Relations and Functions Test

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Relations and Functions Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle f:\left ( 3,4 \right )\rightarrow \left ( 0,1 \right )$$ is defined by $$\displaystyle f\left ( x \right )=x-\left [ x \right ]$$ where $$\displaystyle [x] $$denotes the greatest integer function then $$ f^{-1}(x)$$ is

A
$$\displaystyle \frac{1}{x-\left [ x \right ]}$$

B
$$\displaystyle [x]-x$$

C
$$\displaystyle x-3$$

D
$$\displaystyle x+3$$

Solution

Note that $$\displaystyle f\left ( x \right )=x-\left [ x \right ] =\left\{ x \right\}$$. Hence every number $$x \in \left( {3,4} \right)$$ is mapped to its fractional part.
Hence, the inverse of the function can be written by adding the integer part of the number i.e., $$3$$ to each of the fraction part.
Hence the function $${f^{ - 1}}\left( x \right) = x + 3$$.
Question 2
1 / -0

Let $$\displaystyle f(x)=\frac{ax}{x+1}$$, where $$\displaystyle x\neq -1$$. Then for what value of $$\displaystyle a$$ is $$\displaystyle f( f(x))=x$$ always true

A
$$\displaystyle \sqrt{2}$$

B
$$\displaystyle -\sqrt{2}$$

C
$$1$$

D
$$-1$$

Solution

$$\displaystyle f\left( {f\left( x \right)} \right) = \dfrac{{a\dfrac{{ax}}{{x + 1}}}}{{\dfrac{{ax}}{{x + 1}} + 1}} = \dfrac{{\dfrac{{{a^2}x}}{{x + 1}}}}{{\dfrac{{ax + x + 1}}{{x + 1}}}} = \dfrac{{{a^2}x}}{{ax + x + 1}}$$

Since, $$\displaystyle f\left( {f\left( x \right)} \right) =x$$, we have,

$$\displaystyle\dfrac{{{a^2}x}}{{ax + x + 1}}=x$$.

Simplifying the equation we get,

$${a^2}x = \left( {a + 1} \right){x^2} + x$$

$$\therefore \left( {a + 1} \right){x^2} + \left( {1 - {a^2}} \right)x = 0$$

or $$\left( {a + 1} \right)x\left( {x + 1 - a} \right) = 0$$

Hence the only possible value is $$a=-1$$
Question 3
1 / -0

Let $$ f:R \rightarrow R$$ and $$g:R \rightarrow R$$ be defined by $$f(x)=x^2+2x-3,g(x)=3x-4$$ then $$(gof) (x)=$$

A
$$3x^2+6x-13$$

B
$$3x^2-6x-13$$

C
$$3x^2+6x+13$$

D
$$-3x^2+6x-13$$

Solution

Given that, $$f(x)=x^2+2x-3$$ and $$g(x)=3x-4$$

$$(gof) (x)=g\left(f(x)\right)$$

$$=3(x^2+2x-3)-4$$

$$(gof) (x)=3x^2+6x-13$$

Hence, option A.
Question 4
1 / -0

If $$f(x)=ax+b$$ and $$g(x)=cx+d$$, then $$f(g(x))=g(f(x))$$ implies

A
$$f(a)=g(c)$$

B
$$f(b)=g(b)$$

C
$$f(d)=g(b)$$

D
$$f(c)=g(a)$$

Solution

$$f(g(x))=a(cx+d)+b$$

$$=acx+ad+b$$...(i)

$$g(f(x))=c(ax+b)+d$$

$$=ac(x)+bc+d$$ ...(ii)

$$f(g(x))=g(f(x))$$

From (i) and (ii)

$$ac(x)+bc+d=acx+ad+b$$

$$cb+d=ad+b$$

$$g(b)=f(d)$$
Question 5
1 / -0

If $$f$$ and $$g$$ are two functions such that $$\displaystyle \left ( fg \right )\left ( x \right )=\left ( gf \right )\left ( x \right )$$ for all $$x$$. Then $$f $$ and $$g$$ may be defined as

A
$$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$$

B
$$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$$

C
$$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$$

D
$$\displaystyle f\left ( x \right )=x^{m}, g\left ( x \right )=x^{n}$$ where $$m, n$$ are unequal integers

Solution

A$$)f(x)=$$$$\sqrt{x},g(x)=cosx then (fg)(x) = \sqrt{\cos x} and (gf)(x) = \cos \sqrt{x}$$.They are not equal.

B)$$f(x)=x^{3},g(x)=x+1 then (fg)(x) = (x+1)^{3} and (gf)(x)= x^{3}+1$$.They are not equal.

C)$$f(x)=x1,g(x)=x^{2}+1 then (fg)(x)=x^{2} and (gf)(x)=x^{2}-2x$$. They are not equal.

D)$$f(x)=x^{m}, g(x)=x^{n} then (fg)(x)=x^{n+m} and (gf)(x)=x^{n+m}$$.Hence $$(fg)(x)= (gf)(x)$$
Question 6
1 / -0

If $$f:R\rightarrow R$$ such that $$f(x)=log_3 x$$ then $$f^{-1} x$$ is equal to

A
$$log x^3$$

B
$$3^x$$

C
$$3^{-x}$$

D
$$3^{1/x}$$

Solution

$$f(x)$$ is invertible for all positive $$x$$.

$$y=log_{3}(x)$$

Taking antilogarithm on both sides, we get

$$3^{y}=x$$.

