{"id":32246,"date":"2023-01-05T10:06:57","date_gmt":"2023-01-05T04:36:57","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=32246"},"modified":"2023-01-05T10:06:57","modified_gmt":"2023-01-05T04:36:57","slug":"ap-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6d-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-maths-2b-solutions-chapter-6-ex-6d-in-telugu\/","title":{"rendered":"AP Inter 2nd Year Maths 2B Solutions Chapter 6 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 Ex 6(d)"},"content":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Chapter 6 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 Exercise 6(d) will help students to clear their doubts quickly.<\/p>\n

AP Inter 2nd Year Maths 2B Solutions Chapter 6 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 Exercise 6(d)<\/h2>\n

\u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 6(\u0c21\u0c3f)<\/span><\/p>\n

I. \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c28\u0c41 \u0c38\u0c3e\u0c27\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u222b\\(\\frac{1}{\\sqrt{2 x-3 x^2+1}}\\)dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u222b\\(\\frac{\\sin \\theta}{\\sqrt{2-\\cos ^2 \\theta}}\\) d\u03b8
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u222b\\(\\frac{\\cos x}{\\sin ^2 x+4 \\sin x+5}\\) dx (Mar. 07)
\n\u0c38\u0c3e\u0c27\u0c28:
\nt = sin x \u21d2 dt = cos x dx
\nI = \u222b\\(\\frac{d t}{t^2+4 t+5}\\) = \u222b\\(\\frac{d t}{(t+2)^2+1}\\)
\n= tan-1<\/sup>(t + 2) + C
\n= tan-1<\/sup>(sin x + 2) + C<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\u222b\\(\\frac{d x}{1+\\cos ^2 x}\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\u222b\\(\\frac{d x}{2 \\sin ^2 x+3 \\cos ^2 x}\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u222b\\(\\frac{d x}{2 \\sin ^2 x+3 \\cos ^2 x}\\) = \u222b\\(\\frac{\\sec ^2 x d x}{2 \\tan ^2 x+3}\\)
\nt = tan x \u21d2 dt = sec2<\/sup>x dx
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\u222b\\(\\frac{1}{1+\\tan x}\\) dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n\u222b\\(\\frac{1}{1-\\cot x}\\) dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

II. \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c28\u0c41 \u0c38\u0c3e\u0c27\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u222b\\(\\sqrt{1+3 x-x^2}\\) dx (May ’11)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u222b\\(\\sqrt{1+3 x-x^2}\\) = \u222b\\(\\sqrt{1-\\left(x^2-3 x\\right)}\\) dx
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u222b\\(\\frac{9 \\cos x-\\sin x}{4 \\sin x+5 \\cos x} d x\\) (Mar. 08)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u222b\\(\\frac{2 \\cos x+3 \\sin x}{4 \\cos x+5 \\sin x} d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n2 cos x + 3 sin x = A(4 cos x + 5 sin x) + B(-4 sin x + 5 c\u00f3s x) \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c02\u0c21\u0c3f.
\nsin x, cos x \u0c17\u0c41\u0c23\u0c15\u0c3e\u0c32\u0c28\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c02\u0c21\u0c3f
\n4A + 5B = 2
\n5A – 4B = 3
\n\"AP
\n= \\(\\frac{23}{41}\\)x – \\(\\frac{2}{41}\\). log |4 cos x + 5 sin x| + C<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\u222b\\(\\frac{1}{1+\\sin x+\\cos x}\\)dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\u222b\\(\\frac{1}{3 x^2+x+1}\\)dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\u222b\\(\\frac{d x}{\\sqrt{5-2 x^2+4 x}}\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n5 – 2x2<\/sup> + 4x
\n\"AP<\/p>\n

