11类NCERT解决方案-第2章关系和功能–练习2.3

问题1.以下哪个关系是函数?说明原因。如果它是一个函数，请确定其域和范围。

(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)}

解决方案：

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

Here, each element in domain is having unique/distinct image. So, the given relation is a function.

Domain = {2, 5, 8, 11, 14, 17}

Range of the function = {1}

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

Here, each element in domain is having unique/distinct image. So, the given relation is a function.

Domain = {2, 4, 6, 8, 10, 12, 14}

Range of function = {1, 2, 3, 4, 5, 6, 7}

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

 This relation is not a function since an element 1 corresponds to two elements/images i.e, 3 and 5.

Hence, this relation is not a function.

问题2。找到以下实函数的范围和范围：

(i)f(x)= – | x |

(ii)f(x)=√(9 – x ² )

解决方案：

(i) Given,

f(x) = –|x|, x ∈ R

We know that,  |x| = 

x	if x >= 0
-x	if x < 0

Here f(x) = -x = 

-x	x >= 0
x	x < 0

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

It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.

Therefore, the range of f is given by (–∞, 0]. 

(ii) f(x) = √(9 – x²)

As √(9 – x²) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x² ≥ 0.

|x| <=3

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

For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.

Therefore, the range of f(x) is {x: 0 ≤ x ≤ 3} or we can say Range of f = [0, 3].

问题3.函数f由f(x)= 2x – 5定义。记下

(i)f(0)，(ii)f(7)，(iii)f(-3)

解决方案：

Given, function, f(x) = 2x – 5.

(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11

问题4.将t摄氏度转换为华氏温度的函数“ t”由t(C)= 9C / 5 + 32定义。

求(i)t(0)(ii)t(28)(iii)t(–10)(iv)当t(C)= 212时的C值

解决方案 ：

Here in ques , it is given that :

t(C) = 9C / 5 +32 

So, (i) t(0) = 9(0) / 5 + 32 

                = 0 + 32

                = 32

      (ii) t(28) = 9(28) / 5 + 32 

                    Taking LCM and solving ,

                    = ( 252 +160 ) / 5 

                    = 412 / 5

       (iii) t(-10) = 9(-10) / 5 + 32

                       = -18 + 32

                       = 14

        (iv) Here , in this ques we have to find the value of C.

              Given that , t(C) = 212,

              9C / 5 + 32 = 212

              9C / 5 = 180

              9C = 180 X 5

              C = 100

               The value of C is 100.

问题5.找到以下每个功能的范围。

(i)f(x)= 2 – 3x，x∈R，x> 0。

(ii)f(x)= x2 + 2，x是一个实数。

(iii)f(x)= x，x是一个实数。

解决方案：

(i) Given f (x) = 2 – 3x, x ∈ R, x > 0

∵ x > 0 ⇒ -3x < 0 (Multiplying both sides by -3) 

            ⇒ 2 – 3x < 2 + 0 ⇒ f (x) < 2

∴ Hence, The range of f (x) is (-∞ , 2).

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

We know x²≥ 0 ⇒ x²+ 2 ≥ 0 + 2

⇒ x² + 2 > 2 ∴ f (x) ≥ 2

∴ Hence, The range of f (x) is [2, ∞).

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

Let y = f (x) = x ⇒ y = x

∴ Range of f (x) = Domain of f (x)

∴ Hence, Range of f (x) is R. (f(x) takes all real values)