{"id":118922,"date":"2022-04-29T17:23:19","date_gmt":"2022-04-29T11:53:19","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=118922"},"modified":"2022-05-02T11:34:58","modified_gmt":"2022-05-02T06:04:58","slug":"chapter-2-polynomials-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-2-polynomials-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 2 – Polynomials Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

Exercise 2.1<\/h2>\n

1. The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.<\/h2>\n

\"\"
\nSolutions:
\nGraphical method to find zeroes:-
\nTotal number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.<\/h3>\n

(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.<\/h3>\n

(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.<\/h3>\n

(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.<\/h3>\n

(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.<\/h3>\n

(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.
\n(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.<\/h3>\n

Exercise 2.2<\/h2>\n

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.<\/h2>\n

Solutions:
\n(i) x2\u20132x \u20138
\n\u21d2x2\u2013 4x+2x\u20138 = x(x\u20134)+2(x\u20134) = (x-4)(x+2)
\nTherefore, zeroes of polynomial equation x2\u20132x\u20138 are (4, -2)
\nSum of zeroes = 4\u20132 = 2 = -(-2)\/1 = -(Coefficient of x)\/(Coefficient of x2)
\nProduct of zeroes = 4\u00d7(-2) = -8 =-(8)\/1 = (Constant term)\/(Coefficient of x2)<\/h3>\n

(ii) 4s2\u20134s+1
\n\u21d24s2\u20132s\u20132s+1 = 2s(2s\u20131)\u20131(2s-1) = (2s\u20131)(2s\u20131)
\nTherefore, zeroes of polynomial equation 4s2\u20134s+1 are (1\/2, 1\/2)
\nSum of zeroes = (\u00bd)+(1\/2) = 1 = -(-4)\/4 = -(Coefficient of s)\/(Coefficient of s2)
\nProduct of zeros = (1\/2)\u00d7(1\/2) = 1\/4 = (Constant term)\/(Coefficient of s2\u00a0)<\/h3>\n

(iii) 6×2\u20133\u20137x
\n\u21d26×2\u20137x\u20133 = 6×2\u00a0\u2013 9x + 2x \u2013 3 = 3x(2x \u2013 3) +1(2x \u2013 3) = (3x+1)(2x-3)
\nTherefore, zeroes of polynomial equation 6×2\u20133\u20137x are (-1\/3, 3\/2)
\nSum of zeroes = -(1\/3)+(3\/2) = (7\/6) = -(Coefficient of x)\/(Coefficient of x2)
\nProduct of zeroes = -(1\/3)\u00d7(3\/2) = -(3\/6) = (Constant term) \/(Coefficient of x2\u00a0)<\/h3>\n

(iv) 4u2+8u
\n\u21d2 4u(u+2)
\nTherefore, zeroes of polynomial equation 4u2\u00a0+ 8u are (0, -2).
\nSum of zeroes = 0+(-2) = -2 = -(8\/4) = = -(Coefficient of u)\/(Coefficient of u2)
\nProduct of zeroes = 0\u00d7-2 = 0 = 0\/4 = (Constant term)\/(Coefficient of u2\u00a0)<\/h3>\n

(v) t2\u201315
\n\u21d2 t2\u00a0= 15 or t = \u00b1\u221a15
\nTherefore, zeroes of polynomial equation t2\u00a0\u201315 are (\u221a15, -\u221a15)
\nSum of zeroes =\u221a15+(-\u221a15) = 0= -(0\/1)= -(Coefficient of t) \/ (Coefficient of t2)
\nProduct of zeroes = \u221a15\u00d7(-\u221a15) = -15 = -15\/1 = (Constant term) \/ (Coefficient of t2\u00a0)<\/h3>\n

(vi) 3×2\u2013x\u20134
\n\u21d2 3×2\u20134x+3x\u20134 = x(3x-4)+1(3x-4) = (3x \u2013 4)(x + 1)
\nTherefore, zeroes of polynomial equation3x2\u00a0\u2013 x \u2013 4 are (4\/3, -1)
\nSum of zeroes = (4\/3)+(-1) = (1\/3)= -(-1\/3) = -(Coefficient of x) \/ (Coefficient of x2)
\nProduct of zeroes=(4\/3)\u00d7(-1) = (-4\/3) = (Constant term) \/(Coefficient of x2\u00a0)<\/h3>\n

