Continuity and Differentiability Test

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Continuity and Differentiability Test - 40

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Question 1
1 / -0

If $$u=f(x,y)$$ is a differentiable function of $$x$$ and $$y$$, where $$x$$ and $$y$$ are differentiable functions of $$t$$ then:

A
$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { \partial x }{ \partial t } +\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t } $$

B
$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { dx }{ dt } -\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t } $$

C
$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { dx }{ dt } +\cfrac { \partial f }{ \partial y } .\cfrac { dy }{ dt } $$

D
$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { \partial x }{ \partial t } -\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t } $$

Solution

Given $$u=f(x,y)$$ is a a differentiable function of $$x$$ and $$y$$, where $$x, y$$ are differentiable functions of $$t$$.

We have to find $$\dfrac{du}{dt}$$

Since $$x, y$$ are differentiable functions of $$t$$

Let $$x=g(t), y=h(t)$$

Thus by chain rule we get

$$\dfrac{du}{dt}=\dfrac{\partial f}{\partial x}\cdot \dfrac{dx}{dt}+\dfrac{\partial f}{\partial y}\cdot \dfrac{dy}{dt}$$
Question 2
1 / -0

The set of points, where the function $$f(x) = x|x|$$, is differentiable, is given by

A
$$(-\infty, \infty)$$

B
$$(-\infty, 0) \cup (0, \infty)$$

C
$$(0, \infty)$$

D
$$[0, \infty)$$
Question 3
1 / -0

The function $$\displaystyle f(x) = (x^2 - 1) |x^2 - 3x + 2| + cos (|x|)$$, is not differentiable at x=

A
1

B
-1

C
2

D
0

Solution
Question 4
1 / -0

Let $$f:\left[ {0,2} \right] \to R$$ a function which is continuous on $$\left[ {0,2} \right]$$ and is differentiable on $$\left( {0,2} \right)$$ with $$f\left( 0 \right) = 1$$. Let $$F\left( x \right) = \int\limits_0^{{x^2}} {f\left( {\sqrt t } \right)} dt,$$ for $$x \in \left[ {0,2} \right]$$, if $$F'\left( x \right) = f'\left( x \right),\forall x \in \left( {0,2} \right),$$ then $$F(2)$$ equals to

A
$${e^2} - 1$$

B
$${e^4} - 1$$

C
$$e-1$$

D
$${e^4}$$

Solution

From Newton-Leibnitz's formula,
$$\dfrac{d}{dx}\left[ \displaystyle \int_{\phi(x)}^{\psi(x)} f(t)dt\right]=f\left\{\psi(x)\right\} \left\{ \dfrac{d}{dx}\psi(x)\right\}-f\left\{ \phi(x) \right\} \left\{ \dfrac{d}{dx}\phi(x)\right\} $$

Given that,
$$F(x)=\displaystyle \int_{0}^{x^2}f(\sqrt{t})dt$$
$$\therefore F'(x)=f(x)\cdot \dfrac d{dx}(x^2)-f(0)\cdot \dfrac d{dx}(0)$$
$$\Rightarrow F'(x)=2xf(x)\qquad...(i)$$

Also given that,
$$ F'(x)=f'(x)$$
$$\Rightarrow 2xf(x)=f'(x)$$ [ from $$(i)$$]
$$\Rightarrow \dfrac{f'(x)}{f(x)}=2x$$
Integrating both sides w.r.t. $$x$$,
$$\Rightarrow \displaystyle \int \dfrac{f'(x)}{f(x)}dx=\displaystyle \int 2xdx$$
$$\Rightarrow \ln\{f(x)\}=x^2+c $$
$$\Rightarrow f(x)=e^{{x^2}+c}$$
$$\Rightarrow f(x)=ke^{x^2}$$

Now, $$f(0)=1$$
$$\Rightarrow k=1$$

Hence,$$f(x)=e^{x^2}$$

$$\therefore F(2)=\displaystyle \int_{0}^{4}e^tdt=\big[e^t\big]_{0}^{4}=e^4-1$$
Question 5
1 / -0

Directions For Questions

Let $$F : R \rightarrow R$$ be a thrice differentiable function. Suppose that $$F(1)=0, F(3)=-4$$ and $$F'(x) < 0$$ for all $$x \in (1,3)$$. Let $$f(x)=xF(x)$$ for all $$x \in R$$.

