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Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th Relations and Functions

70. Let A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2} and f; g: A→B be the functions defined by f(x) = x² − x, x ∈ A and g(x) = 2x - 1/2 - 1. x∈ A. Are f and g equal? Justify your answer.

(Hint: One may note that two functions f: A → B and g: A → B such that f(a) = g(a) & mn For E; a ∈A, are called equal functions).

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Vishal Baghel | Contributor-Level 10

4 months ago

It is given that A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2}

Also, it is given that $f, g : A \to B$ are defined by $f (x) = x^{2} - x, x \in A$ and $g (x) = 2 x - \frac{1}{2} - 1, x \in A$ .

It is observed that:

$\begin{array}{l} f (- 1) = (1^{2}) - (- 1) = 1 + 1 = 2 \\ g (- 1) = 2 (- 1) - \frac{1}{2} - 1 = 2 (\frac{3}{2}) - 1 = 3 - 1 = 2 \\ \Rightarrow f (- 1) = g (- 1) \\ f (0) = {(0)}^{\land} 2 - 0 = 0 \\ g (0) = 2 (0) - \frac{1}{2} - 1 = 2 (\frac{1}{2}) - 1 = 1 - 1 = 0 \\ \Rightarrow f (0) = g (0) \\ f (1) = {(1)}^{\land} 2 - 1 = 1 - 1 = 0 \\ g (1) = 2 a - \frac{1}{2} - 1 = 2 (\frac{1}{2}) - 1 = 1 - 1 = 0 \\ \Rightarrow f (1) = g (1) \\ f (2) = {(2)}^{\land} 2 - 2 = 4 - 2 = 2 \\ g (2) = 2 (2) - \frac{1}{2} - 1 = 2 (\frac{3}{2}) - 1 = 3 - 1 = 2 \\ \Rightarrow f (2) = g (2) \\ ∴ f (a) = g (a) \forall a \in A \end{array}$

Hence, the functions f and g are equal.

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Let S = {1, 2, 3 ,....,20}

R₁ = {(a, b) : a divide b}

R₂ = {(a, b) : a is integral multiple of b} a, b Î s

n(R₁ – R₂) = ?

alok kumar singh

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

n (R₁) = 66

R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

$f (x) = {\begin{matrix} e^{- x}, & x < 0 \\ l n x, & x > 0 \end{matrix}$

$g (x) = {\begin{matrix} e^{x}, & x < 0 \\ x, & x > 0 \end{matrix}$

The gof : A->R is

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$g o f (x) = {\begin{matrix} f (x), & f (x) < 0 \\ f (x), & f (x) > 0 \end{matrix}$

$= {\begin{array}{l} e^{l n x} = x & (0, 1) & e^{- x} \\ (- \infty, 0) & l n x & (1, \infty) \end{array}$

If A = {1, 2, 3, … 100}, R = {(x, y) | 2x = 3y, x, y ∈ A} is symmetric relation on A and the number of elements in R is n, the smallest integer value of n is

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$?$ R is symmetric relation

⇒ (y, x) ∈ R V (x, y) ∈ R

(x, y) ∈ R ⇒ 2x = 3y and (y, x) ∈ R ⇒ 3x = 2y

Which holds only for (0, 0)

Which does not belongs to R.

∴ Value of n = 0

Let : f(0, R)->∞ be increasing function such that $\underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)} = 1$ then $\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1)$ is equal to

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f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

$\frac{f (x)}{f (x)} < \frac{f (5 x)}{f (x)} < \frac{f (7 x)}{f (x)}$

$\underset{x \to \infty}{l i m} \frac{f (x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)}$

-> $1 < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < 1$ $\underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} = 1$

$\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1) = 0$

Let A = ${1, 2, 3, . . . . ., 10} a n d f : A \to A$ be defined as

$f (k) = {\begin{matrix} k + 1 & i f k i s o d d \\ k & i f k i s e v e n \end{matrix}$

Then the number of possible functions g : A ® A such that gof = f is:

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Given f (k) = ${\begin{cases} k + 1, k i s o d d \\ k, k i s e v e n \end{cases}$

$? g : A \to A$ such that g (f (x) = f (x)

Case I : If x is even then g (x) = x . (i)

Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ × 1 = 10⁵

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