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Limits Continuity and Differentiability question

2021 · 26 Aug · Shift 1 · Q42
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  5. /2021 · 26 Aug · Shift 1 · Q42

Limits Continuity and Differentiability question

2021 · 26 Aug · Shift 1 · Q42

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
Let a, b ∈\in∈ R, b ∈\in∈ 0, Define a function f(x)={asin⁡π2(x−1),for x≤0tan⁡2x−sin⁡2xbx3,for x>0f(x) = \left\{ {\begin{matrix} {a\sin {\pi \over 2}(x - 1),} & {for\,x \le 0} \\ {{{\tan 2x - \sin 2x} \over {b{x^3}}},} & {for\,x \gt 0} \\ \end{matrix} } \right.f(x)={asin2π​(x−1),bx3tan2x−sin2x​,​forx≤0forx>0​. If f is continuous at x = 0, then 10 −-− ab is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 14

  1. Given function
f(x)={asin⁡(π2(x−1)),x≤0tan⁡2x−sin⁡2xbx3,x>0f(x)= \begin{cases} a\sin\left(\frac{\pi}{2}(x-1)\right), & x\le 0 \\ \dfrac{\tan 2x-\sin 2x}{bx^3}, & x>0 \end{cases}f(x)=⎩⎨⎧​asin(2π​(x−1)),bx3tan2x−sin2x​,​x≤0x>0​

with a,b∈Ra,b\in \mathbb Ra,b∈R and b≠0b\ne 0b=0.

We need continuity at x=0x=0x=0.

  1. Value at x=0x=0x=0 from the left piece

Since the first branch is defined for x≤0x\le 0x≤0,

f(0)=asin⁡(π2(0−1))=asin⁡(−π2)=−a.f(0)=a\sin\left(\frac{\pi}{2}(0-1)\right)=a\sin\left(-\frac{\pi}{2}\right)=-a.f(0)=asin(2π​(0−1))=asin(−2π​)=−a.

So for continuity, we need

lim⁡x→0+tan⁡2x−sin⁡2xbx3=−a.\lim_{x\to 0^+} \frac{\tan 2x-\sin 2x}{bx^3}=-a.x→0+lim​bx3tan2x−sin2x​=−a.
  1. Compute the right-hand limit

Let t=2xt=2xt=2x. Then as x→0+x\to 0^+x→0+, t→0t\to 0t→0.

Using standard expansions near 000:

tan⁡t=t+t33+O(t5),\tan t = t+\frac{t^3}{3}+O(t^5),tant=t+3t3​+O(t5), sin⁡t=t−t36+O(t5).\sin t = t-\frac{t^3}{6}+O(t^5).sint=t−6t3​+O(t5).

Therefore,

tan⁡t−sin⁡t=(t+t33)−(t−t36)+O(t5)=t32+O(t5).\tan t-\sin t = \left(t+\frac{t^3}{3}\right)-\left(t-\frac{t^3}{6}\right)+O(t^5) =\frac{t^3}{2}+O(t^5).tant−sint=(t+3t3​)−(t−6t3​)+O(t5)=2t3​+O(t5).

Now substitute t=2xt=2xt=2x:

tan⁡2x−sin⁡2x=(2x)32+O(x5)=4x3+O(x5).\tan 2x-\sin 2x = \frac{(2x)^3}{2}+O(x^5)=4x^3+O(x^5).tan2x−sin2x=2(2x)3​+O(x5)=4x3+O(x5).

Hence,

\lim_{x\to 0^+}\frac{\tan 2x-\sin 2x}{bx^3} =\lim_{x\to 0^+}\frac{4x^3+O(x^5)}{bx^3}= rac{4}{b}.
  1. Apply continuity condition

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

−a=4b.-a=\frac{4}{b}.−a=b4​.

Multiply by bbb:

ab=−4.ab=-4.ab=−4.
  1. Find 10−ab10-ab10−ab
10−ab=10−(−4)=14.10-ab=10-(-4)=14.10−ab=10−(−4)=14.
  1. Comparison with stored answer

Derived answer = 141414.

Stored correct answer = 141414.

They match.

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