{"id":3195,"date":"2023-11-29T06:27:36","date_gmt":"2023-11-29T00:57:36","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=3195"},"modified":"2023-11-30T12:22:11","modified_gmt":"2023-11-30T06:52:11","slug":"samacheer-kalvi-11th-maths-solutions-chapter-9-ex-9-5","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-11th-maths-solutions-chapter-9-ex-9-5\/","title":{"rendered":"Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.5"},"content":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.5<\/h2>\n

Question 1.
\nProve that f(x) = 2x2<\/sup> + 3x – 5 is continuous at all points in R.
\nSolution:
\nPolynomial functions are continuous at every points of R.
\n\"Samacheer<\/p>\n

Question 2.
\nExamine the continuity of the following:
\n(i) x + sin x
\nSolution:
\nf(x) = x + sin x
\nThe Domain of the function (-\u221e, \u221e)
\n\u2234 f(x) is continuous in (-\u221e, \u221e)(i.e.,) for all x \u2208 R<\/p>\n

(ii) x2<\/sup> cos x
\nSolution:
\nf(x) = x2<\/sup> cos x
\nThe Domain of the function (-\u221e, \u221e)
\n\"Samacheer
\nf(x) is continuous in R<\/p>\n

(iii) ex<\/sup> tan x
\nSolution:
\nThe Domain of the function in R – {(2n + 1) \u03c0\/2}
\n\u2234 The functions is continuous for all x \u2208 R – (2n + 1) \\(\\frac{\\pi}{2}\\), n \u2208 Z<\/p>\n

(iv) e2x<\/sup> + x2<\/sup>
\nf(x) = e2x<\/sup> + x2<\/sup> = 1 + 2x + \\(\\frac{(2 x)^{2}}{2 !}\\) + …………. + x2<\/sup>
\nSolution:
\n\u2234 The functions is continuous for all x \u2208 R<\/p>\n

(v) x.ln x
\nSolution:
\nThus f(x) is continuous for (0, \u221e)<\/p>\n

(vi) \\(\\frac{\\sin x}{x^{2}}\\)
\nSolution:
\nThus f(x) is continuous for all x \u2208 R – {0}<\/p>\n

(vii) \\(\\frac{x^{2}-16}{x+4}\\)
\nSolution:
\nf(x) = \\(\\frac{x^{2}-16}{x+4}=\\frac{(x-4)(x+4)}{x+4}\\)
\nThe function f(x) is continuous for all x \u2208 R – {-4}<\/p>\n

(viii) |x + 2| + |x – 1|
\nSolution:
\nf(x) is continuous for x \u2208 R<\/p>\n

(ix) \\(\\frac{|x-2|}{|x+1|}\\)
\nSolution:
\nThe function is continuous for all x \u2208 R – {-1}<\/p>\n

(x) cot x + tan x
\nSolution:
\nThe function is continuous for all x \u2208 R – \\(\\frac{n \\pi}{2}\\), n \u2208 z.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nFind the points of discontinuity of the function f, where,
\n(i)
\n\"Samacheer
\nSolution:
\nf(3) = 12 + 5 = 17
\n\"Samacheer
\n\u2234 f(x) is discontinuous at x = 3<\/p>\n

(ii)
\n\"Samacheer
\nSolution:
\nf(x) = 4
\n\"Samacheer
\n\u2234 f(x) is continuous for all x \u2208 R<\/p>\n

(iii)
\n\"Samacheer
\nSolution:
\nf(x) = 8 – 3 = 5
\n\"Samacheer
\n\u2234 f(x) is continuous for all x \u2208 R<\/p>\n

(iv)
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\u2234 f(x) is continuous for all x \u2208 [0, \u03c0\/2]<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nAt the given points x0<\/sub> discover whether the given function is continuous or discontinuous citing the reasons for your answer.
\n(i)
\n\"Samacheer
\nSolution:
\nGiven f(x0<\/sub>) = 1
\n\"Samacheer
\n\u2234 f(x) is continuous at x0<\/sub> = 1<\/p>\n

(ii)
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\u2234 f(x) is not continuous at x0<\/sub> = 3<\/p>\n

Question 5.
\nShow that the function \"Samacheer is continuous on (-\u221e, \u221e)
\nSolution:
\n\"Samacheer
\nGiven that f(1) = 3
\n\u2234 f(x) is continuous for all x \u2208 R<\/p>\n

Question 6.
\nFor what value of \u03b1 is this function f(x) = \"Samacheer continuous at x = 1?
\nSolution:
\n\"Samacheer
\n\u2235 f(x) is continuous at x = 1, \u03b1 = 4<\/p>\n

Question 7.
\nLet \"Samacheer Graph the function. Show that f(x) continuous on (-\u221e, \u221e)
\nSolution:
\n\"Samacheer
\n\"Samacheer
\n\u2234 f(x) is continuous in (-\u221e, \u221e)<\/p>\n

Question 8.
\nIf f and g are continuous function with f(3) = 5 and \"Samacheer find g(3).
\nSolution:
\nSince f and g are continuous
\n\"Samacheer
\n2f(3) – g(3) = 4
\n2(5) – g(3) = 4
\n10 – 4 = g(3)
\ng(3) = 6<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nFind the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f.
\n(i)
\n\"Samacheer
\n\u2234 f(x) is not continuous at x = 1
\nSolution:
\n\"Samacheer
\nf(x) is not continuous at x = 1<\/p>\n

(ii)
\n\"Samacheer
\nSolution:
\n\"Samacheer
\n\u2234 f(x) is not continuous at x = 0<\/p>\n

Question 10.
\nA function f is defined as follows:
\n\"Samacheer
\nIs the function continuous?
\nSolution:
\n\"Samacheer
\nFrom (i), (ii) and (iii)
\nf(x) is continuous at x = 0, 1, 3<\/p>\n

Question 11.
\nWhich of the following functions f has removable discontinuity at x = x0<\/sub>? If the discontinuity is removable, find a function g that agrees with f for x \u2260 x0<\/sub> and is continuous on R.
\n(i) f(x) = \\(\\frac{x^{2}-2 x-8}{x+2}\\), x0<\/sub> = -2
\nSolution:
\n\"Samacheer<\/p>\n

(ii) f(x) = \\(\\frac{x^{3}+64}{x+4}\\), x0<\/sub> = -4
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

(iii) f(x) = \\(\\frac{3-\\sqrt{x}}{9-x}\\), x0<\/sub> = 9
\nSolution:
\n\"Samacheer<\/p>\n

Question 12.
\nFind the constant b that makes g continuous on (-\u221e, \u221e)
\n\"Samacheer
\nSolution:
\nSince g(x) is continuous,
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nConsider the function f(x) = x sin \\(\\frac{\\pi}{x}\\). What value must we give f(0) in order to make the function continuous everywhere?
\nSolution:
\n\"Samacheer
\nso to make the function f(x) is continuous at f(0) = 0<\/p>\n

Question 14.
\nThe function f(x) = \\(\\frac{x^{2}-1}{x^{3}-1}\\) is not defined at x = 1. What value must we give f(1) in order to make f(x) continuous at x = 1?
\nSolution:
\n\"Samacheer<\/p>\n

Question 15.
\nState how continuity is destroyed at x = x0<\/sub> for each of the following graphs.
\n\"Samacheer
\nSolution:
\n\"\"<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.5 Question 1. Prove that f(x) = 2×2 + 3x – 5 is continuous at …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[7],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3195"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=3195"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3195\/revisions"}],"predecessor-version":[{"id":60025,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/3195\/revisions\/60025"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=3195"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=3195"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=3195"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}