1015 + Interview Questions in FUNCTIONS IN GENERAL AWARENESS Page 21 InterviewSolution

1001.	Let `f(x)`be real valued and differentiable function on `R`such that `f(x+y)=(f(x)+f(y))/(1-f(x)dotf(y))``f(0)`is equalsa. b. c. d. none of theseA. odd functionB. even functionC. odd and even function simultaneouslyD. neither even nor odd
Answer» Correct Answer - A Putting x = y= 0, we get `f(0)=(f(0)+f(0))/(1-[f(0)]^(2))` `rArr" "f(0)[f^(2)(0)+1]=0rArr f(0)=0( "since "f^(2)(0) ne-1).` Now putting `y=-x`, we get `f(0)=(f(x)+f(-x))/(1-f(x).f(-x))` `rArr" "f(x)+f(-x)=0` `rArr" "f(-x)=-f(x)rArr f(x)` is an odd function.

Discussion

1002.	Find the domain and range of each of the following real valued functions : f(x)= (ax+b)/(bx-a)f(x) = \(\frac{ax+b}{bx-a}\)
Answer» f(x) = (ax+b)/(bx-a) Clearly, f(x) is defined for all real values of x, Except for the case when bx – a = 0 or x = \(\frac{a}{b}\). When x = \(\frac{a}{b}\), f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – \(\{\frac{a}{b}\}\) Let f(x) = y ⇒ ax + b = y(bx – a) ⇒ ax + b = bxy – ay ⇒ ax – bxy = –ay – b ⇒ x(a – by) = –(ay + b) ∴ x = \(-\frac{(ay+b)}{a-by}\) Clearly, when a – by = 0 or y = \(\frac{a}{b}\) , x will be undefined as the division result will be indeterminate. Hence, f(x) cannot take the value \(\frac{a}{b}\). Thus, Range of f = R – \(\{\frac{a}{b}\}\)

1002.

Find the domain and range of each of the following real valued functions : f(x)= (ax+b)/(bx-a)f(x) = \(\frac{ax+b}{bx-a}\)

Answer»

f(x) = (ax+b)/(bx-a)

Clearly,

f(x) is defined for all real values of x,

Except for the case when bx – a = 0 or x = \(\frac{a}{b}\).

When x = \(\frac{a}{b}\),

f(x) will be undefined as the division result will be indeterminate.

Thus,

Domain of f = R – \(\{\frac{a}{b}\}\)

Let f(x) = y

⇒ ax + b = y(bx – a)

⇒ ax + b = bxy – ay

⇒ ax – bxy = –ay – b

⇒ x(a – by) = –(ay + b)

∴ x = \(-\frac{(ay+b)}{a-by}\)

Clearly,

when a – by = 0 or y = \(\frac{a}{b}\) ,

x will be undefined as the division result will be indeterminate.

Hence,

f(x) cannot take the value \(\frac{a}{b}\).

Thus,

Range of f = R – \(\{\frac{a}{b}\}\)

Discussion

1003.	find the domain and range of the function ` f : R to R : f (x) =x^(2) +1`
Answer» Correct Answer - dom (f ) =R and range ( f) `={ y in R : y ge 1}` Dom (f )= R. Also `y=x^(2) +1 rArr =sqrt(y-1)` , x is defined when `y- ge 0, i.e., y ge 1` `:.` range `(f) {y in R : y ge 1}`

Discussion

1004.	Let `g(x)=1+x-[x] and f(x)={(-1",",x lt 0),(0",",x=0),(1",",x gt 0):}`, then for all `x, f[g(x)]` is equal toA. xB. 1C. f(x)D. g(x)
Answer» Correct Answer - B `g(x)=1+x-[x]` is grater than 1 since `x[x] gt 0` `f[g(x)]," since " f(x) =1 " for all " x gt 0`

Discussion

1005.	The range of `f(x) =a^(x) ," where " a gt 0 ` isA. `]-oo , 0]`B. `]-oo , 0)`C. `[0,oo)`D. `(0,oo)`
Answer» Correct Answer - D Clearly `a^(x) gt 0` whetever many be the value of x `:. ` range (f) = `(0,oo)`

