{"id":49999,"date":"2021-05-21T10:56:13","date_gmt":"2021-05-21T05:26:13","guid":{"rendered":"https:\/\/samacheerguru.com\/?p=49999"},"modified":"2021-05-24T15:09:54","modified_gmt":"2021-05-24T09:39:54","slug":"samacheer-kalvi-11th-maths-guide-chapter-7-ex-7-4-in-tamil","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-11th-maths-guide-chapter-7-ex-7-4-in-tamil\/","title":{"rendered":"Samacheer Kalvi 11th Maths Guide Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.4"},"content":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide<\/a> Pdf Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.4 Textbook Questions and Answers, Notes.<\/p>\n

TN Board 11th Maths Solutions Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.4<\/h2>\n

Question 1.
\n(0, 0), (1, 2), (4, 3) \u0b8e\u0ba9\u0bcd\u0bb1 \u0b89\u0b9a\u0bcd\u0b9a\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0bb3\u0bcd\u0bb3\u0bbf\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bca\u0ba3\u0bcd\u0b9f \u0bae\u0bc1\u0b95\u0bcd\u0b95\u0bcb\u0ba3\u0ba4\u0bcd\u0ba4\u0bbf\u0ba9\u0bcd \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\u0bae\u0bc1\u0b95\u0bcd\u0b95\u0bcb\u0ba3\u0ba4\u0bcd\u0ba4\u0bbf\u0ba9\u0bcd \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bbe\u0ba9\u0ba4\u0bc1
\n= \\(\\frac{1}{2}\\) [x1<\/sub>y2<\/sub> – x2<\/sub>y1<\/sub> + x2<\/sub>y3<\/sub> – x3<\/sub>y2<\/sub> + x3<\/sub>y1<\/sub> – x1<\/sub>y3<\/sub>]
\n\u0b87\u0b95\u0bcd\u0b95\u0bcb\u0bb5\u0bc8\u0baf\u0bc8 \u0b85\u0ba3\u0bbf\u0b95\u0bcd\u0b95\u0bcb\u0bb5\u0bc8 \u0bb5\u0b9f\u0bbf\u0bb5\u0ba4\u0bcd\u0ba4\u0bbf\u0bb2\u0bcd<\/p>\n

= \\(\\frac{1}{2}\\) \\(\\left|\\begin{array}{lll}
\nx_{1} & y_{1} & 1 \\\\
\nx_{2} & y_{2} & 1 \\\\
\nx_{3} & y_{3} & 1
\n\\end{array}\\right|\\) \u0b8e\u0ba9 \u0b8e\u0bb4\u0bc1\u0ba4\u0bb2\u0bbe\u0bae\u0bcd.<\/p>\n

\u2234 x1<\/sub>y1<\/sub> (0, 0), x2<\/sub>y2<\/sub> (1, 2) x3<\/sub>y3<\/sub> (4, 3) \u0b8e\u0ba9 \u0b95\u0bca\u0b9f\u0bc1\u0b95\u0bcd\u0b95\u0baa\u0bcd\u0baa\u0b9f\u0bcd\u0b9f\u0bc1\u0bb3\u0bcd\u0bb3\u0ba4\u0bc1.<\/p>\n

\u2234 \u0bae\u0bc1\u0b95\u0bcd\u0b95\u0bcb\u0ba3\u0ba4\u0bcd\u0ba4\u0bbf\u0ba9\u0bcd \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bc1 = \\(\\frac{1}{2}\\) \\(\\left|\\begin{array}{lll}
\n0 & 0 & 1 \\\\
\n1 & 2 & 1 \\\\
\n4 & 3 & 1
\n\\end{array}\\right|\\)
\n= \\(\\left|\\begin{array}{lll}
\n0 & 0 & 1 \\\\
\n1 & 2 & 1 \\\\
\n4 & 3 & 1
\n\\end{array}\\right|\\) [0 (2 – 3) – 0 (1 – 4)+ 1 (3 – 8)]
\n= \\(\\frac{1}{2}\\) (-5)
\n= \\(-\\frac{5}{2}\\) = \\(\\frac{5}{2}\\) = 2.5 \u0b85\u0bb2\u0b95\u0bc1\u0b95\u0bb3\u0bcd<\/p>\n

