{"id":4235,"date":"2023-12-22T15:45:05","date_gmt":"2023-12-22T10:15:05","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=4235"},"modified":"2023-12-23T17:30:44","modified_gmt":"2023-12-23T12:00:44","slug":"maths-1b-limits-and-continuity-important-questions-short-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/maths-1b-limits-and-continuity-important-questions-short-answer-type\/","title":{"rendered":"TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type"},"content":{"rendered":"
Students must practice these Maths 1B Important Questions<\/a> TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type to help strengthen their preparations for exams.<\/p>\n Question 1. Question 2. <\/p>\n Question 3. Question 4. Question 5. Question 6. <\/p>\n Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. <\/p>\n Question 13. Question 14. Question 15. Question 16. Question 17. <\/p>\n Question 18. Question 19. Question 20. Question 21. Question 22. <\/p>\n Question 23. Question 24. Question 25. Question 26. Question 27. Question 28. <\/p>\n Question 29. Question 30. Question 31. Question 32. Question 33. <\/p>\n Question 34. Question 35. Question 36. Question 37. <\/p>\n Question 38. Question 39. Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type to help strengthen their preparations for exams. TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type Question 1. Is the function f, defined by f(x) = , continuous … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/4235"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=4235"}],"version-history":[{"count":5,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/4235\/revisions"}],"predecessor-version":[{"id":5711,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/4235\/revisions\/5711"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=4235"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=4235"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=4235"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}TS Inter 1st Year Maths 1B Limits and Continuity Important Questions Short Answer Type<\/h2>\n
\nIs the function f, defined by f(x) = \\(\\left\\{\\begin{array}{l}
\nx^2 \\text { if } x \\leq 1 \\\\
\nx \\text { if } x>1
\n\\end{array}\\right.\\), continuous on R. [May ’15 (AP), ’11]
\nSolution:
\nWe find the limit at a = 1
\n
\n\u2234 f is continuous at x = 1
\nHence f is continuous on R.<\/p>\n
\nIs f defined by f(x) = \\(\\begin{cases}\\frac{\\sin 2 x}{x} & \\text { if } x \\neq 0 \\\\ 1, & \\text { if } x=0\\end{cases}\\), continuous on ‘0’? [May ’12, ’10, ’04; Mar. ’05]
\nSolution:
\nGiven, f(x) = \\(\\begin{cases}\\frac{\\sin 2 x}{x} & \\text { if } x \\neq 0 \\\\ 1, & \\text { if } x=0\\end{cases}\\)
\nTake a = 0
\n
\n\u2234 f is discontinuous at x = 0.<\/p>\n
\nCheck the continuity of the following function at 2.
\nf(x) = \\(\\begin{cases}\\frac{1}{2}\\left(x^2-4\\right) & \\text { if } 0<x<2 \\\\ 0, & \\text { if } x=2 \\\\ 2-8 x^{-3}, & \\text { if } x>2\\end{cases}\\). [Mar. ’19 (TS): Mar. ’17 (AP): May ’15 (TS), ’08]
\nSolution:
\n
\n<\/p>\n
\nCheck the continuity of f given by f(x) = \\(f(x)= \\begin{cases}\\frac{x^2-9}{x^2-2 x-3} & \\text { if } 0<x<5 \\text { and } x \\neq 3 \\\\ 1.5 & \\text { if } x=3\\end{cases}\\) at the point 3. [Mar. ’15 (AP), ’14, ’13, ’02; May ’04]
\nSolution:
\n<\/p>\n
\nProve that the functions sin x and cos x are continuous on R. [May ’08]
\nSolution:
\n(i) Let f(x) = sin x and a \u2208 R
\n\\(\\lim _{x \\rightarrow a} f(x)=\\lim _{x \\rightarrow a} \\sin x\\) = sin a = f(a)
\n\u2234 \\(\\lim _{x \\rightarrow a} f(x)\\) = f(a)
\n\u2234 f is continuous at x = a
\n\u2234 Since a is arbitrary, f is continuous on R.
