{"id":40594,"date":"2024-02-17T07:42:36","date_gmt":"2024-02-17T02:12:36","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=40594"},"modified":"2024-02-17T16:04:22","modified_gmt":"2024-02-17T10:34:22","slug":"plus-one-maths-chapter-wise-questions-answers-chapter-2","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-one-maths-chapter-wise-questions-answers-chapter-2\/","title":{"rendered":"Plus One Maths Chapter Wise Questions and Answers Chapter 2 Relations and Functions"},"content":{"rendered":"

Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 2 Relations and Functions<\/h2>\n

Plus One Maths Relations and Functions Three Mark Questions and Answers<\/h3>\n

Question 1.
\nLet A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by R = {(a, b): a, b \u2208 A, b is exactly divisible by a}<\/p>\n

    \n
  1. Write R in roster form. (1)<\/li>\n
  2. Find the domain of R. (1)<\/li>\n
  3. Find the range of R. (1)<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 4), (2, 2), (4,4), (6,6), (3,3), (3,6)}<\/li>\n
    2. Domain of R = {1, 2, 3, 4, 6}<\/li>\n
    3. Range of R = {1, 2, 3, 4, 6}<\/li>\n<\/ol>\n

      Question 2.
      \nDetermine the domain and range of the relation R defined by R = {(x, x + 5) : x \u2208 {0, 1, 2, 3, 4, 5}}
      \nAnswer:
      \nR = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9),(5, 10)}
      \nDomain of R = {0, 1, 2, 3, 4, 5}
      \nRange of R = {5, 6, 7, 8, 9, 10}<\/p>\n

      Question 3.
      \nA function f is defined as f(x) = 2x – 5, Write down the values of f(0), f(7), f(-3).
      \nAnswer:
      \nGiven; f(x) = 2x – 5
      \nf(0) = -5;
      \nf(7) = 2(7) – 5 = 14 – 5 = 9
      \nf(-3) = 2(-3) – 5 = -6 – 5 = -11<\/p>\n

      Question 4.
      \nFind the range of the following functions.<\/p>\n

        \n
      1. f(x) = 2 – 3x, x \u2208 R, x>0 (1)<\/li>\n
      2. f(x) = x2<\/sup> + 2, x is a real number. (1)<\/li>\n
      3. f(x) = x, x is a real number. (1)<\/li>\n<\/ol>\n

        Answer:<\/p>\n

          \n
        1. Given; f(x) = 2 – 3x is a first degree polynomial function, therefore the range is R.<\/li>\n
        2. Given; f(x) = x2<\/sup> + 2, The range of x2 is [0, \u221e) , then the range of f(x) = x2<\/sup> + 2 is [2, \u221e)<\/li>\n
        3. Given; f(x) = x is the identity function, therefore the range is R.<\/li>\n<\/ol>\n

          Plus One Maths Relations and Functions Four Mark Questions and Answers<\/h3>\n

          Question 1.
          \nLet A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that<\/p>\n

            \n
          1. A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C) (2)<\/li>\n
          2. A \u00d7 C is a subset of B \u00d7 D (2)<\/li>\n<\/ol>\n

            Answer:
            \n1. A \u00d7 (B \u2229 C) ={1, 2} \u00d7 \u03a6 = \u03a6
            \nA \u00d7 B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
            \nA \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6)}
            \n(A \u00d7 B) \u2229 (A \u00d7 C) = \u03a6
            \nHence; A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n

            2. A \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6)}
            \nB \u00d7 D = {1, 2, 3, 4} \u00d7 {5, 6, 7, 8}
            \n= {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3,6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
            \nHence A \u00d7 C is a subset of B \u00d7 D.<\/p>\n

            Question 2.
            \nThe arrow diagram given below shows a relation R from P to Q. Write the relation in roster form, set-builder form. Find its domain and range.
            \n\"Plus
            \nAnswer:
            \nR – {(9, -3), (9, 3), (4, -2), (4, 2), (25, -5), (25, 5)}
            \nR = {{x, y) : y2<\/sup> = x}
            \nDomain of R = {9, 4, 25}
            \nRange of R = {5, 3, 2, -2, -3, -5}<\/p>\n

            Question 3.
            \nFind the domain of the following.<\/p>\n

              \n
            1. f(x) = \\(\\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\\) (2)<\/li>\n
            2. f(x) = \\(\\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\\) (2)<\/li>\n<\/ol>\n

              Answer:
              \n1. Given; f(x) = \\(\\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\\)
              \nThe function is not defined at points where the denominator becomes zero.
              \nx2<\/sup> – 8x +12 = 0 \u21d2 (x – 6)(x – 2) = 0 \u21d2 x = 2, 6
              \nTherefore domain of fis R – {2, 6}.<\/p>\n

              2. Given; f(x) = \\(\\frac{x^{2}+3 x+5}{x^{2}-5 x+4}\\)
              \nThe function is not defined at points where the denominator becomes zero.
              \nx2<\/sup> – 5x + 4 = 0 \u21d2 (x – 4)(x -1) = 0 \u21d2 x = 1, 4
              \nTherefore domain of f is R – {1, 4}.<\/p>\n

              Question 4.
              \nLet f(x) = \\(=\\sqrt{x}\\) and g(x) = x be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \\(\\left(\\frac{f}{g}\\right)(x)\\).
              \nAnswer:
              \n(f + g)(x) = f(x) + g(x) = \\(=\\sqrt{x}\\) + x
              \n(f – g)(x) = f(x) – g(x) = \\(=\\sqrt{x}\\) – x
              \n(fg)(x) = f(x) \u00d7 g(x) = \\(=\\sqrt{x}\\) \u00d7 x = \\(x^{\\frac{3}{2}}\\)
              \n\"Plus<\/p>\n

