Functions Test 3

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Functions Test 3

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Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

Given a function $$f : A \rightarrow B$$; where $$A = \left \{1, 2, 3, 4, 5\right \}$$ and $$B = \left \{6, 7, 8\right \}$$.

...view full instructions

Find number of all such functions $$y = f(x)$$ which are one-one?

A
$$0$$

B
$$3^{5}$$

C
$$^{5}P_{3}$$

D
$$5^{3}$$

Solution

One-one functions are those in which each element in the domain has a unique image in the range of the function.

Here $$B$$ is the co-domain of the function which has lesser number of elements than the domain itself. This shows that it is not possible for the function to be one-one.
Question 2
1 / -0

The number of real linear functions $$f(x)$$ satisfying $$f\left\{ f(x) \right\} =x+f(x)$$

A
$$0$$

B
$$4$$

C
$$5$$

D
$$2$$

Solution

Let $$f(x) = ax+b $$
$$f(f(x))=a(ax+b)+b = a^2x+ab+b$$

Given, $$f(f(x))=x+f(x)$$
$$\implies a^2x+ab+b = x+ ax+b$$
$$\implies (a^2-a-1)x+ab=0$$
Since above equation is valid for all values of x, we have
$$a^2-a-1 =0$$ and $$ ab=0$$

$$a^2-a-1 =0$$
$$a=\dfrac{ 1\pm \sqrt{1+4}}{2}$$
$$\implies a=\dfrac{ 1\pm \sqrt{5}}{2}$$

$$ab=0$$
Since $$a$$ is non-zero,
$$\implies b=0$$

$$f(x) =\dfrac{ 1+ \sqrt{5}}{2}x$$ and $$f(x) =\dfrac{ 1- \sqrt{5}}{2}x$$ satisfy the condition.
Hence the answer is option (D)
Question 3
1 / -0

Let $$f:N\times N\rightarrow N-\left\{ 1 \right\} $$ be defined as $$f(m,n)=m+n$$, then function $$f$$ is ______.

A
One-One and onto

B
Many- One and not onto

C
One- One and not onto

D
Many- One and onto

Solution

Given $$f(m,n)=m+n$$
Range of this function is set of all natural numbers except $$1$$
[since least value of $$m$$ and $$n$$ is $$1$$]
Therefore the range and co-domain are equal
Therefore it is onto
For every $$(m,n)$$, there exists only one $$m+n$$
Therefore it is one-one also
Hence, $$f$$ is one-one and onto.
Question 4
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
6

D
4

Solution

The total number of functions are $$2^{4}=16$$. From the total number of functions, subtract those functions which do not have 2 as the range =1. Subtract those functions which do not have 1 as the range.
So, the answer is $$16-2=14$$
Question 5
1 / -0

$$f:R \to R,f\left( x \right) = \dfrac {{x^2} + 2x + c}{{x^2} + 4x + c}$$ is onto only if

A
$$c < 2$$

B
$$c \ge 0$$

C
$$c < 0$$

D
$$|c| \le 1$$

Solution

$$f:R\rightarrow R,f\left( x \right) =\cfrac { { x }^{ 2 }+2x+c }{ { x }^{ 2 }+4x+c } \\ \Rightarrow y=\cfrac { { x }^{ 2 }+2x+c }{ { x }^{ 2 }+4x+c } \\ \Rightarrow { x }^{ 2 }y+4xy+cy={ x }^{ 2 }+2x+c\\ \Rightarrow { x }^{ 2 }(y-1)+2x(2y-1)+(cy-c)=0\\ \because x\epsilon R\quad \quad \therefore 4{ (2y-1) }^{ 2 }-4c{ (y-1) }^{ 2 }\ge 0\\ { (2y-1) }^{ 2 }-c{ (y-1) }^{ 2 }\ge 0\\ 4{ y }^{ 2 }+1-4y-c{ y }^{ 2 }+2cy-1\ge 0\\ { y }^{ 2 }(4-c)-y(4-2c)\ge 0\\ \because y\epsilon R\quad \quad { (4-2c) }^{ 2 }-4(4-c)\times 0<0\\ { (4-2c) }^{ 2 }<0\\ { (2c-4) }^{ 2 }<0\\ 2c-4<0\\ c<2$$
Question 6
1 / -0

Suppose that $$g\left( x \right) =1+\sqrt { x }$$ and $$f\left( g\left( x \right) \right) =3+2\sqrt { x } +x$$, then $$f\left( x \right)$$ is

A
$$1+2{ x }^{ 2 }$$

B
$$2+{ x }^{ 2 }$$

C
$$1+x$$

D
$$2+x$$

Solution

$$g(x)=1+\sqrt{x}$$
$$(g(x))^{2}=1+2\sqrt{x}+x$$
$$2+(g(x))^{2}=3+2\sqrt{x}+x$$
$$\therefore f(x)=2+x^{2}$$
Question 7
1 / -0