\n(ii) Let g(x) = cos x and a \u2208 R
\n\\(\\lim _{x \\rightarrow a} g(x)=\\lim _{x \\rightarrow a} \\cos x\\) = cos a = g(a)
\n\u2234 \\(\\lim _{x \\rightarrow a} g(x)\\) = g(a)
\n\u2234 g is continuous at x = a
\n\u2234 since a is arbitrary, g is continuous on R.<\/p>\n
\nFind real constants a, b so that the function f is given by f(x) = \\(\\begin{cases}\\sin x & \\text { if } x \\leq 0 \\\\ x^2+\\mathbf{a} & \\text { if } 0<x<1 \\\\ \\mathbf{b x}+3 & \\text { if } 1 \\leq x \\leq 3 \\\\ -3 & \\text { if } x>3\\end{cases}\\) is continuous on R. [Mar. ’18 (AP & TS); May ’13]
\nSolution:
\nGiven, f(x) = \\(\\begin{cases}\\sin x & \\text { if } x \\leq 0 \\\\ x^2+\\mathbf{a} & \\text { if } 0<x<1 \\\\ \\mathbf{b x}+3 & \\text { if } 1 \\leq x \\leq 3 \\\\ -3 & \\text { if } x>3\\end{cases}\\)
\nSince f is continuous on R.
\nf is continuous at 0, 3.
\n
\nSince f is continuous at x = 3 then
\nLHL = RHL
\n3b + 3 = -3
\n3b = -3 – 3
\nb = -2
\n\u2234 a = 0, b = -2<\/p>\n
\nShow that f(x) = \\(\\left\\{\\begin{array}{cl}
\n\\frac{\\cos a x-\\cos b x}{x^2} & \\text { if } x \\neq 0 \\\\
\n\\frac{1}{2}\\left(b^2-a^2\\right) & \\text { if } x=0
\n\\end{array}\\right.\\) where a and b are real constants, is continuous at ‘0’. [Mar. ’17 (TS), ’13(old), ’09; May ’14; B.P.]
\nSolution:
\n
\n<\/p>\nSome More Maths 1B Limits and Continuity Important Questions Short Answer Type<\/h3>\n
\nFind \\(\\lim _{x \\rightarrow 3} \\frac{x^3-6 x^2+x}{x^2-9}\\)
\nSolution:
\n<\/p>\n
\nFind \\(\\lim _{x \\rightarrow 3} \\frac{x^3-3 x^2}{x^2-5 x+6}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 3} \\frac{x^4-81}{2 x^2-5 x-3}\\)
\nSolution:
\nGiven, \\(\\lim _{x \\rightarrow 3} \\frac{x^4-81}{2 x^2-5 x-3}\\)
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 3} \\frac{x^2-8 x+15}{x^2-9}\\). [Mar. ’16 (AP & TS)]
\nSolution:
\n<\/p>\n
\nIf f(x) = \\(-\\sqrt{25-x^2}\\) then find \\(\\lim _{x \\rightarrow 1} \\frac{f(x)-f(1)}{x-1}\\)
\nSolution:
\n
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{\\sin a x}{\\sin b x}\\), b \u2260 0, a \u2260 b. [Mar. ’18 (TS)]
\nSolution:
\nGiven, \\(\\lim _{x \\rightarrow 0} \\frac{\\sin a x}{\\sin b x}\\)
\nNow dividing the numerator and denominator by x, we get
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{e^{3 x}-1}{x}\\). [Mar. ’18 (AP); May ’15 (TS)]
\nSolution:
\n<\/p>\n
\nEvaluate \\(\\lim _{x \\rightarrow 1} \\frac{\\log _e x}{x-1}\\).
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 3} \\frac{e^x-e^3}{x-3}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{e^{\\sin x}-1}{x}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 1} \\frac{(2 x-1)(\\sqrt{x}-1)}{2 x^2+x-3}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{\\log _e(1+5 x)}{x}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{(1+x)^{\\frac{1}{8}}-(1-x)^{\\frac{1}{8}}}{x}\\)
\nSolution:
\n
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{1-\\cos x}{x}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{\\sec x-1}{x^2}\\)
\nSolution:
\n
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{1-\\cos m x}{1-\\cos n x}\\), n \u2260 0.
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{x\\left(e^x-1\\right)}{1-\\cos x}\\). [May ’14]
\nSolution:
\nGiven, \\(\\lim _{x \\rightarrow 0} \\frac{x\\left(e^x-1\\right)}{1-\\cos x}\\)
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{\\log \\left(1+x^3\\right)}{\\sin ^3 x}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 0} \\frac{x \\tan 2 x-2 x \\tan x}{(1-\\cos 2 x)^2}\\)
\nSolution:
\nGiven, \\(\\lim _{x \\rightarrow 0} \\frac{x \\tan 2 x-2 x \\tan x}{(1-\\cos 2 x)^2}\\)
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow \\infty} \\frac{x^2-\\sin x}{x^2-2}\\). [May ’16 (AP)]
\nSolution:
\n
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 3} \\frac{x^2+3 x+2}{x^2-6 x+9}\\). [Mar. ’19 (AP)]
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow \\infty} \\frac{3 x^2+4 x+5}{2 x^3+3 x-7}\\). [May ’15 (AP)]
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow \\infty} \\frac{6 x^2-x+7}{x+3}\\).
