{"id":31610,"date":"2023-01-03T11:21:10","date_gmt":"2023-01-03T05:51:10","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=31610"},"modified":"2023-01-03T11:29:32","modified_gmt":"2023-01-03T05:59:32","slug":"ap-inter-2nd-year-maths-2b-solutions-chapter-7-ex-7c-in-telugu","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-inter-2nd-year-maths-2b-solutions-chapter-7-ex-7c-in-telugu\/","title":{"rendered":"AP Inter 2nd Year Maths 2B Solutions Chapter 7 \u0c28\u0c3f\u0c36\u0c4d\u0c1a\u0c3f\u0c24 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c41 Ex 7(c)"},"content":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Chapter 7 \u0c28\u0c3f\u0c36\u0c4d\u0c1a\u0c3f\u0c24 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c41 Exercise 7(c) will help students to clear their doubts quickly.<\/p>\n

AP Inter 2nd Year Maths 2B Solutions Chapter 7 \u0c28\u0c3f\u0c36\u0c4d\u0c1a\u0c3f\u0c24 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c41 Exercise 7(c)<\/h2>\n

\u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 – 7 (\u0c38\u0c3f)<\/span><\/p>\n

I. \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\\(\\int_0^{\\pi \/ 2} \\sin ^{10} x d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\frac{9}{10}\\).\\(\\frac{7}{8}\\).\\(\\frac{5}{6}\\).\\(\\frac{3}{4}\\).\\(\\frac{1}{2}\\).\\(\\frac{\\pi}{2}\\) = \\(\\frac{63 \\pi}{512}\\)<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\\(\\int_0^{\\pi \/ 2} \\cos ^{11} x d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nn \u0c2c\u0c47\u0c38\u0c3f \u0c38\u0c02\u0c16\u0c4d\u0c2f
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\\(\\int_0^{\\pi \/ 2}\\) cos7<\/sup> x sin2<\/sup> x dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nI = \\(\\int_0^{\\pi \/ 2}\\) cos7<\/sup> x sin2<\/sup> x dx
\n= \\(\\int_0^{\\pi \/ 2}\\)sinm<\/sup>x.cosn<\/sup> x dx
\nm – \u0c38\u0c30\u0c3f \u0c38\u0c02\u0c16\u0c4d\u0c2f
\nn – \u0c2c\u0c47\u0c38\u0c3f \u0c38\u0c02\u0c16\u0c4d\u0c2f
\n= \\(\\frac{n-1}{m+n}\\).\\(\\frac{n-3}{m+n-2}\\)….\\(\\frac{2}{m+3}\\).\\(\\frac{1}{m+1}\\)
\n= \\(\\frac{7-1}{9}\\) \u00d7 \\(\\frac{7-3}{9}\\) \u00d7 \\(\\frac{7-5}{5}\\) \u00d7 \\(\\frac{1}{2+1}\\)
\n= \\(\\frac{6}{9}\\) . \\(\\frac{4}{7}\\) . \\(\\frac{2}{5}\\) . \\(\\frac{1}{3}\\) = \\(\\frac{16}{315}\\)<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\\(\\int_0^{\\pi \/ 2} \\sin ^4 x \\cos ^4 x d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\int_0^{\\pi \/ 2} \\sin ^m x \\cdot \\cos ^n x d x\\)
\nm, n \u0c32\u0c41 \u0c38\u0c30\u0c3f\u0c38\u0c02\u0c16\u0c4d\u0c2f\u0c32\u0c41
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\\(\\int_0^\\pi\\) sin3<\/sup> x cos6<\/sup> x dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n= \\(\\int_0^\\pi\\) sin2<\/sup> x cos6<\/sup> x dx
\n= \\(\\int_0^\\pi\\)(1 – cos2<\/sup>x) . cos6<\/sup>x. sin x dx
\ncos x = t cos \u03c0 = -1
\n-sin x dx = dt cos 0 = 1
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\\(\\int_0^{2 \\pi}\\) sin2<\/sup> x cos4<\/sup> x dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nsin2<\/sup> x . cos4<\/sup> x dx \u0c38\u0c30\u0c3f \u0c2a\u0c4d\u0c30\u0c2e\u0c47\u0c2f\u0c2e\u0c41
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n\\(\\int_{-\\pi \/ 2}^{\\pi \/ 2} \\sin ^2 \\theta \\cos ^7 \\theta d \\theta\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nsin2<\/sup> \u03b8 . cos7<\/sup> \u03b8 \u0c38\u0c30\u0c3f \u0c2a\u0c4d\u0c30\u0c2e\u0c47\u0c2f\u0c2e\u0c41
\nf(\u03b8) = sin2<\/sup> \u03b8 . cos7<\/sup> \u03b8 d\u03b8
\nf(-\u03b8) = sin2<\/sup>(-\u03b8) . cos7<\/sup>(-\u03b8)
\n= f(\u03b8)
\n= 2\\(\\int_0^{\\pi \/ 2}\\) sin2<\/sup> \u03b8 . cos7<\/sup> \u03b8 d\u03b8
\n\\(\\int_0^{\\pi \/ 2}\\) sinm<\/sup>x.cosn<\/sup>x dx
\nn \u0c2c\u0c47\u0c38\u0c3f \u0c38\u0c02\u0c16\u0c4d\u0c2f, n = 7
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n\\(\\int_{-\\pi \/ 2}^{\\pi \/ 2}\\) sin3<\/sup>\u03b8 cos3<\/sup>\u03b8 d\u03b8
\n\u0c38\u0c3e\u0c27\u0c28:
\nf(\u03b8) = sin3<\/sup>\u03b8 . cos3<\/sup>\u03b8
\nf(-\u03b8) = sin3<\/sup>(-\u03b8) cos3<\/sup>(-\u03b8)
\n= sin3<\/sup>\u03b8 cos3<\/sup> \u03b8 = -f(\u03b8)
\nf(\u03b8) \u0c2c\u0c47\u0c38\u0c3f \u0c2a\u0c4d\u0c30\u0c2e\u0c47\u0c2f\u0c2e\u0c41
\n\u2234 \\(\\int_{-\\pi \/ 2}^{\\pi \/ 2}\\)sin3<\/sup>\u03b8 . cos3<\/sup>\u03b8 d\u03b8 = 0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 9.
\n\\(\\int_0^a x\\left(a^2-x^2\\right)^{\\frac{7}{2}} d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = a sin \u03b8 \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41. a = a sin \u03b8
\ndx = a cos \u03b8 d\u03b8 \u03b8 = \u03c0\/2
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 10.
\n\\(\\int_0^2 x^{3 \/ 2} \\sqrt{2-x} d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = 2 cos2<\/sup> \u03b8
\ndx = 4 cos \u03b8 sin \u03b8 d\u03b8
\n\"AP<\/p>\n

