Cambridge A Level Mathematics 9709 — 2019 Feb/March Paper 2 · Variant 2

1 Solve the equation sec2 tan2 5 tan 4 for Show all necessary working. [4] 1 + 1 = 1 + 0Å < 1 < 180Å. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Use identity 2 2 sec 1 tan θ θ = + B1 Attempt solution of quadratic equation to find two values of tanθ M1 Obtain 1 2 tan , 3 θ = − A1 Obtain 71.6 and 153.4 and no others between 0 and 180 A1 4

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2 Given that x satisfies the equation 2x 3 2x , find the value of + = −1 4x 6x . [4] −3 − ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Solve non-modular equation 2 2 (2 3) (2 1) x x + = − or linear equation with signs of 2x different M1 Obtain 1 2 x = − A1 Substitute negative value into expression and show correct evaluation of modulus at least once M1 Obtain 5 3 2 − = with no errors seen A1 4

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