Chapter 8 Continuity Ex 8.1 Solutions

Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 8 Continuity Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1

(ii) f(x) = on R
Solution:
f(x) = ; x ∈ R
f(x) is a rational function and is continuous for all x ∈ R, except at the points where denominator becomes zero.
Here, denominator x – 3 = 0 when x = 3.
∴ Function f is continuous for all x ∈ R, except at x = 3, where it is not defined.

(ii) f(x) = for x ≠ 1
= 20, for x = 1, at x = 1
Solution:

∴ f(x) is continuous at x = 1

(ii) f(x) = for x ≠ 2
= -24 for x = 2, at x = 2
Solution:

(iii) f(x) = 4x + 1 for x ≤
= , for x > , at x =
Solution:

(iv) f(x) = for 0 ≤ x < 3
= , for 3 ≤ x ≤ 6, at x = 3
Solution:

(ii) If f(x) = , for x ≠ 0
= k for x = 0
is continuous at x = 0, find k.
Solution:
Function f is continuous at x = 0

(iii) For what values of a and b is the function
f(x) = ax + 2b + 18 for x ≤ 0
= x²+ 3a – b for 0 < x ≤ 2 = 8x – 2 for x > 2,
continuous for every x?
Solution:
Function f is continuous for every x.
∴ Function f is continuous at x = 0 and x = 2
As f is continuous at x = 0.
∴
∴
∴ a(0) + 2b + 18 = (0)2 + 3a – b
∴ 3a – 3b = 18
∴ a – b = 6 …..(i)
Also, Function f is continous at x = 2
∴
∴
∴ (2)²+ 3a – b = 8(2) – 2
∴ 4 + 3a – b = 14
∴ 3a – b = 10 …..(ii)
Subtracting (i) from (ii), we get
2a = 4
∴ a = 2
Substituting a = 2 in (i), we get
2 – b = 6
∴ b = -4
∴ a = 2 and b = -4

(iv) For what values of a and b is the function
f(x) = for x < 2
= ax²– bx + 3 for 2 ≤ x < 3
= 2x – a + b for x ≥ 3
continuous in its domain.
Solution:
Function f is continuous for every x on R.
∴ Function f is continuous at x = 2 and x = 3.
As f is continuous at x = 2.
∴

∴ 2 + 2 = a(2)²– b(2) + 3
∴ 4 = 4a – 2b + 3
∴ 4a – 2b = 1 …..(i)
Also function f is continuous at x = 3
∴
∴
∴ a(3)²– b(3) + 3 = 2(3) – a + b
∴ 9a – 3b + 3 = 6 – a + b
∴ 10a – 4b = 3 …..(ii)
Multiplying (i) by 2, we get
8a – 4b = 2 …..(iii)
Subtracting (iii) from (ii), we get
2a = 1
∴ a =
Substituting a = in (i), we get
4() – 2b = 1
∴ 2 – 2b = 1
∴ 1 = 2b
∴ b =
∴ a = and b =