Relations and Functions MCQ Quiz - Objective Question with Answer for Relations and Functions - Download Free PDF

Concept:

• Let f(x) = 3 + 2x, which is a linear function of the form f(x) = ax + b.
• Let g_n(x) be the function obtained by composing f with itself n times: g_n(x) = f ∘ f ∘ ... ∘ f (n times)(x).
• If g_n(x) passes through a fixed point (α, β) for all n ∈ ℕ, then the point satisfies the equation g_n(α) = β for all n.
• This implies that the point (α, β) is a fixed point of all functions g_n.
• We find this by solving for the fixed point that satisfies f(α) = α, which guarantees that the point stays fixed under repeated function compositions.

Calculation:

Let f(x) = 3 + 2x

We want a point (α, β) such that g_n(α) = β for all n ∈ ℕ

⇒ fⁿ(α) = β for all n

Let’s compute a few values:

g₁(x) = f(x) = 3 + 2x

g₂(x) = f(f(x)) = f(3 + 2x) = 3 + 2(3 + 2x) = 3 + 6 + 4x = 9 + 4x

g₃(x) = f(g₂(x)) = f(9 + 4x) = 3 + 2(9 + 4x) = 3 + 18 + 8x = 21 + 8x

So pattern is: g_n(x) = A_n + B_nx

Let’s deduce the recurrence relations:

Initial: A₁ = 3, B₁ = 2

⇒ A_n = 3 + 2A_n-1

⇒ B_n = 2B_n-1

B_n = 2ⁿ

Let us find fixed point: f(α) = α

⇒ 3 + 2α = α

⇒ α = -3

Let β = g_n(α) = g_n(-3)

g_n(x) = A_n + B_nx = A_n - 3 × B_n

So β = A_n - 3 × B_n

We want this constant for all n

Let’s check for small values:

g₁(x) = 3 + 2x ⇒ g₁(-3) = 3 - 6 = -3

g₂(x) = 9 + 4x ⇒ g₂(-3) = 9 - 12 = -3

g₃(x) = 21 + 8x ⇒ g₃(-3) = 21 - 24 = -3

So, β = -3 always

∴ 2α + 3β = -3× 2 + (-3)× 3 = -15

Concept:

Convert decimal to binary

Conversion steps:

• Divide the number by 2.
• Get the integer quotient for the next iteration.
• Get the remainder for the binary digit.
• Repeat the steps until the quotient is equal to 0.

Calculation:

∴ (45)₁₀ = 101101

Calculation:

Given,

The function is \( f(xy) = f(x + y) \) for all real values of x  and y, and f(5) = 10 .

We are tasked with finding:

\( f(20) + f(-20) \)

Using the given functional equation, we have:

For \( f(0 \cdot 5) = f(0 + 5) \), we get:

\( f(0) = f(5) = 10 \)

For \( f(0 \cdot 20) = f(0 + 20) \), we get:

\( f(0) = f(20) = 10 \)

For \( f(0 \cdot -20) = f(0 + (-20)) \), we get:

\( f(0) = f(-20) = 10 \)

Thus,

\( f(20) + f(-20) = 10 + 10 = 20 \)

Hence, the correct answert is Option 3.

Calculation:

Given,

The function is such that: \( f(xy) = f(x + y) \) for all real values of x  and y , and \( f(5) = 10 \).

We need to find the value of \( f(0) \).

Substitute \( x = 5 \) and \( y = 0 \) in the given functional equation \( f(xy) = f(x + y) \):

\( f(5 \cdot 0) = f(5 + 0) \)

\( f(0) = f(5) \)

Since \( f(5) = 10 \), we conclude:

\( f(0) = 10 \)

Hence, the correct answer is Option 4. 

Calculation:

Given,

The function satisfies the functional equation:

\( f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)} \)

Also, we are given that:

\( f(2) = 3 \)

We use the functional equation to calculate f(4) . Substituting x = 4 and y = 2 , we get:

\( \frac{f(4)}{f(2)} = f(2) \)

\( \frac{f(4)}{3} = 3 \quad \Rightarrow f(4) = 3 \times 3 = 9 \)

\( \frac{f(2)}{f(2)} = f(1) \quad \Rightarrow f(1) = 1 \)

\( f(1) \times f(4) = 1 \times 9 = 9 \)

Hence, the correct answer is Option 3.

Concept:

• If a function repeats over at a constant period we say that is a periodic function.
• It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.
• The period of sin x and cos x is 2π

Calculation:

To Find: Period of 4cos3 x - 3cos x

As we know 4cos3 x - 3cos x = cos 3x

Period of cos x is 2π

Therefore, the Period of cos 3x is \(\rm \frac {2\pi}{3}\)

Concept:

If f(x)  is odd function then f(-x) = -f(x)

Calculation:

Given: f(x) = x2 + 4x + 4

Replace x by -x,

⇒ f(-x) = (-x)² + 4(-x) + 4

= x2 - 4x + 4                       (∵ (-x)² = x²)

⇒ f(-x) ≠ ± f(x)

Hence function is neither odd nor even.

Concept:

• Inverse Function:

Let f: A → B be one-one and onto (bijective) function. Then f^-1 exists which is a function f^-1: B → A, which maps each element b ∈ B with an element

a ∈ A such that f(a) = b is called the inverse function of f: A → B.

• Methods to find inverse:

Let f : A → B be a bijective function.

Step – I Put f (x) = y

Step – II Solve the equation y = f (x) to obtain x in terms of y.

Interchange x and y to obtain the inverse of the given function f.

Calculation:

Given: \(f(x) = \frac{1 + 2x}{x + 7}\)

Let y = f(x) = 2x + 1 / x + 7

⇒ xy + 7y = 2x + 1

⇒ 7y – 1 = 2x – xy

⇒ x(2 - y) = 7y – 1

⇒ x = (7y - 1) / (2 - y)

⇒ f^-1 (x) = \(\frac{7x - 1}{2 - x}\)

Concept:

• If a function repeats over at a constant period we say that is a periodic function.
• It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.
• The period of sin x and cos x is 2π

Calculation:

To Find: Period of 3sin x - 4sin3 x

As we know 3sin x - 4sin3 x = sin 3x

Period of sin x is 2π

Therefore, Period of sin 3x is \(\rm \frac {2\pi}{3}\)

Concept:

Logarithm properties:  

Product rule: The log of a product equals the sum of two logs.

\(\rm {\log _a}\left( {mn} \right) = \;{\log _a}m + \;{\log _a}n\)

Quotient rule: The log of a quotient equals the difference of two logs.

\(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\)

Power rule: In the log of power the exponent becomes a coefficient.

\(\rm {\log _a}{m^n} = n{\log _a}m\)

Formula of Logarithms:

If \(\rm lo{g_a}x = b \) then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: \(\rm \log_{3}{(x^{4} - x^3)} - \log_{3} (x - 1) = 3\)

\(\rm \Rightarrow \log_{3} \left[{\frac{(x^{4} - x^3)}{(x - 1)}} \right ] = 3\)        (∵ \(\rm {\log _a}\frac{m}{n} = \;{\log _a}m - \;{\log _a}n\))

\(\rm \Rightarrow \log_{3} \left[{\frac{x^3(x-1)}{(x - 1)}} \right ] = 3\)

\(\rm \Rightarrow \log_{3} x^3 = 3\)

\(\Rightarrow \rm 3\log_3 x = 3\)               (∵ \(\rm {\log _a}{m^n} = n{\log _a}m\)) 

\(\Rightarrow \rm \log_3 x = 1 \\\therefore x=3\)

Concept:

Range: The range of a function is the set of all possible values it can produce, i.e., all values of y for which x is defined.

Note:

The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.

Calculation:

Let, y = f(x) = \(\rm \frac{x+1}{x-3}\)

⇒y(x - 3) = x + 1

⇒yx - 3y - x = 1

⇒ x(y - 1) - 3y = 1

⇒ x(y - 1) = 1 + 3y

⇒\(\rm x = \frac{1+3y}{y-1}\)

It is clear that x is not defined when y - 1 = 0, i.e, when  y = 1

∴ Range (f) = R - {1}

Hence, option (2) is correct.

