# Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.4

Last Updated : 30 Apr, 2021

### (i) f(x) = x3 + 1 and g (x) = x + 1

Solution:

Given, f(x) = x3 + 1 and g(x) = x + 1 and f(x): R â†’ R and g(x): R â†’ R

We know, (f + g)(x) = f(x) + g(x) â‡’ (f + g) (x) = x3 + 1 + x + 1 = x3 + x + 2

So, (f + g)(x) = x3 + x + 2

As, (f â€“ g)(x) = f(x) â€“ g(x) â‡’ (f â€“ g)(x) = x3 + 1 â€“ (x + 1) = x3 + 1 â€“ x â€“ 1 = x3 â€“x

So, (f â€“ g)(x) = x3 â€“ x

Also, (cf)(x) = c Ã— f(x) â‡’ (cf)(x) = c(x3 + 1) = cx3 + c

So, (cf)(x) = cx3 + c

Since, (fg)(x) = f(x)g(x) â‡’ (fg)(x) = (x3 + 1)(x + 1) = (x + 1) (x2 â€“ x + 1) (x + 1) = (x + 1)2 (x2 â€“ x + 1)

So, (fg)(x) = (x + 1)2(x2 â€“ x + 1)

Now, (1/f)(x) = 1/f (x) â‡’ 1/f (x) = 1/(x3 + 1)

Since 1/f(x) is undefined when f(x) = 0 or when x = â€“ 1,

Hence, 1/f: R â€“ {â€“1} â†’ R is given by 1/f (x) = 1 / (x3 + 1)

Lastly, (f/g)(x) = f(x)/g(x) â‡’ (f/g) (x) = (x3 + 1)/(x + 1)

Since (x3 + 1)/(x + 1) is undefined when g(x) = 0 or when x = â€“1.

As x3 + 1 = (x + 1)(x2 â€“ x + 1), we have (f/g)(x) = = x2 â€“ x + 1

Hence, f/g: R â€“ {â€“1} â†’ R is given by (f/g)(x) = x2 â€“ x + 1

### (ii) f(x) =  and g(x) =

Solution:

Since real square root is defined only for non-negative numbers, f(x): [1, âˆž) â†’ R+ and g(x): [â€“1, âˆž) â†’ R+

We know, (f + g)(x) = f(x) + g(x) =

Domain of (f + g) = Domain of f âˆ© Domain of g = [1, âˆž) âˆ© [â€“1, âˆž) = [1, âˆž)

Hence, f + g: [1, âˆž) â†’ R is given by (f + g)(x) =

As, (f â€“ g)(x) = f(x) â€“ g(x) = â€“

Domain of (f â€“ g) = Domain of f âˆ© Domain of g = [1, âˆž) âˆ© [â€“1, âˆž) = [1, âˆž)

Hence, f â€“ g: [1, âˆž) â†’ R is given by (f – g)(x) = â€“

Since, (cf)(x) = c Ã— f(x) = c

Domain of (cf) = Domain of f = [1, âˆž)

Hence, cf: [1, âˆž) â†’ R is given by (cf)(x) = c

Also, (fg)(x) = f(x)g(x) =

Domain of (fg) = Domain of f âˆ© Domain of g = [1, âˆž) âˆ© [â€“1, âˆž) = [1, âˆž)

Hence, fg: [1, âˆž) â†’ R is given by (fg)(x) =

Now, (1/f) (x) = 1/f(x) =

Domain of (1/f) = Domain of f = [1, âˆž)

Since  is also undefined when x â€“ 1 = 0 or x = 1.

Hence, 1/f: (1, âˆž) â†’ R is given by (1/f)(x) =

Lastly, (f/g) (x) = f(x)/g(x) =

Domain of (f/g) = Domain of f âˆ© Domain of g = [1, âˆž) âˆ© [â€“1, âˆž) = [1, âˆž)

Hence, f/g: [1, âˆž) â†’ R is given by (f/g) (x) =

### (i) f + g

Solution:

Given, f(x) = 2x + 5 and g(x) = x2 + x

Both f(x) and g(x) are defined for all x âˆˆ R.

So, domain of f = domain of g = R

We know, (f + g)(x) = f(x) + g(x) â‡’ (f + g)(x) = 2x + 5 + x2 + x

= x2 + 3x + 5

Since (f + g)(x) Is defined for all real numbers x.

Hence, (f + g)(x) = x2 + 3x + 5 and the domain of (f + g) is R.

