# CLASS-11RELATION & FUNCTION - ALGEBRAIC OPERATION ON FUNCTION

Algebraic Operations On Functions

If ‘f ’ and ‘g’ are real valued functions of ‘x’ with domain set A, B respectively, then both ‘f ’ and ‘g’ are defined in A ∩ B. Now we define (f + g), (f – g), f(g), (f/g) as follows –

(i) (f + g) (x) = f(x) + g(x);  Domain A ∩ B

(ii) (f – g) (x) = f(x) – g(x); Domain A ∩ B

(iii) (fg) (x) = f(x). g(x); Domain A ∩ B

f               f(x)

(iv) [-----] (x) =  --------;  Domain = {x ǀ ∈ A ∩ B}

g               g(x)

(v(f + k)x = f(x) + kk is a constant Domain A

(vi) (kf) x = kf (x)

(vii) fⁿ (x) = {f(x)}ⁿ

x - ǀ x ǀ

Example.1)  If f(x) = -----------, then find f(- 1)

ǀ x ǀ

- 1 - ǀ - 1 ǀ          - 1 - 1

Ans.)  f(- 1) = --------------- = ------------  =  - 2    (Ans.)

ǀ - 1 ǀ                 1

x                           f(a)

Example.2) If  f(x) = -------, then prove that -------- = f(a²)

x – 1                       f(a + 1)

f(a)

Ans.)  -----------

f(a + 1)

a

---------

a - 1

=>  ----------------------

a + 1

------------

(a + 1) – 1

a           (a + 1)

=>   -------- ÷ ----------

(a – 1)           a

a              a

=>  --------- X --------

(a – 1)        (a + 1)

a²

=>  ---------- =  f(a²)            (Proved)

(a² - 1)

1 + x

Example.3)  If f(x) = ---------, show that f [f(tan θ)] = - cot θ

1 – x

1 + x

Ans.) f(x) = ----------

1 – x

1 + tan θ

So, f(tan θ) = ------------

1 – tan θ

1 + f(tan θ)

Now, f [f (tan θ)]  = ---------------

1 – f(tan θ)

1 + tan θ                     1 – tan θ + 1 + tan θ

1 + ------------               ----------------------

1 – tan θ                           1 – tan θ

=  ------------------------  = -----------------------------

1 + tan θ                        1 – tan θ – 1 – tan θ

1 - ------------                  ---------------------

1 – tan θ                              1 – tan θ

2              (- 2 tan θ)

=   ------------ ÷ -------------

(1 – tan θ)        (1 – tan θ)

2              (1 - tan θ)

= ------------ X ------------

(1 – tan θ)         (- 2 tan θ)

( - 1)

= ----------

tan θ

=   - cot θ           (Proved)

Example.4) If f(x) = cos (log x), then prove

1                      x²

f(x²). f(y²) - ------ [f (x²y²) + f(------)]  = 0

2                      y²

Ans.)  f(x) = cos (log x)

=>  f(x²) = cos (log x²) = cos (2 log x) ……………………(i)

Similarly,  f(y²) =  cos (2 log y)…………………………(ii)

x²                   x²

f(------) = cos (log ------)

y²                   y²

= cos (log x² - log y²) = cos (2 log x – 2 log y) ………….(iii)

f(x²y²) = cos (log x²y²)

= cos(log x² + log y²)

= cos (2 log x + 2 log y) ………………….(iv)

Now,

1                      x²

f(x²). f(y²) - ------ [f (x²y²) + f(-----)]  = 0

2                      y²

By substitute all the values, obtained from (i), (ii), (iii), and (iv), we get –

1

=> cos (2 log x). cos (2 log y) - ----- [cos (2 log x + 2 log y)

2

+ cos (2 log x – 2 log y)]

[Applying C, D formulae of Trigonometry, we get -]

1

=> cos (2 log x). cos (2 log y) - ----- X 2 cos (2 log x) . cos ( 2 log y)

2

=> cos (2 log x). cos (2 log y) - cos (2 log x). cos (2 log y)

=>   0              (Proved)