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
\n(i) 1\/4 , -1<\/h2>\n

Solution:
\nFrom the formulas of sum and product of zeroes, we know,
\nSum of zeroes = \u03b1+\u03b2
\nProduct of zeroes = \u03b1 \u03b2
\nSum of zeroes = \u03b1+\u03b2 = 1\/4
\nProduct of zeroes = \u03b1 \u03b2 = -1<\/h3>\n

\u2234 If \u03b1 and \u03b2 are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
\nx2\u2013(\u03b1+\u03b2)x +\u03b1\u03b2 = 0
\nx2\u2013(1\/4)x +(-1) = 0
\n4×2\u2013x-4 = 0
\nThus,4×2\u2013x\u20134 is the\u00a0quadratic polynomial.
\n(ii)\u221a2, 1\/3<\/h2>\n

Solution:
\nSum of zeroes = \u03b1 + \u03b2 =\u221a2
\nProduct of zeroes = \u03b1 \u03b2 = 1\/3
\n\u2234 If \u03b1 and \u03b2 are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
\nx2\u2013(\u03b1+\u03b2)x +\u03b1\u03b2 = 0
\nx2\u00a0\u2013(\u221a2)x + (1\/3) = 0
\n3×2-3\u221a2x+1 = 0
\nThus, 3×2-3\u221a2x+1\u00a0is the\u00a0quadratic polynomial.<\/h3>\n

(iii) 0, \u221a5<\/h2>\n

Solution:
\nGiven,
\nSum of zeroes = \u03b1+\u03b2 = 0
\nProduct of zeroes = \u03b1 \u03b2 = \u221a5<\/h3>\n

\u2234 If \u03b1 and \u03b2 are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly
\nas:-
\nx2\u2013(\u03b1+\u03b2)x +\u03b1\u03b2 = 0
\nx2\u2013(0)x +\u221a5= 0
\nThus, x2+\u221a5\u00a0is the\u00a0quadratic polynomial.<\/h3>\n

(iv) 1, 1<\/h2>\n

Solution:
\nGiven,
\nSum of zeroes = \u03b1+\u03b2 = 1
\nProduct of zeroes = \u03b1 \u03b2 = 1
\n\u2234 If \u03b1 and \u03b2 are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
\nx2\u2013(\u03b1+\u03b2)x +\u03b1\u03b2 = 0
\nx2\u2013x+1 = 0
\nThus , x2\u2013x+1is the\u00a0quadratic polynomial.<\/h3>\n

(v) -1\/4, 1\/4<\/h2>\n

Solution:
\nGiven,
\nSum of zeroes = \u03b1+\u03b2 = -1\/4
\nProduct of zeroes = \u03b1 \u03b2 = 1\/4
\n\u2234 If \u03b1 and \u03b2 are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
\nx2\u2013(\u03b1+\u03b2)x +\u03b1\u03b2 = 0
\nx2\u2013(-1\/4)x +(1\/4) = 0
\n4×2+x+1 = 0
\nThus,4×2+x+1 is the\u00a0quadratic polynomial.<\/h3>\n

(vi) 4, 1<\/h2>\n

Solution:
\nGiven,
\nSum of zeroes = \u03b1+\u03b2 =4
\nProduct of zeroes = \u03b1\u03b2 = 1
\n\u2234 If \u03b1 and \u03b2 are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
\nx2\u2013(\u03b1+\u03b2)x+\u03b1\u03b2 = 0
\nx2\u20134x+1 = 0
\nThus, x2\u20134x+1 is the\u00a0quadratic polynomial.<\/h3>\n

Exercise 2.3<\/h2>\n

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
\n(i)\u00a0p(x) = x3-3×2+5x\u20133 , g(x) = x2\u20132<\/h2>\n

Solution:
\nGiven,
\nDividend = p(x) = x3-3×2+5x\u20133
\nDivisor = g(x) = x2\u2013 2<\/h3>\n

\"\"
\nTherefore, upon division we get,
\nQuotient = x\u20133
\nRemainder = 7x\u20139<\/h3>\n

(ii) p(x) = x4-3×2+4x+5 , g(x) = x2+1-x<\/h2>\n

Solution:
\nGiven,
\nDividend = p(x) = x4\u00a0\u2013 3×2\u00a0+ 4x +5
\nDivisor = g(x) = x2\u00a0+1-x<\/h3>\n

