Relations and Functions Test

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Relations and Functions Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

From $$ ] \dfrac{- \pi}{2} , \dfrac{- \pi}{2}[ $$ which of the following is one - one onto function defined in R

A
$$ f(x) = \tan x $$

B
$$ f(x) = \sin x $$

C
$$ f(x) = \cos x $$

D
$$ f(x) = e^{x} + r^{-x}$$
Question 2
1 / -0

Which of the following in one -one function defined from R to R

A
$$ f(x) = |x| $$

B
$$ f(x) = \cos x $$

C
$$ f(x) =e^{x}$$

D
$$ f(x) = x^{2} $$
Question 3
1 / -0

Subtraction is an operation on $$Z$$, which is

A
commutative and associative

B
associative but not commutative

C
neither commutative and nor associative

D
commutative but not associative

Solution

Let $$a, b \in z$$
Commutativity: As we know,
$$a-b \neq b-a$$
Associativity : $$(a-b)-c\neq a-(b-c)$$
So, option (c) is correct.
Question 4
1 / -0

Which of the following functions is not injective ?

A
$$f:R \rightarrow R, f(x)=2x+7$$

B
$$f:[0,\pi]\rightarrow[-1,1],f(x)=\cos x$$

C
$$f:\left [ -\dfrac{\pi}{2},\dfrac{\pi}{2} \right ]\rightarrow R, f(x)=2 \sin x +3$$

D
$$f:R\rightarrow [-1,1],f(x)=\sin x$$

Solution

A:
$$f:R\rightarrow R, f(x)=2x+7$$
$$\dfrac{dy}{dx}=2>0 \rightarrow$$ one-one (Injective)

B:
$$f:[0,\pi]\rightarrow [-1,1],f(x)=\cos x$$
$$\dfrac{dy}{dx}=\sin x=(+ve) \rightarrow$$ one-one (Injective)

C:
$$f: \left[-\dfrac{\pi}2, \dfrac{\pi}2\right], f(x) = 2\sin x+3$$
$$\dfrac{dy}{dx}= 2\cos x=+ve \rightarrow$$ one-one (Injective)

D:
$$f:R\rightarrow [-1,1],f(x)=\sin x$$
$$\dfrac{dy}{dx}=\cos x=+ve$$ & $$-ve \rightarrow$$ many one

Hence, option D.
Question 5
1 / -0

Let $$\displaystyle \mathrm{f}:\mathrm{R}\rightarrow \left[0,\frac{\pi}{2}\right)$$ be defined by $$\mathrm{f}(\mathrm{x})=\mathrm{t}\mathrm{a}\mathrm{n}^{-1}(\mathrm{x}^{2}+\mathrm{x}+\mathrm{a})$$. Then the set of values of '$$\mathrm{a}$$' for which $$\mathrm{f}$$ is onto is

A
$$[0,\infty)$$

B
$$[2, 1]$$

C
$$ \displaystyle \left\{ \frac{1}{4}\right\}$$

D
$$ \displaystyle \left[ \frac{1}{4}, \infty\right)$$

Solution

$$f(x)=\tan^{-1}(x^2+x+a)$$
For, $$f(x)$$ to be onto, codomain should be exactly equal to range.
That is, range of funnction $$ \displaystyle f(x) = \left[ 0,\frac { \pi }{ 2 } \right) $$
So, $$0\leqslant \tan^{-1}(x^2+x+a)<\dfrac{\pi}{2}$$
Now, $$ \displaystyle { x }^{ 2 }+x+a={ \left( x+\dfrac { 1 }{ 2 } \right) }^{ 2 }+a-\dfrac { 1 }{ 4 } $$
The above expression will take all real values from $$ \left[0, \infty \right)$$, only if $$ \displaystyle a = \dfrac{1}{4}$$
Hence, only for $$ a= \dfrac{1}{4}$$, the function is onto.
Question 6
1 / -0

Let $$\displaystyle {f}({x})=\frac{{a}{x}+{b}}{{c}{x}+{d}}$$, then $$fof(x)={x}$$, provided

