Question

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

• A. 1
• B. 4
• C. 5
• D. 7

Answer 1

Answer: (A)

1

Answer 2

Since the given expression is equal to a + bx, so we have to neglect the terms x² and higher powers of x.

$\frac{(1 - 3x)^{1/2} + (1 - x)^{5/3}}{\sqrt{4 - x}}$

= [(1 – 3x)^1/2 + (1 – x)^5/3] [4 – x]^–1/2

= $\left\lbrack 1 + \frac{1}{2}( - 3x) + 1 + \frac{5}{3}(–x) \right\rbrack$ (4^–1/2) $\left( 1 - \frac{x}{4} \right)^{–1/2}$

=$\left\lbrack 1 + \left( - \frac{1}{2} \right)\left( - \frac{x}{4} \right) \right\rbrack = \frac{1}{2}\left( 2 - \frac{19}{6}x \right)\left( 1 + \frac{x}{8} \right)$

= $\frac{1}{2}\left( 2 + \frac{x}{4} - \frac{19x}{6} \right)$=$\frac { 1 } { 2 }$ $\left( 2 - \frac{35x}{12} \right)$= 1–$\frac{35x}{24}$

\ 1–$\frac{35x}{24}$= a + bx

Ž a = 1 and b = – $\frac{35}{24}$.

Hence (1) is correct answer.