Limits and Continuity Test 6

Result

Limits and Continuity Test 6

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$$=

A
$$2$$

B
$$1$$

C
$$0$$

D
$$3$$

Solution

$$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$$
$$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{1-\tan x}{(\frac{\pi}{4}-x)(1+\tan x)}$$.................(dividing the num and deno by cosx)
$$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\tan\left(\frac{\pi}{4}-x\right) }{(\frac{\pi}{4}-x)} = \lim_{h\to 0}\frac{\tan h}{h} = 1$$
Question 2
1 / -0

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$$=

A
$$\displaystyle \frac{1}{3}$$

B
$$\displaystyle \frac{1}{4}$$

C
$$\displaystyle \frac{1}{6}$$

D
$$\displaystyle \frac{1}{8}$$

Solution

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$$
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\sin (\dfrac { \pi }{ 2 } -h) }{ (\pi -2(\dfrac { \pi }{ 2 } -h))^{ 2 } } $$......(x-->h-pi/2)
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\cos { h } }{ 4h^{ 2 } }$$

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

$$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=$$

A
$$1$$

B
$$-1$$

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

D
$$0$$

Solution

$$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}$$

$$=\displaystyle \lim_{x\rightarrow \infty }\frac{\displaystyle 1+\frac{\sin x}{x}}{\displaystyle 1+ \frac{\cos x}{x}}$$

$$=\displaystyle \frac{\displaystyle 1+\frac{\text{Any finite number between -1 and 1}}{\infty}}{\displaystyle 1+ \frac{\text{Any finite number between -1 and 1}}{\infty}}=\frac{1+0}{1+0}=1$$
Question 4
1 / -0

$$ \displaystyle f(x)=\left\{ \begin{matrix} \dfrac { 3 }{ { x }^{ 2 } } \sin { 2{ x }^{ 2 } } ,\; x<0 \\ \dfrac { { x }^{ 2 }+2x+C }{ 1-3{ x }^{ 2 } } ,\; x\ge 0,\; x\neq \frac { 1 }{ \sqrt { 3 } } \\ 0,\; x=\frac { 1 }{ \sqrt { 3 } } \end{matrix} \right. $$
$$\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=$$

A
$$\frac{C}{4}$$

B
2C

C
$$\frac{C}{3}$$

D
C

Solution
Question 5
1 / -0

Let $$f(x)=\begin{cases}x^{2}-1,0 < x < 2\\2x+3,2 \leq x < 3\end{cases},$$
The quadratic equation whose roots are $$\displaystyle \lim_{x\rightarrow 2^{-}}f(x)$$ and $$\displaystyle \lim_{x\rightarrow 2^{+}}f(x)$$ is

A
$$x^{2}-10x+21=0$$

B
$$x^{2}-6x+9=0$$

C
$$x^{2}-14x+49=0$$

D
$$x^{2}+6x+9=0$$

Solution

For $$0 < x < 2$$,
$$f(x) = {x}^{2} - 1 $$
$$\lim_{ x \rightarrow {2}^{-}} f(x) = 3 $$
For, $$2 < x < 3$$
$$f(x) = 2x + 3$$
$$\lim_{ x \rightarrow {2}^{+}} f(x) = 7 $$
Thus the equation with roots, 3 and 7 is
$$ {x}^{2} - 10x + 21 = 0 $$
Question 6
1 / -0

The value of $$\sqrt{e}$$ upto four decimal places is?

A
$$1.6484$$

B
$$1.5237$$

C
$$1.2589$$

D
$$1.9190$$

Solution

$$(e)^{\frac {1}{2}}=1+\dfrac {1}{2}+\dfrac {(\dfrac {1}{2})^2}{2!}+\dfrac {(\dfrac {1}{2})^3}{3!}+\dfrac {(\dfrac {1}{2})^4}{4!}$$
$$=1+0\cdot 5+0\cdot 125+0\cdot 02083+0\cdot 00260$$
$$=1\cdot 6484$$
Question 7
1 / -0

If $$f(x) = \displaystyle \frac {x^2+6x}{\sin x}$$ , then $$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=$$

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

$$\lim_{x\rightarrow 0 }^{ - }f\left( x \right) =\dfrac { { x }^{ 2 }+6x }{ \sin x } \\= \lim_{ h\rightarrow 0 } f\left( h \right) $$
$$=\dfrac { { h }^{ 2 }+6(0-h) }{ \sin(0-h) }= \lim_{h\rightarrow 0 }f\left( h \right) =\dfrac { { h }-6 }{ -\sin(h)/h } = \dfrac { -6 }{ -1 } =6$$
Hence, the answer is "C".
Question 8
1 / -0

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}$$=

A
$$a+b$$

B
$$a-b$$

C
$$-(a+b)$$

D
$$ab$$

Solution

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}$$

$$=\displaystyle a.\lim_{x\to 0}\frac{\log(1+ax)}{ax}-b.\lim_{x\to 0}\frac{\log(1+bx)}{bx}=a-b$$
Question 9
1 / -0

Let $$f$$ be a continuous function on [1,3]. lf $$f$$ takes only rational values for all $$x$$ and $$f(2)=10$$ then $$f(1.5)$$ is equal to

A
$$8$$

B
$$\displaystyle \frac{f(1)+f(3)}{3}$$

C
$$20$$

D
$$10$$

Solution

Since $$f$$ is continuous, so it must take all real values between $$f\left(1\right)$$ and $$f\left(3\right)$$.
But since $$f$$ takes only rational values so $$f$$ must be a constant function.
Hence $$f\left(1.5\right)=f\left(2\right)=10$$
Question 10
1 / -0

$$\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { 3\sin(2{ x }^{ 2 }) }{ { x }^{ 2 } } } =A$$
then the value of $$A$$ is

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

$$\displaystyle \lim_{x\rightarrow { 0^- }} \dfrac { 3\sin({ 2x }^{ 2 }) }{ { x }^{ 2 } } =\lim_{x\rightarrow { 0^- }}\dfrac { 2*3\sin({ 2x }^{ 2 }) }{ { 2x }^{ 2 } } = 6$$

Hence, answer is option $$C$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Limits and Continuity Test 6

Result

Limits and Continuity Test 6

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2$$

$$1$$

$$0$$

$$3$$

Solution

Question 2

$$\displaystyle \frac{1}{3}$$

$$\displaystyle \frac{1}{4}$$

$$\displaystyle \frac{1}{6}$$

$$\displaystyle \frac{1}{8}$$

Solution

Question 3

$$1$$

$$-1$$

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

$$0$$

Solution

Question 4

$$\frac{C}{4}$$

2C

$$\frac{C}{3}$$

C

Solution

Question 5

$$x^{2}-10x+21=0$$

$$x^{2}-6x+9=0$$

$$x^{2}-14x+49=0$$

$$x^{2}+6x+9=0$$

Solution

Question 6

$$1.6484$$

$$1.5237$$

$$1.2589$$

$$1.9190$$

Solution

Question 7

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 8

$$a+b$$

$$a-b$$

$$-(a+b)$$

$$ab$$

Solution

Question 9

$$8$$

$$\displaystyle \frac{f(1)+f(3)}{3}$$

$$20$$

$$10$$

Solution

Question 10

$$2$$

$$4$$

$$6$$

$$8$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.