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Given as

F(x) = (x – a)²(x – b)²

Let us find the f(a + b).

Now, we have

f (a + b) = (a + b – a)² (a + b – b)²

f (a + b) = (b)² (a)²

∴ f (a + b) = a²b²

Given as

f (x) = 1 / (1 – x)

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

Firstly, let us solve for the f {f (x)}.

f {f (x)} = f {1/(1 – x)}

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

= 1/(-x/(1 – x))

= (1 – x)/-x

= (x – 1)/x

∴ f {f (x)} = (x – 1)/x

Then, we shall solve for the f [f {f (x)}]

f [f {f (x)}] = f [(x-1)/x]
= 1/[1 – (x-1)/x]

= 1/[(x – (x-1))/x]

= 1/[(x – x + 1)/x]

= 1/(1/x)

∴ f [f {f (x)}] = x

Thus proved.

Given as

y = f (x) = (ax – b) / (bx – a) ⇒ f (y) = (ay – b) / (by – a)

Let us prove that the x = f (y).

Now, we have,

y = (ax – b) / (bx – a)

On cross-multiplying,

y(bx – a) = ax – b

bxy – ay = ax – b

bxy – ax = ay – b

x(by – a) = ay – b

x = (ay – b) / (by – a) = f (y)

∴ x = f (y)

Thus proved.

f = {(1, 1), (2, 4), (3, 4), (4, 3)}

g = {(1, 1), (3, 27), (4, 64)}

f(1) = 1, g(1) = 1

f(2) = 4, g(3) = 27

f(3) = 4, g(4) = 64

f(4) = 3

(gof) (x) = g(f(x))

(gof) (1) = g(f(1)) = g(1) = 1

(gof) (2) = g(f(2)) = g(4) = 64

(gof) (3) = g(f(3)) = g(4) = 64

(gof) (4) = g(f(4)) = g(3) = 27

∴ gof = {(1, 1), (2, 64), (3, 64), (4, 27)}

Answer : (b) \(\sqrt{x}\)

Given f (x) = sin² x and g(f (x)) = | sin x | 

⇒ g(sin²x) = | sin x | 

⇒ g(sin² x) = \(\sqrt{sin^2 x}\) 

∴  g (x) = \(\sqrt{x}\)

f(x) = 3x⁴ – 5x² + 7

∴ f(x – 1) = 3(x – 1)⁴ – 5(x – 1)² + 7

= 3(x⁴ – ⁴C₁ x³ + ⁴C₂ x² – ⁴C₃ x + ⁴C₄ ) – 5(x² – 2x + 1) + 7

= 3(x⁴ – 4x³ + 6x² – 4x + 1) – 5(x² – 2x + 1) + 7

= 3x⁴ – 12x³ + 18x² – 12x + 3 – 5x² + 10x – 5 + 7

= 3x⁴ – 12x³ + 13x² – 2x + 5

f(x) = 3x + a, f(1) = 7

∴ 3(1) + a = 7

∴ a = 7 – 3 = 4

∴ f(x) = 3x + 4

∴ f(4) = 3(4) + 4 = 12 + 4 = 16

 f(x) = \(\sqrt{x-x^2} + \sqrt {5-x}\)

x – x² ≥ 0 

∴ x² – x ≤ 0 

∴ x(x – 1) ≤ 0 

∴ 0 ≤ x ≤ 1 …..(i) 

5 – x ≥ 0 

∴ x ≤ 5 …..(ii) 

Intersection of intervals given in (i) and (ii) 

gives 

Solution set = [0, 1] 

∴ Domain of f = [0, 1]

Given,

f(x) = x + 1 and 

g(x) = 2x – 3 

Clearly,

Both f(x) and g(x) exist for all real values of x. 

Hence,

Domain of f = Domain of g = R 

Range of f = Range of g = R 

i. f + g 

We know,

(f + g)(x) = f(x) + g(x) 

⇒ (f + g)(x) = x + 1 + 2x – 3 

∴ (f + g)(x) = 3x – 2 

Domain of f + g = Domain of f ∩ Domain of g 

⇒ Domain of f + g = R ∩ R 

∴ Domain of f + g = R 

Thus,

f + g : R → R is given by 

(f + g)(x) = 3x – 2 

ii. f – g 

We know 

(f – g)(x) = f(x) – g(x) 

⇒ (f – g)(x) = x + 1 – (2x – 3) 

