CUET Mathematics 2024 - =|arrayccc1 & x & 1 \\ - x & 1 & x \\ -1 & - x & 1array| (A) =2(1- ^2 x) (B) =2(2- ^2 x) (C) Minimum value of is 2 (D) Maximum value of is 4 Choose the correct answer from the options given below : | PYQs + Solutions | AfterBoards
Skip to main contentSkip to question navigationSkip to solution
IPMAT Indore Free Mocks Topic Tests

CUET Mathematics 2024 PYQs

View all CUET PYQs →
Collection of AfterBoards IPMAT JIPMAT Past Year Papers with Solutions

CUET Mathematics 2024

Algebra
>
Matrices & Determinants

Easy

Δ=∣1cos⁡x1−cos⁡x1cos⁡x−1−cos⁡x1∣\Delta=\left|\begin{array}{ccc}1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1\end{array}\right|
(A) Δ=2(1−cos⁡2x)\Delta=2\left(1-\cos ^{2} x\right)
(B) Δ=2(2−sin⁡2x)\Delta=2\left(2-\sin ^{2} x\right)
(C) Minimum value of Δ\Delta is 22
(D) Maximum value of Δ\Delta is 44
Choose the correct answer from the options given below :

✅ Correct Option: 4
We need to calculate the determinant Δ=∣1cos⁡x1−cos⁡x1cos⁡x−1−cos⁡x1∣\Delta=\left|\begin{array}{ccc}1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1\end{array}\right|
Expanding the determinant using cofactor expansion along the first row:
Δ=1⋅∣1cos⁡x−cos⁡x1∣−cos⁡x⋅∣−cos⁡xcos⁡x−11∣+1⋅∣−cos⁡x1−1−cos⁡x∣\Delta = 1 \cdot \begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix}
Calculating each 2×2 determinant:
∣1cos⁡x−cos⁡x1∣=1⋅1−cos⁡x⋅(−cos⁡x)=1+cos⁡2x\begin{vmatrix} 1 & \cos x \\ -\cos x & 1 \end{vmatrix} = 1 \cdot 1 - \cos x \cdot (-\cos x) = 1 + \cos^2 x
∣−cos⁡xcos⁡x−11∣=(−cos⁡x)⋅1−cos⁡x⋅(−1)=−cos⁡x+cos⁡x=0\begin{vmatrix} -\cos x & \cos x \\ -1 & 1 \end{vmatrix} = (-\cos x) \cdot 1 - \cos x \cdot (-1) = -\cos x + \cos x = 0
∣−cos⁡x1−1−cos⁡x∣=(−cos⁡x)⋅(−cos⁡x)−1⋅(−1)=cos⁡2x+1\begin{vmatrix} -\cos x & 1 \\ -1 & -\cos x \end{vmatrix} = (-\cos x) \cdot (-\cos x) - 1 \cdot (-1) = \cos^2 x + 1
Substituting back into the original equation:
Δ=1⋅(1+cos⁡2x)−cos⁡x⋅0+1⋅(1+cos⁡2x)\Delta = 1 \cdot (1 + \cos^2 x) - \cos x \cdot 0 + 1 \cdot (1 + \cos^2 x)
Δ=1+cos⁡2x+1+cos⁡2x\Delta = 1 + \cos^2 x + 1 + \cos^2 x
Δ=2+2cos⁡2x\Delta = 2 + 2\cos^2 x
Using the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x:
Δ=2+2(1−sin⁡2x)\Delta = 2 + 2(1 - \sin^2 x)
Δ=2+2−2sin⁡2x\Delta = 2 + 2 - 2\sin^2 x
Δ=4−2sin⁡2x\Delta = 4 - 2\sin^2 x
Δ=2(2−sin⁡2x)\Delta = 2(2 - \sin^2 x)
Since 0≤sin⁡2x≤10 \leq \sin^2 x \leq 1 for all real values of xx: \newline - Maximum value: when sin⁡2x=0\sin^2 x = 0, giving Δmax=2(2−0)=4\Delta_{max} = 2(2 - 0) = 4 \newline - Minimum value: when sin⁡2x=1\sin^2 x = 1, giving Δmin=2(2−1)=2\Delta_{min} = 2(2 - 1) = 2
The correct options are (B), (C), and (D).

Related questions:

CUET Mathematics 2024

If AA is a square matrix and II is an identity matrix such that A2=AA^{2}=A, then A(I−2A)3+2A3A(I-2 A)^{3}+2 A^{3} is equal to :

CUET Mathematics 2024

For a square matrix An×nA_{n \times n}
(A) ∣adj⁡A∣=∣A∣n−1|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}
(B) ∣A∣=∣adj⁡A∣n−1|\mathrm{A}|=|\operatorname{adj} \mathrm{A}|^{\mathrm{n}-1}
(C) A(adj⁡A)=∣A∣\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}|
(D) ∣A−1∣=1∣ A∣\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}

CUET Mathematics 2024

If AA and BB are symmetric matrices of the same order, then AB−BAA B-B A is a :

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question