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We know that

f (x₁) = f (x₂)

So we get

3x₁ = 3x₂ => x₁ = x₂

Here, f is one-one

Consider the number 2 in co domain N which has no natural image as 2

Here, f is onto.

We know that

f (x₁) = f (x₂)

So we get

x₁² = x₂²

It can be written as

(x₁ – x₂) (x₁ + x₂) = 0

We get

x₁ – x₂ = 0 where x₁ = x₂

Hence, f is one one.

Consider 2 is the co-domain of N where √2 ∉ N

f (√2) = (√2)² = 2

Hence, f is into.

We know that

f (1) = 1² = 1

f (-1) = -1² = 1

Here, f is many one

Consider – 1 in the codomain R which has no elements in R with square as – 1

Hence, – 1 ∈ R which has no pre image in R

f is many one into.

Given, f = {(1, 6), (2, 5), (4, 3), (5, 2), (8, –1), (10, –3)} 

g = {(2, 0), (3, 2), (5, 6), (7, 10), (8, 12), (10, 16)} 

(1) Domain of f ={1,2,4,5,8,10} 

Domain of g ={2,3,5,7,8,10} 

Domain of (f + g) = {x : x ∈D f ∩ Dg } 

Where Df = Domain of function f, Dg = Domain of function g 

Domain of (f + g) = {2,5,8,10}.

(2) Domain of quotient function f/ g = {x : x ∈D f ∩ Dg and g (x) ≠ 0} 

Domain of (f/g) = {2,5,8,10}.

Given; f(x) = \(\frac{x-1}x\) 

F(x) = 1 – 1/x 

To find f(1/x) replacing x by 1/x 

F(1/x) = 1 – 1/(1/x) 

F(1/x) = 1 – x

Given, A = {a, b, c}, B = {u, v, w} and

f = A → B and g: B → A defined by

f = {(a, v), (b, u), (c, w)} and

g = {(u, b), (v, a), (w, c)}

For both f and g, different elements of domain have different

images

∴ f and g are one – one

Again, for each element in co – domain of f and g, there is a pre – image in the domain

∴ f and g are onto

Thus, f and g are bijective.

Now,

gof = {(a, a), (b, b), (c, c^.)} and

fog = {(u, u), (v, v), (w, w)}

[x]² – 5[x] + 6 = 0 

∴ ([x] – 3)([x] – 2) = 0 

∴ [x] = 3 or 2 

If [x] = 2, then 2 ≤ x < 3 

If [x] = 3, then 3 ≤ x < 4 

∴ Solution set = [2, 4)

f(x) = 2 |x| + 3x

(i) f(2) = 2|2| + 3(2) = 2 (2) + 6 ….. [∵ |x| = x, x > 0]

= 10

(ii) f(-5) = 2 |-5| + 3(-5)

= 2(5) – 15 …..[∵ |x| = -x, x < 0]

= 10 – 15

= -5

f(x) = 4[x] – 3

(i) f(7.2) = 4 [7.2] – 3

= 4(7) – 3 ………[∵ 7 ≤ 7.2 < 8, [7.2] = 7]

= 25

(ii) f(0.5) = 4[0.5] – 3 = 4(0) – 3 ………[∵ 0 ≤ 0.5 < 1, [0.5] = 0]

= -3

(iii) f(-5/2) = f(-2.5)

= 4[-2.5] – 3

= 4(-3) – 3 …….[∵-3 ≤ -2.5 ≤ -2, [-2.5] = -3]

= -15

(iv) f(2π) = 4[2π] – 3

= 4[6.28] – 3 …..[∵ π = 3.14]

= 4(6) – 3 …….[∵ 6 ≤ 6.28 < 7, [6.28] = 6]

= 21

Given as

f(x) = (x + 1)/(x – 1)

Let us prove that the f [f (x)] = x.

f [f (x)] = f [(x+1)/(x-1)]

= [(x+1)/(x-1) + 1]/[(x+1)/(x-1) – 1]

= [[(x+1) + (x-1)]/(x-1)]/[[(x+1) – (x-1)]/(x-1)]

= [(x+1) + (x-1)]/[(x+1) – (x-1)]

= (x+1+x-1)/(x+1-x+1)

= 2x/2

= x

∴ f [f (x)] = x

Thus proved.

f(x) = 2{x} + 5x

(i) {-1} = -1 – [-1] = -1 + 1 = 0

∴ f(-1) = 2 {-1} + 5(-1)

= 2(0) – 5

= -5

(ii) {1/4} = 1/4-0 = 1/4

f(1/4) = 2(1/4) + 5(1/4)

= 2(1/4) + 5/4

 = 7/4

= 1.75

(iii) {-1.2} = -1.2 – [-1.2] = -1.2 + 2 = 0.8

f(-1.2) = 2{-1.2} + 5(-1.2)

= 2(0.8) + (-6)

= -4.4

(iv) {-6} = -6 – [-6] = -6 + 6 = 0

f(-6) = 2{-6} + 5(-6)

= 2(0) – 30

= -30

(i) |x + 4| ≥ 5

The solution of |x| ≥ a is x ≤ -a or x ≥ a

∴ |x + 4| ≥ 5 gives

∴ x + 4 ≤ -5 or x + 4 ≥ 5

∴ x ≤ -5 – 4 or x ≥ 5 – 4

∴ x ≤ -9 or x ≥ 1

∴ The solution set = (-∞, – 9] ∪ [1, ∞)

(ii) |x – 4| + |x – 2| = 3 …..(i)

Case I: x < 2 Equation (i) reduces to

4 – x + 2 – x = 3 …….[x < 2 < 4, x – 4 < 0, x – 2 < 0]

∴ 6 – 3 = 2x

∴ x = 3/2

Case II: 2 ≤ x < 4

Equation (i) reduces to

4 – x + x – 2 = 3

∴ 2 = 3 (absurd)

There is no solution in [2, 4)

Case III: x ≥ 4

Equation (i) reduces to

x – 4 + x – 2 = 3

∴ 2x = 6 + 3 = 9

∴ x = 9/2

∴ x = 3/2, 9/2 are solutions.

