Continuity and Differentiability Test

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Continuity and Differentiability Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$y={ \left( \tan ^{ -1 }{ x } \right) }^{ 2 }$$ and $${ \left( { x }^{ 2 }+1 \right) }^{ 2 }\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2x\left( { x }^{ 2 }+1 \right) \dfrac { dy }{ dx } =k$$, then the value of $$k$$ is

A
$$3$$

B
$$2$$

C
$$1$$

D
$$0$$

Solution

$$y=(\text{tan}^{-1}x)^{2}$$
$$\dfrac{d{y}}{d{x}}=\dfrac{2\text{tan}^{-1}x}{1+x^{2}}$$
$$(1+x^{2})\dfrac{d{y}}{d{x}}=2\text{tan}^{-1}x$$
$$(1+x^{2})\dfrac{d^{2}{y}}{d{x^{2}}}+2{x}\dfrac{d{y}}{d{x}}=2\dfrac{1}{1+x^{2}}$$
$$(1+x^{2})^{2}\dfrac{d^{2}{y}}{d{x^{2}}}+2{x}(1+x^{2})\dfrac{d{y}}{d{x}}=2\implies k=2$$
Question 2
1 / -0

Which of the following is not differentiable in the interval $$(-1,2)$$?

A
$$\displaystyle\int\limits_{x}^{2x}(\log x)^2 dx$$

B
$$\displaystyle\int\limits_{x}^{2x}\large{\frac{\sin x}{x}} dx$$

C
$$\displaystyle\int\limits_{x}^{2x}\large{\frac{1-t+t^2}{1+t+t^2}} dt$$

D
none of these

Solution

For $$x\in (-1,2)$$, $$\displaystyle\int\limits_{x}^{2x}f(x) dx$$ is differentiable in $$(-1,2)$$ if $$f(x)$$ is continuous in $$(-1,2)$$.
For the given functions $$f(x) =\log x$$ is not defined for $$x\leq 0$$ so it is not continuous in $$(-1,2)$$.
So, option $$(A)$$ is not differentiable in $$(-1,2)$$.
But remaining two functions are continuous in $$(-1,2)$$ hence option $$(C)$$ and $$(D)$$ are differentiable in $$(-1,2)$$.
Question 3
1 / -0

Let $$f:(0,\infty ) \to R$$ be a differentiable function such that $$f'(x) = 2 - {{f(x)} \over x}$$ for all $$x \in (0,\infty )$$ and $$f(1) \ne 1$$

A
$$\mathop {\lim }\limits_{x \to 0} f'\left( {{1 \over x}} \right) = 1$$

B
$$\mathop {\lim }\limits_{x \to 0} xf\left( {{1 \over x}} \right) = 2$$

C
$$\mathop {\lim }\limits_{x \to 0} {x^2}f'\left( x \right) = 0$$

D
$$\left| {f(x)} \right| \le 2{\rm{ \ for\ all\ x}} \in {\rm{(0,2)}}$$

Solution

$$f'(x)=2-\dfrac{f(x)}{x}\implies x f'(x)+f(x)=2{x}$$
$$d(x f(x))=2{x} d{x}$$
Integrating on both sides
$$x f(x)=x^{2}+c\implies f(x)=x+\dfrac{c}{x}$$
$$f'(x)=1-\dfrac{c}{x^2}$$
$$\displaystyle \lim _{x\to 0} f'\bigg(\dfrac{1}{x}\bigg)=\lim _{x\to 0} (1-c x^2)=1$$
$$\displaystyle\lim _{x\to 0} x f\bigg(\dfrac{1}{x}\bigg)=\lim _{x\to 0} (1+x^2)=1$$
$$\displaystyle\lim _{x\to 0} x^2 f'(x)=\lim_{x\to 0} (x^{2}-c)=-c$$
$$AM\ge GM$$
$$\implies |f(x)|\ge 2\sqrt{c}$$
Question 4
1 / -0

Let $$f(x)=\begin{cases}
(x-1) \sin \left( \large{\frac{1}{x-1}}\right) \ &\mbox{if}& \ x \ne 1\\
0, \ &\mbox{if}& \ x \ne 1
\end{cases}$$.
Then which one of the following is true?

