SCORE BOOSTER JAMB PREPARATION SERIES II

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Evan Norman
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1 BOOST YOUR JAMB SCORE WITH PAST Polynomials QUESTIONS Part II ALGEBRA by H. O. Aliu J. K. Adewole, PhD (Editor) 1) If 9x 2 + 6xy + 4y 2 is a factor of 27x 3 8y 3, find the other factor. (UTME 2014) 3x 2y Quotient 9x 2 + 6xy + 4y 2 27x 3 8y 3 27x x 2 y + 12xy 2 18x 2 y 12xy 2 8y 3 18x 2 y 12xy 2 8y 3... The Quotient 3x 2y is the other factor. 2) Factorise completely (UTME 2010) Factorise 3) Find the remainder when x 3 2x 2 + 3x 3 is divided by x (UTME 2011) This can be solved using the Long Division x - 2 x x 3 2x 2 + 3x 3 x 3 + x 2x 2 + 2x 3 2x 2 2. Remainder 2x 1 4) Factorize completely 9y 2 16x 2. (UTME 2011) Recall; --- Product of sum --- Difference of two squares 9y 2 16x 2 (3y 4x)( 3y + 4x) ( 3y + 4x) (3y 4x) 5) Find the remainder when 2x 3 11x 2 + 8x 1 is divided by x + 3. (UTME 2012) This can be solved using the Long division or Remainder Theorem. Remainder Theorem If f(x) is divided by ax + b, then the remainder is f. The polynomial f(x) 2x 3 11x 2 + 8x 1 is divided by x + 3, So the remainder is f i.e f (-3). x x - 3 The remainder is therefore obtained by substituting 3 for x in the polynomial. f (x) 2x 3 11x 2 + 8x 1 f (-3) 2( 3 3 ) 11( 3 2 ) + 8( 3) 1 2( 27) 11(9) ) If x 4 is a factor of x 2 x k, then k is.. (UTME 2013) Questions like this can be solved by combining the Factor and the Remainder Theorem. The Factor Theorem If f(a) 0, then x a is a factor of f(x). Dec

2 Since the polynomial has a factor, the remainder is zero (0). Remainder, f(4) 0 Substituting 4 for x and equating to zero gives the unknown value of the term k. f(4) k 0 k 16 4 k 12 7) The remainder when 6p 3 p 2 47p + 30 is divided by p 3 is (UTME 2013) Using the Remainder Theorem, f(p) 6p 3 p 2 47p + 30 Remainder, f( ) or f (3). f(3) 6(3 3 ) (3 2 ) 47(3) (27) ) Factorise 2y 2 15xy + 18x 2. (UTME 2014) Factoring by grouping Since 2y 2 18x x 2 y 2 12xy 3xy, and 12xy 3xy 15xy We can now substitute 12xy 3xy for 15xy and group 2y 2 15xy + 18x 2 2y 2 12xy 3xy + 18x 2 Grouping the above produce (2y 2 12xy) (3xy 18x 2 ) --- Factorise 2y(y 6x) 3x (y 6x) W can now factor out a common factor to get (2y 3x) (y 6x) f(y) y 3 + 4y 2 + ky 6 Since, y 1 is a factor of the polynomial the remainder f(1) 0 f(1) (1 2 ) + k(1) (1 2 ) + k(1) k 6 0 k 6 5 k 1 Change of Subject of Formulae 10) Make Q the subject of the formula if (UTME 2010) 11) Make R the subject of the formula if (UTME 2012) ---- Cross multiply -----Take the square root of both sides 12) Make n the subject of the formula if. (UTME 2012) 9) Find the value of k if y 1 is a factor of y 3 + 4y 2 + ky 6. (UTME 2014) Using the Factor Theorem, Cross multiply Dec

