{"id":17702,"date":"2020-01-04T11:51:41","date_gmt":"2020-01-04T06:21:41","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17702"},"modified":"2021-07-02T15:31:19","modified_gmt":"2021-07-02T10:01:19","slug":"2nd-puc-maths-question-bank-chapter-1-miscellaneous-ex","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-1-miscellaneous-ex\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 1 Relations and Functions Miscellaneous Exercise"},"content":{"rendered":"

Students can Download Maths Chapter 1 Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Miscellaneous Exercise<\/h2>\n

Question 1.
\nLet f : R \u2192 R be defined as f (x) = 10x + 7. Find the function g : R \u2192 R such that
\ngof = fog = IR<\/sub>.
\nAnswer:
\nSince gof = fog = IR<\/sub> ,
\n(gof) (x) = (fog) (x) = x \u2200 x \u2208 R
\ng (f (x)) = f (g (x)) = x
\ng (10x + 7) = x and
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nLet f : W \u2192 W be defined as f(n) = – 1, if n is odd and f(n) = n – 1, if n is even. Show that
\nf is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
\nAnswer:
\nLet m, n \u2208 W and m and n be 2 distinct elements.<\/p>\n

Case I:
\nWhen both m and n are even
\nf(m) = m + 1\u00a0\u00a0\u00a0\u00a0 f (n) = n + 1
\nm \u2260 n \u21d2 m + 1 \u2260 n + 1
\n\u21d2 f (m) \u2260 f (n)<\/p>\n

Case II:
\nWhen both m and n are odd
\nf (m) = m – 1 f(n) = n – 1
\nm \u2260 n \u21d2 m -1 \u2260 n – 1
\n\u21d2 f (m) \u2260 f (n)<\/p>\n

Case III:
\nWhen m is odd and n is even
\nf(m) = m – 1\u00a0 \u00a0 \u00a0 \u00a0f (n) = n + 1
\nf(m) \u2260 f (n)
\nso, in all cases m \u2260 n \u21d2 f (m) \u2260 f (n)
\nHence, f is a one-one function.
\nIf n \u2208 W is any element
\nf (n – 1) = n if n is odd (\u2235 n – liseven)
\nf(n + 1) = n if n is even (\u2235 n + 1 is odd)
\nso, every element of W is the f – image of some element in W.
\nHence f is onto
\nThus, f is both one-one and onto.
\ni.e. f is one-one correspondence. Consequently,
\nf is invertible.
\nf (n – 1) = n if n is odd
\nf (n + 1) = n if n is even
\nn – 1 = f-1<\/sup> (n) if n is odd
\nn + 1 = f-1<\/sup> (n) if n is even
\n\\(\\begin{aligned}&\\text { i.e. } f^{-1}(n)=\\left\\{\\begin{array}{ll}{n-1} & {\\text { if } \\quad n \\text { is odd }} \\\\{n+1} & {\\text { if } \\quad n \\text { is even }}\\end{array}\\right.\\\\&\\text { thus } f^{-1}=f\\end{aligned}\\)<\/p>\n

Question 3.
\nIf f : R \u2192 R is defined by f(x) = x2<\/sup> – 3x +2, find f (f(x)).
\nAnswer:
\nf (x) = x2<\/sup> – 3x + 2
\n\u2234 f (f (x)) = f (y) where
\ny = x2<\/sup> – 3x + 2
\n= y2<\/sup> – 3y + 2
\n= (x2<\/sup> – 3x + 2)2<\/sup> – 3 (x2<\/sup> – 3x + 2) + 2
\n= x4<\/sup> + 9x2<\/sup> + 4 – 6x3<\/sup> + 4x2<\/sup> – 12x – 3x2<\/sup> + 9x – 6 + 2
\n= x4<\/sup> – 6x3<\/sup> + 10x2<\/sup> – 3x.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nShow that the function f : R \u2192 {x \u2208 R : – 1 < x < 1} defined by
\n\\(\\mathbf{f}(\\mathbf{x})=\\frac{\\mathbf{x}}{\\mathbf{1}+|\\mathbf{x}|}, \\mathbf{x} \\in \\mathbf{R}\\) is one\u00a0 one and onto function.
\nAnswer:
\n\"2nd
\n\"2nd
\nhence f is one-one
\nsimilarly we can prove that f is one-one.
\nWhen x1<\/sub> < 0 and x2<\/sub>> 0 and x, < 0 and x2<\/sub> > 0
\nTo prove f is onto
\n\"2nd<\/p>\n