Domain of function as $$f\left( x \right) = \dfrac{{^3\sqrt {{x^2} - 5x + 6} }}{{{x^2} - x - 6}}$$ is

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

B
$$R - \left\{ { - 3,2} \right\}$$

C
$$R - \left\{ { - 2,3} \right\}$$

D
$$R$$

Solution

$$f(x)=(\frac{\sqrt[3]{x^2-5x+6}}{x^2-x-6})$$

$$Cube\>root\>expression\>is\>fine\>for\>any\>value\>of\>the\>term\>inside\>the\>root.$$

$$But\>the\>denomiantor\>must\>not\>be\>equal\>to\>zero.$$

$$x^2-x-6\neq0$$

$$(x-3)(x+2)\neq0$$

$$x\neq3\>or\>x\neq-2$$
$$\therefore$$ domain: R- {-2,3}
Question 8
1 / -0

Read the following information and answer the three items that follow :
Let $$f(x) = x^2 + 2x - 5 $$ and $$g(x) = 5x + 30$$
Consider the following statements:
1. $$f[g(x)]$$ is a polynomial of degree 3.
2. $$g[g(x)]$$ is a polynomial of degree 2.
Which of the above statements is/are correct ?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Neither 1 nor 2

Given,

$$f(x)=x^2+2x-5$$

$$g(x)=5x+30$$

(i)

$$f[g(x)]$$

$$=f[5x+30]$$

$$=(5x+30)^2+2(5x+30)-5$$

upon solving the above equation, we get,

$$f[g(x)]=25x^2+310x+955$$

degree $$=2$$

(ii)

$$g[g(x)]$$

$$=g[5x+30]$$

$$=5(5x+30)+30$$

$$=25x+150+30$$

$$=25x+180$$

$$g[g(x)]=25x+180$$

degree $$=1$$
Question 9
1 / -0

Let $$f(x)=\cfrac { 1 }{ 1-x } $$. Then $$\left\{ f\circ \left( f\circ f \right) \right\} (x)$$

A
$$x$$ for all $$x\in R$$

B
$$x$$ for all $$x\in R-\left\{ 1 \right\} $$

C
$$x$$ for all $$x\in R-\left\{ 0,1 \right\} $$

D
none of these

Solution

$$f(x)=\dfrac{1}{1-x}$$ for $$x\in R-\{1\}$$

$$\{fof\}(x)=\dfrac{1}{1-\dfrac{1}{1-x}}=\dfrac{1}{\dfrac{1-x-1}{1-x}}=\dfrac{x-1}{x}=1-\dfrac{1}{x}$$ for $$x\in R - \{0,1\}$$

$$\{fofof\}(x)=\dfrac{1}{1-fof(x)}=\dfrac{1}{1-\dfrac{x-1}{x}}=\dfrac{1}{\dfrac{x-x+1}{x}}=x$$ for $$x\in R-\{0,1\}$$
Question 10
1 / -0

Read the following information and answer the three items that follow :
Let $$f(x) = x^2 + 2x - 5 $$ and $$g(x) = 5x + 30$$
What are the roots of the equation $$g[f(x)] = 0$$ ?

A
$$1, -1$$

B
$$-1, -1$$

C
$$1, 1$$

D
$$0, 1$$

Solution

Given,

$$f(x)=x^2+2x-5$$

$$g(x)=5x+30$$

$$g[f(x)]=0$$

$$g[x^2+2x-5]=0$$

$$5(x^2+2x-5)+30=0$$

$$5x^2+10x-25+30=0$$

$$5x^2+10x+5=0$$

$$x^2+2x+1=0$$

$$(x+1)^2=0$$

$$\therefore x=-1,-1$$

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Re-Attempt Weekly Quiz Competition

Functions Test 3

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Functions Test 3

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SHARING IS CARING

Question 1

$$0$$

$$3^{5}$$

$$^{5}P_{3}$$

$$5^{3}$$

Solution

Question 2

$$0$$

$$4$$

$$5$$

$$2$$

Solution

Question 3

One-One and onto

Many- One and not onto

One- One and not onto

Many- One and onto

Solution

Question 4

14

16

6

4

Solution

Question 5

$$c < 2$$

$$c \ge 0$$

$$c < 0$$

$$|c| \le 1$$

Solution

Question 6

$$1+2{ x }^{ 2 }$$

$$2+{ x }^{ 2 }$$

$$1+x$$

$$2+x$$

Solution

Question 7

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

$$R - \left\{ { - 3,2} \right\}$$

$$R - \left\{ { - 2,3} \right\}$$

$$R$$

Solution

Question 8

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 9

$$x$$ for all $$x\in R$$

$$x$$ for all $$x\in R-\left\{ 1 \right\} $$

$$x$$ for all $$x\in R-\left\{ 0,1 \right\} $$

none of these

Solution

Question 10

$$1, -1$$

$$-1, -1$$

$$1, 1$$

$$0, 1$$

Solution

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