Class 11: Maths Chapter 8 solutions. Complete Class 11 Maths Chapter 8 Notes.
Contents
Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 8- Continuity
Maharashtra Board 11th Maths Chapter 8, Class 11 Maths Chapter 8 solutions
Ex 8.1
∴ 4 + sin x and 3 – cos x are continuous for all x ∈ R.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = π.
But f(π) is not defined.
∴ f(x) has a removable discontinuity at x = π.
Question 9.
Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous.
Solution:
f(x) is continuous at x = 0 …..(given)
∴ f(x) = x2 + px + q for x ≥ 3 as well as x ≤ 1
Thus, f(x) = x2 + px + q; x ≤ 1
= x + 1; 1 < x < 3 = x2 + px + q; x > 3
f(x) is continuous for all x ∈ R.
∴ f(x) is continuous at x = 1 and x = 3.
As f(x) is continuous at x = 1,
Subtracting (i) from (ii), we get
2p = -6
∴ p = -3
Substituting p = -3 in (i), we get
-3 + q = 1
∴ q = 4
∴ p = -3 and q = 4
Question 15.
Show that there is a root for the equation 2x3 – x – 16 = 0 between 2 and 3.
Solution:
Let f(x) = 2x3 – x – 16
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(2) = 2(2)3 – 2 – 16 = -2 < 0
f(3) = 2(3)3 – 3 – 16 = 35 > 0
∴ f(2) < 0 and f(3) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 2 and 3 such that f(c) = 0.
∴ There is a root of the given equation between 2 and 3.
Question 16.
Show that there is a root for the equation x3 – 3x = 0 between 1 and 2.
Solution:
Let f(x) = x3 – 3x
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = (1)3 – 3(1) = -2 < 0
f(2) = (2)3 – 3(2) = 2 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 1 and 2 such that f(c) = 0.
There is a root of the given equation between 1 and 2.
Question 17.
Let f(x) = ax + b (where a and b are unknown)
= x2 + 5 for x ∈ R
Find the values of a and b, so that f(x) is continuous at x = 1.
Solution:
f(x) = x2 + 5, x ∈ R
∴ f(1) = 1 + 5 = 6
If f(x) = ax + b is continuous at x = 1, then
f(1) = limx→1(ax+b) = a + b
∴ 6 = a + b where, a, b ∈ R
∴ There are infinitely many values of a and b.
Question 18.
Activity: Suppose f(x) = px + 3 for a ≤ x ≤ b
= 5x2 – q for b < x ≤ c
Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the boxes.
Solution:
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Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 8- Continuity
Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 2) :
- Chapter 1- Complex Numbers
- Chapter 2- Sequences and Series
- Chapter 3- Permutations and Combination
- Chapter 4- Methods of Induction and Binomial Theorem
- Chapter 5- Sets and Relations
- Chapter 6- Functions
- Chapter 7- Limits
- Chapter 8- Continuity
- Chapter 9- Differentiation
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