JEE Main202628 January 2026Evening ShiftMathematicsSequences and SeriesActual
Let the arithmetic mean of 1 a and 1 ~b be 5 16 , a >2 . If is such that a , 4, , ~b are in A.P., then the equation x²-a x+2( -2 b)=0 has:
Options
- Aone root in (0,2) and another in (-4,-2)
- Bone root in (1,4) and another in (-2,0)
- Cboth roots in the interval (-2,0)
- Dcomplex roots of magnitude less than 2
Correct answer
B. one root in (1,4) and another in (-2,0)
Step-by-step solution
Given AM of 1 a and 1 b is 5 16 with a > 2 . Since a, 4, , b are in AP: 4 - a = - 4 = b - This gives = 8 - a and b = 12 - 2a . From AM condition: 1 a + 1 b 2 = 5 16 leads to a+b ab = 5 8 Substituting: 20-3a 96-28a+2a^2 = 5 8 160 - 24a = 480 - 140a + 10a^2 10a^2 - 116a + 320 = 0 or 5a^2 - 58a + 160 = 0 Using the quadratic formula gives a = 4 or a = 8 . For a = 4 : = 4, b = 4 . The equation becomes 4x^2 - 4x - 8 = 0 or x^2 - x - 2 = 0 with roots 2 and -1 . One root lies in (1,4) and the other in (-2,0) .