{"id":3167,"date":"2023-10-04T12:16:42","date_gmt":"2023-10-04T12:16:42","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3167"},"modified":"2023-10-04T12:23:00","modified_gmt":"2023-10-04T12:23:00","slug":"week-1-ss2-second-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-1-ss2-second-term-further-mathematics-notes\/","title":{"rendered":"Week 1 – SS2 Second Term Further Mathematics Notes"},"content":{"rendered":"

SECOND TERM E-LEARNING NOTE
\n\t\t\t<\/strong>
\n\u00a0SUBJECT: FURTHER MATHEMATICS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0CLASS: SS2
\n<\/strong>SCHEME OF WORK
\n<\/strong>
\n\u00a0WEEK \u00a0\u00a0\u00a0\u00a0TOPIC
\n<\/strong><\/p>\n

    \n
  1. Differentiation: Limits of Function and First Principle, Differentiation of Polynomial\n<\/li>\n
  2. Differentiation (Continued): Rules of Differentiation\n<\/li>\n
  3. Differentiation of Transcendents: Derivative of Trigonometric Functions and Exponential Functions.\n<\/li>\n
  4. Application of Differentiation: Rate of Change, Equation of Motion, Maximum and Minimum Points and Values of Functions.\n<\/li>\n
  5. Conic Sections: Equation of Circles, General Equation of Circles, Finding Centre and Radius, Equation and Length of Tangents to a Circle.\n<\/li>\n
  6. Conic Sections: The Parabola, Hyperbola and Ellipse\n<\/li>\n
  7. Review of First Half Term\n<\/li>\n
  8. Statistics Probability: Sample Space, Event Space, Combination of Events, Independents and Dependent Events.\n<\/li>\n
  9. Permutation and Combination\n<\/li>\n
  10. Dynamics: Newton’s Laws of Motion\n<\/li>\n
  11. Work, Energy, Power, Impulse and Momentum\n<\/li>\n
  12. Revision and Examination.\n<\/li>\n<\/ol>\n

    \u00a0REFERENCES
    \n<\/strong>Further Mathematics Project 2 and 3.
    \n\t\t\t<\/strong>
    \n\u00a0WEEK ONE
    \n\t\t\t\t<\/strong>TOPIC : LIMITS OF FUNCTIONS AND DIFFERENTIATION FROM THE FIRST PRINCIPLE
    \n<\/strong>The followings are the properties of limits:<\/p>\n

      \n
    1. lim k = k i e\n<\/li>\n<\/ol>\n

      x2\"\"\/<\/sup>a
      \nThe limit of a constant is the constant itself<\/p>\n

        \n
      1. \n
        lim [f(x) + f 2<\/sub> (x) + f3 (x) <\/em><\/sub>+ \u2026 fn<\/sub>(x)]\n<\/div>\n

        = lim f1<\/sub>(x) + lim f2 <\/sub>(x) + lim f3 <\/sub>(x) +limfn<\/sub>(x)\n<\/li>\n<\/ol>\n

        x\"\"\/a x\"\"\/a x\"\"\/ax\"\"\/a
        \ni.e
        \nThe limit of the sum of a finite number of functions is equal to the sum of their respective limits
        \nlim [f1<\/sub>(x) \u2013 f2<\/sub>(x)] = limf1<\/sub>(x) \u2013 limf2<\/sub>(x)
        \nx\"\"\/a x\"\"\/a x\"\"\/a x\"\"\/a
        \ni.e
        \n The limit of the difference of two functions is equal to the difference of their limits.<\/p>\n

          \n
        1. lim [f1<\/sub>(x) f2 <\/sub>(x) f3<\/sub> (x) + \u2026.. fn<\/sub>(x)]\n<\/li>\n<\/ol>\n

          \"\"\/xa
          \n = lim f1<\/sub>(x) lim f2<\/sub> (x) lim f3<\/sub> (x) lim f (x)
          \nx\"\"\/a x\"\"\/a x\"\"\/a\u00a0\u00a0\u00a0\u00a0x\"\"\/a
          \ni.e
          \n The limit of the product of infinite number of functions is equal to the product of their respective limits.<\/p>\n

            \n
          1. \n
            x a[]= lim f1<\/sub>(x)\n\t\t\t\t<\/div>\n

            lim f2<\/sub>(x)
            \nProvided lim f2<\/sub> (x) \"\"\/ 0 i.e
            \nThe limit of the quotient function is equal to the quotient of their limits provided the limit of the divisor is not equal to zero\n<\/li>\n

