Let the arithmetic mean of 1/a and 1/b be 5/16, a > 2

Q. Let the arithmetic mean of 1/a and 1/b be 5/16, a > 2. If α is such that a, 4, α, b are in A.P., then the equation αx2 − ax + 2(α − 2b) = 0 has :

A. one root in (1, 4) and another in (−2, 0)
B. one root in (0, 2) and another in (−4, −2)
C. both roots in the interval (−2, 0)
D. complex roots of magnitude less than 2

Correct Answer: one root in (1, 4) and another in (−2, 0)

Explanation

Given that the arithmetic mean of 1/a and 1/b is 5/16,

(1/a + 1/b)/2 = 5/16

⇒ 1/a + 1/b = 5/8 ⇒ (a + b)/ab = 5/8 …… (1)

Since a, 4, α, b are in A.P.,

4 − a = α − 4 = b − α

From this,

α = 8 − a b = 2α − 4 = 12 − 2a

Substitute a and b in equation (1) and simplify. Using a > 2, the values of a, b and α are obtained consistently.

Now consider the quadratic equation:

f(x) = αx2 − ax + 2(α − 2b)

Evaluating f(x) at x = −2 and x = 0 shows a change of sign, hence one root lies in (−2, 0).

Similarly, evaluating f(x) at x = 1 and x = 4 also shows a change of sign, hence the other root lies in (1, 4).

Therefore, the given equation has

one root in (1, 4) and another in (−2, 0)

Related JEE Main Questions

Considering the principal values of inverse trigonometric functions, the value of the expression tan ( 2 sin−1 ( 2/√13 ) − 2 cos−1 ( 3/√10 ) ) is equal to :
Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line (x − 1)/1 = y/2 = (z − 1)/1. Then the distance of Q from the line (x − 9)/3 = (y − 9)/2 = (z − 5)/(−2) is
Let y = y(x) be the solution of the differential equation x dy/dx − y = x² cot x, x ∈ (0, π). If y(π/2) = π/2, then 6y(π/6) − 8y(π/4) is equal to :
Let the circle x² + y² = 4 intersect x-axis at the points A(a, 0), a > 0 and B(b, 0). Let P(2 cos α, 2 sin α), 0 < α < π/2 and Q(2 cos β, 2 sin β) be two points such that (α − β) = π/2. Then the point of intersection of AQ and BP lies on :
A → B (first reaction) C → D (second reaction). Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is ______ × 10−1 hour−1 (nearest integer).
JEE Main Chapter-wise Previous Year Questions

Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

Scroll to Top