Exercise: 1.3
1. Given,
𝑓𝑔(2)
𝑓(𝑥) → 2𝑥 + 3
= (9 − 1)2 + 3
𝑔(𝑥) → 𝑥 2 − 1
= 67
= 𝑓(22 − 1)
6.
= 𝑓(3)
ℎ(𝑥) → 𝑥 + 2
[𝑥 > 0]
𝑘(𝑥) → √𝑥
[𝑥 > 0]
=2×3+3
a)
=6+3
𝑥 → ℎ(√𝑥)
=9
2. Given,
𝑥 → ℎ𝑘(𝑥)
𝑓(𝑥) = 𝑥 2 − 1
b)
𝑔(𝑥) = 2𝑥 + 3
𝑔𝑓(5)
Given,
𝑥 → √𝑥 + 2
= 𝑔(52 − 1)
⇒ 𝑥 → 𝑘(𝑥 + 2)
= 𝑔(24)
⇒ 𝑥 → 𝑘ℎ(𝑥)
= 2 × 24 + 3
3.
𝑥 → √𝑥 + 2
7.
𝑓(𝑥) = 3𝑥 + 1
10
= 51
𝑔(𝑥) = 2−𝑥
𝑓(𝑥) = (𝑥 + 2)2 − 1
∴ 𝑔𝑓(𝑥) = 𝑔(3𝑥 + 1)
𝑓 2 (3) = 𝑓 𝑓(3)
𝑥∈𝑅
𝑥≠2
10
⇒ 5 = 2−(3𝑥+1)
= 𝑓((3 + 2)2 − 1)
10
⇒ 5 = 2−3𝑥−1
= 𝑓(24)
10
= (24 + 2)2 − 1
⇒ 5 = 1−3𝑥
= 675
⇒ 5 − 15𝑥 = 10
−5
4.
Given,
𝑓(𝑥) = 1 + √𝑥 − 2
10
𝑔(𝑥) = 𝑥 − 1
⇒ 15 = 𝑥
𝑥≥2
1
∴ 𝑥 = −3
𝑥>0
𝑔𝑓(18) = 𝑔(1 + √18 − 2)
8.
𝑔(𝑥) = 𝑥 2 + 2
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ℎ(𝑥) = 3𝑥 − 5
= 𝑔(5)
10
𝑔ℎ(𝑥) = 𝑔(3𝑥 − 5)
= 5 −1
⇒ 51 = (3𝑥 − 5)2 + 2
=1
⇒ 51 = 9𝑥 2 − 30𝑥 + 24 + 2
5.
Given,
𝑓(𝑥)
2
= (𝑥 − 1) + 3
2𝑥+4
⇒ 9𝑥 2 − 30𝑥 + 27 − 51 = 0
[𝑥 > −1]
[𝑥 > 5]
⇒ 9𝑥 2 − 30𝑥 − 24 = 0
𝑔(𝑥)
= 𝑥−5
𝑓𝑔(7)
= 𝑓 ( 7−5 )
⇒ 3𝑥 2 − 12𝑥 + 2𝑥 − 8 = 0
= 𝑓(9)
⇒ 3𝑥(𝑥 − 4) + 2(𝑥 − 4) = 0
2×7+4
⇒ 3𝑥 2 − 10𝑥 − 8 = 0
⇒ (𝑥 − 4)(3𝑥 + 2) = 0
∴𝑥 −4 = 0
3𝑥 + 2 = 0
9.
