**Question 1.If f(x) = 3x + 5, g(x) = 6x – 1, then find(i) (f + g) (x)(ii) (f – g) (2)(iii) (fg) (3)(iv) (f/g) (x) and its domainSolution:**

f(x) = 3x + 5, g (x) = 6x – 1

(i) (f + g) (x) = f (x) + g (x)

= 3x + 5 + 6x – 1

= 9x + 4

(ii) (f – g) (2) = f(2) – g(2)

= [3(2) + 5] – [6(2) – 1]

= 6 + 5 – 12 + 1

= 0

(iii) (fg) (3) = f (3) g(3)

= [3(3) + 5] [6(3) – 1]

= (14) (17)

= 238

**Question 2.Let f: (2, 4, 5} → {2, 3, 6} and g: {2, 3, 6} → {2, 4} be given by f = {(2, 3), (4, 6), (5, 2)} and g = {(2, 4), (3, 4), (6, 2)}. Write down gof.Solution:**

f = {(2, 3), (4, 6), (5, 2)}

∴ f(2) = 3, f(4) = 6, f(5) = 2

g ={(2, 4), (3, 4), (6, 2)}

∴ g(2) = 4, g(3) = 4, g(6) = 2

gof: {2, 4, 5} → {2, 4}

(gof) (2) = g(f(2)) = g(3) = 4

(gof) (4) = g(f(4)) = g(6) = 2

(gof) (5) = g(f(5)) = g(2) = 4

∴ gof = {(2, 4), (4, 2), (5, 4)}

**Question 3.**

(ii) (gof) (x) = g(f(x))

= g(2x^{2} + 3)

= 5(2x + 3) – 2

= 10x^{2} + 13

(iii) (fof) (x) = f(f(x))

(iv) (gog) (x) = g(g(x))

= g(5x – 2)

= 5(5x – 2) – 2

= 25x – 12

**Question 4.Verify that f and g are inverse functions of each other, where**

Replacing x by g(x), we get

Here, f[g(x)] = x and g[f(x)] = x.

∴ f and g are inverse functions of each other.

**Question 5.Check if the following functions have an inverse function. If yes, find the inverse function.(vi) f(x) = Solution:**

(i) f(x) = 5x

^{2}= y (say)

For two values (x

_{1}and x

_{2}) of x, values of the function are equal.

∴ f is not one-one.

∴ f does not have an inverse.

(ii) f(x) = 8 = y (say)

For every value of x, the value of the function f is the same.

∴ f is not one-one i.e. (many-one) function.

∴ f does not have the inverse.

(vi) f(x) = x + 7, x < 0

= 8 – x, x ≥ 0

We observe from the graph that for two values of x, say x_{1} , x_{2} the values of the function are equal.

i.e. f(x_{1}) = f(x_{2})

∴ f is not one-one (i.e. many-one) function.

∴ f does not have inverse.

**Question 6.If f(x) = , then find(i) f(3)(ii) f(2)(iii) f(0)Solution:**

f(x) = x

^{2}+ 3, x ≤ 2

= 5x + 7, x > 2

(i) f(3) = 5(3) + 7

= 15 + 7

= 22

(ii) f(2) = 2^{2} + 3

= 4 + 3

= 7

(iii) f(0) = 0^{2} + 3 = 3

**Question 7.If f(x) = , then find(i) f(-4)(ii) f(-3)(iii) f(1)(iv) f(5)Solution:**

f(x) = 4x – 2, x ≤ -3

= 5, -3 < x < 3

= x

^{2}, x ≥ 3

(i) f(-4) = 4(-4) – 2

= -16 – 2

= -18

(ii) f(-3) = 4(-3) – 2

= -12 – 2

= -14

(iii) f(1) = 5

(iv) f(5) = 5^{2} = 25

**Question 8.If f(x) = 2 |x| + 3x, then find(i) f(2)(ii) f(-5)Solution:**

f(x) = 2 |x| + 3x

(i) f(2) = 2|2| + 3(2)

= 2 (2) + 6 ….. [∵ |x| = x, x > 0]

= 10

(ii) f(-5) = 2 |-5| + 3(-5)

= 2(5) – 15 …..[∵ |x| = -x, x < 0]