A
$$d = -a$$

B
$$d = a$$

C
$$a = b = c = d = 1$$

D
$$a = b = 1$$

Solution

Given,

$$f(x)=\dfrac{ax+b}{cx+d}$$

$$f(f(x))=\dfrac{a\left ( \dfrac{ax+b}{cx+d}\right )+b}{c\left ( \dfrac{ax+b}{cx+d} \right )+d}=x$$

$$=\dfrac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=x$$

$$a(ax+b)+b(cx+d)=x[c(ax+b)+d(cx+d)]$$

$$(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x$$

matching the coefficients of $$x$$, we get,

$$a^2+bc=bc+d^2$$

$$a^2=d^2$$

$$d=\pm a$$

matching the coefficients of $$x^2$$, we get,

$$ac+cd=0$$

$$\therefore d=-a$$

matching constant terms, we get,

$$ab+bd=0$$

$$d=-a$$
Question 7
1 / -0

Let $$\displaystyle \mathrm{f}(\mathrm{x})=\frac{\alpha \mathrm{x}}{\mathrm{x}+1}$$ , $$\mathrm{x}\neq 1$$. then for what value of $$\alpha$$ is $$\mathrm{f}(\mathrm{f}(\mathrm{x}))=\mathrm{x}$$

A
$$\sqrt{2}$$

B
$$-\sqrt{2}$$

C
1

D
-1

Solution

$$f\left( {\frac{{\alpha x}}{{x + 1}}} \right) = \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} \right)}}{{\frac{{\alpha x}}{{x + 1}} + 1}}$$

$$ = \frac{{{\alpha ^2}x}}{{\alpha x + x + 1}}$$

As, it is given that $$f\left( x \right) = x$$, so,

$$\frac{{{\alpha ^2}x}}{{\alpha x + x + 1}} = x$$

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

$${\alpha ^2} = \alpha x + x + 1$$

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

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

$$\alpha \left[ {\alpha - \left( {x + 1} \right)} \right] + 1\left[ {\alpha - \left( {x + 1} \right)} \right] = 0$$

$$\left( {\alpha + 1} \right)\left[ {\alpha - \left( {x + 1} \right)} \right] = 0$$

$$\alpha + 1 = 0\;,\;\alpha - \left( {x + 1} \right) = 0$$

$$\alpha = - 1\;,\;\alpha = x + 1$$

Therefore, the value of $$\alpha $$ is $$ - 1$$.
Question 8
1 / -0

If $$n \geq 2$$ then the number of surjections that can be defined from $$\{1, 2, 3, ....... n\}$$ onto $$\{1, 2\}$$ is

A
$$2n$$

B
$$^nP_2$$

C
$$2^n$$

D
$$2^{n}-2$$

Solution

Let $$A=\{1,2,3,...,n\} $$ and $$B=\{1,2\} $$ .
Therefore, total number of mappings from $$A$$ to $$B$$ is $$2^n$$ of which two functions $$f(x)=1$$ for all $$x\in A$$ and $$g(x)=2$$ for all $$x\in A$$ are not surjective.
Thus, the total number of surjections from $$A$$ to $$B$$ is $$2^n-2$$.
Question 9
1 / -0

If $$f\left( x \right)=\sqrt { 3\left| x \right| -x-2 } $$ and $$g(x)=\sin(x)$$, then the domain of the definition of $$f\circ g\left( x \right) $$ is

A
$$\displaystyle \left\{ 2n\pi +\frac { \pi }{ 2 } \right\} $$

B
$$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) $$

C
$$\displaystyle \left\{ 2n\pi +\frac { 7\pi }{ 6 } \right\} $$

D
$$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) \bigcup _{ n,m\in I }^{ }{ \left( 2n\pi +\frac { \pi }{ 2 } \right) } $$

Solution

We have $$\displaystyle f\left( x \right) =\sqrt { 3\left| x \right| -x-2 } $$ and $$g\left( x \right) =\sin { x } $$

$$\therefore f\circ g\left( x \right) =\sqrt { 3\left| \sin { x } \right| -\sin { x } -2 } $$ which is defined,
if $$3\left| \sin { x } \right| -\sin { x } -2\ge 0$$

