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

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 for all real values of x  and y, and f(5) = 10 .

We are tasked with finding:

Using the given functional equation, we have:

For , we get:

For , we get:

For , we get:

Thus,

Hence, the correct answert is Option 3.

Calculation:

Given,

The function is such that: for all real values of x  and y , and .

We need to find the value of .

Substitute and in the given functional equation :

Since , we conclude:

Hence, the correct answer is Option 4. 

Calculation:

Given,

The function satisfies the functional equation:

Also, we are given that:

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

Hence, the correct answer is Option 3.

Calculation:

Given,

The function satisfies the equation for all positive real values of x and y, and .

We are tasked with finding .

Using the functional equation, for , we have:

Since , we can calculate

Next, to find , we use the functional equation again:

Since , we can calculate

Hence, the correct answer is Option 4.

Concept:

 Period of a function:

• 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 

Concept:

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

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: 

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) = 

Concept:

 Period of a function:

• 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 

Concept:

Logarithm properties:  

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

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

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

Formula of Logarithms:

If  then x = ab (Here a ≠ 1 and a > 0)

Calculation:

Given: 

        (∵ )

               (∵ ) 

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) = 

⇒y(x - 3) = x + 1

⇒yx - 3y - x = 1

⇒ x(y - 1) - 3y = 1

⇒ x(y - 1) = 1 + 3y

⇒

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

Calculation:

Given: f(x) = ln (x + ),__(i)

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

⇒  f(-x) = ln (-x + ),

⇒  f(-x) = ln (-x + ),

Multiply and divide by (x + ) inside the ln function,

⇒  f(-x) =  ,

⇒  f(-x) =  ,

⇒  f(-x) =  ,

⇒  f(-x) =  ,

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,  that implies that .

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

Therefore, 0\) that implies  2.\)

Thus, domain of the given function is