Relations and Functions Test

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Relations and Functions Test - 40

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Question 1
1 / -0

Assertion(A): If $$X=\left \{ x:-1\leq x\leq 1 \right \}$$ and $$f:X\rightarrow X$$ defined by $$f(x)=\sin \pi x; \forall x\in A$$ is not invertible function

Reason (R): For a function $$f$$ to have inverse, it should be a bijection

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true but R is not correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

$$f(-\pi )=f(0)=f(\pi )=0$$
$$\therefore f$$ is not a bijection
$$\therefore \sin\pi x$$ is not invertible.
Question 2
1 / -0

Let $$S$$ be set of all rational numbers. The functions $$f:R\rightarrow R,\ g:R\rightarrow R$$ are defined as
$$f(x)=\begin{cases}
0, & x \in S \\
1, & x \notin S
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\in S \\
0 & x\notin S
\end{cases}$$
then, $$(fog) (\pi)+(gof)(e)=$$

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$2$$

Solution

$$f(x)=\begin{cases}
0, & x \text{ is rational} \\
1, & x \text{ is irrational}
\end{cases}$$
$$g(x)=\begin{cases}
-1 & x\text { is rational} \\
0& x\text { is irrational }
\end{cases}$$
$$fog(\pi )=f(g(\pi ))=f(0)=0$$
so $$gof(e)=g(f(e))=g(1)=-1$$
$$\therefore fog(\pi )+gof(e)=-1$$
Question 3
1 / -0

If $$n\geq 1$$ is any integer, $$\mathrm{d}(n)$$ denotes the number of positive factors of $$n$$, then for any prime number $$\mathrm{p},\ \mathrm{d}(\mathrm{d}(\mathrm{d}(\mathrm{p}^{7})))=$$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

For a number being $$p^{n}$$ the total number of positive factors excluding $$1$$ and the number itself will be $$n-1$$.
Therefore total number of positive factors will be $$n-1+2$$
$$=n+1$$
Hence for $$p^{7}$$ we will have $$8$$ factors.
Now for $$8=2^{3}$$.
Hence $$8$$ will have $$4$$ factors.
Now $$4=2^{2}$$, hence the number of factors will be $$3$$.
Question 4
1 / -0

Let $$\displaystyle f\left( x \right)={ x }^{ 2 }-x+1,x\ge \left( \frac { 1 }{ 2 } \right) $$ then the solution of the equation $$f(x)=f^{-1}(x)$$ is

A
$$x=1$$

B
$$x=2$$

C
$$\displaystyle x=\frac{1}{2}$$

D
None of these

Solution

$$\displaystyle y={ x }^{ 2 }-x+1\Rightarrow { x }^{ 2 }-x+\left( 1-y \right) =0$$
Here, $$a=1,b=-1$$ and $$c=1-y$$

$$\displaystyle x=\frac { 1\pm \sqrt { 1-4\left( 1-y \right) } }{ 2 } \quad \quad \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] $$

$$\displaystyle \because x>\frac { 1 }{ 2 } ,\quad \therefore x=\frac { 1 }{ 2 } +\sqrt { y-\frac { 3 }{ 4 } } $$

$$\displaystyle \Rightarrow f^{ -1 }\left( x \right) =\frac { 1 }{ 2 } +\sqrt { x-\frac { 3 }{ 4 } } $$

Now, $$\displaystyle { x }^{ 2 }-x+1=\frac { 1 }{ 2 } +\sqrt { x-\frac { 3 }{ 4 } } $$
Since the graphs of the original and inverse functions can intersect only on the straight line $$y=x,$$ therefore
$$x=f\left( x \right) \quad \quad \Rightarrow x={ x }^{ 2 }-x+1$$

$$\Rightarrow { x }^{ 2 }-2x+1=0$$

$$\Rightarrow { \left( x-1 \right) }^{ 2 }=0$$
$$\Rightarrow x=1$$
Question 5
1 / -0

lf $$f:[-6,6]\rightarrow \mathbb{R}$$ is defined by $$f(x)=x^{2}-3$$ for $$x\in \mathbb{R}$$ then
$$(fofof)(-1)+(fofof)(0)+(fofof)(1)=$$