\"\"
\nTherefore, upon division we get,
\nQuotient = x2\u00a0+ x\u20133
\nRemainder = 8<\/h3>\n

(iii) p(x) =x4\u20135x+6, g(x) = 2\u2013x2<\/h2>\n

Solution:
\nGiven,
\nDividend = p(x) =x4\u00a0\u2013 5x + 6 = x4\u00a0+0x2\u20135x+6
\nDivisor = g(x) = 2\u2013x2\u00a0= \u2013x2+2<\/h3>\n

\"\"<\/p>\n

Therefore, upon division we get,
\nQuotient = -x2-2
\nRemainder = -5x + 10<\/h3>\n

2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
\n(i) t2-3, 2t4\u00a0+3t3-2t2-9t-12<\/h2>\n

Solutions:
\nGiven,
\nFirst polynomial = t2-3
\nSecond polynomial = 2t4\u00a0+3t3-2t2\u00a0-9t-12<\/h3>\n

\"\"
\nAs we can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor of 2t4\u00a0+3t3-2t2\u00a0-9t-12.<\/h3>\n

(ii)x2+3x+1 , 3×4+5×3-7×2+2x+2<\/h2>\n

Solutions:
\nGiven,
\nFirst polynomial = x2+3x+1
\nSecond polynomial = 3×4+5×3-7×2+2x+2<\/h3>\n

\"\"
\nAs we can see, the remainder is left as 0. Therefore, we say that, x2\u00a0+ 3x + 1 is a factor of 3×4+5×3-7×2+2x+2.<\/h3>\n

(iii) x3-3x+1, x5-4×3+x2+3x+1<\/h2>\n

Solutions:
\nGiven,
\nFirst polynomial = x3-3x+1
\nSecond polynomial = x5-4×3+x2+3x+1<\/h3>\n

\"\"
\nAs we can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is not a factor of x5-4×3+x2+3x+1 .<\/h3>\n

3. Obtain all other zeroes of 3×4+6×3-2×2-10x-5, if two of its zeroes are \u221a(5\/3) and \u2013 \u221a(5\/3).<\/h2>\n

Solutions:
\nSince this is a polynomial equation of degree 4, hence there will be total 4 roots.
\n\u221a(5\/3) and \u2013 \u221a(5\/3)\u00a0are zeroes of polynomial f(x).
\n\u2234\u00a0(x \u2013\u221a(5\/3)) (x+\u221a(5\/3)\u00a0= x2-(5\/3) = 0
\n(3×2\u22125)=0,\u00a0is a factor of given polynomial f(x).
\nNow, when we will divide f(x) by (3×2\u22125) the quotient obtained will also be a factor of f(x) and the remainder will be 0.<\/h3>\n

\"\"
\nTherefore, 3×4\u00a0+6×3\u00a0\u22122×2\u00a0\u221210x\u20135 = (3×2\u00a0\u20135)(x2+2x+1)
\nNow, on further factorizing (x2+2x+1) we get,
\nx2+2x+1\u00a0= x2+x+x+1 = 0
\nx(x+1)+1(x+1) = 0
\n(x+1)(x+1) = 0
\nSo, its zeroes are given by:\u00a0x= \u22121\u00a0and\u00a0x = \u22121.
\nTherefore, all four zeroes of given polynomial equation are:
\n\u221a(5\/3),- \u221a(5\/3) , \u22121 and \u22121.
\nHence, is the answer.<\/h3>\n

4. On dividing x3-3×2+x+2\u00a0by a polynomial g(x), the quotient and remainder were x\u20132 and \u20132x+4, respectively. Find g(x).<\/h2>\n

Solution:
\nGiven,
\nDividend, p(x) = x3-3×2+x+2
\nQuotient = x-2
\nRemainder = \u20132x+4
\nWe have to find the value of Divisor, g(x) =?
\nAs we know,
\nDividend = Divisor \u00d7 Quotient + Remainder
\n\u2234 x3-3×2+x+2 = g(x)\u00d7(x-2) + (-2x+4)
\nx3-3×2+x+2-(-2x+4) = g(x)\u00d7(x-2)
\nTherefore, g(x) \u00d7 (x-2) = x3-3×2+3x-2
\nNow, for finding g(x) we will divide x3-3×2+3x-2 with (x-2)<\/h3>\n

\"\"
\nTherefore,\u00a0g(x) = (x2\u2013x+1)<\/h3>\n

5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
\n(i) deg p(x) = deg q(x)
\n(ii) deg q(x) = deg r(x)
\n(iii) deg r(x) = 0<\/h2>\n

