1. Example:
Let f(x) = and g(x) =
. Describe the function
(f+g)(x) and find its domain.
Solution :
,
1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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2. Example:
Let f(x) = and g(x) = then find
(f + g)x and (fg)x.
Solution :
and
−3, x < −1
| x + 1| + x − 2 , x<0 2x − 1, − 1 ≤ x < 0
f(x) + g(x) = | x + 1| + x + 3, 0 ≤ x <1 =
1 − x + x + 3, 2x + 4, 0 ≤ x < 1
x ≥1 4,
x ≥1
2
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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3. Example:
Let f(x) = and g(x) =
then find the value of
(i) (f + g)(3.5) (ii) f(g(3)) (iii) (fg)(2) (iv) (f −g)(4)
Solution :
(f + g)(3.5) f(g(3))
= f((3 – 3))
= f(3.5) + g(3.5)
= f(0)
= (3.5 – 4) + (3.5 – 3)
= (0)2 – 4(0) + 3
=0 =3
3
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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5. Example:
If the functions f(x) and g(x) are defined on R → R such that
f(x) and g(x) =
then (f − g) (x) is
(A) one-one and onto (B) neither one-one nor onto
(C) one-one but not onto (D) onto but not one-one
(IITJEE 2005)
Solution :
∴ (f − g) : R → R
− x if x ∈ rational
=
x if x ∈ irrational
5
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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6. Since each branch is linear function, function is one – one
Any value from y = –x(x is rational) does not match with
that of y = x(x is irrational)
Also when x is rational y = –x is rational
And when x is irrational y = x is irrational
Thus function takes all real values of x
Hence function is onto
6
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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7. Example:
If f(x) = , x > 0, n ≥ 2, n ∈ N. Then show that (fof) (x)
= x. Find also the inverse of f(x).
Solution : f(x) = (a – xn)1/n
⇒ f(f(x)) = (a – (f(x))n)1/n
⇒ f(f(x)) = [a – ((a – xn)1/n)n]1/n
⇒ f(f(x)) = [a – ((a – xn)]1/n
⇒ f(f(x)) = [a – a + xn]1/n
⇒ f(f(x)) = (xn)1/n
⇒ f(f(x)) = x
Since f(f(x)) = x ,
f– 1 (x) = f(x)
⇒ f– 1 (x) = (a – xn)1/n
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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8. Example:
If f(x) = , then find the value of f(f(x)), f(f(f(x))),
Also find the value of
Solution :
f(x) = ⇒ fof(x) = ⇒ fof(x) =
=
=
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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9. Now fofof(x) = f(f(f(x)))
= = = =x
Finding Now
Now this function depends upon the value of n.
For n = 3, 6, 9, ……. or n = 3k
=x
9
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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10. For n = 3k + 1 ( n = 1, 4, 7, ………)
=
For n = 3k + 2 ( n = 2, 5, 8, ………)
=
10
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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11. Example:
Let f(x) = ax + b and g(x) = cx + d, a ≠ 0, c ≠ 0. Assume a =
1, b = 2. If (fog) (x) = (gof) (x) for all x, what can you say
about c and d ?
Solution :
(fog) (x) = f(g(x)) = a(cx + d) + b
(gof) (x) = f(f(x)) = c(ax + b) + d
Given that, (fog) (x) = (gof) (x) and at a = 1, b = 2
⇒ cx + d + 2 = cx + 2c + d
Comparing constant term, d + 2 = 2c + d
⇒ c = 1 and d is arbitrary
11
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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12. Example:
Let f(x) = Then for what value of α is f(f(x)) = x?
(A) (B) − (C) 1 (D) – 1(IITJEE 2001)
Solution :
x≠ −1
αx
α ÷
f ( f ( x) ) = x ⇒ x + 1 = x ⇒
αx
+1
x +1
⇒ ( α + 1) x2 + 1 − α 2 x = 0 ( )
This is an identity in x
⇒ α + 1 = 0 and 1 − α 2 = 0
⇒ α =− 1
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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13. Example:
If f be the greatest integer function and g be the modulus
function, then find the value of (gof) −(fog)
Solution :
f(x) = [x] and g(x) = |x|
5 5
(gof) − ÷ − (fog) − ÷
3 3
=
=
=
= =2–1 =1
13
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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14. Example:
Let f(x) = , where [.] denotethe
greatest integer function. Then find the value of f(f(−2.3))
Solution :
f(x) = .
First find f(−2.3)
For x < –1, f(x) = 1 + |x|
⇒ f(−2.3) = 1 + |−2.3|
⇒ f(−2.3) = 3.3
Now f(f(−2.3)) = f(3.3)
For x ≥ –1, f(x) = [x]
⇒ f(f(−2.3)) = f(3.3) = [3.3] = 3
14
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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15. Example:
If g(x) = x2 + x − 2 and , then find f(x)
Solution :
……(i)
Also from g(x) = x2 + x − 2
⇒ g(f(x)) = (f(x))2 + f(x) – 2 ……(ii)
Comparing (i) and (ii)
(f(x))2 + f(x) − 2 = 4x2 − 10x + 4
⇒ f(x)2 + f(x) − (4x2 – 10x + 6) = 0
∴ f(x) =
−1 ± 16x2 − 40x + 25
=
2
15
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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16. −1 ± (4x − 5)
=
2
= 2x – 3 or 2 – 2x
16
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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17. Example:
Suppose that g(x) = 1 + and f(g(x)) = 3 + 2 + x, then
find function f(x)
Solution :
g(x) = 1 + and f(g(x)) = 3 + 2 + x …….(i)
⇒ f(1 + )=3+2 +x
Put 1 + =y
⇒ x = (y − 1)2 ⇒ f(y) = 3 + 2(y − 1) + (y − 1)2
= 2 + y2
⇒ f(x) = 2 + x2
17
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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18. Example:
If f(x) = sin x + cos x, g(x) = x2 − 1, then g(f(x)) is invertible
in the domain
(A) (B) (C) (D)[0,π]
(IITJEE 2004)
Solution :
f(x) = sin x + cos x, g(x) = x2 − 1
⇒ g(f(x)) = (sin x + cos x)2 − 1
= sin2x + cos2x + 2sin x cos x – 1
= sin 2x
18
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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19. We know that sin θ is invertible when − π/2 ≤ θ ≤ π/2
⇒ g(f(x)) is invertible in −
π π
⇒− ≤ x≤
4 4
19
S.C.O. 42, Sec. 20 – C chd. 3258041, 4280095
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