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

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.

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.

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.

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).

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.

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).

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.

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

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