KVPY 2019 SB SX Exam Previous Papers

Question : 81 of 120

Marks: +1, -0

The points C and D on a semicircle with AB as diameter are such that AC = 1, CD = 2, and DB = 3. Then the length of AB lies in the interval

Solution:

BC =√{x^2 -1}, AD=√ {x^2- 9}

by Ptolemy's theorem
AB.CD + AC.BD = AD.BC
⇒

2x + 3 =√ {x^2- 9}√ {x^2- 9}

⇒

4x^2 + 12x + 9 = x^4 -10x^2 + 9

⇒

x^4 -14x^2 -12x = 0

⇒ x^3 -14x -12 = 0

Let

\f(x) = x^3 -14x -12

⇒

\f '(x) = 3x^2 -14

⇒ f (x) has only one
positive root

∈ (0,√{14/3})

f(4.1) < 0 and f(4.2) > 0 ⇒ x ∈ (4.1, 4.2)

Go to Question:

Prev Question

Next Question