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 total number of solutions of [x]^2=x+2{x}, where [.] and {.} denote the greatest integer function and fractional part, respectively, is equal to

Solution

(b) [x]^2=x+2{x} ⇒[x]^2=x+3{x} ⇒{x}=([x]^2-[x])/3 ⇒x=-1/3,0,1,8/3

(b) [x]^2=x+2{x} ⇒[x]^2=x+3{x} ⇒{x}=([x]^2-[x])/3 ⇒x=-1/3,0,1,8/3

**Q2.**The domain of definition of the function y=1/log_10〖(1-x)〗 +√(x+2)=

Solution

(c) y=1/log_10(1-x) +√(x-2) y=f(x)+g(x) Then, the domain of given function is D_f∩D_g Now, for the domain of f(x)=1/log_10〖(1-x)〗 , We know it is defined only when 1-x>0 and 1-x≠1

(c) y=1/log_10(1-x) +√(x-2) y=f(x)+g(x) Then, the domain of given function is D_f∩D_g Now, for the domain of f(x)=1/log_10〖(1-x)〗 , We know it is defined only when 1-x>0 and 1-x≠1

**Q3.**Given the function f(x)=(a^x +〖 a〗^(-x))/2 (where a>2). Then f(x+y)+f(x-y)=

Solution

(d) We have fog(x)=f(g(x))=sin〖(log_e|x|)〗 log_e|x| has range R, for which sin(log_e|x| )Ïµ[-1,1] ∴R_1={u:-1≤u≤1} Also gof(x)=g(f(x))=log_e|sinx | ∵0≤|sinx |≤1

(d) We have fog(x)=f(g(x))=sin〖(log_e|x|)〗 log_e|x| has range R, for which sin(log_e|x| )Ïµ[-1,1] ∴R_1={u:-1≤u≤1} Also gof(x)=g(f(x))=log_e|sinx | ∵0≤|sinx |≤1

**Q4.**Let f(x)=([a]^2-5[a]+4) x^3-(6{a}^2-5{a}+1)x-(tanx )×sgn x be an even function for all x∈R, then the sum of all possible values of 'a' is (where [⋅] and {⋅} denote greatest integer function and fractional part functions, respectively)

Solution

(d) f(x)=〖Î±x〗^3-Î²x-(tanx )sgn x f(-x)=f(x) ⇒-〖Î±x〗^3+Î²x-tan〖x sgn x=〗 〖Î±x〗^3-Î²x-(tanx )(sgn x) ⇒2(-〖Î±x〗^2-Î²)x=0∀x∈R ⇒Î±=0 and Î²=0 ∴〖[a]〗^2-5[a]+4=0 and 6〖{a}〗^2-5{a}+1=0 ⇒(3{x}-1)(2{x}-1)=0 ∴a=1+1/3,1+1/2,4+1/3,4+1/2 Sum of values of a=35/3

(d) f(x)=〖Î±x〗^3-Î²x-(tanx )sgn x f(-x)=f(x) ⇒-〖Î±x〗^3+Î²x-tan〖x sgn x=〗 〖Î±x〗^3-Î²x-(tanx )(sgn x) ⇒2(-〖Î±x〗^2-Î²)x=0∀x∈R ⇒Î±=0 and Î²=0 ∴〖[a]〗^2-5[a]+4=0 and 6〖{a}〗^2-5{a}+1=0 ⇒(3{x}-1)(2{x}-1)=0 ∴a=1+1/3,1+1/2,4+1/3,4+1/2 Sum of values of a=35/3

**Q5.**The range of the function f defined by f(x)=[1/sin{x} ] (where [.] and {.} respectively denote the greatest integer and the fractional part functions) is

Solution

(d) ∵{x}∈[0,1) sinx∈(0,sin1 ) as f(x) is defined if sin〖{x}≠0〗 ⇒1/sin{x} ∈(1/sin1 ,∞)⇒[1/sin{x} ]∈{1,2,3,…}

(d) ∵{x}∈[0,1) sinx∈(0,sin1 ) as f(x) is defined if sin〖{x}≠0〗 ⇒1/sin{x} ∈(1/sin1 ,∞)⇒[1/sin{x} ]∈{1,2,3,…}