A
$$f$$ is differentiable neither at $$x=0$$ nor at $$x=1$$

B
$$f$$ is differentiable at $$x=0$$ and $$x=1$$

C
$$f$$ is differentiable at $$x=0$$ but not at $$x=1$$

D
$$f$$ is differentiable at $$x=1$$ but not at $$x=0$$

Solution

$${f}'(x)=\sin\left ( \cfrac{1}{x-1} \right )+(x-1)\cos\left ( \cfrac{1}{x-1} \right )\left ( \cfrac{-1}{(x-1)^{2}} \right )$$
$${f}'(0)=\sin(-1)+(-1)\cos(-1)(-1)$$
$$=\cos1-\sin1$$
As $$x\rightarrow 1 ,{f}'(1)\rightarrow \infty $$ $$\left ( \because \cfrac{1}{x-1}\rightarrow \infty \right )$$
$$\therefore f(x)$$ is differentiable at $$x=0$$, but not at $$x=1$$
Question 5
1 / -0

Let $$f(x)={ x }^{ 3 }-{ x }^{ 2 }+x+1$$ and $$g\left( x \right) =\begin{cases} max\{ f\left( t \right) ;0\le t\le x\} \quad ;\quad 0\le x\le 1 \\ 3-x\quad \quad \quad \quad \quad \quad \quad \quad ;\quad 1<x\le 2 \end{cases}$$ then

A
$$g(x)$$ is continuous and differentiable in $$(0, 2)$$

B
$$g(x)$$ is discontinuous at finite number of points in $$(0,2)$$

C
$$g(x)$$ is non-drivable at $$2$$ points

D
$$g(x)$$ is continuous but non-differentiable at one points

Solution

$$f(x)=x^{3}-x^{2}+x+1$$
$$g(x)=\begin{cases} max.(F(t):0\le t \le x)\ \ \ \ 0\le x\le 1 \\ 3-x; \ \ \ 1\le x\le 2 \end{cases}$$
here $$F(0)=1, F(1)=2$$
If $$F(x)$$ increasing in $$0$$ to $$1$$ then max will be $$f(t)$$
$$F(t)=3x^{2}-2x+1$$
$$0=9x^{2}-2x+1$$
$$x=\dfrac{2\pm \sqrt{8}i}{6}$$
clearly $$f x^{n}$$ is always increasing
So $$g(x)=x^{3}-x^{2}+x+1, 0\le x\le 1$$
$$g-x, 1\le x\le 1$$
$$\displaystyle\lim_{x\rightarrow 1^{-}}g(x)=2, \lim_{x\rightarrow 1^{+}}g(x)=2$$
$$\displaystyle\lim_{x\rightarrow 1}g(x)=2$$
So clearly $$g(x)\rightarrow$$ continuous
$$g(1)=2$$ at $$x=1$$
clearly $$Fx^{n}$$ is deriable
Question 6
1 / -0

If $$x=a{ t }^{ 2 },y=2at$$, then $$\cfrac { dy }{ dx } =$$_____;$$t\neq 0$$

A
$$at$$

B
$$\cfrac { t }{ 2 } $$

C
$$\cfrac { 2 }{ t } $$

D
$$\cfrac { 1 }{ t } $$

Solution

$$x=at^{2}$$
Differentiating both sides with respect to t
$$\dfrac{dx}{dt}=\dfrac{d}{dt}(at^{2})$$
$$\dfrac{dx}{dt}=a\dfrac{d}{dt}(t^{2})$$
$$\dfrac{dx}{dt}=2at$$ (1)
$$y=2at$$
Differentiating both sides with respect to t
$$\dfrac{dy}{dt}=\dfrac{d}{dt}(2at)$$
$$\dfrac{dy}{dt}=2a\dfrac{d}{dt}(t)$$
$$\dfrac{dy}{dt}=2a$$ (2)
Now
$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$$
From (1) and (2)
$$\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$$
Question 7
1 / -0