𝑥 = −3
𝑓(𝑥) = 𝑥 2 − 3
3
𝑔(𝑥) = 𝑥
𝑥>0
⇒
3 2
=0
⇒ 𝑥(𝑥 − 5) + 1(𝑥 − 5) = 0
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⇒ (𝑥 − 5)(𝑥 + 1) = 0
f
∴ 𝑥 = ±4
𝑥2
⇒ 𝑥 2 − 5𝑥 + 𝑥 − 5 = 0
𝑥
3
100+80𝑥−20𝑥 2
⇒ 𝑥 2 − 4𝑥 − 5 = 0
⇒ 13 = ( ) − 3
9
80
⇒ 20𝑥 2 − 80𝑥 − 100 = 0
∴ 𝑓𝑔(𝑥) = 𝑓 (𝑥)
⇒ 𝑥 2 = 16
10
100
3
9
100
⇒ 𝑥 2 + 𝑥 − 20 = 0
𝑥>0
⇒ 16 = 𝑥 2
2
⇒ 𝑥 2 + 2. 𝑥 . 4 + 16 + 3 − 39 = 0
2
𝑥=4
10
⇒ ( 𝑥 + 4) + 3 = 39
∴𝑥−5=0
𝑥+1=0
𝑥=5
𝑥 = −1
12. Given,
10. Given,
3𝑥+5
𝑔(𝑥) = 𝑥 2 − 1
𝑥≥0
𝑥−1
ℎ(𝑥) = 2𝑥 − 7
𝑥≥0
𝑓(𝑥) = 𝑥−2 , 𝑥 ≠ 2
𝑔(𝑥) =
,𝑥 ∈ 𝑅
2
∴ 𝑔ℎ(𝑥) = 0
3𝑥+5
𝑔𝑓(𝑥) = 𝑔 ( 𝑥−2 )
3𝑥+5
⇒ 12 = 𝑥−22
⇒ 𝑔(2𝑥 − 7) = 0
⇒ (2𝑥 − 7)2 − 1 = 0
−1
⇒ 4𝑥 2 − 28𝑥 + 49 − 1 = 0
3𝑥+5−𝑥+2
⇒ 24 =
𝑥−2
⇒ 4𝑥 2 − 28𝑥 + 48 = 0
⇒ 24𝑥 − 48 = 2𝑥 + 7
⇒ 𝑥 2 − 7𝑥 + 12 = 0
⇒ 22𝑥 = 55
⇒ 𝑥 2 − 3𝑥 − 4𝑥 + 12 = 0
55
⇒ 𝑥 = 22
⇒ (𝑥 − 3)(𝑥 − 4) = 0
5
⇒𝑥=2
∴𝑥 −3 = 0
∴ 𝑥 = 2.5
𝑥=3
13. Given,
11. Given,
𝑓(𝑥) = (𝑥 + 4)2 + 3
10
𝑔(𝑥) = 𝑥
∴ 𝑓𝑔(𝑥) = 39
10
⇒ 𝑓 ( 𝑥 ) = 39
𝑥=4
𝑓(𝑥) → 𝑥 3
𝑔(𝑥) → 𝑥 − 1
𝑥>0
𝑥>0
𝑥−4=0
a)
𝑥 → (𝑥 − 1)3
𝑥 → 𝑓(𝑥 − 1)
𝑥 → 𝑓𝑔(𝑥)
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b)
𝑥 → 𝑥3 − 1
𝑥
𝑓𝑔(𝑥)
= 𝑓 (10 − 2)
𝑥 → 𝑔(𝑥 3 )
20−𝑥 2
𝑥 → 𝑔𝑓(𝑥)
c)
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𝑥 →𝑥−2
=
f
𝑥 →𝑥−1−1
d)
=( 2 ) −9
(20−𝑥)2
4
−9
Here, for domain, 20 − 𝑥 < 0
𝑥 → 𝑔(𝑥 − 1)
⇒ 20 < 𝑥
𝑥 → 𝑔 𝑔(𝑥)
⇒ 𝑥 > 20
𝑥 → 𝑔2 (𝑥)
for range, 𝑓𝑔(𝑥) > −9
16. Given,
𝑥 → 𝑥9
1
𝑥 → (𝑥 3 )3
𝑓(𝑥) = 𝑥
𝑥 → 𝑓(𝑥 3 )
𝑥 ∈ 𝑅, 𝑥 ≠ 0
1
𝑔(𝑥) = 𝑥−1
𝑥≠1
𝑥 → 𝑓𝑓(𝑥)
a)
𝑥 → 𝑓 2 (𝑥)
𝑥
14. Given, 𝑓(𝑥) = 𝑥+2
1
𝑓 𝑔(𝑥) = 𝑓 (𝑥−1)
=
𝑥 ≠ −2
3
𝑓𝑔(𝑥)
𝑔(𝑥) = 𝑥
𝑥
=
1
𝑥−1
=𝑥−1
𝑓𝑔(𝑥) is defined for all values of 𝑥 domain
3
∴ 𝑓𝑔(𝑥) = 𝑓 (𝑥)
=3
1
of 𝑓 𝑔(𝑥) is 𝑥 ∈ 𝑅, 𝑥 ≠ 1
3
𝑥
b)
+2
1
𝑔 𝑓(𝑥) = 𝑔 (𝑥)
3
𝑥
3+2𝑥
𝑥
3
1
= 𝑔 (1 )
𝑥
−1
𝑥
= 𝑥 × 3+2𝑥
1
= ( 1−𝑥 )
3
𝑥
= 3+2𝑥
𝑥
= 1−𝑥
For Domain, 3 + 2𝑥 ≠ 0
