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Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.4
  • Last Updated : 30 Apr, 2021

Question 1. Find f + g, f – g, cf (c ∈ R, c ≠ 0), fg, 1/f and f/g in each of the following:

(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 → R and g(x): R → R

We know, (f + g)(x) = f(x) + g(x) ⇒ (f + g) (x) = x3 + 1 + x + 1 = x3 + x + 2

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

As, (f – g)(x) = f(x) – g(x) ⇒ (f – g)(x) = x3 + 1 – (x + 1) = x3 + 1 – x – 1 = x3 –x



So, (f – g)(x) = x3 – x

Also, (cf)(x) = c × f(x) ⇒ (cf)(x) = c(x3 + 1) = cx3 + c

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

Since, (fg)(x) = f(x)g(x) ⇒ (fg)(x) = (x3 + 1)(x + 1) = (x + 1) (x2 – x + 1) (x + 1) = (x + 1)2 (x2 – x + 1)

So, (fg)(x) = (x + 1)2(x2 – x + 1)

Now, (1/f)(x) = 1/f (x) ⇒ 1/f (x) = 1/(x3 + 1)

Since 1/f(x) is undefined when f(x) = 0 or when x = – 1, 

Hence, 1/f: R – {–1} → R is given by 1/f (x) = 1 / (x3 + 1)  



Lastly, (f/g)(x) = f(x)/g(x) ⇒ (f/g) (x) = (x3 + 1)/(x + 1)

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

As x3 + 1 = (x + 1)(x2 – x + 1), we have (f/g)(x) = \frac{(x+1)(x^2-x+1)}{(x+1)}    = x2 – x + 1

Hence, f/g: R – {–1} → R is given by (f/g)(x) = x2 – x + 1

(ii) f(x) = \sqrt{(x-1)} and g(x) = \sqrt{(x+1)}

Solution:

Since real square root is defined only for non-negative numbers, f(x): [1, ∞) → R+ and g(x): [–1, ∞) → R+

We know, (f + g)(x) = f(x) + g(x) = \sqrt{(x-1)}    \sqrt{(x+1)}

Domain of (f + g) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, f + g: [1, ∞) → R is given by (f + g)(x) = \sqrt{(x-1)}    \sqrt{(x+1)}

As, (f – g)(x) = f(x) – g(x) = \sqrt{(x-1)}    – \sqrt{(x+1)}



Domain of (f – g) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, f – g: [1, ∞) → R is given by (f – g)(x) = \sqrt{(x-1)}    – \sqrt{(x+1)}

Since, (cf)(x) = c × f(x) = c \sqrt{(x-1)}

Domain of (cf) = Domain of f = [1, ∞)

Hence, cf: [1, ∞) → R is given by (cf)(x) = c \sqrt{(x-1)}

Also, (fg)(x) = f(x)g(x) = \sqrt{(x-1)}\sqrt{(x+1)}      \sqrt{x^2 -1}

Domain of (fg) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, fg: [1, ∞) → R is given by (fg)(x) = \sqrt{x^2 -1}

Now, (1/f) (x) = 1/f(x) = \frac{1}{\sqrt{x-1}}

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

Since \frac{1}{\sqrt{x-1}}     is also undefined when x – 1 = 0 or x = 1.

Hence, 1/f: (1, ∞) → R is given by (1/f)(x) = \frac{1}{\sqrt{x-1}}

Lastly, (f/g) (x) = f(x)/g(x) =\frac{\sqrt{x-1}}{\sqrt{x+1}}

Domain of (f/g) = Domain of f ∩ Domain of g = [1, ∞) ∩ [–1, ∞) = [1, ∞)

Hence, f/g: [1, ∞) → R is given by (f/g) (x) =\frac{\sqrt{x-1}}{\sqrt{x+1}}

Question 2. Let f(x) = 2x + 5 and g(x) = x2 + x. Describe and find domain in each:

(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 ∈ R.

