GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Gujarat Board Statistics Class 12 GSEB Solutions Part 2 Chapter 4 Limit Ex 4 Textbook Exercise Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Section A

Answer the following questions by selecting a correct option from the given options:

Question 1.
What is the modulus form of 0.3 neighbourhood of 3?
(a)|x – 0.3|<3
(b) |x- 31 < 0.3
(c) |x + 3|< 0.3 (d) |x – 3| > 0.3
Answer:
(b) |x- 31 < 0.3

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 2.
What is the interval form of 0.02 neighbourhood of -2?
(a) (1.98, 2.02)
(b) (- 1.98, 2.02)
(c) (- 2.02, -1.98)
(d) (- 2.02, 1.98)
Answer:
(c) (- 2.02, – 1.98)

Question 3.
What is the interval form of |x – 5| < 0.25?
(a) (4.75, 5.25)
(b) (- 4.75, + 5.25)
(c) (- 5.25, – 4.75)
(d) (- 5.25, 4.75)
Answer:
(a) (4.75, 5.25)

Question 4.
What is the interval form of |2x + 1| < \(\frac{1}{5}\) ?
(a) \(\left(-\frac{6}{5},-\frac{4}{5}\right)\)
(b) \(\left(-\frac{6}{10},-\frac{4}{10}\right)\)
(c) \(\left(\frac{4}{10}, \frac{6}{10}\right)\)
(d) \(\left(-\frac{6}{10}, \frac{4}{10}\right)\)
Answer:
(b) \(\left(-\frac{6}{10},-\frac{4}{10}\right)\)

Question 5.
What is the modulus form of N (5, 0.02) ?
(a) |x + 5|< 0.02
(b) |x – 0.02| < 5 (c) |x – 5| > 0.02
(d) |x – 5| < 0.02
Answer:
(d) |x – 5| < 0.02

Question 6.
If modulus form of N (a, 0.07) is |x – 10| < k, then what will be the value of k?
(a) a
(b) 0.7
(c) 0.07
(d) 9.93
Answer:
(c) 0.07

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 7.
What is the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 1 ?
(a) 9
(b) 10
(c) \(\frac{4}{3}\)
(d) 8
Answer:
(d) 8

Question 8.
What is the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 2 ?
(a) 5
(b) 25
(c) \(\frac{7}{4}\)
(d) 7
Answer:
(a) 5

Question 9.
What is the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 3?
(a) 10
(b) – 2
(c) 8
(d) Indeterminate
Answer:
(a) 10

Question 10.
What is the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 4?
(a) 192
(b) 324
(c) 36
(d) 108
Answer:
(d) 108

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 11.
What is the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 5?
(a) – 5
(b) 5
(c) 4
(d) – 4
Answer:
(b) 5

Question 12.
If y = 10 – 3x and x → – 3, then y tends to which value ?
(a) 1
(b) 9
(c) 19
(d) 7
Answer:
(c) 19

Section B

Answer the following questions in one sentence:

Question 1.
Express 0.09 neighbourhood of 0 in interval form.
Answer:
0.09 neighbourhood of 0 in interval form is expressed by (0 – 0.09, 0 + 0.09) = (- 0.09, 0.09).

Question 2.
Express 0.001 neighbourhood of – 5 in modulus form.
Answer:
0.001 neighbourhood of – 5 in modulus form is expressed by
|x – (- 5) | < 0.001 = |x + 5] < 0.001.

Question 3.
Express |x – 10| < \(\frac{1}{10}\) in neighbourhood form.
Answer:
|x – 10| < \(\frac{1}{10}\) in neighbourhood form is expressed by N (10, \(\frac{1}{10}\)).

Question 4.
Express |2x| < \(\frac{1}{2}\) in interval form.
Answer:
|2x| < \(\frac{1}{2}\) in interval form is expressed by \(\left(-\frac{1}{4}, \frac{1}{4}\right)\)

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 5.
Express N (50, 0.8) in modulus form.
Answer:
N (50, 0.8) in modulus form is expressed by |x – 50| < 0.8.

Question 6.
If N(a, 0.2) = |x – 7|< b, then find the value of a.
Answer:
N (a, 0.2) = | x – 7| < b
If N (a, 0.2) = |x – 7| < b, then a = 7.