### (ii) f â€“ g

Solution:

As (f â€“ g)(x) = f(x) â€“ g(x) â‡’ (f â€“ g)(x)

= 2x + 5 â€“ (x2 + x)

= 2x + 5 â€“ x2 â€“ x

= 5 + x â€“ x2

Since (f â€“ g)(x) is defined for all real numbers x.

Hence, (f â€“ g)(x) = 5 + x â€“ x2 and the domain of (f â€“ g) is R.

### (iii) fg

Solution:

We know, (fg)(x) = f(x)g(x) â‡’ (fg)(x) = (2x + 5)(x2 + x)

= 2x(x2 + x) + 5(x2 + x)

= 2x3 + 2x2 + 5x2 + 5x

= 2x3 + 7x2 + 5x

Since (fg)(x) is defined for all real numbers x.

Hence, (fg)(x) = 2x3 + 7x2 + 5x and the domain of fg is R.

### (iv) f/g

Solution:

We know, (f/g) (x) = f(x)/g(x) â‡’ (f/g)(x) =

Clearly (f/g)(x) is defined for all real values of x, except for the case when x2 + x = 0.

x2 + x = 0 â‡’ x(x + 1) = 0 â‡’ x = 0 or x + 1 = 0 â‡’ x = 0 or â€“1

When x = 0 or â€“1, (f/g)(x) will be undefined.

Hence, The domain of f/g = R â€“ {â€“1, 0}

### Question 3. If f(x) be defined on [â€“2, 2] and is given by  and g(x) = f(|x|) + |f(x)|. Find g(x).

Solution:

f(|x|) = |x|  â€“ 1, where â€“2 â‰¤ x â‰¤ 2

and, |f(x)| =

Thus, g(x) = f(|x|) + |f(x)| =

### (i) f + g

Solution:

Given, f(x) = and g(x) =

f(x) takes real values only when x + 1 â‰¥ 0 x â‰¥ â€“1, x âˆˆ [â€“1, âˆž) â‡’ Domain of f = [â€“1, âˆž)

g(x) takes real values only when 9 â€“ x2 â‰¥ 0

â‡’ x2 â‰¤ 9 â‡’ x2 â€“ 32 â‰¤ 0 â‡’ (x + 3)(x â€“ 3) â‰¤ 0

â‡’ x â‰¥ â€“3 and x â‰¤ 3

â‡’ Domain of g = [â€“3, 3]

We know, (f + g)(x) = f(x) + g(x) â‡’ (f + g) (x) =  +

Domain of f + g = Domain of f âˆ© Domain of g = [â€“1, âˆž) âˆ© [â€“3, 3] = [â€“1, 3]

Hence, f + g: [â€“1, 3] â†’ R is given by (f + g)(x) = f(x) + g(x) =

### (ii) g â€“ f

Solution:

We know, (g â€“ f)(x) = g(x) â€“ f(x) â‡’ (g â€“ f)(x) =â€“

Domain of g â€“ f = Domain of g âˆ© Domain of f = [â€“3, 3] âˆ© [â€“1, âˆž) = [â€“1, 3]

Hence, g â€“ f: [â€“1, 3] â†’ R is given by (g â€“ f) (x) = g(x) â€“ f(x) = â€“

### (iii) fg

Solution:

We know, (fg)(x) = f(x)g(x) â‡’ (fg)(x) =

Domain of fg = Domain of f âˆ© Domain of g = [â€“1, âˆž) âˆ© [â€“3, 3] = [â€“1, 3]

Hence, fg: [â€“1, 3] â†’ R is given by (fg) (x) = f(x) g(x) =

### (iv) f/g

Solution:

We know, (f/g) (x) = f(x)/g(x) â‡’ (f/g) (x) =

Domain of f/g = Domain of f âˆ© Domain of g = [â€“1, âˆž) âˆ© [â€“3, 3] = [â€“1, 3]

However, (f/g) (x) is defined for all real values of x âˆˆ [â€“1, 3], except for the case when 9 â€“ x2 = 0 or x = Â± 3

When x = Â±3, (f/g) (x) will be undefined as the division result will be indeterminate.

Domain of f/g = [â€“1, 3] â€“ {â€“3, 3} = [â€“1, 3)

Hence, f/g: [â€“1, 3) â†’ R is given by (f/g) (x) = f(x)/g(x) =

### (v) g/f

Solution:

We know, (g/f)(x) = g(x)/f(x) â‡’ (g/f)(x) =

Domain of g/f = Domain of f âˆ© Domain of g = [â€“1, âˆž) âˆ© [â€“3, 3] = [â€“1, 3]

However, (g/f) (x) is defined for all real values of x âˆˆ [â€“1, 3], except for the case when x + 1 = 0 or x = â€“1

When x = â€“1, (g/f) (x) will be undefined as the division result will be indeterminate.

