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MCQMediumJEE 2024Solving Linear Equations (Matrix Method)

JEE Mathematics 2024 Question with Solution

Consider the matrices AA = [2−53m]\begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, BB = [20m]\begin{bmatrix} 20 \\ m \end{bmatrix}, and XX = [xy]\begin{bmatrix} x \\ y \end{bmatrix}. Let the set of all mm, for which the system of equations AX=BAX = 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| \, dm is equal to:

  • A

    450450

  • B

    400400

  • C

    500500

  • D

    350350

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

A=[2−53m],B=[20m],X=[xy]A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, \quad B = \begin{bmatrix} 20 \\ m \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}

Find: 8∫ab∣A∣ dm8\int_a^b |A| \, dm where AX=BAX = B has a negative solution, i.e. x<0x < 0 and y<0y < 0.

From AX=BAX = B, we get the system

2x−5y=202x - 5y = 20

and

3x+my=m3x + my = m

From the first equation,

2x=20+5y2x = 20 + 5y

so

x=10+52yx = 10 + \frac{5}{2}y

Substitute into the second equation:

3(10+52y)+my=m3\left(10 + \frac{5}{2}y\right) + my = m 30+152y+my=m30 + \frac{15}{2}y + my = m y(152+m)=m−30y\left(\frac{15}{2} + m\right) = m - 30

Hence,

y=m−30152+m=2m−602m+15y = \frac{m - 30}{\frac{15}{2} + m} = \frac{2m - 60}{2m + 15}

For y<0y < 0,

2m−602m+15<0\frac{2m - 60}{2m + 15} < 0

which gives

−152<m<30-\frac{15}{2} < m < 30

Also,

x=10+52⋅m−30152+m=25m2m+15x = 10 + \frac{5}{2}\cdot \frac{m - 30}{\frac{15}{2} + m} = \frac{25m}{2m + 15}

For x<0x < 0,

25m2m+15<0\frac{25m}{2m + 15} < 0

which gives

−152<m<0-\frac{15}{2} < m < 0

Therefore,

(a,b)=(−152,0)(a,b) = \left(-\frac{15}{2}, 0\right)

Now,

∣A∣=det⁡(A)=2m+15|A| = \det(A) = 2m + 15

On the interval (−152,0)\left(-\frac{15}{2}, 0\right), we have 2m+15>02m + 15 > 0, so

∣A∣=2m+15|A| = 2m + 15

Therefore,

8∫−1520(2m+15) dm=8[m2+15m]−15208\int_{-\frac{15}{2}}^0 (2m + 15) \, dm = 8\left[m^2 + 15m\right]_{-\frac{15}{2}}^0 =8(0−(2254−2252))= 8\left(0 - \left(\frac{225}{4} - \frac{225}{2}\right)\right) =8⋅2254=450= 8\cdot \frac{225}{4} = 450

Therefore, the correct option is A.

Use direct expressions for $$x$$ and $$y$$

Given: the system AX=BAX=B with

A=[2−53m],B=[20m]A = \begin{bmatrix} 2 & -5 \\ 3 & m \end{bmatrix}, \quad B = \begin{bmatrix} 20 \\ m \end{bmatrix}

Find: the value of 8∫ab∣A∣ dm8\int_a^b |A| \, dm.

From the worked solution,

y=2m−602m+15,x=25m2m+15y = \frac{2m - 60}{2m + 15}, \quad x = \frac{25m}{2m + 15}

Now apply the negativity conditions directly.

For y<0y < 0,

−152<m<30-\frac{15}{2} < m < 30

For x<0x < 0,

−152<m<0-\frac{15}{2} < m < 0

Hence the common interval is

(a,b)=(−152,0)(a,b) = \left(-\frac{15}{2}, 0\right)

Next,

∣A∣=2m+15|A| = 2m + 15

which is positive on (−152,0)\left(-\frac{15}{2},0\right). So,

8∫−1520(2m+15) dm=4508\int_{-\frac{15}{2}}^0 (2m+15) \, dm = 450

Therefore, the correct option is A.

Common mistakes

  • Students may find the interval for only y<0y < 0 and stop there. This is wrong because the question requires both x<0x < 0 and y<0y < 0. Always intersect the ranges obtained from both inequalities.

  • Students may treat ∣A∣|A| as the matrix itself instead of the determinant. Here ∣A∣|A| means det⁡(A)=2m+15\det(A)=2m+15. In determinant notation for matrices, vertical bars denote determinant, not modulus of each entry.

  • Students may forget the absolute value while integrating and use sign-changing behavior incorrectly. First check the sign of 2m+152m+15 on the interval (−152,0)\left(-\frac{15}{2},0\right); since it is positive there, ∣2m+15∣=2m+15|2m+15| = 2m+15 on that interval.

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