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Limits Continuity and Differentiability question

2018 · 15 Apr · Shift 1 · Q45
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  5. /2018 · 15 Apr · Shift 1 · Q45

Limits Continuity and Differentiability question

2018 · 15 Apr · Shift 1 · Q45

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let S = {(λ\lambdaλ, μ\muμ) ∈\in∈ R ×\times× R : f(t) = (|λ\lambdaλ| e|t| −μ-\mu−μ). sin (2|t|), t ∈\in∈ R, is a differentiable function}. Then S is a subset of :
  1. A
    R ×\times× [0, ∞\infty∞)
  2. B
    [0, ∞\infty∞) ×\times× R
  3. C
    R ×\times× (−∞-\infty−∞, 0)
  4. D
    (−∞-\infty−∞, 0) ×\times× R
View written solutionFree

Correct answer: A

We need the set of all (λ,μ)∈R×R(\lambda,\mu)\in\mathbb R\times\mathbb R(λ,μ)∈R×R such that

f(t)=(∣λ∣e∣t∣−μ)sin⁡(2∣t∣),t∈Rf(t)=\big(|\lambda|e^{|t|}-\mu\big)\sin(2|t|),\qquad t\in\mathbb Rf(t)=(∣λ∣e∣t∣−μ)sin(2∣t∣),t∈R

is differentiable on R\mathbb RR.

Since ∣t∣|t|∣t∣ is involved, the only possible issue is at t=0t=0t=0. For t≠0t\neq 0t=0, all pieces are smooth, so fff is differentiable automatically.

1. Write f(t)f(t)f(t) separately for t>0t>0t>0 and t<0t<0t<0

For t>0t>0t>0, ∣t∣=t|t|=t∣t∣=t, so

f(t)=(∣λ∣et−μ)sin⁡(2t).f(t)=\big(|\lambda|e^t-\mu\big)\sin(2t).f(t)=(∣λ∣et−μ)sin(2t).

For t<0t<0t<0, ∣t∣=−t|t|=-t∣t∣=−t, so

f(t)=(∣λ∣e−t−μ)sin⁡(−2t)=−(∣λ∣e−t−μ)sin⁡(2t).f(t)=\big(|\lambda|e^{-t}-\mu\big)\sin(-2t) = -\big(|\lambda|e^{-t}-\mu\big)\sin(2t).f(t)=(∣λ∣e−t−μ)sin(−2t)=−(∣λ∣e−t−μ)sin(2t).

Also,

f(0)=(∣λ∣e0−μ)sin⁡0=0.f(0)=\big(|\lambda|e^0-\mu\big)\sin 0=0.f(0)=(∣λ∣e0−μ)sin0=0.

2. Check differentiability at t=0t=0t=0

We compute the derivative at 000 using one-sided limits.

Right-hand derivative

For h>0h>0h>0,

\frac{f(h)-f(0)}{h}= rac{\big(|\lambda|e^h-\mu\big)\sin(2h)}{h}.

As h→0+h\to 0^+h→0+,

∣λ∣eh−μ→∣λ∣−μ,sin⁡(2h)h→2.|\lambda|e^h-\mu \to |\lambda|-\mu, \qquad \frac{\sin(2h)}{h}\to 2.∣λ∣eh−μ→∣λ∣−μ,hsin(2h)​→2.

Hence,

f+′(0)=2(∣λ∣−μ).f'_+(0)=2(|\lambda|-\mu).f+′​(0)=2(∣λ∣−μ).

Left-hand derivative

For h<0h<0h<0,

\frac{f(h)-f(0)}{h}= rac{-\big(|\lambda|e^{-h}-\mu\big)\sin(2h)}{h}.

As h→0−h\to 0^-h→0−,

∣λ∣e−h−μ→∣λ∣−μ,sin⁡(2h)h→2.|\lambda|e^{-h}-\mu \to |\lambda|-\mu, \qquad \frac{\sin(2h)}{h}\to 2.∣λ∣e−h−μ→∣λ∣−μ,hsin(2h)​→2.

Therefore,

f−′(0)=−2(∣λ∣−μ).f'_-(0)=-2(|\lambda|-\mu).f−′​(0)=−2(∣λ∣−μ).

For differentiability at 000, we need

f+′(0)=f−′(0).f'_+(0)=f'_-(0).f+′​(0)=f−′​(0).

So,

2(∣λ∣−μ)=−2(∣λ∣−μ)2(|\lambda|-\mu)=-2(|\lambda|-\mu)2(∣λ∣−μ)=−2(∣λ∣−μ)

which gives

∣λ∣−μ=0⇒μ=∣λ∣.|\lambda|-\mu=0 \quad\Rightarrow\quad \mu=|\lambda|.∣λ∣−μ=0⇒μ=∣λ∣.

3. Describe the set SSS

Thus,

S={(λ,μ)∈R2:μ=∣λ∣}.S=\{(\lambda,\mu)\in\mathbb R^2: \mu=|\lambda|\}.S={(λ,μ)∈R2:μ=∣λ∣}.

Since ∣λ∣≥0|\lambda|\ge 0∣λ∣≥0 for every real λ\lambdaλ, every point in SSS satisfies

μ≥0.\mu\ge 0.μ≥0.

Hence,

S⊆R×[0,∞).S\subseteq \mathbb R\times [0,\infty).S⊆R×[0,∞).

4. Check options

  • A: R×[0,∞)\mathbb R\times[0,\infty)R×[0,∞) — true.
  • B: [0,∞)×R[0,\infty)\times\mathbb R[0,∞)×R — false, because λ\lambdaλ can be negative.
  • C: R×(−∞,0)\mathbb R\times(-\infty,0)R×(−∞,0) — false, because μ=∣λ∣≥0\mu=|\lambda|\ge 0μ=∣λ∣≥0.
  • D: (−∞,0)×R(-\infty,0)\times\mathbb R(−∞,0)×R — false, because λ\lambdaλ need not be negative.

Therefore, the correct option is A.

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