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Limits Continuity and Differentiability question

2018 · Shift 2 · Q36
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  5. /2018 · Shift 2 · Q36

Limits Continuity and Differentiability question

2018 · Shift 2 · Q36

JEE AdvancedMathematicsLimits Continuity and DifferentiabilityMCQ+3 / −1
Let f1:R→Rf_1 : \mathbb{R} \to \mathbb{R}f1​:R→R, f2:(−π2,π2)→Rf_2 : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \mathbb{R}f2​:(−2π​,2π​)→R, f3:(−1,eπ/2−2)→Rf_3 : (-1, e^{\pi/2} - 2) \to \mathbb{R}f3​:(−1,eπ/2−2)→R and f4:R→Rf_4 : \mathbb{R} \to \mathbb{R}f4​:R→R be functions defined by (i) f1(x)=sin⁡(1−e−x2)f_1(x) = \sin\left(\sqrt{1 - e^{-x^2}}\right)f1​(x)=sin(1−e−x2​)(ii) f2(x)={∣sin⁡x∣tan⁡−1xif x≠01if x=0f_2(x) = \left\{ \begin{array}{ll} \dfrac{|\sin x|}{\tan^{-1} x} & \text{if } x \ne 0 \\ 1 & \text{if } x = 0 \end{array} \right.f2​(x)=⎩⎨⎧​tan−1x∣sinx∣​1​if x=0if x=0​ where the inverse trigonometric function tan⁡−1x\tan^{-1} xtan−1x assumes values in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)(−2π​,2π​)(iii) f3(x)=[sin⁡(log⁡e(x+2))]f_3(x) = [\sin(\log_e(x + 2))]f3​(x)=[sin(loge​(x+2))], where for t∈Rt \in \mathbb{R}t∈R, [t][t][t] denotes the greatest integer less than or equal to ttt (iv) f4(x)={x2sin⁡(1x)if x≠00if x=0f_4(x) = \left\{ \begin{array}{ll} x^2 \sin\left(\dfrac{1}{x}\right) & \text{if } x \ne 0 \\ 0 & \text{if } x = 0 \end{array} \right.f4​(x)=⎩⎨⎧​x2sin(x1​)0​if x=0if x=0​ | LIST-I | LIST-II | |--------|--------| | P. The function f1f_1f1​ is | 1. NOT continuous at x=0x = 0x=0 | | Q. The function f2f_2f2​ is | 2. continuous at x=0x = 0x=0 and NOT differentiable at x=0x = 0x=0 | | R. The function f3f_3f3​ is | 3. differentiable at x=0x = 0x=0 and its derivative is NOT continuous at x=0x = 0x=0 | | S. The function f4f_4f4​ is | 4. differentiable at x=0x = 0x=0 and its derivative is continuous at x=0x = 0x=0 |
  1. A
    P →\to→ 2 ; Q →\to→ 3 ; R →\to→ 1 ; S →\to→ 4
  2. B
    P →\to→ 4 ; Q →\to→ 1 ; R →\to→ 2 ; S →\to→ 3
  3. C
    P →\to→ 4 ; Q →\to→ 2 ; R →\to→ 1 ; S →\to→ 3
  4. D
    P →\to→ 2 ; Q →\to→ 1 ; R →\to→ 4 ; S →\to→ 3
View written solutionFree

Correct answer: D

We analyze each function at the relevant point.

1. Function f1(x)=sin⁡ ⁣(1−e−x2)f_1(x)=\sin\!\left(\sqrt{1-e^{-x^2}}\right)f1​(x)=sin(1−e−x2​)

We check continuity/differentiability at x=0x=0x=0.

Step 1: Continuity at x=0x=0x=0

Since e−x2e^{-x^2}e−x2 is continuous,

1−e−x2→1−1=0(x→0).1-e^{-x^2} \to 1-1=0 \quad (x\to 0).1−e−x2→1−1=0(x→0).

