{"id":19652,"date":"2020-01-13T16:23:11","date_gmt":"2020-01-13T10:53:11","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19652"},"modified":"2021-07-02T15:29:39","modified_gmt":"2021-07-02T09:59:39","slug":"2nd-puc-basic-maths-question-bank-chapter-5-ex-5-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-5-ex-5-2\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.2"},"content":{"rendered":"

Students can Download Basic Maths Exercise 5.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.2<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Partial Fractions Ex 5.2 Five Marks Questions and Answers<\/strong><\/p>\n

I. Resolve the following into partial fractions; (3 and 5 marks)<\/p>\n

Question 1.
\n\\(\\frac{x}{(x+1)(x-4)}\\)
\nAnswer:
\n\"2nd
\nPut
\nx = 4, 4 = A(0) +B (4 + 1) \u21d2 4 = 5B \u21d2 B = \\(\\frac{4}{5}\\)
\nPut x = -1, -1 = A(-1-4) +B(0) \u21d2 -1 = -5A = A=\\(\\frac{1}{5}\\)
\nSubstituting both A & B in equation (1) we get
\n\"2nd<\/p>\n

Question 2.
\n\\(\\frac{x+1}{(x+2)(x-3)}\\)
\nAnswer:
\n\\(\\frac{x+1}{(x+2)(x-3)} =\\frac{A}{x+2}+\\frac{B}{x-3}\\)………………….. (1)
\nx = 3, 3 + 1 = A(0) +B(3 + 2)
\n4 = 5B = B = \\(\\frac{4}{5}\\)
\nPut
\nx = -2, -2 + 1 = A(-2 – 3) + B(0)
\n-1 = A(-5)
\nA = 1\/5;
\n3 = -A \u21d2 A = -3
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\n\\(\\frac{7 x-1}{(1-2 x)(1-3 x)}\\)
\nAnswer:
\nLet
\n\\(\\frac{7 x-1}{(1-2 x)(1-3 x)}=\\frac{A}{1-2 x}+\\frac{B}{1-3 x}\\)
\n\u21d2 (7x – 1) = A(1 \u2013 3x) + B(1 \u2013 2x)
\nPut x = 0, -1 = A + B …………..1
\nPut x=1 6 = -2A-B …………………….2
\nSolving 1 & 2 (adding) -A = 5 \u21d2 A = -5
\nB = -1-A = -1 + 5 = 4.
\n\"2nd<\/p>\n

Question 4.
\n\\(\\frac{3-7 x^{2}}{(1-3 x)(1+2 x)(1+x)}\\)
\nAnswer:
\nLet
\n\\(\\frac{3-7 x^{2}}{(1-3 x)(1+2 x)(1+x)}=\\frac{A}{(1-3 x)}+\\frac{B}{1+2 x}+\\frac{C}{1+x}\\) …………… (1)
\n\u2234 3 – 7x2<\/sup> = A(1 + 2x) (1+x) +B(1 \u2013 3x) (1 + x) +C (1 – 3x) (1 + 2x)
\nPut x = -1, 3 – 7 = A(0) + B(0) +C(1 +3) (1-2)
\n-4 = -4C \u21d2 C = 1
\nPut x = 0, 3 = A + B + C = A + B = 2
\nPut x = 1, 3 – 7 = A(1 + 2) (1 + 1) + B(1 – 3) (1 + 1) +C(1 – 3) (1 + 2)
\n-4 = 6A – 4B – 6C \u2235 C= 1
\n-4 + 6 = 6A – 4B = 2
\n3A – 2B = 1 2A + 2B = 4
\n\u21d2 A + B=2
\n\"2nd
\n\"2nd<\/p>\n

