Mathematics

# Let $f(x)\, =\, 3x^{2}.\, \sin \,\displaystyle \frac{1}{x}\, -\, x\cos\, \displaystyle \frac{1}{x},\, x\, \neq\, 0, f(0)\, =\, 0\, f \left (\, \displaystyle \frac{1}{\pi} \right )\, =\, 0$, then which of the  following is/are not correct.

$f(x)$ is non-differentiable at $x = 0$

$f(x)$ is discontinuous at $x = 0$

$f(x)$ is differentiable at $x = 0$

##### SOLUTION
$f(x)\, =\, 3x^{2}. \sin \displaystyle \frac{1}{x}\, -\, x.\cos\, \displaystyle \frac{1}{x}$
$\Rightarrow\, f(x)\, =\,\displaystyle \int \left ( 3x^{2}.\sin \displaystyle \frac{1}{x}\, -\, \cos\, \displaystyle \frac{1}{x} \right ) dx$
$=\, x^{3}\, \sin\, \displaystyle \frac{1}{x}.\, -\, \int \cos\, \frac{1}{x}\, \left (\, -\, \displaystyle \frac{1}{x^{2}} \right )\, x^{3}\, dx\, -\, \int x \cos \displaystyle \frac{1}{x}dx$
$=\, x^{3}\, \sin \, \displaystyle \frac{1}{x}\, +\, C$
Since $f \left ( \displaystyle \frac{1}{\pi} \right )\, =\, 0\, +\, C \, \Rightarrow\, C\, =\, 0$
$\Rightarrow\, f(x)\left \{\begin{matrix} x^{3}\sin\, \displaystyle \frac{a}{x} & , & x \, \neq\, 0 \\ 0 & , & x = 0 \end{matrix} \right \}$

$F(x)$ is clearly continuous and differentiable at $x = 0$ zero with $f(0) = 0$.

$\displaystyle f\, (0)\, =\, \lim_{h \rightarrow 0} \frac{3h^2\, \sin \frac{1}{h}\, -\, h\, \cos \frac{1}{h}}{h}$ $\displaystyle =\, 3h\sin\frac{1}{h}\, -\, \cos \frac{1}{h}$

This limit doesn't exist, hence $f(x)$ is non-differentiable at $x = 0$.

Also $\displaystyle \lim_{x \rightarrow 0}\, f(x)\, =\, 0$.

Thus $f`(x)$ is continuous at $x = 0$.

Its FREE, you're just one step away

Multiple Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

#### Realted Questions

Q1 Subjective Medium
$\int {x\,{{\cos }^{ - 1}}x\,dx}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
If $\displaystyle\int \dfrac{2^x}{\sqrt{1-4^x}}dx=K\sin^{-1}(2^x)+C$, then the value of K is equal to?

 $\int { \tan { x } dx }$ $\cfrac { 1 }{ 2a } \log { \cfrac { x-a }{ x+a } } ,x>a$ $\int { \cot { x } dx }$ $\cfrac1{2a}\log\cfrac{a+x}{a-x}, x Asked in: Mathematics - Integrals 1 Verified Answer | Published on 17th 09, 2020 Q4 Single Correct Medium$\displaystyle \int_0^{2a} f(x) dx = 0$is • A.$\displaystyle \int_0^{2a} f(2a - x) =\int_0^{2a} f(x)$• B.$\displaystyle \int_0^{2a} f(x) = -\int_0^{2a} f (x)$• C.$\displaystyle \int_0^{2a} f(-x) = \int_0^{2a} f(x)$• D.$\displaystyle \int_0^{2a} f(2a - x) = - \int_0^{2a} f(x)$Asked in: Mathematics - Integrals 1 Verified Answer | Published on 17th 09, 2020 Q5 Single Correct Hard Evaluate:$\displaystyle\int \sqrt[3]{x}\cdot \sqrt[7]{1+\sqrt[3]{x^4}}dx$• A.$\displaystyle \dfrac{21}{32}(1+\sqrt{x^4})^{\dfrac{8}{7}}+c$• B.$\displaystyle \dfrac{21}{32}(1+\sqrt{x^4})^{\dfrac{1}{7}}+c$• C.$\displaystyle \dfrac{21}{32}(1+\sqrt[3]{x})^{\dfrac{8}{7}}+c$• D.$\displaystyle \dfrac{21}{32}(1+\sqrt[3]{x^4})^{\dfrac{8}{7}}+c