**Q6.**The period of the function f(x)=[6x+7]+cos〖Ï€x-6x〗, where [.] denotes the greatest integer function, is

Solution

(c) f(x)=[6x+7]+cos〖Ï€x-6x〗 =[6x]+7+cos〖Ï€x-6x〗 =7+cos〖Ï€x-{6x〗} {6x} has the period 1/6 and cosÏ€x has the period 2, then the period of f(x)= LCM of 2 and 1/6 which is 2 Hence, the period is 2

(c) f(x)=[6x+7]+cos〖Ï€x-6x〗 =[6x]+7+cos〖Ï€x-6x〗 =7+cos〖Ï€x-{6x〗} {6x} has the period 1/6 and cosÏ€x has the period 2, then the period of f(x)= LCM of 2 and 1/6 which is 2 Hence, the period is 2

**Q7.**If f(x+y)=f(x).f(y) for all real x,y and f(0)≠0, then the function g(x)=f(x)/(1+{f(x)}^2 ) is

Solution

(a) Given f(x+y)=f(x)f(y) Put x=y=0, then f(0)=1 Put y=-x, then f(0)=f(x)f(-x)⇒f(-x)=1/f(x) Now, g(x)=f(x)/(1+{f(x)}^2 ) ⇒ g(-x) f(-x)/(1+{f(-x)}^2 )=(1/f(x) )/(1+1/{f(x)}^2 ) =f(x)/(1+{f(x)}^2 )=g(x)

(a) Given f(x+y)=f(x)f(y) Put x=y=0, then f(0)=1 Put y=-x, then f(0)=f(x)f(-x)⇒f(-x)=1/f(x) Now, g(x)=f(x)/(1+{f(x)}^2 ) ⇒ g(-x) f(-x)/(1+{f(-x)}^2 )=(1/f(x) )/(1+1/{f(x)}^2 ) =f(x)/(1+{f(x)}^2 )=g(x)

**Q8.**A real-valued function f(x) satisfies the functional equation f(x-y)=f(x)f(y)-f(a-x)f(a+y), where a is a given constant and f(0)=1.f(2a-x) is equal to

Solution

(b) We have f(x-y)=f(x)f(y)-f(a-x)f(a+y) Putting x=a and y=a-x, we get f(a-(x-a)=f(a)f(x-a)-f(0)f(x) (1) Putting x=0, y=0, we get f(0)=f(0)(f(0))-f(a)f(a) ⇒f(0)=(f(0))^2-(f(a))^2 ⇒1=〖(1)〗^2-(f(a))^2 ⇒f(a)=0 ⇒ f(2a-x)=-f(x)

(b) We have f(x-y)=f(x)f(y)-f(a-x)f(a+y) Putting x=a and y=a-x, we get f(a-(x-a)=f(a)f(x-a)-f(0)f(x) (1) Putting x=0, y=0, we get f(0)=f(0)(f(0))-f(a)f(a) ⇒f(0)=(f(0))^2-(f(a))^2 ⇒1=〖(1)〗^2-(f(a))^2 ⇒f(a)=0 ⇒ f(2a-x)=-f(x)

**Q9.**If f and g are one-one function, then

Solution

(c) (a)f(x)=sinx and g(x)=cosx,x∈[0,Ï€⁄2] Here, both f(x) and g(x) are one-one functions But h(x)=f(x)+g(x)=sinx+cosx is many-one as h(0)=h(Ï€⁄2)=1 (b) h(x)=f(x)g(x)=sinx cosx=sin2x/2 is many-one, as h(0)=h(Ï€⁄2)=0 (c)It is a fundamental property

(c) (a)f(x)=sinx and g(x)=cosx,x∈[0,Ï€⁄2] Here, both f(x) and g(x) are one-one functions But h(x)=f(x)+g(x)=sinx+cosx is many-one as h(0)=h(Ï€⁄2)=1 (b) h(x)=f(x)g(x)=sinx cosx=sin2x/2 is many-one, as h(0)=h(Ï€⁄2)=0 (c)It is a fundamental property