Question 7.
If |x + 4| < 0.04 = (k, – 3.96), then find the value of fc.
Answer:
If |x + 4| < 0.04 = (k, – 3.96), then
a – δ = k; where a = – 4, δ = 0.04
∴ k = – 4 – 0.04 = – 4.04

Question 8.
Find the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 6.
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 6 = 3 (5) + 5 = 20

Question 9.
Find the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 7.
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 8

Question 10.
Find the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 9?
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 10

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 11.
Find the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 11
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 12

Question 12.
Find the value of GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 13 where m is an odd number.
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 14

Question 13.
If GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 15 then find the value
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 15
= 6 = 4 (- 1) + k = 6
∴ k = 6 + 4 = 10

Question 14.
If GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 16, then find the value of k.
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 17

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Section C

Answer the following questions as required:

Question 1.
Define an open interval.
Answer:
If a, b ∈ R and a < b, then the set of all real numbers between a and b not including a and b is called an open interval. It is denoted by (a, b).
Thus, (a, b) = {x|a < x < b, x ∈ R}.

Question 2.
Define the 5 neighbourhood of a.
Answer:
If a ∈ R and δ is non-negative real number, then the open interval (a – δ, a + δ) is called δ neighbourhood of a. It is denoted by N (a, δ). Thus,
N (a, δ) = {x|(a – δ) < x < (a + δ), x ∈ R}
= {x| |x – a| < δ. x ∈ R}

Question 3.
Define the punctured δ neighbourhood of a.
Answer:
If a ∈ R and δ is a non-negative real number, then the open interval (a – δ, a + δ) – {a} is called punctured δ neighbourhood of a. It is denoted by N* (a, δ). Thus,
N* (a, δ) = N(a, δ) – {a}
= {x | (a – δ) < x < (a + δ); x ≠ a, x ∈ R}
= {x | |x – a| < δ; x ≠ a, x ∈ R}

Question 4.
Express the interval form (- 0.5, 0.5) in modulus form.
Answer:
Interval form: {- 0.5, 0.5)
∴ a – δ = – 0.5, a + δ = 0.5
By adding, 2a = 0 ∴ a = 0
Putting a = 0 in a + δ, δ = 0.5
In modulus form: |x – a| < δ
∴ For (- 0.5, 0.5), modulus form is expressed by |x – 0| < 0.5 = |x| < 0.5

Question 5.
Express the interval form (- 8.75, – 7.25) in neighbourhood form.
Answer:
Interval form : (- 8.75, – 7.25)
∴ a – δ = – 8.75, a + δ = – 7.25
By adding, 2a = – 16 . ∴ a = – 8
Putting a = – 8 in a + δ = – 7.25,
– 8 + δ = – 7.25 ∴δ = – 7.25 + 8 = 0.75
δ neighbourhood of a: N (a, δ)
∴(- 8.75, – 7.25) in neighbourhood form
= N (- 8, 0.75)

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 6.
If N (k1, 0.5) = (19.5, k2), then find the values of k1 and k2.
Answer:
N (k1, 0.5) = (19.5, k2)
k1 – <5 = 19.5, k1 + S = k2
Here, δ = 0.5
∴ k1 – 0.5 = 19.5
∴ k1 = 19.5 + 0.5 = 20
Putting k1 – 20 and δ = 0.5 in k1 + δ = k2,
20 + 0.5 = k2
k2 = 20.5
Hence, k1 = 20, k2 = 20.5

Question 7.
Express |3x + 1| < 2 in neighbourhood and interval form. ‘
Answer:
|3x + 1| < 2
∴ |x + \(\frac{1}{3}\)| < \(\frac{1}{3}\)
∴ \(\left|x-\left(-\frac{1}{3}\right)\right|<\frac{2}{3}\)
∴ \(\left|x-\left(-\frac{1}{3}\right)\right|<\frac{2}{3}\),
∴ a = – \(\frac{1}{3}\) δ = \(\frac{2}{3}\)

In neighbourhood form: N (a, δ)
∴ Neighbourhood form of |3x + 1| < 2
= N\(\left(-\frac{1}{3}, \frac{2}{3}\right)\)

In Interval form: (a – δ, a + δ)
Interval form of 13x + 11 < 2
= \(\left(-\frac{1}{3}-\frac{2}{3},-\frac{1}{3}+\frac{2}{3}\right)\)
= \(\left(-1, \frac{1}{3}\right)\)

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 8.
If |x – A1| < 0.09 = (A2, 4.09), then find the values of A1 and A2.
Answer:
|x – A1| < 0.09 = (A2, 4.09)
∴ a – δ = A2 and a + δ = 4.09
Here, δ = 0.09 and a = A1
∴ A1 – 0.09 = A2 and A1 + 0.09 = 4.09
Now, A1 + 0.09 = 4.09
∴ A1 = 4.09 – 0.09 = 4
Putting A1 = 4 in A1 – 0.09 = A2,
4 – 0.09 = A2
∴ A2 = 3.91
Hence, A1 = 4 and A2 = 3.91

