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(i) f(x) = 3

∴ 5 – x = 3

∴ x = 5 – 3 = 2

(ii) f(x) = 5

∴ 5 – x = 5

∴ x = 0

Given, 

A = {6, 10, 11, 15, 12} 

f : A → N : f(n) is the highest prime factor of n 

(1) When n = 6, the highest prime factor of 6 is 3. 

Hence, f(6) = 3. 

(2) When n = 10, the highest prime factor of 10 is 5. 

Hence, f(10) = 5. 

(3) When n = 11, the highest prime factor of 11 is 11 as 11 itself is a prime number. Hence, f(11) = 11. 

(4) When n = 15, the highest prime factor of 15 is 5. 

Hence, f(15) = 5. 

(5) When n = 12, the highest prime factor of 12 is 3. 

Hence, f(12) = 3. 

Hence range of f is { 3,5,11}

Given,

f : R → R and 

f(x) = x² + 1. 

We need to find,

f^-1{17} and f^-1{–3}. 

Let f^-1{17} = x 

⇒ f(x) = 17

⇒ x² + 1 = 17 

⇒ x² – 16 = 0 

⇒ (x – 4)(x + 4) = 0 

∴ x = ±4 

Clearly, 

Both –4 and 4 are elements of the domain R. 

Thus, 

f^-1{17} = {–4, 4} 

Now, 

Let f^-1{–3} = x 

⇒ f(x) = –3 

⇒ x² + 1 = –3 

⇒ x² = –4 

However, 

The domain of f is R and for every real number x, 

The value of x² is non-negative. 

Hence, 

There exists no real x for which x² = –4. 

Thus, 

f^-1{–3} = ∅

For a relation to be a function each element of 1^st set should have different image in the second set(Range) i) 

(i) f = {( - 1, 2), (1, 8), (2, 11), (3, 14)} 

Here, each of the first set element has different image in second set. 

∴f is a function whose domain = { - 1, 1, 2, 3} and range (f) = {2, 8, 11, 14} 

(ii) g = {(1, 1), (1, - 1), (4, 2), (9, 3),(16, 4)} 

Here, some of the first set element has same image in second set. 

∴ g is not a function. 

(iii) h = {(a, b), (b, c), (c, b), (d, c)} 

Here, each of the first set element has different image in second set. 

∴h is a function whose domain = {a, b, c, d} and range (h) = {b, c} 

(range is the intersection set of the elements of the second set elements.)

(i) f(x) = 1/x

As we know that, f(x) is defined for all real values of x, except for the case when x = 0.

∴ Domain of f = R – {0}

(ii) f(x) = 1/(x - 7)

As we know that, f (x) is defined for all real values of x, except for the case when x – 7 = 0 or x = 7.

∴ Domain of f = R – {7}

(iii) f(x) = (3x - 2)/(x + 1)

As we know that, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = –1.

∴ Domain of f = R – {–1}

(iv) f(x) = (2x + 1)/(x² - 9)

As we know that, f (x) is defined for all real values of x, except for the case when x² – 9 = 0.

x² – 9 = 0

x² – 3² = 0

(x + 3)(x – 3) = 0

x + 3 = 0 or x – 3 = 0

x = ± 3

∴ Domain of f = R – {–3, 3}

(v) f(x) = (x² + 2x + 1)/(x² - 8x + 12)

As we know that, f(x) is defined for all real values of x, except for the case when x² – 8x + 12 = 0.

x² – 8x + 12 = 0

x² – 2x – 6x + 12 = 0

x(x – 2) – 6(x – 2) = 0

(x – 2)(x – 6) = 0

x – 2 = 0 or x – 6 = 0

x = 2 or 6

∴ Domain of f = R – {2, 6}

(i) Here, we have f(x): R → R and g(x): R → R

(a) f + g

As we know, (f + g)(x) = f(x) + g(x)

(f + g)(x) = x³ + 1 + x + 1

= x³ + x + 2

Therefore, (f + g) (x): R → R

∴ f + g: R → R is given by (f + g) (x) = x³ + x + 2

(b) f – g

As we know, (f – g) (x) = f(x) – g(x)

(f – g) (x) = x³ + 1 – (x + 1)

= x³ + 1 – x – 1

= x³ – x

Therefore, (f – g) (x): R → R

∴ f – g: R → R is given by (f – g) (x) = x³ – x

(c) cf (c ∈ R, c ≠ 0)

As we know, (cf) (x) = c × f(x)

(cf)(x) = c(x³ + 1)

= cx³ + c

Therefore, (cf) (x) : R → R

∴ cf: R → R is given by (cf) (x) = cx³ + c

(d) fg

As we know, (fg) (x) = f(x) g(x)

(fg) (x) = (x³ + 1) (x + 1)

= (x + 1) (x² – x + 1) (x + 1)

= (x + 1)² (x² – x + 1)

Therefore, (fg) (x): R → R

∴ fg: R → R is given by (fg) (x) = (x + 1)²(x² – x + 1)

(e) 1/f

As we know, (1/f) (x) = 1/f (x) 

1/f (x) = 1 / (x³ + 1)
Observe that 1/f(x) is undefined when f(x) = 0 or when x = – 1.

