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 entire graph of the equation y=x^2+kx-x+9 is strictly above the x-axis if and only if

Solution

(b)

(b)

**Q2.**If f:X→Y, where X and Y are sets containing natural numbers, f(x)=(x+5)/(x+2) then the number of elements in the domain and range of f(x) are respectively

Solution

(a) Let y=(x+5)/(x+2)=1+3/(x+2)⇒x=1 Also, y-1=3/(x+2)⇒x+2=3/(y-1) ⇒x=-2+3/(y-1) ⇒y=2 only as x and y are natural numbers

(a) Let y=(x+5)/(x+2)=1+3/(x+2)⇒x=1 Also, y-1=3/(x+2)⇒x+2=3/(y-1) ⇒x=-2+3/(y-1) ⇒y=2 only as x and y are natural numbers

**Q3.**The domain of f(x)=1/√(|cosx |+cosx ) is

Solution

(c) f(x)=√2 sin〖(x+Ï€/4)+2√2〗 Or f(x)=√2 cos〖(x-Ï€/4)+2√2〗 ⇒Y=[√2,3√2] and X=[-3Ï€/4,Ï€/4] or [Ï€/4,5Ï€/4]

(c) f(x)=√2 sin〖(x+Ï€/4)+2√2〗 Or f(x)=√2 cos〖(x-Ï€/4)+2√2〗 ⇒Y=[√2,3√2] and X=[-3Ï€/4,Ï€/4] or [Ï€/4,5Ï€/4]

**Q4.**Let f(n)=1+1/2+1/3+⋯+1/n, then f(1)+f(2)+f(3)+⋯+f(n) is equal to

Solution

(b) In the sum, f(1)+f(2)+f(3)+⋯+f(n), 1 occurs n times, 1/2 occurs (n-1) times, 1/3 occurs (n-2) times and so on ∴f(1)+f(2)+f(3)+⋯+f(n) =n.1+(n-1).1/2+(n-2).1/3+⋯+1.1/n =n(1+1/2+1/3+⋯+1/n)-(1/2+2/3+3/4+⋯+(n-1)/n) =nf(n)-[(1-1/2)+(1-1/3)+(1-1/4)+⋯+(1-1/n)] =nf(n)-[n-f(n)] =(n+1)f(n)-n

(b) In the sum, f(1)+f(2)+f(3)+⋯+f(n), 1 occurs n times, 1/2 occurs (n-1) times, 1/3 occurs (n-2) times and so on ∴f(1)+f(2)+f(3)+⋯+f(n) =n.1+(n-1).1/2+(n-2).1/3+⋯+1.1/n =n(1+1/2+1/3+⋯+1/n)-(1/2+2/3+3/4+⋯+(n-1)/n) =nf(n)-[(1-1/2)+(1-1/3)+(1-1/4)+⋯+(1-1/n)] =nf(n)-[n-f(n)] =(n+1)f(n)-n

**Q5.**The domain of f(x)=cos^(-1)〖((2-|x|)/4)+[log(3-x) ]^(-1) 〗 Is

Solution

(b) cos^(-1)((2-|x|)/4) exists if -1≤(2-|x|)/4≤1 ⇒-6≤-|x|≤2 ⇒-2≤|x|≤6 ⇒|x|≤6 ⇒-6≤x≤6 The function [log(3-x) ]^(-1)=1/log(3-x) is defined if 3-x>0 and x≠2,

(b) cos^(-1)((2-|x|)/4) exists if -1≤(2-|x|)/4≤1 ⇒-6≤-|x|≤2 ⇒-2≤|x|≤6 ⇒|x|≤6 ⇒-6≤x≤6 The function [log(3-x) ]^(-1)=1/log(3-x) is defined if 3-x>0 and x≠2,

**Q6.**If the function f:[1,∞)→[1,∞) is defined by f(x)=2^x(x-1) , then f^(-1) (x) Is

Solution

(b) Given y=2^x(x-1) ⇒x(x-1)=log_2y ⇒x^2-x-log_2y=0 ⇒x=(1±√(1+4 log_2y ))/2 Only x=(1±√(1+4 log_2y ))/2 lies in the domain ⇒f^(-1) (x)=1/2 [1+√(1+4 log_2x )]

(b) Given y=2^x(x-1) ⇒x(x-1)=log_2y ⇒x^2-x-log_2y=0 ⇒x=(1±√(1+4 log_2y ))/2 Only x=(1±√(1+4 log_2y ))/2 lies in the domain ⇒f^(-1) (x)=1/2 [1+√(1+4 log_2x )]

**Q7.**If f(x+1/2)+f(x-1/2)=f(x) for all x∈R, then the period of f(x) is

Solution

(c) f(x+1/2)+f(x-1/2)=f(x) ⇒ f(x+1)+f(x)=f(x+1/2) ⇒ f(x+1)+f(x-1/2)=0 ⇒ f(x+3/2)=-f(x) ⇒ f(x+3)=-f(x+3/2)=f(x) ∴ f(x) is periodic with period 3

(c) f(x+1/2)+f(x-1/2)=f(x) ⇒ f(x+1)+f(x)=f(x+1/2) ⇒ f(x+1)+f(x-1/2)=0 ⇒ f(x+3/2)=-f(x) ⇒ f(x+3)=-f(x+3/2)=f(x) ∴ f(x) is periodic with period 3

**Q8.**The domain of the function f(x)=1/√( ^( 10) C_(x-1)-3× ^( 10) C_x ) contains the points

Solution

(d) Given function is defined if ^( 10) C_(x-1)>3^( 10) C_x ⇒1/(11-x)>3/x⇒4x>33 ⇒x≥9 but x≤10⇒x=9,10

(d) Given function is defined if ^( 10) C_(x-1)>3^( 10) C_x ⇒1/(11-x)>3/x⇒4x>33 ⇒x≥9 but x≤10⇒x=9,10

**Q9.**The function f:N→N (N is the set of natural numbers) defined by f(n)=2n+3 is

Solution

(b) f:N→N,f(n)=2n+3 Here, the range of the function is {5,6,7,…} or N-{1,2,3,4} Which is a subset of N (co-domain). Hence, function is into. Also, it is clear that f(n) is one-one or injective. Hence, f(n) is injective only

(b) f:N→N,f(n)=2n+3 Here, the range of the function is {5,6,7,…} or N-{1,2,3,4} Which is a subset of N (co-domain). Hence, function is into. Also, it is clear that f(n) is one-one or injective. Hence, f(n) is injective only