Q59 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics

If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected, then the approximate value of $\left(1+\frac{3}{4} x\right)^{\frac{1}{2}}\left(1-\frac{2 x}{3}\right)^{-2}$ is

  • A
    $\frac{41+24 x}{41}$
  • B
    $\frac{41-24 x}{41}$
  • Correct answer
    C
    $\frac{24+41 x}{24}$
  • D
    $\frac{24-41 x}{24}$

Solution

If $x$ is so small that $x^2$ and higher powers of $x$ can be neglected, then $$ (1+x)^n=1+n x $$ So, $\left(1+\frac{3}{4} x\right)^{1 / 2}\left(1-\frac{2}{3} x\right)^{-2}=\left(1+\frac{3}{8} x\right)\left(1+\frac{4 x}{3}\right)$ $=1+\frac{3}{8} x+\frac{4 x}{3} \quad$ [on neglecting the $x^2$ term] $$ =\frac{24+41 x}{24} $$ Hence, option (c) is correct.

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

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