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

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

(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

Solution

(a)

(a)

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

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

(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

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

(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_3tanx 〗+log_tanx 〖3)〗 is

Solution

(a) f(x)=sec^(-1)〖(log_3tanx 〗+log_tanx 〖3)〗 f(x)=sec^(-1)(log_3tanx +1/log_3tanx ) Now for log_3tanx to get defined, tanx∈(0,∞ ) ⇒log_3tanx ∈(-∞,∞) or log_3tanx ∈R Also x+1/x≤-2 or x+1/x≥2 ⇒log_3tanx +1/log_3tanx ≤-2 or log_3tanx +1/log_3tanx ≥2 ⇒sec^(-1)(log_3tanx +1/log_3tanx )≤sec^(-1) (-2) or sec^(-1)(log_3tanx +1/log_3tanx )≥sec^(-1) 2 ⇒ f(x)≤2Ï€/3 or f(x)≥Ï€/3 ⇒ f(x)∈[Ï€/3,Ï€/2)∪(Ï€/2,2Ï€/3]

(a) f(x)=sec^(-1)〖(log_3tanx 〗+log_tanx 〖3)〗 f(x)=sec^(-1)(log_3tanx +1/log_3tanx ) Now for log_3tanx to get defined, tanx∈(0,∞ ) ⇒log_3tanx ∈(-∞,∞) or log_3tanx ∈R Also x+1/x≤-2 or x+1/x≥2 ⇒log_3tanx +1/log_3tanx ≤-2 or log_3tanx +1/log_3tanx ≥2 ⇒sec^(-1)(log_3tanx +1/log_3tanx )≤sec^(-1) (-2) or sec^(-1)(log_3tanx +1/log_3tanx )≥sec^(-1) 2 ⇒ f(x)≤2Ï€/3 or f(x)≥Ï€/3 ⇒ f(x)∈[Ï€/3,Ï€/2)∪(Ï€/2,2Ï€/3]

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

Solution

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

(d) It is given that 2^x+2^y=2 ∀ x,y Ïµ 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

Solution

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

(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?

Solution

(d) f(f(x))={█((f(x))^2,for f(x)≥0@f(x)

(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∈

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}}

(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:[-Ï€/3,2Ï€/3]→[0,4] be a function defined as f(x)=√3 sinx-cosx+2. Then f^(-1) (x) is given by

Solution

(b) y=f(x)=√3 sinx-cosx+2=2 sin〖(x-Ï€/6)+2〗 (1) Since f(x) is one-one and onto, f is invertible. From (1) sin〖(x-Ï€/6)=(y-2)/2〗 ⇒x=sin^(-1)〖(y-2)/2+Ï€/6〗 ⇒f^(-1) (x)=sin^(-1)〖((x-2)/2)+Ï€/6〗

(b) y=f(x)=√3 sinx-cosx+2=2 sin〖(x-Ï€/6)+2〗 (1) Since f(x) is one-one and onto, f is invertible. From (1) sin〖(x-Ï€/6)=(y-2)/2〗 ⇒x=sin^(-1)〖(y-2)/2+Ï€/6〗 ⇒f^(-1) (x)=sin^(-1)〖((x-2)/2)+Ï€/6〗