              Question 5.
              \nLet f(x) = x2<\/sup> and g(x) = 2x + 1 be two functions defined over the set of nonnegative real numbers. Find (f + g)(x), (f – g)(x), (fg)(x) and \\(\\left(\\frac{f}{g}\\right)(x)\\).
              \nAnswer:
              \n(f + g)(x) = f(x) + g(x) = x2<\/sup> + 2x + 1
              \n(f – g)(x) = f(x) – g(x) = x2<\/sup> – 2x – 1
              \nf(fg)(x) = f(x) \u00d7 g(x)
              \n= x2<\/sup>(2x +1) = 2x3<\/sup> + x2<\/sup>
              \n\"Plus<\/p>\n

              Question 6.
              \nA = {1, 2}, B = {3, 4}<\/p>\n

                \n
              1. Write A \u00d7 B<\/li>\n
              2. Write the relation from A to B in roster form. (1)<\/li>\n
              3. Represent all possible functions from A to B (Arrow diagram may be used) (2)<\/li>\n<\/ol>\n

                Answer:
                \n1. A \u00d7 B = {(1, 3), (1, 4), (2, 3), (2, 4)}<\/p>\n

                2. Any Subset of A \u00d7 B (say R={(1, 3),(2, 4)})<\/p>\n

                3.
                \n\"Plus<\/p>\n

                Plus One Maths Relations and Functions Six Mark Questions and Answers<\/h3>\n

                Question 1.
                \nLet A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find<\/p>\n

                  \n
                1. A \u00d7 (B \u2229 C) (1)<\/li>\n
                2. (A \u00d7 B) \u2229 (A \u00d7 C) (2)<\/li>\n
                3. A \u00d7 (B \u222a C) (1)<\/li>\n
                4. (A \u00d7 B) \u222a (A \u00d7 C) (2)<\/li>\n<\/ol>\n

                  Answer:
                  \n1. A \u00d7 (B \u2229 C) = {1, 2, 3} \u00d7 {4}
                  \n= {(1, 4), (2, 4), (3, 4)}<\/p>\n

                  2. (A \u00d7 B) \u2229 (A \u00d7 C)
                  \n= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} \u2229 {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5) , (2, 6), (3, 4), (3, 5), (3, 6)}
                  \n= {(1, 4), (2, 4), (3, 4)}<\/p>\n

                  3. A \u00d7 (B \u222a C) = {1, 2, 3} \u00d7 {3, 4, 5, 6}
                  \n= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}<\/p>\n

                  4. (A \u00d7 B) \u222a (A \u00d7 C)
                  \n= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} \u222a {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
                  \n= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}<\/p>\n

                  Question 2.
                  \nFind the domain and range of the following
                  \n\"Plus
                  \nAnswer:
                  \ni) Given; f(x) = -|x|
                  \nD(f) = R, R(f) = (-\u221e, 0]<\/p>\n

                  ii) Given; f(x) = \\(\\sqrt{9-x^{2}}\\)
                  \nx can take values where 9 – x2<\/sup> > 0
                  \n\u21d2 x2<\/sup> \u2264 9 \u21d2 -3 \u2264 x \u2264 3 \u21d2 x \u2208 [-3, 3]
                  \nTherefore domain of f is [-3, 3]
                  \nPut \\(\\sqrt{9-x^{2}}\\) = y, where y \u2265 0
                  \n\u21d2 9 – x2<\/sup> = y2<\/sup>\u21d2 x2<\/sup> = 9 – y2<\/sup>
                  \n\u21d2 x = \\(\\sqrt{9-x^{2}}\\)
                  \n\u21d2 9 – y2<\/sup> \u2265 0 \u21d2 y2<\/sup> \u2264 9 \u21d2 -3 \u2264 y \u2264 3
                  \nTherefore range of fis [0, 3].<\/p>\n

                  iii) Given; f(x) = |x – 1|
                  \nDomain of f is R
                  \nThe range of |x| is [0, \u221e) , then the range of
                  \nf(x) = |x -1| is [0, \u221e)<\/p>\n

                  iv) Given; f(x) = \\(\\sqrt{x-1}\\)
                  \nx can take values where x – 1 \u2265 0
                  \n\u21d2 x \u2265 1 \u21d2 x \u2208 [1, \u221e]
                  \nTherefore domain of fis [1, \u221e]
                  \nThe range of \\(\\sqrt{x}\\) is [0, \u221e), then the range of
                  \nf(x) = \\(\\sqrt{x-1}\\) is [0, \u221e).<\/p>\n

                  Plus One Maths Relations and Functions Practice Problems Questions and Answers<\/h3>\n

                  Question 1.
                  \nIf (x + 1, y – 2) = (3, 1), find the values of x and y.
                  \nAnswer:
                  \n(x + 1, y – 2) = (3, 1) \u21d2 x + 1 = 3, y – 2 = 1 \u21d2 x = 2, y = 3.<\/p>\n

                  Question 2.
                  \nIf \\(\\left(\\frac{x}{3}+1, y-\\frac{2}{3}\\right)=\\left(\\frac{5}{3}, \\frac{1}{3}\\right)\\), find the values of x and y.
                  \nAnswer:
                  \n\"Plus<\/p>\n

                  Question 3.
                  \nIf G = {7, 8}; H = {2, 4, 5}, find G \u00d7 H and H \u00d7 G.
                  \nAnswer:<\/p>\n