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 α x^ 2 -a x+2(α-2 b)=0 has:

Correct answer: B. one root in (1,4) and another in (-2,0) Given AM of (1)/(a) and (1)/(b) is (5)/(16) with a > 2 . Since a, 4, α, b are in AP: 4 - a = α - 4 = b - α This gives α = 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 : α = 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) .

JEE Main 202628 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2026 · 28 January · Evening Shift · Mathematics · Sequences and Series

Let the arithmetic mean of 1 a and 1 ~b be (5)/(16), a>2 . If α is...

Let the arithmetic mean of

\frac{1}{\mathrm{a}}

and

\frac{1}{\mathrm{~b}}

\frac{5}{16}, \mathrm{a}>2

. If

\alpha

is such that

\mathrm{a}, 4, \alpha, \mathrm{~b}

are in A.P., then the equation

\alpha x^{2}-a x+2(\alpha-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

\frac{1}{a}

and

\frac{1}{b}

\frac{5}{16}

with

a > 2

. Since

a, 4, \alpha, b

are in AP:

4 - a = \alpha - 4 = b - \alpha

This gives

\alpha = 8 - a

and

b = 12 - 2a

.
From AM condition:

\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{5}{16}

leads to

\frac{a+b}{ab} = \frac{5}{8}

Substituting:

\frac{20-3a}{96-28a+2a^2} = \frac{5}{8}

160 - 24a = 480 - 140a + 10a^2

10a^2 - 116a + 320 = 0

5a^2 - 58a + 160 = 0

Using the quadratic formula gives

a = 4

a = 8

.
For

a = 4

\alpha = 4, b = 4

.
The equation becomes

4x^2 - 4x - 8 = 0

x^2 - x - 2 = 0

with roots 2 and

-1

.
One root lies in

(1,4)

and the other in

(-2,0)

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