...view full instructions

If $$\int_{1}^{3} F'(x)dx=-12$$ and $$\int_{1}^{3} x^{3} F' '(x) dx=40$$, then the correct expression(s) is/are

A
$$9f'(3)+f'(1)-32=0$$

B
$$\int_{1}^{3} f(x)dx=12$$

C
$$9f' (3)-f'(1)+32=0$$

D
$$\int_{1}^{3} f(x) dx=-12$$

Solution
Question 6
1 / -0

Which of the following function is differentiable at x=0

A
$$\cos \left( {|x|} \right) + |x|$$

B
$$\cos \left( {|x|} \right) - |x|$$

C
$$\sin \left( {|x|} \right) + |x|$$

D
$$\sin \left( {|x|} \right) - |x|$$

Solution

A function is differentiable at 0 only if it's L.H.L &R.H.L is equal.
If $$f(x)=\sin\mid x \mid-\mid x \mid$$ then given that
$$\sin(-x)=-\sin x$$ it can be expressed as,
$$\therefore f(x)=\begin{Bmatrix}\sin x-x\quad if \,x\ge 0\\ -\sin x+x\quad if \,x<0\\ \end{Bmatrix}$$
$$\Rightarrow f^{'}(x)=\begin{Bmatrix}\cos x-1 \quad if\, x\ge 0 \\ -\cos x+1 \quad if\, x<0\\ \end{Bmatrix}$$
Given that $$\cos 0=1$$ we have $$f_{+}^{'}(0)=2$$ & $$f_{-}^{'}(0)=-2$$
$$\therefore f_{+}^{'}(0) \ne f_{-}^{'}(0)$$
If $$f(x)=\cos \mid x \mid-\mid x\mid$$ , then given that $$\cos(-x)=\cos(x)$$ can be written as:
$$\therefore f(x)=\begin{Bmatrix}\cos x-x\quad if \,x\ge 0\\ \cos x+x\quad if \,x<0\\ \end{Bmatrix}$$
$$\Rightarrow f^{'}(x)=\begin{Bmatrix}-\sin x+1 \quad if\, x\ge 0 \\ -\sin x+1 \quad if\, x<0\\ \end{Bmatrix}$$
Given that $$\sin 0=0$$ we have $$f_{+}^{'}(0)=-1$$ & $$f_{-}^{'}(0)=1$$
$$\therefore f_{+}^{'}(0) \ne f_{-}^{'}(0)$$
If $$f(x)=\cos \mid x \mid+\mid x \mid$$, then
$$\therefore f(x)=\begin{Bmatrix}\cos x+x\quad if \,x\ge 0\\ \cos x-x\quad if \,x<0\\ \end{Bmatrix}$$
$$\Rightarrow f^{'}(x)=\begin{Bmatrix}-\sin x+1 \quad if\, x\ge 0 \\ -\sin x-1 \quad if\, x<0\\ \end{Bmatrix}$$
Given that $$\sin 0=0$$ we have $$f_{+}^{'}(0)=1$$ & $$f_{-}^{'}(0)=-1$$
$$\therefore A\Rightarrow \cos(\mid x \mid)+\mid x \mid$$
Question 7
1 / -0

If $$f : R \rightarrow R$$ is defined by
$$f(x) = \left \{\begin{matrix} \dfrac{x + 2}{x^2 + 3x + 2} & if & x \in R - \{-1, -1\} \\ -1 & if & x = -2 \\ 0 & if &x = -1\end{matrix} \right.$$ then $$f(x)$$ continuous on the set

A
$$R$$

B
$$R - \{-2\}$$

C
$$R - \{-1, -2\}$$

D
$$R-\{-1\}$$

Solution
Question 8
1 / -0

If $$x=a\cos^{3}\theta, y=a\sin^{3}\theta$$, then $$1+ {(\dfrac {dy}{dx})}^{2}$$ is

A
$$\sec^{2}\theta$$

B
$$\tan \theta$$

C
$$1$$

D
$$\tan^{2}\theta$$

Solution

Differentiate $$x = a{\cos ^3}\theta $$ with respect to $$\theta $$,

$$\frac{{dx}}{{d\theta }} = - 3a{\cos ^2}\theta \sin \theta $$

Differentiate $$y = a{\sin ^3}\theta $$ with respect to $$\theta $$,

$$\frac{{dy}}{{d\theta }} = 3a{\sin ^2}\theta \cos \theta $$

Now,

$$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}$$

$$ = - \frac{{3a{{\sin }^2}\theta \cos \theta }}{{3a{{\cos }^2}\theta \sin \theta }}$$

$$ = - \tan \theta $$

Then,

$$1 + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( { - \tan \theta } \right)^2}$$

$$ = 1 + {\tan ^2}\theta $$

$$ = {\sec ^2}\theta $$
Question 9
1 / -0

If $$ f\left( x+y \right) =f\left( x \right) +f\left( y \right)$$ then $$ f\left( x \right)$$ may be

A
$$x$$

B
$$x+1$$

C
$${ x }^{ 2 }+1$$

D
$$\log { x }$$

Solution
Question 10
1 / -0

The function $$f\left( x \right) = \,{\sin ^{ - 1}}\left( {\cos \,x} \right)\,is\,: - $$

