{"id":2175,"date":"2023-10-02T10:33:00","date_gmt":"2023-10-02T10:33:00","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2175"},"modified":"2023-10-02T10:35:06","modified_gmt":"2023-10-02T10:35:06","slug":"week-6-and-7-ss1-second-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-6-and-7-ss1-second-term-further-mathematics-notes\/","title":{"rendered":"Week 6 and 7 – SS1 Second Term Further Mathematics Notes"},"content":{"rendered":"

WEEK SIX \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0DATE\u2026\u2026\u2026\u2026\u2026
\n<\/h3>\n

Revision of half term work <\/strong><\/p>\n

\u00a0
\n\u00a0<\/p>\n

WEEK SEVEN \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0DATE\u2026\u2026\u2026\u2026\u2026
\n<\/h3>\n

CONTENT: <\/strong><\/p>\n

Composite Mapping and Inverse Mapping
\n<\/h4>\n

\u00a0COMPOSITE MAPPING: <\/strong>
\n\tA mapping is composite when the co- domain of the first mapping is the domain of the second mapping.
\n\t\t\t<\/strong>Consider the mapping f;X\u2192 Z and g: Z\u2192Y
\n\t\t\t<\/strong><\/p>\n

\u00a0 X Z Yf \u00a0\u00a0\u00a0\u00a0g
\n\t\t\t\t<\/sup>
\n\t\"\"\/
\n\tThe mapping f takes an element in X and produces an image in Z, and then the mapping g takes an element in Z and produces an image in Y. It can be denoted by gf or gof. <\/p>\n

\u00a0Example 1. The mappings f and g are defined by the diagram below: <\/p>\n

\u00a0 F g
\n\"\"\/<\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0
\n\u00a0Determine
\n(a) f(-3) +f(4) (b) f(2) +g(-5) (c) g[f(-3)] (d) g[(-3)] +g[f(4)] <\/p>\n

\u00a0SOLUTION<\/strong>: <\/p>\n

    \n
  1. F(-3) + F(4) = -5 + 9 = 4\n<\/li>\n
  2. F(2) + F(-5) = 5 +(-10) = -5\n<\/li>\n
  3. G[f(-3)] = g(-5) = -10\n<\/li>\n
  4. G[f(-3)] + g[f(4)] = g(-5) + g(9) = -10 + 18 = 8\n<\/li>\n<\/ol>\n

    \u00a0Example 2: The functions f and g on the set of real numbers are defined by f(x) = 3x-1 and g(x) = 5x+2 respectively. Find (a) F [g(x)] (b) g [f(x)] (c) 2f(x) \u2013 g(x)
    \nSOLUTION: <\/strong><\/p>\n

      \n
    1. f[g(x)] = f(5x+2), 5x+2 will represent x in f(x) f ( 5x +2) = 3 (5x+2) \u2013 1\n<\/li>\n<\/ol>\n

      = 15x +5 <\/p>\n

        \n
      1. g [f(x)] = g( 3x-1)\n<\/li>\n<\/ol>\n

        = g(3x-1) = 5(3x-1) + 2
        \n = 15x -5 +2
        \n = 15x-3 <\/p>\n

          \n
        1. 2f(x) \u2013 g(x) = 2(3x-1) \u2013 (5x+2)\n<\/li>\n<\/ol>\n

          = 6x -2 -5x -2
          \n = x-4
          \nINVERSE MAPPING<\/strong>:
          \nA function has an inverse if it’s both one- one and onto. Consider the function f(x) = x-3 on the set p={ -1, 5, 9} into set Q ={ -2,1,3 }
          \n A F B <\/p>\n

          \u00a0
          \n\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0
          \n\u00a0
          \n\u00a0
          \n\u00a0\"\"\/If we reverse the function, that is making range of F to be the domain of the inverse function.
          \n Therefore, B g A <\/p>\n

          \u00a0
          \n\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 <\/p>\n

          \u00a0
          \n\u00a0
          \n\u00a0 g represents the inverse function of f i.e. f-1<\/sup>, to obtain function f-1<\/sup>, we follow the procedure below: f(x) = x \u2013 3 2 y = x-3
          \n\t 2
          \nMake x the subject of the formula
          \n 2y = x-3 x = 2y + 3 Then, f-1<\/sup>(x) = 2x+3. <\/p>\n

          \u00a0Example: The function f is defined on the ser of real numbers by f(x) = 2x-1, (x\u2260-2\/3)
          \n 3x +2
          \nDetermine (a) f-1<\/sup>(x) (b) f-1<\/sup> (-2) (c) determine the largest domain of f-1<\/sup>(x) <\/p>\n

          \u00a0
          \n\u00a0SOLUTION<\/strong>: <\/p>\n

            \n
          1. F(x) = 2x-1 3x+2 f-1<\/sup>(x), y = 2x-1 3x+2\n<\/li>\n<\/ol>\n

            (3x+2) y = 2x-1
            \n 3xy +2y = 2x-1 3xy-2x= -1-2y x (3y-2)= -1-2y x = -1-2y 3y -2 f-1<\/sup>= -1-2x 3x-2 <\/p>\n

