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(x)=(-1)^[2x/Ï€] ,g(x)=|sinx |-|cosx | and ∅(x)=f(x)g(x) (where [.] denotes the greatest integer function) then the respective fundamental periods of f(x), g(x) and f(x),g(x)and ∅(x) are

Solution

(c) Clearly f(x+Ï€)=f(x),g(x+Ï€)=g(x) and ∅(x+Ï€/2) ={(-1)f(x)}{(-1)g(x)}=∅(x)

(c) Clearly f(x+Ï€)=f(x),g(x+Ï€)=g(x) and ∅(x+Ï€/2) ={(-1)f(x)}{(-1)g(x)}=∅(x)

**Q2.**The range of sin^(-1)〖[x^2+1/2]+cos^(-1)[x^2-1/2] 〗, where [.] denotes the greatest integer function, is

Solution

(b) [x^2+1/2]=[x^2-1/2+1]=1+[x^2-1/2] Thus, from domain point of view, [x^2-1/2]=0,-1⇒[x^2+1/2]=1,0 ⇒ f(x)=sin^(-1) (1)+cos^(-1) (0) or sin^(-1) (0)+cos^(-1) (-1) ⇒ f(x)={Ï€}

(b) [x^2+1/2]=[x^2-1/2+1]=1+[x^2-1/2] Thus, from domain point of view, [x^2-1/2]=0,-1⇒[x^2+1/2]=1,0 ⇒ f(x)=sin^(-1) (1)+cos^(-1) (0) or sin^(-1) (0)+cos^(-1) (-1) ⇒ f(x)={Ï€}

**Q3.**If f(x) and g(x) are periodic functions with period 7 and 11, respectively. Then the period of F(x)=f(x)g(x/5)-g(x)f(x/3) is

Solution

(d) The period of f(x) is 7⇒ The period of f(x/3) is 7/(1⁄3)=21 The period of g(x) is 11⇒ The period of g(x/5) is 11/(1⁄5)=55 Hence, T_1= period of f(x)g(x/5)=7×55=385 and T_2= period of g(x)f(x/3)=11×21=231 ∴ period of F(x)=LCM{T_1,T_2} =LCM{385,231} =7×11×3×5 =1155

(d) The period of f(x) is 7⇒ The period of f(x/3) is 7/(1⁄3)=21 The period of g(x) is 11⇒ The period of g(x/5) is 11/(1⁄5)=55 Hence, T_1= period of f(x)g(x/5)=7×55=385 and T_2= period of g(x)f(x/3)=11×21=231 ∴ period of F(x)=LCM{T_1,T_2} =LCM{385,231} =7×11×3×5 =1155

**Q4.**The exhaustive domain of f(x)=√(x^12-x^9+x^4-x+1) is

Solution

(d) f(x)=√(x^12-x^9+x^4-x+1) We must have x^12-x^9+x^4-x+1≥0 Obviously (1) is satisfied by x∈(-∞,0] Also, x^9 (x^3-1)+x(x^3-1)+1≥0∀x∈[1,∞) Further, x^12-x^9+x^4-x+1=(1-x)+x^4 (1-x^5 )+x^12 is also satisfied by x∈(0,1) Hence, the domain is R

(d) f(x)=√(x^12-x^9+x^4-x+1) We must have x^12-x^9+x^4-x+1≥0 Obviously (1) is satisfied by x∈(-∞,0] Also, x^9 (x^3-1)+x(x^3-1)+1≥0∀x∈[1,∞) Further, x^12-x^9+x^4-x+1=(1-x)+x^4 (1-x^5 )+x^12 is also satisfied by x∈(0,1) Hence, the domain is R

**Q5.**Let E={1,2,3,4} and F={1,2}. Then the number of onto functions from E to F is

Solution

(a) From E to F we can define, in all, 2×2×2×2=16 functions (2 options for each elements of E) out of which 2 are into, when all the elements of E map to either 1 or 2. ∴ No. of onto function = 16 – 2 =14

(a) From E to F we can define, in all, 2×2×2×2=16 functions (2 options for each elements of E) out of which 2 are into, when all the elements of E map to either 1 or 2. ∴ No. of onto function = 16 – 2 =14

**Q6.**If f(x)=sin〖x+cos〖x,g(x)=x^2-1〗 〗, then g(f(x)) is invertible in the domain

Solution

(b) ∵g(f(x))=(sin〖x+cosx 〗 )^2-1, is invertible (ie, bijective) ⇒g(f(x))=sin2x, is bijective We know sinx is bijective only when x∈[-Ï€/2,Ï€/2] Thus, g(f(x)) is bijective if, -Ï€/2≤2x≤Ï€/2 ⇒ -Ï€/4≤x≤Ï€/4

(b) ∵g(f(x))=(sin〖x+cosx 〗 )^2-1, is invertible (ie, bijective) ⇒g(f(x))=sin2x, is bijective We know sinx is bijective only when x∈[-Ï€/2,Ï€/2] Thus, g(f(x)) is bijective if, -Ï€/2≤2x≤Ï€/2 ⇒ -Ï€/4≤x≤Ï€/4

**Q7.**The range of the function f(x)=(e^x-e^|x| )/(e^x+e^|x| )

Solution

(c)

(c)

**Q8.**Let f be a function satisfying of x then f(xy)=f(x)/y for all positive real numbers x and y if f(30)=20, then the value of f(40) is

Solution

(a) f(xy)=(f(x))/y ⇒ f(y)=(f(1))/y (putting x=1) ⇒ f(30)=(f(1))/30 or f(1)=30×f(30)=30×20=600 Now f(40)=(f(1))/40=600/40=15

(a) f(xy)=(f(x))/y ⇒ f(y)=(f(1))/y (putting x=1) ⇒ f(30)=(f(1))/30 or f(1)=30×f(30)=30×20=600 Now f(40)=(f(1))/40=600/40=15

**Q9.**The range of f(x)=sin^(-1)((x^2+1)/(x^2+2)) Is

Solution

(c)

(c)

**Q10.**Domain (D) and range (R) of f(x)=sin^(-1)〖(cos^(-1)[x])〗 where [.] denotes the greatest integer function is

Solution

(a) When [x]=0 we have sin^(-1)〖(cos^(-1) 0)=sin^(-1) 〗 (Ï€⁄2), not defined When [x]=-1 we have sin^(-1)〖(cos^(-1)〖-1〗 )=sin^(-1) 〗 (Ï€), not defined When [x]=1 we have sin^(-1)〖(cos^(-1)1 )=sin^(-1) (0)=0〗 Hence, x∈[1,2) and the range of function is {0}

(a) When [x]=0 we have sin^(-1)〖(cos^(-1) 0)=sin^(-1) 〗 (Ï€⁄2), not defined When [x]=-1 we have sin^(-1)〖(cos^(-1)〖-1〗 )=sin^(-1) 〗 (Ï€), not defined When [x]=1 we have sin^(-1)〖(cos^(-1)1 )=sin^(-1) (0)=0〗 Hence, x∈[1,2) and the range of function is {0}