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Limits Continuity and Differentiability question

2022 · 29 Jul · Shift 1 · Q29
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  5. /2022 · 29 Jul · Shift 1 · Q29

Limits Continuity and Differentiability question

2022 · 29 Jul · Shift 1 · Q29

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→0αex+βe−x+γsin⁡xxsin⁡2x=23\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}x→0lim​xsin2xαex+βe−x+γsinx​=32​, where α,β,γ∈R\alpha, \beta, \gamma \in \mathbf{R}α,β,γ∈R, then which of the following is NOT correct?
  1. A
    α2+β2+γ2=6\alpha^{2}+\beta^{2}+\gamma^{2}=6α2+β2+γ2=6
  2. B
    αβ+βγ+γα+1=0\alpha \beta+\beta \gamma+\gamma \alpha+1=0αβ+βγ+γα+1=0
  3. C
    αβ2+βγ2+γα2+3=0\alpha\beta^{2}+\beta \gamma^{2}+\gamma \alpha^{2}+3=0αβ2+βγ2+γα2+3=0
  4. D
    α2−β2+γ2=4\alpha^{2}-\beta^{2}+\gamma^{2}=4α2−β2+γ2=4
View written solutionFree

Correct answer: C

  1. We are given
limx→0αex+βe−x+γsin⁡xxsin⁡2x=23.\\lim_{x\to 0}\frac{\alpha e^x+\beta e^{-x}+\gamma \sin x}{x\sin^2 x}=\frac23.limx→0​xsin2xαex+βe−x+γsinx​=32​.

We need the limit to be finite and equal to 23\frac2332​.

  1. Expand all functions near x=0x=0x=0:
ex=1+x+x22+x36+O(x4),e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4),ex=1+x+2x2​+6x3​+O(x4), e−x=1−x+x22−x36+O(x4),e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+O(x^4),e−x=1−x+2x2​−6x3​+O(x4), sin⁡x=x−x36+O(x5).\sin x=x-\frac{x^3}{6}+O(x^5).sinx=x−6x3​+O(x5).

Also,

sin⁡2x=x2+O(x4)⇒xsin⁡2x=x3+O(x5).\sin^2 x=x^2+O(x^4) \quad\Rightarrow\quad x\sin^2 x=x^3+O(x^5).sin2x=x2+O(x4)⇒xsin2x=x3+O(x5).
  1. Expand the numerator:
αex+βe−x+γsin⁡x\alpha e^x+\beta e^{-x}+\gamma \sin xαex+βe−x+γsinx =α(1+x+x22+x36)+β(1−x+x22−x36)+γ(x−x36)+O(x4).=\alpha\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)+ \beta\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}\right)+ \gamma\left(x-\frac{x^3}{6}\right)+O(x^4).=α(1+x+2x2​+6x3​)+β(1−x+2x2​−6x3​)+γ(x−6x3​)+O(x4).

Collect coefficients:

=(α+β)+(α−β+γ)x+α+β2x2+α−β−γ6x3+O(x4).= (\alpha+\beta) + (\alpha-\beta+\gamma)x + \frac{\alpha+\beta}{2}x^2 + \frac{\alpha-\beta-\gamma}{6}x^3 +O(x^4).=(α+β)+(α−β+γ)x+2α+β​x2+6α−β−γ​x3+O(x4).
  1. Since denominator is of order x3x^3x3, for the limit to exist and be finite, coefficients of 1,x,x21,x,x^21,x,x2 in numerator must vanish:
  • Constant term:
α+β=0\alpha+\beta=0α+β=0
  • Coefficient of xxx:
α−β+γ=0\alpha-\beta+\gamma=0α−β+γ=0
  • Coefficient of x2x^2x2:
α+β2=0,\frac{\alpha+\beta}{2}=0,2α+β​=0,

which is same as α+β=0\alpha+\beta=0α+β=0.

Thus,

β=−α,\beta=-\alpha,β=−α,

and then

α−(−α)+γ=0⇒2α+γ=0⇒γ=−2α.\alpha-(-\alpha)+\gamma=0 \Rightarrow 2\alpha+\gamma=0 \Rightarrow \gamma=-2\alpha.α−(−α)+γ=0⇒2α+γ=0⇒γ=−2α.
  1. Now use the value of the limit.

Numerator becomes

α−β−γ6x3+O(x4).\frac{\alpha-\beta-\gamma}{6}x^3+O(x^4).6α−β−γ​x3+O(x4).

Since denominator xsin⁡2x∼x3x\sin^2x \sim x^3xsin2x∼x3, we get

\lim_{x\to0}\frac{\alpha e^x+\beta e^{-x}+\gamma\sin x}{x\sin^2x} =\frac{\alpha-\beta-\gamma}{6}= rac23.

So,

α−β−γ=4.\alpha-\beta-\gamma=4.α−β−γ=4.

Using β=−α\beta=-\alphaβ=−α and γ=−2α\gamma=-2\alphaγ=−2α,

α−(−α)−(−2α)=4⇒4α=4⇒α=1.\alpha-(-\alpha)-(-2\alpha)=4 \Rightarrow 4\alpha=4 \Rightarrow \alpha=1.α−(−α)−(−2α)=4⇒4α=4⇒α=1.

Hence,

β=−1,γ=−2.\beta=-1,\qquad \gamma=-2.β=−1,γ=−2.
  1. Check each option.

Option A

α2+β2+γ2=12+(−1)2+(−2)2=1+1+4=6.\alpha^2+\beta^2+\gamma^2=1^2+(-1)^2+(-2)^2=1+1+4=6.α2+β2+γ2=12+(−1)2+(−2)2=1+1+4=6.

So A is correct.

Option B

αβ+βγ+γα+1=(1)(−1)+(−1)(−2)+(−2)(1)+1\alpha\beta+\beta\gamma+\gamma\alpha+1 = (1)(-1)+(-1)(-2)+(-2)(1)+1αβ+βγ+γα+1=(1)(−1)+(−1)(−2)+(−2)(1)+1 =−1+2−2+1=0.=-1+2-2+1=0.=−1+2−2+1=0.

So B is correct.

Option C

αβ2+βγ2+γα2+3=1⋅(−1)2+(−1)⋅(−2)2+(−2)⋅12+3\alpha\beta^2+\beta\gamma^2+\gamma\alpha^2+3 =1\cdot(-1)^2+(-1)\cdot(-2)^2+(-2)\cdot1^2+3αβ2+βγ2+γα2+3=1⋅(−1)2+(−1)⋅(−2)2+(−2)⋅12+3 =1−4−2+3=−2≠0.=1-4-2+3=-2\neq 0.=1−4−2+3=−2=0.

So C is NOT correct.

Option D

α2−β2+γ2=12−(−1)2+(−2)2=1−1+4=4.\alpha^2-\beta^2+\gamma^2=1^2-(-1)^2+(-2)^2=1-1+4=4.α2−β2+γ2=12−(−1)2+(−2)2=1−1+4=4.

So D is correct.

  1. Therefore, the only statement which is NOT correct is
C.\boxed{\text{C}}.C​.

Comparison with stored correct answer: stored answer is C, which matches our result.

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