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JEE Main 201412 AprilOnline
Exam
JEE Main
Year
2014
Date
12 April
Mode
Online
Subject
Mathematics
Chapter
Determinants

JEE Main 2014 · 12 April · Online · Mathematics · Determinants

If f(θ)=| array ccc 1 & θ & 1 - θ & 1 & - θ -1 & θ & 1 array | and...

If \mathrm{f}(\theta)=\left|1cos⁡θ1−sin⁡θ1−cos⁡θ−1sin⁡θ1\begin{array}{ccc}1 & \cos \theta & 1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1\end{array}\right| and A\mathrm{A} and B\mathrm{B} are respectively the maximum and the minimum values of f(θ)f(\theta), then (A,B)(A, B) is equal to:

Options

  1. A(3,−1)(3,-1)
  2. B(4,2−2)(4,2-\sqrt{2})
  3. C(2+2,2−2)(2+\sqrt{2}, 2-\sqrt{2})
  4. D(2+2,−1)(2+\sqrt{2},-1)

Correct answer

C. (2+2,2−2)(2+\sqrt{2}, 2-\sqrt{2})

Step-by-step solution

 Let f(θ)=∣1cos⁡θ1−sin⁡θ1−cos⁡θ−1sin⁡θ1∣=(1+sin⁡θcos⁡θ)−cos⁡θ(−sin⁡θ−cos⁡θ)+1(−sin⁡2θ+1)=1+sin⁡θcos⁡θ+sin⁡θcos⁡θ+=2+2sin⁡θcos⁡θ+cos⁡2θ=2+sin⁡2θ+cos⁡2θ\begin{aligned}&\text{ Let }f(\theta)=\left|\begin{array}{ccc}1 & \cos \theta& 1 \\ -\sin \theta& 1 & -\cos \theta\\ -1 & \sin \theta& 1 \end{array}\right| \\ &=(1+\sin \theta \cos \theta)-\cos \theta(-\sin \theta-\cos \theta)\\ &+1\left(-\sin ^2 \theta+1\right) \\ &=1+\sin \theta \cos \theta+\sin \theta \cos \theta+\\ &=2+2 \sin \theta \cos \theta+\cos2 \theta\\ &=2+\sin 2 \theta+\cos2 \theta \end{aligned}Now, maximum value of (1) is 2+12+12=2+22+\sqrt{1^2+1^2}=2+\sqrt{2} and minimum value of (1) is 2-12+12=\sqrt{1^2+1^2}=2-2. \sqrt{2} \text{. }
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