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Limits Continuity and Differentiability question

2024 · 1 Feb · Shift 1 · Q55
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Limits Continuity and Differentiability question

2024 · 1 Feb · Shift 1 · Q55

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1
Let {x}\{x\}{x} denote the fractional part of xxx and f(x)=cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3,xeq0f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x eq 0f(x)={x}−{x}3cos−1(1−{x}2)sin−1(1−{x})​,xeq0. If L\mathrm{L}L and R\mathrm{R}R respectively denotes the left hand limit and the right hand limit of f(x)f(x)f(x) at x=0x=0x=0, then 32π2( L2+R2)\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)π232​( L2+R2) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 18

  1. Understand the fractional part near x=0x=0x=0

For the fractional part function {x}\{x\}{x}:

  • If x→0+x \to 0^+x→0+, then {x}=x\{x\}=x{x}=x.
  • If x→0−x \to 0^-x→0−, then {x}=x−⌊x⌋=x+1\{x\}=x-\lfloor x\rfloor = x+1{x}=x−⌊x⌋=x+1 (since for −1<x<0-1<x<0−1<x<0, ⌊x⌋=−1\lfloor x\rfloor=-1⌊x⌋=−1).

So we must compute:

  • Right hand limit using t={x}=x→0+t=\{x\}=x\to 0^+t={x}=x→0+
  • Left hand limit using t={x}=x+1→1−t=\{x\}=x+1\to 1^-t={x}=x+1→1−

Given

f(x)=cos⁡−1(1−{x}2) sin⁡−1(1−{x}){x}−{x}3f(x)=\frac{\cos^{-1}(1-\{x\}^2)\,\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}f(x)={x}−{x}3cos−1(1−{x}2)sin−1(1−{x})​

Let t={x}t=\{x\}t={x}. Then

f(x)=cos⁡−1(1−t2) sin⁡−1(1−t)t−t3f(x)=\frac{\cos^{-1}(1-t^2)\,\sin^{-1}(1-t)}{t-t^3}f(x)=t−t3cos−1(1−t2)sin−1(1−t)​
  1. Right hand limit RRR

As x→0+x\to 0^+x→0+, we have t→0+t\to 0^+t→0+. Thus

R=lim⁡t→0+cos⁡−1(1−t2) sin⁡−1(1−t)t−t3R=\lim_{t\to 0^+}\frac{\cos^{-1}(1-t^2)\,\sin^{-1}(1-t)}{t-t^3}R=t→0+lim​t−t3cos−1(1−t2)sin−1(1−t)​

Now use standard limits:

(i) For cos⁡−1(1−t2)\cos^{-1}(1-t^2)cos−1(1−t2)

As u→0+u\to 0^+u→0+,

cos⁡−1(1−u)∼2u\cos^{-1}(1-u)\sim \sqrt{2u}cos−1(1−u)∼2u​

Putting u=t2u=t^2u=t2,

cos⁡−1(1−t2)∼2 t\cos^{-1}(1-t^2)\sim \sqrt{2}\,tcos−1(1−t2)∼2​t

(since t>0t>0t>0).

(ii) For sin⁡−1(1−t)\sin^{-1}(1-t)sin−1(1−t)

As t→0+t\to 0^+t→0+,

sin⁡−1(1−t)→sin⁡−1(1)=π2\sin^{-1}(1-t)\to \sin^{-1}(1)=\frac{\pi}{2}sin−1(1−t)→sin−1(1)=2π​

(iii) Denominator

t−t3=t(1−t2)∼tt-t^3=t(1-t^2)\sim tt−t3=t(1−t2)∼t

Hence,

R=lim⁡t→0+(2 t)(π2)t=π2R=\lim_{t\to 0^+}\frac{(\sqrt2\,t)\left(\frac\pi2\right)}{t}=\frac{\pi}{\sqrt2}R=t→0+lim​t(2​t)(2π​)​=2​π​

So,

R2=π22R^2=\frac{\pi^2}{2}R2=2π2​
  1. Left hand limit LLL

As x→0−x\to 0^-x→0−, we have t={x}=x+1→1−t=\{x\}=x+1\to 1^-t={x}=x+1→1−. Let

t=1−h,h→0+t=1-h,\qquad h\to 0^+t=1−h,h→0+

Then

1−t=h1-t=h1−t=h

And

1−t2=1−(1−h)2=2h−h21-t^2=1-(1-h)^2=2h-h^21−t2=1−(1−h)2=2h−h2

Also,

t−t3=t(1−t2)=(1−h)(2h−h2)=h(1−h)(2−h)t-t^3=t(1-t^2)=(1-h)(2h-h^2)=h(1-h)(2-h)t−t3=t(1−t2)=(1−h)(2h−h2)=h(1−h)(2−h)

Now,

L=lim⁡h→0+cos⁡−1(2h−h2?)??L=\lim_{h\to 0^+}\frac{\cos^{-1}(2h-h^2?)?}{?}L=h→0+lim​?cos−1(2h−h2?)?​

Let us substitute carefully:

cos⁡−1(1−t2)=cos⁡−1(2h−h2)\cos^{-1}(1-t^2)=\cos^{-1}(2h-h^2)cos−1(1−t2)=cos−1(2h−h2)

Wait: since

1−t2=2h−h2,1-t^2=2h-h^2,1−t2=2h−h2,

we get

cos⁡−1(1−t2)=cos⁡−1(2h−h2)\cos^{-1}(1-t^2)=\cos^{-1}(2h-h^2)cos−1(1−t2)=cos−1(2h−h2)

As h→0+h\to 0^+h→0+, this tends to

cos⁡−1(0)=π2\cos^{-1}(0)=\frac\pi2cos−1(0)=2π​

Also,

sin⁡−1(1−t)=sin⁡−1(h)∼h\sin^{-1}(1-t)=\sin^{-1}(h)\sim hsin−1(1−t)=sin−1(h)∼h

And denominator:

t−t3=(1−h)(2h−h2)∼2ht-t^3=(1-h)(2h-h^2)\sim 2ht−t3=(1−h)(2h−h2)∼2h

Thus,

L=lim⁡h→0+(π2)(h)2h=π4L=\lim_{h\to 0^+}\frac{\left(\frac\pi2\right)(h)}{2h}=\frac\pi4L=h→0+lim​2h(2π​)(h)​=4π​

So,

L2=π216L^2=\frac{\pi^2}{16}L2=16π2​
  1. Compute the required value

We need

32π2(L2+R2)\frac{32}{\pi^2}(L^2+R^2)π232​(L2+R2)

Now

L2+R2=π216+π22=π2(116+816)=9π216L^2+R^2=\frac{\pi^2}{16}+\frac{\pi^2}{2} =\pi^2\left(\frac1{16}+\frac8{16}\right) =\frac{9\pi^2}{16}L2+R2=16π2​+2π2​=π2(161​+168​)=169π2​

Therefore,

32π2(L2+R2)=32π2⋅9π216=18\frac{32}{\pi^2}(L^2+R^2)=\frac{32}{\pi^2}\cdot \frac{9\pi^2}{16}=18π232​(L2+R2)=π232​⋅169π2​=18
  1. Comparison with stored answer

Derived answer = 181818.

This matches the stored correct answer.

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