Let the arithmetic mean of $\frac{1}{\mathrm{a}}$ and $\frac{1}{\mathrm{~b}}$ be $\frac{5}{16}, \mathrm{a}>2$. 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:

(A) One root in $(1,4)$ and another in $(-2,0)$

(B) One root in $(0,2)$ and another in $(-4,-2)$

(C) Complex roots of magnitude less than 2

(D) Both roots in the interval $(-2,0)$

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Quadratic Equations and Inequalities

Let the arithmetic mean of

\frac{1}{a}

and

\frac{1}{b}

\frac{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:

Answer: (A)

Let the four terms

a, 4, α, b

in A.P. have common difference

d

, so

a = 4 - d, α = 4 + d, b = 4 + 2 d

. The arithmetic-mean condition gives

\frac{1}{4 - d} + \frac{1}{4 + 2 d} = \frac{5}{8} ⟹ \frac{8 + d}{( 4 - d ) ( 4 + 2 d )} = \frac{5}{8},

which reduces to

5 d^{2} - 6 d - 8 = 0

, so

d = 2

d = - \frac{4}{5}

. The requirement

a = 4 - d > 2

forces

d < 2

, hence

d = - \frac{4}{5}

, giving

a = \frac{24}{5}, α = \frac{16}{5}, b = \frac{12}{5}

. The equation

α x^{2} - a x + 2 (α - 2 b) = 0

becomes

\frac{16}{5} x^{2} - \frac{24}{5} x - \frac{16}{5} = 0

, i.e.

2 x^{2} - 3 x - 2 = 0

, so

(2 x + 1) (x - 2) = 0

and the roots are

x = 2

and

x = - \frac{1}{2}

. One root

2

lies in

(1, 4)

and the other

- \frac{1}{2}

lies in

(- 2, 0)

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option