Relations and Functions Test

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Relations and Functions Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Determine all ordered pairs that satisfy $$(x - y)^{2} + x^{2} = 25$$, where $$x$$ and $$y$$ are integers and $$x \geq 0$$. Find the number of different values of $$y$$ that occur

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

$${ \left( x-y \right) }^{ 2 }+{ x }^{ 2 }=25$$
Now, $${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$$
$$9+16=25$$
$$\therefore $$There are 2 possibilities:
$$I.{ \left( x-y \right) }^{ 2 }=9$$ and $${ x }^{ 2 }=16$$
$$\therefore x=\pm 4$$ and $$x-y=\pm 3$$
$$\left( i \right) .x-y=3\Rightarrow \left( 4,1 \right) $$ and $$\left( -4,-7 \right) $$
$$\left( ii \right) .x-y=-3\Rightarrow \left( 4,7 \right) $$ and $$\left( -4,-1 \right) $$
$$II.{ \left( x-y \right) }^{ 2 }=16$$ and $${ x }^{ 2 }=9$$
$$\therefore x=\pm 3$$ and $$x-y=\pm 4$$
$$\left( i \right) .x-y=4\Rightarrow \left( 3,-1 \right) $$ and $$\left( -3,-7 \right) $$
$$\left( ii \right) .x-y=-4\Rightarrow \left( 3,7 \right) $$ and $$\left( -3,-1 \right) $$
$$\therefore $$ Different values of y are $$1,-1,7,-7$$
$$\therefore 4$$ different values of y occur.
Question 2
1 / -0

Express $$\dfrac {b}{a^{5}} $$ in terms of $$t$$ given that $$a = \sqrt [3]{t}$$ and $$b = t^{2}$$.

A
$$t^{-\dfrac {1}{3}}$$

B
$$t^{\dfrac {1}{3}}$$

C
$$t^{\dfrac {5}{6}}$$

D
$$t^{\dfrac {6}{5}}$$

E
$$t^{\dfrac {10}{3}}$$

Solution

$$b = t^2, a = \sqrt[3] {t}$$
$$\Rightarrow a^5 = \sqrt[3] {t^5} = t^{\frac{5}{3}}$$
$$\therefore \cfrac{b}{a^5} = \cfrac{t^2}{t^{\frac{5}{3}}}$$
$$=t^{2-\frac{5}{3}} = t^{\frac{1}{3}}$$
Question 3
1 / -0

If $$f(x) = x^{2} - x$$ and $$g(x) = 3x + 2$$, what is $$g(f(1)) - 2$$?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

E
$$18$$

Solution

Given: $$f(x)=x^2-x, g(x)=3x+2$$
$$g(f(1))-2=?$$
So, $$f(1)=(1)^2-1=0$$
$$g(0)=3(0)+2=2$$
Therefore, $$g(f(1))-2=2-2=0$$
Question 4
1 / -0

$$f(x)=x-\dfrac{1}{x}$$, then $$f(\dfrac{1}{x})=$$
I. $$f(x)$$
II. $$f(-x)$$
III. $$-f(x)$$
IV. $$\dfrac{1}{f(x)}$$

A
I and II

B
II and III

C
III and IV

D
II only

E
II and IV

Solution

$$f(x)=x-\dfrac{1}{x}$$
$$f\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-x$$ (substituting $$\dfrac{1}{x}$$ at the place of x)
$$=-\left(x-\dfrac{1}{x}\right)$$
$$=-f(x)$$

$$f\left(\dfrac{1}{x}\right)=\dfrac{1}{x}-x$$
$$=-\left(-\dfrac{1}{x}\right)+(-x)=f(-x)$$
$$\therefore II$$ & $$III$$.
Question 5
1 / -0

If $$f(x) = 3x + 5$$ and $$f(g(x)) = 6x - 4$$, what is $$g(x)$$?

A
$$2x - 9$$

B
$$2x - 3$$

C
$$3x - 9$$

D
$$3x - 3$$

E
$$9x + 1$$

Solution

Since $$f(x)=3x+5$$ then $$f(g(x))=3g(x)+5$$ .........(1)
Also it is given that $$f(g(x))=6x-4$$ ..........(2)
Therefore, equate equations (1) and (2) as follows:
$$3g(x)=6x-9$$
$$\therefore 3g(x)=3(2x-3)$$
$$\therefore g(x)=2x-3$$
Hence, $$g(x)=2x-3$$.
Question 6
1 / -0

What positive value(s) of x, less than $$360^o$$, will give a minimum value for 4 - 2 sin x cos x?

