NCERT Solutions Class 11 Mathematics Relation Functions Exercise 2-Misc

Question 1:
The relation f is defined by
f(x) =      x², 0 ≤ x ≤ 3
              3x, 3 ≤ x ≤ 10
The relation g is defined by
g(x) =     x², 0 ≤ x ≤ 2
              3x, 2 ≤ x ≤ 10
Show that f is a function and g is not a function.

Answer:
The relation f is defined as
f(x) =      x², 0 ≤ x ≤ 3
              3x, 3 ≤ x ≤ 10
It is observed that for
0 ≤ x < 3, f(x) = x²
3 ≤ x ≤ 10, f(x) = 3x
Also, at x = 3, f(x) = 3² = 9 or f(x) = 3 * 3 = 9
i.e., at x = 3, f(x) = 9
Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique. Thus, the given relation is a function.
The relation g is defined as
g(x) =     x², 0 ≤ x ≤ 2
              3x, 2 ≤ x ≤ 10
It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 * 2 = 6
Hence, element 2 of the domain of the relation g corresponds to two different images
i.e. 4 and 6.
Hence, this relation is not a function.

Question 2:
If f(x) = x², find. {f(1.1) – f(1)}/(1.1 – 1)

Answer:
Given, f(x) = x²
Now, {f(1.1) – f(1)}/(1.1 – 1) =  {(1.1)² – (1)²}/(1.1 – 1)
                                                   = (1.21 – 1)/0.1
                                                   = 0.21/0.1
                                                   = 2.1   

Question 3:
Find the domain of the function f(x) = (x² + 2x + 1)/(x² – 8x + 12)

Answer:
Given function is
      f(x) = (x² + 2x + 1)/(x² – 8x + 12)
⇒ f(x) = (x² + 2x + 1)/{(x – 6)(x – 2)}
It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.
Hence, the domain of f is R – {2, 6}.

Question 4:
Find the domain and the range of the real function f defined by
f(x) = √(x − 1)
 

Answer:
The given real function is f(x) = √(x − 1)
It can be seen that √(x − 1) is defined for x ≥ 1
Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e.,
the domain of f = [1, ∞)
As x ≥ 1
⇒ (x – 1) ≥ 0
⇒ √(x − 1) ≥ 0
Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e.,
 the range of f = [0, ∞)

Question 5:
Find the domain and the range of the real function f defined by
f (x) = |x – 1|

Answer:
The given real function is f (x) = |x – 1|.
It is clear that |x – 1| is defined for all real numbers.
So,  Domain of f = R
Also, for x ∈ R, |x – 1| assumes all real numbers.
Hence, the range of f is the set of all non-negative real numbers.

Question 6:
Let f = {(x, x²/(1 + x²)): x ∈ R} be a function from R into R. Determine the range of f.

Answer:
Given, f = {(x, x²/(1 + x²)): x ∈ R}
               = {(0, 0), (±0.5, 1/5), (±1, 1/2), (±1.5, 9/13), (±2, 4/5), (3, 9/10), (4, 16/17), ………} 
The range of f is the set of all second elements. It can be observed that all these elements are
greater than or equal to 0 but less than 1. [Denominator is greater numerator]
Thus, range of f = [0, 1)

Question 7:
Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.

Answer:
f, g: R → R is defined as f(x) = x + 1, g(x) = 2x – 3
(f + g) (x) = f(x) + g(x)
                 = (x + 1) + (2x – 3)
                 = 3x – 2
So, (f + g) (x) = 3x – 2
(f – g) (x) = f(x) – g(x)
                 = (x + 1) – (2x – 3)
                 = x + 1 – 2x + 3
                 = -x + 4
So, (f – g) (x) = –x + 4
(f/g)(x) = f(x)/g(x), g(x) ≠ 0
(f/g)(x) = (x + 1)/(2x + 3), 2x – 3 ≠ 0 or 2x ≠ 3
(f/g)(x) = (x + 1)/(2x + 3), x ≠ 2/3
 

Question 8:
Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Answer:
f = {(1, 1), (2, 3), (0, –1), (–1, –3)} and f(x) = ax + b
Now, (1, 1) ∈ f
⇒ f(1) = 1
⇒ a * 1 + b = 1
⇒ a + b = 1
Again, (0, –1) ∈ f
⇒ f(0) = –1
⇒ a * 0 + b = –1
⇒ b = –1
On substituting b = –1 in a + b = 1,
We obtain
     a + (–1) = 1
⇒ a = 1 + 1 = 2
Thus, the respective values of a and b are 2 and –1