JEE Main 2018 — Mathematics Calculus

$$\begin{aligned} &\quad S=\left\{(\lambda, \mu) \in \mathrm{R} \times \mathrm{R}: \mathrm{f}(\mathrm{t})=\left(|\lambda| \mathrm{e}^{\mid t}-\mu\right) \sin (2 \mathrm{t} \mid),\right. \\ &t \in R \\ &f(t)=\left(|\lambda| \mathrm{e}^{|t|}-\mu\right) \sin (2|t|) \\ &\left\{\begin{array}{l} \left(|\lambda| e^t-\mu\right) \sin 2 t \quad t>0 \\ \left(|\lambda| e^{-t}-\mu\right)(-\sin 2 t) t < 0 \end{array}\right. \\ &f^{\prime}(t) \\ &\left\{\begin{array}{l} \left(|\lambda| \mathrm{e}^t\right) \sin 2 t+\left(|\lambda| \mathrm{e}^t-\mu\right)(2 \cos 2 t) t>0 \\ |\lambda| \mathrm{e}^{-t} \sin 2 t+\left(|\lambda| e^{-t}-\mu\right)(-2 \cos 2 t) < t < 0 \end{array}\right. \\ &\text { As, } f(t) \text { is differentiable } \\ &\therefore \text { LHD }=\mathrm{RHD} \text { at } t=0 \\ &|\lambda| \cdot \sin 2(0)+\left(|\lambda| \mathrm{e}^0-\mu\right) 2 \cos (0) \\ &=|\lambda| \mathrm{e}^{-0} \cdot \sin 2(0)-2 \cos (0)\left(|\lambda| \mathrm{e}^{-0}-\mu\right) \\ &\Rightarrow 0+(|\lambda|-\mu) 2=0-2(|\lambda|-\mu) \\ &\quad 4(|\lambda|-\mu)=0 \\ &\quad|\lambda|=\mu \end{aligned}$$​S={(λ,μ)∈R×R:f(t)=(∣λ∣e∣t−μ)sin(2t∣),t∈Rf(t)=(∣λ∣e∣t∣−μ)sin(2∣t∣){(∣λ∣et−μ)sin2tt>0(∣λ∣e−t−μ)(−sin2t)t<0​f′(t){(∣λ∣et)sin2t+(∣λ∣et−μ)(2cos2t)t>0∣λ∣e−tsin2t+(∣λ∣e−t−μ)(−2cos2t)<t<0​ As, f(t) is differentiable ∴ LHD =RHD at t=0∣λ∣⋅sin2(0)+(∣λ∣e0−μ)2cos(0)=∣λ∣e−0⋅sin2(0)−2cos(0)(∣λ∣e−0−μ)⇒0+(∣λ∣−μ)2=0−2(∣λ∣−μ)4(∣λ∣−μ)=0∣λ∣=μ​ So, $S \equiv(\lambda, \mu)=\{\lambda \in R \& \mu \in[0, \infty)\}$S≡(λ,μ)={λ∈R&μ∈[0,∞)} Therefore set $\mathrm{S}$S is subset of $R \times[0, \infty)$R×[0,∞)