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:

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

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

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

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

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 $f(4)$

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

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 $f(16)$

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

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 :

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

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

∴ 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 $\underset{x\to 1}{lim}(f\circ f(x))$.

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

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 $\underset{x\to 1}{lim}(x^4-2x^2)$.

Substitute x = 1 into the function:

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

∴ The value of $\underset{x\to 1}{lim}(f\circ f(x))$ 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 $p\ne 0$, 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:

$\{1,4,7,10\}\ne \mathbb{N}$

∴ 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) = $\frac{1}{f^{\prime }(g(x))}$

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

∴ g'(x) is equal to 1 + {g(x)}⁵. 

The correct answer is Option 4.

Calculation

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

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

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

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

⇒ $6a+4+\frac{2}{7}=0$

$a=-\frac{5}{7}$

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

⇒ $|f(x)|=\frac{1}{28}|4x^2+21x+28|$

Now, $28\sum _{\mathrm{i}=1}^5|\mathrm{f}(6)|=28(|\mathrm{f}(1)|+\mathrm{f}(2)+\dots +\mathrm{f}(5))$

$28\cdot \frac{1}{28}\cdot 675=675$

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.

$f^{\prime }(x)=0$ for x in [1.2, 1.9)

Hence option 3 is correct