Class 11 Mathematics | NCERT Exercise 2.3 | Solved Answers and Questions #eduvictors #class11Maths

Class 11 Mathematics | NCERT Exercise 2.3 | Solved Answers and Questions

Q1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

(iii) {(1,3), (1,5), (2,5)}.

Answer: 

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their

unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.

Q2. Find the domain and range of the following real functions:

(i) f(x) = –|x|

(ii) f(x) = $\sqrt{9 − x^2}$.

Answer: We know that f(x) = -|x| x ∈ R

i.e. |x| = {x, x ≥ 0; x, x < 0}

∴ -|x| = {-x, x ≥ 0; -x, x < 0}

Since f(x) is defined for x ∈ R, the domain of f is R.

It can be observed that the range of f(x) = –|x| is all real numbers except positive real numbers.

∴The range of f is (–∞ , 0].

(ii) f(x) = $\sqrt{9 − x^2} = \sqrt{3^2 − x^2} = \sqrt{(3 − x)(3 + x)}$.

Thus f(x) is defined for all real numbers that are greater than or equal to –3 and less than or equal

to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].

For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

Q3. A function f is defined by f(x) = 2x –5. Write down the values of

(i) f(0), (ii) f(7), (iii) f(–3).

Answer: f(x) = 2x –5

(i) f(0) = 2(0) - 5 = -5 

(ii) f(7) = 2(7) - 5 = 14 - 5 = 9

(iii) f(–3) = 2(-3) - 5 = -11

Q4. The function ‘t’ maps temperature in degree Celsius into temperature in

degree Fahrenheit is defined by t(C) = $\frac{9C}{5} + 32$

Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212

Answer: (i) t(0) = $\frac{9×0}{5} + 32$ = 32

(ii) t(28) = $\frac{9×28}{5} + 32 = \frac{252+160}{5} = \frac{412}{5}$ 

(iii) t(–10) = $\frac{9×-10}{5} + 32 = 9×(-2) + 32= 14$

(iv) The value of C, when t(C) = 212

212 =  $\frac{9C}{5} + 32$

212 - 32 = $\frac{9C}{5}$

$\frac{9C}{5}$ = 180

C = $\frac{180×5}{9}$ = 100

Q5. Find the range of each of the following functions.

(i) f(x) = 2 – 3x, x ∈ R, x > 0.

(ii) f(x) = x² + 2, x is a real number.

(iii) f (x) = x, x is a real number

Answer: (i) f(x) = 2 – 3x, x ∈ R, x > 0.

Here, f(x) is a linear function with a negative slope -3.

Since x > 0, As x increases, f(x) decreases without bound.

When x approaches 0+ (say 0.00001), f(x) approaches 2 (for x=0.00001, f(x) = 1.9997 ≈ 2) 

When x approaches +∞, f(x) approaches -∞

Thus Range = (-∞, 2)

(ii) f(x) = x² + 2, x is a real number.

Here, f(x) is a quadratic function. Thus, x² will always be positive.

If x = 0, f(x) = 2

As x approaches +∞ or −∞, x² grows without bound i.e. +∞.

Thus range of f = [2, +∞) 

Alternately, x² ≥ 0

x² + 2 ≥ 0 + 2

f(x) ≥ 2

Thus Range of f(x) is [2, +∞)

(iii) f(x) = x, x is a real number

It is clear that the range of f is the set of all real numbers.

∴ Range of f = R