(\u2234 \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bc1 \u0b8e\u0baa\u0bcd\u0baa\u0bca\u0bb4\u0bc1\u0ba4\u0bc1\u0bae\u0bcd \u0bae\u0bbf\u0b95\u0bc8\u0baf\u0bc7)<\/p>\n

Question 2.
\n(k, 2), (2, 4) \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd (3, 2) \u0b8e\u0ba9\u0bcd\u0bb1 \u0b89\u0b9a\u0bcd\u0b9a\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0bb3\u0bcd\u0bb3\u0bbf\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bca\u0ba3\u0bcd\u0b9f \u0bae\u0bc1\u0b95\u0bcd\u0b95\u0bcb\u0ba3\u0ba4\u0bcd\u0ba4\u0bbf\u0ba9\u0bcd \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bc1 4 \u0b9a\u0ba4\u0bc1\u0bb0 \u0b85\u0bb2\u0b95\u0bc1\u0b95\u0bb3\u0bcd \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, k- \u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\u0bae\u0bc1\u0b95\u0bcd\u0b95\u0bcb\u0ba3\u0ba4\u0bcd\u0ba4\u0bbf\u0ba9\u0bcd \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bc1 = \\(\\frac{1}{2}\\) \\(\\left|\\begin{array}{lll}
\nx_{1} & y_{1} & 1 \\\\
\nx_{2} & y_{2} & 1 \\\\
\nx_{3} & y_{3} & 1
\n\\end{array}\\right|\\)\u00a0-\u0ba9\u0bcd \u0b8e\u0ba3\u0bcd\u0ba3 \u0bb3\u0bb5\u0bc1 \u0b86\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

\u0b8e\u0ba9\u0bb5\u0bc7 4 = \\(\\frac{1}{2}\\) \\(\\left|\\begin{array}{lll}
\nk & 2 & 1 \\\\
\n2 & 4 & 1 \\\\
\n3 & 2 & 1
\n\\end{array}\\right|\\)<\/p>\n

\u21d2 8 = \u00b1 [k(4 – 2) – 2(2 – 3) + 1(4 – 12)]
\n\u21d2 8 = \u00b1 [k (2) + 2 – 8]
\n\u21d2 8 = \u00b1 (2k – 6)<\/p>\n

\u0ba8\u0bbf\u0bb2\u0bc8 (i)
\n\u21d2 8 = 2k – 6
\n\u21d2 2k = 14 \u21d2 k = 7<\/p>\n

\u0ba8\u0bbf\u0bb2\u0bc8 (ii)
\n-2k = 8 – 6
\n-2k = 2
\nk = -1
\nk = -1, 7.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\n\u0b95\u0bc0\u0bb4\u0bcd\u0b95\u0bcd\u0b95\u0bbe\u0ba3\u0bcd\u0baa\u0bb5\u0bb1\u0bcd\u0bb1\u0bbf\u0bb2\u0bcd \u0b8e\u0bb5\u0bc8 \u0baa\u0bc2\u0b9c\u0bcd\u0b9c\u0bbf\u0baf \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0baa\u0bc2\u0b9c\u0bcd\u0b9c\u0bbf\u0baf\u0bae\u0bb1\u0bcd\u0bb1 \u0b95\u0bcb\u0bb5\u0bc8 \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd \u0b8e\u0ba9\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n(i) \\(\\left[\\begin{array}{lll}
\n1 & 2 & 3 \\\\
\n4 & 5 & 6 \\\\
\n7 & 8 & 9
\n\\end{array}\\right]\\)<\/p>\n