Domain of g/f = [â€“1, 3] â€“ {â€“1} = (â€“1, 3]

Hence, g/f: (â€“1, 3] â†’ R is given by (g/f) (x) = g(x)/f(x) =

### (vi) 2f â€“ âˆš5 g

Solution:

We know, (2f â€“ âˆš5g)(x) = 2f(x) â€“ âˆš5g(x)

â‡’ (2f â€“ âˆš5g)(x) = 2f (x) â€“ âˆš5g(x)

= 2  â€“

= 2  â€“

Domain of 2f â€“ âˆš5g = Domain of f âˆ© Domain of g = [â€“1, âˆž) âˆ© [â€“3, 3] = [â€“1, 3]

Hence, 2f â€“ âˆš5g: [â€“1, 3] â†’ R is given by (2f â€“ âˆš5g) (x) = 2f(x) â€“ âˆš5 g(x) = 2 â€“

### (vii) f2 + 7f

Solution:

We know, (f2 + 7f)(x) = f2(x) + (7f)(x)

â‡’ (f2 + 7f) (x) = f(x).f(x) + 7f(x)

= + 7

= x + 1 + 7

Domain of f2 + 7f is same as domain of f = [â€“1, âˆž)

Hence, f2 + 7f: [â€“1, âˆž) â†’ R is given by (f2 + 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7

### (viii) 5/g

Solution:

We know, (5/g)(x) = 5/g(x) â‡’ (5/g)(x) =

Domain of 5/g = Domain of g = [â€“3, 3]

However, (5/g)(x) is defined for all real values of x âˆˆ [â€“3, 3], except for the case when 9 â€“ x2 = 0 or x = Â± 3

When x = Â±3, (5/g)(x) will be undefined.

Thus, Domain of 5/g = [â€“3, 3] â€“ {â€“3, 3} = (â€“3, 3)

Hence, 5/g: (â€“3, 3) â†’ R is given by (5/g)(x) = 5/g(x) =

### (i) f + g

Solution:

f(x) = loge (1 â€“ x) and g(x) = [x]

We know, f(x) takes real values only when 1 â€“ x > 0 or when 1 > x

x < 1, âˆ´ x âˆˆ (â€“âˆž, 1) â‡’ Domain of f = (â€“âˆž, 1)

g(x) is defined for all real numbers x. â‡’ Domain of g = [x], x âˆˆ R = R

Thus, (f + g)(x) = f(x) + g(x) = loge(1 â€“ x) + [x]

Domain of f + g = Domain of f âˆ© Domain of g = (â€“âˆž, 1) âˆ© R = (â€“âˆž, 1)

Hence, f + g: (â€“âˆž, 1) â†’ R is given by (f + g) (x) = loge(1 â€“ x) + [x]

### (ii) fg

Solution:

We know, (fg)(x) = f(x)g(x) â‡’ (fg)(x) = loge(1 â€“ x) Ã— [x] = [x]loge(1 â€“ x)

Domain of fg = Domain of f âˆ© Domain of g = (â€“âˆž, 1) âˆ© R = (â€“âˆž, 1)

Hence, fg: (â€“âˆž, 1) â†’ R is given by (fg) (x) = [x] loge(1 â€“ x).

### (iii) f/g

Solution:

We know, (f/g)(x) = f(x)/g(x) â‡’ (f/g)(x) =

Domain of f/g = Domain of f âˆ© Domain of g = (â€“âˆž, 1) âˆ© R = (â€“âˆž, 1)

However, (f/g) (x) is defined for all real values of x âˆˆ (â€“âˆž, 1), except when [x] = 0.

We have, [x] = 0 when 0 â‰¤ x < 1 or x âˆˆ [0, 1)

When 0 â‰¤ x < 1, (f/g) (x) will be undefined. Domain of f/g = (â€“âˆž, 1) â€“ [0, 1) = (â€“âˆž, 0)

Hence, f/g: (â€“âˆž, 0) â†’ R is given by (f/g)(x) =

### (iv) g/f

Solution:

We know, (g/f)(x) = g(x)/f(x) â‡’ (g/f)(x) =

However, (g/f)(x) is defined for all real values of x âˆˆ (â€“âˆž, 1), except for the case when loge (1 â€“ x) = 0.