Solutions:
\nAccording to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)\u22600. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula;
\nDividend = Divisor \u00d7 Quotient + Remainder
\n\u2234 p(x) = g(x)\u00d7q(x)+r(x)
\nWhere r(x) = 0 or degree of r(x)< degree of g(x).
\nNow let us proof the three given cases as per division algorithm by taking examples for each.<\/h3>\n

(i) deg p(x) = deg q(x)
\nDegree of dividend is equal to degree of quotient, only when the divisor is a constant term.
\nLet us take an example, p(x) = 3×2+3x+3 is a polynomial to be divided by g(x) = 3.
\nSo, (3×2+3x+3)\/3 = x2+x+1 = q(x)
\nThus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x).
\nHence, division algorithm is satisfied here.<\/h3>\n

(ii) deg q(x) = deg r(x)
\nLet us take an example, p(x) = x2\u00a0+ 3 is a polynomial to be divided by g(x) = x \u2013 1.
\nSo,\u00a0x2\u00a0+ 3 = (x \u2013 1)\u00d7(x) + (x + 3)
\nHence, quotient q(x) = x
\nAlso, remainder r(x) = x + 3
\nThus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x).
\nHence, division algorithm is satisfied here.<\/h3>\n

(iii) deg r(x) = 0
\nThe degree of remainder is 0 only when the remainder left after division algorithm is constant.
\nLet us take an example, p(x) = x2\u00a0+ 1 is a polynomial to be divided by g(x) = x.
\nSo, x2\u00a0+ 1 = (x)\u00d7(x) + 1
\nHence, quotient q(x) = x
\nAnd, remainder r(x) = 1
\nClearly, the degree of remainder here is 0.
\nHence, division algorithm is satisfied here.<\/h3>\n

Exercise 2.4<\/h2>\n

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
\n(i) 2×3+x2-5x+2; -1\/2, 1, -2<\/h2>\n

Solution:
\nGiven, p(x)\u00a0=\u00a02×3+x2-5x+2
\nAnd zeroes for p(x) are = 1\/2, 1, -2<\/h3>\n

\u2234 p(1\/2) = 2(1\/2)3+(1\/2)2-5(1\/2)+2 = (1\/4)+(1\/4)-(5\/2)+2 = 0
\np(1) = 2(1)3+(1)2-5(1)+2 = 0
\np(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0
\nHence, proved 1\/2, 1, -2 are the zeroes of 2×3+x2-5x+2.
\nNow, comparing the given polynomial with general expression, we get;
\n\u2234 ax3+bx2+cx+d = 2×3+x2-5x+2
\na=2, b=1, c= -5 and d = 2
\nAs we know, if \u03b1, \u03b2, \u03b3 are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;
\n\u03b1 +\u03b2+\u03b3 = \u2013b\/a
\n\u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1 = c\/a
\n\u03b1 \u03b2\u03b3 = \u2013 d\/a.
\nTherefore, putting the values of zeroes of the polynomial,
\n\u03b1+\u03b2+\u03b3 = \u00bd+1+(-2) = -1\/2 = \u2013b\/a
\n\u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1 = (1\/2\u00d71)+(1 \u00d7-2)+(-2\u00d71\/2) = -5\/2 = c\/a
\n\u03b1 \u03b2 \u03b3 = \u00bd\u00d71\u00d7(-2) = -2\/2 = -d\/a
\nHence, the relationship between the zeroes and the coefficients are satisfied.<\/h3>\n