So, domain of f = domain of g = R

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

Solution:

As (f – g)(x) = f(x) – g(x) ⇒ (f – g)(x) 

= 2x + 5 – (x2 + x) 

= 2x + 5 – x2 – x 

= 5 + x – x2

Since (f – g)(x) is defined for all real numbers x.

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

(iii) fg

Solution:

We know, (fg)(x) = f(x)g(x) ⇒ (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) ⇒ (f/g)(x) = \frac{2x+5}{x^2+x}

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

x2 + x = 0 ⇒ x(x + 1) = 0 ⇒ x = 0 or x + 1 = 0 ⇒ x = 0 or –1

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

Hence, The domain of f/g = R – {–1, 0}

Question 3. If f(x) be defined on [–2, 2] and is given by \begin{cases}-1 \ \ \ ,-2\lex\le0\\x-1 \ \ \  ,0<x\le2\end{cases} and g(x) = f(|x|) + |f(x)|. Find g(x).

Solution:

f(|x|) = |x|  – 1, where –2 ≤ x ≤ 2

and, |f(x)| = \begin{cases}1, \ \ -2\lex\le0\\-(x-1),\ \ 0\lex\le1\\(x-1), \ \ 1\lex\le2\end{cases}

Thus, g(x) = f(|x|) + |f(x)| = \begin{cases}-x, \ \ -2\lex\le0\\0, \ \ 0\lex\le1\\2, \ \ 1\lex\le2\end{cases}

Question 4. Let f, g be two real functions defined by f(x) = \sqrt{x+1}    and g(x) = \sqrt{9-x^2}    . Then, describe each of the following functions.

(i) f + g

Solution:

Given, f(x) = \sqrt{x+1}    and g(x) = \sqrt{9-x^2}

f(x) takes real values only when x + 1 ≥ 0 x ≥ –1, x ∈ [–1, ∞) ⇒ Domain of f = [–1, ∞)

g(x) takes real values only when 9 – x2 ≥ 0 

⇒ x2 ≤ 9 ⇒ x2 – 32 ≤ 0 ⇒ (x + 3)(x – 3) ≤ 0 

⇒ x ≥ –3 and x ≤ 3 

⇒ Domain of g = [–3, 3]

We know, (f + g)(x) = f(x) + g(x) ⇒ (f + g) (x) = \sqrt{x+1}     + \sqrt{9-x^2}

Domain of f + g = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

Hence, f + g: [–1, 3] → R is given by (f + g)(x) = f(x) + g(x) = \sqrt{x+1}    \sqrt{9-x^2}

(ii) g – f

Solution:

We know, (g – f)(x) = g(x) – f(x) ⇒ (g – f)(x) =\sqrt{9-x^2}    – \sqrt{x+1}

Domain of g – f = Domain of g ∩ Domain of f = [–3, 3] ∩ [–1, ∞) = [–1, 3]

Hence, g – f: [–1, 3] → R is given by (g – f) (x) = g(x) – f(x) = \sqrt{9-x^2}    – \sqrt{x+1}

(iii) fg

Solution:

We know, (fg)(x) = f(x)g(x) ⇒ (fg)(x) = \sqrt{x+1}\sqrt{9-x^2}      

\sqrt{x(9-x^2) + (9-x^2)}

\sqrt{9x-x^3+9-x^2}     

\sqrt{9+9x-x^2-x^3}

Domain of fg = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

Hence, fg: [–1, 3] → R is given by (fg) (x) = f(x) g(x) = \sqrt{x+1}\sqrt{9-x^2}    \sqrt{9+9x-x^2-x^3}

(iv) f/g

Solution:

We know, (f/g) (x) = f(x)/g(x) ⇒ (f/g) (x) = \frac{\sqrt{x+1}}{\sqrt{9-x^2}}        

Domain of f/g = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

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

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

Domain of f/g = [–1, 3] – {–3, 3} = [–1, 3)

