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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Mathematics
Chapter
Limits

JEE Advanced 2019 · Paper 2 · Mathematics · Limits

Let F: R → R be a function. We say that f has: PROPERTY 1 if _ h →...

Let F:R→RF: R \to R be a function. We say that ff has:
PROPERTY 11 if lim⁡h→0f(h)−f(0)∣h∣\lim_{h \to 0} \dfrac{f\left(h\right) - f\left(0\right)}{\sqrt{\left|h\right|}} exists and is finite, and
PROPERTY 22 if lim⁡h→0f(h)−f(0)h2\lim_{h \to 0} \dfrac{f\left(h\right) - f\left(0\right)}{{h}^{2}} exists and is finite.
Then which of the following options is/are correct?

Options

  1. Af(x)=x∣x∣f\left(x\right) = x\left|x\right| has PROPERTY 22
  2. Bf(x)=x2/3f\left(x\right) = {x}^{2 / 3} has PROPERTY 11
  3. Cf(x)=sinxf\left(x\right) = sin x has PROPERTY 22
  4. Df(x)=∣x∣f\left(x\right) = \left|x\right| has PROPERTY 11

Correct answer

B. f(x)=x2/3f\left(x\right) = {x}^{2 / 3} has PROPERTY 11

Step-by-step solution

\left(A\right)f\left(x\right) = \left|x\right|<br/><br/>\lim_{h \to 0} \dfrac{\left|h\right| - 0}{\sqrt{\left|h\right|}} = \lim_{h \to 0} \sqrt{\left|h\right|} = 0<br/>(B)f(x)=x23<br/>\left(B\right)f\left(x\right) = {x}^{\dfrac{2}{3}}
lim⁡h→0h23−0∣h∣=lim⁡h→0∣h∣16=0\lim_{h \to 0} \dfrac{{h}^{\dfrac{2}{3}} - 0}{\sqrt{\left|h\right|}} =\lim_{h \to 0} {\left|h\right|}^{\dfrac{1}{6}} = 0
\left(C\right)f\left(x\right) = x\left|x\right|<br/><br/>\lim_{h \to 0} \dfrac{h\left|h\right| - 0}{{h}^{2}} = \lim_{h \to 0} \dfrac{\left|h\right|}{h} = \left\{1,ifh→0+−1,ifh→0−\begin{array}{ll}1, ifh \to {0}^{+} \\ - 1, if h \to {0}^{-}\end{array}\right)
Does not exist
(D)f(x)=sinx\left(D\right)f\left(x\right) = sin x
lim⁡h→0sinh−0h2=lim⁡h→01hsinhh⇒\lim_{h \to 0} \dfrac{sin h - 0}{{h}^{2}} = \lim_{h \to 0} \dfrac{1}{h}\dfrac{sin h}{h} ⇒ does not exist
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