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Solution

The correct option is A $1$
Let $S = x + \frac{1}{x}$
Arthmetic mean $\geq$ Geometric mean
$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \times \frac{1}{x}}$
$⟹ S \geq 2$
Given $S$ is minimum
$⟹ x + \frac{1}{x} = 2 ⟹ x^{2} - 2 x + 1 = 0 ⟹ x = 1$

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Therealnumberxwhenaddedtoitsinversegivestheminimumvalueofthesumatxequalto

The correct option is A 1Let S=x+1xArthmetic mean ≥ Geometric meanx+1x2≥√x×1x⟹S≥2Given S is minimum⟹x+1x=2⟹x2−2x+1=0⟹x=1

$T h e r e a l n u m b e r x w h e n a d d e d t o i t s i n v e r s e g i v e s t h e m i n i m u m v a l u e o f t h e s u m a t x e q u a l t o$

The correct option is A $1$
Let $S = x + \frac{1}{x}$
Arthmetic mean $\geq$ Geometric mean
$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \times \frac{1}{x}}$
$⟹ S \geq 2$
Given $S$ is minimum
$⟹ x + \frac{1}{x} = 2 ⟹ x^{2} - 2 x + 1 = 0 ⟹ x = 1$