Let a -0-q-1- Then the set S of all positive real numbers b satisfying 1- a 21- b 2-4 ab isA- an empty setB- a singleton setC- 0- -D- a finite set containing more than one element

The correct option is A an empty set(1 + a^2 )(1 + b^2 ) = 4ab ⇒ 1 + a^2 + b^2 + a^2b^2 = 4ab ⇒ nbsp;a^2 + b^2 nbsp; 2ab + nbsp;a^2b^2 + 1 2ab nbsp ...

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Standard XII

Mathematics

Composite Function

Let a >0,q'=1...

Question

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

an empty set

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a singleton set

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a finite set containing more than one element

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(0, \infty)

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Solution

The correct option is A an empty set
Given $(1 + a^{2}) (1 + b^{2}) = 4 a b$
$\Rightarrow 1 + a^{2} + b^{2} + a^{2} b^{2} = 4 a b$
$\Rightarrow a^{2} + b^{2} - 2 a b + a^{2} b^{2} + 1 - 2 a b = 0$
$\Rightarrow (a - b)^{2} + (a b - 1)^{2} = 0$
This is only possible when,
$(a - b) = 0$ and $(a b - 1) = 0$
$∴ a = b$ and $∴ a b = 1$
$\Rightarrow a^{2} = 1$
$\Rightarrow a = 1$
but this is not possible as given in the question.
Hence, $S$ will be an empty set.

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Let a>0,a≠1. Then the set S of all positive real numbers b satisfying (1+a2)(1+b2)=4ab is

The correct option is A an empty setGiven (1+a2)(1+b2)=4ab⇒1+a2+b2+a2b2=4ab⇒a2+b2−2ab+a2b2+1−2ab=0⇒(a−b)2+(ab−1)2=0 This is only possible when, (a−b)=0 and (ab−1)=0 ∴a=b and ∴ab=1⇒a2=1⇒a=1 but this is not possible as given in the question. Hence, S will be an empty set.

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