mathrm{y}<0 ),$ be the interval $($ $\mathrm{a}, \mathrm{b} ).$ Then $8 \int_a^b|\mathrm{~A}| \mathrm{dm}$ is equal to$.............$
  • A
    $324$
  • B
    $450$
  • C
    $234$
  • D
    $110$
✓

Answer

$\begin{aligned} & A=\left(\begin{array}{ll}2 & -5 \\ 3 & \mathrm{~m}\end{array}\right), \mathrm{B}=\left(\begin{array}{c}20 \\ \mathrm{~m}\end{array}\right) \\ & \mathrm{X}=\left(\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right)\end{aligned}$
$ 2 \mathrm{x}-5 \mathrm{y}=20 $
$ 3 \mathrm{x}+\mathrm{my}=\mathrm{m} $
$ \Rightarrow \mathrm{y}=\frac{2 \mathrm{~m}-60}{2 \mathrm{~m}+15} $
$ \mathrm{y}<0 \Rightarrow \mathrm{m} \in\left(\frac{-15}{2}, 30\right) $
$ \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 $
 Now,
$ 8 \int_{-15}^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}$
$=450$
Now,
$ 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}$
$=450$

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