Question 9.
Explain the meaning of x → a.
Answer:
When the value of a variable x is brought very close to a number ‘a’ by increasing or decreasing its value, then it can be said that x tends to a. It is denoted by x → a.
x → a means that value of x is very close to a but x ≠ a.
Taking a = 1, let us understand the meaning of x → 1.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 18

Question 10.
Explain the meaning of x → 0.
Answer:
If by decreasing the positive value of a variable x or by increasing the negative value of variable x, the value x is brought very close to ‘O’, then it can be said that x tends to ‘O’. It is denoted by x → 0.
x → 0 means that the value of x is very close to ‘0’ but x ≠ 0.
Let us understand the meaning of x → 0.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 20

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 11.
Define limit of a function.
Answer:
Limit of a function: When the value of a variable x Is brought closer and closer to a number ‘a’, the value of function f(x) approaches closer and closer to a definite number ‘l’, then we can say that as x tends to ‘a’ f(x) tends to ‘l’ that is x → a, f(x) → 1. This definite number ‘l’ is called limit of a function f(x). Symbolically it can be written as follows:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 21
‘l’ is the limit of f(x). Hence f(x) ≠ l.

Question 12.
State multiplication working rule of limit.
Answer:
f(x) and g (x) are two functions of real variable x and GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 22 f(x) = 1, GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 22 g (x) = m.
If f (x) g ∙ (x) is the product of two functions, then multiplication working rule of limit is as follows:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 23
Thus, limit of product of two functions is equal to the product of limit of these two functions.

Question 13.
State division working rule of limit.
Answer:
f (x) and g (x) are two functions of real variable x and GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 22 f(x) = 1, GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 22 g (x) = m.
If \(\frac{f(x)}{g(x)}\) is the division of the two functions. then division working rule of limit is as follows:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 24
Thus, the limit of division of two functions is equal to the division of their limits, where the limit of the function in denominator is not zero.

Question 14.
State the standard form of limit of a polynomial.
Answer:
Suppose, f(x) = a0 + a1x + a2x2 + …… + anxn is a polynomial. The standard form of limit of a polynomial is as follows:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 25

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Section D

Find the values of the following:

Question 1.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 26
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 27

Which is indeterminant. So x – 1 is common factor of the function. Eliminating this common factor from the numerator and denominator of the function limit of the function is found. We consider this matter, then and only then we will find the limit of the function.

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 2.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 28
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 29

Question 3.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 30
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 31

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 4.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 32
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 33

Question 5.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 34
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 35

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 6.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 36
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 37

Question 7.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 38
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 39

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 8.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 40
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 41

Question 9.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 42
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 43

Question 10.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 44
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 45

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 11.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 46
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 47

Question 12.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 48
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 49

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 13.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 50
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 51

Question 14.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 52
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 53

Question 15.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 54
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 55

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 16.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 56
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 57

Question 17.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 58
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 59

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Section E

I. Answer the following as required:

Question 1.
If y = 5x + 7 then using tabular method, prove that when x → 2, y → 17.
Answer:
y = 5x + 7
Here, x → 2. So taking the values of x very close to 2, the following table is prepared:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 60
It is clear from the table that when x is brought nearer to 2 by increasing or decreasing its value the value of y = 5x + 7 approaches to 17. That is, when x → 2, y → 17.
Hence, it is proved that when x → 2, y → 17.

Question 2.
If y = \(\frac{3 x^{2}+16 x+16}{x+4}\) then using tabular method, prove that when x → – 4, y → – 8.
Answer:
y = \(\frac{3 x^{2}+16 x+16}{x+4}\)
= \(\frac{3 x^{2}+12 x+4 x+16}{x+4}\)
= \(\frac{3 x(x+4)+4(x+4)}{x+4}\)
= \(\frac{(x+4)(3 x+4)}{x+4}\)
= 3x + 4
Here, x → – 4. So taking the values of x very close to – 4, the following table is prepared:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 61
It is clear from the table that when x is brought nearer to – 4 by increasing or decreasing its value, the value of y = 3x + 4 approaches to – 8. That is, when x → – 4, y → – 8.
Hence, it is proved that when x → – 4, y → – 8.