The solution set = {3/2, 9/2}

(iii) x² + 7|x| + 12 = 0

∴ (|x|)² + 7|x| + 12 = 0

∴ (|x| + 3) (|x| + 4) = 0

∴ There is no x that satisfies the equation.

The solution set = { } or Φ

(i) f(1/2)

If, 0 ≤ x ≤ 1, f(x) = x

∴ f (1/2) = 1/2

(ii) f(-2)

If, x < 0, f(x) = x²

f(–2) = (–2)²

= 4

∴ f (–2) = 4

(iii) f (1)

If, x ≥ 1, f (x) = 1/x 

f(1) = 1/1

∴ f(1) = 1

(iv) f (√3)

Now, we have √3 = 1.732 > 1 

If, x ≥ 1, f (x) = 1/x 

∴ f (√3) = 1/√3

(v) f (√-3)

As we know that √-3 is not a real number and the function f(x) is defined only when x ∈ R.

Hence, f(√-3) does not exist.

Answer : (c) \( \sqrt{2+x}-2\) 

Let y = f (x) = (x + 2)² – 2 

⇒ y + 2 = (x + 2)² 

⇒ x + 2 = \(\sqrt{y+2}\) 

⇒ x = \(\sqrt{y+2}\) - 2 

⇒ f^-1 (x) = \( \sqrt{2+x}-2\)

Answer : (c) R – {– 3, 3} 

The function log | x² – 9 | is defined when | x² – 9 | > 0

| x² – 9 | is positive for all real values but 

log | x² – 9 | is not defined when | x² – 9 | = 0, 

i.e., function log | x² – 9 | does not exist when x = –3, 3. 

∴  Domain of function is R – {– 3, 3}.

Correct Answer is (B) [1, ∞)

\(f(x)=\frac{1}{1-x^2}\)

\(\Rightarrow y= \frac{1}{1-x^2}\)

⇒ y - yx² = 1

⇒ y - 1 = yx²

⇒ \(x=\sqrt{\frac{y-1}{y}}\)

⇒ \(\frac{y-1}{y}\geq 0\)

⇒ y ≥ 1

∴ range (f) = [1, ∞)

Correct Answer is (B) = [0, 1)

\(f(x)=\frac{x^2}{1+x^2}\)

⇒ \(y=\frac{x^2}{1+x^2}\)

⇒ y + yx² = x²

⇒ y = x² (1 - y)

⇒ x = \(\sqrt{\frac{y}{1-y}}\)

\(\frac{y}{1-y}\geq 0\)

⇒ y ≥ 0

And

1 - y > 0

⇒ y < 1

Taking intersection we get

range (f) = [0, 1)

Correct Answer is (A) [ - 2, 2] 

\(f(x)=x+\frac{1}{x}\)

For this type

Range is

\(-2\leq y\geq2\)

Correct Answer is (D) (0, ∞)

f(x) = a^x 

when x < 0 

0 <  a^x < 1

When x ≥ 0

a^x > 0

Therefore range of f(x) = a^x = (0, ∞)

Given that f: R⁺ → R⁺ and g: R⁺ → R⁺

Therefore, fog: R⁺ → R⁺ and gof: R⁺ → R⁺

Domains of fog and gof are the same.

Let us find fog and gof also we have to check whether they are equal or not,

Consider that (fog)(x) = f(g(x))

= f(√x)

= √x²

= x

Now consider that (gof)(x) = g(f(x))

= g(x²)

= √x²

= x

Therefore, (fog)(x) = (gof)(x), ∀ x ∈ R⁺

Hence, fog = gof 

(i) Given as, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

f(x) = x and g(x) = |x|

(gof)(x) = g(f(x))

= g(x)

= |x|

Now, (fog)(x) = f(g(x))

= f(|x|)

= |x|

(ii) Given as, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

f(x) = x² + 2x − 3 and g(x) = 3x − 4

(gof)(x) = g(f(x))

= g(x² + 2x − 3)

= 3(x² + 2x − 3) − 4

= 3x² + 6x − 9 − 4

= 3x² + 6x − 13

Now, (fog)(x) = f(g(x))

= f(3x − 4)

= (3x − 4)² + 2(3x − 4) − 3

= 9x² + 16 − 24x + 6x – 8 − 3

= 9x² − 18x + 5

(iii) Given as, f: R → R and g: R → R

Therefore, gof: R → R and fog: R → R

f(x) = 8x³ and g(x) = x^1/3

(gof)(x) = g(f(x))

= g(8x³)

= (8x³)^1/3

= [(2x)³]^1/3

= 2x

Now, (fog)(x) = f(g(x))

= f(x^1/3)

= 8(x^1/3)³

= 8x

Given as f: R → R; f(x) = x² + 8 and g: R → R; g(x) = 3x³ + 1.

Consider that (fog)(2) = f(g(2)) 

= f(3 × 2³ + 1) 

= f(3 × 8 + 1)

= f(25)

= 25² + 8

= 633

(gof)(1) = g(f(1)) 

= g(1² + 8) 

= g(9) 

= 3 × 9³ + 1 

= 2188