{"id":30992,"date":"2023-09-12T14:31:32","date_gmt":"2023-09-12T09:01:32","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=30992"},"modified":"2023-09-13T17:06:50","modified_gmt":"2023-09-13T11:36:50","slug":"ap-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8b-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-maths-2b-solutions-chapter-8-ex-8b-in-telugu\/","title":{"rendered":"AP Inter 2nd Year Maths 2B Solutions Chapter 8 \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41 Ex 8(b)"},"content":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Chapter 8 \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41 Exercise 8(b) will help students to clear their doubts quickly.<\/p>\n

AP Inter 2nd Year Maths 2B Solutions Chapter 8 \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41 Exercise 8(b)<\/h2>\n

\u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 – 8(\u0c2c\u0c3f)<\/span><\/p>\n

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

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\\(\\sqrt{1-x^2}\\) dy + \\(\\sqrt{1-y^2}\\) dx = 0 \u0c38\u0c3e\u0c27\u0c3e\u0c30\u0c23 \u0c38\u0c3e\u0c27\u0c28 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28:
\n\u0c26\u0c24\u0c4d\u0c24 \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02
\n\"AP
\nsin-1<\/sup> y = -sin-1<\/sup> x + c
\n\u0c38\u0c3e\u0c27\u0c28 sin-1<\/sup> x + sin-1<\/sup> y = c, c \u0c38\u0c4d\u0c25\u0c3f\u0c30\u0c30\u0c3e\u0c36\u0c3f<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{2 y}{x}\\)\u0c15\u0c41 \u0c38\u0c3e\u0c27\u0c3e\u0c30\u0c23 \u0c38\u0c3e\u0c27\u0c28 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f.
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{2 y}{x}\\)
\n\\(\\int \\frac{d y}{y}\\) = \\(2 \\int \\frac{d x}{x}\\)
\nlog c + log y = 2 log x
\nlog cy = log x2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28 cy = x2<\/sup>, \u0c38\u0c4d\u0c25\u0c3f\u0c30\u0c3e\u0c02\u0c15\u0c2e\u0c41<\/p>\n

II. \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c28\u0c41 \u0c38\u0c3e\u0c27\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{1+y^2}{1+x^2}\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{1+y^2}{1+x^2}\\)
\n\\(\\int \\frac{d y}{1+y^2}\\) = \\(\\int \\frac{d x}{1+x^2}\\)
\ntan-1<\/sup> y = tan-1<\/sup> x + tan-1<\/sup> c, c \u0c38\u0c4d\u0c25\u0c3f\u0c30\u0c3e\u0c02\u0c15\u0c02<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\\(\\frac{d y}{d x}\\) = ey-x<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{e^y}{e^x}\\)
\n\\(\\frac{d y}{e^y}\\) = \\(\\frac{d x}{e^x}\\)
\n\\(\\int \\frac{d y}{1+y^2}\\) = \\(\\int \\frac{d x}{1+x^2}\\)
\ntan-1<\/sup> y = tan-1<\/sup> x + tan-1<\/sup> c, c \u0c38\u0c4d\u0c25\u0c3f\u0c30\u0c3e\u0c02\u0c15\u0c02<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n(ex<\/sup> + 1) y dy + (y + 1) dx = 0
\n\u0c38\u0c3e\u0c27\u0c28:
\n(ex<\/sup> + 1)y. dy = -(y + 1) dx
\n\\(\\frac{y d y}{y+1}\\) = –\\(\\frac{d x}{e^x+1}\\)
\n\\(\\int\\left(1-\\frac{1}{y+1}\\right) d y\\) = \\(\\int-\\frac{e^{-x} d x}{e^{-x}+1}\\)
\ny – log (y + 1) = log(e-x<\/sup> + 1) + log c
\n\u21d2 y – log (y + 1) = log (e-x<\/sup> + 1)
\n\u21d2 y = log (y + 1) + log c (e-x<\/sup> + 1)
\ny = log c (y + 1) (e-x<\/sup> + 1)
\n\u0c38\u0c3e\u0c27\u0c28
\ney<\/sup> = c(y + 1) (e-x<\/sup> + 1)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\\(\\frac{d y}{d x}\\) = ex-y<\/sup> + x2<\/sup> e-y<\/sup> (Mar. ’06; May ’05)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = ex – y<\/sup> + x2<\/sup> . e-y<\/sup>
\n= \\(\\frac{e^x}{e^y}+\\frac{x^2}{e^y}\\)
\n\\(\\int e^y \\cdot d y\\) = \\(\\int\\left(e^x+x^2\\right) d x\\)
\n\u0c38\u0c3e\u0c27\u0c28
\ney<\/sup> = ex<\/sup> + \\(\\frac{x^3}{3}\\) + c<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\ntan y dx + tan x dy = 0
\n\u0c38\u0c3e\u0c27\u0c28:
\ntan y dx = – tan x dy
\n\\(\\frac{d x}{\\tan x}\\) = \\(\\frac{-d y}{\\tan y}\\)
\n\\(\\frac{\\cos x}{\\sin x} d x\\) = \\(-\\frac{\\cos y}{\\sin y} d y\\)
\nlog sin x = -log sin y + log c
\nlog sin x + log sin y = log c
\nlog (sin x. sin y) = log c
\n\u21d2sin x. sin y = c \u0c05\u0c28\u0c47\u0c26\u0c3f \u0c38\u0c3e\u0c27\u0c28<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\\(\\sqrt{1+x^2}\\) dx + \\(\\sqrt{1+y^2}\\) dy = 0
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\sqrt{1+x^2}\\) dx + \\(\\sqrt{1+y^2}\\) dy
\n\u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\ny – x\\(\\frac{d y}{d x}\\) = 5(y2<\/sup> + \\(\\frac{d y}{d x}\\))
\n\u0c38\u0c3e\u0c27\u0c28:
\ny – 5y2<\/sup> = (x + 5)\\(\\frac{d y}{d x}\\)
\n\\(\\frac{d x}{x+5}\\) = \\(\\frac{d y}{y(1-5 y)}\\)
\n‘\u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{x y+y}{x y+x}\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

