Higher Still Level Paper

Transcription

1 Higher Still Level Paper 7. Given the cube with sides units and B is (,,). Also P and Q are the centres of face OCGD and CBFG. Z G y F (a) Coordinates og G are (,, ) D E P Q C B(,, ) (b) Position vectors of p and q are: O A p = i + j + k and q = i + j + k (c) Size of angle POQ is: p q cos θ = p q = (,, ) (,, ) = p q = p = = = q = θ = cos o θ = + + = 6 Page of

2 Higher Still Level Paper 7. Given the diagram. (a) The eact value of sin( c+ d) sin( c+ d) = sin ccos d + sin dcosc = = = 5 = 5 c d (b)(i) The eact value of sin c : sin c = sin ccosc = 5 5 = 5 (b)(ii) The value ofcos d : cos d = cos d sin d = 9 = 8 = = 5 Hence cos d is eactly the same as sin c Page of

3 Higher Still Level Paper 7. To show that the line y = 6 is a tangent to the circle + y + 6 y 7= sub y = 6- into the circle equation = (6 ) + 6 (6 ) 7 = + = + = ( )( ) = = If a tangent then b ac= = ( ) = = Coordinate is = y = 6 - = (,) Page of

4 Higher Still Level Paper 7. Given the diagram and the function o has the form y = asin( b ) + c (a) a = b = c = - P 6 (b) The eact value of the -coordinate of P is given by: sin = sin = sin = = sin =,5,( + 6),(5 + 6),... =,5,9,5,... =,5,,7,... o From the graph it is clear - cordinate is 5 Page of

5 Higher Still Level Paper 7 5. Given the diagram. The circle touches the parabola with equation y = 8 + y at the points P and Q. (a) Given the gradient at Q is. The coordinates of Q are: Gradient is y'( ) = 8 O P Q = 8 8 = = y coordinate 8 = + = Coordinates for P (,) (b) The coordinates for P: By symmetry of the parabola and circle Gradient is y'( ) = 8 - = 8 8 = = Coordinates for P (,) Page 5 of

6 Higher Still Level Paper 7 (c) The coordinates of C are given when the lines perpendicular to each tangent P and Q intersect: Line perpendicular to P(,) y = ( ) y = y = + 6 Line perpendicular to P(,) y = ( ) y = + y = + 5 coordinate of C is + 6 = + 5 = 6 = 8 y coordinate is given by y = 8+ 6 y = = Coordinates for P (8,) Page 6 of

7 Higher Still Level Paper 7 6. Given the diagram of the garden in the shape of the right-angled isosceles triangle. Red values have been added (a)(i) The eact value of ST: ST = + = = = S m 5 o 5 o 5 o 5 o m o 5 5 o T (ii) Area of the decking is: Area = length breadth = ( ) = ( ) = (b) Dimensions that give maimum area of decking are given by: da = d Maimum is given by da = = d = 5 = = Hence dimensions are 5 length = = 5 breadth = 5 Page 7 of

8 Higher Still Level Paper 7 7. The value of y = sin(+ ) d y = cos( + ) y = cos(9) cos() y =.6 [ ] 8. Given the equation y = log ( )., where >, cuts the ais at the point (a,). The value of a is:. a. = log ( a ).. = log ( a ) = + = a a =. Page 8 of

9 Higher Still Level Paper 7 9. Given the graph of y = a, a > (a) Graph of y = a Green line Reflected in y ais (b) Graph of y = a Pink line Move y = a unit to the right.. Given the diagram of the cubic function and its derived function. Both pass through (,6) and the derived function passes through the points (,) and (,). (a) Given that f '( ) is of the form k( a)( b) : (i) Values of a and b are and. (,6) y = f() y = f () (ii) The value of k is: f '( ) = k( a)( b) 6 = k( )( ) 6 = 8k 6 k = = 8 Page 9 of

10 Higher Still Level Paper 7. (b) Equation of cubic is: y = f() f'( ) = ( )( ) = ( 6+ 8) 9 = + 6 (,6) y = f () 9 ( ) = + 9 = + 6+ c f 6 d for = f( ) = 6 9 Cubic has equation f( ) = Page of

11 Higher Still Level Paper 7. Given and y satisfy the equation y = (a) The value of a if (a,6) lies of the graph equation y = : 6 = 6 = = log = a a a a = log a 5. a = 5. log = (b) The value of b if (-.5,b) lies of the graph equation y = : b = b = b = b = Page of

12 Higher Still Level Paper 7. (c) Given a graph of log y against is drawn. y = log y = log log y = log + log log y = log + log log y = (log ) + log Comparing to log y = P+ Q Gradient is log Page of