{"id":39710,"date":"2023-04-21T16:45:36","date_gmt":"2023-04-21T11:15:36","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=39710"},"modified":"2023-04-21T16:45:36","modified_gmt":"2023-04-21T11:15:36","slug":"ap-inter-2nd-year-maths-2a-solutions-chapter-2-ex-2a-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-maths-2a-solutions-chapter-2-ex-2a-in-telugu\/","title":{"rendered":"AP Inter 2nd Year Maths 2A Solutions Chapter 2 \u0c21\u0c3f\u0c2e\u0c4b\u0c2f\u0c30\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 Ex 2(a)"},"content":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2A Textbook Solutions<\/a> Chapter 2 \u0c21\u0c3f\u0c2e\u0c4b\u0c2f\u0c30\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 Exercise 2(a) will help students to clear their doubts quickly.<\/p>\n

AP Inter 2nd Year Maths 2A Solutions Chapter 2 \u0c21\u0c3f\u0c2e\u0c4b\u0c2f\u0c30\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 Exercise 2(a)<\/h2>\n

\u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 2(\u0c0e)<\/span><\/p>\n

I.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\nn \u0c2a\u0c42\u0c30\u0c4d\u0c23\u0c3e\u0c02\u0c15\u0c02 \u0c05\u0c2f\u0c3f\u0c24\u0c47 (1 + i)2n<\/sup> + (1 – i)2n<\/sup> = \\(2^{n+1} \\cos \\frac{n \\pi}{2}\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c35\u0c3f\u0c32\u0c41\u0c35\u0c32\u0c41 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n(i) (1 + i\u221a3)3<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28:
\n1 + i\u221a3 = r (cos \u03b8 + i sin \u03b8) \u0c05\u0c28\u0c41\u0c15\u0c41\u0c02\u0c26\u0c3e\u0c02.
\nr cos \u03b8 = 1, r sin \u03b8 = \u221a3
\nr2<\/sup>(1) = 1 + 3 = 4
\nr = 2
\n\"AP<\/p>\n

(ii) (1 – i)8<\/sup> [Mar. ’07]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

(iii) (1 + i)16<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\"AP<\/p>\n

(iv) \\(\\left(\\frac{\\sqrt{3}}{2}+\\frac{i}{2}\\right)^5-\\left(\\frac{\\sqrt{3}}{2}-\\frac{i}{2}\\right)^5\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

