JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

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

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 $(- 2, 0)$

Solution

Since $a, 4, α, b$ are in A.P. with common difference $d$ : $a = 4 - d, α = 4 + d, b = 4 + 2 d$ .

Given $\frac{1}{2} (\frac{1}{a} + \frac{1}{b}) = \frac{5}{16}$ , i.e. $\frac{1}{4 - d} + \frac{1}{4 + 2 d} = \frac{5}{8}$ .

Solving: $\frac{( 4 + 2 d ) + ( 4 - d )}{( 4 - d ) ( 4 + 2 d )} = \frac{5}{8} \Rightarrow 8 (8 + d) = 5 (4 - d) (4 + 2 d)$ . This yields $d = 2$ (the value consistent with $a > 2$ ).

So $a = 2, α = 6, b = 8$ . The quadratic becomes $6 x^{2} - 2 x + 2 (6 - 16) = 0 \Rightarrow 6 x^{2} - 2 x - 20 = 0 \Rightarrow 3 x^{2} - x - 10 = 0$ .

Factor: $(3 x + 5) (x - 2) = 0 \Rightarrow x = 2$ or $x = - \frac{5}{3}$ .

$x = 2$ lies in $(1, 4)$ and $x = - \frac{5}{3}$ lies in $(- 2, 0)$ .

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