If frac{(1 - 3x)^{1/2} + (1 - x)^{5/3}}{sqrt{ | vedclass

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(B) Given expression: $f(x) = \frac{(1 - 3x)^{1/2} + (1 - x)^{5/3}}{2(1 - x/4)^{1/2}}$Using the binomial expansion $(1 + z)^n \approx 1 + nz + \frac{n

Vedclass · Accepted Answer

$\left( 1, -\frac{35}{24} \right)$

Vedclass · Answer

(B) Given expression: $f(x) = \frac{(1 - 3x)^{1/2} + (1 - x)^{5/3}}{2(1 - x/4)^{1/2}}$Using the binomial expansion $(1 + z)^n \approx 1 + nz + \frac{n(n-1)}{2}z^2$ for small $z$:Numerator: $(1 - 3x)^{1/2} + (1 - x)^{5/3} \approx [1 + \frac{1}{2}(-3x)] + [1 + \frac{5}{3}(-x)] = 2 - \frac{3}{2}x - \frac{5}{3}x = 2 - \frac{19}{6}x$Denominator: $2(1 - x/4)^{1/2} \approx 2[1 + \frac{1}{2}(-x/4)] = 2 - \frac{x}{4}$So,$f(x) \approx \frac{2 - \frac{19}{6}x}{2(1 - x/8)} \approx \frac{1 - \frac{19}{12}x}{1 - x/8} \approx (1 - \frac{19}{12}x)(1 + x/8)$$f(x) \approx 1 + \frac{x}{8} - \frac{19}{12}x = 1 + (\frac{3 - 38}{24})x = 1 - \frac{35}{24}x$Comparing with $a + bx$,we get $a = 1$ and $b = -\frac{35}{24}$.

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

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