Relations and Functions-Quiz-1

RELATIONS AND FUNCTIONS Quiz-1

Dear Readers,

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:[-10,10]→R, where f(x)=sin⁡x+[x^2⁄a] be an odd function. Then the set of values of parameter a is/are

•  (-10,10)~{0}
•  (0,10)
•  [100,∞)
•  (100,∞)

Solution
(d) Since f (x) is an odd function, [x^2/a]=0 for all x∈[-10,10] ⇒0≤x^2/a<1 for all x∈[-10,10]⇒a>100

Q2.If the graph of the function f(x)=(a^x-1)/(x^n (a^x+1) ) is symmetrical about y-axis, then n equals

•  2
•  2/3
•  1/4
•  -1/3

Solution
(d) f(x)=(a^x-1)/(x^n (a^x+1) ) f(x) is symmetrical about y-axis ⇒ f(x)=f(-x) ⇒(a^x-1)/(x^n (a^x+1) )=(a^(-x)-1)/(〖(-x)〗^n (a^(-x)+1) ) ⇒(a^x-1)/(x^n (a^x+1) )=(1-a^x)/(〖(-x)〗^n (1+a^x ) )⇒x^n=-〖(-x)〗^n

Q3.  The domain of the function f(x)=√(log⁡(1/|sin⁡x | ) ) is

•   R-{-π,π}
•  R-{nπ│n∈Z}
•  R-{2nπ│n∈z}
•  (-∞,∞)

Solution
(b) f(x) is defined for log⁡〖(1/|sin⁡x | )≥0〗 ⇒1/|sin⁡x | ≥1 and |sin⁡x |≠0 ⇒|sin⁡x |≠0 [∵1/|sin⁡x | ≥1 for all x] ⇒x≠nπ,n∈Z Hence, the domain of f(x)=R-{nπ:n∈Z}

Q4. Let f(x)=αx/(x+1),x≠-1. Then for what value of α is f(f(x))=x?

•  √2
•  -√2
•  1
•  -1

Solution
(d) f(x)=αx/(x+1),x≠-1 f(f(x))=x⇒α(αx/(x+1))/(αx/(x+1)+1)=x ⇒(α^2 x)/((α+1) x+1)=x ⇒(α+1) x^2+(1-α^2 )x=0 ⇒α+1=0 and 1-α^2=0 [As true ∀ x≠1∴ Eq. (1) is an identity] ⇒α=-1

Q5. If f: [1,∞)→┤ [2,∞)┤ is given by f(x)=x+1/x , then f^(-1) (x) equals

•  ((x+√(x^2-4)))/2
•  x/(1+x^2 )
•  ((x-√(x^2-4)))/2
•  1+√(x^2-4)

Solution
 (a) f: [1,∞)→┤ [2,∞)┤ f(x)=x+1/x=y ⇒x^2-yx+1=0 ⇒x=(y±√(y^2-4))/2 But given f: [1,∞)→┤ [2,∞)┤ ∴x=(y+√(y^2-4))/2

Q6. The domain of f(x)=sin^(-1)⁡[〖2x〗^2-3], where [.] denotes the greatest integer function, is

•  (-√(3/2),√(3/2))
•  (-√(3/2),-1]∪(-√(5/2),√(5/2))
• (-√(5/2),√(5/2))
•  (-√(5/2),-1]∪[1,√(5/2))

Solution
(d)

Q7. Domain of definition of the function f(x)=√(sin^-⁡〖(2x)+π/6〗 ) for real valued x, is

•  [-1/4,1/2]
•  [-1/2,1/2]
•  (-1/2,1/9)
•  [-1/4,1/4]

Solution
(a) Here, f(x)=√(sin^(-1)⁡〖(2x)+π/6〗 ) , to find domain we must have, sin^(-1)⁡〖(2x)+π/6≥0〗 (but-π/2≤sin^(-1)⁡θ≤π/2) ∴ -π/6≤sin^(-1)⁡〖(2x)〗≤π/2 〖⇒sin〗⁡(-π/6)≤2x≤sin⁡(π/2) ⇒ -1/2≤2x≤1 ⇒x∈[-1/4,1/2]

Q8. Let f:R→R,g:R→R be two given functions such that f is injective and g is surjective, then which of the following is injective?

•  gof
•  Tfog
•  gog
•  None of these

Solution
(d) If f is injective and g is surjective ⇒fog is injective ⇒fof is injective

Q9. Let X={a_1,a_2,…,a_6 } and Y={b_1,b_2,b_3 }. The number of functions f from x to y such that it is onto and there are exactly three elements x in X such that f(x)=b_1 is

•  75
•  90
•  100
•  120

Solution
(d) Image b_1 is assigned to any three of the six pre-images in ^6 C_3 ways Rest two images can be assigned to remaining three pre-images in 2^3-2 ways (as function is onto) Hence number of functions are ^6 C_3×(2^3-2)=20×6=120

Q10. f:N→N where f(x)=x-(-1)^x then f is

•  One-one and into
•  Many-one and into
•  One-one and onto
• Many-one and onto

Solution
(c) f(x)={█(x-1,x is even@x+1,x is odd)┤, where is clearly are one-one and onto