          2. lim k f (x) = k lim f (x)\n<\/li>\n<\/ol>\n

            x\"\"\/ax \"\"\/a
            \ni.e
            \nLimit of the product of a constant and a function is equal to the product of the constant and the limit of the function<\/p>\n

            \u00a0Example 1
            \n<\/strong>Evaluate lim( 7 \u2013 2x + 5x2<\/sup> \u2013 4x3<\/sup>)
            \nSolution
            \nlim {7 \u2013 2x + 5x2 <\/sup>– 4x3<\/sup>)
            \nx\"\"\/ a
            \n= lim 7 \"\"\/ 2 lim x + 5 lim x2 \"\"\/<\/sup> 4 lim x
            \nx\"\"\/ ax \"\"\/ax\"\"\/ a x \"\"\/ a
            \n= 7 \"\"\/ 0 + 0 = 7 <\/p>\n

            \u00a0Example 2
            \n<\/strong>Limx2 <\/sup>+ 5x + 9
            \n\t\t\t<\/sup>x\"\"\/0 2x2<\/sup> \u2013 3x + 15
            \nSolution
            \n<\/strong>\"\"\/\"\"\/lim x2 <\/sup>+ 5x + 9 = limx2 <\/sup> + 5x + 9
            \nx2x2<\/sup> \u2013 3x + 15lim2x2 <\/sup>\u2013 3x+15
            \nlim x2<\/sup> +5lim x+ lim 9
            \n\"\"\/x\"\"\/0 x\"\"\/0 x\"\"\/0
            \n\t\t\t<\/sup>2 lim x2 <\/sup>\u2013 3lim x + lim 15
            \nx\"\"\/0 x\"\"\/0 x\"\"\/0
            \n=
            \n=
            \n<\/sup>=
            \n<\/sup>
            \n\u00a0Example <\/strong>
            \n\t\t\"\"\/Evaluate <\/strong>limx2<\/sup><\/strong>\u2013 25
            \n\t\t\t<\/strong>xx\u2013 5
            \nSolution<\/strong>
            \n\t\t\"\"\/Lim x2<\/sup>\u2013 25 = lim =
            \n\t\t\t<\/sup>xx \u2013 5
            \n = lim (x + 5)
            \nx \u2013 5
            \n= lim x + lim 5
            \nx \u2013 5x \u2013 5
            \n= 5 + 5
            \n = 10<\/p>\n

            \u00a0Example
            \n<\/strong>\"\"\/Evaluate lim 3x3<\/sup>+2x2<\/sup>+x+1
            \nx \u2013 5 x3<\/sup> + 2x+ 5
            \nSolution
            \n<\/strong> We know that lim = 0
            \nx \u2013 0
            \nlim3x3<\/sup> + 2x2 <\/sup> + x + 1
            \nx \u2013 0x3 <\/sup>+ 2x + 5
            \nx3<\/sup>(3 + + + )
            \nlimx x= 0
            \nx\"\"\/0 x3<\/sup> ( + )
            \nx x
            \n<\/sup>lim 3 + 2 lim + lim + lim
            \n =x \u2013 0x \u2013 0x \u2013 0x \u2013 0
            \n lim1 + 2 lim+ lim + 5 lim
            \nx \u2013 0x \u2013 0x \u2013 0x \u2013 0
            \n= <\/strong>3+ 0 + 0 + 0
            \n\t\t1 + 0 + 0
            \n=
            \n= 3 <\/p>\n

            \u00a0EVALUATION
            \n<\/strong>Evaluate lim -> 4 x3<\/sup> +4x 6
            \nEvaluastelim x -> -2 x+6\/ 2x +4<\/p>\n

            \u00a0Differentiation From first Principle
            \n<\/strong>The technique adopted in unit 11.3 in finding the derivative of a function from the consideration of the limiting value is called differentiation from first principle.
            \n<\/strong>
            \n\u00a0Example
            \n<\/strong> Find the derivative of f(x) = x2<\/sup> from first principle.
            \nSolution
            \n<\/strong>f (x) = x2<\/sup>
            \n\t\tf(x + \"\"\/x) = ( x + \"\"\/x)2
            \n<\/sup>= x2 <\/sup>+ 2x\"\"\/x + (\"\"\/x)2
            \n<\/sup>f(x + \"\"\/x) \u2013 f (x) = (x + \"\"\/x)2<\/sup> \u2013 x2
            \n<\/sup>= x2<\/sup>+ 2x\"\"\/x + (\"\"\/x)2<\/sup> – x2
            \n<\/sup> = 2x\"\"\/x + (\"\"\/x)2
            \n<\/sup>= <\/strong>2x +
            \nlim= <\/strong>2x
            \n\t\t\t<\/strong>0
            \n\t\t\t<\/strong>\"\"\/f1<\/sup>(x) = 2x<\/p>\n

