Relations and Functions-Quiz-14

RELATIONS AND FUNCTIONS Quiz-14

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. The domain of f(x)= ln (〖ax〗^3+(a+b) x^2+(b+c)x+c), where a>0,b^2-4ac=0, is (where [⋅] represents greatest integer function).

•  (-1,∞)~{-b/2a}
•  (1,∞)~{-b/2a}
•  (-1,1)~{-b/2a}
•  None of these

Solution
A

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. The domain of the function f(x)=√(In_((|x|-1) ) (x^2+4x+4) ) is

•  [-3,-1]∪[1,2]
•  (-2,-1)∪[2,∞)
•  (-∞,-3]∪(-2,-1)∪(2,∞)
•  None of these

Solution
(c) Case I 0<|x|-1<1⇒1<|x|<2, then x^2+4x+4≤1 ⇒x^2+4x+3≤0 ⇒-3≤x≤-1 So x∈(-2,-1) (1) Case II |x|-1>1⇒|x|>2, then x^2+4x+4≥1 ⇒x^2+4x+3≥0 ⇒x≥-1 or x≤-3 So, x∈(-∞,-3]∪(2,∞) (2) From (1) and (2), x∈(-∞,-3]∪(-2,-1)∪(2,∞)

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 xϵ[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. If f(x)=√(n&x^m ),n∈N, is an even function, then m is

•  Even integer
•  Odd integer
•  Any integer
• f(x)-even is not possible

Solution
(a) Given f(x)=√(n&x^m ),n∈N is an even function where m∈I ⇒ f(x)=f(-x) ⇒√(n&x^m )=√(n&(-x)^m ) ⇒x^m=(-x)^m ⇒m is an even integer ⇒m=2k,k∈I