Question 5.
\n\\(\\frac{x^{2}}{(x+1)(x+2)(x+3)}\\)
\nAnswer:
\nLet \\(\\frac{x^{2}}{(x+1)(x+2)(x+3)}=\\frac{A}{x+1}+\\frac{B}{x+2}+\\frac{C}{x+3}\\)
\nx2<\/sup> = A(x +2) (x + 3) +B(x + 1) (x + 3) + C(x + 1) (x + 2)
\nPut x=-1, (-1)2<\/sup> = A(-1 +2) (-1 + 3) + 0 + 0
\n1 = 2A \u21d2 A =\\(\\frac{1}{2}\\)
\nPut x = -2, (-2)2<\/sup> = A(0) + B(-2 + 1) (-2 + 3) + C(0)
\n4 = -B \u21d2 b = -4
\nPut x = -3, (-3)2<\/sup> = A(0) + B(0) +C(-3 + 1) (-3 + 2)
\n9 = -2C:(-1) \u21d2 9 = 2c \u21d2 c = \\(\\frac{9}{2}\\)
\nEquation 1 be becomes<\/p>\n

Question 6.
\nAnswer:
\n\\(\\frac{2 x^{2}+10 x-3}{(x+1)(x-3)(x+3)}\\)
\nLet
\n\"2nd<\/p>\n

2x2<\/sup> +10x \u2013 3 = A[x – 3) (x + 3) + B(x + 1) (x + 3) + C(x +1) (x \u2013 3)
\nPut x=-1, 2 – 10 – 3 = A(-4) (2) + B(0) + C(0)
\n-11 = -8A = A = \\(\\frac{11}{8}\\)
\nPut x = 3, 18 + 30 \u2013 3 = A(0) + B(4)(6) + C(0)
\n+45 = 24B \u21d2 B = \\(\\frac{45}{24}=\\frac{15}{8}\\)
\nPut x = -3, 18 – 30 – 3 = C(-3 + 1) (-3 -3) = C(-2) (-6)-15 = + 12C
\nC = \\(\\frac{-15}{12}=\\frac{-5}{4}\\)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\n\\(\\frac{3 x+1}{x^{2}-6 x+8}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\n\\(\\frac{x^{2}-10 x+13}{(x+1)\\left(x^{2}-5 x+6\\right)}\\)
\nAnswer:
\nLet
\n\"2nd
\nx2<\/sup> – 10x + 13 = A(x-2)(x – 3) + B(x +1) (x – 3) + C(x + 1) (x – 2)
\nPut x=-1, 1 + 10 + 13 = A (-3) (-4) + 0 + 0
\n24 = 12A \u21d2 A = \\(\\frac{24}{12}\\)=2
\nPut x = 2, 4 – 20 + 13 A(0) + B(3) (-1) + C(0)
\n-3 = -3B \u21d2 B = 1
\nPut x = 3, 9 – 30 + 13 = A(0) + B(0) + C(4) (1)
\n-8 = 47 \u21d2 C= -2
\n\"2nd<\/p>\n

Question 9.
\n\\(\\frac{3 x+20}{x^{2}+4 x}\\)
\nAnswer:
\nLet \\(\\frac{3 x+20}{x^{2}+4 x}=\\frac{3 x+20}{x(x+4)}=\\frac{A}{x}+\\frac{B}{x+4}\\) ……..(1)
\n3x + 20 = A(x + 4) + B(x)
\nPut x=0, 20 = 4A \u21d2 A = 5
\nPut x = -4, -12 + 20 = A(0) + B(-4)
\n8 = -4B \u21d2 B = -2
\n\\(\\frac{3 x+20}{x^{2}+4 x}=\\frac{5}{x}-\\frac{2}{x+4}\\)<\/p>\n