= 10 – 15

= -5

**Question 9.If f(x) = 4[x] – 3, where [x] is greatest integer function of x, then find(i) f(7.2)(ii) f(0.5)(iv) f(2π), where π = 3.14Solution:**

f(x) = 4[x] – 3

(i) f(7.2) = 4 [7.2] – 3

= 4(7) – 3 ………[∵ 7 ≤ 7.2 < 8, [7.2] = 7]

= 25

(ii) f(0.5) = 4[0.5] – 3

= 4(0) – 3 ………[∵ 0 ≤ 0.5 < 1, [0.5] = 0]

= -3

= 4[-2.5] – 3

= 4(-3) – 3 …….[∵-3 ≤ -2.5 ≤ -2, [-2.5] = -3]

= -15

(iv) f(2π) = 4[2π] – 3

= 4[6.28] – 3 …..[∵ π = 3.14]

= 4(6) – 3 …….[∵ 6 ≤ 6.28 < 7, [6.28] = 6]

= 21

**Question 10.If f(x) = 2{x} + 5x, where {x} is fractional part function of x, then find(i) f(-1)(ii) f(1/4)(iii) f(-1.2)(iv) f(-6)Solution:**

f(x) = 2{x} + 5x

(i) {-1} = -1 – [-1] = -1 + 1 = 0

∴ f(-1) = 2 {-1} + 5(-1)

= 2(0) – 5

= -5

(iii) {-1.2} = -1.2 – [-1.2] = -1.2 + 2 = 0.8

f(-1.2) = 2{-1.2} + 5(-1.2)

= 2(0.8) + (-6)

= -4.4

(iv) {-6} = -6 – [-6] = -6 + 6 = 0

f(-6) = 2{-6} + 5(-6)

= 2(0) – 30

= -30

**Question 11.Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.(i) |x + 4| ≥ 5(ii) |x – 4| + |x – 2| = 3(iii) x**

^{2}+ 7|x| + 12 = 0(iv) |x| ≤ 3(v) 2|x| = 5(vi) [x + [x + [x]]] = 9(vii) {x} > 4(viii) {x} = o(ix) {x} = 0.5(x) 2{x} = x + [x]Solution:

(i) |x + 4| ≥ 5

The solution of |x| ≥ a is x ≤ -a or x ≥ a

∴ |x + 4| ≥ 5 gives

∴ x + 4 ≤ -5 or x + 4 ≥ 5

∴ x ≤ -5 – 4 or x ≥ 5 – 4

∴ x ≤ -9 or x ≥ 1

∴ The solution set = (-∞, – 9] ∪ [1, ∞)

(ii) |x – 4| + |x – 2| = 3 …..(i)

Case I: x < 2

Equation (i) reduces to

4 – x + 2 – x = 3 …….[x < 2 < 4, x – 4 < 0, x – 2 < 0]

∴ 6 – 3 = 2x

∴ x = 3/2

Case II: 2 ≤ x < 4

Equation (i) reduces to

4 – x + x – 2 = 3

∴ 2 = 3 (absurd)

There is no solution in [2, 4)

(iv) |x| ≤ 3 The solution set of |x| ≤ a is -a ≤ x ≤ a

∴ The required solution is -3 ≤ x ≤ 3

∴ The solution set is [-3, 3]

(v) 2|x| = 5

∴ |x| = 5/2

∴ x = ±5/2

(vi) [x + [x + [x]]] = 9

∴ [x + [x] + [x] ] = 9 …….[[x + n] = [x] + n, if n is an integer]

∴ [x + 2[x]] = 9

∴ [x] + 2[x] = 9 …..[[2[x] is an integer]]

∴ [x] = 3

∴ x ∈ [3, 4)

(vii) {x} > 4

This is a meaningless statement as 0 ≤ {x} < 1

∴ The solution set = { } or Φ

(viii) {x} = 0

∴ x is an integer

∴ The solution set is Z.

(ix) {x} = 0.5

∴ x = ….., -2.5, -1.5, -0.5, 0.5, 1.5, …..

∴ The solution set = {x : x = n + 0.5, n ∈ Z}

(x) 2{x} = x + [x]

= [x] + {x} + [x] ……[x = [x] + {x}]

∴ {x} = 2[x]

R.H.S. is an integer

∴ L.H.S. is an integer

∴ {x} = 0

∴ [x] = 0

∴ x = 0