3 -- Collect like terms -- Factorise 13) If S find t in terms of S. (UTME 2013) S --- Square both side --- Factorise ---- Squre root both sides 14) If gt 2 k w 0, make g the subject of the formula. (UTME 2014) gt 2 k w 0 gt 2 k + w --- Divide through by t 2 g x 2 x 2 + 4x 4 8 4x 4 8 x x 3 To get y, subt. 3 for x in (i) 3 y 2 y 3 2 y 1 x 1 and y 3 16) Solve for x and y respectively in the simultaneous equations 2x 5y 3, x+ 3y 0. (UTME 2011) 2x 5y i x + 3y ii From ii, x 3y. Substitute 3y for x in (i). 2( 3y) 5y i 6y 5y 3 y 3 Substitute 3 for y in eqn. ii x + 3(3) 0 x x 9 (x, y) (- 9, 3) Simultaneous Equation 15) Solve for x and y if x y 2 and x 2 y (UTME 2010) x y i x 2 y ii Let s solve this simultaneous equations using Substitution method. From eqn. i, y x 2 Substitute x 2 for y in (ii) Eqn. ii becomes x 2 (x 2) 2 8 x 2 (x 2 4x + 4) 8 17) Solve for x and y in the equations below. x 2 y 2 4, x + y 2. (UTME 2012) x 2 y i x + y ii From ii, y 2 x, Subt. this for y in (i). x 2 (2 x) 2 4 x 2 (4 4x + x 2 ) 4 x x x 2 4 x 2 x 2 + 4x x 8 x 2 Substitute 2 for x in (ii) 2 + y 2 Dec

4 y 2 2 y 0 (x, y) (2, 0) Variation 27 18) If y varies directly as the square root of x and y 3 when x 16. Calculate y when x 64. (UTME 2010) x when y 1 is Mathematical relation between x and y. y when x is 64 will be 21) T varies inversely as the cube of R. When R 3, T, find T when R 2. (UTME 2011) 19) If x is inversely proportional to y and x 2½ when y 2, find x if y 4. (UTME 2010) --- Equation connecting the variables law connecting the variables x when y 4 is 22) If y varies directly as and y 4 when n 4, find y when n 1. (UTME 2012) 20) If x varies directly as square root of y and x 81 when y 9, find x when y 1. (UTME 2011) 2 Dec

5 --- Relationship between the variables. 23) U is inversely proportional to the cube of V and U 81 when V 2. Find U when V 3. (UTME 2015) 25) If r varies inversely as the square root of s and t, how does s vary with r and t? (utme 2013) U when V 3 is therefore, --- Law connecting the variables 24 24) P varies jointly as m and u, and varies inversely as q. Given that p 4, m 3 and u 2 when q 1, find the value of p when m 6, u 4 and q. (UTME 2013) Answer: s varies inversely as r 2 and t. 26) y varies directly as w 2. When y 8, w 2, find y when w 3. (UTME 2014) 2 2 x 9 18 variables --- Relationship between 10 27) P varies directly as Q and inversely as R. When Q 36 and R 16, P 27. Find the relation between P, Q and R. (UTME 2014) Dec

6 Progression --- Law connecting the variables --- Relation between P, Q and R. 28) Find the sum to infinity of the following series: (UTME 2010) The series is a geometric series because the successive terms does not have a common difference Common ratio, r for 29) The 3rd term of an arithmetic progression is 9 and the 7th term is 29. Find the 10th term of the progression. (UTME 2010) 3rd term a + 2d (i) 7th term a + 6d (ii) Solve the two equations simultaneously to find the first term (a) and the common difference (d). Using Elimination method, Subtract equ (i) from (ii) a + 2d (i) a + 6d (ii) 4d 20 d 5 Substitute 5 for d in eqn. (i) a + 2( 5) 9 a 10 9 a The 10 th term is a + 9d ( 5) ) Find the sum of the first 18 terms of the series 3, 6, 9, 36. (UTME 2011) ( ) ( ) n 18; a 3; and d 3 ( ) 9(6 + 51) 9 (57) ) The second term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms. (UTME 2011) 2 nd term ar (i) 3 rd term ar (ii) Solve the set of equations to get a9 an9d r which we can then use to compute the sum of the first five terms. Divide (ii) by (i) 2 Substitute 2 for r in equation (i) to obtain a. ar 4 2a 4 a 2 The sum of the first five terms of the GP is, 2(32 1) 2(31) 62 Dec

7 32) The nth term of a sequence is. Find the sum of the 3rd and 4th term. (UTME 2012) The nth term is 3 rd term rd term The addition of the two terms is therefore, ) The sum to infinity of a geometric progression is and the first term is. Find the common ration of the progression. (UTME 2012) for We use this formula though we don t know if r < 1 but since the first term, a is negative and the sum to infinitive, is negative, r is most likely to be less than 1., Substitute 3 for in eqn. (i) the sum of the first term and common ratio is 34) If the sum of the first two terms of a G.P is 3 and the sum of the second and the third terms is 6, find the sum of the first term and the common ratio. (UTME 2013) --- (i) --- (ii) From (i), ar and r Substitutes these values into eqn, (ii) 35) The nth term of the progression, is (UTME n 1,, when n 2, the nth term is 36) The 4th term of an A.P. is 13 while the 10th term is 31. Find the 24th term. (UTME 2014) --- (i) --- (ii) 6d Subtracting (i) from (ii) d 3 Dec