Question 10.
\n\\(\\frac{x+3}{(x-1)\\left(x^{2}-4\\right)}\\)
\nAnswer:
\nLet \\(\\frac{x+3}{(x-1)\\left(x^{2}-4\\right)}=\\frac{x+3}{(x-1)(x-2)(x+2)}=\\frac{A}{x-1}+\\frac{B}{x-2}+\\frac{C}{x+2}\\) ……(1)
\n(x + 3) = A[x -2) (x + 2) +B(x – 1) (x + 2) + C(x – 1) (x – 2)
\nPut x= 1, 4 = A(-1)(3) \u21d2 A=\\(\\frac{-4}{3}\\)
\nPut x = 2, 5 = A(0) + B(1) (4) + C(O)
\nPut 5 = 4B \u21d2 B = \\(\\frac{-4}{3}\\)
\nPut x = -2, -2 + 3 = A(0) + B(0) +C(-3) (-4)
\n1 = 12C \u21d2 C = \\(\\frac{1}{12}\\)
\n\"2nd<\/p>\n

Question 11.
\n\\(\\frac{x+3}{x^{3}-x}\\)
\nAnswer:
\nLet \\(\\frac{x+3}{x^{3}-x}=\\frac{x+3}{x(x-1)(x+1)}=\\frac{A}{x}+\\frac{B}{x-1}+\\frac{C}{x+1}\\) ….. (1)
\nx + 3 = A(x-1) (x + 1) + B(x) (x + 1) + C(x) (x – 1)
\nPut x = 0 3 = -A + 0 + 0 \u21d2 A=-3
\nPut x = 1, 4 = A(0) + B(1) (2) + 0 = 4 = 2B \u21d2 B = 2
\nPut x = -1, 2 = A(0) + B(0) + C(-1) (-2); 2 = 2c \u21d2 c = 1
\n\u2234 \\(\\frac{x-3}{x^{3}-x}=\\frac{-3}{x}+\\frac{2}{x-1}+\\frac{1}{x+1}\\)<\/p>\n

Question 12.
\n\\(\\frac{1+3 x+2 x^{2}}{(1-2 x)\\left(1-x^{2}\\right)}\\)
\nAnswer:
\nLet
\n\"2nd
\n\"2nd
\n1 + 3x + 2x2<\/sup> = A(1 – x)(1 + x) +B(1 – 2x)(1 + x) + C(1 – 2x) (1 – x)
\nPut x = 1, 6 = A(0) + B(-1)(2) +C(0);
\n6 = -2B \u21d2 B = -3
\nPut x=-1, 0) = A(0) + B(0) +C(1 + 2)(2)
\n6C = 0 \u21d2 C = 0
\nPut x = 0, 1 = A + B + C = A – 3 = 1 \u21d2 A = 1 + 3 = 4<\/p>\n

\"KSEEB<\/p>\n

Part – B<\/span><\/p>\n

1. Resolve the following into partial fractions; (5 Marks)<\/p>\n

Question 1.
\n\\(\\frac{4}{(x-3)(x+1)^{2}}\\)
\nAnswer:
\nLet \\(\\frac{4}{(x-3)(x+1)^{2}}=\\frac{A}{x-3}+\\frac{B}{x+1}+\\frac{C}{(x+1)^{2}}\\)
\n\u22344 = A(x + 1)2<\/sup> + B(x + 1)(x \u2013 3) + C(x \u2013 3)
\nPut x = 3, 4 = A(4)2<\/sup> + B(0) + C(0)
\n4 = 16A \u21d2 A = \\(\\frac{1}{4}\\)
\nPut x =-1, 4 = A(0) + B(0) + C(-1-3); Put 4 = -4C \u21d2 C = -1
\nPut x = 0 4 = A- 3B – 3C
\n3B = A – 3C – 4
\n\"2nd<\/p>\n

Question 2.
\n\\(\\frac{9}{(x+1)(x+2)^{2}}\\)
\nAnswer:
\nLet \\(\\frac{9}{(x+1)(x+2)^{2}}=\\frac{A}{x+1}+\\frac{B}{x+2}+\\frac{C}{(x+2)^{2}}\\)
\n9 = A(x + 2)2<\/sup> + B(x + 1)(x + 2) +C(x + 1)
\nPut x = -1, 9 = A(-1 + 2)2<\/sup> + 0 + 0
\nA = 9
\nPut x= -2, 9 = A(0) + B(0) + CC – 10 = -9
\nPut x = 0, 9 = 4A + 2B + C
\n9 = 36 + 2B -9
\n9 = 27 + 2B \\(\\frac{-18}{2}\\) = B B = -9
\n\u2234 \\(\\frac{9}{x-1}-\\frac{9}{x+2}-\\frac{9}{(x+1)^{2}}\\)<\/p>\n

