![Answer:
(i) f(x) = 3x + 2 ;g(x) = 6x –
k
fog = f[g(x)]
= f (6x – k)
= 3(6x – k) + 2
= 18x – 3K + 2
g0f= g [f(x)]
= g (3x + 2)
= 6(3x + 2) – k
= 18x + 12 – k
But given fog = gof.
18x – 3x + 2 = 18x + 12 – k
-3k + 2 = 12 – k
-3 k + k = 12-2
-2k = 10
k = −10/2 = -5
The value of k = -5](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-26-320.jpg)
![(ii) f(x) = 2x – k, g(x) = 4x +
5
Answer:
f(x) = 2x – k ; g(x) = 4x + 5
fog = f[g(x)]
= f(4x + 5)
= 2(4x + 5) – k
= 8x + 10 – k
gof = g [f(x)]
= g(2x – k)
= 4(2x – k) + 5
= 8x – 4k + 5
But fog = gof
8x + 10 – k = 8x – 4k + 5
-k + 4k = 5 – 10
3k = -5
k = −5/3
The value of k = −5/3](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-27-320.jpg)
![(i) f(x) = x – 1, g (x) = 3x + 1, h(x)
= x2
fog (x) = f[g(x)]
= f(3x + 1)
= 3x + 1 – 1
fog = 3x
(fog) o h(x) = fog [h(x)] ,
= fog (x2)
= 3(x2)
(fog) oh = 3x2 …..(1)
goh (x) = g[h(x)]
= g(x2)](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-34-320.jpg)
![= 3(x2) + 1
= 3x2 +1
fo(goh) x = f [goh(x)]
= f[3x2 + 1]
= 3x2 + 1 – 1
= 3x2 ….(2)
From (1) and (2) we get
(fog) oh = fo (goh)
Hence it is verified](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-35-320.jpg)
![(ii) f(x) = x2 ; g (x) = 2x and h(x) = x + 4
(fog) x = f[g(x)]
= f (2x)
= (2x)2
= 4x2
(fog) oh (x) = fog [h(x)]
= fog (x + 4)
= 4(x + 4)2
= 4[x2 + 8x + 16]
= 4x2 + 32x + 64 …. (1)
goh (x) = g[h(x)]
= g(x + 4)
= 2(x + 4)
= 2x + 8
fo(goh) x = fo [goh(x)]
= f[2x + 8]
= (2x + 8)2
= 4×2 + 32x + 64 …. (2)
From (1) and (2) we get
(fog) oh = fo(goh)](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-36-320.jpg)
![(iii) f(x) = x – 4 ; g (x) = x2; h(x) = 3x – 5
fog (x) = f[g(x)]
= f(x2)
= x2 – 4
(fog) oh (x) = fog [h(x)]
= fog (3x – 5)
= (3x – 5)2 – 4
= 9x2 – 30x + 25 – 4
= 9x2 – 30x + 21 ….(1)
goh (x) = g[h(x)]
= g(3x – 5)
= (3x – 5)2
= 9x2 + 25 – 30x
fo(goh)x = f[goh(x)]
= f[9x2 – 30x + 25]
= 9x2 – 30x + 25 – 4
= 9x2 – 30x + 21 ….(2)
From (1) and (2) we get
(fog) oh = fo(goh)](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-37-320.jpg)
![Solution:
f ={(-1, 3), (0, -1), 2, -9)
f(x) = (ax) + b ………… (1)
is the equation of all linear functions.
∴ f(-1) = 3
f(0) = -1
f(2) = -9
f(x) = ax + b
f(-1) = -a + b = 3 …………… (2)
f(0) = b = -1
-a – 1 = 3 [∵ substituting b = – 1 in
(2)]
-a = 4
a = -4
The linear function is -4x – 1. [From
(1)]](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-38-320.jpg)
![Answer:
Given C(t) = 3t
C(at1) = 3at1 …. (1)
C(bt2) = 3 bt2 …. (2)
Add (1) and (2)
C(at1) + C(bt2) = 3at1 + 3bt2
C(at1 + bt2) = 3at1 + 3bt2
= Cat1 + Cbt2 [from (1) and (2)]
∴ C(at1 + bt2) = C(at1 + bt2)
Superposition principle is
satisfied.
