KVPY2019MathematicsProperties of Triangles
The points C and D on a semicircle with A B as diameter are such that A C = 1 , C D = 2 and D B = 3 . Then, the length of A B lies in the interval.
Options
- A[ 4 , 4 . 1 )
- B[ 4 , 1 , 4 . 2 )
- C[ 4 . 2 , 4 . 3 )
- D[ 4 . 2 , ∞ )
Correct answer
B. [ 4 , 1 , 4 . 2 )
Step-by-step solution
Let the diameter A B = x ∵ ∠ A C B = π 2 and ∠ A D B = π 2 ∴ B C = x 2 - 1 On applying Ptolemy theorem, we get A B · C D + A C · B C = A D · B C ⇒ 2 x + 3 = x 2 - 1 x 2 - 9 On squaring both sides, we get 4 x 2 + 9 + 12 x = x 4 - 10 x 2 + 9 ⇒ x 4 - 14 x 2 - 12 x = 0 ⇒ x 3 - 14 x - 12 = 0 Let f x = x 3 - 14 x - 12 ∵ f 4 = - 4 f 4 . 1 = - 0 . 479 f 4 . 2 = 3288 ∵ f 4 . 1 · f 4 . 2 0 ⇒ x ∈ [ 4 , 1 , 4 . 2 )