              \n
            1. f-1<\/sup>(-2) i.e x= -2 in f-1<\/sup>(x) f-1<\/sup>(-2) = -1-2(-2) 3(-2)-2\n<\/li>\n<\/ol>\n

              = -1+4
              \n\t -6\u00a0\u00a0\u00a0\u00a02
              \n =\u00a0\u00a0\u00a0\u00a03
              \n\t\t (c) The largest domain of f-1<\/sup>(x) is all real values of x except 2\/3 \u00a0\u00a0\u00a0\u00a08 <\/p>\n

              \u00a0
              \n\u00a0<\/p>\n

              EVALUATION
              \n<\/h3>\n

              1. \u00a0\u00a0\u00a0\u00a0Given g(x) = x3<\/sup> and h(x) = 4x +1 <\/p>\n

                \n
              1. find the value of g(2) + h(2)\n<\/li>\n
              2. find the value of h[g(2)]\n<\/li>\n
              3. find the value of 3g(-1)-4h(-1)\n<\/li>\n<\/ol>\n

                \u00a02. \u00a0\u00a0\u00a0\u00a0A function g(x)=\u221a(x-2), x \u2265 2, find g-1<\/sup>(x) and g-1<\/sup>(4) <\/p>\n

                \u00a0<\/p>\n

                GENERAL EVALUATION
                \n<\/h3>\n
                  \n
                1. Determine the values of p and q if (x -1) and (x + 2) are factors of 2x3<\/sup> + px2<\/sup> \u2013x+ q\n<\/li>\n
                2. If f(x) = 6x3<\/sup> + 13x2<\/sup> +2x \u2013 5, shows that f(-1)= 0\n<\/li>\n
                3. \n
                  Given that f(x) = x2<\/sup>
                  \n\t\t\t\t and g(x)= \u221a(x- 2), x\u22602 find x2<\/sup> + 2\n<\/div>\n
                    \n
                  1. f-1<\/sup>(x)\n<\/li>\n
                  2. g-1<\/sup>(x)\n<\/li>\n
                  3. F(g(x)\n<\/li>\n
                  4. The value of x for which f(g(x) is not undefined.\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                    \u00a0READING ASSIGNMENT<\/strong>: Read Mapping, Further Mathematics Project 2, and page 32- 41. <\/p>\n

                    \u00a0<\/p>\n

                    WEEKEND ASSIGNMENT
                    \n<\/h3>\n
                      \n
                    1. Given that f(x) = x2<\/sup>+ 4x+ 3, for what values of x is f(x) = f(x+1).\n<\/li>\n<\/ol>\n

                      A. -11 B -5 C -3
                      \n\t 2 2 4 <\/p>\n

                        \n
                      1. Given f(x) = x2<\/sup> -1 and g(x) = 2x+3, determine the formula for gf(x)\n<\/li>\n<\/ol>\n

                        A. 2x2<\/sup> +4x+1 B. 2x2<\/sup> +1 C. x+1 <\/p>\n

                          \n
                        1. Given g(x) = xn<\/sup> and g (3) = 81, determine the value of n.\n<\/li>\n<\/ol>\n

                          A -4 B 27 C 4 <\/p>\n

                            \n
                          1. Given that the image of x under the mapping f(x) \u2192 3x +2 is -10. What is the value of x.\n<\/li>\n<\/ol>\n

                            A -4 B -28 C 0 <\/p>\n

                              \n
                            1. If f is a function defined by f(x) = 2x-3, find ff. A. 4x-6 B 2x+3 C. 4x +6\n<\/li>\n<\/ol>\n

                              \u00a0<\/p>\n

                              \u00a0\u00a0\u00a0\u00a0THEORY
                              \n<\/h3>\n
                                \n
                              1. Given the functions f(x) = 3x2<\/sup> \u2013x+5, g(x) = 6x3<\/sup> + 7×2+7x+15. Simplify, as far as possible, the expressions \u00a0\u00a0\u00a0\u00a0(a) 3f(x) – g(x) (b) f(x) g(x) (c) g(x)\/f(x)\n<\/li>\n<\/ol>\n

                                \u00a0<\/p>\n

                                  \n
                                1. A relation R is defined by g(x) = 2 , x \u2260 2, find g-1<\/sup>(x).\n<\/li>\n<\/ol>\n

                                  x-2 <\/p>\n

                                  \u00a0
                                  \n\t\t\t<\/strong>
                                  \n\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

                                  WEEK SIX \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0DATE\u2026\u2026\u2026\u2026\u2026 Revision of half…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,188],"tags":[],"class_list":["post-2175","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2175","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2175"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2175\/revisions"}],"predecessor-version":[{"id":2176,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2175\/revisions\/2176"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2175"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2175"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2175"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}