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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $b = -\frac{35}{24}$.

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $\|x\| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $\|x\| < 1$
$(D)$ If $\|x\|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

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) = $

Similar Questions

The set of all real values of $x$ for which the expansion of $\left(125 x^2-\frac{27}{x}\right)^{-2/3}$ is valid,is

If $3x = 1 + \frac{5}{8} + \frac{5 \times 9}{8 \times 16} + \frac{5 \times 9 \times 13}{8 \times 16 \times 24} + \dots$,then $x^4 + 4x^3 + 6x^2 + 4x = $

If $x=\frac{2 \cdot 5}{3 \cdot 6}-\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9}\left(\frac{2}{5}\right)+\frac{2 \cdot 5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\left(\frac{2}{5}\right)^2-\ldots \infty$,then $7^2(12 x+55)^3=$

The formula $(a + b)^m = a^m + ma^{m-1}b + \frac{m(m - 1)}{1 \cdot 2}a^{m - 2}b^2 + \dots$ holds when

Vedclass Test Series

Exam Paper Generator

Online Exam Module