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 domain of f(x)= ln (〖ax〗^3+(a+b) x^2+(b+c)x+c), where a>0,b^2-4ac=0, is (where [⋅] represents greatest integer function).

Solution

A

A

**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.**The domain of the function f(x)=√(In_((|x|-1) ) (x^2+4x+4) ) is

Solution

(c) Case I 0<|x|-1<1⇒1<|x|<2, then x^2+4x+4≤1 ⇒x^2+4x+3≤0 ⇒-3≤x≤-1 So x∈(-2,-1) (1) Case II |x|-1>1⇒|x|>2, then x^2+4x+4≥1 ⇒x^2+4x+3≥0 ⇒x≥-1 or x≤-3 So, x∈(-∞,-3]∪(2,∞) (2) From (1) and (2), x∈(-∞,-3]∪(-2,-1)∪(2,∞)

(c) Case I 0<|x|-1<1⇒1<|x|<2, then x^2+4x+4≤1 ⇒x^2+4x+3≤0 ⇒-3≤x≤-1 So x∈(-2,-1) (1) Case II |x|-1>1⇒|x|>2, then x^2+4x+4≥1 ⇒x^2+4x+3≥0 ⇒x≥-1 or x≤-3 So, x∈(-∞,-3]∪(2,∞) (2) From (1) and (2), x∈(-∞,-3]∪(-2,-1)∪(2,∞)

**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.**If f(x)=√(n&x^m ),n∈N, is an even function, then m is

Solution

(a) Given f(x)=√(n&x^m ),n∈N is an even function where m∈I ⇒ f(x)=f(-x) ⇒√(n&x^m )=√(n&(-x)^m ) ⇒x^m=(-x)^m ⇒m is an even integer ⇒m=2k,k∈I

(a) Given f(x)=√(n&x^m ),n∈N is an even function where m∈I ⇒ f(x)=f(-x) ⇒√(n&x^m )=√(n&(-x)^m ) ⇒x^m=(-x)^m ⇒m is an even integer ⇒m=2k,k∈I