JEE Main 12 Jan 2019 Evening Question 29
Question: If $ a= \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \\ \end{bmatrix} ; $ then for all $ \theta \in ( \frac{3\pi }{4},\frac{5\pi }{4} ), $ det lies in the interval [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ ( \frac{3}{2},3 ] $
B) $ [ \frac{5}{2},4 ) $
C) $ ( 1,\frac{5}{2} ] $
D) $ ( 0,\frac{3}{2} ] $
Show Answer
Answer:
Correct Answer: A
Solution:
- Here, det $ (A)= \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \\ \end{vmatrix} $ $ =(1+sin^{2}\theta )-sin\theta (0)+1(sin^{2}\theta +1) $
$ =2(1+sin^{2}\theta ) $ $ \because $ $ \theta \in ( \frac{3\pi }{4},\frac{5\pi }{4} )\Rightarrow \sin \theta \in ( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} ) $
$ \Rightarrow $ $ {{\sin }^{2}}\theta \in [ 0,\frac{1}{2} ) $
$ \therefore $ $ \det (A)\in [2,3) $