A
$$\frac{\pi}{4}$$ only

B
$$\frac{5\pi}{4}$$ only

C
$$\frac{\pi}{2}$$ and $$\frac{5\pi}{2}$$

D
$$\frac{3\pi}{2}$$

E
$$\frac{\pi}{4}$$ and $$\frac{5\pi}{4}$$
Solution
- $$4-2sinxcosx = 4-sin2x$$ , It will be minimum if $$sin2x$$ is maximum
- The maximum value of $$sin2x$$ is $$1$$ and occurs at $$2x = 90,450$$
- Which implies $$x=45 , 225$$
- Therefore the correct option is $$E$$
Question 7
1 / -0

If $$f(x)=\dfrac{x+1}{x-1}$$, what is the value of $$f(f(f(f((\frac{3}{5}))))$$?

A
-4

B
0

C
0.6

D
1.3

E
7

Solution

$$f(x)=\dfrac{x+1}{x-1}$$
$$f\left(\dfrac{3}{5}\right)=\dfrac{\dfrac{3}{5}+1}{\dfrac{3}{5}-1}=\dfrac{\dfrac{8}{5}}{\dfrac{-2}{5}}=-4$$
$$f(f\left(\dfrac{3}{5}\right))=\dfrac{f\left(\dfrac{3}{5}\right)+1}{f\left(\dfrac{3}{5}\right)-1}=\dfrac{-4+1}{-4-1}=\dfrac{-3}{-5}$$
$$f(f(f(\dfrac{3}{5})))=\dfrac{\dfrac{3}{5}+1}{\dfrac{3}{5}-1}=-4$$
$$f(f(f(f(\dfrac{3}{5}))))=\dfrac{3}{5}=0.6$$.
Question 8
1 / -0

In the problem below, $$f(x)={x}^{2}$$ and $$g(x)=4x-2$$
Find the following function: $$(f+g)(x)$$

A
$${x}^{2}+4x-2$$

B
$${x}^{2}-4x-2$$

C
$${x}^{2}-4x+2$$

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

Solution

Given, $$f(x)={x}^{2}$$ and $$g(x)=4x-2$$
We need to find $$(f+g)(x)$$
$$\therefore (f+g)(x) = f(x)+g(x) = {x}^{2}+4x-2$$
Question 9
1 / -0

If $$p - q > 0$$, which of the following is true?

A
If q = 0, then p < 0

B
If q < 0, then p < 0

C
If p > 0, then q > 0

D
If q = 0, then p > 0

E
If q < 1, then p > 1
Solution
- Given $$p-q>0$$
- If $$q=0$$ , then $$p-0>0$$ , which gives $$p>0$$
- Therefore option $$D$$ is correct
Question 10
1 / -0

If $$f(x) = 3x - 2$$ and $$g(x) = 7, f[g(x)] =$$

A
$$21x-2$$

B
$$7$$

C
$$19$$

D
$$7x-2$$

E
$$13$$

Solution

Given, $$f(x)=3x-2$$ and $$g(x)=7$$
So, we get $$f(g(x))=f(7)$$
$$=3(7)-2$$
$$=21-2$$
$$=19$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 32

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Relations and Functions Test - 32

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Question 1

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 2

$$t^{-\dfrac {1}{3}}$$

$$t^{\dfrac {1}{3}}$$

$$t^{\dfrac {5}{6}}$$

$$t^{\dfrac {6}{5}}$$

$$t^{\dfrac {10}{3}}$$

Solution

Question 3

$$0$$

$$1$$

$$2$$

$$3$$

$$18$$

Solution

Question 4

I and II

II and III

III and IV

II only

II and IV

Solution

Question 5

$$2x - 9$$

$$2x - 3$$

$$3x - 9$$

$$3x - 3$$

$$9x + 1$$

Solution

Question 6

$$\frac{\pi}{4}$$ only

$$\frac{5\pi}{4}$$ only

$$\frac{\pi}{2}$$ and $$\frac{5\pi}{2}$$

$$\frac{3\pi}{2}$$

$$\frac{\pi}{4}$$ and $$\frac{5\pi}{4}$$

Solution

Question 7

-4

0

0.6

1.3

7

Solution

Question 8

$${x}^{2}+4x-2$$

$${x}^{2}-4x-2$$

$${x}^{2}-4x+2$$

$${x}^{2}-2$$

Solution

Question 9

If q = 0, then p < 0

If q < 0, then p < 0

If p > 0, then q > 0

If q = 0, then p > 0

If q < 1, then p > 1

Solution

Question 10

$$21x-2$$

$$7$$

$$19$$

$$7x-2$$

$$13$$

Solution

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