Q: Let R denote the set of all real numbers. Let a_ i , b_ i R for i \ 1,2,3\ . Define the functions f: R R , g: R R , and h: R R by aligned & f(x)=a_1+10 x+a_2 x^2+a_3 x^3+x^4, \\ & g(x)=b_1+3 x+b_2 x^2

aligned & h(x)=f(x+1)-g(x+2) \\ & =a_1+10(x+1)+a_2(x+1)^2+a_3(x+1)^3+(x+1)^4-b_1-3(x+2)-b_2(x+2)^2-b_3(x+2)^3-(x+2)^4 aligned Coeff. of x^3 in h(x) aligned & =a_3-b_3-4 \\ & f(x)-g(x) 0 x R \\ & a_1+10 x+a_2 x^2+a_3 x^3+x^4-b_1-3 x-b_2 x^2-b_3 x^3-x^4 0 \\ & x^3 (a_3-b_3 )+x^2 (a_2-b_2 )+7 x+ (a_1-b_1 ) 0 aligned Cubic Eq. will become zero at atleast are value of x So it will be quadratic a _3- b _3=0

Q: Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f: N Z and g: Z N defined by f(n)= cases (n+1) / 2 & if n is odd \\ (4-n) / 2 & if n is even c

aligned & f(n)= cases (n+1) / 2 & if n is odd \\ (4-n) / 2 & if n is even cases \\ & f(n)=\ (1,1),(2,1),(3,2),(4,0),(5,3),(6,-1), \ aligned f ( n ) is many one and onto function aligned & g ( n )= \ array cc 3+2 n & if n 0 \\ -2 n & if n <0 array . \\ & g ( n )=\ (-3,6),(-2,4),(-1,2),(0,3),(1,5),(2,7),(3,9),(4,15), \ aligned g ( n ) is one-one and into function f ( ~g ( n ))=2+ n , n ~N fog is one-one and into g(f(n))= cases 4+n & if n is odd natural number \\ 7-n & if n=2,4 \\ n-4 & if n is eve

Q: Let f: R R be a function such that f(x+y)=f(x)+f(y) for all x, y R , and g: R (0, ) be a function such that g(x+y)=g(x) g(y) for all x, y R . If f ( -3 5 )=12 and g ( -1 3 )=2, then the value of (f (

f(x+y)=f(x)+f(y) ...(1) f ( nx )= nf ( x ) n N ...(2) Now put y=-x in eq.(1) aligned & f ( x )+ f (- x )= f (0) \ f (0)=0\ \\ & f (- x )=- f ( x ) aligned f is odd function from eq. (2) from eq. (2) aligned & f(-n x)= nf (-x) \\ & f (- nx )=-n f( x ) \\ & aligned f ( mx )= mf ( x ) m Z ...(3) from eq. (2) and eq. (3) f ( nx )= nf ( x ) n Z ...(4) Now put x= p q where p , q Z , q 0 f ( np q )= nf ( p q ) n Z put n = q f(p)=q f ( p q ) pf (1)= qf ( p q ) from eq.(4) Let f (1)= a then pa = qf ( p q

Q: Let f: R R and g: R R be functions defined by f(x)= \ array ll x|x| ( 1 x ), & x 0, \\ 0, & x=0, array . and g(x)= cases 1-2 x, & 0 x 1 2 \\ 0, & otherwise cases Let a, b, c, d R . Define the function

f(x)= \ array ccc x|x| 1 x & ; & x 0 \\ 0 & ; & x=0 array g(x)= \ array ccc 1-2 x & ; & 0 x 1 2 \\ 0 & ; & otherwise array . . g ( 1 2 -x )= \ array ccc 2 x & ; & 0 1 2 -x 1 2 \\ 0 & ; & otherwise array = \ array ccc 2 x & ; & 0 x 1 2 \\ 0 & ; & otherwise array \ . g(x)+g ( 1 2 -x )= \ array lll 1 & ; & 0 x 1 2 \\ 0 & ; & otherwise array \ (P) Now a =0, ~b =1, c =0, ~d =0 h ( x )= g ( x )+ g ( 1 2 - x )= \ array lll 1 & ; & 0 x 1 2 \\ 0 & ; & otherwise array . Hence Range of h ( x ) is \ 0,1\ (Q

Q: Let S = 0 , 1 ∪ 1 , 2 ∪ 3 , 4 and T = 0 , 1 , 2 , 3 . Then which of the following statements is(are) true?