∴ C(t) = 3t is a linear function.](https://image.slidesharecdn.com/xstdmaths-numbersandsequenceex1-201128163557/85/X-std-maths-Relations-and-functions-ex-1-5-amp-1-6-39-320.jpg)
X std maths - Relations and functions (ex 1.5 & 1.6)
- 1. PEDAGOGY OF MATHEMATICS – PART II BY Dr. I. UMA MAHESWARI Principal Peniel Rural College of Education,Vemparali, Dindigul District iuma_maheswari@yahoo.co.in
- 2. X STD Ex 1.5
- 23. Solution: (i) f(x) = x – 6, g(x) = x2 fog(x) = f(g(x)) = f(x2) = x2 – 6 …………….. (1) gof(x) = g(f(x)) = g(x – 6) = (x – 6)2 = x2 + 36 – 12x = x2 – 12x + 36 ……………… (2) (1) ≠ (2)
- 24. (iii) f(x) = (x+6)/3 g(x) = 3 – x
- 25. (iv) f(x) = 3 + x, g(x) = x – 4 fog(x) = f(g(x)) = f(x – 4) = 3 + x – 4 = x – 1 ………… (1) gof(x) = g(f(x)) = g(3 + x) = 3 + x – 4 = x – 1 ……………… (2) Here fog(x) = gof(x) (v) f(x) = 4x2 – 1, g(x) = 1 + x fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)2 – 1 = 4(1 + x2 + 2x) – 1 = 4 + 4x2 + 8x – 1 = 4x2 + 8x + 3 ……………. (1) gof(x) = g(f(x)) = g(4x2 – 1) = 1 + 4x2 – 1 = 4x2 …………….. (2) (1) ≠ (2) ∴ fog(x) ≠ gof(x)
- 26. Answer: (i) f(x) = 3x + 2 ;g(x) = 6x – k fog = f[g(x)] = f (6x – k) = 3(6x – k) + 2 = 18x – 3K + 2 g0f= g [f(x)] = g (3x + 2) = 6(3x + 2) – k = 18x + 12 – k But given fog = gof. 18x – 3x + 2 = 18x + 12 – k -3k + 2 = 12 – k -3 k + k = 12-2 -2k = 10 k = −10/2 = -5 The value of k = -5
- 27. (ii) f(x) = 2x – k, g(x) = 4x + 5 Answer: f(x) = 2x – k ; g(x) = 4x + 5 fog = f[g(x)] = f(4x + 5) = 2(4x + 5) – k = 8x + 10 – k gof = g [f(x)] = g(2x – k) = 4(2x – k) + 5 = 8x – 4k + 5 But fog = gof 8x + 10 – k = 8x – 4k + 5 -k + 4k = 5 – 10 3k = -5 k = −5/3 The value of k = −5/3
- 34. (i) f(x) = x – 1, g (x) = 3x + 1, h(x) = x2 fog (x) = f[g(x)] = f(3x + 1) = 3x + 1 – 1 fog = 3x (fog) o h(x) = fog [h(x)] , = fog (x2) = 3(x2) (fog) oh = 3x2 …..(1) goh (x) = g[h(x)] = g(x2)
- 35. = 3(x2) + 1 = 3x2 +1 fo(goh) x = f [goh(x)] = f[3x2 + 1] = 3x2 + 1 – 1 = 3x2 ….(2) From (1) and (2) we get (fog) oh = fo (goh) Hence it is verified
- 36. (ii) f(x) = x2 ; g (x) = 2x and h(x) = x + 4 (fog) x = f[g(x)] = f (2x) = (2x)2 = 4x2 (fog) oh (x) = fog [h(x)] = fog (x + 4) = 4(x + 4)2 = 4[x2 + 8x + 16] = 4x2 + 32x + 64 …. (1) goh (x) = g[h(x)] = g(x + 4) = 2(x + 4) = 2x + 8 fo(goh) x = fo [goh(x)] = f[2x + 8] = (2x + 8)2 = 4×2 + 32x + 64 …. (2) From (1) and (2) we get (fog) oh = fo(goh)
- 37. (iii) f(x) = x – 4 ; g (x) = x2; h(x) = 3x – 5 fog (x) = f[g(x)] = f(x2) = x2 – 4 (fog) oh (x) = fog [h(x)] = fog (3x – 5) = (3x – 5)2 – 4 = 9x2 – 30x + 25 – 4 = 9x2 – 30x + 21 ….(1) goh (x) = g[h(x)] = g(3x – 5) = (3x – 5)2 = 9x2 + 25 – 30x fo(goh)x = f[goh(x)] = f[9x2 – 30x + 25] = 9x2 – 30x + 25 – 4 = 9x2 – 30x + 21 ….(2) From (1) and (2) we get (fog) oh = fo(goh)
- 38. Solution: f ={(-1, 3), (0, -1), 2, -9) f(x) = (ax) + b ………… (1) is the equation of all linear functions. ∴ f(-1) = 3 f(0) = -1 f(2) = -9 f(x) = ax + b f(-1) = -a + b = 3 …………… (2) f(0) = b = -1 -a – 1 = 3 [∵ substituting b = – 1 in (2)] -a = 4 a = -4 The linear function is -4x – 1. [From (1)]
- 39. Answer: Given C(t) = 3t C(at1) = 3at1 …. (1) C(bt2) = 3 bt2 …. (2) Add (1) and (2) C(at1) + C(bt2) = 3at1 + 3bt2 C(at1 + bt2) = 3at1 + 3bt2 = Cat1 + Cbt2 [from (1) and (2)] ∴ C(at1 + bt2) = C(at1 + bt2) Superposition principle is satisfied. ∴ C(t) = 3t is a linear function.