Question 5.
\nShow that the function f : R \u2192 R given by f (x) = x3<\/sup> is injective.
\nAnswer:
\nLet x1<\/sub> \u2260 x2<\/sub> be any 2 reals
\n\u2234\u00a0 x1 <\/sub>– x2<\/sub> \u2260 0.
\nAlso,<\/p>\n

\"2nd<\/p>\n

Question 6.
\nGive examples of two functions f : N \u2192 Z and g : Z \u2192 Z such that g : Z \u2192 Z is injective but \u00a3 is not injective. (Hint: Consider f(x) = x and g (x) = |x|).
\nAnswer:
\nAs g (x) = g (-x) = |x| for all x \u2208 Z ,
\n\u2234 g is not one-one
\ni.e. y is not injective
\nAs f : N \u2192\u00a0 Z and g : Z \u2192 Z gof : N \u2192 Z
\nlet x1<\/sub>, x2<\/sub>, \u2208 N such that gof (x1<\/sub>) = gof (x2<\/sub>)
\n\u21d2 g (x1<\/sub>) = g (x2<\/sub>)
\n\u21d2 | x1 <\/sub>| = | x2<\/sub> | \u21d2 x1 <\/sub>= X2
\n<\/sub>(both x1 <\/sub>,x2<\/sub> >0)
\nHence g o f is injective.<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nGive examples of two functions f : N \u2192 N and g : N \u2192 N such that gof is onto but f is not onto. (Hint: Consider f (x) = x + 1 and
\n\\(g(x)=\\left\\{\\begin{array}{r}{x-1 \\text { if } x>1} \\\\{1 \\text { if } x=1}\\end{array}\\right.\\)
\nAnswer:
\nf(x) = x+ lxl + 1 \u2265 1 + 1 \u2200 x \u2208 N
\n(\u2235 v \u2265 1 x \u2208 N)
\n\u21d2 f(x) \u2265 2 \u2200 x \u2208 N . Rf<\/sub> ^ N as 1 t Rf
\n<\/sub>Hence f is not onto
\ngof : \u2192 \u223c \u2192 N such that
\ngof (x) = g (f (x)) = g (x + 1)
\n= (x+1) -1 [ \u2235 \u2200 x \u2208 N,x+ 1 > 1]
\n\u21d2 gof (x) = x\u00a0\u00a0\u00a0\u00a0\u00a0 V x \u2208 N
\n\u2234 Range of gof = N as gof is the identity f Hence gof is onto.<\/p>\n

Question 8.
\nGiven a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A \u2282 B. Is R an equivalence relation on P(X)? Justify your answer.
\nAnswer:
\nSince A C A \u2200 A \u2208 P (X)
\nR is reflexive.
\nAlso, for A, B, C \u2208 P(X),A R B & B R C
\n\u21d2 A C B and B C C
\n\u21d2 A \u2282 C \u21d2 A R C
\n\u2234 R is transitive.
\nR is not symmetric as A C B \u21d2 B C A
\nso ARB \u21d2 BRA
\nHence R is not an equivalence relation.<\/p>\n

Question 9.
\nGiven a non-empty set X, consider the binary operation * : P(X) * P(X) \u2192 P(X) given by
\nA * B = A \u2229 B \u2200 A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation *.
\nAnswer:
\nLet E \u2208 P (X) be an identity elements, then
\nA * E = E * A = A \u2200 A \u2208 P(X)
\n\u21d2 A \u2229 E = E \u2229 A = A \u2200 A \u2208 P(X)
\n\u21d2 X \u2229 E = X as X \u2208 P (X)
\n\u21d2 X C E
\nAlso E C X as E \u2208 P (X)
\n\u2234 E = X
\nThus, X is the identity element.
\nLet A \u2208 P(X) be invertible, then there exists
\nB \u2208 P(X) such that A * B = B*A = X, the identity element.
\n\u21d2 A \u2229 B = B \u2229 A = X
\nX C A & X C B
\nAlso A, B, \u2282 X as A, B \u2208 P (X)
\nA = X = B
\nX is the only invertible element and – X1<\/sup> = B = X.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nFind the number of all onto functions from the set {1, 2, 3,…, n} to itself.
\nAnswer:
\nLet A = {1, 2, 3,……… n}
\nf : A \u2192 A is an onto function, then range of f
\n\u21d2 f is one-one f is one and onto the number of onto function is n<\/sup>Pn<\/sub> = n!<\/p>\n