Discussion

1006.	If `f(x)=log((1+x)/(1-x))a n dg(x)=((3x+x^3)/(1+3x^2))`, then `f(g(x))`is equal to`f(3x)`(b) `{f(x)}^3`(c) `3f(x)`(d) `-f(x)`
Answer» `f(g(x)) = f((3x+x^3)/(1+3x^2))` `=log((1+(3x+x^3)/(1+3x^2))/(1-(3x+x^3)/(1+3x^2)))` `=log((1+3x^2+3x+x^3)/(1+3x^2-3x-x^3))` `=log((1+x)^3/(1-x)^3)` `=log((1+x)/(1-x))^3` `=3log((1+x)/(1-x))` `=3f(x)` `:. f(g(x)) = 3f(x)`

Discussion

1007.	Find the domain of each of the following real valued functions of real variable :i. f(x) = 1/xii. f(x) = \(\frac{1}{x-7}\)iii. f(x) = \(\frac{3x-2}{x+1}\)iv. f(x) = \(\frac{2x+1}{x^2+9}\) v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\)
Answer» i. f(x) = 1/x Clearly, f(x) is defined for all real values of x, Except for the case when x = 0. When x = 0, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {0} ii. f(x) = \(\frac{1}{x-7}\) Clearly, f(x) is defined for all real values of x, Except for the case when x – 7 = 0 or x = 7. When x = 7, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {7} iii. f(x) = \(\frac{3x-2}{x+1}\) Clearly, f(x) is defined for all real values of x, Except for the case when x + 1 = 0 or x = –1. When x = –1, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {–1} iv. f(x) = \(\frac{2x+1}{x^2+9}\) Clearly, 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 When x = ±3, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {–3, 3} v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\) Clearly, 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 When x = 2 or 6, f(x) will be undefined as the division result will be indeterminate. Thus, Domain of f = R – {2, 6}

1007.

Find the domain of each of the following real valued functions of real variable :i. f(x) = 1/xii. f(x) = \(\frac{1}{x-7}\)iii. f(x) = \(\frac{3x-2}{x+1}\)iv. f(x) = \(\frac{2x+1}{x^2+9}\) v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\)

Answer»

i. f(x) = 1/x

Clearly,

f(x) is defined for all real values of x,

Except for the case when x = 0.

When x = 0,

f(x) will be undefined as the division result will be indeterminate.

Thus,

Domain of f = R – {0}

ii. f(x) = \(\frac{1}{x-7}\)

Clearly,

f(x) is defined for all real values of x,

Except for the case when x – 7 = 0 or x = 7.

When x = 7,

f(x) will be undefined as the division result will be indeterminate.

Thus,

Domain of f = R – {7}

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

Clearly,

f(x) is defined for all real values of x,

Except for the case when x + 1 = 0 or x = –1.

When x = –1,

f(x) will be undefined as the division result will be indeterminate.

Thus,

Domain of f = R – {–1}

iv. f(x) = \(\frac{2x+1}{x^2+9}\)

Clearly,

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

When x = ±3,

f(x) will be undefined as the division result will be indeterminate.

Thus,

Domain of f = R – {–3, 3}

v. f(x) = \(\frac{x^2+2x+1}{x^2-8x+12}\)

Clearly,

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

When x = 2 or 6,

f(x) will be undefined as the division result will be indeterminate.

Thus,

Domain of f = R – {2, 6}

Discussion

1008.	Find the domain of each of the following real valued functions of real variable:(i) f (x) = √(x - 2)(ii) f (x) = 1/(√(x2 - 1))(iii) f (x) = √(9 - x2)(iv) f (x) = √(x - 2)/(3 - x)
Answer» (i) As we know that the square of a real number is never negative. Now, f(x) takes real values only when x – 2 ≥ 0 x ≥ 2 ∴ x ∈ [2, ∞) ∴ The domain (f) = [2, ∞) (ii) As we know that the square of a real number is never negative. Now, f(x) takes real values only when x² – 1 ≥ 0 x² – 1² ≥ 0 (x + 1) (x – 1) ≥ 0 x ≤ –1 or x ≥ 1 ∴ x ∈ (–∞, –1] ∪ [1, ∞) Here, in addition f(x) is also undefined when x² – 1 = 0 because denominator will be zero and the result will be indeterminate. x² – 1 = 0 ⇒ x = ± 1 Therefore, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1} x ∈ (–∞, –1) ∪ (1, ∞) ∴ The domain (f) = (–∞, –1) ∪ (1, ∞) (iii) As we know that the square of a real number is never negative. Now, f(x) takes real values only when 9 – x² ≥ 0 9 ≥ x² x² ≤ 9 x² – 9 ≤ 0 x² – 3² ≤ 0 (x + 3)(x – 3) ≤ 0 x ≥ –3 and x ≤ 3 x ∈ [–3, 3] ∴ The domain (f) = [–3, 3] (iv) As we know the square root of a real number is never negative. Now, f(x) takes real values only when x – 2 and 3 – x are both positive and negative. (a) Here, both x – 2 and 3 – x are positive x – 2 ≥ 0 x ≥ 2 3 – x ≥ 0 x ≤ 3 Thus, x ≥ 2 and x ≤ 3 ∴ x ∈ [2, 3] (b) Here, both x – 2 and 3 – x are negative x – 2 ≤ 0 x ≤ 2 3 – x ≤ 0 x ≥ 3 Thus, x ≤ 2 and x ≥ 3 However, the intersection of these sets is null set. Hence, this case is not possible. Thus, x ∈ [2, 3] – {3} x ∈ [2, 3] ∴ The domain (f) = [2, 3]

