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Class 11 (FYJC / HSC)Commerce Mathematics & Statistics2026-27 Syllabus

Chapter 8 Continuity Miscellaneous Exercise 8 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 8 Continuity Miscellaneous Exercise 8. Step-by-step solved exercises, numerical problems, and digest answers.

11 Solved Questions20 Diagrams698 words

Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 8 Continuity Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

I. Discuss the continuity of the following functions at the point(s) or in the interval indicated against them.

Question 1 Maharashtra Board Solution
If f(x) = 2x2 – 2x + 5 for 0 ≤ x < 2 = for 2 ≤ x < 4 = for 4 ≤ x ≤ 7 on its domain.
Solution & Step-by-Step Answer:
The domain of f is [0, 5) ∪ (5, 7] We observe that x = 5 is not included in the domain as f is not defined at x = 5 a. For 0 ≤ x < 2 f(x) = 2x2 – 2x + 5 It is a polynomial function and is continuous at all point in [0, 2)

b. For 2 < x < 4
f(x) =
It is a rational function and is continuous everwhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 1
But x = 1 does not lie in the interval.
f(x) is continuous at all points in (2, 4)

c. For 4 < x ≤ 7, x ≠ 5
i.e. for x ∈ [4, 5) ∪ (5, 7]
∴ f(x) =
It is a rational function and is continuous everywhere except possibly at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 ∉ [4, 5) ∪ (5, 7]
∴ f is continuous at all points in (4, 7] – {5}.

d. Since the definition of function changes around x = 2, x = 4 and x = 7
∴ there is disturbance in behaviour of the function.
So we examine continuity at x = 2, 4, 7 separately.
Continuity at x = 2:

= 2(2)2– 2(2) + 5
= 8 – 4 + 5
= 9

∴ f is continuous at x = 2

e. Continuity at x = 4:

∴ f is continuous at x = 4

Question 2 Maharashtra Board Solution
f(x) = for x ≠ 0 = (log 3)2 for x = 0 at x = 0
Solution & Step-by-Step Answer:
∴ ∴ f is continuous at x = 0

Question 3 Maharashtra Board Solution
f(x) = for x ≠ 0 = (log 5 – 1) for x = 0 at x = 0
Solution & Step-by-Step Answer:
∴ ∴ f is continuous at x = 0

Question 4 Maharashtra Board Solution
f(x) = for x ≠ 1 = 2 for x = 1, at x = 1
Solution & Step-by-Step Answer:
∴ ∴ f is discontinuous at x = 1

Question 5 Maharashtra Board Solution
f(x) = for x ≠ 3 = 3 for x = 3, at x = 3
Solution & Step-by-Step Answer:

(II) Find k if following functions are continuous at the points indicated against them.

Question 1 Maharashtra Board Solution
f(x) = for x ≠ 2 = k for x = 2 at x = 2
Solution & Step-by-Step Answer:

Question 2 Maharashtra Board Solution
f(x) = for x ≠ 0 = for x = 0, at x = 0
Solution & Step-by-Step Answer:

Question 3 Maharashtra Board Solution
f(x) = , for x ≠ 0 = , for x = 0, at x = 0
Solution & Step-by-Step Answer:

III. Find a and b if following functions are continuous at the point indicated against them.

Question 1 Maharashtra Board Solution
f(x) = x2 + a, for x ≥ 0 = 2 + b, for x < 0 and f(1) = 2, is continuous at x = 0
Solution & Step-by-Step Answer:

Question 2 Maharashtra Board Solution
f(x) = + a, for x > 3 = 5, for x = 3 = 2x2 + 3x + b, for x < 3 is continuous at x = 3
Solution & Step-by-Step Answer:

Question 3 Maharashtra Board Solution
f(x) = + a, for x > 0 = 2, for x = 0 = x + 5 – 2b, for x < 0 is continuous at x = 0
Solution & Step-by-Step Answer: