## Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Ex 1.5

Question 1

Using the functions f and g given below, find fog and gof. Check whether fog = gof.

(i) f(x) = x – 6, g(x) = x^{2}

(ii) f(x) = \(\frac { 2 }{ x } \), g(x) = 2x^{2}– 1

(iii) f(x) = \(\frac { x+6 }{ 3 } \)g(x) = 3 – x

(iv) f(x) = 3 + x, g(x) = x – 4

(v) f(x) = 4x^{2}– 1,g(x) = 1 + x

Solution:

(i) f(x) = x – 6, g(x) = x^{2}

fog(x) = f(g(x)) = f(x^{2}) = x^{2} – 6 …(1)

gof(x) = g(f(x)) = g(x – 6) = (x – 6)^{2}

= x^{2} + 36 – 12x = x^{2} – 12x + 36…(2)

(1) ≠ (2)

∴ fog(x) ≠ gof(x)

(ii) f(x) = \(\frac { 2 }{ x } \),g(x) = 2x^{2} – 1

(iii) f(x) = \(\frac { x+6 }{ 3 } \),g(x) = 3 – x

(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) = 4x^{2} – 1, g(x) = 1 + x

fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)^{2} – 1

= 4(1 + x^{2} + 2x) – 1 = 4 + 4x^{2} + 8x – 1

= 4x^{2} + 8x + 3 …(1)

gof(x) = g(f(x)) = g(4x^{2} – 1)

= 1 + 4x^{2} – 1 = 4x^{2} …(2)

(1) ≠ (2)

∴ fog(x) ≠ gof(x)

Question 2.

Find the value of k, such that fog = gof

(i) f(x) = 3x + 2, g(x) = 6x – k

(ii) f(x) = 2x – k, g(x) = 4x + 5

Solution:

(i) f(x) = 3x + 2, g(x) = 6x – k

fog(x) = f(g(x)) = f(6x – k) = 3(6x – k) + 2

= 18x – 3k + 2 …(1)

gof(x) = g(f(x)) = g(3x + 2) = 6(3x + 2) – k

= 18x + 12 – k …(2)

(1) = (2)

2k = -10

k = -5

(ii) f(x) = 2x – k, g(x) = 4x + 5

fog(x) = f(g(x)) = f(4x + 5) = 2(4x + 5) – k

= 8x + 10 – k …(1)

gof(x) = g(f(x)) = g(2x – k) = 4(2x – k)+ 5

= 8x – 4k + 5 ….(2)

(1) = (2)

3k = -5

k = \(\frac { -5 }{ 3 } \)

Question 3.

if f(x) = 2x – 1, g(x) = \(\frac { x+1 }{ 2 } \), show that fog = gof = x

Solution:

f(x) = 2x – 1, g(x) = \(\frac { x+1 }{ 2 } \), fog = gof = x

Question 4.

(i) If f (x) = x^{2} – 1, g(x) = x – 2 find a, if gof(a) = 1.

(ii) Find k, if f(k) = 2k – 1 and fof (k) = 5.

Solution:

(i) f(x) = x^{2} – 1, g(x) = x – 2

Given gof(a) = 1

gof(x) = g(f(x)

= g(x^{2} – 1) = x^{2} – 1 – 2

= x^{2} – 3

gof(a) ⇒ a^{2} – 3 = 1 =+ a^{2} = 4

a = ± 2

(ii) f(k) = 2k – 1

fo f(k) = 5

f(f(k)m = f(2k – 1) = 5

⇒ 2(2k – 1) – 1 = 5

4 k – 2 – 1 = 5 ⇒ 4k = 8

k = 2

Question 5.

Let A,B,C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g : B → C be defined by g(x) = x^{2}. Find the range of fog and gof

Solution:

f(x) = 2x + 1

g(x) = x^{2}

fog(x) = fg(x)) = f(x^{2}) = 2x^{2} + 1

gof(x) = g(f(x)) = g(2x + 1) = (2x + 1)^{2}

= 4x^{2} + 4x + 1

Range of fog is

{y/y = 2x^{2} + 1, x ∈ N}

Range of gof is

{y/y = (2x + 1)^{2}, x ∈ N}.

Question 6.

