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# RD Chapter 3- Functions Ex-3.2 Interview Questions Answers

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Question 1 : If f (x) = x2 – 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).

Given:

f(x) = x2 – 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) + 12 – 6x – 3 +4

= 4x2 + 4x + 1 – 6x+ 1

= 4x2 – 2x + 2

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

x2 – 3x + 4 = 4x2 – 2x + 2

4x2 – 2x + 2 – x2 + 3x – 4 = 0

3x2 + x – 2 = 0

3x2 + 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.

Question 2 : If f (x) = (x – a)2 (x – b)2, find f (a + b).

F (x) = (x – a)2(x – b)2

Let us find f (a + b).

We have,

f (a + b) = (a + b – a)(a+ b – b)2

f (a + b) = (b)2 (a)2

f(a + b) = a2b2

Question 3 : If y = f (x) = (ax – b) / (bx – a), show that x = f (y).

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.

Question 4 : If f (x) = 1 / (1 – x), show that f [f {f (x)}] = x.

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.

Question 5 : If f (x) = (x + 1) / (x – 1), show that f [f (x)] = x.

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.

Question 6 :
If

Find:
(i) f (1/2)
(ii) f (-2)
(iii) f (1)
(iv) f (√3)
(v) f (√-3)

(i) f(1/2)

When, 0 ≤ x ≤ 1, f(x) = x

f (1/2) = ½

(ii) f(-2)

When, x < 0, f(x) = x2

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.

krishan