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

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Functions Test 33

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Question 1

58x+2 \frac{5}{8}x+285​x+2.

85x+2 \frac{8}{5}x+258​x+2.

58x−2 \frac{5}{8}x-285​x−2.

8x-2

5x-2

Solution

Question 2

λ=β\lambda=\betaλ=β & α=μ=0\alpha=\mu=0α=μ=0

α=μ=1\alpha=\mu=1α=μ=1

α+β=1\alpha+\beta =1α+β=1

cos⁡−1(λβ)=π4\cos^{-1}\left(\dfrac {\lambda}{\beta}\right)=\dfrac {\pi}{4}cos−1(βλ​)=4π​

Solution

Question 3

[−2,3]\left[ { - 2,3} \right][−2,3]

(−2,3)\left( { - 2,3} \right)(−2,3)

[−2,3]\left[ { - 2,3]} \right.[−2,3]

(−2,3]( - 2,3](−2,3]

Solution

Question 4

f(x)=x1−x2f(x)=\frac{x}{\sqrt{1-x^{2}}}f(x)=1−x2​x​

f(x)=x1+x2f(x)=\frac{x}{\sqrt{1+x^{2}}}f(x)=1+x2​x​

x2x^{2}x2

x

Solution

Question 5

f(x)f ( x )f(x)

x

0

1

Solution

Question 6

(−∞,0)( - \infty , 0 )(−∞,0)

(0,1)( 0,1 )(0,1)

(1,∞)( 1 , \infty )(1,∞)

(0,∞)( 0 , \infty )(0,∞)

Solution

Question 7

−f(x) - f\left( x \right)−f(x)

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

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

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

Solution

Question 8

2x2−12x^{2}-12x2−1

(2x−1)2(2x-1)^{2}(2x−1)2

2x2+32x^{2}+32x2+3

4x2−4x+34x^{2}-4x+34x2−4x+3

Solution

Question 9

xxx

111

f(x)f(x)f(x)

g(x)g(x)g(x)

Solution

Question 10

a=c

b=d

ad=bc

ab=cd

Solution

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

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

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

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

$\alpha=\mu=1$

$\alpha+\beta =1$

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

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

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

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

$( - 2,3]$

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

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

$x^{2}$

$f ( x )$

$( - \infty , 0 )$

$( 0,1 )$

$( 1 , \infty )$

$( 0 , \infty )$

$- f\left( x \right)$

$3f\left( x \right)$

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

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

$2x^{2}-1$

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

$2x^{2}+3$

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

$x$

$1$

$f(x)$

$g(x)$