The probability that a randomly chosen one-one function from the set <math> <mrow> <mrow> <mo>{</mo> <mrow> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>,</mo> <mi>c</mi> <mo>,</mo> <mi>d</mi> </mrow> <mo>}</mo> </mrow> </mrow> </math> to the set <math> <mrow> <mrow> <mo>{</mo> <mrow> <mn>1</mn> <mo>,</mo> <mn>2</mn> <mo>,</mo> <mn>3</mn> <mo>,</mo> <mn>4</mn> <mo>,</mo> <mn>5</mn> </mrow> <mo>}</mo> </mrow> </mrow> </math> satisfies f(a) + 2f(b) – f(c)&nbsp; = f(d) is:

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Maths NCERT Exemplar Solutions Class 12th Chapter Three Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th

The probability that a randomly chosen one-one function from the set ${a, b, c, d}$ to the set ${1, 2, 3, 4, 5}$ satisfies f(a) + 2f(b) – f(c) = f(d) is:

Option 1 -

$\frac{1}{24}$

Option 2 -

$\frac{1}{40}$

Option 3 -

$\frac{1}{30}$

Option 4 -

$\frac{1}{20}$

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Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

No of one – one functions ® ⁵P₄ = 120

f(a) + 2f(b) – f(c) = f(d) ${1, 2, 3, 4, 5}$

2f(b) = f(d) + f(c) – f(a)

So, f(d) + f(c) – f(a) should be even.

Only possibilities of f(d) f(c) f(a)

Not possible since E E &n

...more

No of one – one functions ® ⁵P₄ = 120

f(a) + 2f(b) – f(c) = f(d) ${1, 2, 3, 4, 5}$

2f(b) = f(d) + f(c) – f(a)

So, f(d) + f(c) – f(a) should be even.

Only possibilities of f(d) f(c) f(a)

Not possible since E E E E - even, O – Odd

function is one one E O O

O O E

O E O

Case (i)

f(d) f(c) f(a)

E O O

2 1 3 similarly we write cases for f(d) = 4

2 1 5 413

2 3 5 431

2 5 1 435

2 3 1 453

2 5 3 415

4512²²²asdf=fhf=kjhfjkdfh45654hdfkjdhjh

Case (ii)

O O E O O E

1 5 2 1 5 4

5 1 2 5 1 4

3 1 2 1 3 4

1 3 2 3 1 4

3 5 2 5 3 4

5 3 2 &nb

less

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Similar Questions for you

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

alok kumar singh

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

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

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

alok kumar singh

$?$ 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

alok kumar singh

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:

alok kumar singh

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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The probability that a randomly chosen one-one function from the set ${a, b, c, d}$ to the set ${1, 2, 3, 4, 5}$ satisfies f(a) + 2f(b) – f(c) = f(d) is:

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The probability that a randomly chosen one-one function from the set { a , b , c , d } to the set { 1 , 2 , 3 , 4 , 5 } satisfies f(a) + 2f(b) – f(c) = f(d) is:

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The probability that a randomly chosen one-one function from the set ${a, b, c, d}$ to the set ${1, 2, 3, 4, 5}$ satisfies f(a) + 2f(b) – f(c) = f(d) is: