# Class 11 RD Sharma Solutions – Chapter 3 Functions – Exercise 3.3

Last Updated : 30 Apr, 2021

### (i) f (x) = 1/x

Solution:

We are given, f (x) = 1/x.

Here, f (x) is defined for all real values of x, except for the case when x = 0.

Therefore, domain of f = R â€“ {0}

### (ii) f (x) = 1/(xâˆ’7)

Solution:

We are given, f (x) = 1/(xâˆ’7).

Here, f (x) is defined for all real values of x, except for the case when x â€“ 7 = 0 or x = 7.

Therefore, domain of f = R â€“ {7}

### (iii) f (x) = (3xâˆ’2)/(x+1)

Solution:

We are given, f (x) = (3xâˆ’2)/(x+1).

Here, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = â€“1.

Therefore, domain of f = R â€“ {â€“1}

### (iv) f (x) = (2x+1)/(x2âˆ’9)

Solution:

We are given, f (x) = (2x+1)/(x2âˆ’9).

Here, f (x) is defined for all real values of x, except for the case when x2 â€“ 9 = 0.

=> x2 â€“ 9 = 0

=> (x + 3)(x â€“ 3) = 0

=> x + 3 = 0 or x â€“ 3 = 0

=> x = Â± 3

Therefore, domain of f = R â€“ {â€“3, 3}

### (v) f (x) = (x2+2x+1)/(x2â€“8x+12)

Solution:

We are given, f (x) = (x2+2x+1)/(x2â€“8x+12).

Here, f(x) is defined for all real values of x, except for the case when x2 â€“ 8x + 12 = 0.

=> x2 â€“ 8x + 12 = 0

=> x2 â€“ 2x â€“ 6x + 12 = 0

=> x(x â€“ 2) â€“ 6(x â€“ 2) = 0

=> (x â€“ 2)(x â€“ 6) = 0

=> x â€“ 2 = 0 or x â€“ 6 = 0

=> x = 2 or 6

Therefore, domain of f = R â€“ {2, 6}

### Question 2. Find the domain of each of the following real-valued functions of a real variable:

(i) f (x) = âˆš(xâ€“2)

Solution:

We are given, f (x) = âˆš(xâ€“2).

Here, f (x) takes real values only when x â€“ 2 â‰¥ 0 as the square of a real number cannot be negative.

=> x â€“ 2 â‰¥ 0

=> x â‰¥ 2

=> x âˆˆ [2, âˆž)

Therefore, domain of f = [2, âˆž)

### (ii) f (x) = 1/(âˆš(x2â€“1))

Solution:

We are given, f (x) = 1/(âˆš(x2â€“1)).

Here, f (x) takes real values only when x2 â€“ 1 > 0 as the square of a real number cannot be negative and the denominator x2 â€“ 1 cannot be zero.

=> x2 â€“ 1 > 0

=> (x + 1) (x â€“ 1) > 0

=> x < â€“1 or x > 1

=> x âˆˆ (â€“âˆž, â€“1) âˆª (1, âˆž)

Therefore, domain of f = (â€“âˆž, â€“1) âˆª (1, âˆž)

### (iii) f (x) = âˆš(9â€“x2)

Solution:

We are given, f (x) = âˆš(9â€“x2).

Here, f (x) takes real values only when 9 â€“ x2 â‰¥ 0 as the square of a real number cannot be negative.

=> 9 â€“ x2 â‰¥ 0

=> 9 â‰¥ x2

=> x2 â‰¤ 9

=> x2 â€“ 9 â‰¤ 0

=> (x + 3)(x â€“ 3) â‰¤ 0

=> x â‰¥ â€“3 and x â‰¤ 3

=> x âˆˆ [â€“3, 3]

Therefore domain of f = [â€“3, 3]

### (iv) f (x) = âˆš[(xâ€“2)/(3â€“x)]

Solution:

We are given, f (x) = âˆš[(xâ€“2)/(3â€“x)].

Here, f (x) takes real values only when x â€“ 2 and 3 â€“ x are both positive and negative.

Case 1. x â€“ 2 â‰¥ 0 and 3 â€“ x â‰¥ 0

=> x â‰¥ 2 and x â‰¤ 3

Therefore, x âˆˆ [2, 3]

Case 2. x â€“ 2 â‰¤ 0 and 3 â€“ x â‰¤ 0.

=> x â‰¤ 2 and x â‰¥ 3

This case is not possible as the intersection of these sets is null set.

Hence, x âˆˆ [2, 3] â€“ {3}

=> x âˆˆ [2, 3)

Therefore, domain of f = [2, 3)

### (i) f (x) = (ax+b)/(bxâ€“a)

Solution:

We are given, f (x) = (ax+b)/(bxâ€“a).

Here, f(x) is defined for all real values of x, except for the case when bx â€“ a = 0 or x = a/b.

So, domain of f = R â€“ (a/b)

Let f (x) = y. So, (ax+b)/(bxâ€“a) = y.

=> ax + b = y(bx â€“ a)

=> ax + b = bxy â€“ ay

=> ax â€“ bxy = â€“ay â€“ b

=> x(a â€“ by) = â€“(ay + b)

=> x = â€“ (ay+b)/(aâ€“by)

When a â€“ by = 0 or y = a/b. Hence, f(x) cannot take the value a/b.

Therefore, range of f = R â€“ (a/b)

### (ii) f (x) = (axâ€“b)/(cxâ€“d)

Solution:

We are given, f (x) = (axâ€“b)/(cxâ€“d).

Here, f(x) is defined for all real values of x, except for the case when cx â€“ d = 0 or x = d/c.

So, domain of f = R â€“ (d/c)

Let f (x) = y. So, (axâ€“b)/(cxâ€“d) = y

=> ax â€“ b = y(cx â€“ d)

=> ax â€“ b = cxy â€“ dy

=> ax â€“ cxy = b â€“ dy

=> x(a â€“ cy) = b â€“ dy

=> x = (bâ€“dy)/(aâ€“cy)

When a â€“ cy = 0 or y = a/c. Hence, f(x) cannot take the value a/c.

Therefore, range of f = R â€“ (a/c)

### (iii) f (x) = âˆš(xâ€“1)

Solution:

We are given, f (x) = âˆš(xâ€“1).

Here, f(x) takes real values only when x â€“ 1 â‰¥ 0.

=> x â‰¥ 1

=> x âˆˆ [1, âˆž)

So, domain of f = [1, âˆž)

When x â‰¥ 1, we have x â€“ 1 â‰¥ 0. So, âˆš(xâ€“1) â‰¥ 0.

=> f (x) â‰¥ 0

=> f(x) âˆˆ [0, âˆž)

Therefore, range of f = [0, âˆž)

### (iv) f (x) = âˆš(xâ€“3)

Solution:

We are given, f (x) = âˆš(xâ€“3).

Here, f (x) takes real values only when x â€“ 3 â‰¥ 0.

=> x â‰¥ 3

=> x âˆˆ [3, âˆž)

So, domain of f = [3, âˆž)

When x â‰¥ 3, we have x â€“ 3 â‰¥ 0. Hence, âˆš(xâ€“3) â‰¥ 0

=> f (x) â‰¥ 0

=> f(x) âˆˆ [0, âˆž)

Therefore, range of f = [0, âˆž)

### (v) f (x) = (xâ€“2)/(2â€“x)

Solution:

We are given, f (x) = (xâ€“2)/(2â€“x).

Here, f(x) is defined for all real values of x, except for the case when 2 â€“ x = 0 or x = 2.

So, domain of f = R â€“ {2}

And also, f (x) = â€“(2â€“x)/(2â€“x) = â€“1

Hence, when x â‰  2, f(x) = â€“1

Therefore, range of f = {â€“1}

### (vi) f (x) = |xâ€“1|

Solution:

We are given, f (x) = |xâ€“1|.

Clearly, f(x) is defined for all real numbers x.

So, domain of f = R

Let f (x) = y. So, |xâ€“1| = y.

Therefore, y can take only the positive values. So, y â‰¥ 0.

Therefore, range of f = (0, âˆž]

### (vii) f (x) = â€“|x|

Solution:

We are given, f (x) = â€“|x|.

Clearly, f(x) is defined for all real numbers x.

So, domain of f = R

Let f (x) = y. So, y = â€“|x|.

Therefore, y can take only the negative values. So, y â‰¤ 0.

Therefore, range of f = (â€“âˆž, 0]

### (viii) f (x) = âˆš(9â€“x2)

Solution:

We are given, f (x) = âˆš(9â€“x2)

Here, f(x) takes real values only when 9 â€“ x2 â‰¥ 0.

=> 9 â‰¥ x2

=> x2 â‰¤ 9

=> x2 â€“ 9 â‰¤ 0

=> (x + 3)(x â€“ 3) â‰¤ 0

=> x â‰¥ â€“3 and x â‰¤ 3

=> x âˆˆ [â€“3, 3]

So, domain of f = [â€“3, 3]

When, x âˆˆ [â€“3, 3], we have 0 â‰¤ 9 â€“ x2 â‰¤ 9.

=> 0 â‰¤ âˆš(9â€“x2) â‰¤ 3

=> 0 â‰¤ f (x) â‰¤ 3

=> f (x) âˆˆ [0, 3]

Therefore, range of f = [0, 3]

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