II. \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f.<\/span><\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1.
\n\\(\\int_0^1 x^5(1-x)^{5 \/ 2} d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = – sin2<\/sup> \u03b8
\ndx = 2 sin \u03b8 . cos \u03b8 . d\u03b8
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\\(\\int_0^4\\)(16 – x2<\/sup>)5\/2<\/sup> dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = 4 sin \u03b8 4 = 4 sin \u03b8
\ndx = 4 cos \u03b8 d\u03b8 sin \u03b8 = 1
\n\u03b8 = \u03c0\/2
\nI = \\(\\int_0^{\\pi \/ 2}\\) (16 – 16sin2<\/sup>\u03b8)5\/2<\/sup> . 4 cos \u03b8 d\u03b8
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\\(\\int_{-3}^3\\) (9 – x2<\/sup>)3\/2<\/sup> x dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nLet f(x) = \\(\\left(9-x^2\\right)_x^{3 \/ 2}\\)
\nf(x) = (9 – (-x2<\/sup>))3\/2<\/sup> (-x)
\n= (9 – x2<\/sup>)3\/2<\/sup> . x
\n= f(-x)
\n\u2234 f \u0c2c\u0c47\u0c38\u0c3f \u0c2a\u0c4d\u0c30\u0c2e\u0c47\u0c2f\u0c2e\u0c41
\n\u2234 \\(\\left(9-x^2\\right)^{3 \/ 2}. x d x\\) = 0<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\\(\\int_0^5\\) x3<\/sup>(25 – x2<\/sup>)7\/2<\/sup> dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nI = \\(\\int_0^5\\)x3<\/sup>(25 – x2<\/sup>)7\/2<\/sup> dx \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41
\nx = 5 sin \u03b8
\ndx = 5 cos \u03b8 d\u03b8
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\"AP
\n\u0c38\u0c3e\u0c27\u0c28:
\nf(x) = sin8<\/sup>x. cos7<\/sup> x \u0c05\u0c28\u0c41\u0c15\u0c4a\u0c28\u0c41\u0c2e\u0c41
\nf(-x) = sin8<\/sup>(-x) . cos7<\/sup>(-x) = sin8<\/sup>x. cos7<\/sup>x
\n\u2234 f \u0c38\u0c30\u0c3f \u0c2a\u0c4d\u0c30\u0c2e\u0c47\u0c2f\u0c02
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 6.
\n\\(\\int_3^7 \\sqrt{\\frac{7-x}{x-3}} d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = 3 cos2<\/sup> \u03b8 + 7 sin2<\/sup> \u03b8
\ndx = (7 – 3) sin 2\u03b8 d\u03b8
\ndx = 4 sin 2\u03b8 d\u03b8
\n\u0c0e\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41 :
\nx = 3 cos2<\/sup> \u03b8 + 7 sin2<\/sup> \u03b8
\n7 = 3 cos2<\/sup> \u03b8 + 7 sin2<\/sup> \u03b8
\n4 cos2<\/sup> \u03b8 = 0
\ncos \u03b8 = 0
\n\u03b8 = \\(\\frac{\\pi}{2}\\)<\/p>\n