𝑥 ∈ 𝑅,
3
∴ 𝑥 ≠ −2
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Domain = 𝑥 ≠ 1, 𝑥 ≠ 0
3
Domain = 𝑥 ∈ 𝑅, 𝑥 ≠ − 2 , 𝑥 ≠ 0
15. Given,
𝑓(𝑥) = 𝑥 2 − 9
𝑥
𝑔(𝑥) = 10 − 2
𝑓𝑔(𝑥)
𝑥
= 𝑓 (10 − 2)
𝑥 2
= (10 − 2) − 9
17. Given,
𝑥<0
𝑥>6
𝑓(𝑥) = 2𝑥 − 6
𝑔(𝑥) = √𝑥
a)
𝑓 𝑔(𝑥) = 𝑓(√𝑥)
= 2 √𝑥 − 6
The function is defined for only positive
integer 𝐷𝑓 = 𝑥 ∈ 𝑅, 𝑥 ≥ 0
2
= 𝑥2 − 𝑥
Putting minimum value of 𝑥 from Domain
𝑥=0
= 𝑥2
2−𝑥
For domain: 𝑥 2 ≠ 0, 𝑥 ≠ 0
∴ 𝑓 𝑔(𝑥) = 2√0 − 6
𝐷𝑓 = 𝑥 ∈ 𝑅, 𝑥 ≠ 0
= −6
b)
1
Again for range,
2−𝑥
Range: 𝑓 𝑔(𝑥) ∈ 𝑅, 𝑓 𝑔(𝑥) ≥ −6
For range: Let 𝑦 = 𝑥 2
𝑔 𝑓(𝑥)
⇒ 𝑥2𝑦 = 2 − 𝑥
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= 𝑔(2𝑥 − 6)
f
= √2𝑥 − 6 …….. (i)
The function is defined for only positive
integer 2𝑥 − 6 ≥ 0, 𝑥 ≥ 3
Domain 𝑥 ∈ 𝑅; 𝑥 ≥ 3
⇒ 𝑥2𝑦 + 𝑥 − 2 = 0
Range, 𝑏 2 − 4𝑎𝑐 ≥ 0
⇒ 12 − 4 × 𝑦 × (−2) ≥ 0
⇒ 1 + 8𝑦 ≥ 0
1
⇒ 𝑦 ≥ −8
For range, putting minimum value of 𝑥
from domain 𝑥 = 3 in (i)
𝑔 𝑓(𝑥) = √2.3 − 6
1
Range: 𝑓𝑔(𝑥) ∈ 𝑅, 𝑓 𝑔(𝑥) ≥ − 8
b) ii. 𝑔 𝑓(𝑥) = 𝑔(2𝑥 2 − 𝑥)
=0
1
= 2𝑥 2 −𝑥
Range: 𝑔𝑓(𝑥) ∈ 𝑅, 𝑔𝑓(𝑥) ≥ 0
1
= 𝑥(2𝑥−1)
18. Given,
1
= 2𝑥 − 𝑥
𝑥 ∈ 𝑅, 𝑥 < 2
For domain, 𝑥 ≠ 0, 2𝑥 − 1 ≠ 0, 𝑥 ≠ 2
𝑔(𝑥) = (𝑥 − 3)2
𝑥 ∈ 𝑅, 𝑥 > 3
∴ Domain (𝑥) ∈ 𝑅, 𝑥 ∈ 0, 𝑥 ≠ 2
2
𝑓(𝑥)
a) i. Range of 𝑓(𝑥):
Range ≥ −
𝑅≥−
1
For Range,
𝑏 2 −4𝑎𝑐
Let, 𝑦 =
4𝑎
1
2𝑥 2 −𝑥
⇒ 2𝑥 2 𝑦 − 𝑥𝑦 = 1
(−1)2 −4×2×0
4×2
⇒ 2𝑥 2 𝑦 − 𝑥𝑦 − 1 = 0
−1
𝑅≥ 8
For Range, 𝑏 2 − 4𝑎𝑐 ≥ 0
1
∴ Range: 𝑓(𝑥) ∈ 𝑅, 𝑓(𝑥) ≥ − 8
⇒ 𝑦 2 − 4 × 2𝑦 × (−1) ≥ 0
a) ii. Range of 𝑔(𝑥);
⇒ 𝑦 2 + 𝑓𝑦 ≥ 0
1
⇒ 𝑦(𝑦 + 8) ≥ 0
Here, 𝑔(𝑥) = 𝑥
Range: 𝑔(𝑥) ∈ 𝑅, 𝑔(𝑥) ≠ 0
∴ Range 𝑔𝑓(𝑥) ∈ 𝑅, 𝑔𝑓(𝑥) > 0,
𝑔𝑓(𝑥) ≤ −𝑓
1
b) i. 𝑓 𝑔(𝑥) = 𝑓 (𝑥)
1 2
1
= 2 (𝑥) − 𝑥
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19. Given,
𝑓(𝑥)
𝑓(𝑥)
= 2𝑥 + 5, 𝑥 ∈ 𝑅, 𝑥 < 2
=2×3−1
=6−1
𝑔(𝑥) = (𝑥 − 3)2 , 𝑥 ∈ 𝑅, 𝑥 > 3
=5
