Continuity and Differentiability Test

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Continuity and Differentiability Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)={ sin }^{ -1 }\left[ \dfrac { 2x }{ 1+{ x }^{ 2 } } \right] $$,then $$f(x)$$ is differentiable on

A
[-1,1]

B
R-{-1,1}

C
R-(-1,1)

D
None of these

Solution
Question 2
1 / -0

f(X)=|x|+|x-1| is continuous at

A
'0' only

B
0,1 only

C
Every where

D
No where

Solution

$$\textbf{Step-1: Finding values of}$$ $$\mathbf{f(x)}$$
$$f(x)=|x|+|x-1|$$
$$f(x)=-x-|x-1|$$ $$\text{at}$$ $$x\leq{0}$$
$$=x-(x-1)$$ $$\text{at}$$ $$0\leq{x}<1$$
$$=x+(x-1)$$ $$\text{at}$$ $$x\geq{1}$$
$$\text{Hence, we get}$$
$$f(x)=1-2x$$ $$\text{at}$$ $$x\leq{0}$$
$$=1$$ $$\text{at}$$ $$0\leq{x}<1$$
$$=2x-1$$ $$\text{at}$$ $$x\geq{1}$$
$$\textbf{Step-2: Finding limits of the function at constraint values}$$
$$\text{At}$$ $$x=0,$$
$$\lim_{x\to 0} f(x)=1$$
$$\text{Hence,}$$ $$f(x)$$ $$\text{is continuous at}$$ $$x=0$$
$$\text{At}$$ $$x=1,$$
$$\lim_{x\to 1} f(x)=1$$
$$\text{Hence,}$$ $$f(x)$$ $$\text{is continuous at}$$ $$x=1$$
$$\text{Hence,}$$ $$f(x)$$ $$\text{is continuous everywhere}$$

$$\textbf{Hence,Correct option is (C)}$$
Question 3
1 / -0

If $$f\left( x \right) = {\left| x \right|^{\left| {\sin x} \right|}}$$, then $${f'}\left( { - \dfrac{\pi }{4}} \right)$$ is equals

A
$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( { - \dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi }} \right)$$

B
$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } + \dfrac{{2\sqrt 2 }}{\pi }} \right)$$

C
$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{\pi }{4} + \dfrac{{2\sqrt 2 }}{\pi }} \right)$$

D
None of these.
Question 4
1 / -0

f(x) is diffrentiable function and (f(x). g(x)) is differentiable a x=a , then

A
g(x) must be differentiable at x=a

B
if g(x) is discontinuous , then f(a) =0

C
f(a) $$\neq $$ 0 then g(x) must be differentiable

D
nothing can be said
Question 5
1 / -0

The order of the differential equation of all circles whose radius is $$4$$, is?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Circle with radius $$4$$ is represented as follows:-
$$(x - h)^2 + (y - k)^2 = 4^2$$ ..... $$(1)$$

Differentiating w.r.t $$x$$
$$2(x - h) + 2(y - k) \dfrac{dy}{dx} = 0$$

$$\Rightarrow\ (x - h) + (y - k) \dfrac{dy}{dx} = 0$$ .....$$(2)$$

Again differentiating w.r.t $$x$$
$$1 + (y - k)\dfrac{d^2y}{dx^2} + \left (\dfrac{dy}{dx} \right )^2 = 0$$

$$\Rightarrow\ (y - k) = \dfrac{-\left (\dfrac{dy}{dx} \right )^2 - 1}{\dfrac{d^2y}{dx^2}} $$ ....$$(3)$$

From $$(2)$$ we get
$$\Rightarrow\ (x - h) = - (y - k) \dfrac{dy}{dx} $$

$$\Rightarrow\ (x - h) = - \left \{ \dfrac{\left (\dfrac{dy}{dx} \right )^2 - 1 }{\dfrac{d^2y}{dx^2}} \right \} \dfrac{dy}{dx} $$

$$\therefore\ (x - h) = \dfrac{\left (\dfrac{dy}{dx} \right )^3 + \left (\dfrac{dy}{dx} \right )}{\dfrac{d^2 y}{dx^2}} $$ .... $$(4)$$

