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

333

444

555

666

Solution

Question 2

t−13t^{-\dfrac {1}{3}}t−31​

t13t^{\dfrac {1}{3}}t31​

t56t^{\dfrac {5}{6}}t65​

t65t^{\dfrac {6}{5}}t56​

t103t^{\dfrac {10}{3}}t310​

Solution

Question 3

000

111

222

333

181818

Solution

Question 4

I and II

II and III

III and IV

II only

II and IV

Solution

Question 5

2x−92x - 92x−9

2x−32x - 32x−3

3x−93x - 93x−9

3x−33x - 33x−3

9x+19x + 19x+1

Solution

Question 6

π4\frac{\pi}{4}4π​ only

5π4\frac{5\pi}{4}45π​ only

π2\frac{\pi}{2}2π​ and 5π2\frac{5\pi}{2}25π​

3π2\frac{3\pi}{2}23π​

π4\frac{\pi}{4}4π​ and 5π4\frac{5\pi}{4}45π​

Solution

Question 7

-4

0

0.6

1.3

7

Solution

Question 8

x2+4x−2{x}^{2}+4x-2x2+4x−2

x2−4x−2{x}^{2}-4x-2x2−4x−2

x2−4x+2{x}^{2}-4x+2x2−4x+2

x2−2{x}^{2}-2x2−2

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−221x-221x−2

777

191919

7x−27x-27x−2

131313

Solution

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Question Analysis

My Perfomance

Score

$3$

$4$

$5$

$6$

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

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

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

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

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

$0$

$1$

$2$

$3$

$18$

$2x - 9$

$2x - 3$

$3x - 9$

$3x - 3$

$9x + 1$

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

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

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

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

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

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

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

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

${x}^{2}-2$

$21x-2$

$7$

$19$

$7x-2$

$13$