Given, S = 0 , 1 ∪ 1 , 2 ∪ 3 , 4 and T = 0 , 1 , 2 , 3 Now, let domain and co-domain of a function y = f x are S and T respectively. Now solving option, A There are infinitely many elements in domain and four elements in co-domain. ⇒ There are infinitely many functions from S to T . ⇒ Option A is correct B If number of elements in domain is greater than number of elements in co-domain, then number of strictly inc

Q: Let the function f : 0 , 1 → ℝ be defined by f x = 4 x 4 x + 2 . Then the value of f 1 40 + f 2 40 + f 3 40 + ⋯ + f 39 40 - f 1 2 is_________

f x = 4 x 4 x + 2 , f 1 - x = 4 1 - x 4 1 - x + 2 = 2 4 x + 2 ⇒ f x + f 1 - x = 1 f 1 40 + f 2 40 + … . . + f 39 40 - f 1 2 = f 1 40 + f 39 40 + f 2 40 + f 38 40 + . . . . . f 19 40 + f 21 40 + f 20 40 - f 1 2 = 19 .

Q: If the function f : R → R is defined by f x = x x − sin x , then which of the following statements is TRUE?

f x = x x − sin x is odd function ∵ f − x = − f x Now, f x = x 2 − x sin x ∀ x ≥ 0 ⇒ f ' x = 2 x − x cos x − sin x ⇒ f ' x = x − sin x + x 1 − cos x > 0 ∴ Graph of y = f x will be From graph, it's one-one and onto both.

Question 1

Let N denote the set of all positive integers. Consider the sets A = \ 1, 2, 3, 4, 5\ and B = \ 1, 2, 3, 4, 5, 6, 7\ . Let S be the set of all functions f: A B such that f(2) 2 and f(4) 4. Consider th

Accepted Answer

The condition g(f(x)) = 2^x for all x A implies that if f(x_1) = f(x_2), then g(f(x_1)) = g(f(x_2)), which gives 2^ x_1 = 2^ x_2 x_1 = x_2. Thus, f must be an injective (one-to-one) function. Conversely, if f is injective, such a function g can always be constructed. Therefore, T is the set of all injective functions f: A B such that f(2) 2 and f(4) 4. The total number of injective functions from A to B is ^ 7 P_ 5 = 7 6 5 4 3 = 2520. Let E_2 be the set of injective functions where f(2) = 2. The

Question 2

Let R denote the set of all real numbers. Let a_ i , b_ i R for i \ 1,2,3\ . Define the functions f: R R , g: R R , and h: R R by aligned & f(x)=a_1+10 x+a_2 x^2+a_3 x^3+x^4, \ & g(x)=b_1+3 x+b_2 x^2

Accepted Answer

aligned & h(x)=f(x+1)-g(x+2) \ & =a_1+10(x+1)+a_2(x+1)^2+a_3(x+1)^3+(x+1)^4-b_1-3(x+2)-b_2(x+2)^2-b_3(x+2)^3-(x+2)^4 aligned Coeff. of x^3 in h(x) aligned & =a_3-b_3-4 \ & f(x)-g(x) 0 x R \ & a_1+10 x+a_2 x^2+a_3 x^3+x^4-b_1-3 x-b_2 x^2-b_3 x^3-x^4 0 \ & x^3 (a_3-b_3 )+x^2 (a_2-b_2 )+7 x+ (a_1-b_1 ) 0 aligned Cubic Eq. will become zero at atleast are value of x So it will be quadratic a _3- b _3=0

Question 3

Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f: N Z and g: Z N defined by f(n)= cases (n+1) / 2 & if n is odd \ (4-n) / 2 & if n is even c

Accepted Answer

aligned & f(n)= cases (n+1) / 2 & if n is odd \ (4-n) / 2 & if n is even cases \ & f(n)=\ (1,1),(2,1),(3,2),(4,0),(5,3),(6,-1), \ aligned f ( n ) is many one and onto function aligned & g ( n )= \ array cc 3+2 n & if n 0 \ -2 n & if n <0 array . \ & g ( n )=\ (-3,6),(-2,4),(-1,2),(0,3),(1,5),(2,7),(3,9),(4,15), \ aligned g ( n ) is one-one and into function f ( ~g ( n ))=2+ n , n ~N fog is one-one and into g(f(n))= cases 4+n & if n is odd natural number \ 7-n & if n=2,4 \ n-4 & if n is eve