A
discontinuous at $$=0$$

B
continuous at $$=0$$

C
differentiable $$=0$$

D
none of these

Solution

Given:$$f\left(x\right)={\sin}^{-1}{\left(\cos{x}\right)}$$
Continuity at $$x=0$$

We have,LHL at $$x=0$$
$$\displaystyle \lim_{x\rightarrow\,{0}^{-}}{f\left(x\right)}=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{\left(0-h\right)}\right)}$$

$$=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{h}\right)}$$

$$={\sin}^{-1}{1}=\dfrac{\pi}{2}$$

RHL at $$x=0$$
$$\displaystyle \lim_{x\rightarrow\,{0}^{+}}{f\left(x\right)}=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{\left(0+h\right)}\right)}$$

$$=\displaystyle \lim_{h\rightarrow\,0}{\sin}^{-1}{\left(\cos{h}\right)}$$

$$={\sin}^{-1}{1}=\dfrac{\pi}{2}$$

$$f\left(0\right)={\sin}^{-1}{\left(\cos{0}\right)}={\sin}^{-1}{1}=\dfrac{\pi}{2}$$

Differentiability at $$x = 0:$$
LHD at $$x = 0=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{f\left(x\right)-f\left(0\right)}{x-0}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(0-h\right)}\right)}-\dfrac{\pi}{2}}{-h}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(-h\right)}\right)}-\dfrac{\pi}{2}}{-h}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\cos{h}\right)}-\dfrac{\pi}{2}}{-h}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{-}}{\dfrac{{\sin}^{-1}{\left(\sin{\left(\dfrac{\pi}{2}-h\right)}\right)}-\dfrac{\pi}{2}}{-h}}$$

$$=\displaystyle \lim_{h\rightarrow\,0}\dfrac{-h}{-h}=1$$

RHD at $$x = 0=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{f\left(x\right)-f\left(0\right)}{x-0}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(0+h\right)}\right)}-\dfrac{\pi}{2}}{h}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\cos{\left(h\right)}\right)}-\dfrac{\pi}{2}}{h}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\cos{h}\right)}-\dfrac{\pi}{2}}{h}}$$

$$=\displaystyle \lim_{x\rightarrow\,{0}^{+}}{\dfrac{{\sin}^{-1}{\left(\sin{\left(\dfrac{\pi}{2}-h\right)}\right)}-\dfrac{\pi}{2}}{h}}$$

$$=\displaystyle \lim_{h\rightarrow\,0}\dfrac{-h}{h}=-1\\$$
$$\therefore\,LHD\neq\,RHD$$
Hence, the function is not differentiable at $$x = 0$$ but is continuous at $$x = 0$$
Hence option$$(b)$$ is correct.

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Continuity and Differentiability Test - 40

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Continuity and Differentiability Test - 40

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SHARING IS CARING

Question 1

$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { \partial x }{ \partial t } +\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t } $$

$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { dx }{ dt } -\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t } $$

$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { dx }{ dt } +\cfrac { \partial f }{ \partial y } .\cfrac { dy }{ dt } $$

$$\cfrac { du }{ dt } =\cfrac { \partial f }{ \partial x } .\cfrac { \partial x }{ \partial t } -\cfrac { \partial f }{ \partial y } .\cfrac { \partial y }{ \partial t } $$

Solution

Question 2

$$(-\infty, \infty)$$

$$(-\infty, 0) \cup (0, \infty)$$

$$(0, \infty)$$

$$[0, \infty)$$

Question 3

1

-1

2

0

Solution

Question 4

$${e^2} - 1$$

$${e^4} - 1$$

$$e-1$$

$${e^4}$$

Solution

Question 5

$$9f'(3)+f'(1)-32=0$$

$$\int_{1}^{3} f(x)dx=12$$

$$9f' (3)-f'(1)+32=0$$

$$\int_{1}^{3} f(x) dx=-12$$

Solution

Question 6

$$\cos \left( {|x|} \right) + |x|$$

$$\cos \left( {|x|} \right) - |x|$$

$$\sin \left( {|x|} \right) + |x|$$

$$\sin \left( {|x|} \right) - |x|$$

Solution

Question 7

$$R$$

$$R - \{-2\}$$

$$R - \{-1, -2\}$$

$$R-\{-1\}$$

Solution

Question 8

$$\sec^{2}\theta$$

$$\tan \theta$$

$$1$$

$$\tan^{2}\theta$$

Solution

Question 9

$$x$$

$$x+1$$

$${ x }^{ 2 }+1$$

$$\log { x }$$

Solution

Question 10

discontinuous at $$=0$$

continuous at $$=0$$

differentiable $$=0$$

none of these

Solution

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