Let a>0, a ≠ 1 . Then, set S of all positive real numbers b satisfying left(1+a^2right)left(1+b

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1.Set Theory

normal

Let $a>0, a \neq 1$. Then, the set $S$ of all positive real numbers $b$ satisfying $\left(1+a^2\right)\left(1+b^2\right)=4 a b$ is

A

an empty set

B

a singleton set

C

a finite set containing more than one element

D

$(0, \infty)$

(KVPY-2019)

Solution

(a)

Given relation

$\left(1+a^2\right)\left(1+b^2\right)=4 a b$

$\Rightarrow a^2+b^2-2 a b=2 a b-1-a^2 b^2$

$\Rightarrow \quad (a-b)^2=-(1-a b)^2$

$\because a > 0, a \neq 1 \text { and } b \text { is a positive real number}$

$\therefore(a-b)^2 \neq 0 \neq-(1-a b)^2, \text { because }(a-b)^2$

$\text { and }(1-a b)^2 \text { are non-negative real numbers}$

Standard 11

Mathematics

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