Question 3.
\n\\(\\frac{3 x+4}{(x+1)^{2}(x-1)}\\)
\nAnswer:
\nLet \\(\\frac{3 x+4}{(x+1)^{2}(x-1)}=\\frac{A}{x+1}+\\frac{B}{(x+1)^{2}}+\\frac{C}{x-1}\\)
\n3x+4 = A(x + 1)(x – 1) + B(x – 1) + C (x + 1)2<\/sup>
\nPut x = -1, \u21d2 3 + 4 = A(0) + B(-2) + C(0)
\n1 = -2B \u21d2 B = \\(-\\frac{1}{2}\\)
\nPut x = 1, = 3 + 4 = A(0) +B(0) + C(2)2<\/sup>
\n7 = 4C \u21d2 C = 5
\nComparing the coefficients of x2<\/sup> on both sides
\n0 = A +C \u21d2 A = -C = \\(-\\frac{7}{4}\\)
\n\"2nd<\/p>\n

Question 4.
\n\\(\\frac{3 x+2}{(x-2)(x+3)^{2}}\\)
\nAnswer:
\nLet \\(\\frac{3 x+2}{(x-2)(x+3)^{2}}=\\frac{A}{x-2}+\\frac{B}{x+3}+\\frac{C}{(x+3)^{2}}\\)
\n(3x + 2) = A(x + 3)2<\/sup> + B(x – 2) (x + 3) + C(x – 2)
\nPut x = 2, 6 + 2 = A(2 + 3)2<\/sup> + B(0)+C (0)
\n8 = 25A \u21d2 A = \\(\\frac{8}{25}\\)
\nPut X=-3, -9 + 2 = A(0) + B(0) +C(-5)
\n-7 = -5C \u21d2 C = \\(\\frac{7}{5}\\)
\nComparing the co-efficients of x2 on both sides
\n0 = A + B \u21d2 B = -A = \\(\\frac{8}{25}\\)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\n\\(\\frac{2 x^{2}-4 x+1}{(x+2)(x-3)^{2}}\\)
\nAnswer:
\n2x2<\/sup> – 4x + 1 = A[x – 3)2<\/sup> + B(x – 2) (x-3) + C (x – 2)
\nPut x = 2, 8-8 + 1 = A(-1)2<\/sup> = A = 1
\nPut x = 3, 18 \u2013 12 + 1 = A(0) + B(0) + C(3 – 2) = 7 = C
\nComparing coefficient of x2<\/sup> = 2 = A + B = B = 2-A = 2 -1 =1
\n\u2234 \\(\\frac{2 x^{2}-4 x+1}{(x-2)(x-3)^{2}}=\\frac{1}{x-2}+\\frac{1}{x-3}+\\frac{7}{(x-3)^{2}}\\)<\/p>\n

Question 6.
\n\\(\\frac{x^{2}}{(x+1)^{2}(x-5)}\\)
\nAnswer:
\nLet \\(\\frac{x^{2}}{(x+1)^{2}(x-5)}=\\frac{A}{x+1}+\\frac{B}{(x+1)^{2}}+\\frac{C}{x-5}\\)
\nx2<\/sup> = A (x + 1)(x – 5) + B(x – 5) + C(x + 1)2<\/sup>
\nPut x = -1, 1 = A(0) + B(-1-5) + C(0)
\n1 = -6B \u21d2 B = \\(-\\frac{1}{6}\\)
\nPut x = 5, 25 = A(0) + B(0) + C(6)2<\/sup>
\nC = \\(\\frac{25}{36}\\)
\nCompare the coefficients of x2 on both sides we get
\n\"2nd<\/p>\n

