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Class 11 (FYJC / HSC)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.

30 Solved Questions37 Diagrams2532 words

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

(I) Select the correct answer from the given alternatives.

Question 1 Maharashtra Board Solution
f(x) = , for x ≠ = log √2, for x = (A) f is continuous at x = (B) f has a jump discontinuity at x = (C) f has a removable discontinuity (D)
Solution & Step-by-Step Answer:
(A) f is continuous at x = Hint:

Question 2 Maharashtra Board Solution
If f(x) = , for x ≠ is continuous at x = , then f() = (A) (B) (C) (D)
Solution & Step-by-Step Answer:
(D) Hint:

Question 3 Maharashtra Board Solution
If f(x) = , for x ≠ 0 is continuous at x = 0, then f(0) is (A) (B) (C) (D)
Solution & Step-by-Step Answer:
(D) Hint:

Question 4 Maharashtra Board Solution
f(x) = , for x ∈ [-6, -3] (A) f is discontinuous at x = 2 (B) f is discontinuous at x = -4 (C) f is discontinuous at x = 0 (D) f is discontinuous at x = 2 and x = -4
Solution & Step-by-Step Answer:
(B) f is discontinuous at x = -4 Hint: f(x) = , for x ∈ [-6, -3] = Here f(x) is a rational function and is continuous everywhere except at the points Where denominator becomes zero. Here, denominator becomes zero when x = -4 or x = 2 But x = 2 does not lie in the given interval. ∴ x = -4 is the point of discontinuity.
Question 5 Maharashtra Board Solution
If f(x) = ax2 + bx + 1, for |x – 1| ≥ 3 and = 4x + 5, for -2 < x < 4 is continuous everywhere then, (A) a = , b = 3 (B) a = , b = -3 (C) a = , b = 3 (D) a = , b = -3
Solution & Step-by-Step Answer:
(A) a = , b = 3 Hint: f(x) = ax2 + bx + 1, |x – 1| ≥ 3 = 4x + 5; -2 < x < 4 The first interval is |x – 1| ≥ 3 ∴ x – 1 ≥ 3 or x – 1 ≤ -3 ∴ x ≥ 4 or x ≤ -2 ∴ f(x) is same for x ≤ -2 as well as x ≥ 4. ∴ f(x) is defined as: f(x) = ax2 + bx + 1; x ≤ -2 = 4x + 5; -2 < x < 4 = ax2 + bx + 1; x ≥ 4 f(x) is continuous everywhere. ∴ f(x) is continuous at x = -2 and x = 4. As f(x) is continuous at x = -2, ∴ ∴ a(-2)2 + b(-2) + 1 = 4(-2) + 5 ∴ 4a – 2b + 1 = -3 ∴ 4a – 2b = -4 ∴ 2a – b = -2 …..(i) ∵ f(x) is continuous at x = 4, ∴ 4(4) + 5 = a(4)2 + b(4) + 1 16a + 4b + 1 = 21 16a + 4b = 20 4a + b = 5 …..(ii) Adding (i) and (ii), we get 6a = 3 ∴ a = Substituting a = in (ii), we get 4() + b = 5 ∴ 2 + b = 5 ∴ b = 3 ∴ a = , b = 3
Question 6 Maharashtra Board Solution
f(x) = , for x ≠ 0 = k, for x = 0 is continuous at x = 0, then ‘k’ = (A) (B) (C) (D)
Solution & Step-by-Step Answer:
(B) Hint:

Question 7 Maharashtra Board Solution
f(x) = , for x ≠ 0 = k, for x = 0, is continuous at x = 0, then value of ‘k’ is (A) 6 (B) 4 (C) (log 2) (log 4) (D) 3 log 4
Solution & Step-by-Step Answer:
(A) 6 Hint:

Question 8 Maharashtra Board Solution
If f(x) = , for x ≠ 0 is continuous at x = 0 then the value of f(0) is (A) (B) log 2. log 3 (C) (D) None of these
Solution & Step-by-Step Answer:
(B) log 2. log 3 Hint:

Question 9 Maharashtra Board Solution
If f(x) = , for x ≠ 0 and f(0) = k, is continuous at x = 0, then k is (A) e7 (B) e3 (C) e12 (D)
Solution & Step-by-Step Answer:
(C) e12 Hint:

Question 10 Maharashtra Board Solution
If f(x) = for x ∈ (-1, 2), then f is discontinuous at (A) x = -1, 0, 1, 2 (B) x = -1, 0, 1 (C) x = 0, 1 (D) x = 2
Solution & Step-by-Step Answer:
(C) x = 0, 1 Hint: f(x) = , x ∈ (-1, 2) This function is discontinuous at all integer values of x between -1 and 2. ∴ f(x) is discontinuous at x = 0 and x = 1.