Hence, f/g: [–1, 3) → R is given by (f/g) (x) = f(x)/g(x) = \frac{\sqrt{x+1}}{\sqrt{9-x^2}}

(v) g/f

Solution:

We know, (g/f)(x) = g(x)/f(x) ⇒ (g/f)(x) = \frac{\sqrt{9-x^2}}{\sqrt{x+1}}

Domain of g/f = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

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

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

Domain of g/f = [–1, 3] – {–1} = (–1, 3]

Hence, g/f: (–1, 3] → R is given by (g/f) (x) = g(x)/f(x) = \frac{\sqrt{9-x^2}}{\sqrt{x+1}}         

(vi) 2f – √5 g    

Solution:

We know, (2f – √5g)(x) = 2f(x) – √5g(x) 

⇒ (2f – √5g)(x) = 2f (x) – √5g(x)

= 2 \sqrt{x+1}     – \sqrt5\sqrt{9-x^2}        

= 2 \sqrt{x+1}     – \sqrt{45- 5x^2}

Domain of 2f – √5g = Domain of f ∩ Domain of g = [–1, ∞) ∩ [–3, 3] = [–1, 3]

Hence, 2f – √5g: [–1, 3] → R is given by (2f – √5g) (x) = 2f(x) – √5 g(x) = 2 \sqrt{x+1}     \sqrt{45- 5x^2}

(vii) f2 + 7f

Solution:

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

⇒ (f2 + 7f) (x) = f(x).f(x) + 7f(x) 

=\sqrt{x+1}.\sqrt{x+1} + 7 \sqrt{x+1}      

= x + 1 + 7 \sqrt{x+1}

Domain of f2 + 7f is same as domain of f = [–1, ∞)

Hence, f2 + 7f: [–1, ∞) → R is given by (f2 + 7f) (x) = f(x) f(x) + 7f(x) = x + 1 + 7 \sqrt{x+1}

(viii) 5/g  

Solution:

We know, (5/g)(x) = 5/g(x) ⇒ (5/g)(x) = \frac{5}{\sqrt{9-x^2}}

Domain of 5/g = Domain of g = [–3, 3]

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

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

Thus, Domain of 5/g = [–3, 3] – {–3, 3} = (–3, 3)

Hence, 5/g: (–3, 3) → R is given by (5/g)(x) = 5/g(x) = \frac{5}{\sqrt{9-x^2}}

Question 5. If f(x) = loge (1 – x) and g(x) = [x], then determine each of the following functions:

(i) f + g

Solution:

f(x) = loge (1 – x) and g(x) = [x]

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

x < 1, ∴ x ∈ (–∞, 1) ⇒ Domain of f = (–∞, 1)

 g(x) is defined for all real numbers x. ⇒ Domain of g = [x], x ∈ R = R

Thus, (f + g)(x) = f(x) + g(x) = loge(1 – x) + [x]

Domain of f + g = Domain of f ∩ Domain of g = (–∞, 1) ∩ R = (–∞, 1)

Hence, f + g: (–∞, 1) → R is given by (f + g) (x) = loge(1 – x) + [x]

(ii) fg

Solution:

We know, (fg)(x) = f(x)g(x) ⇒ (fg)(x) = loge(1 – x) × [x] = [x]loge(1 – x)

Domain of fg = Domain of f ∩ Domain of g = (–∞, 1) ∩ R = (–∞, 1)

Hence, fg: (–∞, 1) → R is given by (fg) (x) = [x] loge(1 – x).

(iii) f/g

Solution:

We know, (f/g)(x) = f(x)/g(x) ⇒ (f/g)(x) = \frac{log_e(1 - x)}{x}

Domain of f/g = Domain of f ∩ Domain of g = (–∞, 1) ∩ R = (–∞, 1)

However, (f/g) (x) is defined for all real values of x ∈ (–∞, 1), except when [x] = 0.

