Hits: 165

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2009 | Oct-Nov | (P2-9709/22) | Q#6

Hits: 165 Question     i.       Express  in the form , where  and , stating the  exact value of R and and giving the value of  correct to 2 decimal places.    ii.       Hence solve the equation Giving all solutions in the interval . Solution      i.  We are given the expression; We are […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2009 | Oct-Nov | (P2-9709/21) | Q#5

Hits: 77 Question     i.       Express  in terms of .    ii.       Hence find the exact value of . Solution      i.   We are given that  and we are required to express it in terms of . Let us start from . We can write  as; We have the trigonometric identity;    […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2009 | May-Jun | (P2-9709/02) | Q#5

Hits: 142   Question Solve the equation sec x=4 – 2 tan2 x, giving all solutions in the interval . Solution We are required to solve the equation; Since , therefore, we can write . It is evident that it is a quadratic equation in . Let , then we can write; Now we have […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2009 | Oct-Nov | (P1-9709/12) | Q#4

Hits: 305   Question Functions  and  are defined by  for .      i.       Find the range of .    ii.       Sketch the graph of .   iii.       State, with a reason, whether  has an inverse. Solution i.   We have the function; We can write it as; We know that; Hereby; We can find the range of   by substituting extreme possible values of ; Therefore; […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2009 | Oct-Nov | (P1-9709/11) | Q#2

Hits: 268   Question The equation of a curve is . The equation of a line is . On the same diagram, sketch the curve and the line for . Solution First we sketch  for . We can find the points of the graph as follows. Now we sketch the line  for . We can write it as; Slope-Intercept form […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2009 | May-Jun | (P1-9709/01) | Q#4

Hits: 693   Question The diagram shows the graph of  for .      i.       Find the values of ,  and .    ii.      Find the smallest value of  in the interval  for which . Solution i.   We are given that; When we observe given graph of this function, it is evident that y-intercept of the graph, i.e. when , is 3. […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2009 | Oct-Nov | (P1-9709/11) | Q#5

Hits: 2120 Question The diagram shows a semicircle ABC with centre O and radius 6 cm. The point B is such that angle BOA is 90o and BD is an arc of a circle with centre A. Find i.       the length of the arc BD,    ii.       the area of the shaded region. Solution      i.   Expression for length […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2009 | Oct-Nov | (P1-9709/12) | Q#5

Hits: 266 Question i.               Prove the identity . ii.               Solve the equation  for . Solution i.   We have the trigonometric identity; We can rewrite it in two ways; ii.   To solve the equation  for , we can express, as demonstrated in (i), the right hand side of given equation as; Therefore the given equation can be written as; […]

Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2009 | Oct-Nov | (P1-9709/11) | Q#1

Hits: 265 Question Solve the equation  for . Solution To solve this equation for , we can substitute . Hence, Since given interval is  , for  interval can be found as follows; Multiplying the entire inequality with 2; Adding  to entire inequality; Since ; Hence the given interval for  is . To solve  equation for interval , Using calculator we can find the value of . […]