Substituting $$(4)$$ and $$(3)$$ in eq. $$(1)$$
$$\left (\dfrac{\left(\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx}}{\dfrac{d^2y}{dx^2}} \right )^2 + \left (\dfrac{-\left (\dfrac{dy}{dx} \right )^2 - 1}{\dfrac{d^2y}{dx^2}} \right )^2 = 4^2$$

$$ \left [\left (\dfrac{dy}{dx} \right )^3 + \dfrac{dy}{dx} \right ]^2 + \left [\left (\dfrac{dy}{dx} \right )^2 + 1 \right ]^2 = 16 \dfrac{d^2y}{dx^2} $$

$$\therefore$$ Order of D.E of circle whose radius is $$4$$ is $$2$$.
Question 6
1 / -0

Give that f(x) =xg(x) /$$ \left | x \right | $$ , g(0) = 0 and f(x) is continous at x=0. Then the value of f' (0)

A
Does not exist

B
is -1

C
is 1

D
is 0

Solution
Question 7
1 / -0

If $$f(x) = \left\{\begin{matrix} \dfrac{x\log \cos x}{\log(1+x^2)}, & x \neq 0\\ 0, & x=0\end{matrix}\right.$$ then

A
f(x) is not continuous at x=0.

B
f(x) is continuous at x=0.

C
f(x) is continuous at x=0 but not differentiable at x=0.

D
f(x) is differentiable at x=0.

Solution
Question 8
1 / -0

Let $$f:(-1,1)\rightarrow R$$ be a differentiable function satisfying
$$(f'(x))^4=16(f(x))^2$$ for all $$x\in (-1,1)$$
$$f(0)=0$$
The number of such functions is

A
$$2$$

B
$$3$$

C
$$4$$

D
more than $$4$$

Solution

Given:$${\left({f}^{\prime}{\left(x\right)}\right)}^{4}=16{\left(f{\left(x\right)}\right)}^{2}$$ for all $$x\in\,\left(–1 , 1\right)$$
$$f\left(0\right)=0$$
$$\Rightarrow\, {\left({f}^{\prime}{\left(x\right)}\right)}^{2}=\pm\,4f\left(x\right)$$
$$\Rightarrow\, {f}^{\prime}{\left(x\right)}=\pm\,2\sqrt{\pm\,f\left(x\right)}$$

$$(i)$$Case $$1$$
$${f}^{\prime}{\left(x\right)}=2\sqrt{\pm\,f\left(x\right)}$$
$$\displaystyle\int{\dfrac{d\left(f\left(x\right)\right)}{f\left(x\right)}}=\displaystyle\int{2\,dx}$$
$$\Rightarrow\,2\sqrt{f\left(x\right)}=2x+c$$
$$\Rightarrow\,f\left(0\right)=0\Rightarrow\,c=0$$
$$\Rightarrow\,\sqrt{f\left(x\right)}=x$$
$$\Rightarrow\,x\ge 0$$
$$f\left(x\right)={x}^{2},\,\,0\le x<1$$

$$(ii)$$Case $$2$$
$${f}^{\prime}{\left(x\right)}=-2\sqrt{f\left(x\right)}$$
$$\Rightarrow\,\sqrt{f\left(x\right)}=-x\Rightarrow\,x\le 0$$
$$f\left(x\right)={x}^{2};\,\,\,-1<x\le 0$$

$$(iii)$$Case $$3$$
$${f}^{\prime}{\left(x\right)}=2\sqrt{-f\left(x\right)}$$
$$\sqrt{-f\left(x\right)}=x$$
$$-f\left(x\right)={x}^{2}$$
$$f\left(x\right)=-{x}^{2};\,\,\,0\le x< 1$$

$$(iv)$$Case $$4$$
$${f}^{\prime}{\left(x\right)}=-2\sqrt{-f\left(x\right)}$$
$$\sqrt{-f\left(x\right)}=-x$$
$$f\left(x\right)=-{x}^{2};\,\,\,-1< x\le 0$$

$$(v)$$Also, one singular solution of given differential equation is
$$f\left(x\right)=0,\,\,-1<x<1$$