If the function $$f(x)=\dfrac{e^{x^{2}}-\cos x}{x^{2}}$$ for $$x \neq 0$$ continuous at $$x=0$$ then $$f(0)=$$

A
$$\dfrac{1}{2}$$

B
$$\dfrac{3}{2}$$

C
$$2$$

D
$$\dfrac{1}{3}$$

Solution

Here , function is continuous ,

So, left limit and right limit boh are equal and equals to the value of $$f(x)$$ at $$x=0$$

Therefore

$$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}-cosx}{x^2}$$

Solving limit by derivative approach,

=> $$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{e^{x^2}.2x+sinx}{2x}$$

=> $$\displaystyle \lim_{x\rightarrow0} f(x)=\lim_{x\rightarrow0} \dfrac{2e^{x^2}+e^{x^2}.x^2+cosx}{2x}$$

Putting $$x=0$$, we get

=> $$f(0)=\dfrac{2+0+1}{2}=\dfrac{3}{2}$$
Question 8
1 / -0

The function $$f : R /{0} \rightarrow R$$ given by $$f(x) =
\dfrac{1}{x} - \dfrac{2}{e^{2x} -1}$$ can be made continuous at $$x=0$$ by
defining $$f(0)$$ as

A
0

B
1

C
2

D
-1

Solution

Given $$f(x)=\dfrac{1}{x}-\dfrac{2}{e^{2x}-1}$$

For $$f(x)$$ to be continuous at $$x=0$$ , its limit should exists at $$x=0$$ and must be finite.

Hence
$$f(0)=\lim\limits_{x\to0}\dfrac{e^{2x}-1-2x}{x(e^{2x}-1)}$$

$$=\lim\limits_{x\to0}\dfrac{1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}.....-1-2x}{x+2x^2+\frac{4x^3}{2!}.....-x}=\dfrac{\frac{4x^2}{2!}}{2x^2}=1$$......[neglecting higher power of $$x$$]
Question 9
1 / -0

The function $$f(x) =
\ sin^{-1} (\ cosx)$$ is

A
Discontinuous at $$x=0$$

B
continuous at $$x=0$$

C
differentiable at $$x=0$$

D
None of these

Solution

We know that the cosine function, is nothing more than the sin $$\dfrac{\pi}{2}$$ radians out of phase, as proved below,

$$\cos(\theta-\dfrac{\pi}{2})=\cos(\theta)\cos(-\dfrac{\pi}{2})-\sin(\theta)\sin(-\dfrac{\pi}{2})$$

$$\cos(\theta-\dfrac{\pi}{2})=\cos(\theta).0-(-\sin(\theta)\sin(\dfrac{\pi}{2}))$$

$$\cos(\theta-\dfrac{\pi}{2})=\cos(\theta).1=\sin(\theta)$$

So, it $$f(x)=sin^{-1}(cosx)$$ is continuous at $$x=0$$

So we can say that the since function, 90 degree ahead, is the cousin function.

$$arc\sin(\cos(x))=arcsin(sin(x+\dfrac{\pi}{2}))$$

Using the property of inverse that $$f^{-1}(f(x))=x$$, we have

$$arc\sin(\cos(x))=x+\dfrac{\pi}{2}$$
Question 10
1 / -0

Let $$f(x) = \begin{cases} x & x<1 \\ 2-x & 1 \leq x \leq 2 \\ -2+3x-x^2 & x>2 \end{cases}$$
then $$f(x)$$ is