Question 7.
\n\\(\\frac{4-7 x}{(2+3 x)(1+x)^{2}}\\)
\nAnswer:
\nLet \\(\\frac{4-7 x}{(2+3 x)(1+x)^{2}}=\\frac{A}{2+3 x}+\\frac{B}{1+x}+\\frac{C}{(1+x)^{2}}\\)
\n\u21d2 4 – 7x = A(1 + x)2<\/sup> +B(1 + x) (2 + 3x) + C(2 + 3x)
\nPut x=-1, 11 = A(0) + B(0) + C(-1)
\nC = -11
\nPut x = 0 4 = A + 2B + 2C \u21d2 26 = A + 2B …..(1)
\nCompare the coefficients off of x2<\/sup> on both sides
\n0 = A + 3B …….(2)
\nEquations 1 – 2 gives 26 =-B = 0 \u21d2B = -26& A = 78
\n\u2234 \\(\\frac{4-7 x}{(2+3 x)(1+x)^{2}}=\\frac{78}{2+3 x}-\\frac{26}{1+x}-\\frac{11}{(1+x)^{2}}\\)<\/p>\n

Question 8.
\n\\(\\frac{1+2 x}{(x+2)^{2}(x-1)}\\)
\nAnswer:
\nLet \\(\\frac{1+2 x}{(x+2)^{2}(x-1)}=\\frac{A}{x+2}+\\frac{B}{(x+2)^{2}}+\\frac{C}{x-1}\\)
\n1 + 2x = A (x + 2) (x – 1) + B(x – 1) +C(x + 2)2<\/sup>
\n3 = A(0) + B(0) + C(9) \u21d2 C = \\(\\frac{1}{3}\\)
\nPut x = -2, 1 – 4 = B(-3) \u21d2 B = 1
\nComparing the coefficient of x2<\/sup> on both sides we get
\n\"2nd<\/p>\n

Question 9.
\n\\(\\frac{9 x-27}{(2+1)(x-2)^{2}}\\)
\nAnswer:
\nLet \\(\\frac{9 x-27}{(2+1)(x-2)^{2}}=\\frac{A}{x+1}+\\frac{B}{x-2}+\\frac{C}{(x-2)^{2}}\\)
\n9x – 27 = A(x – 2)2<\/sup> + B(x + 1)(x – 2) + C(x + 1)
\nPut x = -1, -36 = A(-3)2<\/sup> + B(0) + C(0)
\n-36 = 9A \u21d2 A=-4
\nPut x = 2, 18 – 27 = A(0) + B(0) +C(2+1)
\n-9 = 3C \u21d2 C =-3
\nComparing the coefficients of x? on both sides
\n0 = A + B = B = -A \u21d2 B =-(-4) = 4
\n\"2nd<\/p>\n

Question 10.
\n\\(\\frac{2 x+5}{(x+2)(x-1)^{2}}\\)
\nAnswer:
\nLet \\(\\frac{2 x+5}{(x+2)(x-1)^{2}}=\\frac{A}{x+2}+\\frac{B}{x-1}+\\frac{C}{(x-1)^{2}}\\)
\n2x + 5 = A(x – 1)2<\/sup> + B(x – 1)(x + 2) +C(x + 2)
\nPut x = -2, -4 + 5 = A(-3)2<\/sup> + 0 + 0
\n1 = 9A \u21d2 A =\\(\\frac{1}{9}\\)
\nPut x= 1, 2 + 5 = A(0) + B(0) + C(1 + 2)
\n7 = 3C \u21d2 C = \\(\\frac{7}{3}\\)
\nComparing the coeff of x2 on both sides
\n0 = A+B \u21d2 B = A = \\(-\\frac{1}{9}\\)
\n\"2nd<\/p>\n