Therefore, 1/f: R – {–1} → R is given by 1/f (x) = 1 / (x³ + 1) 

(f) f/g

As we know, (f/g) (x) = f(x)/g(x) 

(f/g) (x) = (x³ + 1) / (x + 1)
Observe that (x³ + 1) / (x + 1) is undefined when g(x) = 0 or when x = –1.

By using x³ + 1 = (x + 1) (x² – x + 1), we have

(f/g) (x) = [(x + 1) (x² –  x + 1)/(x + 1)]

= x² – x + 1

∴ f/g: R – {–1} → R is given by (f/g) (x) = x² – x + 1

(ii) Now, we have f(x): [1, ∞) → R⁺ and g(x): [–1, ∞) → R⁺ as real square root is defined only for non-negative numbers.

(a) f + g

As we know, (f + g) (x) = f(x) + g(x)

(f + g) (x) = √(x - 1) + √(x + 1)

Domain of (f + g) = Domain of f ∩ Domain of g

Domain of (f + g) = [1, ∞) ∩ [–1, ∞)

Domain of (f + g) = [1, ∞)

∴ f + g: [1, ∞) → R is given by (f + g) (x) = √(x - 1) + √(x + 1)

(b) f – g

As we know, (f – g) (x) = f(x) – g(x)

(f - g) (x) = √(x - 1) – √(x + 1)

Domain of (f – g) = Domain of f ∩ Domain of g

Domain of (f – g) = [1, ∞) ∩ [–1, ∞)

Domain of (f – g) = [1, ∞)

∴ f – g: [1, ∞) → R is given by (f - g) (x) = √(x - 1) – √(x + 1)

(c) cf (c ∈ R, c ≠ 0)

As we know, (cf) (x) = c × f(x)

(cf) (x) = c√(x - 1)

Domain of (cf) = Domain of f

Domain of (cf) = [1, ∞)

∴ cf: [1, ∞) → R is given by (cf) (x) = c√(x - 1)

(d) fg

As we know, (fg) (x) = f(x) g(x)

(fg) (x) = √(x - 1) √(x + 1)

= √(x² - 1)

Domain of (fg) = Domain of f ∩ Domain of g

Domain of (fg) = [1, ∞) ∩ [–1, ∞)

Domain of (fg) = [1, ∞)

∴ fg: [1, ∞) → R is given by (fg) (x) = √(x² - 1)

(e) 1/f

As we know, (1/f) (x) = 1/f(x) 

(1/f) (x) = 1/√(x - 1)
Domain of (1/f) = Domain of f

Domain of (1/f) = [1, ∞)

Observe that 1/√(x - 1) is also undefined when x – 1 = 0 or x = 1.

∴ 1/f: (1, ∞) → R is given by (1/f) (x) = 1/√(x - 1)

(f) f/g

As we know, (f/g) (x) = f(x)/g(x) 

(f/g) (x) = √(x - 1)/√(x + 1)

(f/g) (x) = √[(x - 1)/(x + 1)]

Domain of (f/g) = Domain of f ∩ Domain of g

Domain of (f/g) = [1, ∞) ∩ [–1, ∞)

Domain of (f/g) = [1, ∞)

Thus, f/g: [1, ∞) → R is given by (f/g) (x) = √[(x - 1)/(x + 1)]

Option : (B) 

A function is said to exist when we get a unique value of y for any value of x.. 

If we get 2 values of y for any value of x, then it is not a function.. 

Therefore,

Option B is correct . 

NOTE : To check if a given curve is a function or not, draw the curve and then draw a line parallel to y-axis.. If it intersects the curve at only one point, then it is a function, else not..

Option : (B)

Since A has 3 elements and B has 2..

then number of functions from A to B

= 2\(\times\)2\(\times\)2

= 2³

= 8

Given: f(x) = \(\sqrt{\frac{x-5}{3-x}}\)

Need to find: Where the functions are defined. 

The condition for the function to be defined,

3 - x > 0

⇒ x < 3

So, the domain of the function is the set of all the real numbers lesser than 3. 

The domain of the function, D_f(x) = (-∞, 3). 

The condition for the range of the function to be defined,

x - 5 ≥ 0 _& 3 - x > 0

⇒ x ≥ 5 _& x < 3

Both the conditions can’t be satisfied simultaneously. That means there is no range for the function f(x).