1008.

Find the domain of each of the following real valued functions of real variable:(i) f (x) = √(x - 2)(ii) f (x) = 1/(√(x2 - 1))(iii) f (x) = √(9 - x2)(iv) f (x) = √(x - 2)/(3 - x)

Answer»

(i) As we know that the square of a real number is never negative.

Now, f(x) takes real values only when x – 2 ≥ 0

x ≥ 2

∴ x ∈ [2, ∞)

∴ The domain (f) = [2, ∞)

(ii) As we know that the square of a real number is never negative.

Now, f(x) takes real values only when x² – 1 ≥ 0

x² – 1² ≥ 0

(x + 1) (x – 1) ≥ 0

x ≤ –1 or x ≥ 1

∴ x ∈ (–∞, –1] ∪ [1, ∞)

Here, in addition f(x) is also undefined when x² – 1 = 0 because denominator will be zero and the result will be indeterminate.

x² – 1 = 0 ⇒ x = ± 1

Therefore, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}

x ∈ (–∞, –1) ∪ (1, ∞)

∴ The domain (f) = (–∞, –1) ∪ (1, ∞)

(iii) As we know that the square of a real number is never negative.

Now, f(x) takes real values only when 9 – x² ≥ 0

9 ≥ x²

x² ≤ 9

x² – 9 ≤ 0

x² – 3² ≤ 0

(x + 3)(x – 3) ≤ 0

x ≥ –3 and x ≤ 3

x ∈ [–3, 3]

∴ The domain (f) = [–3, 3]

(iv) As we know the square root of a real number is never negative.

Now, f(x) takes real values only when x – 2 and 3 – x are both positive and negative.

(a) Here, both x – 2 and 3 – x are positive

x – 2 ≥ 0

x ≥ 2

3 – x ≥ 0

x ≤ 3

Thus, x ≥ 2 and x ≤ 3

∴ x ∈ [2, 3]

(b) Here, both x – 2 and 3 – x are negative

x – 2 ≤ 0

x ≤ 2

3 – x ≤ 0

x ≥ 3

Thus, x ≤ 2 and x ≥ 3

However, the intersection of these sets is null set. Hence, this case is not possible.

Thus, x ∈ [2, 3] – {3}

x ∈ [2, 3]

∴ The domain (f) = [2, 3]

Discussion

1009.	Let `f(x)=a^(x)(a gt 0)` be written as `f(x)=f_(1)(x)+f_(2)(x), " where " f_(1)(x)` is an function and `f_(2)(x)` is an odd function. Then `f_(1)(x+y)+f_(1)(x-y)` equalsA. `2f_(1)(x+y)f_(2)(x-y)`B. `2f_(1)(x+y)f_(1)(x-y)`C. `2f_(1)(x)f_(2)(y)`D. `2f_(1)(x)f_(1)(y)`
Answer» Correct Answer - D Given, function `f(x)=a^(x),a gt 0` is written as sum of an even and odd function `f_(1)(x) and f_(2)(x)` respectively. Clearly, `f_(1)(x) =(a^(x)+a^(-x))/(2) and f_(2)(x)=(a^(x)-a^(-x))/(2)` So, `f_(1)(x+y)+f_(1)(x-y)` `=(1)/(2)[a^(x+y)+a^(-(x+y))]+(1)/(2)[a^(x-y)+a^(-(x-y))]` `=(1)/(2)[a^(x)a^(y)+(1)/(a^(x)a^(y))+(a^(x))/(a^(y))+(a^(y))/(a^(x))]` `=(1)/(2)[a^(x)(a^(y)+(1)/(a^(y)))+(1)/(a^(x))((1)/(a^(y))+a^(y))]` ` =(1)/(2)(a^(x)+(1)/(a^(x)))(a^(y)+(1)/(a^(y)))` ` =2((a^(x)+a^(-x))/(2))((a^(y)+a^(-y))/(2))= 2f_(1)(x)*f_(1)(y)`