Question 11.
\n\\(\\frac{x}{(1+2 x)^{2}(1-3 x)}\\)
\nAnswer:
\nLet \\(\\frac{x}{(1+2 x)^{2}(1-3 x)}=\\frac{A}{1+2 x}+\\frac{B}{(1+2 x)^{2}}+\\frac{C}{1-3 x}\\)
\nX = A(1 + 2x) (1 \u2013 3x) + B(1 \u2013 3x) + C(1 + 2x)2
\nPut x = 0 we get A + B + C = 0
\nComparing co efficient of x2 both sides
\n0 = -6A + 40 \u21d2 6A = 4C \u21d23A = 20
\nWe have A+B+C = 0
\n\u21d2 3A + 3B + 3C = 0
\n2C + 3B + 3C = 0
\n5C = -3B
\n: 3A = 2C
\nComparing the coefficient of x both sides<\/p>\n

\"2nd<\/p>\n

\"KSEEB<\/p>\n

Part – C<\/span><\/p>\n

III. Resolve into partial fractions: (5 Marks)<\/p>\n

Question 1.
\n\\(\\frac{2 x^{2}-7 x+1}{x^{2}-3 x-4}\\)
\nAnswer:
\n\\(\\frac{2 x^{2}-7 x+1}{x^{2}-3 x-4}\\)
\nThis is an improper fraction
\n\u2234 Convert into proper fraction by actual division.
\n\"2nd
\n9- x = A(x + 1) + B(x – 4)
\nPut x = 4, 9-4 = A(4 + 1) + B(0)
\n5 = 5A \u21d2 A = 1
\nPut x=-1, 9-(-1) = A(0) +B(-1-4)
\n10 = -5B \u21d2 B = -2
\n\"2nd<\/p>\n

Question 2.
\n\\(\\frac{4 x^{2}-3 x+5}{(2-x)(1+x)}\\)
\nAnswer:
\n\"2nd
\nx + 13 = A(1 + x) + B(2 – x)
\nPut x = 2, 15 = A[1 + 2) + B(0)
\n15 = 3A \u21d2 A = \\(\\frac{15}{3}\\) = 5
\nPut x=-1, -1 + 13 = A(0) +B(2-(-1))
\n12 = 3B \u21d2 B = 4
\n\"2nd
\nQuestion 3.
\n\\(\\frac{x^{3}+7 x^{2}+17 x+11}{x^{2}+5 x+6}\\)
\nAnswer:
\n\"2nd
\n(x – 1) = A(x + 3) + B(x + 2)
\nPut x = -2, -3 = A(-2 + 3) + B(0)
\n-3 = A \u21d2 A=-3
\nPut X=-3, -4 = A(0) + B(-3 + 2)
\n-4 = -B \u21d2 B = 4
\n\"2nd<\/p>\n

Question 4.
\n\\(\\frac{2 x^{2}+3 x+2}{x^{2}-x-2}\\)
\nAnswer:
\n\"2nd
\n\u2234 5x + 6 = A (x + 1) +B(x – 2)
\nPut x = 2, 16 = A(3) \u21d2 A = \\(\\frac{16}{3}\\)
\nPut x = -1, 1 = A(0) + B(-3)=B = \\(\\frac{1}{-3}\\)
\n1 2x+3x+2-21 16 1 x – x – 2 =2+3(x-2) 3(x+1)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\n\\(\\frac{2 x^{3}+x^{2}-x-3}{x(x-1)(2 x+3)}\\)
\nAnswer:
\n\"2nd
\n\u2234 2x – 3 = A[x -1) (2x + 3) +B(x) (2x + 3) + C(x) (x – 1)
\nPut x = 0 -3 = A(-1) (3) + B(0) + C(0)
\n-3 = -3A \u21d2 A = 1
\nPut x = 1, -1 = A(0) + B(1) (2 + 3) + C(0)
\n-1 = 5B \u21d2 B = \\(-\\frac{1}{5}\\)
\nCompare the co-efficient of x2<\/sup> both sides
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 5.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 5 Partial Fractions Ex 5.2 Part – A 2nd PUC …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19652"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=19652"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19652\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19652"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19652"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19652"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}