Hence

1−e−x2→0,\sqrt{1-e^{-x^2}} \to 0,1−e−x2​→0,

and therefore

f1(x)=sin⁡(1−e−x2)→sin⁡0=0.f_1(x)=\sin\left(\sqrt{1-e^{-x^2}}\right) \to \sin 0=0.f1​(x)=sin(1−e−x2​)→sin0=0.

Also,

f1(0)=sin⁡(1−e0)=sin⁡0=0.f_1(0)=\sin\left(\sqrt{1-e^0}\right)=\sin 0=0.f1​(0)=sin(1−e0​)=sin0=0.

So f1f_1f1​ is continuous at x=0x=0x=0.

Step 2: Differentiability at x=0x=0x=0

Use the derivative definition:

f1′(0)=lim⁡h→0f1(h)−f1(0)h=lim⁡h→0sin⁡(1−e−h2)h.f_1'(0)=\lim_{h\to 0}\frac{f_1(h)-f_1(0)}{h} =\lim_{h\to 0}\frac{\sin\left(\sqrt{1-e^{-h^2}}\right)}{h}.f1′​(0)=h→0lim​hf1​(h)−f1​(0)​=h→0lim​hsin(1−e−h2​)​.

For small hhh,

e−h2=1−h2+o(h2)  ⟹  1−e−h2=h2+o(h2).e^{-h^2}=1-h^2+o(h^2) \implies 1-e^{-h^2}=h^2+o(h^2).e−h2=1−h2+o(h2)⟹1−e−h2=h2+o(h2).

Thus

1−e−h2∼∣h∣.\sqrt{1-e^{-h^2}}\sim |h|.1−e−h2​∼∣h∣.

Also sin⁡u∼u\sin u\sim usinu∼u as u→0u\to 0u→0, so

sin⁡(1−e−h2)∼∣h∣.\sin\left(\sqrt{1-e^{-h^2}}\right)\sim |h|.sin(1−e−h2​)∼∣h∣.

Therefore

sin⁡(1−e−h2)h∼∣h∣h,\frac{\sin\left(\sqrt{1-e^{-h^2}}\right)}{h}\sim \frac{|h|}{h},hsin(1−e−h2​)​∼h∣h∣​,

whose right-hand limit is 111 and left-hand limit is −1-1−1.

Hence f1f_1f1​ is not differentiable at x=0x=0x=0, though it is continuous there.

So,

P→2.P \to 2.P→2.

2. Function

f2(x)={∣sin⁡x∣tan⁡−1x,x≠0,1,x=0.f_2(x)=\begin{cases} \dfrac{|\sin x|}{\tan^{-1}x}, & x\ne 0,\\[4pt] 1, & x=0. \end{cases}f2​(x)=⎩⎨⎧​tan−1x∣sinx∣​,1,​x=0,x=0.​

We check continuity at x=0x=0x=0.

As x→0x\to 0x→0,

∣sin⁡x∣∼∣x∣,tan⁡−1x∼x.|\sin x|\sim |x|, \qquad \tan^{-1}x\sim x.∣sinx∣∼∣x∣,tan−1x∼x.

Thus

∣sin⁡x∣tan⁡−1x∼∣x∣x.\frac{|\sin x|}{\tan^{-1}x} \sim \frac{|x|}{x}.tan−1x∣sinx∣​∼x∣x∣​.

So,

  • as x→0+x\to 0^+x→0+, limit =1=1=1,
  • as x→0−x\to 0^-x→0−, limit =−1=-1=−1.

The two-sided limit does not exist. Therefore f2f_2f2​ is not continuous at x=0x=0x=0.

So,

Q→1.Q \to 1.Q→1.

3. Function

f3(x)=[sin⁡(ln⁡(x+2))],x∈(−1,eπ/2−2).f_3(x)=[\sin(\ln(x+2))], \qquad x\in (-1,e^{\pi/2}-2).f3​(x)=[sin(ln(x+2))],x∈(−1,eπ/2−2).

We check continuity at x=0x=0x=0.