8 Substitute 3 for d in eqn. (i) 4 the 24 th term is ) What is the common ratio of the G.P. ( ) ( ) (UTME 2011) Rationalize, Binary Operations 38) If x * y x + y 2, find the value of (2*3)*5 (UTME 2010) x * y x + y 2 2* (2*3)*5 11*5 11* (2*3)* ) If p and q are two non-zero numbers and 18(p + q) (18 + p)q, which of the following must be true? (UTME 2010) 18(p + q) (18 + p)q 18p + 18q 18q + pq 18q pq P 18 This is the only condition that must be true for the right hand side of the equation to be equal to the left hand side. 40) A binary operation on real numbers is defined by x y xy + x + y for two real numbers x and y. Find the value of 3 ⅔. (UTME 2011) x y xy + x + y 3 ⅔ (3 ⅔) ( ⅔) ⅔ 1 ⅔ ⅓ 41) The binary operation * is defined on the set of integers such that p*q pq + p q. Find 2*(3*4). (UTME 2012) p*q pq + p q 3*4 (3 4) *4 11 2*(3*4) 2*11 2*11 (2 11) *(3*4) 13 42) A binary operation on the set of real numbers is defined by m*n for all m, n R. If the identity element is 2, find the inverse of -5. (UTME 2012) The Inverse Element (Binary Operation) If, S being a non-empty set which is closed under the binary operation *, and we can find an element such that, where e is the identity element in S under *, then is called the inverse of x in S. The inverse of x is denoted as. Dec

9 So, to find the inverse of 5 in set R under the given binary operation * Let the inverse of 5 i.e be represented with y --- divide through by the inverse of, denoted in the set R under * is 43) If a binary operation * is defined by x*y x + 2y, find 2*(3*4) (UTME 2013) x*y x + 2y 3*4 3 + (2 x 4) *(3*4) 2*11 2* (2 x 11) *(3*4) 11 44) A binary operation * is defined by x*y x y. If x * 2 12 x, find the possible values of x. (UTME 2014) x*y x y x * 2 12 x Let y 2, So x*y x*2 x 2 x 2 12 x x x Factorise (x + 4)(x 3) 0 x or x 3 0 x 4, or x 3 x 3, 4 Matrices and Determinants 45) If, find the value of x. (UTME 2010) Determinant of a matrix of Order 2 x 2 If A * +, then Determinant of A (Det A), 15 7x x x 21 / ) Evaluate (utme 2010) Determinant of matrix of Order 3 x 3 If A [ ], + + Det A + + [ ] [ ] [ ] + 2[(6 1) (3 9)] 0 [(4 1) (3 8)] + 5 [(4 9) (6 8)] Dec

10 + 2 (6 27) (36 48) + 2 ( 21) ( 12) ) If P * +, what is P -1? (UTME 2010) Inverse of a matrix For a 2 x 2 matrix, the inverse of matrix A denoted by A 1 is If A, then A 1 * + A 1 * + * + * + * + A 1 * + To evaluate the determinant of the matrix + + [ ] [ ] [ ] +4(9 + 1) 2 (6 1) 1 (2 + 3) ) The inverse of matrix N * + is (UTME 2011) A 1 * + N 1 * + N 1 * + * + N 1 * + 48) If, find the value of x. (UTME 2011) Equate the determinants of the two matrices and this gives an algebraic linear equation from which we can obtain x. 51) If, find the value of x. (UTME 2012) Equate the determinants and solve for x. 6 49) Evaluate (UTME 2011) 52) Given that is a unit matrix of order 3. Find (UTME 2011) Dec

