composite function class 10 unit 1

 Here are a few things that you need to consider before checking the solutions:

1. If A, B, and C be three non-empty sets such that f is the function from A->B and g is the function from B->C, then gof(x) or g[f(x)] is a composite function defined from A->C.
2. When mathematical operations are performed in two or more functions, it is known as combination of functions.
3. Composition of functions is written with a small circle ( o ).
4. Disintegrating the composite function to get its constituent functions is known as decomposition of functions.

1. In each of the following conditions find fog and gof.

a) f(x) = 2x and g(x) = x²
Solution:
fog(x) = f[g(x)]= f(x²)= 2(x²)= 2x²
And,gof(x) = g[f(x)]= g[2x]= (2x)²= 4x²

c) f(x) = x² + 2 and g(x) = 2x + 1
Solution:fog(x) = f[g(x)]= f[2x + 1]
= (2x + 1)² + 2
= (4x² + 4x + 1) + 2= 4x² + 4x + 3
And,gof(x) = g[f(x)]
= g[x² + 2]
= 2(x² + 2) + 1
= 2x² + 5

e) f(x) = 2x² – 3 and g(x) = 3x – 2Solution:fog(x) = f[g(x)]= f[3x – 2]= 2(3x – 2)² – 3= 2(9x² – 12x + 4) – 3= 18x² – 24x + 5And,gof(x) = g[f(x)]= g[2x² – 3]= 3(2x² – 3) – 2= 6x² – 11

2. In each of the following conditions, find fog(3) and gof(2).

a) f:x -> x² + 5 and g:x -> x³ +6
Solution:Given,f(x) = x² + 5 and g(x) = x³ + 6Now,fog(x) = f[g(x)]= f[x³ + 6]= (x³ + 6)² + 5fog(3) = (3³ + 6)² + 5= (27 +6)² + 5= (33)² + 5= 1094And,gof(x) = g[f(x)]= g[x² + 5]= (x² + 5)³ + 6gof(2) = (2² + 5)³ + 6= (4 + 5)³ + 6= 9³ + 6= 735

c) f:x -> x³ + 3x and g:x -> x² + 2x
Solution:Given,f(x) = x³ + 3x and g(x) = x² + 2xNow,fog(x) = f[g(x)]= f[x² + 2x]= (x² + 2x)³ + 3(x² + 2x)fog(3) = (3² + 2×3)³ + 3(3² + 2×3)= (9 +6)³ + 3(9 + 6)= (15)³ + 3(15)= 3420And,gof(x) = g[f(x)]= g[x³ + 3x]= (x³ + 3x)² + 2(x³ + 3x)gof(2) = (2³ + 3×2)² + 2(2³ + 3×2)= (8 + 6)² + 2(8 + 6)= 14² + 2(14)= 224

e) f:x -> x² + 2x + 3 and g:x -> x² + 4x + 5
Solution:Given,f(x) = x² + 2x + 3 and g(x) = x² + 4x + 5Now,fog(x) = f[g(x)]= f[x² + 4x + 5]= (x² + 4x + 5)² + 2(x² + 4x + 5) + 3fog(3) = (3² + 4×3 + 5)² + 2(3² + 4×3 + 5) + 3= (9 + 12 + 5)² + 2(9 + 12 + 5) + 3= 29² + 2(29) + 3= 902And,gof(x) = g[f(x)]= g[x² + 2x + 3]= (x² +2x + 3)² + 4(x² + 2x + 3) + 5gof(2) = (2² + 2×2 + 3)² + 4(2² + 2×2 + 3) + 5= (4 + 4 + 3)² + 4(4 + 4 + 3) + 5= (11)² + 4(11) + 5= 170

3a) If f:x -> 3x + b and ff(2) = 12, find the value of b.
Solution:Given,f(x) = 3x + bff(2) = 12We know,ff(x) = fof(x) = f[f(x)]= f[3x + b]= 3(3x + b) + bNow,fof(2) = 12or, 3(3×2 + b) + b = 12or, 3(6 + b) + b = 12or, 18 + 3b + b = 12or, 4b = 12 – 18or, 4b = – 6So, b = – 3/2

4a) If f(x) = 4x, g(x) = x + 1 and fog(x) = 20, find the value of x.
Solution:Given,fog(x) = 20or, 20 = fog(x)or, 20 = f[g(x)]or, 20 = f[x + 1]or, 20 = 4(x + 1)or, 20/4 = 4/4 (x + 1)or, 5 = x + 1So, x = 4

