Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Limits Continuity and Differentiability question

2021 · 16 Mar · Shift 2 · Q35
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Limits Continuity and Differentiability
  5. /2021 · 16 Mar · Shift 2 · Q35

Limits Continuity and Differentiability question

2021 · 16 Mar · Shift 2 · Q35

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let α∈\alpha\inα∈ R be such that the function f(x)={cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3,xe0α,x=0f(x) = \left\{ {\begin{matrix} {{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x e 0} \\ {\alpha ,} & {x = 0} \\ \end{matrix} } \right.f(x)={{x}−{x}3cos−1(1−{x}2)sin−1(1−{x})​,α,​xe0x=0​ is continuous at x = 0, where {x} = x −-− [ x ] is the greatest integer less than or equal to x. Then :
  1. A
    no such α\alphaα exists
  2. B
    α\alphaα = 0
  3. C
    α\alphaα=π4{\pi \over 4}4π​
  4. D
    α\alphaα=π2{\pi \over {\sqrt 2 }}2​π​
View written solutionFree

Correct answer: A

  1. Interpret the function carefully

The function is

f(x)={cos⁡−1(1−{x}2) sin⁡−1(1−{x}){x}−{x}3,x≠0,α,x=0,f(x)= \begin{cases} \dfrac{\cos^{-1}(1-\{x\}^2)\,\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x\neq 0,\\[6pt] \alpha, & x=0, \end{cases}f(x)=⎩⎨⎧​{x}−{x}3cos−1(1−{x}2)sin−1(1−{x})​,α,​x=0,x=0,​

where {x}=x−[x]\{x\}=x-[x]{x}=x−[x] is the fractional part of xxx.

We need continuity at x=0x=0x=0, so we must check lim⁡x→0f(x).\lim_{x\to 0} f(x).limx→0​f(x). If this limit exists, then α\alphaα must equal that limit.

  1. Behavior of fractional part near x=0x=0x=0

The key fact is:

  • If x→0+x\to 0^+x→0+, then [x]=0[x]=0[x]=0, so {x}=x.\{x\}=x.{x}=x.
  • If x→0−x\to 0^-x→0−, then [x]=−1[x]=-1[x]=−1, so {x}=x+1→1−.\{x\}=x+1 \to 1^-.{x}=x+1→1−.

Thus the left-hand and right-hand limits may behave very differently.

  1. Right-hand limit as x→0+x\to 0^+x→0+

For x>0x>0x>0 sufficiently small, {x}=x\{x\}=x{x}=x. Hence

f(x)=cos⁡−1(1−x2) sin⁡−1(1−x)x−x3.f(x)=\frac{\cos^{-1}(1-x^2)\,\sin^{-1}(1-x)}{x-x^3}.f(x)=x−x3cos−1(1−x2)sin−1(1−x)​.

So

f(x)=cos⁡−1(1−x2)x⋅sin⁡−1(1−x)1−x2.f(x)=\frac{\cos^{-1}(1-x^2)}{x}\cdot \frac{\sin^{-1}(1-x)}{1-x^2}.f(x)=xcos−1(1−x2)​⋅1−x2sin−1(1−x)​.

Now use standard limits:

(i) First factor

For small t>0t>0t>0, cos⁡−1(1−t)∼2t.\cos^{-1}(1-t)\sim \sqrt{2t}.cos−1(1−t)∼2t​. Putting t=x2t=x^2t=x2,

Hence

(ii) Second factor

As x→0+x\to 0^+x→0+, 1−x→1−,1-x\to 1^-,1−x→1−, so

Also,

Therefore

Thus

  1. Left-hand limit as x→0−x\to 0^-x→0−

For x<0x<0x<0 sufficiently close to 000, we have {x}=x+1.\{x\}=x+1.{x}=x+1. Let u={x}=x+1.u=\{x\}=x+1.u={x}=x+1. Then as x→0−x\to 0^-x→0−,

Now

f(x)=cos⁡−1(1−u2) sin⁡−1(1−u)u−u3.f(x)=\frac{\cos^{-1}(1-u^2)\,\sin^{-1}(1-u)}{u-u^3}.f(x)=u−u3cos−1(1−u2)sin−1(1−u)​.

Take the limit as u→1−u\to 1^-u→1−.

Compute each part:

  • 1−u2→0,1-u^2 \to 0,1−u2→0, so cos⁡−1(1−u2)→cos⁡−1(0)=π2.\cos^{-1}(1-u^2)\to \cos^{-1}(0)=\frac{\pi}{2}.cos−1(1−u2)→cos−1(0)=2π​.
  • 1−u→0+,1-u\to 0^+,1−u→0+, so sin⁡−1(1−u)→0.\sin^{-1}(1-u)\to 0.sin−1(1−u)→0.
  • u−u3=u(1−u2)→1⋅(1−1)=0.u-u^3=u(1-u^2)\to 1\cdot(1-1)=0.u−u3=u(1−u2)→1⋅(1−1)=0.

