{"id":564133,"date":"2021-12-23T09:26:43","date_gmt":"2021-12-23T09:26:43","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=564133"},"modified":"2021-12-28T06:42:21","modified_gmt":"2021-12-28T06:42:21","slug":"maharashtra-board-solutions-class-11-arts-science-maths-part-2-chapter-8-continuity","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-11-arts-science-maths-part-2-chapter-8-continuity\/","title":{"rendered":"Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 8- Continuity"},"content":{"rendered":"\n

Class 11: Maths Chapter 8 solutions. Complete Class 11 Maths Chapter 8 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 8- Continuity<\/h2>\n\n\n\n

Maharashtra Board 11th Maths Chapter 8, Class 11 Maths Chapter 8 solutions<\/p>\n\n\n\n

Ex 8.1<\/h4>\n\n\n\n
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\u2234 4 + sin x and 3 \u2013 cos x are continuous for all x \u2208 R.
\u2234 f(x) is continuous for both the given intervals.
Let us test the continuity at x = \u03c0.<\/p>\n\n\n\n

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But f(\u03c0) is not defined.
\u2234 f(x) has a removable discontinuity at x = \u03c0.<\/p>\n\n\n\n

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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.<\/strong><\/p>\n\n\n\n

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Solution:<\/strong><\/p>\n\n\n\n

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f(x) is continuous at x = 0 \u2026..(given)<\/p>\n\n\n\n

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\u2234 f(x) = x2<\/sup> + px + q for x \u2265 3 as well as x \u2264 1
Thus, f(x) = x2<\/sup> + px + q; x \u2264 1
= x + 1; 1 < x < 3 = x2<\/sup> + px + q; x > 3
f(x) is continuous for all x \u2208 R.
\u2234 f(x) is continuous at x = 1 and x = 3.
As f(x) is continuous at x = 1,
<\/p>\n\n\n\n

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Subtracting (i) from (ii), we get
2p = -6
\u2234 p = -3
Substituting p = -3 in (i), we get
-3 + q = 1
\u2234 q = 4
\u2234 p = -3 and q = 4<\/p>\n\n\n\n

Question 15.
Show that there is a root for the equation 2x3<\/sup> \u2013 x \u2013 16 = 0 between 2 and 3.
Solution:
Let f(x) = 2x3<\/sup> \u2013 x \u2013 16
f(x) is a polynomial function and hence it is continuous for all x \u2208 R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(2) = 2(2)3<\/sup> \u2013 2 \u2013 16 = -2 < 0
f(3) = 2(3)3<\/sup> \u2013 3 \u2013 16 = 35 > 0
\u2234 f(2) < 0 and f(3) > 0
\u2234 By intermediate value theorem,
there has to be point \u2018c\u2019 between 2 and 3 such that f(c) = 0.
\u2234 There is a root of the given equation between 2 and 3.<\/p>\n\n\n\n

Question 16.
Show that there is a root for the equation x3<\/sup> \u2013 3x = 0 between 1 and 2.
Solution:<\/strong>
Let f(x) = x3<\/sup> \u2013 3x
f(x) is a polynomial function and hence it is continuous for all x \u2208 R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = (1)3<\/sup> \u2013 3(1) = -2 < 0
f(2) = (2)3<\/sup> \u2013 3(2) = 2 > 0
\u2234 f(1) < 0 and f(2) > 0
\u2234 By intermediate value theorem,
there has to be point \u2018c\u2019 between 1 and 2 such that f(c) = 0.
There is a root of the given equation between 1 and 2.<\/p>\n\n\n\n

Question 17.
Let f(x) = ax + b (where a and b are unknown)
= x2<\/sup> + 5 for x \u2208 R<\/strong>
Find the values of a and b, so that f(x) is continuous at x = 1.
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Solution:
<\/strong>f(x) = x2<\/sup> + 5, x \u2208 R
\u2234 f(1) = 1 + 5 = 6
If f(x) = ax + b is continuous at x = 1, then
f(1) = limx\u21921(ax+b) = a + b
\u2234 6 = a + b where, a, b \u2208 R
\u2234 There are infinitely many values of a and b.<\/p>\n\n\n\n

Question 18.
Activity: Suppose f(x) = px + 3 for a \u2264 x \u2264 b
= 5x2<\/sup> \u2013 q for b < x \u2264 c
Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the boxes.<\/strong>
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Solution:
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Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 8- Continuity<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 11-Arts & Science Maths (Part 2): Chapter 8- Continuity PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 11 Arts & Science Maths (Part 2) :<\/strong><\/h2>\n\n\n\n