If $$\sin { x } >0,$$ then $$2\sin { x } -2\ge 0\quad $$
$$\Rightarrow \sin { x } \ge 1$$
$$\displaystyle \Rightarrow \sin { x } =1\Rightarrow x=\frac { \pi }{ 2 } $$

If $$\sin { x } <0,$$ then $$\displaystyle -4\sin { x } -2\ge 0\Rightarrow -1\le -\frac { 1 }{ 2 } $$

$$\displaystyle \therefore x\in \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) \bigcup _{ n,m\in I }^{ }{ \left( 2n\pi +\frac { \pi }{ 2 } \right) } n,m\in I$$
Question 10
1 / -0

If $$f:R\rightarrow \left [\dfrac {\pi}{6}, \dfrac {\pi}{2}\right ), f(x)=\sin^{-1}\left (\dfrac {x^2-a}{x^2+1}\right )$$ is a onto function, then set of values of $$a$$ is

A
$$\left \{-\dfrac {1}{2}\right \}$$

B
$$\left [-\dfrac {1}{2}, -1\right )$$

C
$$(-1, \infty)$$

D
none of these

Solution

Given, $$f(x)$$ is onto.
$$\therefore \displaystyle \frac {\pi}{6}\leq \sin^{-1}\left (\frac {-x^2-a}{x^2+1}\right ) < \frac {\pi}{2}$$
$$\Rightarrow \displaystyle \frac {1}{2}\leq \frac {x^2-a}{x^2+1} < 1$$
$$\Rightarrow \displaystyle \frac {1}{2}\leq 1-\frac {(a+1)}{x^2+1} < 1, \forall x\epsilon R $$
$$\Rightarrow a+1 > 0$$
$$\Rightarrow a\in (-1, \infty)$$
Hence, (c) is the correct answer.

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Relations and Functions Test - 71

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Question 1

$$ f(x) = \tan x $$

$$ f(x) = \sin x $$

$$ f(x) = \cos x $$

$$ f(x) = e^{x} + r^{-x}$$

Question 2

$$ f(x) = |x| $$

$$ f(x) = \cos x $$

$$ f(x) =e^{x}$$

$$ f(x) = x^{2} $$

Question 3

commutative and associative

associative but not commutative

neither commutative and nor associative

commutative but not associative

Solution

Question 4

$$f:R \rightarrow R, f(x)=2x+7$$

$$f:[0,\pi]\rightarrow[-1,1],f(x)=\cos x$$

$$f:\left [ -\dfrac{\pi}{2},\dfrac{\pi}{2} \right ]\rightarrow R, f(x)=2 \sin x +3$$

$$f:R\rightarrow [-1,1],f(x)=\sin x$$

Solution

Question 5

$$[0,\infty)$$

$$[2, 1]$$

$$ \displaystyle \left\{ \frac{1}{4}\right\}$$

$$ \displaystyle \left[ \frac{1}{4}, \infty\right)$$

Solution

Question 6

$$d = -a$$

$$d = a$$

$$a = b = c = d = 1$$

$$a = b = 1$$

Solution

Question 7

$$\sqrt{2}$$

$$-\sqrt{2}$$

1

-1

Solution

Question 8

$$2n$$

$$^nP_2$$

$$2^n$$

$$2^{n}-2$$

Solution

Question 9

$$\displaystyle \left\{ 2n\pi +\frac { \pi }{ 2 } \right\} $$

$$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) $$

$$\displaystyle \left\{ 2n\pi +\frac { 7\pi }{ 6 } \right\} $$

$$\displaystyle \left( 2n\pi +\frac { 7\pi }{ 6 } ,2n\pi +\frac { 11\pi }{ 6 } \right) \bigcup _{ n,m\in I }^{ }{ \left( 2n\pi +\frac { \pi }{ 2 } \right) } $$

Solution

Question 10

$$\left \{-\dfrac {1}{2}\right \}$$

$$\left [-\dfrac {1}{2}, -1\right )$$

$$(-1, \infty)$$

none of these

Solution

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