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

Question 1 Maharashtra Board Solution
f(x) = , for 3 ≤ x ≤ 6, x ≠ 5 = 10, for x = 5 = , for 6 < x ≤ 9
Solution & Step-by-Step Answer:

Question 2 Maharashtra Board Solution
f(x) = 2x2 – 2x + 5, for 0 ≤ x ≤ 2 = , for 2 < x < 4 = , for 4 ≤ x ≤ 7 and x ≠ 5 = 7, for x = 5
Solution & Step-by-Step Answer:
The domain of f(x) is [0, 7]. (i) For 0 ≤ x ≤ 2 f(x) = 2x2 – 2x + 5 It is a polynomial function and is Continuous at all point in [0, 2).

(ii) 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).

(iii) For 4 ≤ x ≤ 7, x ≤ 5
f(x) =
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 does not lie in the interval.
∴ f(x) is continuous at all points in (4, 7] – {5}.

(iv) For continuity at x = 2:

∴ f(x) is continuous at x = 2.

(v) For continuity at x = 4:

∴ f(x) is continuous at x = 4.

(vi) For continuity at x = 5.
f(5) = 7

∴ f(x) is discontinuous at x = 5.
Thus, f(x) is continuous at all points on its domain except at x = 5.

Question 3 Maharashtra Board Solution
f(x) = , for x ≠ 0 f(0) = , at x = 0 on ≤ x ≤
Solution & Step-by-Step Answer:
The domain of f(x) is [, ] (i) For [, ] – {0}: f(x) = It is a rational function and is continuous everywhere except at points where its denominator becomes zero. Denominator becomes zero when cos x = 1, i.e., x = 0 But x = 0 does not lie in the interval. ∴ f(x) is continuous at all points in [, ] – {0}

(ii) For continuity at x = 0:



∴ f(x) is discontinuous at x = 0.

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

Question 5 Maharashtra Board Solution
f(x) = , for x ≠ -1 = 0, for x = -1, at x = -1
Solution & Step-by-Step Answer:

Question 6 Maharashtra Board Solution
f(x) = [x + 1] for x ∈ [-2, 2) Where [*] is greatest integer function.
Solution & Step-by-Step Answer:
f(x) = [x + 1], x ∈ [-2, 2) ∴ f(x) = -1, x ∈ [-2, -1) = 0, x ∈ [-1, 0) = 1, x ∈ [0, 1) = 2, x ∈ [1, 2) ∴ ∴ f(x) is discontinuous at x = -1. Similarly, f(x) is discontinuous at the points x = 0 and x = 1.

Question 7 Maharashtra Board Solution
f(x) = 2x2 + x + 1, for |x – 3| ≥ 2 = x2 + 3, for 1 < x < 5
Solution & Step-by-Step Answer:
|x – 3| ≥ 2 ∴ x – 3 ≥ 2 or x – 3 ≤ -2 ∴ x ≥ 5 or x ≤ 1 ∴ f(x) = 2x2 + x + 1, x ≤ 1 = x2 + 3, 1 < x < 5 = 2x2 + x + 1, x ≥ 5 Consider the intervals x < 1, i.e., (-∞, 1) 1 < x < 5, i.e., (1, 5) x > 5, i.e., (5, ∞) In all these intervals, f(x) is a polynomial function and hence is continuous at all points. For continuity at x = 1: ∴ f(x) is discontinuous at x = 5. ∴ f(x) is continuous for all x ∈ R, except at x = 5.

III. Identify discontinuities if any for the following functions as either a jump or a removable discontinuity on their respective domains.

