Limits and Continuity Test 8

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Limits and Continuity Test 8

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Question 1
1 / -0

Let $$f : R\rightarrow R$$ be any function, Define
$$g:R\rightarrow R$$ by $$g(x)=|f(x)|\forall x$$, then

A
$$g$$ is continuous if $$f$$ is not continuous

B
$$g$$ is not continuous if $$f$$ is not continuous

C
$$g$$ is continuous if $$f$$ is continuous

D
$$g$$ is differentiable if $$f$$ is differentiable

Solution

From the property of the modulus function, it is clear that it is continuous along the entire number line.
Thus, the function defined by $$g(x) = |f(x)|$$ is continuous if $$f(x)$$ is a continuous function.
Question 2
1 / -0

$$\displaystyle \lim_{x\rightarrow \infty }\frac{2x+7\sin x}{4x+3\cos x}=$$

A
$$1$$

B
$$-1$$

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

D
$$-\dfrac{1}{2}$$

Solution

We divide both the numerator and denominator by x.
Since, from the property of trigonometric functions we know that $$sin x$$ and $$cos x$$ can only have their values between $$-1$$ and $$1$$
Thus, the expression transforms to,
$$lim _{x \rightarrow \infty } \dfrac{2 + 7\dfrac{sin x}{x}}{4 + 3\dfrac{cos x}{x}} $$
Now applying the limit,
$$= \dfrac{2 +0}{4+0} $$
$$= \dfrac{1}{2} $$
Hence, option 'C' is correct.
Question 3
1 / -0

The function $$y=3\sqrt{x}-|x-1|$$ is continuous at

A
$$x<0$$

B
$$x\geq 1$$

C
$$0\leq x\leq 1$$

D
$$x\geq 0$$

Solution

Since, $$|x|$$ is a function that is continuous along the entire number line, therefore, $$|x-1|$$ is continuous for all $$x$$
Now, we know that a negative value cannot go inside the square root sign. So $$x >= 0$$
Thus the function $$f(x)$$ defined by $$ 3{x}^{0.5} - |x -1| $$ is continuous for all $$x >= 0$$
Question 4
1 / -0

lf $$\mathrm{f}(\mathrm{x})=\left\{\begin{matrix}1+x &x\leq 1 \\ 3-ax^{2}& x>1\end{matrix}\right.$$ is continuous at $${x}=1$$ then $${a}=({a}>0)$$

A
$$1$$

B
$$2$$

C
$$0$$

D
$$3$$

Solution

For the function to be continuous at $$x = 1$$
The LHL should be equal to RHL
LHL
$$\lim_{ x \rightarrow {1}^{-}} f(x) $$
$$= \lim_{ x \rightarrow {1}^{-}} 1 + x $$
$$= 2$$
RHL
$$\lim_{x \rightarrow {1}^{+}} f(x) $$
$$=\lim_{ x \rightarrow {1}^{+}} 3 - a{x}^{2} $$
$$= 3-a$$
Since, LHL = RHL
So, $$2 = 3-a$$
Hence, $$a = 1 $$
Question 5
1 / -0

The function $$\displaystyle \mathrm{f}({x})=\frac{1+\sin x-\cos x}{1-\sin x-\cos x}$$ is not defined at $${x}=0$$. The value of $$\mathrm{f}(\mathrm{0})$$ so that $$\mathrm{f}({x})$$ is continuous at $${x}=0$$ is

A
$$1$$

B
$$-1$$

C
$$0$$

D
$$2$$

Solution

Since, on applying the limit, the function is of $$ \dfrac{0}{0} $$ form,
we apply the L-Hospital's Rule and differentiate both the numerator and the denominator individually.
Thus, the question transforms to,
$$\displaystyle\lim_{x \rightarrow 0} \dfrac{cos x +sin x}{sin x - cos x} $$

Now applying the limit, we get
$$\Rightarrow \dfrac{0+1}{0-1} = -1$$
Thus, for the function to be continuous at $$x = 0$$, $$f(0) = -1$$
Question 6
1 / -0

$$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}=$$

A
$${e}$$

B
$$e^{2}$$

C
$$-1$$

D
1

Solution

$$\displaystyle Lt_{x\rightarrow 0^+}(sinx)^{\tan x}$$

$$\log { k } =Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x } } $$

$$k={ e }^{ Lt_{ x\rightarrow 0^+ }\tan { x } \log { \sin { x } } }$$

$$ k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \log { \sin { x } } }{ \cot { x } } }$$

It is of the form $$\displaystyle \frac{\infty}{\infty}$$, so applying L-Hospital's rule

$$k={ e }^{ Lt_{ x\rightarrow 0^+ }\displaystyle \frac { \cot { x } }{ -\csc ^{ 2 }{ x } } }$$

$$k={ e }^{ 0 }=1$$
Question 7
1 / -0

lf the function $$\displaystyle \mathrm{f}({x})=\frac{e^{x^{2}}-\cos {x}}{x^{2}}$$ for $$x \neq 0$$ is continuous at $${x}=0$$ then $$\mathrm{f}(\mathrm{0})=$$

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

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

C
$$2$$

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

Solution

$${ f }({ x })=\dfrac { e^{ x^{ 2 } }-{ c }{ o }{ s }{ x } }{ x^{ 2 } } $$

Given f(x) is continuous at $$x=0$$

$$\displaystyle \lim _{ x\rightarrow 0 } f(x)=f(0)$$

$$\displaystyle=\lim _{ x\rightarrow 0 } \dfrac { e^{ x^{ 2 } }-{ c }{ o }{ s }{ x } }{ x^{ 2 } } $$

It is of the form $$\displaystyle \dfrac{0}{0}$$ , so applying L-Hospital's rule

$$\displaystyle =\lim _{ x\rightarrow 0 } \dfrac { 2xe^{ x^{ 2 } }+\sin { x } }{ 2x } $$

$$\displaystyle \lim_{x\to 0} e^{x^2}+\lim_{x\to 0} \dfrac{\sin x}{2x}$$

$$\displaystyle =1+\dfrac{1}{2}=\dfrac{3}{2}$$
Question 8
1 / -0

$$\displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}=$$

A
$$\mathrm{e}$$

B
$$\mathrm{e}^{4}$$

C
$$\mathrm{e}^{-1}$$

D
$$\mathrm{e}^{-4}$$

Solution

Let, $$\text{L}= \displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}$$
Put $$y=\pi-x\Rightarrow x\to \pi\Leftrightarrow y\to 0$$
$$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1- 4 \tan \mathrm{(\pi-y)} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{(\pi-y)}}$$
$$\quad \quad = \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{-\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{y}}[\because \tan (\pi-y)=-\tan y]$$
Clearly form of the limit is $$1^{\infty}$$
$$\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{\dfrac{1}{4\tan y}\times(-4)}=e^{-4}$$
Question 9
1 / -0

$$\displaystyle \lim_{n\rightarrow \infty }(\pi n)^{2/n}=$$

A
0

B
1

C
2

D
3

Solution

Let $$l = \displaystyle \lim_{n\rightarrow \infty }(\pi n)^{2/n}$$
Take log both sides,
$$\displaystyle \log l = 2\lim_{n\to \infty}\frac{\log(\pi n)}{n}=2\lim_{n\to \infty}\frac{\log(n)+\log(\pi)}{n}$$
Clearly form of the limit is $$\dfrac{\infty}{\infty}$$
Applying L'Hospital's rule,
$$\Rightarrow \displaystyle \log l =2\lim_{n\to \infty}\frac{1/n}{1} =0$$
$$\Rightarrow l =e^0=1$$
Question 10
1 / -0

$$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} \displaystyle\frac { { x }^{ 5 }-32 }{ x-2 } , & x\neq 2 \end{matrix} \\ \begin{matrix} k, & x=2 \end{matrix} \end{cases}$$ is continuous at $$x=2$$, then the value of $$k$$ is

A
$$10$$

B
$$15$$

C
$$35$$

D
$$80$$

Solution

Function $$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} \dfrac { { x }^{ 5 }-32 }{ x-2 } , & x\neq 2 \end{matrix} \\ \begin{matrix} k, & x=2 \end{matrix} \end{cases}$$ is continuous at $$x=2$$
We know that $$\displaystyle \lim _{ x\rightarrow 2 }{ f\left( x \right) } =\lim _{ x\rightarrow 2 }{ \left( \frac { { x }^{ 5 }-32 }{ x-2 } \right) } =\lim _{ x\rightarrow 2 }{ \left( \frac { { x }^{ 5 }-{ 2 }^{ 5 } }{ x-2 } \right) } $$
We also know that $$\displaystyle \lim _{ x\rightarrow a }{ \left( \frac { { x }^{ n }-{ a }^{ n } }{ x-a } \right) } =n{ a }^{ n-1 }$$
Therefore, $$\displaystyle \lim _{ x\rightarrow 2 }{ f\left( x \right) } =5\times { 2 }^{ 4 }=80$$
Since $$f(2)=k$$ and $$f(x)$$ is continuous at $$x=2,$$ therefore $$\displaystyle \lim _{ x\rightarrow 2 }{ f\left( x \right) } =f\left( 2 \right) \Rightarrow k=80.$$

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Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 8

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Limits and Continuity Test 8

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SHARING IS CARING

Question 1

$$g$$ is continuous if $$f$$ is not continuous

$$g$$ is not continuous if $$f$$ is not continuous

$$g$$ is continuous if $$f$$ is continuous

$$g$$ is differentiable if $$f$$ is differentiable

Solution

Question 2

$$1$$

$$-1$$

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

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

Solution

Question 3

$$x<0$$

$$x\geq 1$$

$$0\leq x\leq 1$$

$$x\geq 0$$

Solution

Question 4

$$1$$

$$2$$

$$0$$

$$3$$

Solution

Question 5

$$1$$

$$-1$$

$$0$$

$$2$$

Solution

Question 6

$${e}$$

$$e^{2}$$

$$-1$$

1

Solution

Question 7

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

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

$$2$$

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

Solution

Question 8

$$\mathrm{e}$$

$$\mathrm{e}^{4}$$

$$\mathrm{e}^{-1}$$

$$\mathrm{e}^{-4}$$

Solution

Question 9

0

1

2

3

Solution

Question 10

$$10$$

$$15$$

$$35$$

$$80$$

Solution

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