So this is of type (π/2)⋅00\dfrac{(\pi/2)\cdot 0}{0}0(π/2)⋅0​, so we simplify asymptotically.

Let h=1−u→0+.h=1-u \to 0^+.h=1−u→0+. Then u=1−hu=1-hu=1−h. Now:

(i) Numerator factors

First, 1−u=h,1-u= h,1−u=h, so sin⁡−1(1−u)=sin⁡−1(h)∼h.\sin^{-1}(1-u)=\sin^{-1}(h)\sim h.sin−1(1−u)=sin−1(h)∼h.

Next, 1−u2=1−(1−h)2=1−(1−2h+h2)=2h−h2→0,1-u^2=1-(1-h)^2=1-(1-2h+h^2)=2h-h^2\to 0,1−u2=1−(1−h)2=1−(1−2h+h2)=2h−h2→0, thus cos⁡−1(1−u2)→cos⁡−1(0)=π2.\cos^{-1}(1-u^2) \to \cos^{-1}(0)=\frac{\pi}{2}.cos−1(1−u2)→cos−1(0)=2π​. So asymptotically, cos⁡−1(1−u2)∼π2.\cos^{-1}(1-u^2)\sim \frac{\pi}{2}.cos−1(1−u2)∼2π​.

(ii) Denominator

u−u3=u(1−u2)=(1−h)(2h−h2)∼2h.u-u^3=u(1-u^2)=(1-h)(2h-h^2)\sim 2h.u−u3=u(1−u2)=(1−h)(2h−h2)∼2h.

Hence,

f(x)∼(π2)(h)2h=π4.f(x)\sim \frac{\left(\frac{\pi}{2}\right)(h)}{2h}=\frac{\pi}{4}.f(x)∼2h(2π​)(h)​=4π​.

Therefore

  1. Compare left-hand and right-hand limits

We obtained: lim⁡x→0+f(x)=π2,\lim_{x\to 0^+} f(x)=\frac{\pi}{\sqrt{2}},limx→0+​f(x)=2​π​, lim⁡x→0−f(x)=π4.\lim_{x\to 0^-} f(x)=\frac{\pi}{4}.limx→0−​f(x)=4π​.

Since π2≠π4,\frac{\pi}{\sqrt{2}} \neq \frac{\pi}{4},2​π​=4π​, the two-sided limit at x=0x=0x=0 does not exist.

Therefore, there is no real number α\alphaα that can make fff continuous at x=0x=0x=0.

  1. Evaluate options
  • A: no such α\alphaα exists ✅
  • B: α=0\alpha=0α=0 ❌
  • C: α=π4\alpha=\dfrac{\pi}{4}α=4π​ ❌ (only left-hand limit)
  • D: α=π2\alpha=\dfrac{\pi}{\sqrt{2}}α=2​π​ ❌ (only right-hand limit)
  1. Final answer

The correct option is A.\boxed{\text{A}}.A​.

This matches the stored correct answer.

PreviousNext

More from Limits Continuity and Differentiability

  • Let f : R → R and g : R → R be defined as f(x)={x+a,∣x−1∣,​x<0x≥0​ and g(x)={x+1,(x−1)2+b,​x<0x≥0​…2021 · Numerical
  • The value of x→0+lim​x−x3cos−1(x−[x]2).sin−1(x−[x]2)​, where [ x ] denotes the greatest integer ≤ x is :2021 · MCQ
  • If the function f(x)=x4cos(sinx)−cosx​ is continuous at each point in its domain and f(0)=k1​, then k is ​.2021 · Numerical
  • The value of the limit θ→0lim​sin(2πsin2θ)tan(πcos2θ)​ is equal to :2021 · MCQ
  • The value of n→∞lim​n2[r]+[2r]+...+[nr]​, where r is a non-zero real number and [r] denotes the greatest integer less than or equal to r, is equal to :2021 · MCQ
  • If x→0lim​3x3sin−1x−tan−1x​ is equal to L, then the value of (6L + 1) is2021 · MCQ
  • If f(x)={∣x∣1​ax2+b​;∣x∣≥1;∣x∣<1​ is differentiable at every point of the domain, then the values of a and b are respectively :2021 · MCQ
  • Let f:R→R be a function defined as f(x)=⎩⎨⎧​2xsin(a+1)x+sin2x​bbx5/2x+bx3​−x​​​if x<0if x=0if x>0​…2021 · MCQ