JEE Advanced 2012 Paper 1
Exam JEE Advanced
Year 2012
Paper Paper 1
Subject Mathematics
Chapter Trigonometric Equations JEE Advanced 2012 · Paper 1 · Mathematics · Trigonometric Equations
Let θ, ∈[0,2 π] be such that 2 θ(1- )= ^ 2 θ × ( (θ)/(2)+ (θ)/(2))...
Let
θ , φ ∈ [ 0 , 2 π ] \theta, \varphi \in[0,2 \pi] θ , φ ∈ [ 0 , 2 π ] be such that
2 cos θ ( 1 − sin φ ) = sin 2 θ × ( tan θ 2 + cot θ 2 ) cos φ − 1 , tan ( 2 π − θ ) > 0 2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta \times \left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1, \tan (2 \pi-\theta)>0 2 cos θ ( 1 − sin φ ) = sin 2 θ × ( tan 2 θ + cot 2 θ ) cos φ − 1 , tan ( 2 π − θ ) > 0 and
− 1 < sin θ < − 3 2 -1 < \sin \theta < -\frac{\sqrt{3}}{2} − 1 < sin θ < − 2 3 , then
φ \varphi φ cannot satisfy
Options A 0 < φ < π 2 0 < \varphi < \frac{\pi}{2} 0 < φ < 2 π B π 2 < φ < 4 π 3 \frac{\pi}{2} < \varphi < \frac{4 \pi}{3} 2 π < φ < 3 4 π C 4 π 3 < φ < 3 π 2 \frac{4 \pi}{3} < \varphi < \frac{3 \pi}{2} 3 4 π < φ < 2 3 π D 3 π 2 < φ < 2 π \frac{3 \pi}{2} < \varphi < 2 \pi 2 3 π < φ < 2 π Correct answer A. 0 < φ < π 2 0 < \varphi < \frac{\pi}{2} 0 < φ < 2 π
Step-by-step solution
As
tan ( 2 π − θ ) > 0 \tan (2 \pi-\theta)>0 tan ( 2 π − θ ) > 0 and
− 1 < sin θ < − 3 2 , θ ∈ [ 0 , 2 π ] -1 < \sin \theta < -\frac{\sqrt{3}}{2}, \theta \in[0,2 \pi] − 1 < sin θ < − 2 3 , θ ∈ [ 0 , 2 π ]
Hence
3 π 2 < θ < 5 π 3 \frac{3 \pi}{2} < \theta < \frac{5 \pi}{3} 2 3 π < θ < 3 5 π
Now
2 cos θ ( 1 − sin φ ) = sin 2 θ ( tan θ 2 + cot θ 2 ) cos φ − 1 2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1 2 cos θ ( 1 − sin φ ) = sin 2 θ ( tan 2 θ + cot 2 θ ) cos φ − 1 ⇒ 2 cos θ ( 1 − sin φ ) = 2 sin θ cos φ − 1 \Rightarrow 2 \cos \theta(1-\sin \varphi)=2 \sin \theta \cos \varphi-1 ⇒ 2 cos θ ( 1 − sin φ ) = 2 sin θ cos φ − 1 ⇒ 2 cos θ + 1 = 2 sin ( θ + φ ) \Rightarrow 2 \cos \theta+1=2 \sin (\theta+\varphi) ⇒ 2 cos θ + 1 = 2 sin ( θ + φ )
As
θ ∈ ( 3 π 2 , 5 π 3 ) , 1 < 2 sin ( θ + φ ) < 2 \quad \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right), 1 < 2 \sin (\theta+\varphi) < 2 θ ∈ ( 2 3 π , 3 5 π ) , 1 < 2 sin ( θ + φ ) < 2
As
θ + φ ∈ ( π 6 , 5 π 6 ) \theta+\varphi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right) θ + φ ∈ ( 6 π , 6 5 π ) or
( θ + φ ) ∈ ( 13 π 6 , 17 π 6 ) (\theta+\varphi) \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right) ( θ + φ ) ∈ ( 6 13 π , 6 17 π )
We have
φ ∈ ( − 3 π 2 , − 2 π 3 ) ∪ ( 2 π 3 , 7 π 6 ) \varphi \in\left(-\frac{3 \pi}{2},-\frac{2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right) φ ∈ ( − 2 3 π , − 3 2 π ) ∪ ( 3 2 π , 6 7 π )
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