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Definition of Functions

If the functi...

Question

If the function $F : R \to R$ defined by $f (x) = \frac{3^{x} + 3^{- x}}{2},$ then show that $f (x + y) + f (x - y) = 2 f (x) f (y)$

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Solution

$f (x) = \frac{1}{2} [3^{x} + \frac{1}{3^{x}}]$

$f (x + y) = \frac{1}{2} [3^{x} {.3}^{y} + \frac{1}{3^{x} {.3}^{y}}]$

$f (x - y) = \frac{1}{2} [\frac{3^{x}}{3^{y}} + \frac{3^{y}}{3^{x}}]$

$f (x + y) + f (x - y) = \frac{1}{2} [3^{x} {.3}^{y} + \frac{3^{x}}{3^{y}} + \frac{1}{3^{x} {.3}^{y}} + \frac{3^{y}}{3^{x}}]$

$= \frac{1}{2} [3^{x} [3^{y} + \frac{1}{3^{y}}] + \frac{1}{3^{x}} [\frac{1}{3^{y}} + 3^{y}]]$

$= \frac{1}{2} [(3^{x} + \frac{1}{3^{x}}) (3^{y} + \frac{1}{3^{y}})]$

$= 2. f (x) . f (y)$

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Similar questions

Q. If the function

E : R \to

f (x) \frac{3^{x} + 3^{- x}}{2}

f (x + y) + f (x - y) = 2 f (x) f (y)

.

Q. If a function

f (x)

satisfies the relation

f (x + y) + f (x - y) = 2 f (x) f (y) \forall x, y \in R

f (0) \neq 0

Q. Assertion :If

f (x + y) + f (x - y) = 2 f (x) f (y) \forall x, y \in R

f (0) \neq 0

f (x)

is an even function. Reason: If

f (- x) = f (x)

f (x)

f : R \to R

f (x + y) - k x y = f (x) + 2 y^{2} \forall x, y \in R

f (1) = 2; f (2) = 8

where k is some constant, then

f (x + y) . f (\frac{1}{x + y}) = (x + y \neq 0)

Q. Classify the following functions as injection, surjection or bijection :
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : N → N given by f(x) = x³
(iv) f : Z → Z given by f(x) = x³
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x² + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x³ + 1
(x) f : R → R, defined by f(x) = x³ − x
(xi) f : R → R, defined by f(x) = sin²x + cos²x
(xii) f : Q − {3} → Q, defined by

f (x) = \frac{2 x + 3}{x - 3}

(xiii) f : Q → Q, defined by f(x) = x³ + 1
(xiv) f : R → R, defined by f(x) = 5x³ + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x²
(xvii) f : R → R, defined by f(x) =

\frac{x}{x^{2} + 1}

[NCERT EXEMPLAR]

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