Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

If the minimum and the maximum values of the function  f:\left [ \frac{\pi }{2},\frac{\pi }{2} \right ]\rightarrow R, defined by  f(\theta )=\begin{vmatrix} -\sin ^{2}\theta & -1-\sin ^{2}\theta & 1\\ -\cos ^{2}\theta &-1-\cos ^{2}\theta & 1\\ 12 & 10 &-2 \end{vmatrix} are m and M respectively, then the ordered pair (m,M) is equal to :
Option: 1 (0,2\sqrt{2})  
Option: 2 (-4,0)
Option: 3 (-4,4)
Option: 4 (0,4)
 

Answers (1)

best_answer

\\\mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\left(\mathrm{C}_{1}-\mathrm{C}_{2}\right) \\ \mathrm{f}(\theta)=\left|\begin{array}{ccc} \sin ^{2} \theta & -1-\sin ^{2} \theta & 0 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 0 \\ 12 & 10 & -4 \end{array}\right| \\ =-4\left[\left(1+\cos ^{2} \theta\right) \sin ^{2} \theta-\cos ^{2} \theta\left(1+\sin ^{2} \theta\right)\right] \\ =-4\left[\sin ^{2} \theta+\cos ^{2} \theta-\cos ^{2} \theta-\cos ^{2} \theta \sin ^{2} \theta\right]

\\\mathrm{f}(\theta)=4 \cos 2 \theta \\ \theta \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \\ 2 \theta \in\left[\frac{\pi}{2}, \pi\right] \\ \mathrm{f}(\theta) \in[-4,0] \\ (\mathrm{m}, \mathrm{M})=(-4,0)

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 result date announced; Steps to check NTA session 2 scores
anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question