(ii) \\(\\left[\\begin{array}{rrr}
\n2 & -3 & 5 \\\\
\n6 & 0 & 4 \\\\
\n1 & 5 & -7
\n\\end{array}\\right]\\)<\/p>\n

(iii) \\(\\left[\\begin{array}{ccc}
\n0 & a-b & k \\\\
\nb-a & 0 & 5 \\\\
\n-k & -5 & 0
\n\\end{array}\\right]\\)
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n(i) A = \\(\\left[\\begin{array}{lll}
\n1 & 2 & 3 \\\\
\n4 & 5 & 6 \\\\
\n7 & 8 & 9
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0b95.<\/p>\n

|A| = \\(\\left|\\begin{array}{lll}
\n1 & 2 & 3 \\\\
\n4 & 5 & 6 \\\\
\n7 & 8 & 9
\n\\end{array}\\right|\\)
\n= 1(45 – 48) – 2(36 – 42) + 3(32 – 35)
\n= -3 – 2(-6) + 3(-3)
\n= -3 + 12 – 9 = -12 + 12 = 0
\n\u2234 A \u0b92\u0bb0\u0bc1 \u0baa\u0bc2\u0b9c\u0bcd\u0b9c\u0bbf\u0baf\u0b95\u0bcd\u0b95\u0bcb\u0bb5\u0bc8 \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

(ii) A =\\(\\left[\\begin{array}{ccc}
\n2 & -3 & 5 \\\\
\n6 & 0 & 4 \\\\
\n1 & 5 & -7
\n\\end{array}\\right]\\) \u0b8e\u0ba9\u0bcd\u0b95<\/p>\n

|Al = \\(\\left|\\begin{array}{ccc}
\n2 & -3 & 5 \\\\
\n6 & 0 & 4 \\\\
\n1 & 5 & -7
\n\\end{array}\\right|\\)
\n= 2(0 – 20) + 3(-42 – 4) + 5(30 – 0)
\n= – 40 + 3 \u00d7 (-46) + 150
\n= – 40 – 138 + 150
\n= – 178 + 150 = -28 \u2260 0
\n\u2234 A \u0b92\u0bb0\u0bc1 \u0baa\u0bc2\u0b9c\u0bcd\u0b9c\u0bbf\u0baf\u0bae\u0bb1\u0bcd\u0bb1 \u0b95\u0bcb\u0bb5\u0bc8 \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

(iii) B = \\(\\left[\\begin{array}{ccc}
\n0 & a-b & k \\\\
\nb-a & 0 & 5 \\\\
\n-k & -5 & 0
\n\\end{array}\\right]\\)<\/p>\n

|B| = \\(\\left|\\begin{array}{ccc}
\n0 & a-b & k \\\\
\nb-a & 0 & 5 \\\\
\n-k & -5 & 0
\n\\end{array}\\right|\\)
\n= 0 (0 + 25) – (a – b) (0 + 5k) + k [-5(b – a) – 0]
\n= -5k(a – b) – 5k(b – a)
\n= + 5k (b – a) – 5k (b – a) = 0
\nB \u0b92\u0bb0\u0bc1 \u0baa\u0bc2\u0b9c\u0bcd\u0b9c\u0bbf\u0baf\u0b95\u0bcd \u0b95\u0bcb\u0bb5\u0bc8 \u0b85\u0ba3\u0bbf\u0baf\u0bbe\u0b95\u0bc1\u0bae\u0bcd.<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\n\u0baa\u0bbf\u0ba9\u0bcd\u0bb5\u0bb0\u0bc1\u0bb5\u0ba9 \u0baa\u0bc2\u0b9c\u0bcd\u0b9c\u0bbf\u0baf\u0b95\u0bcd \u0b95\u0bcb\u0bb5\u0bc8 \u0b85\u0ba3\u0bbf\u0b95\u0bb3\u0bcd \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, a \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd b \u0b86\u0b95\u0bbf\u0baf\u0bb5\u0bb1\u0bcd\u0bb1\u0bbf\u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n(i) A = \\(\\left[\\begin{array}{cc}
\n7 & 3 \\\\
\n-2 & a
\n\\end{array}\\right]\\)<\/p>\n

