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JEE Main 201323 AprilOnline
Exam
JEE Main
Year
2013
Date
23 April
Mode
Online
Subject
Mathematics
Chapter
Determinants

JEE Main 2013 · 23 April · Online · Mathematics · Determinants

Statement-1: The system of linear equations aligned & x+( α)y+(...

Statement-1: The system of linear equations x+(sin⁡α)y+(cos⁡α)z=0x+(cos⁡α)y+(sin⁡α)z=0x−(sin⁡α)y−(cos⁡α)z=0\begin{aligned}& x+(\sin \alpha)y+(\cos \alpha)z=0 \\ & x+(\cos \alpha)y+(\sin \alpha)z=0 \\ & x-(\sin \alpha)y-(\cos \alpha)z=0 \end{aligned}has a non-trivial solutionforonlyonevalueofαfor only one value of\alpha lying in the interval (0,π2)\left(0, \frac{\pi}{2}\right). Statement-2: The equation in \alpha\left|cos⁡αsin⁡αcos⁡αsin⁡αcos⁡αsin⁡αcos⁡α−sin⁡α−cos⁡α\begin{array}{ccc} \cos \alpha& \sin \alpha& \cos \alpha\\ \sin \alpha& \cos \alpha& \sin \alpha\\ \cos \alpha& -\sin \alpha& -\cos \alpha \end{array}\right|=0 has only one solution lying in the interval $\left(0, \frac{\pi}{2}\right)

Options

  1. AStatement-1 is true, Statement-2 is true, Statement-2 is not correct explantion for Statement-1.
  2. BStatement-1 is true, Statement-2 is true, Statement-2 is a correct explantion for Statement-1.
  3. CStatement-1 is true, Statement- 2 is false.
  4. DStatememt-1 is false, Statement-2 is true.

Correct answer

C. Statement-1 is true, Statement- 2 is false.

Step-by-step solution

⇒α=64x3×16x=43Δ1=∣1sin⁡αcos⁡α1cos⁡αsin⁡α1−sin⁡αcos⁡α∣=∣0sin⁡α−cos⁡αcos⁡α−sin⁡α0cos⁡α+sin⁡αsin⁡α−cos⁡α1−sin⁡αcos⁡α∣=(sin⁡α−cos⁡α)2−(cos⁡2α−sin⁡2α)=sin⁡2α+cos⁡2α−2sin⁡α⋅cos⁡α−cos⁡2α=2sin⁡2α−2sin⁡α⋅cos⁡α=2sin⁡2α(sin⁡α−cos⁡α)\begin{aligned}& \Rightarrow \alpha=\frac{64 x}{3 \times 16 x}=\frac{4}{3}\\ & \Delta_1=\left|\begin{array}{ccc}1 & \sin \alpha& \cos \alpha\\ 1 & \cos \alpha& \sin \alpha\\ 1 & -\sin \alpha& \cos \alpha \end{array}\right|\\ & =\left|\begin{array}{ccc}0 & \sin \alpha-\cos \alpha& \cos \alpha-\sin \alpha\\ 0 & \cos \alpha+\sin \alpha& \sin \alpha-\cos \alpha\\ 1 & -\sin \alpha& \cos \alpha \end{array}\right|\\ & =(\sin \alpha-\cos \alpha)^2-\left(\cos ^2 \alpha-\sin ^2 \alpha\right)\\ & =\sin ^2 \alpha+\cos ^2 \alpha-2 \sin \alpha \cdot \cos \alpha-\cos ^2 \alpha\\ & =2 \sin ^2 \alpha-2 \sin \alpha \cdot \cos \alpha\\ & =2 \sin ^2 \alpha(\sin \alpha-\cos \alpha) \end{aligned} Now, sin⁡α−cos⁡α=0\sin \alpha-\cos \alpha=0 for only α=π4 in (0,π2)∴Δ1=2(sin⁡α)×0=0,\begin{aligned}\alpha& =\frac{\pi}{4} \text{ in } \left(0, \frac{\pi}{2}\right)\\ \therefore \quad \Delta_1 & =2(\sin \alpha) \times0=0, \end{aligned}since value of \sin \alpha is finite for \alpha \in \left(0, \frac{\pi}{2}\right)Hence non-trivivial solution for only one value of \alpha in \left(0, \frac{\pi}{2}\right) ∣cos⁡αsin⁡αcos⁡αsin⁡αcos⁡αsin⁡αcos⁡α−sin⁡α−cos⁡α∣=0⇒∣0sin⁡αcos⁡α0cos⁡αsin⁡α2cos⁡α−sin⁡α−cos⁡α∣=0⇒2cos⁡α(sin⁡2α−cos⁡2α)=0∴cos⁡α=0 or sin⁡2α−cos⁡2α=0\begin{aligned}& \left|\begin{array}{ccc} \cos \alpha& \sin \alpha& \cos \alpha\\ \sin \alpha& \cos \alpha& \sin \alpha\\ \cos \alpha& -\sin \alpha& -\cos \alpha \end{array}\right|=0 \\ & \Rightarrow\left| \begin{array}{ccc}0 & \sin \alpha& \cos \alpha\\ 0 & \cos \alpha& \sin \alpha\\ 2 \cos \alpha& -\sin \alpha& -\cos \alpha \end{array} \right|= 0 \\ & \Rightarrow 2 \cos \alpha\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ & \therefore \cos \alpha=0 \text{ or } \sin ^2 \alpha-\cos ^2 \alpha=0 \end{aligned}But \cos \alpha=0notpossibleforanyvalueofnot possible for any value ofα∈(0,π2)∴sin⁡2α−cos⁡2α=0⇒sin⁡α=−cos⁡α,\begin{aligned}& \alpha \in \left(0, \frac{\pi}{2}\right)\\ & \therefore \sin ^2 \alpha-\cos ^2 \alpha=0 \Rightarrow \sin \alpha=-\cos \alpha, \end{aligned}whichisalsonotpossibleforanyvalueofwhich is also not possible for any value of\alpha \in \left(0, \frac{\pi}{2}\right)$Hence, there is no solution.
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