A
differentiable at $$x=1$$

B
differentiable at $$x=2$$

C
differentiable at $$x=1$$ and $$x=2$$

D
not differentiable at $$x=10$$

Solution

Differentiability check at 1
$$\lim _{ h\rightarrow 0 }{ \cfrac { f(1+h)-f(1) }{ h } } \\ \Rightarrow \cfrac { 2-(1+h)-(2-1) }{ h } \\ \Rightarrow \cfrac { -h }{ h } =-1\\ \lim _{ h\rightarrow 0 }{ \cfrac { f(1)-f(1-h) }{ h } } \\ \Rightarrow \cfrac { 1-(1-h) }{ h } \\ \Rightarrow \cfrac { h }{ h } =1\\ $$
Not differentiable at 1.
Differentiability check at 2
$$\lim _{ h\rightarrow 0 }{ \cfrac { f(2+h)-f(2) }{ h } } \\ \Rightarrow \cfrac { (-2+3(2+h)-{ (2+h) }^{ 2 }) }{ h } -0\\ \Rightarrow \cfrac { -2+6+3h-4-{ h }^{ 2 }-4h }{ h } \\ \Rightarrow \cfrac { -{ h }^{ 2 }-h }{ h } \\ \Rightarrow \cfrac { -h(h+1) }{ h } =-1\\ \lim _{ h\rightarrow 0 }{ \cfrac { f(2)-f(2-h) }{ h } } \\ \Rightarrow \cfrac { 0-(2-(2-h)) }{ h } \\ \Rightarrow \cfrac { -2+2-h }{ h } \Rightarrow -1$$
Differentiable at 2.

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Re-Attempt Weekly Quiz Competition

Continuity and Differentiability Test - 23

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Continuity and Differentiability Test - 23

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SHARING IS CARING

Question 1

$$3$$

$$2$$

$$1$$

$$0$$

Solution

Question 2

$$\displaystyle\int\limits_{x}^{2x}(\log x)^2 dx$$

$$\displaystyle\int\limits_{x}^{2x}\large{\frac{\sin x}{x}} dx$$

$$\displaystyle\int\limits_{x}^{2x}\large{\frac{1-t+t^2}{1+t+t^2}} dt$$

none of these

Solution

Question 3

$$\mathop {\lim }\limits_{x \to 0} f'\left( {{1 \over x}} \right) = 1$$

$$\mathop {\lim }\limits_{x \to 0} xf\left( {{1 \over x}} \right) = 2$$

$$\mathop {\lim }\limits_{x \to 0} {x^2}f'\left( x \right) = 0$$

$$\left| {f(x)} \right| \le 2{\rm{ \ for\ all\ x}} \in {\rm{(0,2)}}$$

Solution

Question 4

$$f$$ is differentiable neither at $$x=0$$ nor at $$x=1$$

$$f$$ is differentiable at $$x=0$$ and $$x=1$$

$$f$$ is differentiable at $$x=0$$ but not at $$x=1$$

$$f$$ is differentiable at $$x=1$$ but not at $$x=0$$

Solution

Question 5

$$g(x)$$ is continuous and differentiable in $$(0, 2)$$

$$g(x)$$ is discontinuous at finite number of points in $$(0,2)$$

$$g(x)$$ is non-drivable at $$2$$ points

$$g(x)$$ is continuous but non-differentiable at one points

Solution

Question 6

$$at$$

$$\cfrac { t }{ 2 } $$

$$\cfrac { 2 }{ t } $$

$$\cfrac { 1 }{ t } $$

Solution

Question 7

$$\dfrac{1}{2}$$

$$\dfrac{3}{2}$$

$$2$$

$$\dfrac{1}{3}$$

Solution

Question 8

0

1

2

-1

Solution

Question 9

Discontinuous at $$x=0$$

continuous at $$x=0$$

differentiable at $$x=0$$

None of these

Solution

Question 10

differentiable at $$x=1$$

differentiable at $$x=2$$

differentiable at $$x=1$$ and $$x=2$$

not differentiable at $$x=10$$

Solution

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