Question 4

Let f: R R be a function such that f(x+y)=f(x)+f(y) for all x, y R , and g: R (0, ) be a function such that g(x+y)=g(x) g(y) for all x, y R . If f ( -3 5 )=12 and g ( -1 3 )=2, then the value of (f (

Accepted Answer

f(x+y)=f(x)+f(y) ...(1) f ( nx )= nf ( x ) n N ...(2) Now put y=-x in eq.(1) aligned & f ( x )+ f (- x )= f (0) \ f (0)=0\ \ & f (- x )=- f ( x ) aligned f is odd function from eq. (2) from eq. (2) aligned & f(-n x)= nf (-x) \ & f (- nx )=-n f( x ) \ & aligned f ( mx )= mf ( x ) m Z ...(3) from eq. (2) and eq. (3) f ( nx )= nf ( x ) n Z ...(4) Now put x= p q where p , q Z , q 0 f ( np q )= nf ( p q ) n Z put n = q f(p)=q f ( p q ) pf (1)= qf ( p q ) from eq.(4) Let f (1)= a then pa = qf ( p q

Question 5

Let f: R R and g: R R be functions defined by f(x)= \ array ll x|x| ( 1 x ), & x 0, \ 0, & x=0, array . and g(x)= cases 1-2 x, & 0 x 1 2 \ 0, & otherwise cases Let a, b, c, d R . Define the function

Accepted Answer

f(x)= \ array ccc x|x| 1 x & ; & x 0 \ 0 & ; & x=0 array g(x)= \ array ccc 1-2 x & ; & 0 x 1 2 \ 0 & ; & otherwise array . . g ( 1 2 -x )= \ array ccc 2 x & ; & 0 1 2 -x 1 2 \ 0 & ; & otherwise array = \ array ccc 2 x & ; & 0 x 1 2 \ 0 & ; & otherwise array \ . g(x)+g ( 1 2 -x )= \ array lll 1 & ; & 0 x 1 2 \ 0 & ; & otherwise array \ (P) Now a =0, ~b =1, c =0, ~d =0 h ( x )= g ( x )+ g ( 1 2 - x )= \ array lll 1 & ; & 0 x 1 2 \ 0 & ; & otherwise array . Hence Range of h ( x ) is \ 0,1\ (Q

Question 6

Let S = 0 , 1 &#8746; 1 , &#160; 2 &#8746; 3 , &#160; 4  and T = 0 , &#160; 1 , &#160; 2 , &#160; 3 . Then which of the following statements is(are) true?

Accepted Answer

Given, S = 0 , &#160; 1 &#8746; 1 , &#160; 2 &#8746; 3 , &#160; 4  and  T = 0 , &#160; 1 , &#160; 2 , &#160; 3 Now, let domain and co-domain of a function y = f x are S and T  respectively. Now solving option, A There are infinitely many elements in domain and four elements in co-domain. &#8658; There are infinitely many functions from S to T . &#8658; Option A is correct B If number of elements in domain is greater than number of elements in co-domain, then number of strictly inc

Question 7

Let the function  f : 0 , 1 &#8594; &#8477; be defined by  f x = 4 x 4 x + 2 . Then the value of  f 1 40 + f 2 40 + f 3 40 + &#8943; + f 39 40 - f 1 2   is_________

Accepted Answer

f x = 4 x 4 x + 2 , &#160;&#160; f 1 - x = 4 1 - x 4 1 - x + 2 = 2 4 x + 2 &#8658; f x + f 1 - x = 1 f 1 40 + f 2 40 + &#8230; . . + f 39 40 - f 1 2 = f 1 40 + f 39 40 + f 2 40 + f 38 40 + . . . . . f 19 40 + f 21 40 + f 20 40 - f 1 2 = 19 .

Question 8

If the function f : R &#8594; R is defined by f x = x x &#8722; sin x , then which of the following statements is TRUE?

Accepted Answer

f x = x x − sin x is odd function ∵ f − x = − f x Now, f x = x 2 − x sin x ∀ x ≥ 0 ⇒ f ' x = 2 x − x cos x − sin x ⇒ f ' x = x − sin x + x 1 − cos x > 0 ∴ Graph of y = f x will be From graph, it's one-one and onto both.

Functions — Previous Year Questions & Solutions