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 x is real, then the value of the expression (x^2+14x+9)/(x^2+2x+3) lies between

Solution

(c) (x^2+14x+9)/(x^2+2x+3)=y ⇒x^2+14x+9=x^2 y+2xy+3y ⇒x^2 (y-1)+2x(y-7)+(3y-9)=0 Since x is real, ∴4(y-7)^2-4(3y-9)(y-1)>0 ⇒4(y^2+49-14y)-4(〖3y〗^2+9-12y)>0

(c) (x^2+14x+9)/(x^2+2x+3)=y ⇒x^2+14x+9=x^2 y+2xy+3y ⇒x^2 (y-1)+2x(y-7)+(3y-9)=0 Since x is real, ∴4(y-7)^2-4(3y-9)(y-1)>0 ⇒4(y^2+49-14y)-4(〖3y〗^2+9-12y)>0

**Q2.**The domain of the function f(x)=x/√(sin〖(ln x)-cos(ln x) 〗 )(n∈Z) Is

Solution

(b) For the domain sin(lnx )>cos〖(lnx)〗 and x>0

(b) For the domain sin(lnx )>cos〖(lnx)〗 and x>0

**Q3.**If af(x+1)+bf(1/(x+1))=x,x≠-1,a≠b, then f(2) is equal to

Solution

(a) af(x+1)+bf(1/(x+1))=(x+1)-1 (1) Replacing x+1 by 1/(x+1), we get ∴ af(1/(x+1))+bf(x+1)=1/(x+1)-1 (2) (1)×a-(2)×b⇒(a^2-b^2 )f(x+1)=a(x+1)-a-b/(x+1)+b Putting x=1,(a^2-b^2 )f(2)=2a-a-b/2+b=a+b/2 =(2a+b)/2

(a) af(x+1)+bf(1/(x+1))=(x+1)-1 (1) Replacing x+1 by 1/(x+1), we get ∴ af(1/(x+1))+bf(x+1)=1/(x+1)-1 (2) (1)×a-(2)×b⇒(a^2-b^2 )f(x+1)=a(x+1)-a-b/(x+1)+b Putting x=1,(a^2-b^2 )f(2)=2a-a-b/2+b=a+b/2 =(2a+b)/2

**Q4.**The range of f(x)=[|sinx |+|cosx |], where [.] denotes the greatest integer function, is

Solution

(c) y=|sinx |+|cosx | ⇒y^2=1+|sin2x | ⇒1≤y^2≤2 ⇒y∈[1,√2] ⇒ f(x)=1∀x∈R

(c) y=|sinx |+|cosx | ⇒y^2=1+|sin2x | ⇒1≤y^2≤2 ⇒y∈[1,√2] ⇒ f(x)=1∀x∈R

**Q5.**The number of roots of the equation x sinx=1,x∈[-2Ï€,0)∪(0,2Ï€], is

Solution

(c) x sinx=1 (1) ⇒y=sinx=1/x Root of equation (1) will be given by the point(s) of intersection of the graphs y=sinx and y=1/x. Graphically, it is clear that we get four roots.

(c) x sinx=1 (1) ⇒y=sinx=1/x Root of equation (1) will be given by the point(s) of intersection of the graphs y=sinx and y=1/x. Graphically, it is clear that we get four roots.

**Q6.**If f(2x+3y,2x-7y)=20x, then f(x,y) equals

Solution

(b) Let 2x+3y=m and 2x-7y=n ⇒y=(m-n)/10 and x=(7m-3n)/20 ⇒f(m,n)=7m+3n ⇒ f(x,y)=7x+3y

(b) Let 2x+3y=m and 2x-7y=n ⇒y=(m-n)/10 and x=(7m-3n)/20 ⇒f(m,n)=7m+3n ⇒ f(x,y)=7x+3y

**Q7.**The range of f(x)=sin^(-1)(√(x^2+x+1)) is

Solution

(c) For the function to get defined 0≤x^2+x+1≤1, But x^2+x+1≥3/4⇒√3/2≤√(x^2+x+1)≤1 ⇒Ï€/3≤sin^(-1)〖(√(x^2+x+1))≤Ï€/2 〗

(c) For the function to get defined 0≤x^2+x+1≤1, But x^2+x+1≥3/4⇒√3/2≤√(x^2+x+1)≤1 ⇒Ï€/3≤sin^(-1)〖(√(x^2+x+1))≤Ï€/2 〗

**Q8.**The function f(x)=sin(log(x+√(1+x^2 )) ) is

Solution

(b) f(x)=sin(log(x+√(1+x^2 )) ) ⇒f(-x)=sin[log(-x+√(1+x^2 )) ] ⇒ f(-x)=sinlog((√(1+x^2 )-x) ((√(1+x^2 )+x))/((√(1+x^2 )+x) )) ⇒ f(-x)=sinlog[1/((x+√(1+x^2 )) )] ⇒ f(-x)=sin[-log(x+√(1+x^2 )) ] ⇒ f(-x)=〖-sin〗[log(x+√(1+x^2 )) ] ⇒ f(-x)=-f(x) ⇒ f(x) is an odd function

(b) f(x)=sin(log(x+√(1+x^2 )) ) ⇒f(-x)=sin[log(-x+√(1+x^2 )) ] ⇒ f(-x)=sinlog((√(1+x^2 )-x) ((√(1+x^2 )+x))/((√(1+x^2 )+x) )) ⇒ f(-x)=sinlog[1/((x+√(1+x^2 )) )] ⇒ f(-x)=sin[-log(x+√(1+x^2 )) ] ⇒ f(-x)=〖-sin〗[log(x+√(1+x^2 )) ] ⇒ f(-x)=-f(x) ⇒ f(x) is an odd function

**Q9.**The function f(x)=sec^(-1)x/√(x-[x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x∈

Solution

(b) The function sec^(-1)x is defined for all x∈R-(-1,1) and the function 1/√(x-[x] ) is defined for all x∈R-Z So the given function is defined for all x∈R-{(-1,1)∪{n│n∈Z}}

(b) The function sec^(-1)x is defined for all x∈R-(-1,1) and the function 1/√(x-[x] ) is defined for all x∈R-Z So the given function is defined for all x∈R-{(-1,1)∪{n│n∈Z}}

**Q10.**Let f(x)=(x+1)^2-1,x≥1, Then the set {x:f(x)=f^(-1) (x)} is

Solution

(c) Since f(x)=(x+1)^2-1 is continuous function, solution of f(x)=f^(-1) (x) lies on the line y=x ⇒f(x)=f^(-1) (x)=x ⇒(x+1)^2-1=x ⇒x^2+x=0 ⇒x=0 or -1 ⇒ The required set is {0,-1}

(c) Since f(x)=(x+1)^2-1 is continuous function, solution of f(x)=f^(-1) (x) lies on the line y=x ⇒f(x)=f^(-1) (x)=x ⇒(x+1)^2-1=x ⇒x^2+x=0 ⇒x=0 or -1 ⇒ The required set is {0,-1}