Operations on Functions MCQ Quiz - Objective Question with Answer for Operations on Functions - Download Free PDF

Calculation:

Given,

The function is \( f(xy) = f(x + y) \) for all real values of x  and y, and f(5) = 10 .

We are tasked with finding:

\( f(20) + f(-20) \)

Using the given functional equation, we have:

For \( f(0 \cdot 5) = f(0 + 5) \), we get:

\( f(0) = f(5) = 10 \)

For \( f(0 \cdot 20) = f(0 + 20) \), we get:

\( f(0) = f(20) = 10 \)

For \( f(0 \cdot -20) = f(0 + (-20)) \), we get:

\( f(0) = f(-20) = 10 \)

Thus,

\( f(20) + f(-20) = 10 + 10 = 20 \)

Hence, the correct answert is Option 3.

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.

Concept:

For two functions f(x) and g(x), the function [(f × g)(x)] is defined as f(x) × g(x).

Calculation:

f(x) = x + 5 ⇒ f(5) = 5 + 5 = 10.

\(\rm g(x)= \dfrac{1}{2x+5} \Rightarrow g(5)= \dfrac{1}{2\times5+5}=\dfrac{1}{15}\).

∴ [(f × g)(5)] = f(5) × g(5) = \(\rm 10\times\dfrac{1}{15}=\dfrac{2}{3}\).

Given 

f"(x) = -f(x)      ---(i)

g(x) = f'(x)

\(F(x) = {\left\{ {f \left( {\frac{x}{2}} \right)} \right\}^2} + {\left\{ {g \left( {\frac{x}{2}} \right)} \right\}^2}\)      ---(ii)

F(5) = 5

from equation (1)

we can say

f(x) = a sin x

f'(x) = a cos x = g(x)

f"(x) = -a sin x = - f(x)

\(f\left(\frac{x}{2}\right)= a\ sin\left(\frac{x}{2}\right)\)     ---(iii)

\(g\left(\frac{x}{2}\right)= a\ cos\left(\frac{x}{2}\right)\)      ---(iv)

from equation (ii), (iii) & (iv)

\(F(x) = {\left\{ {a\sin \left( {\frac{x}{2}} \right)} \right\}^2} + {\left\{ {a\cos \left( {\frac{x}{2}} \right)} \right\}^2}\)

\(F(x)=a^2sin^2\frac{x}{2}+a^2cos^2\frac{x}{2}\)

\(F(x) = {a^2}\left\{ {{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2}} \right\}\)

F(x) = a²

f(5) = a² = 5

f(10) = a² = 5

Concept:

Real valued function: A function f : A → B is called a real-valued function if B is a subset of R (set of all real numbers).

If A and B both are subsets of R, then f is called a real function.

Calculation:

f(x) = 2x3 + 7x2 - 3

∴ f(x - 1) = 2(x - 1)3 + 7(x - 1)2 - 3

⇒ 2(x³ - 3x² + 3x - 1) + 7(x² + 1 - 2x) - 3

⇒ 2x3 - 6x2 + 6x - 2 + 7x2 + 7 - 14x - 3

⇒ 2x3 + x2 - 8x + 2

Concept:

Trigonometric Formulae:

\(\rm \sin 2x = \frac{2\tan x}{1\ +\ \tan^2x}\)

\(\rm \cos 2x = \frac{1\ -\ \tan^2x}{1\ +\ \tan^2x}\)

\(\rm \tan 2x = \frac{2\tan x}{1\ -\ \tan^2x}\)

Calculation:

We have f(x) = \(\rm {2x\over {1\ +\ x^2}}\).

Substituting x = tan θ, we get:

⇒ f(tan θ) = \(\rm {2\tan\theta\over {1\ +\ \tan^2 \theta}}\) = sin 2θ.