III. \u0c15\u0c3f\u0c02\u0c26\u0c3f \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c2e\u0c41\u0c32\u0c28\u0c41 \u0c38\u0c3e\u0c27\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\\(\\frac{d y}{d x}\\) = \\(\\frac{1+y^2}{\\left(1+x^2\\right) x y}\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\nlog (1 + y2<\/sup>)= log x2<\/sup> – log (1 + x2<\/sup>) + log c
\nlog (1 + x2<\/sup>) + log (1 + y2<\/sup>) = log x2<\/sup> + log c
\n\u0c38\u0c3e\u0c27\u0c28 (1 + x2<\/sup>) (1 + y2<\/sup>) = cx2<\/sup><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\\(\\frac{d y}{d x}\\) + x2<\/sup> = x2<\/sup> . e3y<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{d y}{d x}\\) + x2<\/sup> = x2<\/sup> . e3y<\/sup>
\n\\(\\frac{d y}{d x}\\) = x2<\/sup>. e3y<\/sup> – x2<\/sup>
\n= x2<\/sup> (e3y<\/sup> – 1)
\n\\(\\int \\frac{d y}{e^{3 y}-1}\\) = \\(\\int x^2 d x\\)
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n(xy2<\/sup> + x) dx + (yx2<\/sup> + y) dy = 0 (Mar. ’07)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx(y2<\/sup> + 1) dx + y (x2<\/sup> + 1) dy = 0
\nx(y2<\/sup> + 1) dx + y (x2<\/sup> + 1 ) dy = 0
\n(1 + x2<\/sup>) (1 + y2<\/sup>) \u0c24\u0c4b \u0c2d\u0c3e\u0c17\u0c3f\u0c02\u0c1a\u0c17\u0c3e
\n\\(\\frac{x d x}{1+x^2}\\) + \\(\\frac{y d y}{1+y^2}\\) = 0
\n\u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e
\n\\(\\int \\frac{x d x}{1+x^2}\\) + \\(\\int \\frac{y d y}{1+y^2}\\) = 0
\n\\(\\frac{1}{2}\\)[(log (1 + x2<\/sup>) + log (1 + y2<\/sup>)] = log c
\nlog (1 + x2<\/sup>) (1 + y2<\/sup>) = 2 log c = log c2<\/sup>
\n\u0c38\u0c3e\u0c27\u0c28 (1 + x2<\/sup>) (1 + y2<\/sup>) = k. \u0c07\u0c15\u0c4d\u0c15\u0c21 k = c2<\/sup><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\\(\\frac{d \\mathbf{y}}{\\mathbf{d x}}\\) = 2y tanh x
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{d \\mathbf{y}}{\\mathbf{d x}}\\) = 2y tanh x
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{y}}\\) = 2 tanh x dx
\n\u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32\u0c3e \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e
\n\\(\\int \\frac{d y}{y}\\) = 2\\(\\int \\tanh x d x\\)
\nlog y = 2 log |cosh x | + log c
\nlny = 2ln cosh x + lnc
\ny = c cos2<\/sup>h x<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\nsin-1<\/sup> \\(\\left(\\frac{\\mathbf{d y}}{\\mathbf{d x}}\\right)\\) = x + y (May ’07)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = sin (x + y)
\nx + y = t
\n1 + \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = \\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\)
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\) – 1 = sin t
\n\\(\\frac{\\mathrm{dt}}{\\mathrm{dx}}\\) = 1 + sin t
\n\\(\\frac{d t}{1+\\sin t}\\) = dx