Question 1 Maharashtra Board Solution
f(x) = x2 + x – 3, for x ∈ [-5, -2) = x2 – 5, for x ∈ (-2, 5]
Solution & Step-by-Step Answer:
f(-2) has not been defined. = (-2)2 + (-2) – 3 = 4 – 2 – 3 = -1 = (-2)2 – 5 = 4 – 5 = -1 ∴ ∴ But f(-2) has not been defined. ∴ f(x) has a removable discontinuity at x = -2.
Question 2 Maharashtra Board Solution
f(x) = x2 + 5x + 1, for 0 ≤ x ≤ 3 = x3 + x + 5, for 3 < x ≤ 6
Solution & Step-by-Step Answer:
= (3)2 + 5(3) + 1 = 9 + 15 + 1 = 25 = (3)3 + 3 + 5 = 27 + 3 + 5 = 35 ∴ ∴ does not exist. ∴ f(x) is discontinuous at x = 3. ∴ f(x) has a jump discontinuity at x = 3.
Question 3 Maharashtra Board Solution
f(x) = , for x ∈ [0, 3) = , for x ∈ [3, 6]
Solution & Step-by-Step Answer:
∴ f(x) is continuous at x = 3.

IV. Discuss the continuity of the following functions at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous.

Question 1 Maharashtra Board Solution
f(x) =
Solution & Step-by-Step Answer:
f(x) = = ∴ f(x) is not defined at x = 4 and x = -3. ∴ The domain of function f = R – {-3, 4}. For x ≠ -3 and 4, f(x) is discontinuous at x = 4 and x = -3. This discontinuity is removable. ∴ f(x) can be redefined as f(x) = , for x ≠ 4, x ≠ -3 = -5, for x ∈ R – {-3, 4}, x = -3 = 2, for x ∈ R – {-3, 4}, x = 4

Question 2 Maharashtra Board Solution
f(x) = x2 + 2x + 5, for x ≤ 3 = x3 – 2x2 – 5, for x > 3
Solution & Step-by-Step Answer:
∴ f(x) is discontinuous at x = 3. This discontinuity is irremovable.

V. Find k if the 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:
f(x) is continuous at x = 2. …..(given)

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

VI. Find a and b if the following functions are continuous at the points or on the interval indicated against them.

Question 1 Maharashtra Board Solution
f(x) = , for x < 0 = , for x = 0 = , for x < 0
Solution & Step-by-Step Answer:

Question 2 Maharashtra Board Solution
f(x) = ax2 + bx + 1, for |2x – 3| ≥ 2 = 3x + 2, for < x <
Solution & Step-by-Step Answer:

VII. Find f(a), if f is continuous at x = a where,

Question 1 Maharashtra Board Solution
f(x) = , for x ≠ 1 and at a = 1.
Solution & Step-by-Step Answer:

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

VIII. Solve using intermediate value theorem.

Question 1 Maharashtra Board Solution
Show that 5x – 6x = 0 has a root in [1, 2].
Solution & Step-by-Step Answer:
Let f(x) = 5x – 6x 5x and 6x are continuous functions for all x ∈ R. ∴ 5x – 6x is also continuous for all x ∈ R. i.e., f(x) 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) = 51 – 6(1) = -1 < 0 f(2) = (5)2 – 6(2) = 13 > 0 ∴ f(1) < 0 and f(2) > 0 ∴ By intermediate value theorem, there has to be a point ‘c’ between 1 and 2 such that f(c) = 0. ∴ There is a root of the given equation in [1, 2].
Question 2 Maharashtra Board Solution
Show that x3 – 5x2 + 3x + 6 = 0 has at least two real roots between x = 1 and x = 5.
Solution & Step-by-Step Answer:
Let f(x) = x3 – 5x2 + 3x + 6 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. Here, we have been asked to show that f(x) has at least two roots between x = 1 and x = 5. f(1) = (1)3 – 5(1)2 + 3(1) + 6 = 5 > 0 f(2) = (2)3 – 5(2)2 + 3(2) + 6 = 8 – 20 + 6 + 6 = 0 ∴ x = 2 is a root of f(x). Also, f(3) = (3)3 – 5(3)2 + 3(3) + 6 = 27 – 45 + 9 + 6 = -3 < 0 f(4) = (4)3 – 5(4)2 + 3(4) + 6 = 64 – 80 + 12 + 6 = 2 > 0 ∴ f(3) < 0 and f(4) > 0 ∴ By intermediate value theorem, there has to be a point ‘c’ between 3 and 4 such that f(c) = 0. ∴ There are two roots, x = 2 and a root between x = 3 and x = 4. Thus, there are at least two roots of the given equation between x = 1 and x = 5.