If S = { x ∈ R : 3 ( 3 + 2 ) x + ( 3 − 2 ) x = 1 0 3 } then number of elements in set S is

3 − 2 = ( 3 + 2 ) ( 3 − 2 ) ( 3 + 2 ) = 1 3 + 2 Let 3 + 2 = t t x + 1 t x = 1 0 3 Let t x = y y + 1 y = 1 0 3 y = 3 or 1 3 ( 3 + 2 ) x = 3 or 1 3 x l o g ( 3 + 2 ) = l n 3 or –ln3 x = l n 3 l n ( 3 + 3 ) or − l n 3 3 + 2 two real values of x f ( x ) = { e − x , x 0 g ( x ) = { e x , x 0

If $S = {x \in R : 3 {(\sqrt[]{3} + \sqrt[]{2})}^{x} + {(\sqrt[]{3} - \sqrt[]{2})}^{x} = \frac{10}{3}}$ then number of elements in set S is

Option 1 - Zero

Option 2 - 1

Option 3 - 2

Option 4 - 3

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Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 3

Detailed Solution:

$\sqrt[]{3} - \sqrt[]{2} = \frac{(\sqrt[]{3} + \sqrt[]{2}) (\sqrt[]{3} - \sqrt[]{2})}{(\sqrt[]{3} + \sqrt[]{2})} = \frac{1}{\sqrt[]{3} + \sqrt[]{2}}$

Let $\sqrt[]{3} + \sqrt[]{2} = t$

$t^{x} + \frac{1}{t^{x}} = \frac{10}{3}$

Let $t^{x} = y$ $y + \frac{1}{y} = \frac{10}{3}$

y = 3 or $\frac{1}{3}$

${(\sqrt[]{3} + \sqrt[]{2})}^{x} = 3$ or $\frac{1}{3}$

$x l o g (\sqrt[]{3} + \sqrt[]{2}) = l n 3$ or –ln3

$x = \frac{l n 3}{l n (\sqrt[]{3} + \sqrt[]{3})}$ or $\frac{- l n 3}{\sqrt[]{3} + \sqrt[]{2}}$

two real values of x

$f (x) = {\begin{matrix} e^{- x}, & x < 0 \\ l n x, & x > 0 \end{matrix}$

$g (x) = {\begin{matrix} e^{x}, & x < 0 \\ x, & x > 0 \end{matrix}$

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