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

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

$f(g(x))=x$ where $g(x)$ is the inverse of $f(x)$.

Now differentiating both the sides with respect to $x$, we get

$f^{\prime }(g(x))g^{\prime }(x)=1$

${\displaystyle \frac{g^{\prime }(x)}{1+g(x{)}^5}}=1$

$g^{\prime }(x)=1+g(x{)}^5$ .

Given:

f(x) = log (x - 1) - log (x - 2)

Calculation:

⇒ f(x) is defined for all x satisfying

⇒ x - 1 > 0 and x - 2 > 0 i.e. x > 2

⇒  Domain (f) = (2, ∞)

⇒ g(x) is defined for all x satisfying

⇒  $\frac{x-1}{x-2}>0$ ⇒ x ∈(-∞, 1) U (2, ∞) 

⇒  Domain (g) = (-∞, 1) U (2, ∞)

∴ The f(x) and g(x) are equal for all x belonging to their common domain i.e. (2, ∞).

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.

$\mathrm{g}(\mathrm{x})={\displaystyle \frac{1}{2\mathrm{x}+5}}\Rightarrow \mathrm{g}(5)={\displaystyle \frac{1}{2\times 5+5}}={\displaystyle \frac{1}{15}}$.

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

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)

$f\left(\frac{x}{2}\right)=asin\left(\frac{x}{2}\right)$     ---(iii)

$g\left(\frac{x}{2}\right)=acos\left(\frac{x}{2}\right)$      ---(iv)

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

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

$F(x)=a^{2}si{n}^2\frac{x}{2}+a^{2}co{s}^2\frac{x}{2}$

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

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:

$\sin 2\mathrm{x}=\frac{2\tan \mathrm{x}}{1+{\tan }^2\mathrm{x}}$

$\cos 2\mathrm{x}=\frac{1-{\tan }^2\mathrm{x}}{1+{\tan }^2\mathrm{x}}$

$\tan 2\mathrm{x}=\frac{2\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}$

Calculation:

We have f(x) = $\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}$.

Substituting x = tan θ, we get:

⇒ f(tan θ) = $\frac{2\tan \theta }{1+{\tan }^2\theta }$ = 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 

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

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

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

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

So, x = 2 satisfies the equation 

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

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

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

Hence, x = ± 2 satisfies the equation 

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

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.

Concept:

For any two non-empty sets A and B, we have:

I. A X B = {(a, b) | a ∈ A and b ∈ B}

II. B X A = {(b, a) | a ∈ A and b ∈ B}

Calculation:

Given:  A = {1, 2} and B = {8, 9} .

We know that, A X B = {(a, b) | a ∈ A and b ∈ B}

⇒ A X B = {(a,8), (a, 9), (b, 8), (b, 9)}

We know that, B X A = {(b, a) | a ∈ A and b ∈ B}

⇒ B X A = {(1, a), (1, b), (2, a), (2, b)}

⇒ (A X B) ∪ (B X A) = {(1,8), (1, 9), (2, 8), (2, 9), (8, 1), (8, 2), (9, 1), (9, 2)}