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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

If the equation |2 x+ ^2 a|+|2 x+3+2 a|=0 has exactly one solution...

If the equation ∣2x+sin⁡2a∣+∣2x+3+2sin⁡a∣=0\left|2 x+\sin ^2 a\right|+|2 x+3+2 \sin a|=0 has exactly one solution x=λx=\lambda (where ' aa ' is a constant) then find the value of 4λ24 \lambda^2.

Correct answer

1

Step-by-step solution

∣2x+sin⁡2a∣+∣2x+3+2sin⁡a∣=0⇒2x+sin⁡2a=0 and 2x+3+2sin⁡a=0 Hence, sin⁡2a−2sin⁡a−3=0⇒⇒sin⁡a=−1∴2x=−1⇒x=λ=−1/2∴4λ2=1\begin{array}{rlrlrl} & \left|2 x+\sin ^2 a\right|+|2 x+3+2 \sin a| & =0 & & \\ \Rightarrow & 2 x+\sin ^2 a & =0 & \text { and } & 2 x+3+2 \sin a & =0 \\ \text { Hence, } & \sin ^2 a-2 \sin a-3 & =0 \Rightarrow & \Rightarrow & \sin a=-1 \\ \therefore & 2 x & =-1 \Rightarrow & & & x=\lambda=-1 / 2 \\ \therefore & 4 \lambda^2 & =1 & & & \end{array}
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