Operations on Functions MCQ Quiz in বাংলা - Objective Question with Answer for Operations on Functions - বিনামূল্যে ডাউনলোড করুন [PDF]

Last updated on Mar 15, 2025

পাওয়া Operations on Functions उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Operations on Functions MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।

Latest Operations on Functions MCQ Objective Questions

Top Operations on Functions MCQ Objective Questions

Operations on Functions Question 1:

Comprehension:

Consider the following for the two (02) items that follow:

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

What is f(0) equal to?

  1. 0
  2. 1
  3. 5
  4. 10

Answer (Detailed Solution Below)

Option 4 : 10

Operations on Functions Question 1 Detailed Solution

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. 

Operations on Functions Question 2:

Comprehension:

Consider the following for the two (02) items that follow:
The function f(x) satisfies  \(f(x), f \left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}.\)  ​for all positive real values of x and y, and f(2) = 3

What is f(1)f(4) equal to?

  1. 4
  2. 8
  3. 9
  4. 18

Answer (Detailed Solution Below)

Option 3 : 9

Operations on Functions Question 2 Detailed Solution

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.

Operations on Functions Question 3:

Comprehension:

Consider the following for the two (02) items that follow:
The function f(x) satisfies  \(f(x), f \left( \frac{x}{y} \right) = \frac{f(x)}{f(y)}.\)  ​for all positive real values of x and y, and f(2) = 3

What is f(16) equal to?

  1. 18
  2. 27
  3. 54
  4. 81

Answer (Detailed Solution Below)

Option 4 : 81

Operations on Functions Question 3 Detailed Solution

Calculation:

Given,

The function satisfies the equation \( \frac{f(x)}{f(y)} = \frac{x}{y} \) for all positive real values of x and y, and \( f(2) = 3 \).

We are tasked with finding \( f(16) \).

Using the functional equation, for \( f(4) \), we have:

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

Since \( f(2) = 3 \), we can calculate \( f(4) \)

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

Next, to find \( f(16) \), we use the functional equation again:

\( \frac{f(16)}{f(4)} = f\left( \frac{16}{4} \right) = f(4) \)

Since \( f(4) = 9 \), we can calculate \( f(16) \)

\( \frac{f(16)}{9} = 9 \quad \Rightarrow \quad f(16) = 81 \)

Hence, the correct answer is Option 4.

Operations on Functions Question 4:

Comprehension:

Consider the following for the two (02) items that follow :
Let the function f(x) = x2- 1

What is the area bounded by the function f(x) and the x-axis?

  1. 1/3 square unit
  2. 2/3 square unit
  3. 4/3 square units
  4. 2 square units

Answer (Detailed Solution Below)

Option 3 : 4/3 square units

Operations on Functions Question 4 Detailed Solution

Calculation:

qImage68484d518d641eb957ac2a1e

 

 

The function is \( y = x^2 - 1 \), and we need to find the area bounded by the curve and the x-axis between x = -1  and x = 1 .

The required area is given by the definite integral of the function from  -1  to 1 :

\( \text{Area} = \int_{-1}^{1} (x^2 - 1) \, dx \)

\( \int (x^2 - 1) \, dx = \frac{x^3}{3} - x \)

Evaluate the integral from  -1 to  1 :

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

\( = \left(\frac{1}{3} - 1\right) - \left(\frac{-1}{3} + 1\right) = \left(\frac{1}{3} - \frac{3}{3}\right) - \left(\frac{-1}{3} + \frac{3}{3}\right) = \frac{2}{3} - \frac{2}{3} = \frac{4}{3} \)

∴ The area is \( \frac{4}{3} \) square units.

Hence, the correct answer is Option 3.

Operations on Functions Question 5:

Comprehension:

Consider the following for the two (02) items that follow :
Let the function f(x) = x2- 1

What is \(\lim_{x \to 1} \{f \circ f(x)\}\) equal to?

  1. -1
  2. 0
  3. 1
  4. 2

Answer (Detailed Solution Below)

Option 1 : -1

Operations on Functions Question 5 Detailed Solution

Calculation:

Given,

The function is \( f(x) = x^2 - 1 \) and we are tasked with finding \( \lim_{x \to 1} (f \circ f(x)) \).

\( f \circ f(x) = f(f(x)) \)

Given \( f(x) = x^2 - 1 \), we substitute \( f(x) \) into itself:

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

Now, apply the function \( f \) to \( (x^2 - 1) \):

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

This simplifies to:

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

Thus, \( f \circ f(x) = x^4 - 2x^2 \).

Now, we need to find \( \lim_{x \to 1} (x^4 - 2x^2) \).

Substitute x = 1 into the function:

\( f \circ f(1) = 1^4 - 2(1^2) = 1 - 2 = -1 \)

∴ The value of \( \lim_{x \to 1} (f \circ f(x)) \) is -1.

The correct answer is Option (1).

Operations on Functions Question 6:

Comprehension:

Consider the following for the two (02) items that follow:
Let f = {(1, 1), (2, 4), (3, 7), (4, 10)}

Consider the following statements:
I. f is one-one function.
II. f is onto function if the codomain is the set of natural numbers.
Which of the statements given above is/are correct?

  1. I only
  2. II only
  3. Both I and II
  4. Neither I not II

Answer (Detailed Solution Below)

Option 1 : I only

Operations on Functions Question 6 Detailed Solution

Calculation:

Given,

The function is \(f(x)=px+q\), a linear polynomial.

Since \(f(x)=px+q\) is linear with \(p\neq0\), it is one-one (injective): different \(x\) give different \(f(x)\).

The actual outputs given are \(\{1,4,7,10\}\), so the range of \(f\) is \(\{1,4,7,10\}\).