⇒ (f – g)(x) = x + 1 – 2x + 3 

∴ (f – g)(x) = –x + 4 

Domain of f – g = Domain of f ∩ Domain of g 

⇒ Domain of f – g = R ∩ R 

∴ Domain of f – g = R 

Thus, 

f – g : R → R is given by 

(f – g)(x) = –x + 4

iii. \(\frac{f}{g}\) 

We know,

(\(\frac{f}{g}\))(x) = \(\frac{f(x)}{g(x)}\)

∴ (\(\frac{f}{g}\))(x) = \(\frac{x+1}{2x-3}\) 

Clearly,

(\(\frac{f}{g}\))(x) is defined for all real values of x,

Except for the case when 2x – 3 = 0 or x = \(\frac{3}{2}\).

When  x = \(\frac{3}{2}\),(\(\frac{f}{g}\))(x) will be undefined as the division result will be indeterminate. 

Thus,

Domain of \(\frac{f}{g}\) = R – \(\{\frac{3}{2}\}\)

Thus, 

\(\frac{f}{g}\) : R – \(\{\frac{3}{2}\}\)  → R is given by

(\(\frac{f}{g}\))(x) = \(\frac{x+1}{2x-3}\) 

Given,

f(x) = x² – 3x + 4.

We need to find x satisfying 

f(x) = f(2x + 1). 

We have,

f(2x + 1) = (2x + 1)² – 3(2x + 1) + 4 

⇒ f(2x + 1) = (2x)² + 2(2x)(1) + 1² – 6x – 3 + 4 

⇒ f(2x + 1) = 4x² + 4x + 1 – 6x + 1 

∴ f(2x + 1) = 4x² – 2x + 2 

Now, 

f(x) = f(2x + 1) 

⇒ x² – 3x + 4 = 4x² – 2x + 2 

⇒ 3x² + x – 2 = 0 

⇒ 3x² + 3x – 2x – 2 = 0 

⇒ 3x(x + 1) – 2(x + 1) = 0 

⇒ (x + 1)(3x – 2) = 0 

⇒ x + 1 = 0 or 3x – 2 = 0 

⇒ x = –1 or 3x = 2 

∴ x = –1 or \(\frac{2}{3}\)

Thus,

The required values of x are –1 and \(\frac{2}{3}\).

Given,

f(x) = {(x²,when x<0),(x, when 0≤x≤1),(1/x, when x>1)

i. f(\(\frac{1}{2}\))

When 0 ≤ x ≤ 1, 

f(x) = x

∴  f(\(\frac{1}{2}\)) = \(\frac{1}{2}\)

ii. f(–2) 

When x < 0, 

f(x) = x² 

⇒ f(–2) = (–2)² 

∴ f(–2) = 4

iii. f(1) 

When x ≥ 1,

f(x) = \(\frac{1}{x}\)

⇒ f(1) = \(\frac{1}{1}\)

∴ f(1) = 1

iv. f(\(\sqrt 3\))

We have,

\(\sqrt 3\) ≈ 1.732 > 1

When x ≥ 1,

f(x) = \(\frac{1}{x}\)

∴ f(\(\sqrt 3\)) = \(\frac{1}{\sqrt 3}\)

v. f(\(\sqrt {-3}\))

We know, 

\(\sqrt {-3}\) is not a real number and the function f(x) is defined only when x ∈ R.

Thus,

 f(\(\sqrt {-3}\)) does not exist.

g(x) = \(\frac{x+4}{x-2}\) 

Function g is defined everywhere except at x = 2.

∴ Domain of g = R – {2}

Let y = g(x) = \(\frac{x+4}{x-2}\) 

∴ (x – 2) y = x + 4

∴ x(y – 1) = 4 + 2y

∴ For every y, we can find x, except for y = 1.

∴ y = 1 ∉ range of function g

∴ Range of g = R – {1}

2[2x – 5] – 1 = 7 

∴ [2x – 5] = 7+1/2 = 4

∴ [2x] – 5 = 4 

∴ [2x] = 9 

∴ 9 ≤ 2x < 10

∴ 9/2 ≤ x < 5

∴ Solution set =[9/2, 5]

[x – 2] + [x + 2] + {x} = 0 

∴ [x] – 2 + [x] + 2 + {x} = 0 

∴ [x] + x = 0 …..[{x} + [x] = x] 