Discussion

1010.	The range of the function `f(x)=(x^2+x+2)/(x^2+x+1),x in R ,`isA. `(1,infty)`B. `(1,11//7)`C. `(1,7//3]`D. `(1,7//5)`
Answer» Correct Answer - C Let `y=f(x)=(x^(2)+x+2)/(x^(2)+x+1), x in R` `therefore y=(x^(2)+x+2)/(x^(2)+x+1)` `y=1+(1)/(x^(2)+x+1) " "[i.e. y gt 1]` ...(i) ` rArr yx^(2)+yx+y=x^(2)+x+2` ` rArr x^(2) (y-1) + x(y-1)+(y-2)=0, AA x in R` Since, x is real, `D ge 0` `rArr (y-1)^(2)-4(y-1)(y-2) ge 0` ` rArr (y-1) {(y-1) -4(y-2)} ge 0` `rArr (y-1)(-3y+7) ge 0` `rArr 1 le y le (7)/(3) " ...(ii)" ` From Eqs. (i) and (ii), Range ` in (1,(7)/(3)]`

Discussion

1011.	Find the domain of each of the following real valued functions of realvariable:`f(x)=sqrt(x-2)`(ii) `f(x)=1/(sqrt(x^2-1))`(iii) `f(x)=sqrt(9-x^2)`(iv) `f(x)=sqrt((x-2)/(3-x))`
Answer» (i) `f(x) = sqrt(x-2)` As, anything under square root can not be negative, `:. x - 2 ge 0 => x ge 2` So, domain of `f(x)` is `x in [2,oo]`. (ii) `f(x) = 1/sqrt(x^2-1)` As, anything under square root can not be negative and denominator can not be zero, `:. x^2-1 gt 0=> (x-1)(x+1) gt 0` So, domain of `f(x)` is `x in (-oo,-1) uu (1,oo)`. (iii) `f(x) = sqrt(9-x^2)` As, anything under square root can not be negative, `:. 9-x^2 ge 0 =>(3-x)(3+x) ge 0 =>(x-3)(x+3) le 0` So, domain of `f(x)` is `x in [-3,3]`. (iv) `f(x) = sqrt((x-2)/(3-x))` As, anything under square root can not be negative and denominator can not be zero, `:. x-2 ge 0 and 3-x gt 0` `=> x ge 2 and x lt 3` So, domain of `f(x)` is `x in [2,3)`.

Discussion

1012.	The domain of definition of the function`f(x)=sqrt(sin^(-1)(2x)+pi/6)`for real-valued `x`isA. `[-(1)/(4),(1)/(2)]`B. `[-(1)/(2),(1)/(2)]`C. `(-(1)/(2),(1)/(9))`D. `[-(1)/(4),(1)/(4)]`
Answer» Correct Answer - A Here, `f(x)=sqrt(sin^(-1)(2x)+(pi)/(6))`, to find domain we must have, `sin^(-1)(2x)+(pi)/(6) ge 0 " "["but "-(pi)/(2) le sin^(-1)theta le (pi)/(2)]` ` -(pi)/(6) le sin^(-1) (2x) le (pi)/(2)` `sin(-(pi)/(6)) le 2x le "sin"(pi)/(2) rArr (-1)/(2) le 2x le (1)/(2)` `(-1)/(4) le x le (1)/(2)` `because x in [(-1)/(4), (1)/(2)]`

Discussion

1013.	Let A = {p, q, r, s} and B = {1, 2, 3}. Which of the following relations from A to B is not a function? i. R1 = {(p, 1), (q, 2), (r, 1), (s, 2)} ii. R2 = {(p, 1), (q, 1), (r, 1), (s, 1)} iii. R3 = {(p, 1), (q, 2), (p, 2), (s, 3)} iv. R4 = {(p, 2), (q, 3), (r, 2), (s, 2)}
Answer» Given, A = {p, q, r, s} and B = {1, 2, 3} i. R₁ = {(p, 1), (q, 2), (r, 1), (s, 2)} Every element of set A has an ordered pair in the relation R₁ and no two ordered pairs have the same first component but different second components. Hence, The given relation R₁ is a function. ii. R₂= {(p, 1), (q, 1), (r, 1), (s, 1)} Every element of set A has an ordered pair in the relation R₂, and no two ordered pairs have the same first component but different second components. Hence, The given relation R₂ is a function. iii. R₃ = {(p, 1), (q, 2), (p, 2), (s, 3)} Every element of set A has an ordered pair in the relation R₃. However, Two ordered pairs (p, 1) and (p, 2) have the same first component but different second components. Hence, The given relation R₃ is not a function. iv. R₄ = {(p, 2), (q, 3), (r, 2), (s, 2)} Every element of set A has an ordered pair in the relation R₄, and no two ordered pairs have the same first component but different second components. Hence, the given relation R₄ is a function.

