Relations and Functions-Quiz-2

RELATIONS AND FUNCTIONS Quiz-2

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 f is a function such that f(0)=2,f(1)=3 and f(x+2)=2f(x)-f(x+1) for every real x, then f(5) is

•  7
•  13
•  1
•  15

Solution
(b) Put x=0⇒f(2)=2f(0)-f(1)=2×2-3=1 Put x=1⇒f(3)=6-1=5 Put x=2⇒f(4)=2f(2)-f(3)=2×1-5=-3 Put x=3⇒f(5)=2f(3)-f(4)=2(5)-(-3)=13

Q2.The domain of the function f(x)=log_2 (〖-log〗_(1⁄2) (1+1/x^(1⁄4) )-1) is

•  (0,1)
•  (0,1]
•  [1,∞)
•  (1,∞)

Solution
(a)

Q3.  The domain of f(x)=√(〖2{x}〗^2-3{x}+1), where {.} denotes the fractional part in [-1,1], is

•   [-1,1]~(1/2,1)
•  [-1,-1/2]∪[0,1/2]∪{1}
•  [-1,1/2]
•  [-1/2,1]

Solution
(b) We must have 〖2{x}〗^2-3{x}+1≥0⇒{x}≥1 or {x}≤1⁄2 Thus, we have 0≤{x}≤1⁄2⇒x∈[n,n+1/2],n∈I

Q4. Let f:R→R and g:R→R be two one-one and onto functions such that they are the mirror images if each other about the line y=a. If h(x)=f(x)+g(x), then h(x) is

•  One-one and onto.
•  Only one-one and not onto.
•  Only onto but not one-one.
•  Neither one-one nor onto.

Solution
(d) y=f(x) and y=g(x) are mirror image of each other about line y=a ⇒ for some x=b,g(b)-a=a-f(b) ⇒f(b)+g(b)=2a ⇒h(b)f(b)+g(b)=2a (constant) Hence h(x) is constant function Thus it is neither one-one nor onto

Q5. The range of f(x)=sec^(-1)⁡〖(log_3⁡tan⁡x 〗+log_tan⁡x ⁡〖3)〗 is

•  [π/3,π/2)∪(π/2,2π/3]
•  [0,π/2)
•  (2π/3,π]
•  None of these

Solution
 (a) f(x)=sec^(-1)⁡〖(log_3⁡tan⁡x 〗+log_tan⁡x ⁡〖3)〗 f(x)=sec^(-1)⁡(log_3⁡tan⁡x +1/log_3⁡tan⁡x ) Now for log_3⁡tan⁡x to get defined, tan⁡x∈(0,∞ ) ⇒log_3⁡tan⁡x ∈(-∞,∞) or log_3⁡tan⁡x ∈R Also x+1/x≤-2 or x+1/x≥2 ⇒log_3⁡tan⁡x +1/log_3⁡tan⁡x ≤-2 or log_3⁡tan⁡x +1/log_3⁡tan⁡x ≥2 ⇒sec^(-1)⁡(log_3⁡tan⁡x +1/log_3⁡tan⁡x )≤sec^(-1) (-2) or sec^(-1)⁡(log_3⁡tan⁡x +1/log_3⁡tan⁡x )≥sec^(-1) 2 ⇒ f(x)≤2π/3 or f(x)≥π/3 ⇒ f(x)∈[π/3,π/2)∪(π/2,2π/3]

Q6. The domain of definition of the function f(x) given by the equation 2^x+2^y=2 is

•  0x≤1
•  0≤x≤1
• -∞
•  -∞

Solution
(d) It is given that 2^x+2^y=2 ∀ x,y ϵ R ⇒2^y=2-2^x ⇒y=log_2⁡(2-2^x )

Q7. The domain of f(x) is (0, 1), then, domain of f(e^x )+f(ln|x|) is

•  (-1,e)
•  (1,e)
•  (-e,-1)
•  (-e,1)

Solution
(c) f(x) is defined for x∈(0,,1) ⇒f(e^x )+f(ln⁡|x| )

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) f(f(x))={█((f(x))^2,for f(x)≥0@f(x)

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:[-π/3,2π/3]→[0,4] be a function defined as f(x)=√3 sin⁡x-cos⁡x+2. Then f^(-1) (x) is given by

•  sin^(-1)⁡〖((x-2)/2)-ð/6〗
•  sin^(-1)⁡〖((x-2)/2)+π/6〗
•  2π/3+cos^(-1)⁡((x-2)/2)
• None of these

Solution
(b) y=f(x)=√3 sin⁡x-cos⁡x+2=2 sin⁡〖(x-π/6)+2〗 (1) Since f(x) is one-one and onto, f is invertible. From (1) sin⁡〖(x-π/6)=(y-2)/2〗 ⇒x=sin^(-1)⁡〖(y-2)/2+π/6〗 ⇒f^(-1) (x)=sin^(-1)⁡〖((x-2)/2)+π/6〗