Hence, there are more than $$4$$ function possible
$${f}_{1}\left( x \right) =\begin{cases} {x}^{2};0\le x< 1 \\ {-x}^{2}; -1<x<0 \end{cases}$$
$${f}_{2}\left( x \right) =\begin{cases} -{x}^{2};0\le x< 1 \\ {x}^{2}; -1<x<0 \end{cases}$$
$${f}_{3}\left( x \right) = {x}^{2},\,-1<x<1$$
$${f}_{4}\left( x \right) =-{x}^{2},\,-1<x<1$$
$${f}_{5}\left( x \right) =0,\,-1<x<1$$...
Hence there are more than $$4$$ solutions
Question 9
1 / -0

Let $$f(x)$$ be a function satisfying $$f(x+y) = f(x)+f(y)$$ and $$f(x) = xg(x)$$ $$\forall x, ~y \in$$ R, where $$g(x)$$ is a continuous function then, which of the following is true?

A
$$f'(x)=g'(0)$$

B
$$f'(x)=g (0)$$

C
$$f(x)=g (0)$$

D
$$f(x)=g'(0)$$

Solution
Question 10
1 / -0

If the function $$f:[0,8] \rightarrow R$$ is differentiable, then for$$0<a, b<2, \int_{0}^{8} f(t) d t$$ is equal to

A
$$3\left[\alpha^{3} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$$

B
$$3\left[\alpha^{3} f(\alpha)+\beta^{3} f(\beta)\right]$$

C
$${3}\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]$$

D
$$3\left[\alpha^{2} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$$

Solution

$$\operatorname{Let} g(x)=\int_{0}^{x^{3}} f(t) d t\\$$
$$ \text { Now } \int_{0}^{8} f(t) d t=g(2) =\dfrac{g(2)-g(1)}{2-1}+\dfrac{g(1)-g(0)}{1-0} \\$$
$$=g^{\prime}(\alpha)+g^{\prime}(\beta) \\$$
$$=3\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right] $$

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 49

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Continuity and Differentiability Test - 49

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SHARING IS CARING

Question 1

[-1,1]

R-{-1,1}

R-(-1,1)

None of these

Solution

Question 2

'0' only

0,1 only

Every where

No where

Solution

Question 3

$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( { - \dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } - \dfrac{{2\sqrt 2 }}{\pi }} \right)$$

$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{4}{\pi } + \dfrac{{2\sqrt 2 }}{\pi }} \right)$$

$${\left( {\dfrac{\pi }{4}} \right)^{1/\sqrt 2 }}\left( {\dfrac{{\sqrt 2 }}{2}{\text{ln}}\dfrac{\pi }{4} + \dfrac{{2\sqrt 2 }}{\pi }} \right)$$

None of these.

Question 4

g(x) must be differentiable at x=a

if g(x) is discontinuous , then f(a) =0

f(a) $$\neq $$ 0 then g(x) must be differentiable

nothing can be said

Question 5

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 6

Does not exist

is -1

is 1

is 0

Solution

Question 7

f(x) is not continuous at x=0.

f(x) is continuous at x=0.

f(x) is continuous at x=0 but not differentiable at x=0.

f(x) is differentiable at x=0.

Solution

Question 8

$$2$$

$$3$$

$$4$$

more than $$4$$

Solution

Question 9

$$f'(x)=g'(0)$$

$$f'(x)=g (0)$$

$$f(x)=g (0)$$

$$f(x)=g'(0)$$

Solution

Question 10

$$3\left[\alpha^{3} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$$

$$3\left[\alpha^{3} f(\alpha)+\beta^{3} f(\beta)\right]$$

$${3}\left[\alpha^{2} f\left(\alpha^{3}\right)+\beta^{2} f\left(\beta^{3}\right)\right]$$

$$3\left[\alpha^{2} f\left(\alpha^{2}\right)+\beta^{2} f\left(\beta^{2}\right)\right]$$

Solution

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