Question 11.
\nLet S = {a, b, c} and T = {1, 2, 3}. Find F1 <\/sup>of the following functions F from S to T, if it exists.
\n(i) F = {(a, 3), (b, 2), (c, 1)}
\n(ii) F = {(a, 2),(i, 1), (c, 1)}
\nAnswer:
\n(i) Range of {1, 2, 3} = T
\n\u21d2 F is onto Also F is onto as different elements of S have different F – images.
\n\u2234 F-1<\/sup> exist and F-1<\/sup> = {(3, a), (2, b), (1, c)}
\n\u2234 F1<\/sup> T \u2192 S as defined by
\nF1<\/sup> (3) = a, F-1<\/sup> (2) = b, F1<\/sup> (1) = c.<\/p>\n

(ii) F (b) = 1, F (c) = 1 hence F (n1<\/sub>) = F (n2<\/sub>) \u2260 n1<\/sub>, = n2
\n<\/sub>\u21d2 F is not one-one .: hence not invertible.<\/p>\n

Question 12.
\nConsider the binary operations * : R x R \u2192 R and o : R x R \u2192 R defined as a* b = |a-b| and a o b = a, \u2200 a, b \u2208 R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that \u2200 a, b, c \u2208 R, a *(b o c) = (a * b) o (a * b). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over? Justify your answer.
\nAnswer:
\n\u2200 a, b \u2208 R
\na * b = |a – b| = |b – a| = b * a
\nhence commutative \u2200 3, 5 ,1 \u2208 R
\n(3 * 5) * 7
\n|3 – 5| * 7 = 2 * 7= |2 – 7| = 5
\n3* (5 * 7) = 3* |5 – 7| =3 * 2= |3 – 2| = 1
\na* (b * c) & (a * b) *c
\n\u2234 * is not associative.
\na o b = a, b o a = b
\nhence a o b \u2260 boa
\nhence not commutative
\n(a o b)o c : a o c = a
\na o (b o c) = a o b = a
\nhence o is associative
\na*(b o c) (a * b)=|a – b|
\n(a * b) o (a * c) = |a – b| o |a – c| = |a – b|
\nhence a* (b o c) = (a * b) o (a * c)
\nhence o is distributive over *.<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nGiven a non-empty set X, let * : P(X) \u00d7 P(X) \u2192 P(X) be defined as A * B = (A – B) \u222a (B – A), \u2200 A, B \u2208 P(X). Show that the empty set \u03c6) is the identity for the operation * and all the elements A of P(X) are invertible with A-1<\/sup> = A.
\n(Hint: (A – \u03d5) \u222a (\u03d5 – A) = A and (A – A) \u222a (A – A) = A * A = \u03d5
\nAnswer:
\n\u03d5 \u2208 P (X) be an identity element.
\nTo prove it let E \u2208 P (X) be the identity element such that A * E = E * A = A
\n\u2200 A \u2208 P (X)
\n(A – E) \u222a (E – A) = A \u21d2 E = \u03c6
\ni.e. (A – \u03d5) \u222a (E – \u03d5)) = A
\nA * \u03d5 = \u03d5 * A = A
\n* A = A, hence \u03d5 is the identity element.
\nLet Be P(X) be the inverse of A
\n\u2234 A* B = B * A = \u03c6 \u2200 A e P(X)
\n(A – B) \u222a (B – A) = 0 \u21d2 B = A
\nbecause A – B = \u03d5 \u00a0 \u00a0B – A = \u03d5\u21d2 A = B
\n\u2234 \u2200 A \u2208 P(X),\u00a0 A * A = \u03d5
\nA is the invertible element of A
\n\u2234 A-1<\/sup> = A<\/p>\n

