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Relations and Functions

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2 months ago

Relations and Functions Class 12th

The total number of functions, f : {1,2,3,4} -> {1,2,3,4,5,6} such that f(1) + f(2) = f(3), is equal to:

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Vishal Baghel

Contributor-Level 10

Case 1 : If f (3) = 3 then f (1) and f (2) take 1 or 2

No. of ways = 2.6 = 12

Case 2 : If f (3) = 5 then f (1) and f (2) take 2 or 3

OR 1 and 4

No. of ways = 2.6.2 = 24

Similarly for all other cases

Total no. of ways 90.

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2 months ago

Relations and Functions Class 12th

Let $f, g : N - {1}$ N be functions defined by f(a) = a, where a is the maximum of the powers of those primes p such that p^a divides a, and g(a) = a + 1, for all $a \in N$ - {1}. Then, the function f + g is

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alok kumar singh

Contributor-Level 10

f (a) = a, a is max of powers of prime P such that P^a divides a.

f (2) = 1; g (2) = 3

f (3) = 1; g (3) = 4

f (4) = 2; g (4) = 5

f (5) = 1; g (5) = 6

f (2) + g (2) = 4

f (3) + g (3) = 5

f (4) + g (4) = 7

f (5) + g (5) = 7

So, f (x) + g (x) is many-one-and an into function. We won't get 'I' in the range.

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2 months ago

Relations and Functions Class 12th

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a₀ = a₁ = 0 and a_n+2 = 3a_n+1 – 2a_n + 1, $\forall n \geq 0 .$

Then $a_{25} a_{23} - 2 a_{25} a_{22} - 2 a_{23} a_{24} + 4 a_{22} a_{24}$ is equal to

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alok kumar singh

Contributor-Level 10

a₀ = 0, a₁ = 0

a_n+2 = 3a_n+1 – 2a_n + 1

a₂₅ a₂₃ – 2a₂₅a₂₂ – a₂₃a₂₄ + 4a₂₂a₂₄ =?

a₂ = 3a₁ – 2a₀ + 1

a₃ = 3a₂ – 2a₁ + 1

a₄ = 3a₃ – 2a₂ + 1

a₅ = 3a₄ – 2a₃ + 1

a_n+2 = 3a_n+1 – 2a_n + 1

$(+) \Rightarrow (a_{2} + a_{3} + a_{4} + . . . . + a_{n + 1} + a_{n + 2})$

-> a_n+2 = 2 (a₂ + a₃ + …. + a_n + a_n+1) –2 (a₁ + a₂ + ….+ a_n) + n + 1

a_n+2 = 2a_n+1 + n + 1

a₂₅ a₂₃ -2a₂₅ a₂₂ -a₂₃ a₂₄ + 4a₂₂ a₂₄

= a₂₅ (a₂₃ – 2a₂₂) -2a₂₄ (a₂₃ – 2a₂₂)

As a_n+2 = 2a_n+1 + n + 1

-> a_n+2 – 2a_n+1 = n + 1

-> a_n+1 -2a_n = n

-> 24 * 22 = 528

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2 months ago

Relations and Functions Class 12th

Let f(x) be a quadratic polynomial with leading coefficient 1 such that f(0) = p, p $\neq 0$ , and f(1) = $\frac{1}{3} .$ If the equations f(x) = 0 and fo fo fo f(x) = 0 have a common real root , then f(3) is equal to

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Payal Gupta

Contributor-Level 10

Let f (x) = (x -α) (x - β)

It is given that f (0) = p = p

and

$f (1) = \frac{1}{3} \Rightarrow (1 - α) (1 - β) = \frac{1}{3}$

Now, let us assume that is the common root of f (x) = 0 and fofofof (x) = 0

fofofof (x) = 0

$f (- 3) = (- 3 - \frac{7}{6}) (3 - 3) = 25$

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New answer posted

2 months ago

Relations and Functions Class 12th

The number of one-one functions f:{a, b, c, d} ® {0, 1, 2, ……., 10} such that 2f (a) – f(b) + 3f(c) + f(d) = 0 is ___________.

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alok kumar singh

Contributor-Level 10

f(b) = 2f(a) + 3f(c) + f(d)

Value of f(c) Value of f(a) Number of functions

1 7

...more

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2 months ago

Relations and Functions Class 12th

The sum of all the real roots of the equation $(e^{2 x} - 4) (6 e^{2 x} - 5 e^{x} + 1) = 0 i s$

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Payal Gupta

Contributor-Level 10

$e^{2 x} - 4 = 0 o r 6 e^{2 x} - 5 e^{x} + 1 = 0$

$\Rightarrow x = l n 2 x = l n (\frac{1}{3}), l n (\frac{1}{2})$

= $- l n 3, - l n 2$

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2 months ago

Relations and Functions 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:

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alok kumar singh

Contributor-Level 10

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 &

...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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New question posted

2 months ago

Relations and Functions 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:

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New answer posted

2 months ago

Relations and Functions Class 12th

Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is

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Payal Gupta

Contributor-Level 10

P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)

6x = 5 = 0 $x = \frac{5}{6}$

P (atmost 2H)

$P (O H, 5 T) + P (1 H, 4 T) + P (2 H, 3 T)$

$= \frac{1}{6^{5}} (1 + 25 + 250) = \frac{276}{6^{5}} = \frac{4^{6}}{6^{4}}$

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New answer posted

2 months ago

Relations and Functions Class 12th

The mean and standard of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to:

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Vishal Baghel

Contributor-Level 10

$\bar{x} = 15, σ = 2 \Rightarrow σ^{2} = 4$

$∴ x_{1} + x_{2} + . . . . + x_{50} = 15 * 50 = 750$

$4 = \frac{x_{1}^{2} + x_{2}^{2} + . . . . . + x_{50}^{2}}{50} - 225$

Let a be the correct observation and b is the incorrect observation then a + b = 70 and

$16 = \frac{75 - b + a}{50}$

$= \frac{50 * 229 + 6 0^{2} - 1 0^{2}}{50} - 256 = 43$

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