We have, [x] = 0 when 0 ≤ x < 1 or x ∈ [0, 1)

When 0 ≤ x < 1, (f/g) (x) will be undefined. Domain of f/g = (–∞, 1) – [0, 1) = (–∞, 0)

Hence, f/g: (–∞, 0) → R is given by (f/g)(x) = \frac{log_e(1 - x)}{x}

(iv) g/f

Solution:

We know, (g/f)(x) = g(x)/f(x) ⇒ (g/f)(x) = \frac{x}{log_e (1 - x)}

However, (g/f)(x) is defined for all real values of x ∈ (–∞, 1), except for the case when loge (1 – x) = 0. 

⇒ loge (1 – x) = 0 ⇒ 1 – 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 = (–∞, 1) – {0} = (–∞, 0) ∪ (0, 1)

Hence, g/f: (–∞, 0) ∪ (0, 1) → R is given by (g/f)(x) = \frac{x}{log_e (1 - x)}

(v) (f + g)(–1)

Solution:

(f + g) (x) = loge(1 – x) + [x], x ∈ (–∞, 1)

Substituting x = –1 in the above equation, we get

(f + g)(–1) = loge(1 – (–1)) + [–1] 

= loge(1 + 1) + (–1) 

= loge2 – 1

Hence, (f + g)(–1) = loge2 – 1

(vi) (fg)(0)

Solution:

We have, (fg)(x) = [x]loge(1 – x), x ∈ (–∞, 1)

Substituting x = 0 in the above equation, we get

(fg)(0) = [0]loge(1 – 0) 

= 0 × loge1 = 0

Hence, (fg) (0) = 0

(vii) (f/g)(1/2) 

Solution:

(f/g)(x) =  \frac{log_e(1 - x)}{x}    , x ∈ (–∞, 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) =\frac{x}{log_e (1 - x)}    , x ∈ (–∞, 0) ∪ (0, ∞)

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

(g/f)(1/2) = \frac{x}{log(1 - x)}

\frac{\frac{1}{2}}{log_e(1 -\frac{1}{2})}     

\frac{0.5}{log_e(\frac{1}{2})}     

= \frac{0}{loge(1/2)}     = 0

Hence, (g/f)(1/2) = 0 

Question 6. If f, g, h are real functions defined by f(x) = \sqrt{x+1}    , g(x) = 1/x and h(x) = 2x2 – 3, then find the values of (2f + g – h)(1) and (2f + g – h)(0).

Solution:

Since, f(x) is defined for x + 1 ≥ 0 ⇒ x ≥ – 1 

 ⇒ x ∈ [– 1, ∞] = Domain of f(x)

Now, (2f+g –h)(x) = 2f(x) + g(x) – h(x) 

= 2 \sqrt{x+1}     + 1/x  – (2x2 – 3)

 = 2 \sqrt{1+1}     + 1/1 – 2(1)2 + 3

(2f + g – h)(x) = 2√2 + 2

Since (2f+g –h)(0) does not lie in the domain x ∈ [– 1, ∞] – 0, it does not exist.

Question 7. The function f(x) is defined by: f(x) = \begin{cases}1-x , \ \ x<0\\1 , \ \ x=0\\1+x , \ \ x>0\end{cases}. 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 ⇒ R be defined respectively as f(x) = x + 1 and g(x) = 2x – 3. Find f + g, f – g and f/g.

Solution:

We know, (f + g)(x) = f(x) + g(x) =  (x + 1 + 2x – 3) = 3x – 2.

Now, (f – g)(x) = f(x) – g(x) = (x + 1) – (2x – 3) = 4 – x.

f/g(x) = f(x)/g(x) = \frac{x+1}{2x-3}

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

Solution:

(f + g)(x) = f(x) + g(x) = √x + x 

(f – g)(x) = f(x) – g(x) = √x – x

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

f/g(x) = f(x)/g(x) = √x/x = 1/√x 

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 – g)(x) = f(x) – g(x) = x2 – 2x – 1

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

f/g(x) = f(x)/g(x) = \frac{x^2}{2x+1}

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