Question ID: #949
Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}$, $a \gt 2$.
If $a$ is such that $a, 4, \alpha, b$ are in A.P., then the equation $\alpha x^{2}-ax+2(\alpha-2b)=0$ has:
- (1) One root in (1,4) and another in (-2,0)
- (2) One root in (0,2) and another in (-4, -2)
- (3) Complex roots of magnitude less than 2
- (4) Both roots in the interval (-2, 0)
Solution:
Given AM of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$:
$$\frac{\frac{1}{a} + \frac{1}{b}}{2} = \frac{5}{16} \Rightarrow \frac{a+b}{ab} = \frac{5}{8}$$
Given $a, 4, \alpha, b$ are in A.P.
Let the common difference be $d$.
$4 = a + d \Rightarrow a = 4 – d$
$\alpha = 4 + d$
$b = 4 + 2d$
Substitute $a$ and $b$ into the AM relation:
$$\frac{(4-d) + (4+2d)}{(4-d)(4+2d)} = \frac{5}{8}$$
$$\frac{8+d}{16 + 8d – 4d – 2d^2} = \frac{5}{8}$$
$$\frac{8+d}{16 + 4d – 2d^2} = \frac{5}{8}$$
$$8(8+d) = 5(16 + 4d – 2d^2)$$
$$64 + 8d = 80 + 20d – 10d^2$$
$$10d^2 – 12d – 16 = 0$$
$$5d^2 – 6d – 8 = 0$$
Solving for $d$:
$$5d^2 – 10d + 4d – 8 = 0$$
$$5d(d-2) + 4(d-2) = 0$$
$$(5d+4)(d-2) = 0 \Rightarrow d = 2 \text{ or } d = -0.8$$
Assuming integer terms as per standard problem types, take $d=2$.
Find the coefficients:
$a = 4 – 2 = 2$
$\alpha = 4 + 2 = 6$
$b = 4 + 4 = 8$
The quadratic equation is $\alpha x^2 – ax + 2(\alpha – 2b) = 0$:
$$6x^2 – 2x + 2(6 – 16) = 0$$
$$6x^2 – 2x + 2(-10) = 0$$
$$6x^2 – 2x – 20 = 0$$
$$3x^2 – x – 10 = 0$$
Roots:
$$3x^2 – 6x + 5x – 10 = 0$$
$$3x(x-2) + 5(x-2) = 0$$
$$(3x+5)(x-2) = 0$$
$$x = 2, \quad x = -\frac{5}{3} \approx -1.67$$
Root $x=2 \in (1, 4)$.
Root $x=-1.67 \in (-2, 0)$.
Ans. (1)
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