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] and {x} represent the integral and fractional parts of x, respectively, then the value of ∑_(r=1)^2000▒({x+r})/2000 is

Solution

(c) ∑_(r=1)^2000▒({x+r})/2000=∑_(r=1)^2000▒({x})/2000=2000 ({x})/2000={x}

(c) ∑_(r=1)^2000▒({x+r})/2000=∑_(r=1)^2000▒({x})/2000=2000 ({x})/2000={x}

**Q2.**If f:[0,∞]→[0,∞] and f(x)=x/(1+x), then f is

Solution

(b) Here, f:[0,∞]→[0,∞)ie, domain is [0,∞) and codomain is [0,∞). For one-one f(x)=x/(1+x) ⇒ f^' (x)=1/(1+x)^2 >0,∀ x∈[0,∞) ∴f(x) is increasing in its domain. Thus, f(x) is one-one in its domain. For onto (we find range) f(x)=x/(1+x) ie,y=x/(1+x)⇒y+yx=x ⇒x=y/(1-y)⇒y/(1-y)≥0 as x≥0 ∴0≤y≠1 ie, Range ≠ Codomain ∴f(x) is one-one but not onto.

(b) Here, f:[0,∞]→[0,∞)ie, domain is [0,∞) and codomain is [0,∞). For one-one f(x)=x/(1+x) ⇒ f^' (x)=1/(1+x)^2 >0,∀ x∈[0,∞) ∴f(x) is increasing in its domain. Thus, f(x) is one-one in its domain. For onto (we find range) f(x)=x/(1+x) ie,y=x/(1+x)⇒y+yx=x ⇒x=y/(1-y)⇒y/(1-y)≥0 as x≥0 ∴0≤y≠1 ie, Range ≠ Codomain ∴f(x) is one-one but not onto.

**Q3.**If f(x)={█(x,x is rational@1-x, x is irrational)┤ then f(f(x)) is

Solution

(a) f(f(x))={█(f(x),f(x) is rational@1-f(x),f(x) is irrational)┤ ⇒ f(f(x))={█(x,x is rational@1-(1-x)=x,x is irrational)┤

(a) f(f(x))={█(f(x),f(x) is rational@1-f(x),f(x) is irrational)┤ ⇒ f(f(x))={█(x,x is rational@1-(1-x)=x,x is irrational)┤

**Q4.**Let h(x)=|kx+5|, the domain of f(x) is [-5,7], the domain of f(h(x)) is [-6,1] and the range of h(x) is the same as the domain of f(x), then the value of k is

Solution

(b) -5≤|kx+5|≤7 ⇒-12≤kx≤2 where -6≤x≤1 ⇒-6≤k/2 x≤1 where -6≤x≤1 ∴k=2 [∵ the range of h(x)= the domain of f(x)]

(b) -5≤|kx+5|≤7 ⇒-12≤kx≤2 where -6≤x≤1 ⇒-6≤k/2 x≤1 where -6≤x≤1 ∴k=2 [∵ the range of h(x)= the domain of f(x)]

**Q5.**f(x)={█(x,if x is rational@0,if x is irrational)┤ and f(x)={█(0,if x is rational@x,if x is irrational )┤. Then, f-g is

Solution

(d) Let Ï•(x)=f(x)-g(x) ={█(x,x∈Q@-x,x∉Q)┤ For one-one Take any straight line parallel to x-axis which will intersect Ï•(x) only at one point. ⇒Ï•(x) is one-one. Foe onto As, Ï•(x)={█(x,x∈Q@-x,x∉Q)┤, which shows y=x and y=-x for irrational values ⇒y∉ real numbers. ∴ Range=Codomain ⇒ Ï•(x) is onto. Thus, f-g is one-one and onto.

(d) Let Ï•(x)=f(x)-g(x) ={█(x,x∈Q@-x,x∉Q)┤ For one-one Take any straight line parallel to x-axis which will intersect Ï•(x) only at one point. ⇒Ï•(x) is one-one. Foe onto As, Ï•(x)={█(x,x∈Q@-x,x∉Q)┤, which shows y=x and y=-x for irrational values ⇒y∉ real numbers. ∴ Range=Codomain ⇒ Ï•(x) is onto. Thus, f-g is one-one and onto.