III. \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c15\u0c28\u0c41\u0c32\u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\u222b\\(\\frac{x+1}{\\sqrt{x^2-x+1}}\\) dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\u222b(6x + 5)\\(\\sqrt{6-2 x^2+x}\\) dx (May 06)
\n\u0c38\u0c3e\u0c27\u0c28:
\n6x + 5 = A(1 – 4x) + B \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c02\u0c21\u0c3f
\nx \u0c17\u0c41\u0c23\u0c15\u0c3e\u0c32\u0c28\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c02\u0c21\u0c3f
\n6 = -4A \u21d2 A = \\(\\frac{-3}{2}\\)
\n\u0c38\u0c4d\u0c25\u0c3f\u0c30\u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c02\u0c21\u0c3f.
\nA + B = 5
\nB = 5 – A = 5 + \\(\\frac{3}{2}\\) = \\(\\frac{13}{2}\\)
\n\"AP
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\u222b\\(\\frac{d x}{4+5 \\sin x}\\) (Mar. 05)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\u222b\\(\\frac{1}{2-3 \\cos 2 x}\\) dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nt = tan x \u21d2 dt = sec2<\/sup> x dx
\n= (1 + tan2<\/sup>x) dx
\n= (1 + t2<\/sup>) dx
\ndx = \\(\\frac{d t}{1+t^2}\\)
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\u222bx \\(\\sqrt{1+x-x^2}\\) dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = A(1 – 2x) + B \u0c05\u0c28\u0c41\u0c15\u0c41\u0c02\u0c26\u0c3e\u0c02
\nx \u0c17\u0c41\u0c23\u0c15\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c02\u0c21\u0c3f
\n1 = -2A \u21d2 A = –\\(\\frac{1}{2}\\)
\n\u0c38\u0c4d\u0c25\u0c3f\u0c30\u0c2a\u0c26\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c02\u0c21\u0c3f
\n0 = A + B \u21d2 B = -A = \\(\\frac{1}{2}\\)
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\u222b\\(\\frac{d x}{(1+x) \\sqrt{3+2 x-x^2}}\\) (May 05)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n\u222b\\(\\frac{d x}{4 \\cos x+3 \\sin x}\\) (Mar. 06)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n\u222b\\(\\frac{1}{\\sin x+\\sqrt{3} \\cos x}\\) dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 9.
\n\u222b\\(\\frac{d x}{5+4 \\cos 2 x}\\) (Mar. 11)
\n\u0c38\u0c3e\u0c27\u0c28:
\nt = tan x \u21d2 dt = sec2<\/sup>x dx
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 10.
\n\u222b\\(\\frac{2 \\sin x+3 \\cos x+4}{3 \\sin x+4 \\cos x+5}\\) dx (Mar. 11) (A.P.Mar. 17) (A.P.Mar. 16) (T.S.Mar. 16)
\n\u0c38\u0c3e\u0c27\u0c28:
\n2 sin x + 3 cos x + 4 = A(3 sin x + 4 cos x + 5) + B(3 cos x – 4 sin x) + C \u0c05\u0c28\u0c41\u0c15\u0c4b\u0c66\u0c21\u0c3f.
\nsin x, \u0c17\u0c41\u0c23\u0c15\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c41 3A – 4B = 2
\ncos x, \u0c17\u0c41\u0c23\u0c15\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c41 4A + 3B = 2
\n\"AP
\n\"AP
\n(1) \u0c32\u0c4b \u0c2a\u0c4d\u0c30\u0c24\u0c3f\u0c15\u0c4d\u0c37\u0c47\u0c2a\u0c3f\u0c02\u0c1a\u0c17\u0c3e
\nI = \\(\\frac{18}{25}\\) . x + \\(\\frac{1}{25}\\) log |3 sin x + 4 cos x + 5| – \\(\\frac{4}{5\\left(3+\\tan \\frac{x}{2}\\right)}\\) + C<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 11.
\n\u222b\\(\\sqrt{\\frac{5-x}{x-2}}\\)dx, x \u2208 (2, 5)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\u0c38\u0c1c\u0c3e\u0c24\u0c40\u0c2f \u0c2a\u0c26\u0c3e\u0c32 \u0c17\u0c41\u0c23\u0c15\u0c3e\u0c32\u0c41 \u0c38\u0c2e\u0c3e\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e
\n-2A = -1 \u21d2 A = \\(\\frac{1}{2}\\)
\n7A + B = 5
\n\"AP
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 12.
\n\u222b\\(\\sqrt{\\frac{1+x}{1-x}}\\) dx, x \u2208 (-1, 1).
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 13.
\n\u222b\\(\\frac{d x}{(1+x) \\sqrt{3+2 x-x^2}}\\), x \u2208 (-1, 3).
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 14.
\n\u222b\\(\\frac{d x}{(x+2) \\sqrt{x+1}}\\), x \u2208 (-1, \u221e).
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 15.
\n\u222b\\(\\frac{d x}{(2 x+3) \\sqrt{x+2}}\\), x \u2208 I \u2282 (-2, \u221e) \\ \\(\\left\\{\\frac{-3}{2}\\right\\}\\).
\n\u0c38\u0c3e\u0c27\u0c28:
\nx + 2 = t2<\/sup> \u21d2 dx = 2t dt \u0c2e\u0c30\u0c3f\u0c2f\u0c41
\n2x + 3 = 2(t2<\/sup> – 2) + 3 = 2t2<\/sup> – 1
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 16.
\n\u222b\\(\\frac{1}{(1+\\sqrt{x}) \\sqrt{x-x^2}}\\) dx, x \u2208 (0, 1).
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u222b\\(\\frac{1}{(1+\\sqrt{x}) \\sqrt{x-x^2}}\\) dx
\n\u0c2a\u0c4d\u0c30\u0c24\u0c3f\u0c15\u0c4d\u0c37\u0c47\u0c2a\u0c3f\u0c02\u0c1a\u0c17\u0c3e x = t2<\/sup> \u21d2 dx = 2t dt
\n\"AP
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 17.
\n\u222b\\(\\frac{d x}{(x+1) \\sqrt{2 x^2+3 x+1}}\\), x \u2208 I \u2282 R \\ [-1, \\(\\frac{-1}{2}\\)]
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 18.
\n\u222b\\(\\sqrt{e^x-4}\\) dx, x \u2208 [loge<\/sub> 4, \u221e)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 19.
\n\"AP
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 20.
\n\u222b\\(\\frac{d x}{1+x^4}\\), x \u2208 R.
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP
\n\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2B Textbook Solutions Chapter 6 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 Exercise 6(d) will help students to clear their doubts quickly. AP Inter 2nd Year Maths 2B Solutions Chapter 6 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 Exercise 6(d) \u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 6(\u0c21\u0c3f) I. \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c28\u0c41 \u0c38\u0c3e\u0c27\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u222bdx \u0c38\u0c3e\u0c27\u0c28: \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2. \u222b d\u03b8 \u0c38\u0c3e\u0c27\u0c28: \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3. \u222b dx (Mar. 07) \u0c38\u0c3e\u0c27\u0c28: … Read more<\/a><\/p>\n","protected":false},"author":4,"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\/32246"}],"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\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=32246"}],"version-history":[{"count":4,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/32246\/revisions"}],"predecessor-version":[{"id":32514,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/32246\/revisions\/32514"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=32246"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=32246"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=32246"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}