Functions Test 48

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Functions Test 48

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Weekly Quiz Competition

Question 1
1 / -0

Find the maximum value of $$g(f(x))$$ if:
$$f(x) = x + 4$$ and
$$g(x) = 6 - x^{2}$$

A
$$-6$$

B
$$-4$$

C
$$2$$

D
$$4$$

E
$$6$$

Solution

Given, $$f(x)=x+4, g(x)=6-x^2$$
$$\therefore g(f(x)) = g(x+4) = 6-{(x+4)}^{2}$$
The minimum value of $${(x+4)}^{2}$$ is $$0$$ which occurs at $$x=-4$$.
So, the maximum value of $$g(f(x))$$ is $$6-0 = 6$$.
Question 2
1 / -0

$$f(x)\, =\, \sqrt{\log_{x}(\cos2 \pi x)}$$ Domain of the function is :

A
$$(0, \dfrac14) \cup (\dfrac34, 1) \cup (\dfrac54,\dfrac74) \cup \{x : x \in N, x \leq 2\}$$

B
$$(0, \dfrac14) \cup (\dfrac34, 1) \cup \{x : x \in N, x \leq 2\}$$

C
$$(0, \dfrac14) \cup (\dfrac14, 1) \cup \{x : x \in N, x \leq 2\}$$

D
$$(0, \dfrac14) \cup (\dfrac34, \dfrac32) \cup \{x : x \in N, x \leq 2\}$$

Solution

$$f\left( x \right)=\sqrt { \log _{ x }{ \left( \cos { 2\pi x } \right) } } $$
i) $$log$$ is defined when its base $$x>0$$ and $$x\neq 1$$

ii) and $$\displaystyle\cos { 2\pi x } >0 \Rightarrow 2\pi x>(2n+1)\frac { \pi }{ 2 } $$$$\displaystyle\Rightarrow x>\frac { \left( 2n+1 \right) }{ 4 } $$

iii) Square roots is defined when $$\log _{ x }{ \left( \cos { 2\pi x } \right) } \ge 0$$
$$\Rightarrow\cos { 2\pi x } \ge1 \Rightarrow 2\pi x\ge2n\pi \Rightarrow x\ge n$$

Therefore from (i),(ii) and (iii) we get

$$(0, \dfrac14) \cup (\dfrac34, 1) \cup \{x : x \in N, x \leq 2\}$$
Question 3
1 / -0

If f(x)=x+5 and g(x)=$$\displaystyle \sqrt{x^{2}-9}$$ then find the domain of gof(x)

A
(-8,-2)

B
$$\displaystyle \left ( -\infty ,-8 \right )\cup \left ( -2,\infty \right )$$

C
$$\displaystyle (-\infty ,-8]\cup [-2,\infty )$$

D
$$\displaystyle (-(-\infty ,-8]\cup [-2,\infty )$$

Solution

$$ gof(x) = g(f(x)) = \sqrt {{(x+5)}^{2} -9} = \sqrt {x^2 + 25 + 10x -9} $$
           $$ = \sqrt {x^2 + 16 + 10x} = \sqrt {(x+8)(x+2)} $$

Now, for the square root to have a real value, both $$ (x+8)\ and\ (x+2) $$ should be positive or equal to zero or both should be negative or equal to zero.

Case 1: When both
$$ x+8 \ge 0 $$   and $$ x + 2 \ge 0 $$
$$ => x \ge -8 $$ and $$ x\ge -2 $$

$$ => x \ge -2 $$ is the solution for this case

Case 2: When both
$$ x+8 \le 0 $$   and $$ x + 2 \le 0 $$
$$ => x \le -8 $$ and $$ x\le -2 $$

$$ => x \le -8 $$ is the solution for this case.

So, combining these both, we get the domain, $$ x \le -8\ and\ x \ge -2$$
Question 4
1 / -0

The set of real values of $$x$$ satisfying the equality $$\left [\displaystyle \frac{3}{x}\, \right]\, +\, \left[\dfrac{4}{x}\, \right]\, =\, 5$$ (where $$[. ]$$ denotes the greatest integer function) belongs to the interval $$\left( a,\, \displaystyle \frac{b}{c}\, \right]$$ where $$a, b, c$$ $$\in$$ $$N$$ and $$\displaystyle \frac{b}{c}$$ is in its lowest form. Find the value of $$a + b + c + abc$$.

A
$$25$$

B
$$18$$

C
$$20$$

D
$$10$$

Solution

Given, $$\left[ \dfrac { 3 }{ x } \, \right] \, +\, \left[ \dfrac { 4 }{ x } \, \right] \, =\, 5,$$

$$ 2 \leq\dfrac { 3 }{ x } <3,\quad 3 \leq \dfrac { 4 }{ x } <4, $$

$$ x \leq\dfrac { 3 }{ 2 } ,\quad x>1\quad and\quad x\leq\dfrac { 4 }{ 3 } ,\quad x>1$$

$$ i.e,\quad x \leq\dfrac { 4 }{ 3 } ,\quad x>1 \implies x \epsilon (1,\dfrac { 4 }{ 3 } ]$$

$$\therefore a=1,\quad b=4,\quad c=3$$

$$ a + b + c + abc=1+4+3+1.3.4=20$$
Question 5
1 / -0

Find $$g(x)$$, if $$f(x) = 5x^{2} + 4$$ and $$f(g(3)) = 84$$

A
$$3x - 10$$

B
$$4x - 7$$

C
$$6x - 17$$

D
$$x^{2} - 5$$

E
$$x^{2} - 3$$

Solution

Given, $$f(x)=5x^2+4$$, $$f(g(3))=84$$
$$\Rightarrow f(g(3)) = 5{g(3)}^{2} +4 = 84$$
$$\Rightarrow 5{g(3)}^{2}=80$$
$$\Rightarrow g(3) = \sqrt{16} = 4$$
$$\Rightarrow g(x) = {x}^{2}-5$$ satisfies the relation $$g(3) = 4$$
Therefore, the correct option is $$D$$.
Question 6
1 / -0