Question 14.
\nDefine a binary operation * on the set {0, 1, 2, 3, 4, 5} as
\n\\(\\mathbf{a} * \\mathbf{b}=\\left\\{\\begin{array}{ll}{\\mathbf{a}+\\mathbf{b},} & {\\text { if } \\mathbf{a}+\\mathbf{b}<\\mathbf{6}} \\\\{\\mathbf{a}+\\mathbf{b}-\\mathbf{6}} & {\\text { if } \\mathbf{a}+\\mathbf{b} \\geq \\mathbf{6}}\\end{array}\\right.\\)
\nShow that zero is the identity for this operation and each element a \u00b10 of the set is invertible with 6-a being the inverse of a.
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n\n
*<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
0<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n0<\/td>\n<\/tr>\n
2<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n0<\/td>\n1<\/td>\n<\/tr>\n
3<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
4<\/td>\n4<\/td>\n5<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
5<\/td>\n5<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

From the composition table it is obvious that a
\n* 0 = a \u2200 a \u2208 {0, 1\/2, 3,4, 5}
\n\u2234 0 is the identity element
\nAlso \u2200 a \u2208 {0, 1, 2, 3, 4, 5} \u2203 (6 – a) \u2208 {0,1,2, 3,4,5} such that
\na * (6 – a) = (6 – a) * a = 0, a \u2260 0
\nhence 6 – a is the inverse of a .
\nHowever when a = 0, 6 – a g {0, 1, 2, 3,4, 5} hence 6 – a is the inverse of a when a \u2260 0.<\/p>\n

Question 15.
\nLet A = {- 1, 0, 1, 2}, B = {- 4, – 2, 0, 2} and f, g : A \u2192 B be functions defined by
\nf (x) = x2<\/sup> – x, x \u2208<\/span> A and g (x) = -1, x \u2208<\/span> A, Are f and g equal? Justify your answer.
\n(Hint: One may note that two functions f : A \u2192 B and g : A \u2192 B such that f(a) =
\n\\(2\\left|x-\\frac{1}{2}\\right|-1, x \\in A\u00a0 A\\), are called equal functions).
\nAnswer:
\nf (-1) = (-1)2<\/sup> + 1=2
\nf (0) = 02<\/sup> – 0 = 0
\nf(1) = (1)2<\/sup>– 1 = 0
\nf(2) = (2)2<\/sup> -2 = 2<\/p>\n

\"2nd<\/p>\n

Question 16.
\nLet A = {1, 2,3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
\n(A) 1
\n(B) 2
\n(C) 3
\n(D) 4
\nAnswer:
\nR1<\/sub> = {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (3, 1), (2, 1)} is the only relation which is reflexive symmetric but not transitive and is such that {1, 2], {1, 3} \u2208 R1<\/sub> Correct answer is \u201cA\u201d<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nLet A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
\n(A) 1
\n(B) 2
\n(C )3
\n(D) 4
\nAnswer:
\n(i) {(1, 1), (1, 2), (2, 1), (2,2), (2, 3)}
\n(ii) {(1,1), (2,2), (3,3), (1,2), (2,1), (1,3), (3,1)} (2,3), (3,2)}
\nThere are two equivalence relation
\nCorrect answer is \u201cB\u201d.<\/p>\n

Question 18.
\nLet f : R \u2192 R be the Signum Function defined as
\n\\(\\mathbf{f}(\\mathbf{x})=\\left\\{\\begin{array}{ll}{\\mathbf{1},} {\\mathbf{x}>\\mathbf{0}} \\\\{\\mathbf{0},} & {\\mathbf{x}=\\mathbf{0}} \\\\{-\\mathbf{1},} & {\\mathbf{x}<\\mathbf{0}}\\end{array}\\right.\\)
\nAnswer:
\nfog (n) = f g(n) = f (0) = 0, 0 \u2264 n < 1
\nf (1) = n = 1
\ngof (n) = g [1] = 1
\n\"2nd
\nhence fog and gof does not coincide.<\/p>\n

Question 19.
\nNumber of binary operations on the set {a, b} are
\n(A) 10
\n(B) 16
\n(C) 20
\n(D) 8
\nAnswer:
\nNumber of Binary operation on a set {a, b} is
\nn (A) = 2\u00a0 \u00a0 \u2234 n(A\u00d7A) = 4
\nnumber of binary operation is 24<\/sup> = 16.
\ncorrect answer is B.<\/p>\n