A
$$f(4\sqrt{2})$$

B
$$f(3\sqrt{2})$$

C
$$f(2\sqrt{2})$$

D
$$f(\sqrt{2})$$

Solution

$$fofof(-1)=fof(-2)=f(1)=-2$$
$$fofof(1)=fof(-2)=f(1)=-2$$
$$fofof(0)=fof(-3)=f(0)=33$$
$$\therefore fofof(-1)+fofof(1)+fofof(0)=29$$
$$=32-3$$
$$=(4\sqrt{2})^{2}-3$$
$$\Rightarrow f(4\sqrt{2})=(4\sqrt{2})^{2}-3$$

Question 6

1 / -0

lf $$f$$ : $$R\rightarrow R$$ is defined by
$$f(x)=\left\{\begin{array}{l}x+4 & x<-4\\3x+2 & -4\leq x<4\\x-4 & x\geq 4\end{array}\right.$$
then the correct matching of list I to List II is.

List - I	List - II
$$\mathrm{A}) f(-5)+f(-4)=$$	$$\mathrm{i}) 14$$
$$\mathrm{B}) f(\|f(-8)\|)=$$	ii $$) 4$$
$$\mathrm{C}) f(f(-7)+f(3))=$$	$$\mathrm{i}\mathrm{i}\mathrm{i})-11$$
$$\mathrm{D}) f(f(f(f(0)))+1=$$	$$\mathrm{i}\mathrm{v})-1$$
	v) $$1$$
	vi) $$0$$

A-iii , B-vi , C-ii , D- v

A-iii , B-iv , C-ii , D- vi

A-iv , B-iii , C-ii , D- i

A-ii , B-vi , C-v , D- ii

Solution

$$f(-5)=-5+4=-1 ; f(-4)=3(-4)+2=-10$$
$$\therefore f(-5)+f(4)=-11$$
$$f(|f(-8)|)=f((-8+4))=f(4)=4-4=0$$
$$f(f(-7)) tf(3) = f(-7+4)+3(3)+2$$
$$=3(-3)+2+3(3)+2$$
$$=4$$
$$f(f(f(f(0))))+1=f(f(f(2)))+1=f(f(8))+1$$
$$=f(4)+1=0+1=1$$

Question 7
1 / -0

If $$f:R\rightarrow R$$ is defined by $$f(x)=x^{2}-10x+21 $$ then $$ f^{-1}(-3)$$ is

A
$$\left \{ -4,6 \right \}$$

B
$$\left \{ 4,6 \right \}$$

C
$$\left \{ -4, 4, 6 \right \}$$

D
Not Invertible

Solution

Let $$f^{-1}(-3)=t$$
$$\Rightarrow f(t)=-3$$
$$t^{2}-10t+21=-3$$
$$t^{2}-10t+24=0$$
$$t^{2}-6t-4t+24=0$$
$$t(t-6)-4(t-6)=0$$
$$\Rightarrow t=6,4 = f^{-1}(-3)$$
Question 8
1 / -0

lf $$g(f(x)) =|\sin \mathrm{x}|,f(g(x)) =(\sin\sqrt{\mathrm{x}})^{2}$$, then

A
$${f}({x})=\sin^{2} {x},{g}({x})=\sqrt{{x}}$$

B
$${f}({x})=\sin x,g({x})=|{x}|$$

C
$${f}({x})={x}^{2},{g}({x})=\sin\sqrt{{x}}$$

D
$${f}, {g}$$ cannot be determined

Solution

$$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$$ .......(i)
$$f(g(x))=\sin^{2}\sqrt{x}$$ .......(ii)

Comparing $$i$$ and $$ii$$

$$\Rightarrow$$$$g(f(x))=|\sin x|=\sqrt{\sin^{2}x}$$

$$\Rightarrow$$$$g(\sin^{2}x)=\sqrt{\sin^{2}x}=g(f(x))$$

And
$$\Rightarrow$$$$f(g(x))=\sin^{2}\sqrt{x}$$

$$\Rightarrow$$$$f(\sqrt{x})=\sin^{2}\sqrt{x}=f(g(x))$$

Therefore
$$\Rightarrow$$$$g(x)=\sqrt{x}$$

$$\Rightarrow$$$$f(x)=\sin^{2}x$$
Question 9
1 / -0

lf $$ f(x)=x-x^{2}+x^{3}-x^{4}+\ldots..\infty$$ when $$|x|<1$$, then the ascending order of the following is
a) $$f(1/2)$$
b) $$f^{-1}(1/2)$$
c) $$ f(-1/2)$$
d) $$f^{-1}(-1/2)$$