Directions For Questions

Let $$f: A\rightarrow B, g: B\rightarrow C$$ and $$h:C\rightarrow D$$ be three functions, then the function $$gof:A \rightarrow C$$ defined by gof(x) g[f(x)] for all $$x \epsilon A$$ is called the composition of f and g.

...view full instructions

If $$h(x) = x^2, g(x)= x^2 -3$$ and f(x)= x -2, what can you say about ho(gof) and (hog)of?

A
(hog)of $$\neq $$ ho(gof)

B
ho(gof) = (hog)of

C
(hog)of = 4 ho(gof)

D
ho(gof)=$$(x^2-4x-1)^2$$

Solution
Question 7
1 / -0

Which of the following functions are not identical?

A
$$f(x)=\frac{x}{x^2}$$ and $$g(x) = \frac{1}{x}$$

B
$$f(x)=\frac{x^2}{x}$$ and $$g(x) =$$

C
$$f(x)=In \,x^4$$ and g(x)= 4 In Xx

D
f(x) = In {(x-1)(x-2)} and g(x) = In (x-2)+In (x-3)
Question 8
1 / -0

The function $$f(x)={x}^{2}+bx+c$$, where $$b$$ and $$c$$ real constants, describes

A
one-to-one mapping

B
onto mapping

C
not one-to-one onto mapping

D
neither one-to-one nor onto mapping

Solution

$$f\left( x \right) =x^{2}+bx+c$$ when $$f\left( x_{1} \right) =f\left( x_2 \right)$$
Since this a quadratic function and quadratic function is neither one-one nor onto.
so option D is correct.
Question 9
1 / -0

Directions For Questions

Let $$f: A\rightarrow B, g: B\rightarrow C$$ and $$h:C\rightarrow D$$ be three functions, then the function $$gof:A \rightarrow C$$ defined by gof(x) g[f(x)] for all $$x \epsilon A$$ is called the composition of f and g.

...view full instructions

If f(x)=x+2and $$g(x)=x^2-3$$, then which is true?

A
fog $$\neq $$ gog

B
2fog = gof

C
fog = gof

D
fog = 2 gof

Solution
Question 10
1 / -0

If $$f(x) = 4x^{2} - 1$$ and $$g(x) = 8x + 7, g\circ f(2) =$$

A
$$15$$

B
$$23$$

C
$$127$$

D
$$345$$

E
$$2115$$

Solution

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Re-Attempt Weekly Quiz Competition

Functions Test 48

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Functions Test 48

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Rank

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SHARING IS CARING

Question 1

$$-6$$

$$-4$$

$$2$$

$$4$$

$$6$$

Solution

Question 2

$$(0, \dfrac14) \cup (\dfrac34, 1) \cup (\dfrac54,\dfrac74) \cup \{x : x \in N, x \leq 2\}$$

$$(0, \dfrac14) \cup (\dfrac34, 1) \cup \{x : x \in N, x \leq 2\}$$

$$(0, \dfrac14) \cup (\dfrac14, 1) \cup \{x : x \in N, x \leq 2\}$$

$$(0, \dfrac14) \cup (\dfrac34, \dfrac32) \cup \{x : x \in N, x \leq 2\}$$

Solution

Question 3

(-8,-2)

$$\displaystyle \left ( -\infty ,-8 \right )\cup \left ( -2,\infty \right )$$

$$\displaystyle (-\infty ,-8]\cup [-2,\infty )$$

$$\displaystyle (-(-\infty ,-8]\cup [-2,\infty )$$

Solution

Question 4

$$25$$

$$18$$

$$20$$

$$10$$

Solution

Question 5

$$3x - 10$$

$$4x - 7$$

$$6x - 17$$

$$x^{2} - 5$$

$$x^{2} - 3$$

Solution

Question 6

(hog)of $$\neq $$ ho(gof)

ho(gof) = (hog)of

(hog)of = 4 ho(gof)

ho(gof)=$$(x^2-4x-1)^2$$

Solution

Question 7

$$f(x)=\frac{x}{x^2}$$ and $$g(x) = \frac{1}{x}$$

$$f(x)=\frac{x^2}{x}$$ and $$g(x) =$$

$$f(x)=In \,x^4$$ and g(x)= 4 In Xx

f(x) = In {(x-1)(x-2)} and g(x) = In (x-2)+In (x-3)

Question 8

one-to-one mapping

onto mapping

not one-to-one onto mapping

neither one-to-one nor onto mapping

Solution

Question 9

fog $$\neq $$ gog

2fog = gof

fog = gof

fog = 2 gof

Solution

Question 10

$$15$$

$$23$$

$$127$$

$$345$$

$$2115$$

Solution

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