∴ x = 0

[x/2] + [x/3] = 5x/6

L.H.S. = an integer 

R.H.S. = an integer 

∴ x = 6k, where k is an integer

-2 < [x] ≤ 7 

∴ -2 < x < 8 

∴ Solution set = (-2, 8)

g(x) = \(\frac{18-2x^2}7\) 

g(x) = 0

∴ 18 – 2x² = 0

∴ x² = 9

∴ x = ±3

Given f(x) = 2x + 1

⇒ f = {(1, 2(1) + 1), (2, 2(2) + 1), (3, 2(3) + 1), (4, 2(4) + 1)}

= {(1, 3), (2, 5), (3, 7), (4, 9)}

Also given g(x) = x² − 2

⇒ g = {(3, 32 − 2), (5, 52 − 2), (7, 72 − 2), (9, 92 − 2)}

= {(3, 7), (5, 23), (7, 47), (9, 79)}

So, f and g are bijections and,

Hence, f⁻¹: B → A and g⁻¹: C → B exist.

Therefore, f⁻¹= {(3, 1), (5, 2), (7, 3), (9, 4)} 

And g⁻¹= {(7, 3), (23, 5), (47, 7), (79, 9)}

Now, (f⁻¹og⁻¹): C → A

f⁻¹og⁻¹ = {(7, 1), (23, 2), (47, 3), (79, 4)} ...(1)

Also, f: A → B and g: B → C,

⇒ gof: A → C, (gof)⁻¹ : C → A

Therefore, f⁻¹og⁻¹ and (gof)⁻¹ have same domains.

(gof)(x) = g(f(x))

=g(2x + 1)

=(2x +1 )² − 2

⇒ (gof)(x) = 4x² + 4x + 1 − 2

⇒ (gof)(x) = 4x² + 4x − 1

Now, (gof)(1) = g(f(1)) 

= 4 + 4 − 1 

= 7,

(gof)(2) = g(f(2))

= 4 + 4 – 1 = 23,

(gof)(3) = g(f(3))

= 4 + 4 – 1 = 47 and 

(gof)(4) = g(f(4))

= 4 + 4 − 1 = 79

Therefore, gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

⇒ (gof) – 1 = {(7, 1), (23, 2), (47, 3), (79, 4)} …(2)

From equations (1) and (2), we get:

(gof)⁻¹ = f⁻¹og⁻¹

g(x) = 6x² + x – 2

g(x) = 0

∴ 6x + x – 2 = 0

∴ (2x – 1) (3x + 2) = 0

∴ 2x – 1 = 0 or 3x + 2 = 0

∴ x = 1/2 or x = -2/3

|x² – 9| + |x² – 4| = 5 

∴ |(x – 3) (x + 3)| + |(x – 2) ( x + 2)| = 5 ………(i) 

Case I: x < -3 

Also, x < -2, x < 2, x < 3

∴ (x – 3) (x + 3) > 0 and (x – 2) (x + 2) > 0 

Equation (i) reduces to  x² – 9 + x² – 4 = 5 

∴ 2x² = 18 

∴ x = -3 or 3 (both rejected as x < -3)

Case II: -3 ≤ x < -2 

As x < -2, x < 3 

∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0 

Equation (i) reduces to -(x² – 9) + x² – 4 = 5 

∴ 5 = 5 (true) 

-3 ≤ x < -2 is a solution ….(ii)

Case III: -2 ≤ x < 2 As x > -3, x < 3 

∴ (x – 3) (x + 3) < 0, 

(x – 2) (x + 2) < 0 

Equation (i) reduces to  9 – x² + 4 – x2 = 5 

∴ 2x² = 13 – 5 

∴ x² = 4 

∴ x = -2 is a solution …..(iii)

Case IV: 2 ≤ x < 3 As x > -3, x > -2 

∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0 

Equation (i) reduces to  9 – x² + x² – 4 = 5 

∴ 5 = 5 (true) 

∴ 2 ≤ x < 3 is a solution ……(iv)

Case V: 3 ≤ x As x > -3, x > -2, x > 2 

∴ (x + 3) (x – 3) > 0, (x – 2) (x + 2) > 0 

Equation (i) reduces to x² – 9 + x² – 4 = 5 

∴ 2x = 18 

∴ x = 9 

∴ x = 3 …..(v) 

(x = -3 rejected as x ≥ 3) From (ii), (iii), (iv), (v), we get 

∴ Solution set = [-3, -2] ∪ [2, 3]

(i) |x| ≤ 3 The solution set of |x| ≤ a is -a ≤ x ≤ a

∴ The required solution is -3 ≤ x ≤ 3

∴ The solution set is [-3, 3]

(ii) 2|x| = 5

∴ |x| = 5/2

∴ x = ± 5/2

(iii) [x + [x + [x]]] = 9 

∴ [x + [x] + [x] ] = 9 …….[[x + n] = [x] + n, if n is an integer]

∴ [x + 2[x]] = 9

∴ [x] + 2[x] = 9 …..[[2[x] is an integer]]

∴ [x] = 3

∴ x ∈ [3, 4)

One – One Function: – A function f: A → B is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Bijection Function: – A function f: A → B is said to be a bijection function if it is one – one as well as onto function.

