If $\frac{(1 - 3x)^{\frac{1}{2}} + {(1 - x)}^{\frac{5}{3}}}{\sqrt{4 - x}}$ is approximately equal to $a + bx$ for small values of $x$, then $(a,b)$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
$\frac{{(1 - 3x)}^{\frac{1}{2}} + {(1 - x)}^{\frac{5}{3}}}{2\left( 1 - \frac{x}{4} ight)^{\frac{1}{2}}}$
$= \frac{\left\lbrack \begin{array}{r}
\left\lbrack 1 + \frac{1}{2}( - 3x) + \frac{1}{2}\left( - \frac{1}{2} ight)\frac{1}{2}( - 3x)^{2} + ... ightbrack + \
\left\lbrack 1 + \frac{5}{3}( - x) + \frac{5}{3} \bullet \frac{2}{3} \bullet \frac{1}{2}( - x)^{2} + ... ightbrack
\end{array} ightbrack}{2\left\lbrack 1 + \frac{1}{2}\left( - \frac{x}{4} ight) + \frac{1}{2}\left( - \frac{1}{2} ight)\frac{1}{2}\left( - \frac{x}{4} ight)^{2} + ... ightbrack}$
$= \frac{2\left\lbrack 1 - \frac{19}{12}x - \frac{41}{144}x^{2} - ... ightbrack}{2\left\lbrack 1 - \frac{x}{8} - \frac{1}{128}x^{2} - ... ightbrack}$
$= \left\lbrack 1 - \frac{19}{12}x - \frac{41}{144}x^{2} - ... ightbrack\left\lbrack 1 - \frac{x}{8} - \frac{1}{128}x^{2}... ightbrack^{- 1}$
$= 1 - \frac{35}{24}x + ...$
On neglecting higher powers of $x$, we get
$a + bx = 1 - \frac{35}{24}x$
$\Longrightarrow \ a = 1,\ b = - \frac{35}{24}$

Q. If $\frac{(1 - 3x)^{\frac{1}{2}} + {(1 - x)}^{\frac{5}{3}}}{\sqrt{4 - x}}$ is approximately equal to $a + bx$ for small values of $x$, then $(a,b)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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