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow \\infty} \\frac{x^2+5 x+2}{2 x^2-5 x+1}\\). [May ’14; Mar. ’17 (AP)]
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow 2}\\left[\\frac{1}{x-2}-\\frac{4}{x^2-4}\\right]\\)
\nSolution:
\n
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow-\\infty} \\frac{5 x^3+4}{\\sqrt{2 x^4+1}}\\)
\nSolution:
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow \\infty} \\frac{2+\\cos ^2 x}{x+2007}\\)
\nSolution:
\nGiven, \\(\\lim _{x \\rightarrow \\infty} \\frac{2+\\cos ^2 x}{x+2007}\\)
\nWe know that
\n-1 \u2264 cos x \u2264 1
\n0 \u2264 cos2<\/sup>x \u2264 1
\n2 + 0 \u2264 2 + cos2<\/sup>x \u2264 2 + 1
\n2 \u2264 2 + cos2<\/sup>x \u2264 3
\n\\(\\frac{2}{x+2007} \\leq \\frac{2+\\cos ^2 x}{x+2007} \\leq \\frac{3}{x+2007}\\)
\n<\/p>\n
\nCompute \\(\\lim _{x \\rightarrow-\\infty} \\frac{6 x^2-\\cos 3 x}{x^2+5}\\)
\nSolution:
\nWe know that
\n-1 \u2264 cos x \u2264 1
\n-1 \u2265 cos 3x \u2265 1
\n1 \u2265 -cos 3x \u2265 -1
\n-1 \u2264 -cos 3x \u2264 1
\n<\/p>\n
\nShow that f, given by f(x) = \\(\\frac{\\mathbf{x}-|\\mathbf{x}|}{\\mathbf{x}}\\) (x \u2260 0), is continuous on R – {0}.
\nSolution:
\n
\n\u2234 f is discontinuous at x = 0.
\n\u2234 Hence f is continuous on R – {0}.<\/p>\n
\nIf f is a function defined by f(x) = \\(\\begin{cases}\\frac{x-1}{\\sqrt{x}-1} & \\text { if } x>1 \\\\ 5-3 x & \\text { if }-2 \\leq x \\leq 1 \\\\ \\frac{6}{x-10} & \\text { if } x<-2\\end{cases}\\) then discuss the continuity of ‘f’.
\nSolution:
\n
\n<\/p>\n
\nIf f, given by f(x) = \\(\\begin{cases}\\mathbf{k}^2 x-k & \\text { if } \\mathbf{k} \\geq 1 \\\\ 2 & \\text { if } x<1\\end{cases}\\) is a continuous function on R, then find the values of k. [Mar. ’15 (TS)]
\nSolution:
\nGiven, \\(\\begin{cases}\\mathbf{k}^2 x-k & \\text { if } \\mathbf{k} \\geq 1 \\\\ 2 & \\text { if } x<1\\end{cases}\\)
\n\u2234 f is continuous on R
\n\u2234 f is continuous at x = 1
\nat x = 1, LHL = RHL = f(1) ………(1)
\n
\nFrom (1), LHL = RHL
\n\u21d2 2 = k2<\/sup> – k
\n\u21d2 k2<\/sup> – k – 2 = 0
\n\u21d2 k2<\/sup> – 2k + k – 2 = 0
\n\u21d2 k(k – 2) + 1(k – 2) = 0
\n\u21d2 (k – 2)(k + 1) = 0
\n\u21d2 k – 2 = 0 (or) k + 1 = 0
\n\u21d2 k = 2 (or) k = -1<\/p>\n
\nCheck the continuity of ‘f’ given by f(x) = \\(\\left\\{\\begin{array}{rlr}
\n4-x^2, & \\text { if } & x \\leq 0 \\\\
\n\\mathbf{x}-5, & \\text { if } & 0 4 x^2-9, & \\text { if } & 1<x<2 \\\\
\n3 x+4, & \\text { if } & x \\geq 2
\n\\end{array}\\right.\\) at points x = 0, 1, 2. [Mar. ’16 (TS)]
\nSolution:
\n
\n<\/p>\n","protected":false},"excerpt":{"rendered":"