5a) If f(x) = x + 2 and g(x+1) = x + 1 then find gof(4).
Solution:Given,f(x) = x + 2g(x + 1) = x + 1Now,g(x + 1) = 1(x + 1)or, g(x) = 1xSo, g(x) = xAnd,gof(x) = g[f(x)]= g[x + 2]= x + 2So, gof(4) = 4 + 2 = 6

6a) If f(4x – 15) = 8x – 27, find fof(2).Solution:Given,f(4x – 15) = 8x – 27or, f(4x – 15) = 8x – 30 + 3or, f(4x – 15) = 2(4x – 15) + 3So, f(x) = 2x + 3Now,fof(x) = f[f(x)]= f[2x + 3]= 2(2x + 3) + 3= 4x + 9So, fof(2) = 4×2 + 9 = 8 + 9 = 17

7a) If f(3x – 1) = -6x +3, find fof(2).Solution:Given,f(3x – 1) = -6x + 3or, f(3x – 1) = – (6x – 3)or, f(3x – 1) = – (6x – 2 – 1)or, f(3x – 1) = – 2(3x – 1) + 1So, f(x) = -2x + 1Now,fof(x) = f[f(x)]= f(-2x + 1)= -2(-2x + 1) + 1= 4x – 2 + 1= 4x – 1So, fof(2) = 4(2) – 1 = 8 -1 = 7

8a) If f(x) = x + 2, g(x) = x² and k(x) = 3x +1, find the value of.
(i) fok(2)Solution:fok(x) = f[k(x)]= f[3x +1]= (3x +1) + 2= 3x + 1 + 2= 3x + 3So, fok(2) = 3(2) + 3 = 6+3 = 9
(ii) kok(2)Solution:kok(x) = k[k(x)]= k[3x +1]= 3(3x +1) + 1= 9x + 3 + 1= 9x + 4So, kok(2) = 9(2) +4 = 18+4 = 22
(iii) f(kok)(-5)Solution:kok(x) = 9x + 4 (from ii)f(kok)(x) = f[kok(x)]= f[9x +4]= (9x +4) + 2= 9x + 4 + 2= 9x + 6So, f(kok)(-5) = 9(-5) + 6 = -45+6 = -39
(iv) f(gok)(x)Solution:(gok)(x) = g[k(x)]= g[3x +1]= (3x +1)²= 9x² + 6x +1Now,f(gok)(x) = f[gok(x)]= f[9x² + 6x +1]= (9x² + 6x +1) + 2= 9x² + 6x +1 +2= 9x² + 6x +3

9a) If f(x) = 2x +3, gof(x) = 2x -1 then find fog(2) and gof(2).
Solution:Given,f(x) = 2x +3gof(x) = 2x -1
We know,gof(x) = g[f(x)]or, 2x -1 = g[2x +3]or, 2x – (4-3) = g[2x +3]or, 2x – 4 + 3 = g[2x +3]or, [2x +3] – 4 = g[2x +3]or, x -4 = g(x)So, g(x) = x – 4
Now,fog(x) = f[g(x)]= f[x -4]= 2(x -4) +3= 2x -8 + 3= 2x -5So, fog(2) = 2(2) – 5 = 4-5 = -1
And,gof(2) = 2(2) -1= 4 -1= 3

10a) If f(x+1) = x +3, g(x+1) = x+1 and h(x) = x² then find (hog)of.
Solution:Given,f(x+1) = x+3g(x+1) = x +1h(x) = x²

We know,f(x+1) = x+3or, f(x+1) = [x+1] +2so, f(x) = x +2
And,g(x+1) = [x+1]so, g(x) = x
Now,(hog)(x) = h[g(x)]= h[x]= x²
Again,(hog)of = (hog)[f(x)]= (hog)(x +2)= (x +2)²

11a) If f(x) = x² +x +2 and g(x) = 2x² -7 then prove that: 2fog(2) = f(2).g(2).
Solution:Given,f(x) = x² + x +2g(x) = 2x² -7To prove: 2 fog(2) = f(2).g(2)
Taking LHS= 2 fog(2)= 2 f[g(2)]= 2 f[2(2)² – 7]= 2 f[8 -7]= 2 f[1]= 2 [(1)² + 1 +2]= 2 [1 + 3]= 2×4= 8
Taking RHS= f(2).g(2)= [(2)² + 2 + 2] × [2(2)² – 7]= [4 + 2 + 2] × [8 – 7]= [8] × 1= 8
Since, LHS = RHS, 2 fog(2) = f(2).g(2)#proved.