II.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\nx2<\/sup> – 2x + 4 = 0 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02 \u0c2e\u0c42\u0c32\u0c3e\u0c32\u0c41 \u03b1, \u03b2 \u0c32\u0c41 \u0c05\u0c2f\u0c3f\u0c24\u0c47 n \u2208 N \u0c15\u0c41 \u03b1n<\/sup> + \u03b2n<\/sup> = \\(2^{n+1} \\cos \\left(\\frac{n \\pi}{3}\\right)\\) \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f. [Mar. ’14, ’11]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\ncos \u03b1 + cos \u03b2 + cos \u03b3 = 0 = sin \u03b1 + sin \u03b2 + sin \u03b3 \u0c05\u0c2f\u0c3f\u0c24\u0c47
\n(i) cos 3\u03b1 + cos 3\u03b2 + cos 3\u03b3 = 3 cos(\u03b1 + \u03b2 + \u03b3)
\n(ii) sin 3\u03b1 + sin 3\u03b2 + sin 3\u03b3 = 3 sin(a + \u03b2 + \u03b3)
\n(iii) cos (\u03b1 + \u03b2) + cos (\u03b2 + \u03b3) + cos (\u03b3 + \u03b1) = 0 \u0c05\u0c28\u0c3f \u0c30\u0c41\u0c1c\u0c41\u0c35\u0c41 \u0c1a\u0c47\u0c2f\u0c02\u0c21\u0c3f. [May, Mar ’08]
\n\u0c38\u0c3e\u0c27\u0c28:
\ncos \u03b1 +cos \u03b2 + cos \u03b3 = 0
\nsin \u03b1 + sin \u03b2 + sin \u03b3 = 0
\n\u2234 (cos \u03b1 + cos \u03b2 + cos \u03b3) + i(sin \u03b1 + sin \u03b2 + sin \u03b3) = 0
\ni.e., (cos \u03b1 + i sin \u03b1) + (cos \u03b2 + i sin \u03b2) +(cos \u03b3 + i sin \u03b3) = 0
\na + b + c = 0 \u0c05\u0c2f\u0c3f\u0c24\u0c47 a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc
\n\u0c15\u0c3e\u0c2c\u0c1f\u0c4d\u0c1f\u0c3f (cos \u03b1 + i sin \u03b1)3<\/sup> + (cos \u03b2 + i sin \u03b2)3<\/sup> + (cos \u03b3 + i sin \u03b3)3<\/sup> = 3(cos \u03b1 + i sin \u03b1) (cos \u03b2 + i sin \u03b2) + (cos \u03b3 + i sin \u03b3)
\ni.e., cos 3\u03b1 + i sin 3\u03b1 + cos 3\u03b2 + i sin 3\u03b2 + cos 3\u03b3 + i sin 3\u03b3 = 3[cos(\u03b1 + \u03b2 + \u03b3) + i sin (\u03b1 + \u03b2 + \u03b3)] – (1)
\n\u0c35\u0c3e\u0c38\u0c4d\u0c24\u0c35, \u0c38\u0c02\u0c15\u0c40\u0c30\u0c4d\u0c23 \u0c2d\u0c3e\u0c17\u0c3e\u0c32\u0c28\u0c41 \u0c2a\u0c4b\u0c32\u0c4d\u0c1a\u0c17\u0c3e,
\n(i) cos 3\u03b1 + cos 3\u03b2 + cos 3\u03b3 = 3 cos(\u03b1 + \u03b2 + \u03b3)
\n(ii) sin 3\u03b1 + sin 3\u03b2 + sin 3\u03b3 = 3 sin(\u03b1 + \u03b2 + \u03b3)
\n(iii) a = cos \u03b1 + i sin \u03b1 \u0c05\u0c2f\u0c3f\u0c24\u0c47 \\(\\frac{1}{a}\\) = cos \u03b1 – i sin \u03b1
\nb = cos \u03b2 + i sin \u03b2 \u0c05\u0c2f\u0c3f\u0c24\u0c47 \\(\\frac{1}{b}\\) = cos \u03b2 – i sin \u03b2
\nc = cos \u03b3 + i sin \u03b3 \u0c05\u0c2f\u0c3f\u0c24\u0c47 \\(\\frac{1}{c}\\) = cos \u03b3 – i sin \u03b3
\n\u0c07\u0c2a\u0c4d\u0c2a\u0c41\u0c21\u0c41 \\(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\) = (cos \u03b1 + cos \u03b2 + cos \u03b3) – i(sin \u03b1 + sin \u03b2 + sin \u03b3)
\n= 0 – i(0)
\n= 0
\n\u21d2 \\(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=0\\)
\n\u21d2 bc + ca + ab = 0
\n\u21d2 [cos(\u03b2 + \u03b3)+ i sin(\u03b2 + \u03b3)] + [cos(\u03b3 + \u03b1) + i sin(\u03b3 + \u03b1)] + [cos(\u03b1 + \u03b2) + i sin(\u03b1 + \u03b2)] = 0
\n\u21d2 [cos(\u03b1 + \u03b2) + cos(\u03b2 + \u03b3) + cos(\u03b3 + \u03b1)] + i[sin(\u03b1 + \u03b2) sin(\u03b2 + \u03b3) + sin(\u03b3 + \u03b1)] = 0