(ii) B = \\(\\left[\\begin{array}{ccc}
\nb-1 & 2 & 3 \\\\
\n3 & 1 & 2 \\\\
\n1 & -2 & 4
\n\\end{array}\\right]\\)
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n(i) A = \\(\\left[\\begin{array}{cc}
\n7 & 3 \\\\
\n-2 & a
\n\\end{array}\\right]\\)<\/p>\n

|A| = \\(\\left|\\begin{array}{cc}
\n7 & 3 \\\\
\n-2 & a
\n\\end{array}\\right|\\) = 0 \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd
\n7a + 6 = 0
\n7a = – 6
\na = \\(-\\frac{6}{7}\\)<\/p>\n

(ii) B = \\(\\left[\\begin{array}{ccc}
\nb-1 & 2 & 3 \\\\
\n3 & 1 & 2 \\\\
\n1 & -2 & 4
\n\\end{array}\\right]\\)<\/p>\n

|B| = \\(\\left|\\begin{array}{ccc}
\nb-1 & 2 & 3 \\\\
\n3 & 1 & 2 \\\\
\n1 & -2 & 4
\n\\end{array}\\right|\\) = 0 \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd
\n(b – 1) (4 + 4) – 2(12 – 2) + 3(-6 – 1) = 0
\n(b – 1) (8) – 20 – 21 = 0
\n8b – 8 – 41 = 0
\n8b – 49 = 0
\n8b = 49
\nb = \\(\\frac{49}{8}\\)<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\ncos 2\u03b8 = 0 \u0b8e\u0ba9\u0bbf\u0bb2\u0bcd, \\(\\left|\\begin{array}{rrr}
\n0 & \\cos \\theta & \\sin \\theta \\\\
\n\\cos \\theta & \\sin \\theta & 0 \\\\
\n\\sin \\theta & 0 & \\cos \\theta
\n\\end{array}\\right|^{2}\\)\u00a0-\u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\nA = \\(\\left|\\begin{array}{ccc}
\n0 & \\cos \\theta & \\sin \\theta \\\\
\n\\cos \\theta & \\sin \\theta & 0 \\\\
\n\\sin \\theta & 0 & \\cos \\theta
\n\\end{array}\\right| \\times\\left|\\begin{array}{rrr}
\n0 & \\cos \\theta & \\sin \\theta \\\\
\n\\cos \\theta & \\sin \\theta & 0 \\\\
\n\\sin \\theta & 0 & \\cos \\theta
\n\\end{array}\\right|\\) \u0b8e\u0ba9\u0bcd\u0b95 .
\ncos 2\u03b8 = 2 cos2<\/sup>\u03b8 – 1 = 0
\n2 cos2<\/sup>\u03b8 = 1
\ncos2<\/sup>\u03b8 = \\(\\frac{1}{2}\\)
\ncos \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\), sin \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\)
\n\"Tamilnadu<\/p>\n

Question 6.
\n\\(\\left|\\begin{array}{cc}
\n\\log _{3} 64 & \\log _{4} 3 \\\\
\n\\log _{3} 8 & \\log _{4} 9
\n\\end{array}\\right| \\times\\left|\\begin{array}{ll}
\n\\log _{2} 3 & \\log _{8} 3 \\\\
\n\\log _{3}^{c} 4 & \\log _{3} 4
\n\\end{array}\\right|\\) \u0b8e\u0ba9\u0bcd\u0bb1 \u0baa\u0bc6\u0bb0\u0bc1\u0b95\u0bcd\u0b95\u0bb2\u0bbf\u0ba9\u0bcd \u0bae\u0ba4\u0bbf\u0baa\u0bcd\u0baa\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95.
\n\u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 :
\n\\(\\left|\\begin{array}{cc}
\n\\log _{3} 64 & \\log _{4} 3 \\\\
\n\\log _{3} 8 & \\log _{4} 9
\n\\end{array}\\right| \\times\\left|\\begin{array}{ll}
\n\\log _{2} 3 & \\log _{8} 3 \\\\
\n\\log _{3}^{c} 4 & \\log _{3} 4
\n\\end{array}\\right|\\)<\/p>\n