Now, f: N → N given by f(x) = x³

Check for Injectivity:

Let x,y be elements belongs to N i.e x, y ∈ N such that

So, from definition

⇒ f(x) = f(y)

⇒ x³ = y³

⇒ x³ – y³ = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to N i.e y ∈ N be arbitrary, then

⇒ f(x) = y

⇒ x³ = y

⇒ x = 3√y 

⇒ 3√y not belongs to X for non–perfect square value of y.

Since f attain only cubic number like 1,8,27….

Therefore no non – perfect square value of y has a pre–image in domain N.

Hence, f is not Onto function.

Thus, Not Bijective also.

One – One Function: – A function f: A → B is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Bijection Function: – A function f: A → B is said to be a bijection function if it is one – one as well as onto function.

Now, f : N → N given by f(x) = x³

Check for Injectivity:

Let x,y be elements belongs to N i.e x, y ∈ N such that

⇒ f(x) = f(y)

⇒ x³ = y³

⇒ x³ – y³ = 0

⇒ (x – y)(x² + y² + xy) = 0

As x, y ∈ N therefore x² + y² + xy >0

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to N i.e y ∈ N be arbitrary, then

⇒ f(x) = y

⇒ x³ = y

⇒ x = 3√y

⇒ 3√y not belongs to N for non–perfect cube value of y.

Since f attain only cubic number like 1,8,27….,

Therefore no non – perfect cubic values of y in N (co – domain) has a pre–image in domain N.

Hence, f is not onto function

Thus, Not Bijective also

One – One Function: – A function f: A → B is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Bijection Function: – A function f: A → B is said to be a bijection function if it is one – one as well as onto function.

Now, f : Z → Z given by f(x) = x²

Check for Injectivity:

Let x₁, – x₁ be elements belongs to Z i.e x₁, - x₁ ∈ Z such that

So, from definition

⇒ x₁ ≠ – x₁

⇒ (x₁)² = ( – x₁)²

⇒ f(x₁)² = f( – x₁)²

Hence f is not One – One function

Check for Surjectivity:

Let y be element belongs to Z i.e y ∈ Z be arbitrary, then

⇒ f(x) = y

⇒ x² = y

⇒ x = ± √y 

⇒ √y not belongs to Z for non–perfect square value of y.

Therefore no non – perfect square value of y has a pre–image in domain Z.

Hence, f is not Onto function.

Thus, Not Bijective also.

One – One Function: – A function f: A → B is said to be a one – one functions or an injection if different elements of A have different images in B.

So, f: A → B is One – One function

⇔ a≠b

⇒ f(a)≠f(b) for all a, b ∈ A

⇔ f(a) = f(b)

⇒ a = b for all a, b ∈ A

Onto Function: – A function f: A → B is said to be a onto function or surjection if every element of A i.e, if f(A) = B or range of f is the co – domain of f.

So, f: A → B is Surjection iff for each b ∈ B, there exists a ∈ B such that f(a) = b

Bijection Function: – A function f: A → B is said to be a bijection function if it is one – one as well as onto function.

Now, f: N → N given by f(x) = x²

Check for Injectivity:

Let x,y be elements belongs to N i.e x, y ∈ N such that

So, from definition

⇒ f(x) = f(y)

⇒ x² = y²

⇒ x² – y² = 0

⇒ (x – y)(x + y) = 0

As x, y ∈ N therefore x + y > 0

⇒ x – y = 0

⇒ x = y

Hence f is One – One function

Check for Surjectivity:

Let y be element belongs to N i.e y ∈ N be arbitrary, then

⇒ f(x) = y

⇒ x² = y

⇒ x = √y 

⇒ √y not belongs to N for non–perfect square value of y.

Therefore no non – perfect square value of y has a pre–image in domain N.

Hence, f is not Onto function.

Thus, Not Bijective also.