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 f:N→N defined by f(x)=x^2+x+1,x∈N, then f is

Solution

(c) Let x,y∈N such that f(x)=f(y) Then f(x)=f(y) ⇒x^2+x+1=y^2+y+1 ⇒(x-y)(x+y+1)=0 ⇒x=y or x=(-y-1)∉N ∴ f is one-one Also, f(x) does not take all positive integral values. Hence f is into

(c) Let x,y∈N such that f(x)=f(y) Then f(x)=f(y) ⇒x^2+x+1=y^2+y+1 ⇒(x-y)(x+y+1)=0 ⇒x=y or x=(-y-1)∉N ∴ f is one-one Also, f(x) does not take all positive integral values. Hence f is into

**Q2.**If f(3x+2)+f(3x+29)=0 ∀ x∈R, then the period of f(x) is

Solution

(d) f(3x+2)+f(3x+29)=0 (1) Replacing x by x+9, we get f(3(x+9)+2)+f(3(x+9)+29=0 ⇒ f(3x+29)+f(3x+56)=0 (2) From (1) and (2), we get f(3x+2)=f(3x+56) ⇒ f(3x+2)=f(3(x+18)+2) ⇒ f(x) is periodic with period 54

(d) f(3x+2)+f(3x+29)=0 (1) Replacing x by x+9, we get f(3(x+9)+2)+f(3(x+9)+29=0 ⇒ f(3x+29)+f(3x+56)=0 (2) From (1) and (2), we get f(3x+2)=f(3x+56) ⇒ f(3x+2)=f(3(x+18)+2) ⇒ f(x) is periodic with period 54

**Q3.**Which of the following functions is periodic?

Solution

(a) f(x)={x} is periodic with period 1 f(x)=sin〖1/x〗 for x≠0,f(0)=0 is non-periodic as g(x)=1/x is non-periodic Also f(x)=x cosx is non-periodic as g(x)=x is non-periodic

(a) f(x)={x} is periodic with period 1 f(x)=sin〖1/x〗 for x≠0,f(0)=0 is non-periodic as g(x)=1/x is non-periodic Also f(x)=x cosx is non-periodic as g(x)=x is non-periodic

**Q5.**The domain of the function f(x)=sin^(-1)(3-x)/In(|x|-2) Is

Solution

(b) f(x)=sin^(-1)(3-x)/log〖(|x|-2)〗 Let g(x)=sin^(-1)(3-x) ⇒-1≤3-x≤1 The domain of g(x) is [2, 4] And let h(x)=log(|x|-2) ⇒|x|-2>0 or |x|>2 ⇒x<-2 or x>2 ⇒(-∞,-2)∪(2,∞) We know that (f⁄g)(x) f(x)/(g(x))∀x∈D_1∩D_2-{x∈R:g(x)=0} ∴ the domain of f(x)=(2,4]-{3}=(2,3)∪(3,4]

(b) f(x)=sin^(-1)(3-x)/log〖(|x|-2)〗 Let g(x)=sin^(-1)(3-x) ⇒-1≤3-x≤1 The domain of g(x) is [2, 4] And let h(x)=log(|x|-2) ⇒|x|-2>0 or |x|>2 ⇒x<-2 or x>2 ⇒(-∞,-2)∪(2,∞) We know that (f⁄g)(x) f(x)/(g(x))∀x∈D_1∩D_2-{x∈R:g(x)=0} ∴ the domain of f(x)=(2,4]-{3}=(2,3)∪(3,4]

**Q6.**If f(x) is an even function and satisfies the relation x^2 f(x)-2f(1/x)=g(x) where g(x) is an odd function, then f(5) equals

Solution

(a) x^2 f(x)-2f(1/x)=g(x) and 2f(1/x)-4x^2 f(x)=2x^2 g(1/x) (Replacing x by 1/x) ⇒ -3x^2 f(x)=g(x)+2x^2 g(1/x) (Eliminating f(1/x)) ⇒ f(x)=-((g(x)=2x^2 g(1/x))/(3x^2 )) ∵ g(x) and x^2 are odd and even functions, respectively So, f(x) is an odd function But f(x) is given even ⇒ f(x)=0∀x Hence, f(5)=0

(a) x^2 f(x)-2f(1/x)=g(x) and 2f(1/x)-4x^2 f(x)=2x^2 g(1/x) (Replacing x by 1/x) ⇒ -3x^2 f(x)=g(x)+2x^2 g(1/x) (Eliminating f(1/x)) ⇒ f(x)=-((g(x)=2x^2 g(1/x))/(3x^2 )) ∵ g(x) and x^2 are odd and even functions, respectively So, f(x) is an odd function But f(x) is given even ⇒ f(x)=0∀x Hence, f(5)=0

**Q7.**If f(x)=sin〖([x]Ï€)〗/(x^2+x+1), where [.] denotes the greatest integer function, then

Solution

(c) f(x)=sin〖[x]Ï€〗/(x^2+x+1) Let [x]=n∈ integer ⇒sin〖[x]Ï€=0〗 ⇒ f(x)=0 ⇒f(x) is constant function

(c) f(x)=sin〖[x]Ï€〗/(x^2+x+1) Let [x]=n∈ integer ⇒sin〖[x]Ï€=0〗 ⇒ f(x)=0 ⇒f(x) is constant function

**Q8.**If x satisfies |x-1|+|x-2|+|x-3|≥6, then

Solution

(c)

(c)

**Q9.**The period of function 2^({x})+sinÏ€ 〖x+3〗^{x⁄2} +cos2Ï€x (where {x} denotes the fractional part of x) is

Solution

(a) The period of sinÏ€x and cos2Ï€x is 2 and 1, respectively The period of 2^({x}) is 1 The period of 3^({x⁄2}) is 2 Hence, the period of f(x) is LCM of 1 and 2=2

(a) The period of sinÏ€x and cos2Ï€x is 2 and 1, respectively The period of 2^({x}) is 1 The period of 3^({x⁄2}) is 2 Hence, the period of f(x) is LCM of 1 and 2=2

**Q10.**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)