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
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?
Answer (Detailed Solution Below)
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:
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?
Answer (Detailed Solution Below)
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:
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?
Answer (Detailed Solution Below)
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:
Let the function f(x) = x2- 1
What is the area bounded by the function f(x) and the x-axis?
Answer (Detailed Solution Below)
Operations on Functions Question 4 Detailed Solution
Calculation:
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:
Let the function f(x) = x2- 1
What is \(\lim_{x \to 1} \{f \circ f(x)\}\) equal to?
Answer (Detailed Solution Below)
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:
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?
Answer (Detailed Solution Below)
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:
Let f = {(1, 1), (2, 4), (3, 7), (4, 10)}
If f(x) = px + q then what is the value of (p + q) ?
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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 _______.
Answer (Detailed Solution Below)
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