Hence,

$$f^{-1}(x)=3^{x}$$

Hence, option 'B' is correct.
Question 7
1 / -0

If $$f:\left( 3,6 \right) \rightarrow \left( 1,3 \right) $$ is a function defined by $$\displaystyle f\left( x \right)=x-\left[ \frac { x }{ 3 } \right] $$ $$($$ where $$\left[ . \right] $$ denotes the greatest integer function $$),$$ then $$\displaystyle f^{ -1 }\left( x \right)=$$

A
$$x-1$$

B
$$x+1$$

C
$$x$$

D
none of these

Solution

Given that $$\displaystyle f\left( x \right) =x-\left[ \frac { x }{ 3 } \right] $$

where $$\displaystyle f:\left( 3,6 \right) \rightarrow \left( 1,3 \right) $$

Now, inside the given domain, we will always have $$\displaystyle \left[ \frac { x }{ 3 } \right] =1$$

$$\displaystyle \therefore f\left( x \right) =x-1$$ throughout its domain

$$\displaystyle \Rightarrow y=x-1$$

$$\displaystyle \Rightarrow x=y+1={ f }^{ -1 }\left( y \right) $$

$$\displaystyle \therefore { f }^{ -1 }\left( x \right) =x+1$$
Question 8
1 / -0

Let $$S$$ be a set containing $$n$$ elements. Then the total number of binary operations on $$S$$ is

A
$$n^n$$

B
$${2^n}^{2}$$

C
$$n^{n^2}$$

D
$$n^2$$

Solution

No of binary operations imply
$$S\times S\rightarrow S$$
$$=n(S)^{[n(S\times S)]}$$
$$=n^{n\times n}$$
$$=n^{n^2}$$
Question 9
1 / -0

Let $$f: R\rightarrow R$$ be defined by $$f(x)=3x-4$$ then $$f^{-1}(x)$$ is

A
$$\dfrac{1}{3}(x+4)$$

B
$$\dfrac{1}{3}(x-4)$$

C
$$3x+4$$

D
not defined

Solution

$$f(x)$$ is invertible for all real values of $$x$$
Hence
$$y=3x-4$$
$$y+4=3x$$
$$\dfrac{y+4}{3}=x$$
$$f^{-1}(x)=\dfrac{x+4}{3}$$
Question 10
1 / -0

A function $$y=f\left ( x \right )$$ is invertible only when

A
$$y=f\left ( x \right )$$ is monotonic increasing

B
$$y=f\left ( x \right )$$ is bijective

C
$$y=f\left ( x \right )$$ is monotonic decreasing

D
invertible

Solution

Invertible function is defined as, the function $$f$$ applied to an input $$x$$ gives a result of $$y$$, then applying its inverse function $$g$$ to $$y$$ gives the result $$x$$, and vice versa. i.e., $$f\left ( x \right )=y$$ $$\Rightarrow g\left ( y \right )$$ = $$x$$.
And bijective function has the same definition as that of an Invertible function

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 44

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Relations and Functions Test - 44

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SHARING IS CARING

Question 1

$$\displaystyle \frac{1}{x-\left [ x \right ]}$$

$$\displaystyle [x]-x$$

$$\displaystyle x-3$$

$$\displaystyle x+3$$

Solution

Question 2

$$\displaystyle \sqrt{2}$$

$$\displaystyle -\sqrt{2}$$

$$1$$

$$-1$$

Solution

Question 3

$$3x^2+6x-13$$

$$3x^2-6x-13$$

$$3x^2+6x+13$$

$$-3x^2+6x-13$$

Solution

Question 4

$$f(a)=g(c)$$

$$f(b)=g(b)$$

$$f(d)=g(b)$$

$$f(c)=g(a)$$

Solution

Question 5

$$\displaystyle f\left ( x \right )=\sqrt{x}, g\left ( x \right )=\cos x$$

$$\displaystyle f\left ( x \right )=x^{3}, g\left ( x \right )=x+1$$

$$\displaystyle f\left ( x \right )=x-1, g\left ( x \right )=x^{2}+1$$

$$\displaystyle f\left ( x \right )=x^{m}, g\left ( x \right )=x^{n}$$ where $$m, n$$ are unequal integers

Solution

Question 6

$$log x^3$$

$$3^x$$

$$3^{-x}$$

$$3^{1/x}$$

Solution

Question 7

$$x-1$$

$$x+1$$

$$x$$

none of these

Solution

Question 8

$$n^n$$

$${2^n}^{2}$$

$$n^{n^2}$$

$$n^2$$

Solution

Question 9

$$\dfrac{1}{3}(x+4)$$

$$\dfrac{1}{3}(x-4)$$

$$3x+4$$

not defined

Solution

Question 10

$$y=f\left ( x \right )$$ is monotonic increasing

$$y=f\left ( x \right )$$ is bijective

$$y=f\left ( x \right )$$ is monotonic decreasing

invertible

Solution

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