{"id":11049,"date":"2024-03-08T10:14:10","date_gmt":"2024-03-08T04:44:10","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=11049"},"modified":"2024-03-11T17:52:26","modified_gmt":"2024-03-11T12:22:26","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-6-ex-6c","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-6-ex-6c\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(c)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions<\/a> Chapter 6 Binomial Theorem Ex 6(c) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(c)<\/h2>\n

I.
\nQuestion 1.
\nFind an approximate value of the following corrected to 4 decimal places.
\ni) \\(\\sqrt[5]{242}\\)
\nii) \\(\\sqrt[7]{127}\\)
\niii) \\(\\sqrt[5]{32.16}\\)
\niv) \\(\\sqrt{199}\\)
\nv) \\(\\sqrt[3]{1002}-\\sqrt[3]{998}\\)
\nvi) \\((1.02)^{3 \/ 2}-(0.98)^{3 \/ 2}\\)
\nSolution:
\ni) \\(\\sqrt[5]{242}\\) = (243 – 1)\\(\\frac{1}{5}\\)<\/sup>
\n= (243)\\(\\frac{1}{5}\\)<\/sup> (1 – \\(\\frac{1}{243}\\))\\(\\frac{1}{5}\\)<\/sup>
\n= 3 \\(\\left[1-\\frac{1}{5} \\cdot \\frac{1}{24.3}+\\frac{\\frac{1}{5}\\left(\\frac{1}{5}-1\\right)}{2 !}\\left(\\frac{1}{243}\\right)^2-\\ldots .\\right]\\)
\n= 3 [1 – 0.000823 + ……………]
\n= 3 (0.999177)
\n\u21d2 \\(\\sqrt[5]{242}\\) = 2.997531.<\/p>\n

ii) \\(\\sqrt[7]{127}\\)<\/p>\n

\"TS<\/p>\n

= 2 (1 – 0.0011161 + ……………)
\n= 2 (0.99888) = 1.9977.<\/p>\n

iii) \\(\\sqrt[5]{32.16}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

iv) \\(\\sqrt{199}\\)
\n= (196 + 3)1\/2<\/sup>
\n= (196)1\/2<\/sup> (1 + \\(\\frac{3}{196}\\))1\/2<\/sup>
\n= 14 (1 + 0.0153)1\/2<\/sup>
\n= 14 [1 + \\(\\frac{0.0153}{2}\\) + \\(\\frac{\\frac{1}{2}\\left(\\frac{1}{2}-1\\right)}{2 !}(0.0153)^2\\) + ……………..]
\n= 14 [1 + 0.00765]
\n= 14 (1.00765) = 14.1071.<\/p>\n

v) \\(\\sqrt[3]{1002}-\\sqrt[3]{998}\\)<\/p>\n

\"TS<\/p>\n

vi) \\((1.02)^{3 \/ 2}-(0.98)^{3 \/ 2}\\)
\n= (1 + 0.02)3\/2<\/sup> – (1 – 0.02)3\/2<\/sup><\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nIf |x| is so small that x2<\/sup> and higher powers of x may be neglected, then find approximate values of the following.
\ni) \\(\\frac{(4+3 x)^{\\frac{1}{2}}}{(3-2 x)^2}\\)
\nii) \\(\\frac{\\left(1-\\frac{2 x}{3}\\right)^{\\frac{3}{2}}(32+5 x)^{\\frac{1}{5}}}{(3-x)^3}\\)
\niii) \\(\\sqrt{4-x}\\left(3-\\frac{x}{2}\\right)^{-1}\\)
\niv) \\(\\frac{\\sqrt{4+x}+\\sqrt[3]{8+x}}{(1+2 x)+(1-2 x)^{\\frac{-1}{3}}}\\)
\nv) \\(\\frac{(8+3 x)^{\\frac{2}{3}}}{(2+3 x) \\sqrt{4-5 x}}\\)
\nSolution:
\ni) \\(\\frac{(4+3 x)^{\\frac{1}{2}}}{(3-2 x)^2}\\)<\/p>\n

\"TS<\/p>\n

ii) \\(\\frac{\\left(1-\\frac{2 x}{3}\\right)^{\\frac{3}{2}}(32+5 x)^{\\frac{1}{5}}}{(3-x)^3}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

iii) \\(\\sqrt{4-x}\\left(3-\\frac{x}{2}\\right)^{-1}\\)<\/p>\n

\"TS<\/p>\n

iv) \\(\\frac{\\sqrt{4+x}+\\sqrt[3]{8+x}}{(1+2 x)+(1-2 x)^{\\frac{-1}{3}}}\\)<\/p>\n

\"TS<\/p>\n

v) \\(\\frac{(8+3 x)^{\\frac{2}{3}}}{(2+3 x) \\sqrt{4-5 x}}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nSuppose s and t are positive and t is very small when compared to s. Then find an approximate value of \\(\\left(\\frac{s}{s+t}\\right)^{\\frac{1}{3}}-\\left(\\frac{s}{s-t}\\right)^{\\frac{1}{3}}\\).
\nSolution:
\n\\(\\left(\\frac{s}{s+t}\\right)^{\\frac{1}{3}}-\\left(\\frac{s}{s-t}\\right)^{\\frac{1}{3}}\\)<\/p>\n

\"TS<\/p>\n

Question 4.
\nSuppose p, q are positive and p is very small when compared to q. Then find an approximate value of \\(\\left(\\frac{q}{q+p}\\right)^{\\frac{1}{2}}+\\left(\\frac{q}{q-p}\\right)^{\\frac{1}{2}}\\).
\nSolution:
\n\\(\\left(\\frac{q}{q+p}\\right)^{\\frac{1}{2}}+\\left(\\frac{q}{q-p}\\right)^{\\frac{1}{2}}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\nBy neglecting x4<\/sup> and higher powers of x, find an approximate value of \\(\\sqrt[3]{x^2+64}-\\sqrt[3]{x^2+27}\\).
\nSolution:
\n\\(\\sqrt[3]{x^2+64}-\\sqrt[3]{x^2+27}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 6.
\nExpand 3\u221a3 in increasing powers of \\(\\frac{2}{3}\\).
\nSolution:
\n3\u221a3 = 3\\(\\frac{2}{3}\\)<\/sup>
\n= \\(\\left(\\frac{1}{3}\\right)^{\\frac{-3}{2}}\\)<\/p>\n

\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(c) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(c) I. Question 1. Find an approximate value of the following corrected to 4 decimal places. i) ii) iii) … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11049"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=11049"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11049\/revisions"}],"predecessor-version":[{"id":11065,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/11049\/revisions\/11065"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=11049"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=11049"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=11049"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}