Functions Test 31

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Functions Test 31

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Weekly Quiz Competition

Question 1
1 / -0

If $$f : A \rightarrow B $$ defined as $$f(x) = x^2+2x+\frac{1}{1+(x+1)^2}$$ is onto function, then set B is equal to

A
$$[0, \infty)$$

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

C
$$[2, \infty)$$

D
$$(-\infty, \infty)$$

Solution

$$f:A\rightarrow B,\quad f(x)={ x }^{ 2 }+2x+\cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } \\ \because \cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } >0\\ \therefore f(x)={ x }^{ 2 }+2x+\cfrac { 1 }{ 1+{ (x+1) }^{ 2 } } \\ a=1,\quad b=2\\ D=4-4\times 1\times (\cfrac { 1 }{ +ve } )\quad \ge 0\\ \because x\epsilon R$$
Question 2
1 / -0

The domain of the function $$f(x)=\sin \left (\dfrac{1}{x} \right )$$ is

A
$$x\in R$$

B
$$x\in R-\{0\}$$

C
$$x\in (-1,1)$$

D
$$x\in [-1,1]$$

Solution

$$f\left(x\right)=\sin\dfrac{1}{x}$$
Domain of $$\sin(x)$$ is $$x\in R \Rightarrow \dfrac{1}{x}\in R$$

$$\because \ at \ x=0$$

$$\left (\dfrac {1} {x} \right) becomes \ \infty$$

$$\therefore \ domain \ of \ f\left(x\right)=x\in R-\left\{0\right\}$$

$$\therefore$$ The option is $$B$$
Question 3
1 / -0

If $$g(f(x) ) = |\sin x |$$ and $$f(g(x))=(\sin\sqrt x)^2$$ , then

A
$$f(x) = \sin^2x. g(x) =\sqrt x$$

B
$$f(x) = \sin x , g(x) =|x| $$

C
$$f(x) = x^2, g(x) = \sin \sqrt x$$

D
f and g can not be determined

Solution

$$g(f(x))=|sinx|=\sqrt{(sinx)^2}\\f(g(x))=(sin\sqrt x )^2=sin^2\sqrt x\\\therefore f(x)=sin^2x \>and \>g(x)=\sqrt x$$
Question 4
1 / -0

The domain of the function $$f(x) = \dfrac{arc \, cot \, X}{\sqrt{X^2 - [X^2]}}$$, where [X] denotes the greatest integer not greater than x, is :

A
$$R$$

B
$$R - \{0\}$$

C
$$R - \{\pm \sqrt{n} : n \in 1^+ \cup \{0\} \}$$

D
$$R - \{n : n \in 1\}$$

Solution

$$f(x)=\cfrac { { cot }^{ -1 }X }{ \sqrt { { x }^{ 2 }-\left[ { x }^{ 2 } \right] } } $$
The real value of $$x$$ is always greater than or equal to greater integer function. But as integer function $$ {x}^{2}$$ will get equal to $$\left[ { x }^{ 2 } \right].$$
Hence $$X\in R\{n:n\in 1\}.$$
Question 5
1 / -0

The set onto which the derivative of the function $$f(x)=x(\log x-1)$$ maps the range $$[1,\infty )$$ is

A
$$\left[1,\infty \right)$$

B
$$\left( e,\infty \right)$$

C
$$\left[e,\infty \right)$$

D
$$\left( 0,0 \right)$$

Solution

Given $$f (x) = x [\log x-1] x \in [+, \infty]$$
$$f (x) = x \left[ \dfrac{d}{dx} (\log x-1) \right]+ (\log x-1) \dfrac{d}{dx} (x)$$
$$=x. \dfrac{1}{x}+ (\log x-1).1$$
$$=1 -1 \log x-1$$
$$f(x)= \log x$$
$$\therefore $$ The set of $$f(x)= \log x$$ that maps
$$[1, \infty )$$ is $$[e, \infty )$$
Question 6
1 / -0

Let $$E=\{1, 2, 3, 4\}$$ and $$F=\{1, 2\}$$ then the number of onto functions from E to F is

A
$$14$$

B
$$16$$

C
$$12$$

D
$$8$$

Solution

Number of onto function from $$E$$ to $$F.$$

$$E=\{1,2,3,4\}$$ $$F=\{1,2\} $$

Number of function from $$E$$ to $$F=2\times 2\times 2\times 2=16$$

We have to exclude functions where $$f(x)=1$$ & $$ f(x)=2$$.

$$\therefore$$ Total number of onto function $$=16-2=14$$.
Question 7
1 / -0

Let $$f\left( x \right) ={ x }^{ 2 },g\left( x \right) ={ 2 }^{ x }$$, then solution set of $$fog\left( x \right) =gof\left( x \right) $$ is