            \u00a0Example
            \n<\/strong>Find the derivative of y = x3<\/sup> from first principle
            \nSolution<\/strong>
            \n\t\t y = x3
            \n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y + \"\"\/y = (x + \"\"\/x)3
            \n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= x3<\/sup> + 3x3\"\"\/<\/sup>x + 3x (\"\"\/x)2<\/sup>+ (\"\"\/x)3
            \n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\/<\/sup>y = (x +\"\"\/x)3<\/sup> \u2013 x3
            \n<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 3x2\"\"\/<\/sup> + 3x(\"\"\/x)2<\/sup>+ (\"\"\/x)3
            \n<\/sup> = 3x2<\/sup> + 3x\"\"\/x + (\"\"\/x)2<\/sup>
            \n\t\tlim = 3x2
            \n<\/sup>x\"\"\/0
            \nHence = 3x2<\/sup><\/p>\n

            \u00a0Example
            \n<\/strong> Find the derivative of y = from first principle.
            \nSolution
            \n<\/strong>y =
            \n\u00a0\u00a0\u00a0\u00a0 y + \"\"\/y = –
            \n\"\"\/y =
            \n =
            \n=
            \n= =
            \n=
            \nlim = –
            \n=
            \nHence = – <\/p>\n

            \u00a0Example
            \n<\/strong>Find the derivative of y = c, where c is a constant, from first principle.
            \nSolution<\/strong>
            \n\t\t y = c
            \n\u00a0\u00a0\u00a0\u00a0y+ = c ( since c is a constant)
            \n\u00a0\u00a0\u00a0\u00a0= <\/strong>0
            \n\u00a0\u00a0\u00a0\u00a0 = 0
            \nlim = 0
            \n= = 0 <\/p>\n

            \u00a0Hence the derivative of a constant is zero.
            \n<\/strong>EVALUATION
            \n<\/strong>Find the derivative of the following using first principle.
            \n1. y=3x2 <\/sup>+ 4 (2) y= x3<\/sup> -2x2<\/sup> + 2x -5<\/p>\n

            \u00a0GENERAL EVALUATION
            \n<\/strong>1) <\/strong>Evaluate (i) lim x-> 0 x4<\/sup> + 5x \/ x2<\/sup> + 3 (ii) lim x-> 2 3x + 7
            \n2) Differentiate from the first principle y= 2x2<\/sup> +3x + 5
            \n3) Find the gradient function of y = x2<\/sup> +3x +1 (4) Differentiate y =5x4<\/sup> +7x3<\/sup> + 6x2<\/sup> \u2013 9x +4<\/p>\n

            \u00a0READING ASSIGNMENT:<\/strong>New further Maths Project 2 page 113- 120<\/p>\n

            \u00a0WEEKEND ASSIGNMENT
            \n<\/strong>1<\/strong>) Evaluate limx-> 1<\/sub> 4x2<\/sup> + 3x a) 4 b) 3 c) 7 d) 0
            \n2) Evaluate limx-> 0<\/sub> x2<\/sup> + 9 a) 3 b) 9 c) 6 d) 1
            \n3) Evaluate limx-> 0 <\/sub>( x+3) ( 3x-3) a) 27 b) 6 c) 9 d) -9
            \n4) Differentiate 8x2<\/sup> + 10 a) 8x b) 16x c) 10 d) 18x
            \n5) Find the derivative of y = b where b is a constant a) 0 b) bx c) x d) 1<\/p>\n

            \u00a0THEORY
            \n<\/strong>1) Evaluate limx-> -2<\/sub> 3x3<\/sup> +4 \/ x2<\/sup> +4 (2) Differentiate from the first principle y = 7x3<\/sup> + 5x2<\/sup> \u2013 6x +5 <\/p>\n","protected":false},"excerpt":{"rendered":"

            SECOND TERM E-LEARNING NOTE \u00a0SUBJECT: FURTHER MATHEMATICS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0CLASS: SS2 SCHEME OF WORK \u00a0WEEK \u00a0\u00a0\u00a0\u00a0TOPIC Differentiation: Limits…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,253],"tags":[],"class_list":["post-3167","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3167","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=3167"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3167\/revisions"}],"predecessor-version":[{"id":3168,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3167\/revisions\/3168"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3167"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3167"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3167"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}