Functions Test 33

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Functions Test 33

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Weekly Quiz Competition

Question 1
1 / -0

If f(g(x))=5x+2 and g(x)=8x then f(x)=

A
$$ \frac{5}{8}x+2$$.

B
$$ \frac{8}{5}x+2$$.

C
$$ \frac{5}{8}x-2$$.

D
8x-2

E
5x-2

Solution
Question 2
1 / -0

if $$f$$ is a bijective function such that $$f(x)=\dfrac {\lambda x+\mu}{ax+\beta}$$ and if $$f^{-1}(x)=f(x)$$, then

A
$$\lambda=\beta$$ & $$\alpha=\mu=0$$

B
$$\alpha=\mu=1$$

C
$$\alpha+\beta =1$$

D
$$\cos^{-1}\left(\dfrac {\lambda}{\beta}\right)=\dfrac {\pi}{4}$$

Solution

undefined
Question 3
1 / -0

If $$f\left( x \right) = (1 - x)$$ , $$x \in \left[ { - 3,3} \right]$$ , then the domain of $$f\left( {f\left( x \right)} \right)$$ is

A
$$\left[ { - 2,3} \right]$$

B
$$\left( { - 2,3} \right)$$

C
$$\left[ { - 2,3]} \right.$$

D
$$( - 2,3]$$

Solution

$$f\left(x\right)=1-x$$
$$f\left(f\left(x\right)\right)=f\left(1-x\right)=1-1+x=x$$
And $$x\in\left[-3,3\right]$$
Domain of $$f\left(x\right)$$ is $$\left[-3,3\right]$$
Domain of $$f\left(f\left(x\right)\right)=$$Domain of $$f\left(1-x\right)$$
Domain of $$f\left(f\left(x\right)\right)=\left[1-3,3\right]=\left[-2,3\right]$$
Question 4
1 / -0

If $$f(x)=\frac{x}{\sqrt{1-x^{2}}}$$ and g(x) = $$f(x)=\frac{x}{\sqrt{1+x^{2}}}$$ , then (fog)(x) =

A
$$f(x)=\frac{x}{\sqrt{1-x^{2}}}$$

B
$$f(x)=\frac{x}{\sqrt{1+x^{2}}}$$

C
$$x^{2}$$

D
x

Solution

Given:-
$$f{\left( x \right)} = \cfrac{x}{\sqrt{1 - {x}^{2}}}$$
$$g{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
To find:-
$$fog{\left( x \right)} = ?$$
$$fog{\left( x \right)}$$
$$= f{\left( g{\left( x \right)} \right)}$$
$$= f{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - {\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}^{2}}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{1 - \left( \cfrac{{x}^{2}}{1 + {x}^{2}} \right)}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\sqrt{\left( \cfrac{1 + {x}^{2} - {x}^{2}}{1 + {x}^{2}} \right)}}$$
$$= \cfrac{\left( \cfrac{x}{\sqrt{1 + {x}^{2}}} \right)}{\left( \cfrac{1}{1 + {x}^{2}} \right)}$$
$$= \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
$$\Rightarrow fog{\left( x \right)} = \cfrac{x}{\sqrt{1 + {x}^{2}}}$$
Hence the correct answer is $$\cfrac{x}{\sqrt{1 + {x}^{2}}}$$.
Question 5
1 / -0

If $$ f ( x ) = \left( a - x ^ { n } \right) ^ { 1 / n }$$ where $$ a > 0$$ and } $$n$$ is a positive integer then$$( f o f ) ( x )$$ is

A
$$f ( x )$$

B
x

C
0

D
1

Solution

$$f(x)=(a-x^n)^{1/n}$$
$$\therefore f(f(x))=(a-(f(x))^n))^{1/n}$$
$$\therefore f(f(x))=(a-((a-x^n)^{1/n)^n)^{1/n}}$$
$$\therefore f(f(x))=(a-((a-x^n)))^{1/n}$$
$$\therefore f(f(x))=(a-a+x^n)^{1/n}$$
$$\therefore f(f(x))=(x^n)^{1/n}$$
$$\therefore(f(x))=x$$
Question 6
1 / -0

The function $$f ( x ) = \sqrt { \log _ { x ^ { 2 } } x }$$ is defined for $$x$$:

A
$$( - \infty , 0 )$$

B
$$( 0,1 )$$

C
$$( 1 , \infty )$$

D
$$( 0 , \infty )$$

Solution

$$\sqrt{\log_{{x}^{2}}{x}}$$ is defined when $${x}^{2}>0$$ and $${x}^{2}\neq 1$$ and $$x>0$$
$$\Rightarrow\,\left(0,\infty\right)-\left\{1\right\}=\left(1,\infty\right)$$
Question 7
1 / -0

if $$f\left( x \right) = \log \left( {\dfrac{{1 +x}}{{1 - x}}} \right)$$ and $$g\left( x \right) = \dfrac{{3x + {x^3}}}{{1 + 3{x^2}}}$$ then $$\left( {f(g(x)))} \right)$$ is equal to

