NABTEB Mathematics 2024 Theory — Question 12

Model answer

(a) Since V lies on line OP, ∠VOQ = ∠POQ = 56°. Using the exterior angle theorem on triangle OVQ, the exterior angle ∠SVP (=109°) equals the sum of the two opposite interior angles: ∠SVP = ∠VOQ + ∠VQO. So 109° = 56° + ∠VQO → ∠VQO = 53°. Since OQ is a radius and PR is a tangent to the circle at Q, OQ ⊥ PR, so ∠OQP = 90°. (i) ∠TQP = ∠OQP − ∠VQO = 90° − 53° = 37°. (ii) By the tangent-chord angle theorem, the angle between tangent QP and chord QT equals the angle in the alternate segment: ∠QTS = ∠TQP = 37°. (b) Factorising: 2n²−3n−2 = (2n+1)(n−2); 2n²+3n+1 = (2n+1)(n+1); n²−1 = (n−1)(n+1); n²−4 = (n−2)(n+2). Expression = [(2n+1)(n−2)/((2n+1)(n+1))] × [(n−1)(n+1)/((n−2)(n+2))] = [(n−2)/(n+1)] × [(n−1)(n+1)/((n−2)(n+2))] = (n−1)(n+1)/[(n+1)(n+2)] [the (n−2) terms cancel] = (n−1)/(n+2) [the (n+1) terms cancel].