A
R

B
$$\left\{ 0 \right\} $$

C
$$\left\{ 0,2 \right\} $$

D
None of these

Solution
Question 8
1 / -0

If $$D$$ is the domain of the function $${\sec ^{ - 1}}\left( {\log x} \right)$$, then $$D$$ contains

A
$$\left( { - \infty ,\frac{1}{e}} \right)$$

B
$$e-1$$

C
$$1$$

D
None of these

Solution

$$f(x)=\sec^{-1}(\log x)$$
$$x > 0; x\neq 1$$
$$-\infty < \log x \le -1\cup 1\le \log x \le \infty$$
$$x \in \left (0, \dfrac {1}{e}\right) \cup [e, \infty]$$
$$D$$ is correct
Question 9
1 / -0

Let $$f(x+\dfrac{1}{x})=x^2+\dfrac{1}{x^2}(x\neq 0)$$, then $$f(x)=$$

A
$$x^2$$

B
$$x^2-1$$

C
$$x^2-2$$

D
N.O.T

Solution

We are given
$$f(x+ \dfrac{1}{2})= x^{2}+ \dfrac{1}{x^{2}}$$ (where $$x \neq 0$$)

$$= x^{2}+ \dfrac{1}{x^{2}}+2-2$$.

$$=x^{2}+ \dfrac{1}{x^{2}}+ 2 (x^{2}) \left( \dfrac{1}{x^{2}} \right)-2$$.

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

So, simply put $$x+ \dfrac{1}{x} \rightarrow x$$

we get $$f(x) = x^{2}-2$$
Question 10
1 / -0

If f(x)=sin log $$\left( {\sqrt {4 - {x^2}} /\left( {1 - x} \right)} \right)$$ then the domain and range f are (respectively)

A
$$[-1,1],(-1,1)$$

B
$$(-2,1),(-1,1)$$

C
$$(1,2),[-1,1]$$

D
$$(-2,1),[-1,1]$$

Solution

consider the given function ,

$$f\left( x \right)=\sin {{\log }_{e}}\left( \dfrac{\sqrt{4-{{x}^{2}}}}{1-x} \right)$$

Now ,
for $$\log_e\dfrac{\sqrt{4x^2}}{1-x}$$ to be defined $$ \dfrac{\sqrt{4-{{x}^{2}}}}{1-x}>0 $$
$$ \dfrac{\sqrt{4-{{x}^{2}}}}{1-x}>0 $$

$$ \sqrt{4-{{x}^{2}}}>0 $$, $$x\not=1$$ and $$1-x>0$$

$$ 4-{{x}^{2}}>0 $$ $$1>x$$

$$ 4>{{x}^{2}} $$ $$x<1$$

$$ x^2-4<0$$

$$ (x+2)(x-2)<0 $$
$$-2<x<2$$

combining the two we get
$$-2<x<1$$
$$x\epsilon(-2,1)$$
Hence , the domain of the function is $$(-2,1)$$

Hence, this is the range of the function $$[-1, 1]$$

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Re-Attempt Weekly Quiz Competition

Functions Test 31

Result

Functions Test 31

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SHARING IS CARING

Question 1

$$[0, \infty)$$

$$(-\infty, 0)$$

$$[2, \infty)$$

$$(-\infty, \infty)$$

Solution

Question 2

$$x\in R$$

$$x\in R-\{0\}$$

$$x\in (-1,1)$$

$$x\in [-1,1]$$

Solution

Question 3

$$f(x) = \sin^2x. g(x) =\sqrt x$$

$$f(x) = \sin x , g(x) =|x| $$

$$f(x) = x^2, g(x) = \sin \sqrt x$$

f and g can not be determined

Solution

Question 4

$$R$$

$$R - \{0\}$$

$$R - \{\pm \sqrt{n} : n \in 1^+ \cup \{0\} \}$$

$$R - \{n : n \in 1\}$$

Solution

Question 5

$$\left[1,\infty \right)$$

$$\left( e,\infty \right)$$

$$\left[e,\infty \right)$$

$$\left( 0,0 \right)$$

Solution

Question 6

$$14$$

$$16$$

$$12$$

$$8$$

Solution

Question 7

R

$$\left\{ 0 \right\} $$

$$\left\{ 0,2 \right\} $$

None of these

Solution

Question 8

$$\left( { - \infty ,\frac{1}{e}} \right)$$

$$e-1$$

$$1$$

None of these

Solution

Question 9

$$x^2$$

$$x^2-1$$

$$x^2-2$$

N.O.T

Solution

Question 10

$$[-1,1],(-1,1)$$

$$(-2,1),(-1,1)$$

$$(1,2),[-1,1]$$

$$(-2,1),[-1,1]$$

Solution

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