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) + k; k 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)