Question

A real value of x satisfies the equation  \[\frac{3 - 4ix}{3 + 4ix} = a - ib (a, b \in \mathbb{R}), if a^2 + b^2 =\]

पर्याय

• 1

• -1

• 2

• -2

Answer 1

\[a - ib = \frac{3 - 4ix}{3 + 4ix}\]

\[ = \frac{3 - 4ix}{3 + 4ix} \times \frac{3 - 4ix}{3 - 4ix}\]

\[ = \frac{9 + 16 x^2 i^2 - 24xi}{9 - 16 x^2 i^2}\]

\[ = \frac{\left( 9 - 16 x^2 \right) - i\left( 24x \right)}{9 + 16 x^2}\]

\[ \Rightarrow \left| a - ib \right|^2 = \left| \frac{\left( 9 - 16 x^2 \right) - i\left( 24x \right)}{9 + 16 x^2} \right|^2 \]

\[ \Rightarrow a^2 + b^2 = \frac{\left( 9 - 16 x^2 \right)^2 + \left( 24x \right)^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = \frac{81 + 256 x^4 - 288 x^2 + 576 x^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = \frac{81 + 256 x^4 + 288 x^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = \frac{\left( 9 + 16 x^2 \right)^2}{\left( 9 + 16 x^2 \right)^2}\]

\[ = 1\]

Hence, the correct option is (a).