â‡’ loge (1 â€“ x) = 0 â‡’ 1 â€“ x = 1

or x = 0

When x = 0, (g/f) (x) will be undefined as the division result will be indeterminate.

Domain of g/f = (â€“âˆž, 1) â€“ {0} = (â€“âˆž, 0) âˆª (0, 1)

Hence, g/f: (â€“âˆž, 0) âˆª (0, 1) â†’ R is given by (g/f)(x) =

### (v) (f + g)(â€“1)

Solution:

(f + g) (x) = loge(1 â€“ x) + [x], x âˆˆ (â€“âˆž, 1)

Substituting x = â€“1 in the above equation, we get

(f + g)(â€“1) = loge(1 â€“ (â€“1)) + [â€“1]

= loge(1 + 1) + (â€“1)

= loge2 â€“ 1

Hence, (f + g)(â€“1) = loge2 â€“ 1

### (vi) (fg)(0)

Solution:

We have, (fg)(x) = [x]loge(1 â€“ x), x âˆˆ (â€“âˆž, 1)

Substituting x = 0 in the above equation, we get

(fg)(0) = [0]loge(1 â€“ 0)

= 0 Ã— loge1 = 0

Hence, (fg) (0) = 0

### (vii) (f/g)(1/2)

Solution:

(f/g)(x) =  , x âˆˆ (â€“âˆž, 0)

However, 1/2 is not in the domain of f/g.

Hence, (f/g)(1/2) does not exist.

### (viii) (g/f)(1/2)

Solution:

We have, (g/f)(x) =, x âˆˆ (â€“âˆž, 0) âˆª (0, âˆž)

Substituting x=1/2 in the above equation, we get

(g/f)(1/2) =

= = 0

Hence, (g/f)(1/2) = 0

### Question 6. If f, g, h are real functions defined by f(x) = , g(x) = 1/x and h(x) = 2x2 â€“ 3, then find the values of (2f + g â€“ h)(1) and (2f + g â€“ h)(0).

Solution:

Since, f(x) is defined for x + 1 â‰¥ 0 â‡’ x â‰¥ â€“ 1

â‡’ x âˆˆ [â€“ 1, âˆž] = Domain of f(x)

Now, (2f+g â€“h)(x) = 2f(x) + g(x) â€“ h(x)

= 2 + 1/x  â€“ (2x2 â€“ 3)

= 2 + 1/1 â€“ 2(1)2 + 3

(2f + g â€“ h)(x) = 2âˆš2 + 2

Since (2f+g â€“h)(0) does not lie in the domain x âˆˆ [â€“ 1, âˆž] â€“ 0, it does not exist.

### Question 7. The function f(x) is defined by: f(x) = . Draw the graph of f(x).

Solution:

The graph of f(x) for x < 0 lies to the left of origin.

The graph of f(x) for x > 0 lies to the right of origin.

The graph of f(x) for x = 0 is represented by the point (0,1).

### Question 8. Let f, g: R â‡’ R be defined respectively as f(x) = x + 1 and g(x) = 2x â€“ 3. Find f + g, f – g and f/g.

Solution:

We know, (f + g)(x) = f(x) + g(x) =  (x + 1 + 2x â€“ 3) = 3x â€“ 2.

Now, (f â€“ g)(x) = f(x) â€“ g(x) = (x + 1) â€“ (2x â€“ 3) = 4 â€“ x.

f/g(x) = f(x)/g(x) =

### Question 9. Let f and g be defined respectively as f(x) = âˆšx and g(x) = x. Find f + g, f – g, fg, and f/g.

Solution:

(f + g)(x) = f(x) + g(x) = âˆšx + x

(f â€“ g)(x) = f(x) â€“ g(x) = âˆšx â€“ x

(fg)(x) = f(x)g(x) = âˆšx.x = x1/2 + 1 = x3/2

f/g(x) = f(x)/g(x) = âˆšx/x = 1/âˆšx

### Question 10. Let f(x) = x2 and g(x) = 2x + 1 be two real functions. Find f + g, f – g, fg, and f/g.

Solution:

(f + g)(x) = f(x) + g(x) = x2 + 2x + 1 = (x + 1)2

(f â€“ g)(x) = f(x) â€“ g(x) = x2 â€“ 2x â€“ 1

(fg)(x) = f(x)g(x) = x2(2x + 1) = 2x3 + x2

f/g(x) = f(x)/g(x) =

Previous
Next