\n\u0c35\u0c3e\u0c38\u0c4d\u0c24\u0c35 \u0c2d\u0c3e\u0c17\u0c3e\u0c32\u0c28\u0c41 \u0c2a\u0c4b\u0c32\u0c4d\u0c1a\u0c17\u0c3e,
\ncos(\u03b1 + \u03b2) + cos(\u03b2 + \u03b3) + cos(\u03b3 + \u03b1) = 0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\nn \u0c2a\u0c42\u0c30\u0c4d\u0c23\u0c3e\u0c02\u0c15\u0c02, z = cis \u03b8, (\u03b8 \u2260 (2n + 1) \u03c0\/2) \u0c05\u0c2f\u0c3f\u0c24\u0c47 \\(\\frac{z^{2 n}-1}{z^{2 n}+1}\\) = i tan n\u03b8 \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28:
\nz = cis \u03b8 = cos \u03b8 + i sin \u03b8, \u03b8 \u2260 (2n + 1)\u03c0\/2
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n(1 + x)n<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + …… an<\/sub>xn<\/sup>, n \u2208 N \u0c05\u0c2f\u0c3f\u0c24\u0c47
\n(i) a0<\/sub> – a2<\/sub> + a4<\/sub> – a6<\/sub> + ……. = \\(2^{n \/ 2} \\cos \\left[\\frac{n \\pi}{4}\\right]\\)
\n(ii) a1<\/sub> – a3<\/sub> + a5<\/sub> – a7<\/sub> + ….. = \\(2^{n \/ 2} \\sin \\left[\\frac{n \\pi}{4}\\right]\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u0c07\u0c1a\u0c4d\u0c1a\u0c3f\u0c28\u0c26\u0c3f (1 + x)n<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + …… an<\/sub>xn<\/sup>
\nx = i \u0c05\u0c2f\u0c3f\u0c24\u0c47
\n\"AP
\n\u0c35\u0c3e\u0c38\u0c4d\u0c24\u0c35 \u0c2e\u0c30\u0c3f\u0c2f\u0c41 \u0c15\u0c32\u0c4d\u0c2a\u0c3f\u0c24 \u0c2d\u0c3e\u0c17\u0c2e\u0c41\u0c32\u0c28\u0c41 \u0c2a\u0c4b\u0c32\u0c4d\u0c1a\u0c17\u0c3e,
\n(i) a0<\/sub> – a2<\/sub> + a4<\/sub> – a6<\/sub> + ……. = \\(2^{n \/ 2} \\cos \\left[\\frac{n \\pi}{4}\\right]\\)
\n(ii) a1<\/sub> – a3<\/sub> + a5<\/sub> – a7<\/sub> + ….. = \\(2^{n \/ 2} \\sin \\left[\\frac{n \\pi}{4}\\right]\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2A Textbook Solutions Chapter 2 \u0c21\u0c3f\u0c2e\u0c4b\u0c2f\u0c30\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 Exercise 2(a) will help students to clear their doubts quickly. AP Inter 2nd Year Maths 2A Solutions Chapter 2 \u0c21\u0c3f\u0c2e\u0c4b\u0c2f\u0c30\u0c4d \u0c38\u0c3f\u0c26\u0c4d\u0c27\u0c3e\u0c02\u0c24\u0c02 Exercise 2(a) \u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 2(\u0c0e) I. \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. n \u0c2a\u0c42\u0c30\u0c4d\u0c23\u0c3e\u0c02\u0c15\u0c02 \u0c05\u0c2f\u0c3f\u0c24\u0c47 (1 + i)2n + (1 – i)2n = \u0c05\u0c28\u0c3f \u0c1a\u0c42\u0c2a\u0c02\u0c21\u0c3f. \u0c38\u0c3e\u0c27\u0c28: \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/39710"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=39710"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/39710\/revisions"}],"predecessor-version":[{"id":39723,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/39710\/revisions\/39723"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=39710"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=39710"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=39710"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}