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Matrices and Determinants question

2020 · 5 Sep · Shift 1 · Q30
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  5. /2020 · 5 Sep · Shift 1 · Q30

Matrices and Determinants question

2020 · 5 Sep · Shift 1 · Q30

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
If the minimum and the maximum values of the function f:[π4,π2]→Rf:\left[ {{\pi \over 4},{\pi \over 2}} \right] \to Rf:[4π​,2π​]→R, defined by f(θ)=∣−sin⁡2θ−1−sin⁡2θ1−cos⁡2θ−1−cos⁡2θ11210−2∣f\left( \theta \right) = \left| {\begin{matrix} { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \\ { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \\ {12} & {10} & { - 2} \\ \end{matrix} } \right|f(θ)=​−sin2θ−cos2θ12​−1−sin2θ−1−cos2θ10​11−2​​ are m and M respectively, then the ordered pair (m,M) is equal to :
  1. A
    (0,22)\left( {0,2\sqrt 2 } \right)(0,22​)
  2. B
    (-4, 0)
  3. C
    (-4, 4)
  4. D
    (0, 4)
View written solutionFree

Correct answer: B

  1. Let
D(θ)=∣−sin⁡2θ−1−sin⁡2θ1−cos⁡2θ−1−cos⁡2θ11210−2∣.D(\theta)=\begin{vmatrix} -\sin^2\theta & -1-\sin^2\theta & 1\\ -\cos^2\theta & -1-\cos^2\theta & 1\\ 12 & 10 & -2 \end{vmatrix}.D(θ)=​−sin2θ−cos2θ12​−1−sin2θ−1−cos2θ10​11−2​​.

We need the minimum and maximum values of f(θ)=D(θ)f(\theta)=D(\theta)f(θ)=D(θ) on [π4,π2]\left[\frac\pi4,\frac\pi2\right][4π​,2π​].

  1. Use the identity sin⁡2θ+cos⁡2θ=1.\sin^2\theta+\cos^2\theta=1.sin2θ+cos2θ=1. Set

so that a+b=1a+b=1a+b=1, with a∈[12,1],b=1−a.a\in\left[\frac12,1\right],\qquad b=1-a.a∈[21​,1],b=1−a. Then

D=∣−a−1−a1−b−1−b11210−2∣.D=\begin{vmatrix} -a & -1-a & 1\\ -b & -1-b & 1\\ 12 & 10 & -2 \end{vmatrix}.D=​−a−b12​−1−a−1−b10​11−2​​.
  1. Subtract row 2 from row 1: R1→R1−R2.R_1\to R_1-R_2.R1​→R1​−R2​. Then
D=∣−(a−b)−(a−b)0−b−1−b11210−2∣.D=\begin{vmatrix} -(a-b) & -(a-b) & 0\\ -b & -1-b & 1\\ 12 & 10 & -2 \end{vmatrix}.D=​−(a−b)−b12​−(a−b)−1−b10​01−2​​.

Factor out −(a−b)-(a-b)−(a−b) from the first row:

D=−(a−b)∣110−b−1−b11210−2∣.D=-(a-b)\begin{vmatrix} 1 & 1 & 0\\ -b & -1-b & 1\\ 12 & 10 & -2 \end{vmatrix}.D=−(a−b)​1−b12​1−1−b10​01−2​​.
  1. Now compute the remaining determinant by expanding along the first row:
∣110−b−1−b11210−2∣=1∣−1−b110−2∣−1∣−b112−2∣.\begin{vmatrix} 1 & 1 & 0\\ -b & -1-b & 1\\ 12 & 10 & -2 \end{vmatrix} =1\begin{vmatrix}-1-b & 1\\ 10 & -2\end{vmatrix}-1\begin{vmatrix}-b & 1\\ 12 & -2\end{vmatrix}.​1−b12​1−1−b10​01−2​​=1​−1−b10​1−2​​−1​−b12​1−2​​.

Compute each minor:

∣−1−b110−2∣=(−1−b)(−2)−10=2+2b−10=2b−8,\begin{vmatrix}-1-b & 1\\ 10 & -2\end{vmatrix}=(-1-b)(-2)-10=2+2b-10=2b-8,​−1−b10​1−2​​=(−1−b)(−2)−10=2+2b−10=2b−8, ∣−b112−2∣=(−b)(−2)−12=2b−12.\begin{vmatrix}-b & 1\\ 12 & -2\end{vmatrix}=(-b)(-2)-12=2b-12.​−b12​1−2​​=(−b)(−2)−12=2b−12.

Hence

(2b−8)−(2b−12)=4.(2b-8)-(2b-12)=4.(2b−8)−(2b−12)=4.

Therefore, D=−4(a−b).D=-4(a-b).D=−4(a−b).

  1. Since a−b=sin⁡2θ−cos⁡2θ=−cos⁡2θ,a-b=\sin^2\theta-\cos^2\theta=-\cos2\theta,a−b=sin2θ−cos2θ=−cos2θ, we get

So f(θ)=4cos⁡2θ.f(\theta)=4\cos2\theta.f(θ)=4cos2θ.

  1. Now θ∈[π4,π2]\theta\in\left[\frac\pi4,\frac\pi2\right]θ∈[4π​,2π​], so 2θ∈[π2,π].2\theta\in\left[\frac\pi2,\pi\right].2θ∈[2π​,π]. On this interval, cos⁡2θ\cos2\thetacos2θ decreases from 000 to −1-1−1. Hence f(θ)=4cos⁡2θf(\theta)=4\cos2\thetaf(θ)=4cos2θ varies from 0down to−4.0 \quad \text{down to} \quad -4.0down to−4. Thus, M=0,m=−4.M=0,\qquad m=-4.M=0,m=−4.

  2. Therefore,

(m,M)=(−4,0).(m,M)=(-4,0).(m,M)=(−4,0).
  1. Checking options:
  • A: incorrect
  • B: correct
  • C: incorrect
  • D: incorrect
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