Relations and Functions-Quiz-5

RELATIONS AND FUNCTIONS Quiz-5

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. If x is real, then the value of the expression (x^2+14x+9)/(x^2+2x+3) lies between

•  4 and 5
•  5 and -4
•  -5 and 4
•  None of these

Solution
(c) (x^2+14x+9)/(x^2+2x+3)=y ⇒x^2+14x+9=x^2 y+2xy+3y ⇒x^2 (y-1)+2x(y-7)+(3y-9)=0 Since x is real, ∴4(y-7)^2-4(3y-9)(y-1)>0 ⇒4(y^2+49-14y)-4(〖3y〗^2+9-12y)>0

Q2.The domain of the function f(x)=x/√(sin⁡〖(ln x)-cos⁡(ln x) 〗 )(n∈Z) Is

•  (e^2nπ,e^((3n+1/2)π))
•  (e^(2n+1⁄4)π,e^(2n+5⁄4)π )
•  (e^(2n+1⁄4)π,e^((3n-3⁄(4)) π) )
•  None of these

Solution
(b) For the domain sin⁡(ln⁡x )>cos⁡〖(ln⁡x)〗 and x>0

Q3.  If af(x+1)+bf(1/(x+1))=x,x≠-1,a≠b, then f(2) is equal to

•   (2a+b)/2(a^2-b^2 )
•  a/(a^2-b^2 )
•  (a+2b)/(a^2-b^2 )
•  None of these

Solution
(a) af(x+1)+bf(1/(x+1))=(x+1)-1 (1) Replacing x+1 by 1/(x+1), we get ∴ af(1/(x+1))+bf(x+1)=1/(x+1)-1 (2) (1)×a-(2)×b⇒(a^2-b^2 )f(x+1)=a(x+1)-a-b/(x+1)+b Putting x=1,(a^2-b^2 )f(2)=2a-a-b/2+b=a+b/2 =(2a+b)/2

Q4. The range of f(x)=[|sin⁡x |+|cos⁡x |], where [.] denotes the greatest integer function, is

•   {0}
•  {0, 1}
•  {1}
•  None of these

Solution
(c) y=|sin⁡x |+|cos⁡x | ⇒y^2=1+|sin⁡2x | ⇒1≤y^2≤2 ⇒y∈[1,√2] ⇒ f(x)=1∀x∈R

Q5. The number of roots of the equation x sin⁡x=1,x∈[-2π,0)∪(0,2π], is

•  2
•  3
•  4
•  1

Solution
 (c) x sin⁡x=1 (1) ⇒y=sin⁡x=1/x Root of equation (1) will be given by the point(s) of intersection of the graphs y=sin⁡x and y=1/x. Graphically, it is clear that we get four roots.

Q6. If f(2x+3y,2x-7y)=20x, then f(x,y) equals

•  7x-3y
•  7x+3y
• 3x-7y
•  x-ky

Solution
(b) Let 2x+3y=m and 2x-7y=n ⇒y=(m-n)/10 and x=(7m-3n)/20 ⇒f(m,n)=7m+3n ⇒ f(x,y)=7x+3y

Q7. The range of f(x)=sin^(-1)⁡(√(x^2+x+1)) is

•  (0,π/2]
•  (0,π/3]
•  [π/3,π/2]
•  [π/6,π/3]

Solution
(c) For the function to get defined 0≤x^2+x+1≤1, But x^2+x+1≥3/4⇒√3/2≤√(x^2+x+1)≤1 ⇒π/3≤sin^(-1)⁡〖(√(x^2+x+1))≤π/2 〗

Q8. The function f(x)=sin⁡(log⁡(x+√(1+x^2 )) ) is

•  Even function
•  TOdd function
•  Neither even nor odd
•  Periodic function

Solution
(b) f(x)=sin⁡(log⁡(x+√(1+x^2 )) ) ⇒f(-x)=sin⁡[log⁡(-x+√(1+x^2 )) ] ⇒ f(-x)=sin⁡log⁡((√(1+x^2 )-x) ((√(1+x^2 )+x))/((√(1+x^2 )+x) )) ⇒ f(-x)=sin⁡log⁡[1/((x+√(1+x^2 )) )] ⇒ f(-x)=sin⁡[-log⁡(x+√(1+x^2 )) ] ⇒ f(-x)=〖-sin〗⁡[log⁡(x+√(1+x^2 )) ] ⇒ f(-x)=-f(x) ⇒ f(x) is an odd function

Q9. The function f(x)=sec^(-1)⁡x/√(x-[x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x∈

•  R
•  R-{(-1,1)∪{n│n∈Z}}
•  R^+-(0,1)
•  R^+-{n│n∈N}

Solution
(b) The function sec^(-1)⁡x is defined for all x∈R-(-1,1) and the function 1/√(x-[x] ) is defined for all x∈R-Z So the given function is defined for all x∈R-{(-1,1)∪{n│n∈Z}}

Q10. Let f(x)=(x+1)^2-1,x≥1, Then the set {x:f(x)=f^(-1) (x)} is

•  {0,-1,(-3+i√3)/2,(-3-i√3)/2 }
•  {0,1,-1}
•  {0,-1}
• Empty

Solution
(c) Since f(x)=(x+1)^2-1 is continuous function, solution of f(x)=f^(-1) (x) lies on the line y=x ⇒f(x)=f^(-1) (x)=x ⇒(x+1)^2-1=x ⇒x^2+x=0 ⇒x=0 or -1 ⇒ The required set is {0,-1}