Step 1: Value at x=0x=0x=0

f3(0)=[sin⁡(ln⁡2)].f_3(0)=[\sin(\ln 2)].f3​(0)=[sin(ln2)].

Since 0<ln⁡2<10<\ln 2<10<ln2<1, we have

0<sin⁡(ln⁡2)<1.0<\sin(\ln 2)<1.0<sin(ln2)<1.

Hence

[sin⁡(ln⁡2)]=0.[\sin(\ln 2)]=0.[sin(ln2)]=0.

So f3(0)=0f_3(0)=0f3​(0)=0.

Step 2: Behavior near x=0x=0x=0

The function sin⁡(ln⁡(x+2))\sin(\ln(x+2))sin(ln(x+2)) is continuous near x=0x=0x=0, and at x=0x=0x=0 its value is strictly between 000 and 111. Therefore, for all xxx sufficiently close to 000,

0<sin⁡(ln⁡(x+2))<1.0<\sin(\ln(x+2))<1.0<sin(ln(x+2))<1.

Hence its greatest integer value remains

[sin⁡(ln⁡(x+2))]=0.[\sin(\ln(x+2))]=0.[sin(ln(x+2))]=0.

So near x=0x=0x=0, f3(x)=0f_3(x)=0f3​(x)=0 identically.

Therefore f3f_3f3​ is continuous at x=0x=0x=0, and in fact differentiable there with derivative

f3′(0)=0.f_3'(0)=0.f3′​(0)=0.

Also, since f3f_3f3​ is constant in a neighborhood of 000, its derivative is 000 in that neighborhood, so the derivative is continuous at 000.

Thus,

R→4.R \to 4.R→4.

4. Function

f4(x)={x2sin⁡(1x),x≠0,0,x=0.f_4(x)=\begin{cases} x^2\sin\left(\dfrac1x\right), & x\ne 0,\\[4pt] 0, & x=0. \end{cases}f4​(x)=⎩⎨⎧​x2sin(x1​),0,​x=0,x=0.​

We check differentiability and continuity of derivative at x=0x=0x=0.

Step 1: Differentiability at x=0x=0x=0

f4′(0)=lim⁡h→0f4(h)−f4(0)h=lim⁡h→0h2sin⁡(1/h)h=lim⁡h→0hsin⁡(1/h)=0,f_4'(0)=\lim_{h\to 0}\frac{f_4(h)-f_4(0)}{h} =\lim_{h\to 0}\frac{h^2\sin(1/h)}{h} =\lim_{h\to 0} h\sin(1/h)=0,f4′​(0)=h→0lim​hf4​(h)−f4​(0)​=h→0lim​hh2sin(1/h)​=h→0lim​hsin(1/h)=0,

since ∣hsin⁡(1/h)∣≤∣h∣→0|h\sin(1/h)|\le |h|\to 0∣hsin(1/h)∣≤∣h∣→0.

So f4f_4f4​ is differentiable at 000.

Step 2: Derivative for x≠0x\ne 0x=0

For x≠0x\ne 0x=0,

f4′(x)=2xsin⁡(1x)−cos⁡(1x).f_4'(x)=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right).f4′​(x)=2xsin(x1​)−cos(x1​).

Now check continuity of derivative at 000.

As x→0x\to 0x→0,

2xsin⁡(1x)→0,2x\sin\left(\frac1x\right)\to 0,2xsin(x1​)→0,

but

cos⁡(1x)\cos\left(\frac1x\right)cos(x1​)

does not have a limit. Hence f4′(x)f_4'(x)f4′​(x) has no limit as x→0x\to 0x→0.

Therefore f4′f_4'f4′​ is not continuous at x=0x=0x=0.

So,

S→3.S \to 3.S→3.

5. Final matching

We obtained:

P→2,Q→1,R→4,S→3.P\to 2,\qquad Q\to 1,\qquad R\to 4,\qquad S\to 3.P→2,Q→1,R→4,S→3.

This corresponds to Option D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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