A
$$ - f\left( x \right)$$

B
$$3f\left( x \right)$$

C
$${\left( {f\left( x \right)} \right)^3}$$

D
$$f\left( {3x} \right)$$

Solution

$$f(x)=\log\bigg(\dfrac{1+x}{1-x}\bigg)\ and\ g(x)=\dfrac{3x+x^{3}}{1+3x^{2}}$$

$$f(g(x))=\log\bigg(\dfrac{1+\dfrac{3x+x^{3}}{1+3x^{2}}}{1-\dfrac{3x+x^{3}}{1+3x^{2}}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}}}{\dfrac{1+3x^{2}-3x-x^{3}}{1+3x^{2}}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{1+3x^{2}+3x+x^{3}}{1+3x^{2}-3x-x^{3}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{(1+x)^3}{(1-x)^{3}}\bigg)$$

$$f(g(x))=\log\bigg(\dfrac{1+x}{1-x}\bigg)^{3}$$

$$f(g(x))=3\log\bigg(\dfrac{1+x}{1-x}\bigg)$$

$$f(g(x))=3f(x)$$
Question 8
1 / -0

If $$f:R \rightarrow R, f(x)=2x-1$$ and $$g; R \rightarrow R, g(x)=x^{2}+2$$, then $$(gof)(x)$$ equals-

A
$$2x^{2}-1$$

B
$$(2x-1)^{2}$$

C
$$2x^{2}+3$$

D
$$4x^{2}-4x+3$$

Solution

$$f(x)=2x-1$$
$$g(x)=x^2+2$$
$$g(f(x))=(2x-1)^2+2=4x^2-4x+3$$.
Question 9
1 / -0

Let $$g(x)=1+x-[x]\quad $$ and $$f(x)=\begin{cases} -1\quad if\quad x<0 \\ 0\quad \quad if\quad x=0 \\ 1\quad \quad if\quad x>0 \end{cases}$$ , then $$\forall \:x,fog(x)$$ equals

A
$$x$$

B
$$1$$

C
$$f(x)$$

D
$$g(x)$$

Solution

given g(x)=1+x=[x] 1+{x} {x} donates the fractional part of x so,
g(x) is always positive
so, input to f(g(x)) is always positive and
hence, fog(x)=1
Question 10
1 / -0

$$f:c \to c$$ is defined as $$f(x) = \dfrac{{ax + b}}{{cx + d}},bd \ne 0$$ then $$f$$ is a constant function when,

A
a=c

B
b=d

C
ad=bc

D
ab=cd

Solution

f($$x$$)=$$\frac{ax+b}{cx+d}$$ is a constant function,
then lets say it equal to same constant m.
$$m(cx+d)=ax+b$$
$$a=mc $$
$$b=md $$
$$\frac{a}{c}$$ =$$\frac{b}{d}=m$$
$$\frac{a}{b}$$ =$$\frac{c}{d}$$
$$ad=bc$$
C is correct.

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Re-Attempt Weekly Quiz Competition

Functions Test 33

Result

Functions Test 33

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SHARING IS CARING

Question 1

$$ \frac{5}{8}x+2$$.

$$ \frac{8}{5}x+2$$.

$$ \frac{5}{8}x-2$$.

8x-2

5x-2

Solution

Question 2

$$\lambda=\beta$$ & $$\alpha=\mu=0$$

$$\alpha=\mu=1$$

$$\alpha+\beta =1$$

$$\cos^{-1}\left(\dfrac {\lambda}{\beta}\right)=\dfrac {\pi}{4}$$

Solution

Question 3

$$\left[ { - 2,3} \right]$$

$$\left( { - 2,3} \right)$$

$$\left[ { - 2,3]} \right.$$

$$( - 2,3]$$

Solution

Question 4

$$f(x)=\frac{x}{\sqrt{1-x^{2}}}$$

$$f(x)=\frac{x}{\sqrt{1+x^{2}}}$$

$$x^{2}$$

x

Solution

Question 5

$$f ( x )$$

x

0

1

Solution

Question 6

$$( - \infty , 0 )$$

$$( 0,1 )$$

$$( 1 , \infty )$$

$$( 0 , \infty )$$

Solution

Question 7

$$ - f\left( x \right)$$

$$3f\left( x \right)$$

$${\left( {f\left( x \right)} \right)^3}$$

$$f\left( {3x} \right)$$

Solution

Question 8

$$2x^{2}-1$$

$$(2x-1)^{2}$$

$$2x^{2}+3$$

$$4x^{2}-4x+3$$

Solution

Question 9

$$x$$

$$1$$

$$f(x)$$

$$g(x)$$

Solution

Question 10

a=c

b=d

ad=bc

ab=cd

Solution

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