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 range of f(x)=(x+1)(x+2)(x+3)(x+4)+5 for x∈[-6,6] is

Solution

(a) Let g(x)=(x+1)(x+2)(x+3)(x+4) The rough graph of g(x) is given as ∴g(x)=(x+1)(x+2)(x+3)(x+4) =(x+1)(x+4)(x+2)(x+3) =(x^2+5x+4)(x^2+5x+6) =t(t+2)=〖(t+1)〗^2-1, Where t=x^2+5x Now g_min=-1, for which x^2+5x-1 has real roots in [-6,6] Also g(6)=7×8×9×10=5040 Hence, the range of g(x) is [-1,5040] for x∈[-6,6] Then, the range of f(x) is [4,5045]

(a) Let g(x)=(x+1)(x+2)(x+3)(x+4) The rough graph of g(x) is given as ∴g(x)=(x+1)(x+2)(x+3)(x+4) =(x+1)(x+4)(x+2)(x+3) =(x^2+5x+4)(x^2+5x+6) =t(t+2)=〖(t+1)〗^2-1, Where t=x^2+5x Now g_min=-1, for which x^2+5x-1 has real roots in [-6,6] Also g(6)=7×8×9×10=5040 Hence, the range of g(x) is [-1,5040] for x∈[-6,6] Then, the range of f(x) is [4,5045]

**Q2.**If f(x)=log_e((x^2+e)/(x^2+1)), then the range of f(x) is

Solution

(d) f(x)=ln((x^2+e)/(x^2+1))= ln((x^2+1+e-1)/(x^2+1))= ln(1+(e-1)/(x^2+1))

(d) f(x)=ln((x^2+e)/(x^2+1))= ln((x^2+1+e-1)/(x^2+1))= ln(1+(e-1)/(x^2+1))

**Q3.**Let f(x)=√(|x|-{x}) (where {.} denotes the fractional part of x) and X,Y are its domain and range, respectively, then

Solution

(c) f(x)=√(|x|-{x}) is defined if |x|≥{x} ⇒x∈(-∞-1/2]∪[0,∞)⇒Y∈[0,∞)

(c) f(x)=√(|x|-{x}) is defined if |x|≥{x} ⇒x∈(-∞-1/2]∪[0,∞)⇒Y∈[0,∞)

**Q4.**If X and Y are two non-empty sets where f:X→Y is function is defined such that f(C)={f(x):x∈C}for C⊆X And f^(-1) (D)={x:f(x)∈D}for D⊆Y, For any A⊆X and B⊆Y, then

Solution

(c) The given data is shown in the figure below Since, f^(-1) (D)=x ⇒ f(x)=D Now, if B⊂X,f(B)⊂D ⇒ f^(-1) (f(B))=B

(c) The given data is shown in the figure below Since, f^(-1) (D)=x ⇒ f(x)=D Now, if B⊂X,f(B)⊂D ⇒ f^(-1) (f(B))=B

**Q5.**Possible values of a such that the equation x^2+2ax+a=√(a^2+x-1/16)-1/16,x≥-a, has two distinct real roots are given by

Solution

(d) The equation is x^2+2ax+1/16=-a+√(a^2+x-1/16) ⇒ f(x)=f^(-1) (x) Which has the solution if x^2+2ax+1/16=x ⇒x^2+(2a-1)x+1/16=0 For real and distinct roots (2a-1)^2-4 1/16≥0 ⇒2a-1≤(-1)/2 or 2a-1≥1/2⇒a≤1/4 or a≥3/4

(d) The equation is x^2+2ax+1/16=-a+√(a^2+x-1/16) ⇒ f(x)=f^(-1) (x) Which has the solution if x^2+2ax+1/16=x ⇒x^2+(2a-1)x+1/16=0 For real and distinct roots (2a-1)^2-4 1/16≥0 ⇒2a-1≤(-1)/2 or 2a-1≥1/2⇒a≤1/4 or a≥3/4

**Q6.**If f(x) is a polynomial satisfying f(x)f(1⁄x)=f(x)+f(1⁄x) and f(3)=28, then f(4) is equal to

Solution

(b) f(x)=x^n+1 ⇒f(3)=3^n+1=28 ⇒3^n=27 ∴n=3 ⇒f(4)=4^3+1=65

(b) f(x)=x^n+1 ⇒f(3)=3^n+1=28 ⇒3^n=27 ∴n=3 ⇒f(4)=4^3+1=65

**Q7.**If f(x+f(y))=f(x)+y ∀ x,y∈R and f(0)=1, then the value of f(7) is

Solution

(a) f(x+f(y))=f(x)+y,f(0)=1 Putting y=0, we get f(x+f(0))=f(x)+0 ⇒ f(x+1)=f(x)∀x∈R Thus, f(x) is the period with 1 as one of its period ⇒f(7)=f(6)=f(5)=⋯=f(1)=(0)=1

(a) f(x+f(y))=f(x)+y,f(0)=1 Putting y=0, we get f(x+f(0))=f(x)+0 ⇒ f(x+1)=f(x)∀x∈R Thus, f(x) is the period with 1 as one of its period ⇒f(7)=f(6)=f(5)=⋯=f(1)=(0)=1

**Q8.**If f(x)=cos〖(log_ex)〗, then f(x)f(y)-1/2 [f(x/y)+f(xy)] has the value

Solution

(d) f(x)=cos(logx ) ⇒f(x)f(y)-1/2 [f(x/y)+f(xy)] =cos(logx ) cos(logy )-1/2 [cos(log〖x-logy 〗 ) ]+cos(log〖x+logy 〗 ) =cos(logx ) cos(logy )-1/2 [2 cos(logx ) cos(logy ) ] = 0

(d) f(x)=cos(logx ) ⇒f(x)f(y)-1/2 [f(x/y)+f(xy)] =cos(logx ) cos(logy )-1/2 [cos(log〖x-logy 〗 ) ]+cos(log〖x+logy 〗 ) =cos(logx ) cos(logy )-1/2 [2 cos(logx ) cos(logy ) ] = 0

**Q9.**If f is periodic, g is polynomial function and f(g(x)) is periodic and g(2)=3, g(4)=7 then g(6) is

Solution

(c) From the given data g(x) must be linear function Hence, g(x)=ax+b Also g(2)=2a+b=3 and g(4)=4a+b=7 Solving, we get a=2 and b=-1 Hence, g(x)=2x-1 Then, g(6)=11

(c) From the given data g(x) must be linear function Hence, g(x)=ax+b Also g(2)=2a+b=3 and g(4)=4a+b=7 Solving, we get a=2 and b=-1 Hence, g(x)=2x-1 Then, g(6)=11

**Q10.**A function F(x) satisfies the functional equation x^2 F(x)+F(1-x)=2x-x^4 for all real x. F(x) must be

Solution

(b) x^2 F(x)+F(1-x)=2x-x^4 (1) Replacing x by 1-x, we get ⇒(1-x)^2 F(1-x)+F(x)=2(1-x)-〖(1-x)〗^4 (2) Eliminating F(1-x) from (1) and (2), we get F(x)=1-x^2

(b) x^2 F(x)+F(1-x)=2x-x^4 (1) Replacing x by 1-x, we get ⇒(1-x)^2 F(1-x)+F(x)=2(1-x)-〖(1-x)〗^4 (2) Eliminating F(1-x) from (1) and (2), we get F(x)=1-x^2