Relations and Functions Test - 12

Given that: n(A) = m and n (B) = n

\(\therefore \) n(A × B) = n(A) . n(B) = mn

So, the total number of non-empty relations from A to B \(2^{mn} - 1\)

We have \([x]^2\) – 5[x] + 6 = 0

⇒ \([x]^2\) – 3[x] 2[x] + 6 = 0

⇒ [x]([x] – 3) – 2([x] – 3) = 0

 ⇒ ([x] – 3)([x] – 2) = 0 ⇒ [x] = 2, 3

If 3 \(\leq\) x < 4, then [x]=3

So, x \(\in\) [2, 4).

Given that: f (x) = \(\frac{1}{1 - 2cosx}\) 

We know that – 1 \(\leq\) cos x ≤ 1

⇒ -2 ≤ - 2 cos x ≤ 2

⇒ - 2 + 1 ≤ 1 – 2 cos x ≤ 2 + 1

⇒ - 1 ≤ 1 – 2 cos x ≤ 3

⇒ \(\frac{1}{1 - 2 cos x} \geq \frac{1}{3}\) and \(\frac{1}{1 - 2 cos x} \leq -1\)

So, the range of f(x) = \((-\infty, -1] \cup[\frac{1}{3}, \infty)\)

Given that: \(f(x) = \sqrt{1 + x^2}\)

\(\implies f(xy) = \sqrt{1 + x^2y^2}\)

and f(x).f(y) \(= \sqrt{1 + x^2}.\sqrt{1 + y^2}\)

\( = \sqrt{1 + x^2 + y^2 + x^2y^2}\)

\(\therefore \sqrt{1 + x^2y^2} \leq \sqrt{1 + x^2 + y^2 + x^2y^2}\)

⇒ f(xy) ≤ f(x).f(y)

Let \(f(x) = \sqrt{a^2 - x^2}\)

f(x) is defined if \(a^2 - x^2 \geq 0\)

\(\implies x^2 - a^2 \leq 0 \\ \implies x^2 \leq a^2\)

⇒ \(|x| \leq a \)

⇒ -a ≤ x ≤ a

\(\therefore\)  Domain of f(x) = [-a, a]

Given that: f(x) = ax + b

⇒ f(- 1) = a(- 1) + b

⇒ - 5 = - a + b

⇒ a – b = 5 ….(i)

f(3) = 3a + b

⇒ 3 = 3a + b

⇒ 3a + b = 3 …..(ii)

On solving eqn. (i) and (ii), we get a = 2, b = - 3

Given that: \(f(x) = \sqrt{4 - x} + \frac{1}{\sqrt{x^2 - 1}}\)

f(x) is defined if

4 – x ≥ 0 or \(x^2\) – 1 > 0

⇒ - x ≥ - 4 or (x – 1)(x + 1) > 0

⇒ x ≤ 4 or x < - 1 and x > 1

\(\therefore\) Domain of f(x) is (- ∞, - 1) \(\cup\)(1, 4]

Given that: f(x) = \(\frac{4 - x}{x - 4}\)

We know that f(x) is defined if x – 4 ≠ 0 ⇒ x ≠ 4

So, the domain of f(x) is = R – {4}

Let \(f(x) = y = \frac{4 - x}{x - 4}=-1\)

\(\therefore\) Range of f(x) =  { - 1)

Given that: \(f(x) = \sqrt{x - 1}\)

f(x) is defined if x – 1 ≥ 0 ⇒ x ≥ 1

\(\therefore \) Domain of f(x) = [1, ∞)

 \(Let,f(x) = y \)

\(\implies \sqrt{x - 1} =y\)

\(\implies y^2 = x - 1\)

\(\implies x = y^2 + 1\)

If x \(\in\)  [1, ∞) then y \(\in\) R

\(\therefore \) Range of f(x) = [0, ∞)

Given that: f(x) = \(\frac{x^2 + 2x + 1}{x^2 - x - 6}\)

F(x) is defined if \(x^2 - x - 6 \neq 0\)

⇒ \(x^2 - 3x + 2x - 6 \neq 0\)

⇒ (x – 3)(x + 2) ≠ 0 ⇒ x ≠ - 2, x ≠ 3

So, the domain of f(x) = R – {- 2, 3}

Given that: f(x) = 2 – |x – 5|

Here, f(x) is defined for x \(\in\) R

\(\therefore\) Domain of f(x) = R

Now, |x – 5| ≥ 0

⇒ - |x – 5| ≤ 0

⇒ 2 - |x – 5| ≤ 2

⇒ f(x) ≤ 2

\(\therefore\) Range of f(x) = (- ∞, 2]

Given that: f(x) = \(3x^2 - 1 \) and g(x) = 3 + x

f(x) = g(x)

⇒ \(3x^2 - 1 = 3 + x\)

⇒ \(3x^2 - x - 4 = 0\)

⇒ \(3x^2 - 4x + 3x - 4 = 0\)

⇒ x(3x – 4) + 1(3x – 4) = 0

⇒ (x + 1)(3x – 4) = 0

⇒ x + 1 = 0 or 3x – 4 = 0

⇒ x = - 1, or \(x = \frac{4}{3}\)

\(\therefore\) Domain = \(\{-1, \frac{4}{3}\}\)

Given, n(A) = 4, n(B) = 3 We know that, the total number of relations from two finite sets A to B is given by

\(= 2^{n(A).n(B)} \)

= \(2^{4 × 3}\) = \(2^{12} \)

Let A = { 1,2 } and {A} = {1, 2} 

⇒ P(A) = {{ 1 },{ 2 },{ 1, 2}, φ}, 

Then, A ∪ P(A) ≠ P(A) 

and {A} ∩ P(A) = {{1,2} ∩ {1},{2},{1,2}, φ }

= { 1,2 } =  A

B′ = S − B = S − {2,3,5,6} = {0,1,4,7,8,9}

A′ = S − A = S − {1,2,3,4} = {0, 5, 6, 7, 8, 9}

\(\therefore\) B′ − A′ = {0,1,4,7,8,9} - {0,5,6,7,8,9}

= {1,4}