The codomain is the set of natural numbers \(\mathbb{N}\). For \(f\) to be onto, its range must equal its codomain, but here:

\(\{1,4,7,10\}\neq\mathbb{N}\)

∴ f is one-one but not onto.

Hence, the correct answer is Option 1.

Operations on Functions Question 7:

Comprehension:

Consider the following for the two (02) items that follow:
Let f = {(1, 1), (2, 4), (3, 7), (4, 10)}

If f(x) = px + q then what is the value of (p + q) ?

  1. -1
  2. 0
  3. 1
  4. 5

Answer (Detailed Solution Below)

Option 3 : 1

Operations on Functions Question 7 Detailed Solution

Calculation:

Given,

The function is \(f = \{(1, 1), (2, 4), (3, 7), (4, 10)\} \), and we are also given that the function is \( f(x) = px + q \), where p and q  are constants.

Using the points f(1) = 1  and f(2) = 4 , we create the following system of equations:

- From f(1) = 1 , we have \(p(1) + q = 1 \), so p + q = 1 .......(1).

- From f(2) = 4 , we have p(2) + q = 4 , so 2p + q = 4  ..........( 2).

Subtract Equation 1 from Equation 2:

\( (2p + q) - (p + q) = 4 - 1 \)

\( p = 3 \)

Now, substitute p = 3 into Equation 1:

\( 3 + q = 1 \)

\( q = -2 \)

\( p + q = 3 + (-2) = 1 \)

∴ The value of p + q  is 1.

The correct answer is Option (3).

Operations on Functions Question 8:

If g is the inverse of a function f and \(f^{′}(x)=\frac{1}{1+x^5}\), then g′(x) is equal to

  1. 1 + x5
  2. 5x4
  3. \(\frac{1}{1+\{g(x)\}^5}\)
  4. 1 + {g(x)}5 
  5. None of these 

Answer (Detailed Solution Below)

Option 4 : 1 + {g(x)}5 

Operations on Functions Question 8 Detailed Solution

Calculation:

Given, g is the inverse of f

⇒ g-1(x) = f(x)

⇒ f(g(x)) = x

Differentiating wrt x, we get:

f '(g(x))g'(x) = 1

⇒ g'(x) = \(\frac{1}{f'(g(x))}\)

⇒ g'(x) = 1 + {g(x)}5 [∵ \(f^{′}(x)=\frac{1}{1+x^5}\)]

∴ g'(x) is equal to 1 + {g(x)}5

The correct answer is Option 4.

Operations on Functions Question 9:

Let f : R → R be a function defined by f(x) = (2 + 3a)x2 \(\left(\frac{a+2}{a-1}\right)\) x + b, a ≠ 1. If f(x + y) = f(x) + f(y) + 1 - \(\frac{2}{7}\)xy, then the value of \(28 \sum_{i=1}^{5}|f(i)|\) is:

  1. 715 
  2. 735 
  3. 545 
  4. 675 

Answer (Detailed Solution Below)

Option 4 : 675 

Operations on Functions Question 9 Detailed Solution

Calculation

\(f(x)=(3 a+2) x^{2}+\left(\frac{a+2}{a-1}\right) x+b\)

\(\mathrm{f}\left(\mathrm{x}+\frac{1}{2}\right)=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})+1-\frac{2}{7} \mathrm{xy} \quad ...(1)\)

In (1) Put x = y = 0 f(0) = 2f(0) + 1 ⇒ f(0) = –1

So, f(0) = 0 + 0 + b = –1 b = –1

In (1) Put y = –x f(0) = f(x) + f(–x) + 1 + \(\frac{2}{7}\) x2

–1 = 2(3a + 2)x2 + 2b + 1 \(\frac{2}{7}\) x2

\(-1=\left(2(3 a+2)+\frac{2}{7}\right) x^{2}+1-2\)

⇒ \(6 a+4+\frac{2}{7}=0\)

\(a=-\frac{5}{7}\)

So f(x) = \(-\frac{1}{7} x^{2}-\frac{3}{4} x-1\)

⇒ \(|f(x)|=\frac{1}{28}\left|4 x^{2}+21 x+28\right|\)

Now, \(28 \sum_{\mathrm{i}=1}^{5}|\mathrm{f}(6)|=28(|\mathrm{f}(1)|+\mathrm{f}(2)+\ldots+\mathrm{f}(5))\)

\(28 \cdot \frac{1}{28} \cdot 675=675\)

Hence option 4 is correct

Operations on Functions Question 10:

Function f : [1.2, 1.9] → R, f(x) = [x], where [x] denotes the greatest int less than or equal to x. Then _______.

  1. f'(x) = 1
  2. f is not differentiative
  3. f'(x) = 0
  4. f is not continuous function

Answer (Detailed Solution Below)

Option 3 : f'(x) = 0

Operations on Functions Question 10 Detailed Solution

Given:

f: [1.2, 1.9] → R, f(x) = [x], where [x] denotes the greatest integer less than or equal to x.

Concept Used:

The greatest integer function [x] is discontinuous at integer values.

The derivative of a constant function is 0.

Calculation:

Given:

f: [1.2, 1.9] → R, f(x) = [x], where [x] denotes the greatest integer less than or equal to x.

In the interval [1.2, 1.9), the function f(x) = [x] is defined as:

f(x) = 1, for all x in [1.2, 1.9)

This is because the greatest integer less than or equal to any number in this interval is 1.

Since f(x) is a constant function in the given interval, its derivative is 0.

\(f'(x) = 0\) for x in [1.2, 1.9)

Hence option 3 is correct

Get Free Access Now
Hot Links: teen patti download teen patti master 2023 teen patti lotus teen patti party