11 A unit matrix of size n is the n x n square matrix with ones on the main diagonal and zeros elsewhere. It is denoted by. [ ] A unit matrix of order 3 or 3 x 3 identity matrix. + 1 [(1 1) (0 0)] 0 [(0 1) (1 0)] + 0 [(0 0) (1 0)] +1 (1 0) (1) 1 53) If * + * +, find 2P + Q. (UTME 2011) Addition and Subtraction of matrices If * +, 2P 2 * + 2P * * Scalar multiplication of vector 2P + Q * + * + * + * + * + * + 55) Find y, if (UTME 2014) This is simultaneous equation presented in matrix form. Write out the equation and solve for x and y. 5x 6y (i) x 2 2x 7y (ii) x 5 Using Elimination method to solve the equations Multiply equation (i) by 2 and (ii) by 5 10x 12y (iii) 10x 35y (iv) Subtract the new equation (iii) from (iv) 35y y 69 y 3 This is the required answer Substitute 3 for y in equation (i) To get y 5x 6(3) 7 5x x x 5 54) Find the inverse of * + (UTME 2013) Inverse of * + * + * + * + * + Alternatively, Use the matrix or Determinant method for solving simultaneous equation. Matrix method for solving simultaneous equation The simultaneous equations can be written in matrix form as --- (i) --- (ii) * + * + [ ] Dec

12 A. X Y X And * + * + X * + * + [ ] The values of x and y (matrix X) is the vector multiplication of the inverse of A and the solution set (Y). So, for The inverse of is 6 57) Find the value of (UTME 2014) Inequalities 58) For what range of values of xis. (UTME 2010) --- Subtract from both side --- Subtract from both side x * + [ ] [ ] --- Vector multiplication of matrices Multiply both sides by 6 x 5 and y 3 56) If, find x. (UTME 2011) 59) Solve the inequalities (UTME 2010) The inequality has to be considered in two stages STAGE 1 Dec

13 @ y 0, x & The graph of the inequality is therefore, STAGE 2 (or <x) These two solutions sets can be combined as Combining the two inequalities gives 60) Which of the following diagrams represents the solution of the inequalities. (UTME 2011) The straight line y 0, or ; (2, 0) x 0, ; (0, 2) A y-intercept form graph with intercept So, plot the two points (2, 0) and (0, 2) The graph of the solution sets that therefore satisfy the two inequalities is The quadratic inequality has these simple factors & So, gives Let The zeros of the function are & Plot the x 0, ; and 61) Solve the inequalities. (UTME 2011). --- Add 18 to both sides --- Subtract from both sides --- Divide through by --- The sign changes when you divide with a negative number Dec

14 62)Solve the inequalities. (UTME 2011) A Quadratic Inequality --- Subtract 15 from both side from both sides --- Subtract and --- Factorise and group When solving a quadratic inequality, you need to take more options into consideration. (a) For a less than - < problem, one factor must be positive while the other negative. (b) For a greater than - > problem, there are two ways that this product can be greater than zero (positive). Both factors are positive or both factors are negative. You must check out both possibilities. 64) Divide through by The sign changes The solution can either be -Both positive and or The only condition that makes both through is OR -Both Negative and or The only condition that makes both through is The set of solution is 63) The value of y for which is... (UTME 2012) or The graph above is correctly represented by (UTME 2013) This is a graphical solution of a quadratic equation and the equations can be obtained from the roots (the two values of x at the point where the curve strikes the x-axis) 65) Find the range of values of m which satisfy. (UTME 2014) There are two ways to solving this quadratic inequality. This product could be less than zero (negative) if and Dec

15 and This is not possible because m cannot be less than 3 and also greater than Divide through by The sign changes Or and and This is the possible solution and can be combined as 68) Solve for x: Ix - 2 I< 3 (UTME 2013) 66) The shaded region above is represented by the equation.. (UTME 2012) 69) What is the solution of? (UTME 2014) --- Multiply both sides by The equation of a straight line is y mx + c For this inequality, it wsill be y mx + c The gradient, m can be obtained from the two points given i.e (0, 4) & (1, 0) The intercept c on the y axis is the value of y at x 0, and is 4. The inequality is therefore Alternatively, The Equation of the line can be obtained directly as --- Cross multiply Also, can not satisfy the condition of the inequality as this produce a positive number greater than 1 but will. 70) Evaluate the inequality. (UTME 2014) 67) Evaluate 3(x+2) > 6(x+3). (UTME 2013) from both sides --- Subtract 6 and 6x both sides --- Subtract and from Dec

16 --- Divide through by The sign changes Dec

Day 6: 6.4 Solving Polynomial Equations Warm Up: Factor. 1. x 2-2x x 2-9x x 2 + 6x + 5

Day 6: 6.4 Solving Polynomial Equations Warm Up: Factor. 1. x 2-2x - 15 2. x 2-9x + 14 3. x 2 + 6x + 5 Solving Equations by Factoring Recall the factoring pattern: Difference of Squares:...... Note: There