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:[-10,10]→R, where f(x)=sinx+[x^2⁄a] be an odd function. Then the set of values of parameter a is/are

Solution

(d) Since f (x) is an odd function, [x^2/a]=0 for all x∈[-10,10] ⇒0≤x^2/a<1 for all x∈[-10,10]⇒a>100

(d) Since f (x) is an odd function, [x^2/a]=0 for all x∈[-10,10] ⇒0≤x^2/a<1 for all x∈[-10,10]⇒a>100

**Q2.**If the graph of the function f(x)=(a^x-1)/(x^n (a^x+1) ) is symmetrical about y-axis, then n equals

Solution

(d) f(x)=(a^x-1)/(x^n (a^x+1) ) f(x) is symmetrical about y-axis ⇒ f(x)=f(-x) ⇒(a^x-1)/(x^n (a^x+1) )=(a^(-x)-1)/(〖(-x)〗^n (a^(-x)+1) ) ⇒(a^x-1)/(x^n (a^x+1) )=(1-a^x)/(〖(-x)〗^n (1+a^x ) )⇒x^n=-〖(-x)〗^n

(d) f(x)=(a^x-1)/(x^n (a^x+1) ) f(x) is symmetrical about y-axis ⇒ f(x)=f(-x) ⇒(a^x-1)/(x^n (a^x+1) )=(a^(-x)-1)/(〖(-x)〗^n (a^(-x)+1) ) ⇒(a^x-1)/(x^n (a^x+1) )=(1-a^x)/(〖(-x)〗^n (1+a^x ) )⇒x^n=-〖(-x)〗^n

**Q3.**The domain of the function f(x)=√(log(1/|sinx | ) ) is

Solution

(b) f(x) is defined for log〖(1/|sinx | )≥0〗 ⇒1/|sinx | ≥1 and |sinx |≠0 ⇒|sinx |≠0 [∵1/|sinx | ≥1 for all x] ⇒x≠nÏ€,n∈Z Hence, the domain of f(x)=R-{nÏ€:n∈Z}

(b) f(x) is defined for log〖(1/|sinx | )≥0〗 ⇒1/|sinx | ≥1 and |sinx |≠0 ⇒|sinx |≠0 [∵1/|sinx | ≥1 for all x] ⇒x≠nÏ€,n∈Z Hence, the domain of f(x)=R-{nÏ€:n∈Z}

**Q4.**Let f(x)=Î±x/(x+1),x≠-1. Then for what value of Î± is f(f(x))=x?

Solution

(d) f(x)=Î±x/(x+1),x≠-1 f(f(x))=x⇒Î±(Î±x/(x+1))/(Î±x/(x+1)+1)=x ⇒(Î±^2 x)/((Î±+1) x+1)=x ⇒(Î±+1) x^2+(1-Î±^2 )x=0 ⇒Î±+1=0 and 1-Î±^2=0 [As true ∀ x≠1∴ Eq. (1) is an identity] ⇒Î±=-1

(d) f(x)=Î±x/(x+1),x≠-1 f(f(x))=x⇒Î±(Î±x/(x+1))/(Î±x/(x+1)+1)=x ⇒(Î±^2 x)/((Î±+1) x+1)=x ⇒(Î±+1) x^2+(1-Î±^2 )x=0 ⇒Î±+1=0 and 1-Î±^2=0 [As true ∀ x≠1∴ Eq. (1) is an identity] ⇒Î±=-1

**Q5.**If f: [1,∞)→┤ [2,∞)┤ is given by f(x)=x+1/x , then f^(-1) (x) equals

Solution

(a) f: [1,∞)→┤ [2,∞)┤ f(x)=x+1/x=y ⇒x^2-yx+1=0 ⇒x=(y±√(y^2-4))/2 But given f: [1,∞)→┤ [2,∞)┤ ∴x=(y+√(y^2-4))/2

(a) f: [1,∞)→┤ [2,∞)┤ f(x)=x+1/x=y ⇒x^2-yx+1=0 ⇒x=(y±√(y^2-4))/2 But given f: [1,∞)→┤ [2,∞)┤ ∴x=(y+√(y^2-4))/2

**Q6.**The domain of f(x)=sin^(-1)[〖2x〗^2-3], where [.] denotes the greatest integer function, is

Solution

(d)

(d)

**Q7.**Domain of definition of the function f(x)=√(sin^-〖(2x)+Ï€/6〗 ) for real valued x, is

Solution

(a) Here, f(x)=√(sin^(-1)〖(2x)+Ï€/6〗 ) , to find domain we must have, sin^(-1)〖(2x)+Ï€/6≥0〗 (but-Ï€/2≤sin^(-1)Î¸≤Ï€/2) ∴ -Ï€/6≤sin^(-1)〖(2x)〗≤Ï€/2 〖⇒sin〗(-Ï€/6)≤2x≤sin(Ï€/2) ⇒ -1/2≤2x≤1 ⇒x∈[-1/4,1/2]

(a) Here, f(x)=√(sin^(-1)〖(2x)+Ï€/6〗 ) , to find domain we must have, sin^(-1)〖(2x)+Ï€/6≥0〗 (but-Ï€/2≤sin^(-1)Î¸≤Ï€/2) ∴ -Ï€/6≤sin^(-1)〖(2x)〗≤Ï€/2 〖⇒sin〗(-Ï€/6)≤2x≤sin(Ï€/2) ⇒ -1/2≤2x≤1 ⇒x∈[-1/4,1/2]

**Q8.**Let f:R→R,g:R→R be two given functions such that f is injective and g is surjective, then which of the following is injective?

Solution

(d) If f is injective and g is surjective ⇒fog is injective ⇒fof is injective

(d) If f is injective and g is surjective ⇒fog is injective ⇒fof is injective

**Q9.**Let X={a_1,a_2,…,a_6 } and Y={b_1,b_2,b_3 }. The number of functions f from x to y such that it is onto and there are exactly three elements x in X such that f(x)=b_1 is

Solution

(d) Image b_1 is assigned to any three of the six pre-images in ^6 C_3 ways Rest two images can be assigned to remaining three pre-images in 2^3-2 ways (as function is onto) Hence number of functions are ^6 C_3×(2^3-2)=20×6=120

(d) Image b_1 is assigned to any three of the six pre-images in ^6 C_3 ways Rest two images can be assigned to remaining three pre-images in 2^3-2 ways (as function is onto) Hence number of functions are ^6 C_3×(2^3-2)=20×6=120

**Q10.**f:N→N where f(x)=x-(-1)^x then f is

Solution

(c) f(x)={█(x-1,x is even@x+1,x is odd)┤, where is clearly are one-one and onto

(c) f(x)={█(x-1,x is even@x+1,x is odd)┤, where is clearly are one-one and onto