Relations and Functions Test

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Relations and Functions Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Let f: R → R be defined by f(x) = \(3x ^2\) - 5 and g: R → R by g(x) = \(x\over x^2+1\). Then gof is

A
\(3x^2-5\over 9x^4-30x^2+26\)

B
\(3x^2-5\over 9x^4-6x^2+26\)

C
\(3x^2\over x^4+2x^2-4\)

D
\(3x^2\over 9x^4+30x^2-2\)

Solution

Given that

f(x) = \(3x^2-5\) and g(x) = \(x\over x^2+1\)

gof = g{f(x)} = g(\(3x^2-5\))

= \(3x^2-5\over (3x^2-5)^2+1\) = \(3x^2-5\over 9x^4-30x^2+25+1\)

= \(3x^2-5\over 9x^4-30x^2+26\)
Question 2
1 / -0

Which of the following functions from Z into Z are bijections?

A
f(x) = \(x^3\)

B
f(x) = x + 2

C
f(x) = 2x + 1

D
f(x) = \(x^2\) + 1

Solution

Here,

f(x) = x + 2 ⇒ f(\(x_1\)) = f(\(x_2\))

\(x_1\) + 2 = \(x_2\) + 2 ⇒ \(x_1\) = \(x_2\)

Let y = x + 2

x = y - 2

f(y-2) = x+2

=(y-2)+2

=y

\(\implies f(x)=y\)

Hence, f(x) is one-one and onto.
Question 3
1 / -0

Let f: R → R be the functions defined by f(x) = \(x ^3\) + 5. Then \(f ^{-1}\)(x) is

A
\((x+5)^{1\over3}\)

B
\((x-5)^{1\over3}\)

C
\((5-x)^{1\over3}\)

D
\((5+x)^{1\over3}\)

Solution

Given that, f(x) = \(x^3\) + 5

Let y = \(x^3\) + 5 ⇒ \(x^3\) = y - 5

x = \((y-5)^{1\over3}\) ⇒ \(f^{-1}(x)=\,(x-5)^{\frac{1}{3}}\)
Question 4
1 / -0

Let f: A → B and g: B → C be the bijective functions. Then \((gof) ^{-1}\) is

A
\(f^{-1}og^{-1}\)

B
\(fog\)

C
\(g^{-1}of^{-1}\)

D
\(gof\)

Solution

Given that,

f: A → B and g: B → C be the bijective functions.

\((gof)^{-1}\) = \(f^{-1}og^{-1}\)
Question 5
1 / -0

Let f: R -\(\Big\{{3\over5}\Big\}\) → R be defined by f(x) = \(3x+2\over 5x-3\). Then

A
\(f^{-1}(x) =f(x)\)

B
\(f^{-1}(x) =-f(x)\)

C
\( ( f o f ) x = -x\)

D
\(f^{-1}(x) ={1\over19}f(x)\)

Solution

Given that,

f(x) = \(3x+2\over 5x-3\)

Let y = \(3x+2\over 5x-3\)

3x + 2 = 5xy - 3y

⇒ x(3 - 5y) = -3y - 2

x = \(3y+2\over 5y-3\) ⇒ \(f^{-1}\)(x) = \(3x+2\over 5x-3\)

Therefore, \(f^{-1}(x) =f(x)\)
Question 6
1 / -0

Let f: [0, 1] → [0, 1] be defined by \(f(x) = \begin{cases} x, & \quad \text{if } x \text{ is rational}\\ 1-x, & \quad \text{if } x \text{ is irrational} \end{cases}\) Then (\(fof\)) x is

A
Constant

B
1 + x

C
x

D
None of these

Solution

Given that

f: [0, 1] → [0, 1] be defined by

\(f(x) = \begin{cases} x, & \quad \text{if } x \text{ is rational}\\ 1-x, & \quad \text{if } x \text{ is irrational} \end{cases}\)

\(\therefore\) (\(fof\))x = f(f(x)) = x
Question 7
1 / -0

Let f: [2, \(\infty\)) → R be the function defined by f(x) = \(x ^2 -4x + 5\), then the range of f is