If f(x) = x^{2} – 1. Find (i)f(x) = x^{2} – 1, (ii)fofof

Solution:

(i) f(x) = x^{2} – 1

fof(x) = f(fx)) = f(x^{2} – 1)

= (x^{2} – 1 )^{2} – 1;

= x^{4} – 2x^{2} + 1 – 1

= x^{4} – 2x^{2}

(ii) fofof = f o f(f(x))

= f o f (x^{4} – 2x^{2})

= f(f(x^{4} – 2x^{2}))

= (x^{4} – 2x^{2})^{2} – 1 =

= x^{8} – 4x^{6} + 4x^{4} – 1

Question 7.

If f: R → R and g : R → R are defined by f(x) = x^{5} and g(x) = x^{4} then check if f,g are one-one and fog is one-one?

Solution:

f(x) = x^{5}

g(x) = x^{4}

fog = fog(x) = f(g(x)) = f(x^{4})

= (x^{4})^{5} = x^{20}

f is one-one, g is not one-one.

∵ g(1) = 1^{4} = 1

g(-1) = ( -1)^{4} = 1

Different elements have same images

fog is not one-one. [∵ fog (1) = fog (-1) = 1]

Question 8.

Consider the functions f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh) in each case.

(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x^{2}

(ii) f(x) = x^{2}, g(x) = 2x and h(x) = x + 4

(iii) f(x) = x – 4, g(x) = x^{2} and h(x) = 3x – 5

Solution:

(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x^{2}

f(x) = x – 1

g(x) = 3x + 1

f(x) = x^{2}

(fog)oh = fo(goh)

LHS = (fog)oh

fog = f(g(x)) = f(3x + 1) = 3x + 1 – 1 = 3x

(fog)oh = (fog)(h(x)) = (fog)(x2) = 3×2 …(1)

RHS = fo(goh)

goh = g(h(x)) = g(x^{2}) = 3x^{2} + 1

fo(goh) = f(3x^{2} + 1) = 3x^{2} + 1 – 1= 3x^{2 }…(2)

LHS = RHS Hence it is verified.

(ii) f(x) = x^{2}, g(x) = 2x, h(x) = x + 4

(fog)oh = fo(goh)

LHS = (fog)oh

fog = f(g(x)) = f(2x) = (2x)^{2} = 4x^{2}

(fog)oh = (fog) h(x) = (fog) (x + 4)

= 4(x + 4)^{2} = 4(x^{2} + 8x+16)

= 4x^{2} + 32x + 64 …(1)

RHS = fo(goh) goh = g(h(x)) = g(x + 4)

= 2(x + 4) = (2x + 8)

fo(goh) = f(goh) = f(2x + 8) = (2x + 8)^{2}

= 4x^{2} + 32x + 64 …(2)

(1) = (2)

LHS = RHS

∴ (fog)oh = fo(goh) It is proved.

(iii) f(x) = x – 4, g(x) = x^{2}, h(x) = 3x – 5

(fog)oh = fo(goh)

LHS = (fog)oh

fog = f(g(x)) = f(x^{2}) = x^{2} – 4

(fog)oh = (fog)(3x – 5) = (3x – 5)^{2} – 4

= 9x^{2} – 30x + 25 -4

= 9x^{2} – 30x + 21 …(1)

∴ RHS = fo(goh)

(goh) = g(h(x)) = g(3x – 5) = (3x – 5)^{2}

= 9x^{2} – 30x + 25

fo(goh) = f(9x^{2} – 30 x + 25)

= 9x^{2} – 30x + 25 – 4

= 9x^{2} – 30x + 21 …(2)

(1) = (2)

LHS = RHS

∴ (fog)oh = fo(goh)

It is proved.

Question 9.

Let f ={(-1, 3),(0, -1),(2, -9)} be a linear function from Z into Z . Find f(x).

Solution:

f ={(-1,3), (o,-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 = -A

The linear function is -4x – 1. [From (1)]

Question 10.

In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at_{1} + bt_{2}) = aC(t_{1}) + bC(t_{2}), where a,b are constants. Show that the circuit C(t) = 31 is linear.

Solution:

Given C(t) = 3t. To prove that the function is linear

C(at_{1}) = 3a(t_{1})

C(bt_{2}) = 3 b(t_{2})

C(at_{1} + bt_{2}) = 3 [at_{1} + bt_{2}] = 3 at_{1} + 3 bt_{2}

= a(3t_{1}) + b(3t_{2}) = a[C(t_{1}) + b(Ct_{62})]

∴ Superposition principle is satisfied.

Hence C(t) = 3t is linear function.