Functions Test 11

Result

Functions Test 11

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

Find the domain of $$x$$ if $$f(x)=\sqrt {x^2-|x|-2}$$

A
$$x\in R-(-2, 2)$$

B
$$x\in R$$

C
$$x\in R-(0, 2)$$

D
None of these

Solution

Case I $$x<0$$
Hence $$f(x)=\sqrt{x^2-(-x)-2}$$
$$=\sqrt{x^2+x-2}$$
$$=\sqrt{(x+2)(x-1)}$$
Therefore the domain will be $$(-\infty,-2]$$....$$(D_{1})$$
Case II $$x>0$$
Hence $$f(x)=\sqrt{x^2-(x)-2}$$
$$=\sqrt{x^2-x-2}$$
$$=\sqrt{(x-2)(x+1)}$$
Therefore the domain will be $$[2,\infty)$$....$$(D_{2})$$
Hence the domain of the function $$f(x)$$ from $$D_{1}$$ and $$ D_{2}$$ will be
$$=(-\infty,-2]\cup[2,\infty)$$
Question 2
1 / -0

Find the domain of $$x^2+2$$

A
R

B
N

C
Z

D
All of the above

Solution

The function is a polynomial function. Hence its domain includes all real numbers which includes the set of all integers, Z and the set of all natural numbers, N. So correct option is D.
Question 3
1 / -0

The domain for, $$f(x)=\cos^{-1}[x]$$ is

A
$$x\in [-1, 2)$$

B
$$x\in [-1, 1]$$

C
$$x\in (-1, 2)$$

D
$$x\in (-1, 1)$$

Solution

As, $$y=\cos^{-1}$$ $$x$$ exists when $$-1 \leq x \leq 1$$.
$$\therefore f(x)=\cos^{-1} [x]$$ exists, when $$-1\leq [x]\leq 1$$
$$\Rightarrow -1\leq x < 2$$ or $$x\in [-1, 2)$$
Question 4
1 / -0

The domain of the function $$\displaystyle f(x)=[\sin x] \cos\left ( \frac {\pi}{[x-1]} \right )$$ is

A
$$\in R-[1, 2)$$

B
$$\in R-(1, 2)$$

C
$$\in R-[1]$$

D
$$\in R-(1, 2]$$

Solution

Given $$\displaystyle f(x)=[\sin x] \cos\left ( \frac {\pi}{[x-1]} \right )$$

$$[\sin x]$$ is always defined

$$\cos \left ( \dfrac {\pi}{[x-1]} \right )$$ is also defined except when $$[x-1]=0$$

$$\Rightarrow 0\leq x-1 < 1$$
$$\Rightarrow 1 \leq x < 2$$
Hence, domain $$\in R-[1, 2)$$.
Question 5
1 / -0

If $$f(x) = \dfrac{2x+5}{x^{2} + x + 5}$$, then $$f\left [ f(- 1 ) \right ]$$ is equal to

A
$$\dfrac{149}{155}$$

B
$$\dfrac{155}{147}$$

C
$$\dfrac{155}{149}$$

D
$$\dfrac{147}{155}$$

Solution

Given expression is $$ f(x)=\dfrac { 2x+5 }{ { x }^{ 2 }+x+5 } $$
$$ \therefore f(-1)=\dfrac { 2\times (-1)+5 }{ (-1)^{ 2 }+(-1)+5 } =\dfrac { 3 }{ 5 } $$
$$ \therefore f\left( f\left( -1 \right) \right) =\dfrac { 2\times \dfrac { 3 }{ 5 } +5 }{ \left( \dfrac { 3 }{ 5 } \right) ^{ 2 }+\dfrac { 3 }{ 5 } +5 } =\dfrac { 155 }{ 149 }$$
Question 6
1 / -0

Which one of the following relation is a function

A

B

C

D
All of these

Solution

As in option C every element in the domain has a unique image.
Question 7
1 / -0

If $$f$$ and $$g$$ are one-one functions from $$R\to R$$, then

A
$$f+g$$ is one-one

B
$$fg$$ is one-one

C
$$fog$$ is one-one

D
none of these

Solution

Let $${x_1},{x_2} \in R$$ be two distinct elements, then $$g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$$, as $$g$$ is one-one function. Similarly $$f\left( {g\left( {{x_1}} \right)} \right) \ne f\left( {g\left( {{x_2}} \right)} \right)$$ as $$f$$ is also one-one function.
Hence, $$f\circ g$$ is one-one function.
Note that $$f+g$$ and $$f\cdot g$$ may not one-one functions even if $$f$$ and $$g$$ are one one. For example consider $$f\left( x \right) = x$$ and $$g\left( x \right) =- x$$, then $$f+g$$ and $$f\cdot g$$ are not one-one.
Question 8
1 / -0

Which of the following functions are one-one?

