**Answer
1** :

Given:

f(x) = x^{2} – 3x + 4.

Let us find x satisfying f (x) = f (2x + 1).

We have,

f (2x + 1) = (2x + 1)^{2} –3(2x + 1) + 4

= (2x)^{ 2} +2(2x) (1) + 1^{2} – 6x – 3 +4

= 4x^{2} + 4x + 1 – 6x+ 1

= 4x^{2} – 2x + 2

Now, f (x) = f (2x + 1)

x^{2} – 3x + 4 = 4x^{2} – 2x + 2

4x^{2} – 2x + 2 – x^{2} + 3x – 4 = 0

3x^{2} + x – 2 = 0

3x^{2} + 3x – 2x – 2 =0

3x(x + 1) – 2(x + 1) = 0

(x + 1)(3x – 2) = 0

x + 1 = 0 or 3x – 2 = 0

x = –1 or 3x = 2

x = –1 or 2/3

∴The values of x are –1 and 2/3.

**Answer
2** :

F (x) = (x – a)^{2}(x – b)^{2}

Let us find f (a + b).

We have,

f (a + b) = (a + b – a)^{2 }(a+ b – b)^{2}

f (a + b) = (b)^{2} (a)^{2}

∴ f(a + b) = a^{2}b^{2}

**Answer
3** :

Given:

y = f (x) = (ax – b) / (bx – a) ⇒ f (y) = (ay – b) / (by – a)

Let us prove that x = f (y).

We have,

y = (ax – b) / (bx – a)

By cross-multiplying,

y(bx – a) = ax – b

bxy – ay = ax – b

bxy – ax = ay – b

x(by – a) = ay – b

x = (ay – b) / (by – a) = f (y)

∴ x= f (y)

Hence proved.

**Answer
4** :

Given:

f (x) = 1 / (1 – x)

Let us prove that f [f {f (x)}] = x.

Firstly, let us solve for f {f (x)}.

f {f (x)} = f {1/(1 – x)}

= 1 / 1 – (1/(1 – x))

= 1 / [(1 – x – 1)/(1 – x)]

= 1 / (-x/(1 – x))

= (1 – x) / -x

= (x – 1) / x

∴ f {f (x)} = (x – 1) / x

Now, we shall solve for f [f {f (x)}]

f [f {f (x)}] = f [(x-1)/x]

= 1 / [1 – (x-1)/x]

= 1 / [(x – (x-1))/x]

= 1 / [(x – x + 1)/x]

= 1 / (1/x)

∴ f [f {f (x)}] = x

Hence proved.

**Answer
5** :

Given:

f (x) = (x + 1) / (x – 1)

Let us prove that f [f (x)] = x.

f [f (x)] = f [(x+1)/(x-1)]

= [(x+1)/(x-1) + 1] / [(x+1)/(x-1) – 1]

= [[(x+1) + (x-1)]/(x-1)] / [[(x+1) – (x-1)]/(x-1)]

= [(x+1) + (x-1)] / [(x+1) – (x-1)]

= (x+1+x-1)/(x+1-x+1)

= 2x/2

= x

∴ f [f (x)] = x

Hence proved.

If

Find:

(i) f (1/2)

(ii) f (-2)

(iii) f (1)

(iv) f (√3)

(v) f (√-3)

**Answer
6** :

**(i)** f(1/2)

When, 0 ≤ x ≤ 1, f(x) = x

∴f (1/2) = ½

**(ii)** f(-2)

When, x < 0, f(x) = x^{2}

f (–2) = (–2)^{2}

= 4

∴ f(–2) = 4

**(iii)** f(1)

When, x ≥ 1, f (x) = 1/x

f (1) = 1/1

∴ f(1)= 1

**(iv)** f(√3)

We have √3 = 1.732 > 1

When, x ≥ 1, f (x) = 1/x

∴f (√3) = 1/√3

**(v)** f(√-3)

We know √-3 is not a real number and the function f(x) isdefined only when x ∈ R.

∴ f(√-3) does not exist.

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