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Limits Continuity and Differentiability question

2021 · 25 Jul · Shift 1 · Q27
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  5. /2021 · 25 Jul · Shift 1 · Q27

Limits Continuity and Differentiability question

2021 · 25 Jul · Shift 1 · Q27

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f : R →\to→ R be defined as f(x)={λ∣x2−5x+6∣μ(5x−x2−6),x<2etan⁡(x−2)x−[x],x>2μ,x=2f(x) = \left\{ {\begin{matrix} {{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x \lt 2} \\ {{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x \gt 2} \\ {\mu ,} & {x = 2} \\ \end{matrix} } \right.f(x)=⎩⎨⎧​μ(5x−x2−6)λ∣x2−5x+6∣​,ex−[x]tan(x−2)​,μ,​x<2x>2x=2​ where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then λ\lambdaλ+μ\muμ is equal to :
  1. A
    e(−-− e + 1)
  2. B
    e(e −-− 2)
  3. C
    1
  4. D
    2e −-− 1
View written solutionFree

Correct answer: A

  1. Given piecewise function
f(x)={λ∣x2−5x+6∣μ(5x−x2−6),x<2etan⁡(x−2)x−[x],x>2μ,x=2f(x)= \begin{cases} \dfrac{\lambda |x^2-5x+6|}{\mu(5x-x^2-6)}, & x<2 \\ e^{\dfrac{\tan(x-2)}{x-[x]}}, & x>2 \\ \mu, & x=2 \end{cases}f(x)=⎩⎨⎧​μ(5x−x2−6)λ∣x2−5x+6∣​,ex−[x]tan(x−2)​,μ,​x<2x>2x=2​

We need continuity at x=2x=2x=2, so

lim⁡x→2−f(x)=f(2)=lim⁡x→2+f(x)=μ.\lim_{x\to 2^-} f(x)=f(2)=\lim_{x\to 2^+} f(x)=\mu.x→2−lim​f(x)=f(2)=x→2+lim​f(x)=μ.
  1. Left-hand limit as x→2−x\to 2^-x→2−

First factorize:

x2−5x+6=(x−2)(x−3).x^2-5x+6=(x-2)(x-3).x2−5x+6=(x−2)(x−3).

Also,

5x−x2−6=−(x2−5x+6)=−(x−2)(x−3).5x-x^2-6=-(x^2-5x+6)=-(x-2)(x-3).5x−x2−6=−(x2−5x+6)=−(x−2)(x−3).

So for x<2x<2x<2,

f(x)=λ∣(x−2)(x−3)∣μ [−(x−2)(x−3)].f(x)=\frac{\lambda |(x-2)(x-3)|}{\mu\,[-(x-2)(x-3)]}.f(x)=μ[−(x−2)(x−3)]λ∣(x−2)(x−3)∣​.

Now when x<2x<2x<2 and close to 222:

  • x−2<0x-2<0x−2<0
  • x−3<0x-3<0x−3<0
  • hence (x−2)(x−3)>0(x-2)(x-3)>0(x−2)(x−3)>0

Therefore,

∣(x−2)(x−3)∣=(x−2)(x−3).|(x-2)(x-3)|=(x-2)(x-3).∣(x−2)(x−3)∣=(x−2)(x−3).

Thus

f(x)=λ(x−2)(x−3)−μ(x−2)(x−3)=−λμ.f(x)=\frac{\lambda (x-2)(x-3)}{-\mu (x-2)(x-3)}=-\frac{\lambda}{\mu}.f(x)=−μ(x−2)(x−3)λ(x−2)(x−3)​=−μλ​.

Hence,

lim⁡x→2−f(x)=−λμ.\lim_{x\to 2^-} f(x)=-\frac{\lambda}{\mu}.x→2−lim​f(x)=−μλ​.

For continuity,

−λμ=μ.-\frac{\lambda}{\mu}=\mu.−μλ​=μ.

So,

λ=−μ2.(1)\lambda=-\mu^2. \qquad (1)λ=−μ2.(1)
  1. Right-hand limit as x→2+x\to 2^+x→2+

For x→2+x\to 2^+x→2+, we have 2<x<32<x<32<x<3, so

[x]=2.[x]=2.[x]=2.

Hence

x−[x]=x−2.x-[x]=x-2.x−[x]=x−2.

Therefore for x>2x>2x>2 near 222,

f(x)=etan⁡(x−2)x−2.f(x)=e^{\frac{\tan(x-2)}{x-2}}.f(x)=ex−2tan(x−2)​.

Let h=x−2h=x-2h=x−2. Then h→0+h\to 0^+h→0+, and

lim⁡x→2+f(x)=elim⁡h→0+tan⁡hh=e1=e.\lim_{x\to 2^+} f(x)=e^{\lim_{h\to 0^+}\frac{\tan h}{h}}=e^1=e.x→2+lim​f(x)=elimh→0+​htanh​=e1=e.

So continuity gives

μ=e.(2)\mu=e. \qquad (2)μ=e.(2)
  1. Find λ\lambdaλ

From (1),

λ=−μ2=−e2.\lambda=-\mu^2=-e^2.λ=−μ2=−e2.

Therefore,

λ+μ=−e2+e=e(1−e).\lambda+\mu=-e^2+e=e(1-e).λ+μ=−e2+e=e(1−e).

So,

λ+μ=e(−e+1).\lambda+\mu=e(-e+1).λ+μ=e(−e+1).
  1. Match with options
e(−e+1)e(-e+1)e(−e+1)

corresponds to Option A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.

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