Relations and Functions Test

Result

Relations and Functions Test - 81

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Weekly Quiz Competition

Question 1
1 / -0

If 27 $$^{*}$$ 3 =243 and 5 $$^{*}$$ 4 = 80 .Then what is the value of 3 $$^{*}$$ 7 = ?

A
84

B
147

C
63

D
23

Solution
Question 2
1 / -0

Consider the function $$f\left( x \right) ={ e }^{ x }$$ and $$g\left( x \right)=\sin ^{ -1 }{ x } $$, then which of the following is/are necessarily true.

A
Domain of $$gof =$$ Domain of $$f$$

B
Range of $$gof\ \subset$$ Range of $$g$$

C
Domain of $$gof$$ is $$\left( -\infty ,0 \right) $$

D
Range of $$gof$$ is $$\left( -\dfrac{\pi}{2} ,0 \right) $$

Solution

$$\text { solution: } \quad f(x)=e^{x} \quad g(x)=\sin ^{-1}(x) \\$$

$$\text { gof }=\sin ^{-1}\left(e^{x}\right) \\$$

$$\therefore-1 \leqslant e^{x} \leqslant 1 \\$$

$$\text { so we need } x \in(-\infty \text { o] } \\$$

$$\text { domain of gof is }(-\infty, 0] \\$$

$$\text { Range of gof is same as range of } g \\$$

$$\text { Ansuer. : option (B) }$$
Question 3
1 / -0

Let N be the set of natural numbers and two functions f and g be defined as
and g(n)=$$n-{ \left( -1 \right) }^{ n }$$ then fog is:

A
one-one but not onto

B
onto but not one-one

C
neither one-one nor into

D
both one-one and onto

Solution
Question 4
1 / -0

Let $$f:R\rightarrow R$$ defined by $$f\left( x \right) =\frac { { e }^{ { x }^{ 2 } }-{ e }^{ -x^{ 2 } } }{ { e }^{ x^{ 2 } }+{ e }^{ { -x }^{ 2 } } } ,$$ then

A
f(x) is one-one but not onto

B
f(x) is neither one-one nor onto

C
f(x) is many one but onto

D
f(x) is one-one and onto

Solution

$$\text { solution: } \quad f(x)=\frac{e^{x^{2}}-e^{-x^{2}}}{e^{x^{2}}+e^{-x^{2}}}=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1} \\$$

$$f^{\prime}(x)=\frac{\left\{\left(e^{x^{2}}\right)^{2}+1\right\}\left\{4 x\left(e^{x^{2}}\right)\right\}-\left\{\left(e^{x^{2}}\right)^{2}-1\right\}\left\{4 x\left(e^{x^{2}}\right)\right\}}{\left(\left(e^{x^{2}}\right)^{2}+1\right)^{2}} \\$$

$$\text { Since } f^{\prime}(x) \text { is nither positive, nor negative } \\$$

$$\text { for all } x \text { so } f(x) \text { is not one - one function } \\$$

$$\qquad f(x)=\frac{\left(e^{x^{2}}\right)^{2}-1}{\left(e^{x^{2}}\right)^{2}+1}$$

Range of $$f(x)$$ is $$[0,1 )$$

since Range of $$f(x)$$ is not equal to co-domain

of $$f(x)$$ so $$f(x)$$ is not onto function.

Answer: option: (B)
Question 5
1 / -0

Which one of the following is one-one?

A
$$f:R\rightarrow R$$ given by $$f(x) =\left| x-1 \right| $$ for all $$x\in R$$

B
$$g:\left[ -\pi /2,\pi /2 \right] \rightarrow given by$$ $$g(x)=\left| sinx \right| $$

C
$$h:\left[ -\pi /2,\pi /2 \right] \in R$$ given by $$h(x) =sin x $$ for all $$x\in \left[ -\pi /2,\pi /2 \right] $$

D
$$\phi :R\rightarrow R$$ given by $$f(x)={ x }^{ 2 }-4$$ for all $$x\in R$$
Question 6
1 / -0

The function $$f:R\rightarrow R$$ defined by $$f\left( x \right) =\frac { { e }^{ \left| x \right| }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } } $$ is

A
One-One and onto

B
One-one but not onto

C
Not one-one but onto

D
Neither one-one nor onto

Solution
Question 7
1 / -0

Let f(x+y)=f f(x) f(y) and f(x) =1+x g(x) G(x), where $$\underset { x\rightarrow 0 }{ lim } g\left( x \right) =a and \underset { x\rightarrow 0 }{ lim } G\left( x \right) =b,$$ then f' (x) is equal to

A
1+ab

B
ab

C
f(x)

D
ab f(x)

Solution
Question 8
1 / -0

Let $$f:R\rightarrow R$$, be defined as $$f(x)={ e }^{ x^{ 2 } }+cosx$$ then f is

A
One-one and onto

B
One-one and into

C
Many-one and onto

D
many-one and into

Solution
Question 9
1 / -0

$$f:N\rightarrow N\quad where\quad f\left( x \right) =x-{ (-1) }^{ x }$$, then 'f' is

A
one-one and into

B
many- one and into

C
one-one and onto

D
many-one and onto

Solution
Question 10
1 / -0

Let (X) be a function satisfying f' (X) = f (X) with f (0) = 1 and g (X) be a function that satisfies f (X) + g (x) = $${ x }^{ 2 },$$ Then the value of the integral $$\int _{ 0 }^{ 1 }{ f } (x)\quad g\quad (x)\quad dx,\quad is$$

A
$$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 5 }{ 2 } $$

B
$$e+\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } $$

C
$$e-\frac { { e }^{ 2 } }{ 2 } -\frac { 3 }{ 2 } $$

D
$$e+\frac { { e }^{ 2 } }{ 2 } +\frac { 5 }{ 2 } $$

Solution