A
\(R\)

B
\([1,\infty)\)

C
\([4,\infty)\)

D
\([5,\infty)\)

Solution

Given that,

f(x) = \(x ^2 -4x + 5\)

Let y = \(x ^2 -4x + 5\)

⇒ y = \(x ^2 -4x + 4+1\) = \((x-2)^2\) + 1

⇒ \((x-2)^2\) = y - 1 ⇒ x - 2 = \(\sqrt{y-1}\)

⇒ x = 2 + \(\sqrt{y-1}\)

\(\therefore\) y - 1 \(\geq\) 0, y \(\geq\) 1

Range = \([1,\infty)\)
Question 8
1 / -0

Let f: N → R be the function defined by f(x) = \(2x-1\over 2\) and g: Q → R be another function defined by g(x) = x + 2. Then (gof)\((\frac{3}{2})\) is

A
1

B
2

C
4

D
None of these

Solution

Given that

f(x) = \(2x-1\over 2\) and g(x) = x + 2

\((gof)(\frac{3}{2})\) = \(g\Big[f\Big({3\over2}\Big)\Big]\) = g\(\Bigg({2\times{3\over2}-1\over2}+2\Bigg)\)

= g(1) = 1 + 2 = 3
Question 9
1 / -0

Let f: R → R be defined by \(f(x) = \begin{cases} 2x :x>3 & \quad \\ x^2:1 < x \leq 3 & \quad \\3x : x \leq 1 \end{cases}\) Then f(-1) + f(2) + f(4) is

A
9

B
14

C
5

D
8

Solution

Given that

\(f(x) = \begin{cases} 2x :x>3 & \quad \\ x^2:1 < x \leq 3 & \quad \\3x : x \leq 1 \end{cases}\)

f(-1) + f(2) + f(4)

= 3(-1) + \((2)^2\) + 2 x 4

= -3 + 4 + 8 = 9
Question 10
1 / -0

Let f: R → R be given by f(x) = tan x. Then \(f ^{-1}\) (1) is

A
\(\pi\over4\)

B
\(\pi\over5\)

C
\(\pi\over6\)

D
\(\pi\over3\)

Solution

Given that,

f(x) = tan x

Let y = tan x ⇒ x = \(tan^{-1}\)y

⇒ \(f^{-1}\)(x) = \(tan^{-1}\)x ⇒ \(f^{-1}\)(1) = \(tan^{-1}\)1

⇒ \(tan^{-1}\)tan\(\pi\over4\) = \(\pi\over4\) [\(\because\) tan\(\pi\over4\) = 1]

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 28

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Relations and Functions Test - 28

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SHARING IS CARING

Question 1

\(3x^2-5\over 9x^4-30x^2+26\)

\(3x^2-5\over 9x^4-6x^2+26\)

\(3x^2\over x^4+2x^2-4\)

\(3x^2\over 9x^4+30x^2-2\)

Solution

Question 2

f(x) = \(x^3\)

f(x) = x + 2

f(x) = 2x + 1

f(x) = \(x^2\) + 1

Solution

Question 3

\((x+5)^{1\over3}\)

\((x-5)^{1\over3}\)

\((5-x)^{1\over3}\)

\((5+x)^{1\over3}\)

Solution

Question 4

\(f^{-1}og^{-1}\)

\(fog\)

\(g^{-1}of^{-1}\)

\(gof\)

Solution

Question 5

\(f^{-1}(x) =f(x)\)

\(f^{-1}(x) =-f(x)\)

\( ( f o f ) x = -x\)

\(f^{-1}(x) ={1\over19}f(x)\)

Solution

Question 6

Constant

1 + x

x

None of these

Solution

Question 7

\(R\)

\([1,\infty)\)

\([4,\infty)\)

\([5,\infty)\)

Solution

Question 8

1

2

4

None of these

Solution

Question 9

9

14

5

8

Solution

Question 10

\(\pi\over4\)

\(\pi\over5\)

\(\pi\over6\)

\(\pi\over3\)

Solution

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