JEE Main 2012 Offline
Exam JEE Main
Year 2012
Mode Offline
Subject Mathematics
Chapter Trigonometric Equations JEE Main 2012 · Offline · Mathematics · Trigonometric Equations
The equation e^ x -e^ - x -4=0 has
The equation
e sin x − e − sin x − 4 = 0 e^{\sin x}-e^{-\sin x}-4=0 e s i n x − e − s i n x − 4 = 0 has
Options A infinite number of real roots B no real roots C exactly one real root D exactly four real roots Correct answer B. no real roots
Step-by-step solution
e sin x − e − sin x = 4 ⇒ e sin x = t e^{\sin x}-e^{-\sin x}=4 \quad \Rightarrow e^{\sin x}=t e s i n x − e − s i n x = 4 ⇒ e s i n x = t t − 1 t = 4 t-\frac{1}{t}=4 t − t 1 = 4 t 2 − 4 t − 1 = 0 ⇒ t = 4 ± 16 + 4 2 ⇒ t = 4 ± 2 5 2 ⇒ t = 2 ± 5 \begin{array}{ll}t^2-4 t-1=0 & \Rightarrow t=\frac{4 \pm \sqrt{16+4}}{2} \\ \Rightarrow t=\frac{4 \pm 2 \sqrt{5}}{2} & \Rightarrow t=2 \pm \sqrt{5}\end{array} t 2 − 4 t − 1 = 0 ⇒ t = 2 4 ± 2 5 ⇒ t = 2 4 ± 16 + 4 ⇒ t = 2 ± 5 e sin x = 2 ± 5 − 1 ≤ sin x ≤ 1 1 e ≤ e sin x ≤ e e^{\sin x}=2 \pm \sqrt{5} \quad \quad-1 \leq \sin x \leq 1 \quad \quad \frac{1}{e} \leq e^{\sin x} \leq e e s i n x = 2 ± 5 − 1 ≤ sin x ≤ 1 e 1 ≤ e s i n x ≤ e e sin x = 2 + 5 e^{\sin x}=2+\sqrt{5} e s i n x = 2 + 5 not possible
e sin x = 2 − 5 e^{\sin x}=2-\sqrt{5} e s i n x = 2 − 5 not possible
∴ \therefore ∴ hence no solution
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