A
a, b, c, d

B
c, d, a, b

C
b, a, d, c

D
d, c, a, b

Solution

$$f(x)=x-x^2+x^3-x^4$$
$$f(x)=\dfrac{x}{1+x}(x)< 1$$
$$f\left ( \dfrac{1}{2} \right )=\dfrac{1}{3}$$
$$f\left ( \dfrac{-1}{2} \right )=-1$$
$$f(x)=\dfrac{x}{1+x}$$
$$y=\dfrac{x}{1+x}$$
$$y+xy=x$$
$$\dfrac{y}{(u-1)}=x$$
$$f^{-1}x=\dfrac{-x}{x-1}$$
$$f^{-1}\left ( \dfrac{1}{2} \right )=1$$
$$f^{-1}\left ( \dfrac{-1}{2} \right )=\dfrac{-1}{3}$$
Question 10
1 / -0

Let $$f$$ be an injective function with domain $$\{x, y, z\}$$and range $$\{1,2,3\}$$ such that exactly one of the follwowing statements is correct and the remaining are false :
$${f}({x})=1,{f}({y})\neq 1$$,
$${f}({z})\neq 2$$,
then the value of $${f}^{-1}(1)$$ is

A
$$x$$

B
$$y$$

C
$$z$$

D
none

Solution

It gives three cases
Case (1) When $$f(x)=1$$ is true.
In this case remaining two are false
$$f(y)=1$$ and $$f(z)=2$$
This means $$x$$ and $$y$$ have the same image, so $$f(x)$$ is not an injective, which is a contradiction.

Case (2) when $$f\left( y \right)\neq 1$$ is true
If $$f\left( y \right)\neq 1$$is true then the remaining statements are false.
$$\therefore f\left( x\right)\neq 1$$ and $$f(z)=2$$
i.e., both $$x$$ and $$y$$ are not mapped to $$1.$$
So, either both associate to $$2$$ or $$3,$$
Thus, it is not injective.

Case (3) When $$f(z) \neq 2$$ is true
If $$f(z) \neq2$$ is true then remaining statements are fales
$$\therefore$$ If $$f(x) \neq1$$ and $$f(y)=1$$
But $$f$$ is injective
Thus, we have $$f(x)=2, f(y)=1$$ and $$f(z)=3$$
Hence, $$\displaystyle f^{ -1 }\left( 1 \right)=y$$

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Relations and Functions Test - 40

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Relations and Functions Test - 40

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SHARING IS CARING

Question 1

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not correct explanation of A

A is true but R is false

A is false but R is true

Solution

Question 2

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 3

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

$$x=1$$

$$x=2$$

$$\displaystyle x=\frac{1}{2}$$

None of these

Solution

Question 5

$$f(4\sqrt{2})$$

$$f(3\sqrt{2})$$

$$f(2\sqrt{2})$$

$$f(\sqrt{2})$$

Solution

Question 6

A-iii , B-vi , C-ii , D- v

A-iii , B-iv , C-ii , D- vi

A-iv , B-iii , C-ii , D- i

A-ii , B-vi , C-v , D- ii

Solution

Question 7

$$\left \{ -4,6 \right \}$$

$$\left \{ 4,6 \right \}$$

$$\left \{ -4, 4, 6 \right \}$$

Not Invertible

Solution

Question 8

$${f}({x})=\sin^{2} {x},{g}({x})=\sqrt{{x}}$$

$${f}({x})=\sin x,g({x})=|{x}|$$

$${f}({x})={x}^{2},{g}({x})=\sin\sqrt{{x}}$$

$${f}, {g}$$ cannot be determined

Solution

Question 9

a, b, c, d

c, d, a, b

b, a, d, c

d, c, a, b

Solution

Question 10

$$x$$

$$y$$

$$z$$

none

Solution

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