Q62 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics

The expansion of $\left(1+x+x^2\right)^{-3 / 2}$ in powers of $x$ is valid if

  • A
    $|\mathrm{x}| < 1$
  • B
    $|x| < \frac{1}{2}$
  • Correct answer
    C
    $\left|x+\frac{1}{2}\right| < \frac{\sqrt{5}}{2}$
  • D
    $-\frac{1}{2}-\frac{\sqrt{5}}{2} < x < 1$

Solution

$$ \left(1+x+x^2\right)^{-3 / 2} $$ for $(1+x)^n(n \notin N)$ to be defined $|x| < 1$. So $\left(1+\mathrm{x}+\mathrm{x}^2\right)^{-3 / 2}$ is defined if $$ \begin{aligned} & \left|x^2+x\right| < 1 \\ & \Rightarrow x^2+2 \cdot x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2 < 1 \\ & \Rightarrow\left(x+\frac{1}{2}\right)^2 < \frac{5}{4} \\ & \therefore\left|x+\frac{1}{2}\right| < \frac{\sqrt{5}}{2} . \end{aligned} $$

Hence, the correct answer is option: 'C'.

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