\u0c26\u0c3f\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41
\nx = 3 cos2<\/sup> \u03b8 + 7 sin2<\/sup> \u03b8
\n3 = 3 sin2<\/sup> \u03b8 + 7 sin2<\/sup> \u03b8
\n4 sin2<\/sup> \u03b8 = 0
\nsin \u03b8 = 0
\n\u03b8 = 0
\n7 – x = 7 – (3 cos2<\/sup>\u03b8 + 7 sin2<\/sup> \u03b8)
\n= (7 – 3) cos2<\/sup>\u03b8
\n= 4 cos2<\/sup>\u03b8
\nx – 3 = 3 cos2<\/sup>\u03b8 + 7 sin2<\/sup> \u03b8 – 3
\n= (7 – 3) sin2<\/sup>\u03b8
\n= 4 sin2<\/sup>\u03b8
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 7.
\n\\(\\int_2^6 \\sqrt{(6-x)(x-2)} d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = 2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8
\ndx = (6 – 2) sin 2\u03b8 d\u03b8
\ndx = 4 sin 2\u03b8 d\u03b8<\/p>\n

\u0c0e\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41 :
\nx = 2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8
\n6 = 2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8
\n4 cos2<\/sup>\u03b8 = 0
\ncos \u03b8 = 0
\n\u03b8 = \\(\\frac{\\pi}{2}\\)<\/p>\n

\u0c26\u0c3f\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41:<\/p>\n

x = 2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8
\n2 = 2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8
\n4 sin2<\/sup>\u03b8 = 0
\n\u03b8 = 0
\n6 – x = 6 – (2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8)
\n= (6 – 2) cos2<\/sup>\u03b8
\n= 4 cos2<\/sup>\u03b8
\nx – 2 = 2 cos2<\/sup>\u03b8 + 6 sin2<\/sup>\u03b8 – 2
\n= (6 – 2)sin2<\/sup>\u03b8
\n= 4 sin2<\/sup>\u03b8
\n\"AP<\/p>\n

\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 8.
\n\\(\\int_0^{\\pi \/ 2}\\)tan5<\/sup>x cos8<\/sup>x dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP<\/p>\n