2nd PUC Maths Relations and Functions Miscellaneous Exercise Additional Questions and Answers<\/h3>\n

One Mark Questions:<\/p>\n

Question 1.
\nf: R \u2192 R defined by \\(\\left(f(x)=\\left(3-x^{3}\\right)^{\\frac{1}{3}}\\right.\\) find fof (x) [CBSE 2010]
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nIf\u00a0 \\(\\mathbf{f}(\\mathbf{x})=\\frac{1}{1-\\mathbf{x}} \\forall \\mathbf{x} \\in \\mathbf{R}, \\mathbf{n} \\neq \\mathbf{1}, \\text { Find } \\mathbf{f}[\\mathbf{f}(\\mathbf{f}(\\mathbf{x}))]\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nLet A = (1, 2, 3}, B = (4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A \u2192 B. State whether f is one-one or not f (1) = 4, f(2) = 5, f (3) = 6 [CBSE 2011]
\nAnswer:
\nf (x1<\/sub>) = f(x2<\/sub>) \u2192 x1<\/sub> = x2<\/sub> \u2200 n \u2208 A
\nhence f is one-one.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\n* : R x R \u2192 R is a binary operation, a* b = 2a + b . Find (2* 3) * 4 [CBSE 2011]
\nAnswer:
\n(2 * 3) *4 = (4 + 7) *4 = 11 * 4 = 26.<\/p>\n

Question 5.
\nIf \\(f(n)=\\frac{4 n+3}{6 n-4}, n \\neq \\frac{2}{3}\\). show that f[f(n)]= n
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Four Marks Questions:<\/p>\n

Question 1.
\nShow that the function f: R \u2192 R defined by \\(f(n)=\\frac{2 n-1}{3}, n \\in R\\) is one -one onto function .Also find the inverse of f.
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nLet f : R \u2192 R defined as f(n) = 10n + 7 . Find the function g : R\u2192 R such the
\ngof = fog = IR
\n<\/sub>Answer:
\n\"2nd<\/p>\n

Question 3.
\nShow that f : A \u2192 B where A = (0,2 \u03c0) and B = [-1,1] such that fin) = sin n is not. invertible\u00a0 [I I T]
\nAnswer:
\n\\(\\text { Let } n_{1}=\\frac{\\pi}{4}, \\text { and } n_{2}=\\frac{3 \\pi}{4}\\)
\n\\(f\\left(n_{1}\\right)=\\frac{1}{\\sqrt{2}} f\\left(n_{2}\\right)=\\frac{1}{\\sqrt{2}}\\)
\nf(n1<\/sub>) = f(n2<\/sub>) \u21d2 n1<\/sub>\u2260 n2
\n<\/sub>hence it is not one-one
\n\u2200 y \u2208 [-1, 1] \u2203 n \u2208 (0, 2\u03c0) such that f (n) = sin n
\nhence f is onto
\nsince f is not one-one.
\nIt is not invertible.<\/p>\n

Question 4.
\nIs the function sin (sin1<\/sup> n) bijective? [I I T] Which is defined on [-1, 1] to [-1, 1].
\nAnswer:
\nLet f (n) = sin (sin-1<\/sup> n)
\nlet n1<\/sub>\u00a0n2<\/sub> \u2208 [-1, 1]
\nf (n1<\/sub>) = f (n2<\/sub>) \u21d2 sin (sin-1<\/sup> n1<\/sub>) = sin [sin1<\/sup> (n2<\/sub>)]
\n\u21d2 n1 <\/sub>= n2<\/sub>
\nhence f is one-one
\nlet y = sin (sin-1<\/sup> n) = n
\n\u2200 y \u2208\u00a0 [-1,1] \u2203 [-1, 1] f(n) = n
\nhence f is onto
\nsince f is one-one and onto it is bijective.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIf f be greatest integer function and g be the absolute function. Find
\n\\(fog\\left( \\cfrac { -3 }{ 2 } \\right) +gof\\left( \\cfrac { 4 }{ 3 } \\right) \\) [CBSE 2007]
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 1 Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Miscellaneous Exercise Question 1. Let f : R …<\/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\/17702"}],"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=17702"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17702\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17702"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17702"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17702"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}