A
$$f:R\rightarrow R$$ given by $$ f(x)={ 2x }^{ 2 }+1$$ for all $$\quad x\in R$$

B
$$g:Z\rightarrow Z$$ given by $$ g(x)={ x }^{ 4 }$$ for all $$\quad x\in R$$

C
$$h:R\rightarrow R$$ given by $$ h(x)={ x }^{ 3 }+4$$ for all $$\quad x\in R$$

D
$$\phi :C\rightarrow C$$ given $$ \phi (z)={ 2z }^{ 6 }+4$$ for all $$\quad x\in R$$

Solution

For all even powered function,
$$f(x_{1})=f(x_{2})$$ where $$x_{1}=\pm(x_{2})$$
Hence, the only possible one-one function is $$f(x)=x^3+4$$
$$f(x)$$ is a strictly increasing function. So, it is one-one.
Question 9
1 / -0

The domain of the function $$\displaystyle f(x)=\sqrt{x^{2}-[x]^{2}},$$ where $$[x]=$$ the greatest integer less than or equal to $$x$$, is

A
$$R$$

B
$$[0,+\infty)$$

C
$$(-\infty,0]$$

D
none of these

Solution

If $$x \ge 0$$, then $$\left[x\right]$$ is an integer part of $$x$$.

Hence $${x^2} \ge {\left[ x \right]^2}$$ and $$\displaystyle f\left( x \right) = \sqrt {{x^2} - {{\left[ x \right]}^2}}$$ is well-defined.

If $$x < 0$$ say $$x=-a$$ where $$a$$ is a positive number,
then $$\displaystyle \left[ x \right] =- \left( {\left[ a \right] + 1} \right)$$.

Hence, $${x^2} < {\left[ x \right]^2}$$ and the function $$\displaystyle f\left( x \right) = \sqrt {{x^2} - {{\left[ x \right]}^2}}$$ is not defined.

Therefore the domain of the function is $$\left[ {0,\infty } \right)$$.
Question 10
1 / -0

If $$f:R\rightarrow R$$ given by $$f(x)={ x }^{ 3 }+({ a+2)x }^{ 2 }+3ax+5$$ is one-one, then $$a$$ belongs to the interval

A
$$(-\infty ,1)$$

B
$$(1 ,\infty)$$

C
$$(1 ,4)$$

D
$$(4 ,\infty)$$

Solution

If $$ f(x) $$ is one one then $$ f'\left( x \right) $$ should be either positive or negative for all $$ x $$.
$$f'\left( x \right) = 3{ x }^{ 2 }+2(a+2)x+3a$$
$$\Rightarrow D = 4{ (a+2) }^{ 2 }-4\times3\times3a<0$$
$$\Rightarrow$$ $${ a }^{ 2 }-5a+4<0$$
$$\Rightarrow$$ $$(a-1)(a-4)<0$$
Hence, $$a\in \left( 1,4 \right) $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Functions Test 11

Result

Functions Test 11

Score

-

Rank

-

SHARING IS CARING

Question 1

$$x\in R-(-2, 2)$$

$$x\in R$$

$$x\in R-(0, 2)$$

None of these

Solution

Question 2

R

N

Z

All of the above

Solution

Question 3

$$x\in [-1, 2)$$

$$x\in [-1, 1]$$

$$x\in (-1, 2)$$

$$x\in (-1, 1)$$

Solution

Question 4

$$\in R-[1, 2)$$

$$\in R-(1, 2)$$

$$\in R-[1]$$

$$\in R-(1, 2]$$

Solution

Question 5

$$\dfrac{149}{155}$$

$$\dfrac{155}{147}$$

$$\dfrac{155}{149}$$

$$\dfrac{147}{155}$$

Solution

Question 6

All of these

Solution

Question 7

$$f+g$$ is one-one

$$fg$$ is one-one

$$fog$$ is one-one

none of these

Solution

Question 8

$$f:R\rightarrow R$$ given by $$ f(x)={ 2x }^{ 2 }+1$$ for all $$\quad x\in R$$

$$g:Z\rightarrow Z$$ given by $$ g(x)={ x }^{ 4 }$$ for all $$\quad x\in R$$

$$h:R\rightarrow R$$ given by $$ h(x)={ x }^{ 3 }+4$$ for all $$\quad x\in R$$

$$\phi :C\rightarrow C$$ given $$ \phi (z)={ 2z }^{ 6 }+4$$ for all $$\quad x\in R$$

Solution

Question 9

$$R$$

$$[0,+\infty)$$

$$(-\infty,0]$$

none of these

Solution

Question 10

$$(-\infty ,1)$$

$$(1 ,\infty)$$

$$(1 ,4)$$

$$(4 ,\infty)$$

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.