Let S=\ (λ, μ) ∈ R × R: f(t)=(|λ| e^t-μ) · (2|t|)., t ∈ R, is a differentiable function \ . Then S is a subest of?

Correct answer: A. R ×[0, ∞) aligned & S=\ (λ, μ) ∈ R × R: f(t)=(|λ| e^ t -μ) (2 t ),. &t ∈ ℝ &f(t)=(|λ| e^ |t| -μ) (2|t|) &\ array l (|λ|e^t-μ) 2 t t>0 (|λ|e^ -t -μ)(- 2 t) t 0 |λ| e^ -t 2 t+(|λ|e^ -t -μ)(-2 2 t) < t < 0 array . As, f(t) is differentiable & LHD =RHD at t=0 &|λ| · 2(0)+(|λ| e^0-μ)^ 2 (0) &=|λ| e^ -0 · 2(0)-2 (0)(|λ| e^ -0 -μ) & 0+(|λ|-μ)^ 2 =0-2(|λ|-μ) & 4(|λ|-μ)=0 & |λ|=μ aligned So, S (λ, μ)=\ λ ∈ ℝ \& μ ∈[0, ∞)\ Therefore set S is subset of R ×[0, ∞)

JEE Main 201815 AprilMorning ShiftOnline

Exam: JEE Main
Year: 2018
Date: 15 April
Shift: Morning Shift
Mode: Online
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2018 · 15 April · Morning Shift · Online · Mathematics · Continuity and Differentiability

Let S=\ (λ, μ) ∈ R × R: f(t)=(|λ| e^t-μ) · (2|t|)., t ∈ R, is a...

Let

S=\left\{(\lambda, \mu) \in R \times R: f(t)=\left(|\lambda| e^t-\mu\right) \cdot \sin (2|t|)\right., t \in R,\text{ is } a\text{ differentiable }\text{ function } \}

. Then

S

is a subest of?

Options

A $R \times[0, \infty)$
B $(-\infty, 0) \times R$
C $[0, \infty) \times R$
D $R \times(-\infty, 0)$

Correct answer

A. $R \times[0, \infty)$

Step-by-step solution

\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 \mathbb{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{\text{ is }\text{ differentiable } }\\ &\therefore \text{ LHD }=\mathrm{RHD} \text{ at } t=0 \\ &|\lambda| \cdot \sin2(0)+\left(|\lambda| \mathrm{e}^0-\mu\right)^{2}\cos (0) \\ &=|\lambda| \mathrm{e}^{-0} \cdot \sin2(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}

So,

S \equiv(\lambda, \mu)=\{\lambda \in

\mathbb{R}

\& \mu \in[0, \infty)\}

Therefore set \mathrm{S}

is subset of

R \times[0, \infty) $

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