## RELATIONS AND FUNCTIONS Quiz-7

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 g(x)=1+x-[x] and f(x)={■(-1,&x<0@0,&x=0@1,&x>0)┤ Then for all x, f(g(x)) is equal to (where [.] represents greatest integer function)
•  x
•  1
•  f(x)
•  g(x)
Solution
(b) g(x)=1+{x},f(x)={■(-1,&x<0@0,&x=0 @1,&x>0)┤ where {x} represents the fractional part function.

Q2.Let f(x)=x+2|x+1|+2|x-1|. If f(x)=k has exactly one real solution, then the value of k is
•  3
•  0 >
•   1
•  2
Solution
(a) Let f(x)=x+2|x+1|+2|x-1| ⇒ f(x)={█(x-2(x+1)-2(x-1),x<-1@x+2(x+1)-2(x-1),-1≤x≤1@x+2(x+1)+2(x-1),x>1)┤ Or f(x)={█(-3x,x<-1@x+4,-1≤x≤1@5x,x>1)┤

Q3.  If f(x)=〖ax〗^7+〖bx〗^3+cx-5,a,b,c are real constants and f(-7)=7, then the range of f(7)+17 cos x is
•   [-34,0]
•  [0,34]
•  [-34,34]
•  None of these
Solution
(a) f(7)+f(-7)=-10 ⇒f(7)=-17 ⇒f(7)+17 cos⁡〖x=-17+17 cos⁡x 〗 which has the range [-34,0]

Q4. If [cos^(-1)⁡x ]+[cos^(-1)⁡x ]=0, where [.] denotes the greatest integer function, then the complete set of values of x is
•  (cos⁡1,1]
•  (cos⁡1,cot⁡1 )
•  (cot⁡1,1]
•  [0,cot⁡1 )
Solution
(c) We have [cos^(-1)⁡x ]≥0∀x∈[-1,1] And [cot^(-1)⁡x ]≥0 ∀ x∈R Hence, [cot^(-1)⁡x ]+[cot^(-1)⁡x ]=0 ⇒[cot^(-1)⁡x ]=[cot^(-1)⁡x ]=0 If [cos^(-1)⁡x ]=0⇒x∈(cos⁡〖1,1] 〗 If [cot^(-1)⁡x ]=0⇒x∈(cot⁡〖1,∞〗 ) ⇒x∈(cot⁡〖1,1] 〗

Q5.The range of f(x)=√((1-cos⁡x ) √((1-cos⁡x ) √((1-cos⁡x ) √(…∞)) ) ) is
•  [0, 1]
•  [0,1/2]
•  [0, 2]
•   None of these
Solution
(c) Given f(x)=√((1-cos⁡x ) √((1-cos⁡x ) √((1-cos⁡x ) √(…∞)) ) ) ⇒ f(x)=(1-cos⁡x )^(1/2) (1-cos⁡x )^(1/4) (1-cos⁡x )^(1/8)…∞ ⇒ f(x)=(1-cos⁡x )^(1/2+1/4+1/8+⋯∞) ⇒ f(x)=(1-cos⁡x )^((1/2)/(1-1/2)) ⇒ f(x)=1-cos⁡x ⇒ The range of f(x) is [0, 2)

Q6. The values of b and c for which the identity f(x+1)-f(x)=8x+3 is satisfied, where f(x)=bx^2+cx+d, are
•  b=2,c=1
•  b=4,c=-1
• b=-1,c=4
•  b=-1,c=1
Solution
(b) ∵f(x+1)-f(x)=8x+3 ⇒{b(x+1)^2+c(x+1)+d}-{bx^2+cx+d}=8x+3 ⇒b{(x+1)^2-x^2 }+c=8x+3 ⇒b(2x+1)+c=8x+3 On comparing co-efficient of x and constant term, we get 2b=8 and b+c=3 Then b=4 and c=-1

Q7. If the graph of y=f(x) is symmetrical about lines x=1 and x=2, then which of the following is true?
•  f(x+1)=f(x)
•  f(x+3)=f(x)
•  f(x+2)=f(x)
•  None of these
Solution
(c) From the given data f(1-x)=f(1+x) (1) And f(2-x)=f(2+x) (2) In (2) replacing x by 1+x, we have f(1-x)=f(3+x) ⇒ f(1+x)=f(3+x) [From (1)] ⇒ f(x)=f(2+x)

Q8. If the function f:[1,∞)→:[1,∞) is defined by f(x)=2^(x(x-1)), then f^(-1) (x) is
•  (1/2)^(x(x-1))
•  T1/2 (1+√(1+4 log_2⁡x ))
•  1/2 (1-√(1+4 log_2⁡x ))
•  Not defined
Solution
(b) y=2^x(x-1) ⇒x^2-x-log_2⁡〖y=0;〗 ⇒x=1/2(1±√(1+4 log_2⁡y )) Since xÏµ[1,∞), we choose x=1/2 (1+√(1+4 log_2⁡y )) Or f^(-1) (x)=1/2 (1+√(1+4 log_2⁡x ))

Q9. Which of the following functions is inverse to itself ?
•  f(x)=(1-x)/(1+x)
•  f(x)=5^log⁡x
•  f(x)=2^x(x-1)
•  None of these
Solution
(a) By checking for different function, we find that for f(x)=(1-x)/(1+x),f^(-1) (x)=f(x)

Q10. The domain of the function f(x)=log_(3+x) (x^2-1) is
•  (-3,-1)∪(1,∞)
•  [-3,-1)∪[1,∞)
•  (-3,-2)∪(-2,-1)∪(1,∞)
• [-3,-2)∪(-2,-1)∪[1,∞)
Solution
(c) f(x) is to be defined when x^2-1>0 and 3+x>0 and 3+x≠1 ⇒x^2>1 and x>-3 and x≠-2 ⇒x<-1 or x>1 and x>-3 and x≠-2 ∴ D_f=(-3,-2)∪(-2,-1)∪(1,∞)

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