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.**Let g(x)=1+x-[x] and f(x)={■(-1,&x<0@0,&x=0@1,&x>0)┤ Then for all x, f(g(x)) is equal to (where [.] represents greatest integer function)

Solution

(b) g(x)=1+{x},f(x)={■(-1,&x<0@0,&x=0 @1,&x>0)┤ where {x} represents the fractional part function.

(b) g(x)=1+{x},f(x)={■(-1,&x<0@0,&x=0 @1,&x>0)┤ where {x} represents the fractional part function.

**Q2.**Let f(x)=x+2|x+1|+2|x-1|. If f(x)=k has exactly one real solution, then the value of k is

Solution

(a) Let f(x)=x+2|x+1|+2|x-1| ⇒ f(x)={█(x-2(x+1)-2(x-1),x<-1@x+2(x+1)-2(x-1),-1≤x≤1@x+2(x+1)+2(x-1),x>1)┤ Or f(x)={█(-3x,x<-1@x+4,-1≤x≤1@5x,x>1)┤

(a) Let f(x)=x+2|x+1|+2|x-1| ⇒ f(x)={█(x-2(x+1)-2(x-1),x<-1@x+2(x+1)-2(x-1),-1≤x≤1@x+2(x+1)+2(x-1),x>1)┤ Or f(x)={█(-3x,x<-1@x+4,-1≤x≤1@5x,x>1)┤

**Q3.**If f(x)=〖ax〗^7+〖bx〗^3+cx-5,a,b,c are real constants and f(-7)=7, then the range of f(7)+17 cos x is

Solution

(a) f(7)+f(-7)=-10 ⇒f(7)=-17 ⇒f(7)+17 cos〖x=-17+17 cosx 〗 which has the range [-34,0]

(a) f(7)+f(-7)=-10 ⇒f(7)=-17 ⇒f(7)+17 cos〖x=-17+17 cosx 〗 which has the range [-34,0]

**Q4.**If [cos^(-1)x ]+[cos^(-1)x ]=0, where [.] denotes the greatest integer function, then the complete set of values of x is

Solution

(c) We have [cos^(-1)x ]≥0∀x∈[-1,1] And [cot^(-1)x ]≥0 ∀ x∈R Hence, [cot^(-1)x ]+[cot^(-1)x ]=0 ⇒[cot^(-1)x ]=[cot^(-1)x ]=0 If [cos^(-1)x ]=0⇒x∈(cos〖1,1] 〗 If [cot^(-1)x ]=0⇒x∈(cot〖1,∞〗 ) ⇒x∈(cot〖1,1] 〗

(c) We have [cos^(-1)x ]≥0∀x∈[-1,1] And [cot^(-1)x ]≥0 ∀ x∈R Hence, [cot^(-1)x ]+[cot^(-1)x ]=0 ⇒[cot^(-1)x ]=[cot^(-1)x ]=0 If [cos^(-1)x ]=0⇒x∈(cos〖1,1] 〗 If [cot^(-1)x ]=0⇒x∈(cot〖1,∞〗 ) ⇒x∈(cot〖1,1] 〗

**Q5.**The range of f(x)=√((1-cosx ) √((1-cosx ) √((1-cosx ) √(…∞)) ) ) is

Solution

(c) Given f(x)=√((1-cosx ) √((1-cosx ) √((1-cosx ) √(…∞)) ) ) ⇒ f(x)=(1-cosx )^(1/2) (1-cosx )^(1/4) (1-cosx )^(1/8)…∞ ⇒ f(x)=(1-cosx )^(1/2+1/4+1/8+⋯∞) ⇒ f(x)=(1-cosx )^((1/2)/(1-1/2)) ⇒ f(x)=1-cosx ⇒ The range of f(x) is [0, 2)

(c) Given f(x)=√((1-cosx ) √((1-cosx ) √((1-cosx ) √(…∞)) ) ) ⇒ f(x)=(1-cosx )^(1/2) (1-cosx )^(1/4) (1-cosx )^(1/8)…∞ ⇒ f(x)=(1-cosx )^(1/2+1/4+1/8+⋯∞) ⇒ f(x)=(1-cosx )^((1/2)/(1-1/2)) ⇒ f(x)=1-cosx ⇒ The range of f(x) is [0, 2)

**Q6.**The values of b and c for which the identity f(x+1)-f(x)=8x+3 is satisfied, where f(x)=bx^2+cx+d, are

Solution

(b) ∵f(x+1)-f(x)=8x+3 ⇒{b(x+1)^2+c(x+1)+d}-{bx^2+cx+d}=8x+3 ⇒b{(x+1)^2-x^2 }+c=8x+3 ⇒b(2x+1)+c=8x+3 On comparing co-efficient of x and constant term, we get 2b=8 and b+c=3 Then b=4 and c=-1

(b) ∵f(x+1)-f(x)=8x+3 ⇒{b(x+1)^2+c(x+1)+d}-{bx^2+cx+d}=8x+3 ⇒b{(x+1)^2-x^2 }+c=8x+3 ⇒b(2x+1)+c=8x+3 On comparing co-efficient of x and constant term, we get 2b=8 and b+c=3 Then b=4 and c=-1

**Q7.**If the graph of y=f(x) is symmetrical about lines x=1 and x=2, then which of the following is true?

Solution

(c) From the given data f(1-x)=f(1+x) (1) And f(2-x)=f(2+x) (2) In (2) replacing x by 1+x, we have f(1-x)=f(3+x) ⇒ f(1+x)=f(3+x) [From (1)] ⇒ f(x)=f(2+x)

(c) From the given data f(1-x)=f(1+x) (1) And f(2-x)=f(2+x) (2) In (2) replacing x by 1+x, we have f(1-x)=f(3+x) ⇒ f(1+x)=f(3+x) [From (1)] ⇒ f(x)=f(2+x)

**Q8.**If the function f:[1,∞)→:[1,∞) is defined by f(x)=2^(x(x-1)), then f^(-1) (x) is

Solution

(b) y=2^x(x-1) ⇒x^2-x-log_2〖y=0;〗 ⇒x=1/2(1±√(1+4 log_2y )) Since xÏµ[1,∞), we choose x=1/2 (1+√(1+4 log_2y )) Or f^(-1) (x)=1/2 (1+√(1+4 log_2x ))

(b) y=2^x(x-1) ⇒x^2-x-log_2〖y=0;〗 ⇒x=1/2(1±√(1+4 log_2y )) Since xÏµ[1,∞), we choose x=1/2 (1+√(1+4 log_2y )) Or f^(-1) (x)=1/2 (1+√(1+4 log_2x ))

**Q9.**Which of the following functions is inverse to itself ?

Solution

(a) By checking for different function, we find that for f(x)=(1-x)/(1+x),f^(-1) (x)=f(x)

(a) By checking for different function, we find that for f(x)=(1-x)/(1+x),f^(-1) (x)=f(x)

**Q10.**The domain of the function f(x)=log_(3+x) (x^2-1) is

Solution

(c) f(x) is to be defined when x^2-1>0 and 3+x>0 and 3+x≠1 ⇒x^2>1 and x>-3 and x≠-2 ⇒x<-1 or x>1 and x>-3 and x≠-2 ∴ D_f=(-3,-2)∪(-2,-1)∪(1,∞)

(c) f(x) is to be defined when x^2-1>0 and 3+x>0 and 3+x≠1 ⇒x^2>1 and x>-3 and x≠-2 ⇒x<-1 or x>1 and x>-3 and x≠-2 ∴ D_f=(-3,-2)∪(-2,-1)∪(1,∞)