(ii) x3-4×2+5x-2\u00a0;2, 1, 1<\/h2>\n

Solution:
\nGiven, p(x) = x3-4×2+5x-2
\nAnd zeroes for p(x) are 2,1,1.
\n\u2234 p(2)= 23-4(2)2+5(2)-2 = 0
\np(1) = 13-(4\u00d712\u00a0)+(5\u00d71)-2 = 0
\nHence proved, 2, 1, 1 are the zeroes of x3-4×2+5x-2
\nNow, comparing the given polynomial with general expression, we get;
\n\u2234 ax3+bx2+cx+d = x3-4×2+5x-2
\na = 1, b = -4, c = 5 and d = -2
\nAs we know, if \u03b1, \u03b2, \u03b3 are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;
\n\u03b1 + \u03b2 + \u03b3 = \u2013b\/a
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = c\/a
\n\u03b1 \u03b2 \u03b3 = \u2013 d\/a.
\nTherefore, putting the values of zeroes of the polynomial,
\n\u03b1 +\u03b2+\u03b3 = 2+1+1 = 4 = -(-4)\/1 = \u2013b\/a
\n\u03b1\u03b2+\u03b2\u03b3+\u03b3\u03b1 = 2\u00d71+1\u00d71+1\u00d72 = 5 = 5\/1= c\/a
\n\u03b1\u03b2\u03b3 = 2\u00d71\u00d71 = 2 = -(-2)\/1 = -d\/a
\nHence, the relationship between the zeroes and the coefficients are satisfied.<\/h3>\n

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, \u20137, \u201314 respectively.<\/h2>\n

Solution:
\nLet us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be \u03b1, \u03b2, \u03b3.
\nAs per the given question,
\n\u03b1+\u03b2+\u03b3 = -b\/a = 2\/1
\n\u03b1\u03b2 +\u03b2\u03b3+\u03b3\u03b1 = c\/a = -7\/1
\n\u03b1 \u03b2\u03b3 = -d\/a = -14\/1
\nThus, from above three expressions we get the values of coefficient of polynomial.
\na = 1, b = -2, c = -7, d = 14
\nHence, the cubic polynomial is x3-2×2-7x+14<\/h3>\n

3. If the zeroes of the polynomial x3-3×2+x+1\u00a0are a \u2013 b, a, a + b, find a and b.<\/h2>\n

Solution:
\nWe are given with the polynomial here,
\np(x) = x3-3×2+x+1
\nAnd zeroes are given as a \u2013 b, a, a + b
\nNow, comparing the given polynomial with general expression, we get;
\n\u2234px3+qx2+rx+s = x3-3×2+x+1
\np = 1, q = -3, r = 1 and s = 1
\nSum of zeroes = a \u2013 b + a + a + b
\n-q\/p = 3a
\nPutting the values q and p.
\n-(-3)\/1 = 3a
\na=1
\nThus, the zeroes are 1-b, 1, 1+b.
\nNow, product of zeroes = 1(1-b)(1+b)
\n-s\/p = 1-b2
\n-1\/1 = 1-b2
\nb2\u00a0= 1+1 = 2
\nb = \u00b1\u221a2
\nHence,1-\u221a2, 1 ,1+\u221a2 are the zeroes of x3-3×2+x+1.<\/h3>\n

4. If two zeroes of the polynomial x4-6×3-26×2+138x-35\u00a0are 2 \u00b1\u221a3,\u00a0find other zeroes.<\/h2>\n

Solution:
\nSince this is a polynomial equation of degree 4, hence there will be total 4 roots.
\nLet f(x) = x4-6×3-26×2+138x-35
\nSince 2 +\u221a3\u00a0and 2-\u221a3\u00a0are zeroes of given polynomial f(x).
\n\u2234 [x\u2212(2+\u221a3)] [x\u2212(2-\u221a3)] = 0
\n(x\u22122\u2212\u221a3)(x\u22122+\u221a3) = 0
\nOn multiplying the above equation we get,
\nx2-4x+1, this is a factor of a given polynomial f(x).
\nNow, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.<\/h3>\n

\"\"
\nSo, x4-6×3-26×2+138x-35 = (x2-4x+1)(x2\u00a0\u20132x\u221235)
\nNow, on further factorizing (x2\u20132x\u221235) we get,
\nx2\u2013(7\u22125)x \u221235\u00a0= x2\u2013 7x+5x+35 = 0
\nx(x \u22127)+5(x\u22127) = 0
\n(x+5)(x\u22127) = 0
\nSo, its zeroes are given by:
\nx= \u22125 and x = 7.
\nTherefore, all four zeroes of given polynomial equation are: 2+\u221a3\u00a0, 2-\u221a3,\u00a0\u22125 and 7.<\/h3>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Class 10 Mathematics NCERT book solutions for Chapter 2 – Polynomials Questions and Answers.<\/p>\n","protected":false},"author":21830,"featured_media":117421,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":{"0":"post-118922","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-education"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118922","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=118922"}],"version-history":[{"count":3,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118922\/revisions"}],"predecessor-version":[{"id":118966,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/118922\/revisions\/118966"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/117421"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=118922"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=118922"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=118922"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}