\n\u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32\u0c3e \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e
\n\\(\\int \\frac{d t}{1+\\sin t}[latex] = [latex]\\int \\mathrm{d} x[latex]
\n[latex]\\int \\frac{1-\\sin t}{\\cos ^2 t} d t\\) = x + c
\n\\(\\int \\sec ^2 t d t\\) – \\(\\int \\tan t \\cdot \\sec t d t\\) = x + c
\ntan t – sec t = x + c
\n\u21d2 tan (x + y) – sec (x + y) = x + c<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\\(\\frac{d y}{d x}\\) + \\(\\frac{y^2+y+1}{x^2+x+1}\\) = 0
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{-d y}{y^2+y+1}\\) = \\(\\frac{d x}{x^2+x+1}\\)
\n\u0c07\u0c30\u0c41\u0c35\u0c48\u0c2a\u0c41\u0c32\u0c3e \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c02 \u0c1a\u0c47\u0c2f\u0c17\u0c3e
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n\\(\\frac{d y}{d x}\\) = tan2<\/sup> (x + y)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{d y}{d x}\\) = tan2<\/sup> (x + y)
\nv = x + y \u0c05\u0c28\u0c41\u0c15\u0c41\u0c02\u0c26\u0c3e\u0c02
\n\\(\\frac{\\mathrm{dv}}{\\mathrm{dx}}\\) = 1 + \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\)
\n= 1 + tan2<\/sup> v = sec2<\/sup> v
\n\\(\\int \\frac{d v}{\\sec ^2 v}\\) = \\(\\int d x\\)
\n= \\(\\int \\cos ^2 v \\cdot d v\\) = x + c
\n\\(\\int \\frac{(1+\\cos 2 v)}{2} d v\\) = x + c
\n\\(\\int(1+\\cos 2 v) d v\\) = 2x + 2c
\nv + \\(\\frac{\\sin 2 v}{2}\\) = 2x + 2c
\n2v + sin 2v = 4x + c’
\n2(x + y) + sin 2(x + y) = 4x + c’
\nx – y – \\(\\frac{1}{2}\\)[2(x + y)] = c<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2B Textbook Solutions Chapter 8 \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41 Exercise 8(b) will help students to clear their doubts quickly. AP Inter 2nd Year Maths 2B Solutions Chapter 8 \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c3e\u0c32\u0c41 Exercise 8(b) \u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 – 8(\u0c2c\u0c3f) I. \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. dy + dx = 0 \u0c38\u0c3e\u0c27\u0c3e\u0c30\u0c23 \u0c38\u0c3e\u0c27\u0c28 \u0c15\u0c28\u0c41\u0c15\u0c4d\u0c15\u0c4b\u0c02\u0c21\u0c3f. \u0c38\u0c3e\u0c27\u0c28: \u0c26\u0c24\u0c4d\u0c24 \u0c05\u0c35\u0c15\u0c32\u0c28 \u0c38\u0c2e\u0c40\u0c15\u0c30\u0c23\u0c02 sin-1 y … 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\/30992"}],"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=30992"}],"version-history":[{"count":5,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/30992\/revisions"}],"predecessor-version":[{"id":31210,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/30992\/revisions\/31210"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=30992"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=30992"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=30992"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}