a) i. Putting minimum value of 𝑥
Range: 𝑓(𝑥) ∈ 𝑅, 𝑓(𝑥) > 5
from domain 𝑥 = 2
for range putting 𝑥 = 7,
𝑓(𝑥)
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=2×2+5
=9
𝑥>3
Range: 𝑔(𝑥) ∈ 𝑅, 𝑔(𝑥) > 47
b)
𝑔𝑓(𝑥)
= (3 − 3)2
= 4𝑥 2 − 4𝑥 + 1 − 2
=0
= 4𝑥 2 − 4𝑥 − 1
Range 𝑔(𝑥) ∈ 𝑅, 𝑔(𝑥) > 0
b)
c)
𝑔𝑓(𝑥)
= 𝑔(2𝑥 − 1)
= 4𝑥 2 − 4𝑥 − 1
𝑔 𝑓(𝑥) = 𝑔(2𝑥 + 5)
= (2𝑥 + 5 − 3)2
∴ 𝑔(𝑥) = 𝑥 2 − 2, 𝑥 > 7
= (2𝑥 + 2)2 ……. (i)
c)
= 𝑔(2𝑥 − 1)
= (2𝑥 − 1)2 − 2
for range putting 𝑥 = 3
𝑔(𝑥)
= 72 − 2
= 47
f
∴ Range, 𝑓(𝑥) < 9
a) ii. 𝑔(𝑥) = (𝑥 − 3)2
𝑔(𝑥)
= 2𝑥 − 1 > 7
Domain of 𝑔 𝑓(𝑥)
⇒𝑥>4
When 𝑔(𝑥) = (𝑥 − 3)2 then 𝑥 > 3
Domain: 𝑥 ∈ 𝑅, 𝑥 > 4
So, 2𝑥 + 5 > 3
For range: putting 𝑥 = 4
⇒ 𝑥 > −1
𝑔𝑓(𝑥)
= 47
So, domain, 𝑔 𝑓(𝑥) ∈ 𝑅, −1 < 𝑥 < 3
Range: i – (i) putting value,
(2 × (−1) + 2)2 < 𝑔 𝑓(𝑥) < (2 × 3 + 2)2
= 4 × 42 − 4 × 4 − 1
∴ Range: 𝑔𝑓(𝑥) ∈ 𝑅, 𝑔𝑓(𝑥) > 47
21. Given,
0 < 𝑔𝑓(𝑥) < 64
𝑓(𝑥) = 3𝑥 + 2, 𝑥 ∈ 𝑅, 𝑥 > 1
Range: 𝑔𝑓(𝑥) ∈ 𝑅, 0 < 𝑔𝑓(𝑥) < 6
𝑔(𝑥) = 𝑥 2 + 1, 𝑥 ∈ 𝑅, 𝑥 > 8
a) i. Range of 𝑓(𝑥):
20. Given,
𝑓(𝑥) = 2𝑥 − 1
𝑥 ∈ 𝑅, 𝑥 > 3
𝑔(𝑥) = 𝑥 2 − 2
𝑥 ∈ 𝑅, 𝑥 > 7
a) i. Range of 𝑓(𝑥):
𝑓(𝑥) = 2𝑥 − 1; 𝑥 > 3
for range putting 𝑥 = 3
𝑓(𝑥) = 3𝑥 + 2, 𝑥 > 1
Putting 𝑥 = 1
𝑓(𝑥)
=3×1+2
=5
Range: 𝑓(𝑥) ∈ 𝑅, 𝑓(𝑥) > 5
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a) ii. Rangel of 𝑔(𝑥)
𝑔(𝑥) = 𝑥 2 + 1, 𝑥 > 8
putting 𝑥 = 8
= 82 + 1
𝑔(𝑥)
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= 65
Range: 𝑔(𝑥) ∈ 𝑅, 𝑔(𝑥) > 65
b)
𝑔 𝑓(𝑥) = 𝑔(3𝑥 + 2)
= (3𝑥 + 2)2 + 1
= 9𝑥 2 + 12𝑥 + 4 + 1
= 9𝑥 2 + 12𝑥 + 5
c)
Domain of 𝑔𝑓(𝑥):
𝑔𝑓(𝑥)
= 𝑔(3𝑥 + 2)
= 9𝑥 2 + 12𝑥 + 5
3𝑥 + 2 > 𝑓
⇒ 3𝑥 > 8 − 2
⇒𝑥>2
∴ Domain: 𝑥 ∈ 𝑅, 𝑥 > 2
Range of 𝑔𝑓(𝑥):
= 9𝑥 2 + 12𝑥 + 5
𝑔𝑓(𝑥)
Putting 𝑥 = 2
= 9 × 22 + 12 × 2 + 5
𝑔𝑓(𝑥)
= 65
∴ Rangle: 𝑔𝑓(𝑥) ∈ 𝑅, 𝑔𝑓(𝑥) > 65
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