JEE Main 2018 · 15 April · Mathematics · Continuity and Differentiability
B. R×[0,∞))R \times \left[0, \infty )\right)R×[0,∞))
Let S={(λ,μ)∈R×R}S = \left\{\left(\lambda, \mu\right) \in R \times R\right\}S={(λ,μ)∈R×R} and
f(t)=(∣λ∣e∣t∣−μ).sin2∣t∣f\left(t\right) = \left(\left|\lambda\right|{e}^{\left|t\right|} - \mu\right) . \sin 2\left|t\right|f(t)=(∣λ∣e∣t∣−μ).sin2∣t∣
RHD=limh→0f(0+h)−0h=limh→0(∣λ∣eh−μ)sin2hh=2(∣λ∣−μ)RHD = \lim_{h \to 0} \dfrac{f\left(0 + h\right) - 0}{h} = \lim_{h \to 0} \left(\left|\lambda\right|{e}^{h} - \mu\right)\dfrac{\sin 2 h}{h} = 2 ( \left|\lambda\right| - \mu )RHD=limh→0hf(0+h)−0=limh→0(∣λ∣eh−μ)hsin2h=2(∣λ∣−μ)
LHD=limh→0f(0−h)−0−h=limh→0(∣λ∣eh−μ)sin2h−h=−2(∣λ∣−μ)LHD = \lim_{h \to 0} \dfrac{f\left(0 - h\right) - 0}{- h} = \lim_{h \to 0} \left(\left|\lambda\right|{e}^{h} - \mu\right)\dfrac{\sin 2 h}{- h} = - 2 ( \left|\lambda\right| - \mu )LHD=limh→0−hf(0−h)−0=limh→0(∣λ∣eh−μ)−hsin2h=−2(∣λ∣−μ)
Now, LHD=RHDLHD = RHDLHD=RHD if ∣λ∣=μ\left|\lambda\right| = \mu∣λ∣=μ
⇒μ≥0and⇒ \mu \ge 0 and⇒μ≥0and\lambda \in R$
