The number of ordered pairs (a, b) of positiv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $\frac{2a-1}{b} = \alpha$ and $\frac{2b-1}{a} = \beta$,where $\alpha, \beta \in \mathbb{Z}^+$.Since $a, b \ge 1$,we have $2a-1 \ge 1$ and $2b-

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let $\frac{2a-1}{b} = \alpha$ and $\frac{2b-1}{a} = \beta$,where $\alpha, \beta \in \mathbb{Z}^+$.Since $a, b \ge 1$,we have $2a-1 \ge 1$ and $2b-1 \ge 1$,so $\alpha, \beta \ge 1$.From the equations,$2a-1 = \alpha b$ and $2b-1 = \beta a$.Substituting $b = \frac{2a-1}{\alpha}$ into the second equation: $2(\frac{2a-1}{\alpha}) - 1 = \beta a$.$4a - 2 - \alpha = \alpha \beta a$,which implies $a(4 - \alpha \beta) = \alpha + 2$.Since $a > 0$ and $\alpha + 2 > 0$,we must have $4 - \alpha \beta > 0$,so $\alpha \beta Possible pairs $(\alpha, \beta)$ are $(1, 1), (1, 2), (1, 3), (2, 1), (3, 1)$.Case $1$: $(\alpha, \beta) = (1, 1) \implies a(4-1) = 1+2 \implies 3a = 3 \implies a = 1$. Then $b = \frac{2(1)-1}{1} = 1$. Pair: $(1, 1)$.Case $2$: $(\alpha, \beta) = (1, 2) \implies a(4-2) = 1+2 \implies 2a = 3$,no integer solution.Case $3$: $(\alpha, \beta) = (1, 3) \implies a(4-3) = 1+2 \implies a = 3$. Then $b = \frac{2(3)-1}{1} = 5$. Pair: $(3, 5)$.Case $4$: $(\alpha, \beta) = (2, 1) \implies a(4-2) = 2+2 \implies 2a = 4 \implies a = 2$. Then $b = \frac{2(2)-1}{2} = 1.5$,not an integer.Case $5$: $(\alpha, \beta) = (3, 1) \implies a(4-3) = 3+2 \implies a = 5$. Then $b = \frac{2(5)-1}{3} = 3$. Pair: $(5, 3)$.The ordered pairs are $(1, 1), (3, 5), (5, 3)$. Total $3$ pairs.

The number of ordered pairs $(a, b)$ of positive integers such that $\frac{2a-1}{b}$ and $\frac{2b-1}{a}$ are both integers is

Similar Questions

If $\tan \alpha$ equals the integral solution of the inequality $4x^2 - 16x + 15 < 0$ and $\cos \beta$ equals the slope of the bisector of the first quadrant,then $\sin(\alpha + \beta)\sin(\alpha - \beta)$ is equal to

Let $f(x)$ be a quadratic polynomial with leading coefficient $1$ such that $f(0)=p, p \neq 0$ and $f(1)=\frac{1}{3}$. If the equations $f(x)=0$ and $f(f(f(f(x))))=0$ have a common real root,then $f(-3)$ is equal to $........$

For all real values of $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

If the minimum value of $f(x) = x^2 + 2bx + 2c^2$ is greater than the maximum value of $g(x) = -x^2 - 2cx + b^2$ for all real $x$,then:

If $\alpha, \beta$ are the irrational roots of the equation $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$,then the roots of the equation $(\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module