1013.

Let A = {p, q, r, s} and B = {1, 2, 3}. Which of the following relations from A to B is not a function? i. R1 = {(p, 1), (q, 2), (r, 1), (s, 2)} ii. R2 = {(p, 1), (q, 1), (r, 1), (s, 1)} iii. R3 = {(p, 1), (q, 2), (p, 2), (s, 3)} iv. R4 = {(p, 2), (q, 3), (r, 2), (s, 2)}

Answer»

Given,

A = {p, q, r, s} and

B = {1, 2, 3}

i. R₁ = {(p, 1), (q, 2), (r, 1), (s, 2)}

Every element of set A has an ordered pair in the relation R₁ and no two ordered pairs have the same first component but different second components.

Hence,

The given relation R₁ is a function.

ii. R₂= {(p, 1), (q, 1), (r, 1), (s, 1)}

Every element of set A has an ordered pair in the relation R₂, and no two ordered pairs have the same first component but different second components.

Hence,

The given relation R₂ is a function.

iii. R₃ = {(p, 1), (q, 2), (p, 2), (s, 3)}

Every element of set A has an ordered pair in the relation R₃.

However,

Two ordered pairs (p, 1) and (p, 2) have the same first component but different second components.

Hence,

The given relation R₃ is not a function.

iv. R₄ = {(p, 2), (q, 3), (r, 2), (s, 2)}

Every element of set A has an ordered pair in the relation R₄, and no two ordered pairs have the same first component but different second components.

Hence,

the given relation R₄ is a function.

Discussion

1014.	Let A = {9, 10, 11, 12, 13} and let f : A → Z be a function given by f(n) = the highest prime factor of n. Find the range of f.
Answer» Given, A = {9, 10, 11, 12, 13} f : A → Z such that f(n) = the highest prime factor of n. A is the domain of the function f. Hence, The range is the set of elements f(n) for all n ∈ A. We have, f(9) = highest prime factor of 9 The prime factorization of 9 = 3² Thus, The highest prime factor of 9 is 3. ∴ f(9) = 3 We have, f(10) = highest prime factor of 10 The prime factorization of 10 = 2 × 5 Thus, The highest prime factor of 10 is 5. ∴ f(10) = 5 We have, f(11) = highest prime factor of 11 We know, 11 is a prime number. ∴ f(11) = 11 We have, f(12) = highest prime factor of 12 The prime factorization of 12 = 2² × 3 Thus, The highest prime factor of 12 is 3. ∴ f(12) = 3 We have, f(13) = highest prime factor of 13 We know, 13 is a prime number. ∴ f(13) = 13 Thus, The range of f is {3, 5, 11, 13}.

1014.

Let A = {9, 10, 11, 12, 13} and let f : A → Z be a function given by f(n) = the highest prime factor of n. Find the range of f.

Answer»

Given,

A = {9, 10, 11, 12, 13}

f : A → Z such that

f(n) = the highest prime factor of n.

A is the domain of the function f.

Hence,

The range is the set of elements f(n) for all n ∈ A.

We have,

f(9) = highest prime factor of 9

The prime factorization of 9 = 3²

Thus,

The highest prime factor of 9 is 3.

∴ f(9) = 3

We have,

f(10) = highest prime factor of 10

The prime factorization of 10 = 2 × 5

Thus,

The highest prime factor of 10 is 5.

∴ f(10) = 5

We have,

f(11) = highest prime factor of 11

We know,

11 is a prime number.

∴ f(11) = 11

We have,

f(12) = highest prime factor of 12

The prime factorization of 12 = 2² × 3

Thus,

The highest prime factor of 12 is 3.

∴ f(12) = 3

We have,

f(13) = highest prime factor of 13

We know,

13 is a prime number.

∴ f(13) = 13

Thus,

The range of f is {3, 5, 11, 13}.

