## RELATIONS AND FUNCTIONS Quiz-9

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 [x] and {x} represent the integral and fractional parts of x, respectively, then the value of ∑_(r=1)^2000▒({x+r})/2000 is
•  x
•  [x]
•  {x}
•  x+2001
Solution
(c) ∑_(r=1)^2000▒({x+r})/2000=∑_(r=1)^2000▒({x})/2000=2000 ({x})/2000={x}

Q2.If f:[0,∞]→[0,∞] and f(x)=x/(1+x), then f is
•  One-one and onto
•  One-one but not onto >
•  Onto but not one-one
•  d) Neither one-one nor onto
Solution
(b) Here, f:[0,∞]→[0,∞)ie, domain is [0,∞) and codomain is [0,∞). For one-one f(x)=x/(1+x) ⇒ f^' (x)=1/(1+x)^2 >0,∀ x∈[0,∞) ∴f(x) is increasing in its domain. Thus, f(x) is one-one in its domain. For onto (we find range) f(x)=x/(1+x) ie,y=x/(1+x)⇒y+yx=x ⇒x=y/(1-y)⇒y/(1-y)≥0 as x≥0 ∴0≤y≠1 ie, Range ≠ Codomain ∴f(x) is one-one but not onto.

Q3.  If f(x)={█(x,x is rational@1-x, x is irrational)┤ then f(f(x)) is
•   x∀x∈R
•  ={█(x,x is irrational@1-x,x is rational)┤
•  {█(x,x is rational@1-x,x is irrational)┤
•  None of these
Solution
(a) f(f(x))={█(f(x),f(x) is rational@1-f(x),f(x) is irrational)┤ ⇒ f(f(x))={█(x,x is rational@1-(1-x)=x,x is irrational)┤

Q4. Let h(x)=|kx+5|, the domain of f(x) is [-5,7], the domain of f(h(x)) is [-6,1] and the range of h(x) is the same as the domain of f(x), then the value of k is
•  1
•  2
•  3
•  4
Solution
(b) -5≤|kx+5|≤7 ⇒-12≤kx≤2 where -6≤x≤1 ⇒-6≤k/2 x≤1 where -6≤x≤1 ∴k=2 [∵ the range of h(x)= the domain of f(x)]

Q5.f(x)={█(x,if x is rational@0,if x is irrational)┤ and f(x)={█(0,if x is rational@x,if x is irrational )┤. Then, f-g is
•  One-one and into
•  Neither one-one nor onto
•  Many one and onto
•   One-one and onto
Solution
(d) Let ϕ(x)=f(x)-g(x) ={█(x,x∈Q@-x,x∉Q)┤ For one-one Take any straight line parallel to x-axis which will intersect ϕ(x) only at one point. ⇒ϕ(x) is one-one. Foe onto As, ϕ(x)={█(x,x∈Q@-x,x∉Q)┤, which shows y=x and y=-x for irrational values ⇒y∉ real numbers. ∴ Range=Codomain ⇒ ϕ(x) is onto. Thus, f-g is one-one and onto.

Q6. The domain of f(x)=log|log⁡x | Is
•  (0,∞)
•  (1,∞)
• (0,1)∪(1,∞)
•  (-∞,1)
Solution
(c) f(x)=log⁡〖|log⁡x |,〗 f(x) is defined if |log⁡x |>0 and x>0 i.e., if x>0 and x≠1 (∵|log⁡x |>0 if x≠1) ⇒x∈(0,1)∪(1,∞)

Q7. If f:R→R is an invertible function such that f(x) and f^(-1) (x) are symmetric about the line y=-x, then
•  f(x) is odd
•  f(x) and f^(-1) (x) may not be symmetric about the line y=x
•  3\f(x) may not be odd
•  None of these
Solution
(a) Since f(x) and f^(-1) (x) are symmetric about the line y=-x If (α,β) lies on y=f(x) then (-β,-α) on y=f^(-1) (x) ⇒(-α,-β) lies on y=f(x) ⇒y=f(x) is odd

Q8. If f(2x+y/8,2x-y/8)=xy, then f(m,n)+(n,m)=0
•  Only when m=n
•  TOnly when m≠n
•  Only when m=-n
•  For all m and n
Solution
(d) Let 2x+y/8=α and 2x-y/8=β, then x=(α+β)/4 and y=4(α-β) Given, f(2x+y/8,2x-y/8)=xy ⇒f(α,β)=α^2-β^2 ⇒f(m,n)+f(n,m)=m^2-n^2+n^2-m^2=0 for all m,n

Q9. The period of the function f(x)=c^(sin^2⁡x+sin^2⁡(x+π/3)+cos⁡x cos⁡(x+π/3) ) is (where c is constant)
•  1
•  π/2
•  π
•  Cannot be determined
Solution
(d) sin^2⁡x+sin^2⁡(x+π/3)+cos⁡x cos⁡(x+π/3) =sin^2⁡x+(sin⁡x/2+(√3 cos⁡x)/2)^2+cos⁡x (cos⁡x/2-(√3 sin⁡x)/2) =sin^2⁡x+sin^2⁡x/4+(3 cos^2⁡x)/4+cos^2⁡x/2 =(5 sin^2⁡x)/4+(5 cos^2⁡x)/4=5⁄4 Hence, f(x)=c^(5⁄4)= constant, which is periodic whose period cannot be determined

Q10. The range of f(x)=[1+sin⁡x ]+[2+sin⁡〖x/2〗 ]+[3+sin⁡〖x/3〗 ]+⋯+[n+sin⁡〖x/n〗 ],∀ x∈[0,π], where [.] denotes the greatest integer function, is
•  {(n^2+n-2)/2,n(n+1)/2}
•  {n(n+1)/2}
•  {(n^2+n-2)/2,n(n+1)/2,(n^2+n+2)/2}
• {n(n+1)/2,(n^2+n+2)/2}
Solution
(d) f(x)=n(n+1)/2+[sin⁡x ]+[sin⁡〖x/2〗 ]+⋯+[sin⁡〖x/n〗 ] Thus, the range of f(x)={n(n+1)/2,n(n+1)/2+1} as x∈[0,π] #### Written by: AUTHORNAME

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