Concept:

If x = a satisfies the equation f(x) = b then f(a) = b

Calculation:

Here, we have to find the real values of x such that 

\(\left( {5 + 2\sqrt 6 } \right)^{\rm x^{2} - 3} + \left( {5 - 2\sqrt 6 } \right)^{\rm x^{2} - 3} = 10\)

Let f(x) = \(\left( {5 + 2\sqrt 6 } \right)^{\rm x^{2} - 3} + \left( {5 - 2\sqrt 6 } \right)^{\rm x^{2} - 3}\)

We can see that if we substitute x = 2 in f(x) then we get

⇒ f(2) = \(\left( {5 + 2\sqrt 6 } \right)^{\rm 4 - 3} + \left( {5 - 2\sqrt 6 } \right)^{\rm 4 - 3} = 10\)

So, x = 2 satisfies the equation 

\(\left( {5 + 2\sqrt 6 } \right)^{\rm x^{2} - 3} + \left( {5 - 2\sqrt 6 } \right)^{\rm x^{2} - 3} = 10\)

Similarly, if we substitute x = - 2 in f(x) then we get

⇒ f(- 2) = \(\left( {5 + 2\sqrt 6 } \right)^{\rm 4 - 3} + \left( {5 - 2\sqrt 6 } \right)^{\rm 4 - 3} = 10\)

Hence, x = ± 2 satisfies the equation 

\(\left( {5 + 2\sqrt 6 } \right)^{\rm x^{2} - 3} + \left( {5 - 2\sqrt 6 } \right)^{\rm x^{2} - 3} = 10\)

Concept:

If p(x) = a₀ + a₁ x + …… + a_nxⁿ , where coefficients of x are real and if a_n ≠ 0 .Then p(x) is a polynomial of degree n.

Calculation:

Given: f(x) = x² + 2x – 5 and g(x) = 5x + 30

⇒ f[g(x)] = f(5x + 30) = (5x + 30)² + 2(5x + 30) – 5 = 25x² + 310x + 955

⇒ f[g(x)] is a polynomial of degree 2.

Hence statement 1 is false.

⇒ g[g(x)] = g(5x + 30) = 5(5x + 30) + 30 = 25x + 180

⇒ g[g(x)] is a polynomial of degree 1.

Concept:

If α and β are the roots of the equation f(x) = ax² + bx + c = 0. Then f(α) = 0 = f(β).

Calculation:

Given: f(x) = x² + 2x – 5 and g(x) = 5x + 30

⇒ g[f(x)] = g(x² + 2x – 5) = 5 (x² + 2x – 5) + 30 = 5x² + 10x + 5 = 0.

⇒ 5x² + 10x + 5 = 0

⇒ x² + 2x + 1 = (x + 1)² = 0

⇒ x = - 1, -1

Explanation:

Given that f is a continuous function on [1,3] and it takes only rational value for all "x"  and f(2) = 10.

If f is continuous, then it must take only take rational values, so f must be a constant function.

Hence f(3/2) = 10.

Calculation:

Given: h(x) = 5f(x) – xg(x), where f(x) = x² + 2x – 5 and g(x) = 5x + 30

⇒ h(x) = 5 × (x² + 2x – 5) – x × (5x + 30)

⇒ 5x2 + 10x – 25 – 5x2 - 30x

⇒ h(x) = -20 x – 25

⇒ h’(x) = - 20.

Calculation:

Given,

The function is \( f(xy) = f(x + y) \) for all real values of x  and y, and f(5) = 10 .

We are tasked with finding:

\( f(20) + f(-20) \)

Using the given functional equation, we have:

For \( f(0 \cdot 5) = f(0 + 5) \), we get:

\( f(0) = f(5) = 10 \)

For \( f(0 \cdot 20) = f(0 + 20) \), we get:

\( f(0) = f(20) = 10 \)

For \( f(0 \cdot -20) = f(0 + (-20)) \), we get:

\( f(0) = f(-20) = 10 \)

Thus,

\( f(20) + f(-20) = 10 + 10 = 20 \)

Hence, the correct answert is Option 3.