## RELATIONS AND FUNCTIONS Quiz-4

In mathematics, a function can be defined as a rule that relates every element in one set, called the domain, to exactly one element in another set, called the range. For example, y = x + 3 and y = x2 – 1 are functions because every x-value produces a different y-value. A relation is any set of ordered-pair numbers..

Q1. If f(x)=(-1)^[2x/π] ,g(x)=|sin⁡x |-|cos⁡x | and ∅(x)=f(x)g(x) (where [.] denotes the greatest integer function) then the respective fundamental periods of f(x), g(x) and f(x),g(x)and ∅(x) are
•  π,π,π
•  π,2π,π
•   π,π,π/2
•  π,π/2,π
Solution
(c) Clearly f(x+π)=f(x),g(x+π)=g(x) and ∅(x+π/2) ={(-1)f(x)}{(-1)g(x)}=∅(x)

Q2.The range of sin^(-1)⁡〖[x^2+1/2]+cos^(-1)⁡[x^2-1/2] 〗, where [.] denotes the greatest integer function, is
•  {π/2,π}
•  {π}
•  {π/2}
•  None of these
Solution
(b) [x^2+1/2]=[x^2-1/2+1]=1+[x^2-1/2] Thus, from domain point of view, [x^2-1/2]=0,-1⇒[x^2+1/2]=1,0 ⇒ f(x)=sin^(-1) (1)+cos^(-1) (0) or sin^(-1) (0)+cos^(-1) (-1) ⇒ f(x)={π}

Q3.  If f(x) and g(x) are periodic functions with period 7 and 11, respectively. Then the period of F(x)=f(x)g(x/5)-g(x)f(x/3) is
•   177
•  222
•  433
•  1155
Solution
(d) The period of f(x) is 7⇒ The period of f(x/3) is 7/(1⁄3)=21 The period of g(x) is 11⇒ The period of g(x/5) is 11/(1⁄5)=55 Hence, T_1= period of f(x)g(x/5)=7×55=385 and T_2= period of g(x)f(x/3)=11×21=231 ∴ period of F(x)=LCM{T_1,T_2} =LCM{385,231} =7×11×3×5 =1155

Q4. The exhaustive domain of f(x)=√(x^12-x^9+x^4-x+1) is
•   [0, 1]
•  [1,∞)
•  (-∞,1]
•  R
Solution
(d) f(x)=√(x^12-x^9+x^4-x+1) We must have x^12-x^9+x^4-x+1≥0 Obviously (1) is satisfied by x∈(-∞,0] Also, x^9 (x^3-1)+x(x^3-1)+1≥0∀x∈[1,∞) Further, x^12-x^9+x^4-x+1=(1-x)+x^4 (1-x^5 )+x^12 is also satisfied by x∈(0,1) Hence, the domain is R

Q5. Let E={1,2,3,4} and F={1,2}. Then the number of onto functions from E to F is
•  14
•  16
•  12
•  8
Solution
(a) From E to F we can define, in all, 2×2×2×2=16 functions (2 options for each elements of E) out of which 2 are into, when all the elements of E map to either 1 or 2. ∴ No. of onto function = 16 – 2 =14

Q6. If f(x)=sin⁡〖x+cos⁡〖x,g(x)=x^2-1〗 〗, then g(f(x)) is invertible in the domain
•  [0,π/2]
•  [-π/4,π/4]
• [-π/2,π/2]
•  [0,π]
Solution
(b) ∵g(f(x))=(sin⁡〖x+cos⁡x 〗 )^2-1, is invertible (ie, bijective) ⇒g(f(x))=sin⁡2x, is bijective We know sin⁡x is bijective only when x∈[-π/2,π/2] Thus, g(f(x)) is bijective if, -π/2≤2x≤π/2 ⇒ -π/4≤x≤π/4

Q7. The range of the function f(x)=(e^x-e^|x| )/(e^x+e^|x| )
•  (-∞,∞)
•  [0,1)
•  (-1,0]
•  (-1,1)
Solution
(c)

Q8. Let f be a function satisfying of x then f(xy)=f(x)/y for all positive real numbers x and y if f(30)=20, then the value of f(40) is
•  15
•  T20
•  40
•  60
Solution
(a) f(xy)=(f(x))/y ⇒ f(y)=(f(1))/y (putting x=1) ⇒ f(30)=(f(1))/30 or f(1)=30×f(30)=30×20=600 Now f(40)=(f(1))/40=600/40=15

Q9. The range of f(x)=sin^(-1)⁡((x^2+1)/(x^2+2)) Is
•  [0,π⁄(2])
•  (0,π⁄6)
•  [π⁄6,π⁄(2])
•  None of these
Solution
(c)

Q10. Domain (D) and range (R) of f(x)=sin^(-1)⁡〖(cos^(-1)⁡[x])〗 where [.] denotes the greatest integer function is
•  D≡x∈[1,2),R∈{0}
•  D≡x∈[0,1],R≡{-1,0,1}
•  D≡x∈[-1,1],R≡{0,sin^(-1)⁡〖(π/2),sin^(-1)⁡(π) 〗 }
• D≡x∈[-1,1],R≡{-π/2,0,π/2}
Solution
(a) When [x]=0 we have sin^(-1)⁡〖(cos^(-1) 0)=sin^(-1) 〗 (π⁄2), not defined When [x]=-1 we have sin^(-1)⁡〖(cos^(-1)⁡〖-1〗 )=sin^(-1) 〗 (π), not defined When [x]=1 we have sin^(-1)⁡〖(cos^(-1)⁡1 )=sin^(-1) (0)=0〗 Hence, x∈[1,2) and the range of function is {0} #### Written by: AUTHORNAME

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