Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

Let $A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$ where $0 \leq \theta \leq 2 \pi$. Then ${det}(A)$ lies in the interval :

Options:

$[2,3]$

$[3,4]$

$[2,4]$

$(2,4)$

Correct Answer:

$[2,4]$

Explanation:

$A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right]$

$R_3 \rightarrow R_3+R_1 \quad A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ 0 & 0 & 2\end{array}\right]$

$|A|=\left|\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ 0 & 0 & 2\end{array}\right|$

$=2\left|\begin{array}{cc}1 & \sin \theta \\ -\sin \theta & 1\end{array}\right|$

taking determinant across R3

$=2\left|\begin{array}{cc}1 & \sin \theta \\ -\sin \theta & 1\end{array}\right|$

$f(\theta)=2\left(1+\sin ^2 \theta\right)=|A|$

0 ≤ sin2θ ≤ 1  ⇒ 1 ≤ (1 + sin2θ) ≤ 2  ⇒  2 ≤ 2 (1+sin2θ) ≤ 4

⇒ [2, 4]