= \\(\\left|\\begin{array}{ll}
\n\\log _{3} 64 \\cdot \\log _{2} 3+\\log _{4} 3 \\cdot \\log _{3} 4 & \\log _{3} 64 \\cdot \\log _{8} 3+\\log _{4} 3 \\cdot \\log _{3} 4 \\\\
\n\\log _{3} 8 \\cdot \\log _{2} 3+\\log _{4} 9 \\cdot \\log _{3} 4 & \\log _{3} 8 \\cdot \\log _{8} 3+\\log _{4} 9 \\cdot \\log _{3} 4
\n\\end{array}\\right|\\) [\u2235 logy<\/sub>x . logx<\/sub> y = 1]<\/p>\n

= \\(\\left|\\begin{array}{lc}
\n6 \\log _{3} 2 \\log _{2} 3+1 & 2 \\log _{3} 88_{8} 3+1 \\\\
\n3 \\log _{3} 2 \\log _{2} 3+1 & 1+2 \\log _{4} 3 \\log _{3} 4
\n\\end{array}\\right|\\) [\u2235 logx<\/sub> x = 1]<\/p>\n

= \\(\\left|\\begin{array}{ll}
\n6+1 & 2+1 \\\\
\n3+2 & 1+2
\n\\end{array}\\right|=\\left|\\begin{array}{ll}
\n7 & 3 \\\\
\n5 & 3
\n\\end{array}\\right|\\)
\n= 21 – 15 = 6<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.4 Textbook Questions and Answers, Notes. TN Board 11th Maths Solutions Chapter 7 \u0b95\u0ba3\u0b99\u0bcd\u0b95\u0bb3\u0bcd, \u0ba4\u0bca\u0b9f\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd \u0bae\u0bb1\u0bcd\u0bb1\u0bc1\u0bae\u0bcd \u0b9a\u0bbe\u0bb0\u0bcd\u0baa\u0bc1\u0b95\u0bb3\u0bcd Ex 7.4 Question 1. (0, 0), (1, 2), (4, 3) \u0b8e\u0ba9\u0bcd\u0bb1 \u0b89\u0b9a\u0bcd\u0b9a\u0bbf\u0baa\u0bcd\u0baa\u0bc1\u0bb3\u0bcd\u0bb3\u0bbf\u0b95\u0bb3\u0bc8\u0b95\u0bcd \u0b95\u0bca\u0ba3\u0bcd\u0b9f \u0bae\u0bc1\u0b95\u0bcd\u0b95\u0bcb\u0ba3\u0ba4\u0bcd\u0ba4\u0bbf\u0ba9\u0bcd \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bc8\u0b95\u0bcd \u0b95\u0bbe\u0ba3\u0bcd\u0b95. \u0ba4\u0bc0\u0bb0\u0bcd\u0bb5\u0bc1 : \u0bae\u0bc1\u0b95\u0bcd\u0b95\u0bcb\u0ba3\u0ba4\u0bcd\u0ba4\u0bbf\u0ba9\u0bcd \u0baa\u0bb0\u0baa\u0bcd\u0baa\u0bbe\u0ba9\u0ba4\u0bc1 […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[7],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/49999"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=49999"}],"version-history":[{"count":0,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/49999\/revisions"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=49999"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=49999"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=49999"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}