Functions Test 9

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Functions Test 9

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Weekly Quiz Competition

Question 1
1 / -0

The domain of $${f}({x})=\sqrt{2-\log_{3}(x-1)}$$ is

A
$$(2,12]$$

B
$$(-\infty,10]$$

C
$$(3,12]$$

D
$$(1,10]$$

Solution

$$f$$ is defined only if term under root is positive.
$$\Rightarrow 2-\log_{3}(x-1)\geq 0$$
$$\log_{3}(x-1)\leq 2$$
$$(x-1)\leq 9$$
$$x\leq 10$$
$$\log (x-1)$$ is defined if $$x-1>0 \Rightarrow x>1$$
$$\therefore$$ Domain of $$f$$ is $$(1,10]$$.
Question 2
1 / -0

The domain of the function defined by $$ \mathrm{f}({x})=^{(7-x)}P_{(x-3)}$$ is

A
$$\{3,7\}$$

B
$$\{3,4,5,6,7\}$$

C
$$\{3,4,5\}$$

D
$$\{1,2,3,4\}$$

Solution

$$f$$ is defined only if $$7-x\geq x-3$$ and $$7-x\geq 0$$.
$$\Rightarrow x\leq 5$$ and $$x\geq 3$$
$$\therefore$$ $$f$$ is defined at $$x=\left \{ 3,4,5 \right \}$$
Question 3
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 4
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 5
1 / -0

The domain of $$f(x) =\left ( {\sin }^{ -1 }x+{ \text{cosec} }^{ -1 }x\right )$$ is

A
$$[-1,1]$$

B
$$(-1, 1)$$

C
$$\{-1, 1\}$$

D
$$(-\infty, \infty)$$
Question 6
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 7

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 8
1 / -0

The domain of $$\mathrm{f}(\mathrm{x})=\log_{\mathrm{x}}(9-\mathrm{x}^{2})$$ is

A
$$(-3,3)$$

B
$$(0,\infty)$$

C
$$(0,1) \cup (1,\infty)$$

D
$$(0,1) \cup (1,3)$$

Solution

For $$f$$ to be defined $$x>0$$ and $$x\neq 1$$
$$9-x^{2}\geq 0 \Rightarrow x\in (-3,3)$$
$$\therefore x\in (0,1)\cup (1,3)$$
Question 9
1 / -0

The domain of $$\displaystyle f(x)=\sqrt{\log\left(\frac{1}{|\sin x|}\right)}$$ is

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

B
$$R-\displaystyle \{\frac{n\pi}{2}:n\in Z\}$$

C
$$\mathrm{R}-\{\mathrm{n}\pi :\mathrm{n}\in \mathrm{Z}\}$$

D
$$ R-\displaystyle \{n\pi+(-1)2\frac{\pi}{2}:n\in Z\}$$

Solution

$$\displaystyle \frac{1}{|\sin x|}\geq 1 $$
$$\therefore \log$$ is always defined except at which denominator is zero,i.e, when $$\sin x=0 \Rightarrow x=n\pi$$
$$\therefore$$ $$f$$ is defined an $$R-n\pi \ \ \forall n$$.
Question 10
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
$$\Rightarrow$$$$g(x)=\sqrt{x}$$

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

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Re-Attempt Weekly Quiz Competition

Functions Test 9

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Functions Test 9

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SHARING IS CARING

Question 1

$$(2,12]$$

$$(-\infty,10]$$

$$(3,12]$$

$$(1,10]$$

Solution

Question 2

$$\{3,7\}$$

$$\{3,4,5,6,7\}$$

$$\{3,4,5\}$$

$$\{1,2,3,4\}$$

Solution

Question 3

$$-1$$

$$0$$

$$1$$

$$2$$

Solution

Question 4

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 5

$$[-1,1]$$

$$(-1, 1)$$

$$\{-1, 1\}$$

$$(-\infty, \infty)$$

Question 6

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

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

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

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

Solution

Question 7

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 8

$$(-3,3)$$

$$(0,\infty)$$

$$(0,1) \cup (1,\infty)$$

$$(0,1) \cup (1,3)$$

Solution

Question 9

$$(-\infty,\infty)$$

$$R-\displaystyle \{\frac{n\pi}{2}:n\in Z\}$$

$$\mathrm{R}-\{\mathrm{n}\pi :\mathrm{n}\in \mathrm{Z}\}$$

$$ R-\displaystyle \{n\pi+(-1)2\frac{\pi}{2}:n\in Z\}$$

Solution

Question 10

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

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

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

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

Solution

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