III. \u0c26\u0c3f\u0c17\u0c41\u0c35 \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\\(\\int_0^1\\)x7\/2<\/sup>(1 – x)5\/2<\/sup> dx
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = sin2<\/sup>\u03b8
\ndx = 2 sin \u03b8 cos \u03b8 d\u03b8
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2.
\n\\(\\int_0^\\pi\\) (1 + cos x)3<\/sup> dx
\n\u0c38\u0c3e\u0c27\u0c28:
\n\\(\\int_0^\\pi\\) (1 + cos x)3<\/sup> dx
\n= \\(\\int_0^\\pi\\)(2 cos2<\/sup>\\(\\frac{x}{2}\\))3<\/sup> dx
\n= \\(\\int_0^\\pi\\) 23<\/sup>. cos6<\/sup> \\(\\frac{x}{2}\\) dx
\n\\(\\frac{x}{2}\\) = t \u21d2 dx = 2 dt<\/p>\n

\u0c0e\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41 : \u21d2 x = \u03c0 \u21d2 \\(\\frac{\\pi}{2}\\) = t
\n\u0c26\u0c3f\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41 : \u21d2 x = 0 \u21d2 0 = t
\n= 8 \\(\\int_0^{\\pi \/ 2}\\) cos6<\/sup> 6(2 dt)
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 3.
\n\\(\\int_4^9 \\frac{d x}{\\sqrt{(9-x)(x-4)}}\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8
\ndx = (9 – 4) sin 2\u03b8 d\u03b8
\ndx = 5 sin 2\u03b8 d\u03b8
\n\u0c0e\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41 :
\nx = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8
\n9 = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8
\n5 cos2<\/sup>\u03b8 = 0
\n\u03b8 = \\(\\frac{\\pi}{2}\\)
\n\u0c26\u0c3f\u0c17\u0c41\u0c35 \u0c39\u0c26\u0c4d\u0c26\u0c41 :
\nx = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8
\n4 = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8
\n5 sin2<\/sup>\u03b8 = o
\n\u03b8 = 0
\n9 – x = 9 – (4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8)
\n= (9 – 4) cos2<\/sup>\u03b8
\n= 5 cos2<\/sup>\u03b8
\nx – 4 = 4 cos2<\/sup>\u03b8 + 9 sin2<\/sup>\u03b8 – 4
\n= (9 – 4)sin2<\/sup>\u03b8
\n= 5 sin2<\/sup>\u03b8
\n\"AP
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 4.
\n\\(\\int_0^5 x^2(\\sqrt{5-x})^7 d x\\)
\n\u0c38\u0c3e\u0c27\u0c28:
\nx = 5 sin2<\/sup> \u03b8
\ndx = 10 sin\u03b8 cos \u03b8 d\u03b8
\n\"AP<\/p>\n

\u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 5.
\n\\(\\int_0^{2 \\pi}\\) (1 + cos x)5<\/sup> (1 – cos x)3<\/sup> dx.
\n\u0c38\u0c3e\u0c27\u0c28:
\n\"AP
\n\"AP<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the\u00a0Intermediate 2nd Year Maths 2B Textbook Solutions Chapter 7 \u0c28\u0c3f\u0c36\u0c4d\u0c1a\u0c3f\u0c24 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c41 Exercise 7(c) will help students to clear their doubts quickly. AP Inter 2nd Year Maths 2B Solutions Chapter 7 \u0c28\u0c3f\u0c36\u0c4d\u0c1a\u0c3f\u0c24 \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c41 Exercise 7(c) \u0c05\u0c2d\u0c4d\u0c2f\u0c3e\u0c38\u0c02 – 7 (\u0c38\u0c3f) I. \u0c15\u0c4d\u0c30\u0c3f\u0c02\u0c26\u0c3f \u0c38\u0c2e\u0c3e\u0c15\u0c32\u0c28\u0c41\u0c32\u0c28\u0c41 \u0c17\u0c23\u0c3f\u0c02\u0c1a\u0c02\u0c21\u0c3f. \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 1. \u0c38\u0c3e\u0c27\u0c28: ….. = \u0c2a\u0c4d\u0c30\u0c36\u0c4d\u0c28 2. \u0c38\u0c3e\u0c27\u0c28: n \u0c2c\u0c47\u0c38\u0c3f \u0c38\u0c02\u0c16\u0c4d\u0c2f \u0c2a\u0c4d\u0c30\u0c36\u0c4d\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\/31610"}],"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=31610"}],"version-history":[{"count":6,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/31610\/revisions"}],"predecessor-version":[{"id":31773,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/31610\/revisions\/31773"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=31610"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=31610"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=31610"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}