**Q6.**The domain of f(x)=log|logx | Is

Solution

(c) f(x)=log〖|logx |,〗 f(x) is defined if |logx |>0 and x>0 i.e., if x>0 and x≠1 (∵|logx |>0 if x≠1) ⇒x∈(0,1)∪(1,∞)

(c) f(x)=log〖|logx |,〗 f(x) is defined if |logx |>0 and x>0 i.e., if x>0 and x≠1 (∵|logx |>0 if x≠1) ⇒x∈(0,1)∪(1,∞)

**Q7.**If f:R→R is an invertible function such that f(x) and f^(-1) (x) are symmetric about the line y=-x, then

Solution

(a) Since f(x) and f^(-1) (x) are symmetric about the line y=-x If (Î±,Î²) lies on y=f(x) then (-Î²,-Î±) on y=f^(-1) (x) ⇒(-Î±,-Î²) lies on y=f(x) ⇒y=f(x) is odd

(a) Since f(x) and f^(-1) (x) are symmetric about the line y=-x If (Î±,Î²) lies on y=f(x) then (-Î²,-Î±) on y=f^(-1) (x) ⇒(-Î±,-Î²) lies on y=f(x) ⇒y=f(x) is odd

**Q8.**If f(2x+y/8,2x-y/8)=xy, then f(m,n)+(n,m)=0

Solution

(d) Let 2x+y/8=Î± and 2x-y/8=Î², then x=(Î±+Î²)/4 and y=4(Î±-Î²) Given, f(2x+y/8,2x-y/8)=xy ⇒f(Î±,Î²)=Î±^2-Î²^2 ⇒f(m,n)+f(n,m)=m^2-n^2+n^2-m^2=0 for all m,n

(d) Let 2x+y/8=Î± and 2x-y/8=Î², then x=(Î±+Î²)/4 and y=4(Î±-Î²) Given, f(2x+y/8,2x-y/8)=xy ⇒f(Î±,Î²)=Î±^2-Î²^2 ⇒f(m,n)+f(n,m)=m^2-n^2+n^2-m^2=0 for all m,n

**Q9.**The period of the function f(x)=c^(sin^2x+sin^2(x+Ï€/3)+cosx cos(x+Ï€/3) ) is (where c is constant)

Solution

(d) sin^2x+sin^2(x+Ï€/3)+cosx cos(x+Ï€/3) =sin^2x+(sinx/2+(√3 cosx)/2)^2+cosx (cosx/2-(√3 sinx)/2) =sin^2x+sin^2x/4+(3 cos^2x)/4+cos^2x/2 =(5 sin^2x)/4+(5 cos^2x)/4=5⁄4 Hence, f(x)=c^(5⁄4)= constant, which is periodic whose period cannot be determined

(d) sin^2x+sin^2(x+Ï€/3)+cosx cos(x+Ï€/3) =sin^2x+(sinx/2+(√3 cosx)/2)^2+cosx (cosx/2-(√3 sinx)/2) =sin^2x+sin^2x/4+(3 cos^2x)/4+cos^2x/2 =(5 sin^2x)/4+(5 cos^2x)/4=5⁄4 Hence, f(x)=c^(5⁄4)= constant, which is periodic whose period cannot be determined

**Q10.**The range of f(x)=[1+sinx ]+[2+sin〖x/2〗 ]+[3+sin〖x/3〗 ]+⋯+[n+sin〖x/n〗 ],∀ x∈[0,Ï€], where [.] denotes the greatest integer function, is

Solution

(d) f(x)=n(n+1)/2+[sinx ]+[sin〖x/2〗 ]+⋯+[sin〖x/n〗 ] Thus, the range of f(x)={n(n+1)/2,n(n+1)/2+1} as x∈[0,Ï€]

(d) f(x)=n(n+1)/2+[sinx ]+[sin〖x/2〗 ]+⋯+[sin〖x/n〗 ] Thus, the range of f(x)={n(n+1)/2,n(n+1)/2+1} as x∈[0,Ï€]