If x is so large that terms containing x^{-3}, x^{-4}, x^{-5}, can be neglected, then the approximate value of \;≤ft(3x-5/4x²+3)^{-4/5} is [AP EAPCET 21 May 2025 S1]

Correct Answer is : (4x3)4/5(1+43x+135x2)≤ft(4x/3)^{4/5} ≤ft(1 + 4/3x + 13/5x²)(34x)4/5(1+3x4+5x213)

≤ft(3/4x)^{4/5} ≤ft(1 - 4/3x - 7/5x²)

≤ft(4x/3)^{4/5} ≤ft(1 + 4/3x + 13/5x²)

≤ft(4x/3)^{4/5} ≤ft(1 + 4/3x - 13/5x²)

≤ft(3/4x)^{4/5} ≤ft(1 - 4/3x + 7/5x²)

Correct Answer is : (4x3)4/5(1+43x+135x2)≤ft(4x/3)^{4/5} ≤ft(1 + 4/3x + 13/5x²)(34x)4/5(1+3x4+5x213)

Test Index

Binomial Theorem

Section: Mathematics

Question : 14 of 33

Marks: +1, -0

x^{-3}, x^{-4}, x^{-5}, \dots

\;\left(\frac{3x-5}{4x^2+3}\right)^{-\frac{4}{5}}

$\left(\frac{3}{4x}\right)^{4/5} \left(1 - \frac{4}{3x} - \frac{7}{5x^2}\right)$
$\left(\frac{4x}{3}\right)^{4/5} \left(1 + \frac{4}{3x} + \frac{13}{5x^2}\right)$
$\left(\frac{4x}{3}\right)^{4/5} \left(1 + \frac{4}{3x} - \frac{13}{5x^2}\right)$
$\left(\frac{3}{4x}\right)^{4/5} \left(1 - \frac{4}{3x} + \frac{7}{5x^2}\right)$

Solution:

We have,

\;\left(\frac{3x-5}{4x^2+3}\right)^{-\frac{4}{5}} = \;\left(\frac{4x^2+3}{3x-5}\right)^{\frac{4}{5}}

\; = \;\left(\frac{4x^2\left(1+\frac{3}{4x^2}\right)}{3x\left(1-\frac{5}{3x}\right)}\right)^{\frac{4}{5}} = \;\left(\frac{4x}{3}\right)^{\frac{4}{5}} \frac{\left(1+\frac{3}{4x^2}\right)^{4/5}}{\left(1-\frac{5}{3x}\right)^{4/5}}

\; \Rightarrow \left(\frac{4x}{3}\right)^{\frac{4}{5}} \left(1+\frac{3}{4x^2}\right)^{\frac{4}{5}} \cdot \left(1-\frac{5}{3x}\right)^{-\frac{4}{5}}

\; = \left(\frac{4x}{3}\right)^{\frac{4}{5}} \left(1 + \frac{4}{5} \cdot \frac{3}{4x^2} + \dots\right)

\; \left(1 + \frac{5}{3x} \cdot \frac{4}{5} + \frac{4}{5} \left(+\frac{9}{5}\right) \left(\frac{5}{3x}\right)\right)^2 \cdot \frac{1}{2!} + \dots

\; = \left(\frac{4x}{3}\right)^{\frac{4}{5}} \left(1+\frac{3}{5x^2}+\dots\right) \left(1+\frac{4}{3x}+\frac{2}{x^2}\right)

\; = \left(\frac{4x}{3}\right)^{\frac{4}{5}} \left(1 + \frac{4}{3x} + \left(\frac{3}{5}+2\right)^{\frac{1}{x^2}}\right)

\; = \left(\frac{4x}{3}\right)^{\frac{4}{5}} \left(1+\frac{4}{3x}+\frac{13}{5x^2}\right)

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