If f x = x - 1 x + 1 , prove that f x - f y 1 + f x . f y = x - y 1 + x y .

Given, f x = x - 1 x + 1 Now, f x - f y 1 + f x . f y = x - 1 x + 1 - y - 1 y + 1 1 + x - 1 x + 1 y - 1 y + 1 = x - 1 y + 1 - x + 1 y - 1 x + 1 y + 1 x + 1 y + 1 + x - 1 y - 1 x + 1 y + 1 = x y + x - y - 1 - x y + x - y + 1 x y + x + y + 1 + x y - x - y + 1 = 2 x - 2 y 2 x y + 2 = x - y 1 + x y Hence, proved.

If f x = x - 1 x + 1 , prove that f x - f y 1 + f x . f y = x - y 1 + x y .

Given, f x = x - 1 x + 1 Now, f x - f y 1 + f x . f y = x - 1 x + 1 - y - 1 y + 1 1 + x - 1 x + 1 y - 1 y + 1 = x - 1 y + 1 - x + 1 y - 1 x + 1 y + 1 x + 1 y + 1 + x - 1 y - 1 x + 1 y + 1 = x y + x - y - 1 - x y + x - y + 1 x y + x + y + 1 + x y - x - y + 1 = 2 x - 2 y 2 x y + 2 = x - y 1 + x y Hence, proved.

If f x + 2 f 1 x = 3 x , x ≠ 0 , and S = x ∈ R : f x = f - x , then S

Contains more than two elements

MEDIUM

Earn 100

If $f (x) = \frac{x - 1}{x + 1}$ , prove that $\frac{f (x) - f (y)}{1 + f (x) . f (y)} = \frac{x - y}{1 + x y}$ .

Important Questions on Relations and Functions

HARD

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let the function $f : [0, 1] \to ℝ$ be defined by $f (x) = \frac{4^{x}}{4^{x} + 2}$ . Then the value of $f (\frac{1}{40}) + f (\frac{2}{40}) + f (\frac{3}{40}) + \dots + f (\frac{39}{40}) - f (\frac{1}{2})$ is_________

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let,

f : R \to R

be a function such that

f (x) = x^{3} + x^{2} f^{'} (1) + x f^{"} (2) + f^{'''} (3), \forall x \in R .

Then

f (2)

equals

EASY

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

f : R \to R

be defined by

f (x) = \{\begin{cases} 2 x; & x > 3 \\ x^{2}; & 1 < x \leq 3 \\ 3 x; & x \leq 1 \end{cases}

Then

f (- 1) + f (2) + f (4)

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

f (x) = \frac{x}{x - 1}, f (3 x)

in terms of

f (x)

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

f (x) + 2 f (\frac{1}{x}) = 3 x, x \neq 0,

and

S = \{x \in R : f (x) = f (- x)\}

, then

S

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Number of functions

f : [0, 1] \to [0, 1]

satisfying

|f (x) - f (y)| = |x - y|

for all

x, y

[0, 1]

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

f : R \to R

be a function which satisfies

f (x + y) = f (x) + f (y), \forall x, y \in R

. If

f (1) = 2

and

g (n) = \sum_{k = 1}^{(n - 1)} f (k),

n \in N

then the value of

n

, for which

g (n) = 20

, is

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

f (x)

is an even function and satisfies the relation

x^{2} f (x) - 2 f (\frac{1}{x}) = g (x)

where

g (x)

is an odd function, then

f (5)

equals

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

f : [0,1] \to R

be such that

f (x y) = f (x) \cdot f (y),

for all

x, y \in [0,1],

and

f (0) \neq 0 .

y = y (x)

satisfies the differential equation,

\frac{d y}{d x} = f (x)

with

y (0) = 1

then

y (\frac{1}{4}) + y (\frac{3}{4})

is equal to:

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

S

be the set of points where the function

, f (x) = |2 - | x - 3 ||, x \in R,

is not differentiable. Then

\sum_{x \in S} f (f (x))

is equal to

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

A

denote the set of all real numbers

x

such that

x^{3} - [x]^{3} = (x - [x])^{3},

where

[x]

is the greatest integer less than or equal to

x .

Then

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

f : x \to y

be such that

f (1) = 2

and

f (x + y) = f (x) f (y)

for all natural numbers

x

and

y

.

\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1)

then

a

is equal to

HARD

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

A function $f (x)$ is given by $f (x) = \frac{5^{x}}{5^{x} + 5}$ , then the sum of the series $f (\frac{1}{20}) + f (\frac{2}{20}) + f (\frac{3}{20}) + \dots + f (\frac{39}{20})$ is equal to:

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

f (x) = \{\begin{array}{l} x \sin (\frac{1}{x}), when x \neq 0 \\ 1, when x = 0 \end{array}\}

and

A = \{x \in R : f (x) = 1\}

. Then,

A

has

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

a + α = 1, b + β = 2

and

a f (x) + α f (\frac{1}{x}) = b x + \frac{β}{x}, x \neq 0,

then the value of the expression

\frac{f (x) + f (\frac{1}{x})}{x + \frac{1}{x}}

is ___________.

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

f : R \to R

be defined as

f (x + y) + f (x - y) = 2 f (x) f (y), f (\frac{1}{2}) = - 1 .

Then the value of

\sum_{k = 1}^{20} \frac{1}{\sin (k) \sin (k + f (k))}

is equal to :

MEDIUM

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

f (x + y) = f (x) + f (y) - x y - 1 \forall x, y \in R

and

f (1) = 1

, then the number of solutions of

f (n) = n, n \in N

HARD

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

\sum_{k = 1}^{10} f (a + k) = 16 (2^{10} - 1),

where the function

f

satisfies

f (x + y) = f (x) f (y)

for all natural numbers

x, y

and

f (1) = 2 .

Then the natural number

a

is:

EASY

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

f : R \to R

satisfy

f (x) f (y) = f (x y)

for all real numbers

x

and

y

.

f (2) = 4

,

then

f (\frac{1}{2}) =

HARD

Mathematics>Algebra>Relations and Functions>Algebra of Real Functions

Let

a, b, c \in R

. If

f (x) = a x^{2} + b x + c

is such that

a + b + c = 3

and

f (x + y) = f (x) + f (y) + x y, \forall x, y \in R

, then

\sum_{n = 1}^{10} f (n)

is equal to:

If fx=x-1x+1, prove that fx-fy1+fx.fy=x-y1+xy.

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

If $f (x) = \frac{x - 1}{x + 1}$ , prove that $\frac{f (x) - f (y)}{1 + f (x) . f (y)} = \frac{x - y}{1 + x y}$ .