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 3.
Using tabular method, prove that GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 62 does not exist.
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 62
Here, f(x) = \(\frac{3}{x+1}\)
x → – 1. So taking the values of x very close to – 1, the following table is prepared:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 68
It is clear from the table that when x is brought nearer to – 1 by increasing or decreasing its value, the value of f(x) does not approach to a particular value. That is, when x → – 1, f(x) does not tend to a particular value.

Hence, it is proved that GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 62 does not exist.

II. Find the values of the following using tabular method:

Question 1.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 64
Answer:
Here, f(x) = \(\frac{x^{2}-3 x-10}{x-5}\)
Putting x = 5 in f(x)
f(x) = \(\frac{0}{0}\). So the common factor of numerator and denominator of f(x) is x – 5. After eliminating this common factor from f(x), we find the limit of f(x).
f(x) = \(\frac{x^{2}-5 x+2 x-10}{x-5}\)
= \(\frac{x(x-5)+2(x-5)}{x-5}\)
= \(\frac{(x-5)(x+2)}{x-5}\)
= x + 2
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 65
Taking the values of x very close to 5, the following table is prepared:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 66
It is clear from the table that when x is brought nearer to 5 by increasing or decreasing its value. The value of f(x) approaches to 7. That is, when x → 5, f(x) → 7.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 67

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 2.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 68
Answer:
f(x) = \(\frac{2 x^{2}+3 x-5}{x-1}\)
Putting x = 1 in f(x)
f(x) = \(\frac{0}{0}\). So the common factor (x – 1) of numerator and denominator of f(x), limit is found by tabular method.
f(x) = \(\frac{2 x^{2}+3 x-5}{x-1}\)
= \(\frac{2 x^{2}+5 x-2 x-5}{x-1}\)
= \(\frac{x(2 x+5)-1(2 x+5)}{x-1}\)
= \(\frac{(2 x+5)(x-1)}{x-1}\)
= 2x + 5
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 69
Taking the values of x very close to 1, the following table is prepared:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 70
It is clear from the table that when x is brought nearer to 1 by increasing or decreasing its value, the value of f(x) approaches to 7. That is. when x → 1, f(x) → 7.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 71

Question 3.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 72
Answer:
Here, f(x) = \(\frac{4 x^{2}+5 x+1}{x+1}\)
Putting x = – 1 in f(x), we get f(x) = \(\frac{0}{0}\).
So after eliminating the common factor (x + 1) of numerator and denominator of f(x), the limit of f(x) is found.
f(x) = \(\frac{4 x^{2}+5 x+1}{x+1}\)
= \(\frac{4 x^{2}+4 x+x+1}{x+1}\)
= \(\frac{4 x(x+1)+1(x+1)}{x+1}\)
= \(\frac{(4 x+1)(x+1)}{(x+1)}\)
= 4x + 1
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 73
Taking the values of x very close to – 1, the following table is prepared:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 74
It is clear from the table that when x is brought nearer to – 1 by increasing or decreasing its value, the value of f(x) approaches to – 3. That is. when x → – 1, f(x) → – 3.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 75

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 4.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 76
Answer:
Here, f(x) = 3x – 1 and x → 0. Taking the values of x very close to 0, the following table is prepared:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 77
It is clear from the table that when x is brought nearer to 0 by increasing or decreasing its value, the value of f(x) approaches to – 1. That is. when x → 0, f(x) → – 1.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 78

III. Find the values of the following:

Question 1.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 79
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 80

Question 2.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 81
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 82

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 3.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 83
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 84

Question 4.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 85
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 86

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 5.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 87
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 88

Question 6.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 89
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 90

Question 7.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 91
Answer:
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 92

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 8.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 93
Answer:
f(x) = 2x2 + 3
∴f(2 + h) = 2(2 + h)2 + 3
= 2 (4 + 4h + h2) + 3
= 8 + 8h + 2h2 + 3
= 2h2 + 8h + 11
f(2) = 2 (22) + 3 = 2 × 4 + 3 = 8 + 3 = 11
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 94

Question 9.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 95
Answer:
f(x) = x2
∴ f(2 + x) = (2 + x)2 = 4 + 4x + x2
f(2 – x) = (2 – x)2 = 4 – 4x + x2
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 96

GSEB Solutions Class 12 Statistics Part 2 Chapter 4 Limit Ex 4

Question 10.
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 97
Answer:
f(x) = x2 + x
∴ f(2) = (2)2 + 2 = 4 + 2 = 6
GSEB Solutions Class 12 Statistics Chapter 4 Limit Ex 4 98

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