Mistake PointsIt is given in the Question that, f(x) is real function. So,

f(x) has real values for value of x other than x = 3

∴ Domain of given function = R - {3}, where R is set of all real numbers

Concept:

Domin of sin-1 x is [-1, 1]

Adding or subtracting the same quantity from both sides of an inequality leaves the inequality symbol unchanged.

Calculation:

Given:  f(x) = sin-1 (x + 1) 

As we know, domin of sin1 x is [-1, 1]

Therefore, -1 ≤ (x + 1) ≤ 1

subtracting 1 in above inequality, 

⇒ -1 - 1 ≤ x + 1 - 1 ≤ 1 - 1

⇒ -2 ≤ x ≤ 0

∴ Domin of sin-1 (x + 1) is [-2, 0] 

Mistake Points[-2, 0] is different from [-2, 0). '[' and ']' indicates that the end number (2 and 0) is also included. '(' and ')' indicates that 2 and 0 are not taken into consideration.

Concept:

• (x + y)(x - y) = x² - y²
• \(\ln { 1\over a} = \ln 1 - \ln a = 0 - \ln a = -\ln a\)

Calculation:

Given: f(x) = ln (x + \(\sqrt{1+\text{x}^2}\)),__(i)

Replace x by (-x) in (i),

⇒  f(-x) = ln (-x + \(\sqrt{1+(\text{-x})^2}\)),

⇒  f(-x) = ln (-x + \(\sqrt{1+\text{x}^2}\)),

Multiply and divide by (x + \(\sqrt{1+\text{x}^2}\)) inside the ln function,

⇒  f(-x) =  \(\ln (-x + \sqrt{1+\text{x}^2}.{ x + \sqrt{1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({- x^2 + {1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({1 \over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(-\ln ({ x + \sqrt{1+\text{x}^2}})\),

From (i),

⇒  f(-x) =  - f(x)\

⇒  f(-x) + f(x) = 0

∴ The correct answer is option (1).

Concept:

Range of a Function:

• The range of a function is the set of all possible output values (f(x)) for the given domain.
• For the function f(x) = x², where x ∈ ℝ, the output is always a non-negative real number.
• This is because the square of any real number is always greater than or equal to zero.
• Minimum value of f(x) occurs at x = 0, which gives f(0) = 0.
• As x increases or decreases, f(x) increases without any upper limit.

Calculation:

Given,

Function: f(x) = x²

Domain: x ∈ ℝ

⇒ f(0) = 0 ∈ range

⇒ For x = ±1, ±2, f(x) = 1, 4, which are > 0

⇒ f(x) ≥ 0 for all real x

⇒ So, the range is [0, ∞)

⇒ f(x) can take any value from 0 to ∞

The range of f(x) is the set of all non-negative real numbers.

Understanding Positive vs Non-negative Real Numbers:

• Positive real numbers are all real numbers greater than zero, i.e., (0, ∞), and do not include 0.
• Non-negative real numbers include all positive real numbers and 0, i.e., [0, ∞).
• The function f(x) = x² is defined for all real numbers, and it gives an output ≥ 0 for any x ∈ ℝ.
• At x = 0, f(x) = 0² = 0, which is part of the range.
• Since 0 is not included in positive real numbers, option 3 is incorrect.
•  

∴ Option 3 is incorrect because it excludes 0, which is part of the range.

Hence Option 1 is the correct answer

Concept:

1. Domain of a  functions:

• The domain of a function is the set of all possible values of the independent variable. That is all the possible inputs for a function.

Calculation:

Observe that the given function is in the form of numerator and denominator. The function will be well defined for all non zero values of the denominator.

Therefore, \({\rm{x}} - 2{\rm{\;}} \ne 0\) that implies that \({\rm{x\;}} \ne 2\).

Similarly square root function is well defined for all non-negative values.

Therefore, \({\rm{x}} - 2 > 0\) that implies \({\rm{x}} > 2.\)

Thus, domain of the given function is \(\left( {2,{\rm{\;}}\infty } \right).\)