## RELATIONS AND FUNCTIONS Quiz-3

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. Let f:N→N defined by f(x)=x^2+x+1,x∈N, then f is
•  One-one onto
•  Many-one onto
•  One-one but not onto
•  None of these
Solution
(c) Let x,y∈N such that f(x)=f(y) Then f(x)=f(y) ⇒x^2+x+1=y^2+y+1 ⇒(x-y)(x+y+1)=0 ⇒x=y or x=(-y-1)∉N ∴ f is one-one Also, f(x) does not take all positive integral values. Hence f is into

Q2.If f(3x+2)+f(3x+29)=0 ∀ x∈R, then the period of f(x) is
•  7
•  8
•  10
•  None of these
Solution
(d) f(3x+2)+f(3x+29)=0 (1) Replacing x by x+9, we get f(3(x+9)+2)+f(3(x+9)+29=0 ⇒ f(3x+29)+f(3x+56)=0 (2) From (1) and (2), we get f(3x+2)=f(3x+56) ⇒ f(3x+2)=f(3(x+18)+2) ⇒ f(x) is periodic with period 54

Q3.  Which of the following functions is periodic?
•   f(x)=x-[x] where [x] denotes the largest integer less than or equal to the real number x
•  f(x)=sin⁡〖1/x〗 for x≠0, f(0)=0
•  f(x)=x cos⁡x
•  None of these
Solution
(a) f(x)={x} is periodic with period 1 f(x)=sin⁡〖1/x〗 for x≠0,f(0)=0 is non-periodic as g(x)=1/x is non-periodic Also f(x)=x cos⁡x is non-periodic as g(x)=x is non-periodic
•   An even function
•  An odd function
•  A periodic function
•  None of these.
Solution
(b)
Q5.The domain of the function f(x)=sin^(-1)⁡(3-x)/In(|x|-2) Is
•  [2,4]
•  (2,3)∪(3,4]
•  [2,∞)
•  (-∞,-3)∪[2,∞)
Solution
(b) f(x)=sin^(-1)⁡(3-x)/log⁡〖(|x|-2)〗 Let g(x)=sin^(-1)⁡(3-x) ⇒-1≤3-x≤1 The domain of g(x) is [2, 4] And let h(x)=log⁡(|x|-2) ⇒|x|-2>0 or |x|>2 ⇒x<-2 or x>2 ⇒(-∞,-2)∪(2,∞) We know that (f⁄g)(x) f(x)/(g(x))∀x∈D_1∩D_2-{x∈R:g(x)=0} ∴ the domain of f(x)=(2,4]-{3}=(2,3)∪(3,4]

Q6. If f(x) is an even function and satisfies the relation x^2 f(x)-2f(1/x)=g(x) where g(x) is an odd function, then f(5) equals
•  0
•  50/75
• 49/75
•  None of these
Solution
(a) x^2 f(x)-2f(1/x)=g(x) and 2f(1/x)-4x^2 f(x)=2x^2 g(1/x) (Replacing x by 1/x) ⇒ -3x^2 f(x)=g(x)+2x^2 g(1/x) (Eliminating f(1/x)) ⇒ f(x)=-((g(x)=2x^2 g(1/x))/(3x^2 )) ∵ g(x) and x^2 are odd and even functions, respectively So, f(x) is an odd function But f(x) is given even ⇒ f(x)=0∀x Hence, f(5)=0

Q7. If f(x)=sin⁡〖([x]π)〗/(x^2+x+1), where [.] denotes the greatest integer function, then
•  f is one-one
•  f is not one-one and non-constant
•  f is a constant function
•  None of these
Solution
(c) f(x)=sin⁡〖[x]π〗/(x^2+x+1) Let [x]=n∈ integer ⇒sin⁡〖[x]π=0〗 ⇒ f(x)=0 ⇒f(x) is constant function

Q8. If x satisfies |x-1|+|x-2|+|x-3|≥6, then
•  0≤x≤4
•  Tx≤-2 or x≥4
•  x≤0 or x≥4
•  None of these
Solution
(c)

Q9. The period of function 2^({x})+sin⁡π 〖x+3〗^{x⁄2} +cos⁡2πx (where {x} denotes the fractional part of x) is
•  2
•  1
•  3
•  None of these
Solution
(a) The period of sin⁡πx and cos⁡2πx is 2 and 1, respectively The period of 2^({x}) is 1 The period of 3^({x⁄2}) is 2 Hence, the period of f(x) is LCM of 1 and 2=2

Q10. 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) #### Written by: AUTHORNAME

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