12a) If f(x) = 3x -2 and fog(x) = 6x – 2, find the value of x such that gof(x) = 8.
Solution:
Given,f(x) = 3x – 2fog(x) = 6x – 2
To find: value of x when gof(x) = 8
We know,fog(x) = f[g(x)]or, 6x – 2 = 3[g(x)] – 2or, 6x – 2 + 2 = 3[g(x)] – 2 + 2or, 6x = 3[g(x)]or, g(x) = $\frac{6x}{3}$So, g(x) = 2
Now,gof(x) = g[f(x)]
or, 8 = g[3x – 2]
or, 8 = 2(3x – 2)
or, 8 = 6x – 4
or, 8 + 4 = 6x
or, 12 = 6x
or, x = $\frac{12}{6}$
So, x = 2

13a) If g(x) = 2x and fog(x) = 6x – 2, then find the value of x such that gof(x) = 10.
Solution:
Given,g(x) = 2xfog(x) = 6x – 2
To find: value of x where gof(x) = 10
We know,fog(x) = f[g(x)]or, 6x – 2 = f[2x]or, 3(2x) – 2 = f[2x]So, f(x) = 3x – 2
Now,gof(x) = g[f(x)]or, 10 = g[3x – 2]or, 10 = 2(3x – 2)or, 2×5 = 2(3x – 2)or, 5 = 3x – 2or, 5 + 2 = 3xor, 3x = 7
So, x = $\frac{7}{3}$

14a) If f(x) = $\frac{6}{x-2}$ where x≠2 and g(x) = kx² – 1, what is the value of k at gf(5) = 7?
Solution:
Given,f(x) = $\frac{6}{x-2}$, x≠2g(x) = kx² – 1
To find: value of k where gf(5) = 7
We know,gf(x) = gof(x) = g[f(x)]
or, gf(x) = g[ $\frac{6}{x-2}$]
or, gf(x) = k ${\left(\frac{6}{x-2}\right)}^2–1$
or, gf(x) = k $\left(\frac{36}{(x-2{)}^2}\right)–1$

Also,gf(5) = 7
or, $k\times \frac{36}{(5-2{)}^2}$  – 1= 7
or, $k\times \frac{36}{3^2}$ = 7 + 1
or, $k\times \frac{36}{9}$ = 8
or, k × 4 = 8
or, k × 4 = 2 × 4
$\therefore k=2$

15a) If p(x) = $\frac{1}{(x+2{)}^3}$ where x≠2, and p(x) = fog(x), find the possible formulas of f(x) and g(x).
Solution:Here,p(x) = ${\displaystyle \frac{1}{(x+2{)}^3}}$
Given,p(x) = fog(x) = f[g(x)]
So, possible formulas of f(x) and g(x) are:
Formula 1g(x) = (x+2)f(x) = $\frac{1}{x^3}$
Formula 2g(x) = $(x+2{)}^3$f(x) = $\frac{1}{x}$
In conclusion, you can form as many formulae as you wish until it satisfies the given composite function.

16a) If f(x) = $\frac{2x+3}{7}$ and g(x)= $\frac{7x–3}{2}$, prove that fog(x) is an identity function.
Solution:Given,f(x) = ${\displaystyle \frac{2x+3}{7}}$g(x) = ${\displaystyle \frac{7x-3}{2}}$
To prove: fog(x) = x
Taking LHS,$fog(x)$
$=f[g(x)]$
$={\displaystyle \frac{7x}{7}}$
$=x$
= RHS#proved
Thus, fog(x) is an identity function.

16c) If f(x) = $\frac{4x+5}{7x+3}$ and g(x) = 15, then prove that fog (x) and gof(x) both are constant functions.
Solution:
A function f(x) = c, where c is any constant number but not a variable is said to be a constant function.
Here,
f(x) = $\frac{4x+5}{7x+3}$g(x) = 15
Now,fog(x) = f[g(x)]
= f[15]
= $\frac{4(15)+5}{7(15)+3}$
= $\frac{60+5}{105+3}$
= $\frac{65}{108}$

And,gof(x) = g[f(x)]
= g[ $\frac{4x+5}{7x+3}$]
= 15

Since, both fog(x) and gof(x) have a constant output, they are constant functions. #proved.

16e) Let f(x) = 2x + 3 and g(x) = x²+2x+1 be two functions. Prove that gof(x) is a quadratic function.
Solution:
An equation in the form f(x) = ax² + bx + c, where a is any number except 0 is called a quadratic function.
Here,f(x) = 2x + 3g(x) = x² + 2x + 1
gof(x) = g[f(x)]
= g[2x + 3]
= (2x + 3)² + 2(2x + 3) +1
= 4x² + 12x + 9 + 4x + 6 + 1
= 4x² + 16x + 16
So, gof(x) is a quadratic function because its degree is 2. #proved.