Discussion

1015.	The function f is defined by f(x) = {(x2,0≤x≤3)(3x,3≤x≤10)\(f(x) =\begin{cases}x^2, 0≤x≤3\\3x,3≤x≤10\end{cases}\)The relation g is defined by \(g(x) =\begin{cases}x^2, 0≤x≤2\\3x,2≤x≤10\end{cases}\) Show that f is a function and g is not a function.
Answer» Given, f(x) = {(x^2,0≤x≤3)(3x,3≤x≤10) and \(g(x) =\begin{cases}x^2, 0≤x≤2\\3x,2≤x≤10\end{cases}\) Let us first show that f is a function. When 0 ≤ x ≤ 3, f(x) = x². The function x² associates all the numbers 0 ≤ x ≤ 3 to unique numbers in R. Hence, The images of {x ∈ Z: 0 ≤ x ≤ 3} exist and are unique. When 3 ≤ x ≤ 10, f(x) = 3x. The function x² associates all the numbers 3 ≤ x ≤ 10 to unique numbers in R. Hence, The images of {x ∈ Z: 3 ≤ x ≤ 10} exist and are unique. When x = 3, Using the first definition, we have f(3) = 3² = 9 When x = 3, Using the second definition, we have f(3) = 3(3) = 9 Hence, The image of x = 3 is also distinct. Thus, As every element of the domain has an image and no element has more than one image, f is a function. Now, Let us show that g is not a function. When 0 ≤ x ≤ 2, g(x) = x². The function x² associates all the numbers 0 ≤ x ≤ 2 to unique numbers in R. Hence, The images of {x ∈ Z: 0 ≤ x ≤ 2} exist and are unique. When 2 ≤ x ≤ 10, g(x) = 3x. The function x² associates all the numbers 2 ≤ x ≤ 10 to unique numbers in R. Hence, The images of {x ∈ Z: 2 ≤ x ≤ 10} exist and are unique. When x = 2, Using the first definition, we have g(2) = 2² = 4 When x = 2, Using the second definition, we have g(2) = 3(2) = 6 Here, The element 2 of the domain is associated with two elements distinct elements 4 and 6. Thus, g is not a function.

1015.

The function f is defined by f(x) = {(x2,0≤x≤3)(3x,3≤x≤10)\(f(x) =\begin{cases}x^2, 0≤x≤3\\3x,3≤x≤10\end{cases}\)The relation g is defined by \(g(x) =\begin{cases}x^2, 0≤x≤2\\3x,2≤x≤10\end{cases}\) Show that f is a function and g is not a function.

Answer»

Given,

f(x) = {(x^2,0≤x≤3)(3x,3≤x≤10) and

\(g(x) =\begin{cases}x^2, 0≤x≤2\\3x,2≤x≤10\end{cases}\)

Let us first show that f is a function.

When 0 ≤ x ≤ 3,

f(x) = x².

The function x² associates all the numbers 0 ≤ x ≤ 3 to unique numbers in R.

Hence,

The images of {x ∈ Z: 0 ≤ x ≤ 3} exist and are unique.

When 3 ≤ x ≤ 10,

f(x) = 3x.

The function x² associates all the numbers 3 ≤ x ≤ 10 to unique numbers in R.

Hence,

The images of {x ∈ Z: 3 ≤ x ≤ 10} exist and are unique.

When x = 3,

Using the first definition, we have

f(3) = 3² = 9

When x = 3,

Using the second definition, we have

f(3) = 3(3) = 9

Hence,

The image of x = 3 is also distinct.

Thus,

As every element of the domain has an image and no element has more than one image, f is a function.

Now,

Let us show that g is not a function.

When 0 ≤ x ≤ 2, g(x) = x².

The function x² associates all the numbers 0 ≤ x ≤ 2 to unique numbers in R.

Hence,

The images of {x ∈ Z: 0 ≤ x ≤ 2} exist and are unique.

When 2 ≤ x ≤ 10, g(x) = 3x.

The function x² associates all the numbers 2 ≤ x ≤ 10 to unique numbers in R.

Hence,

The images of {x ∈ Z: 2 ≤ x ≤ 10} exist and are unique.

When x = 2,

Using the first definition, we have

g(2) = 2² = 4

When x = 2,

Using the second definition, we have

g(2) = 3(2) = 6

Here,

The element 2 of the domain is associated with two elements distinct elements 4 and 6.

Thus,

g is not a function.

Discussion

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