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Limits Continuity and Differentiability question

2021 · 22 Jul · Shift 2 · Q36
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  5. /2021 · 22 Jul · Shift 2 · Q36

Limits Continuity and Differentiability question

2021 · 22 Jul · Shift 2 · Q36

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f : R →\to→ R be defined as f(x)={x3(1−cos⁡2x)2log⁡e(1+2xe−2x(1−xe−x)2),xe0α,x=0f(x) = \left\{ {\begin{matrix} {{{{x^3}} \over {{{(1 - \cos 2x)}^2}}}{{\log }_e}\left( {{{1 + 2x{e^{ - 2x}}} \over {{{(1 - x{e^{ - x}})}^2}}}} \right),} & {x e 0} \\ {\alpha ,} & {x = 0} \\ \end{matrix} } \right.f(x)={(1−cos2x)2x3​loge​((1−xe−x)21+2xe−2x​),α,​xe0x=0​ If f is continuous at x = 0, then α\alphaα is equal to :
  1. A
    1
  2. B
    3
  3. C
    0
  4. D
    2
View written solutionFree

Correct answer: A

We need continuity at x=0x=0x=0, so

α=lim⁡x→0x3(1−cos⁡2x)2 ln⁡ ⁣(1+2xe−2x(1−xe−x)2).\alpha=\lim_{x\to 0} \frac{x^3}{(1-\cos 2x)^2}\,\ln\! \left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right).α=x→0lim​(1−cos2x)2x3​ln((1−xe−x)21+2xe−2x​).

1. Simplify the prefactor

Using 1−cos⁡2x=2sin⁡2x∼2x2(x→0),1-\cos 2x = 2\sin^2 x \sim 2x^2 \quad (x\to 0),1−cos2x=2sin2x∼2x2(x→0), we get

(1−cos⁡2x)2∼(2x2)2=4x4.(1-\cos 2x)^2 \sim (2x^2)^2=4x^4.(1−cos2x)2∼(2x2)2=4x4.

Hence

x3(1−cos⁡2x)2∼x34x4=14x.\frac{x^3}{(1-\cos 2x)^2} \sim \frac{x^3}{4x^4}=\frac{1}{4x}.(1−cos2x)2x3​∼4x4x3​=4x1​.

So the limit becomes

α=lim⁡x→014xln⁡(1+2xe−2x(1−xe−x)2).\alpha=\lim_{x\to 0} \frac{1}{4x}\ln\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right).α=x→0lim​4x1​ln((1−xe−x)21+2xe−2x​).

Thus we need the first-order term of the logarithm.

2. Expand the inside of the logarithm

Let

Expand e−2xe^{-2x}e−2x and e−xe^{-x}e−x

e−2x=1−2x+2x2+O(x3),e^{-2x}=1-2x+2x^2+O(x^3),e−2x=1−2x+2x2+O(x3),

so

2xe−2x=2x−4x2+4x3+O(x4).2xe^{-2x}=2x-4x^2+4x^3+O(x^4).2xe−2x=2x−4x2+4x3+O(x4).

Therefore

A=1+2x−4x2+O(x3).A=1+2x-4x^2+O(x^3).A=1+2x−4x2+O(x3).

Also,

e−x=1−x+x22+O(x3),e^{-x}=1-x+\frac{x^2}{2}+O(x^3),e−x=1−x+2x2​+O(x3),

so

xe−x=x−x2+x32+O(x4).xe^{-x}=x-x^2+\frac{x^3}{2}+O(x^4).xe−x=x−x2+2x3​+O(x4).

Hence

1−xe−x=1−x+x2+O(x3).1-xe^{-x}=1-x+x^2+O(x^3).1−xe−x=1−x+x2+O(x3).

Squaring,

B=(1−x+x2+O(x3))2=1−2x+3x2+O(x3).B=(1-x+x^2+O(x^3))^2=1-2x+3x^2+O(x^3).B=(1−x+x2+O(x3))2=1−2x+3x2+O(x3).

So

AB=1+2x−4x2+O(x3)1−2x+3x2+O(x3).\frac{A}{B}=\frac{1+2x-4x^2+O(x^3)}{1-2x+3x^2+O(x^3)}.BA​=1−2x+3x2+O(x3)1+2x−4x2+O(x3)​.

Using 11−2x+3x2=1+2x+O(x2)\dfrac{1}{1-2x+3x^2}=1+2x+O(x^2)1−2x+3x21​=1+2x+O(x2) for first-order accuracy,

AB=(1+2x+O(x2))(1+2x+O(x2))=1+4x+O(x2).\frac{A}{B}= (1+2x+O(x^2))(1+2x+O(x^2))=1+4x+O(x^2).BA​=(1+2x+O(x2))(1+2x+O(x2))=1+4x+O(x2).

Therefore

ln⁡(AB)=ln⁡(1+4x+O(x2))=4x+O(x2).\ln\left(\frac{A}{B}\right)=\ln(1+4x+O(x^2))=4x+O(x^2).ln(BA​)=ln(1+4x+O(x2))=4x+O(x2).

3. Compute the limit

Thus

α=lim⁡x→014x(4x+O(x2))=1.\alpha=\lim_{x\to 0} \frac{1}{4x}(4x+O(x^2))=1.α=x→0lim​4x1​(4x+O(x2))=1.

4. Final answer

For continuity at x=0x=0x=0,

α=1.\boxed{\alpha=1}.α=1​.

So the correct option is A.

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