Question:
If \(|x|\) is so small that \(x^2\) and higher powers of \(x\) may be neglected, then the approximate value of
\[
\left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}}
\]
is
If \(|x|\) is so small that \(x^2\) and higher powers of \(x\) may be neglected, then the approximate value of
\[
\left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}}
\]
is
Show Hint
For approximation problems:
\[
(1+u)^n \approx 1+nu
\]
when \(|u|\) is very small. Also remember:
\[
\frac{1+a}{1+b}\approx 1+a-b,
\]
neglecting terms involving \(ab,a^2,b^2\), etc.
Updated On: Jul 9, 2026
- \(4+\dfrac{9x}{8}\)
- \(1-\dfrac{7x}{32}\)
- \(1+\dfrac{9x}{32}\)
- \(4-\dfrac{7x}{8}\) \bigskip
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
If \(|u|\ll 1\), then using the Binomial Approximation,
\[
(1+u)^n \approx 1+nu,
\]
neglecting terms involving \(u^2\) and higher powers.
Step 1: Rewrite the given expression. \[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} = \left(\frac{64\left(1+\frac{3x}{64}\right)} {8\left(1+\frac{3x}{8}\right)}\right)^{\frac{2}{3}}. \] \[ = \left[ 8\cdot \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right]^{\frac{2}{3}}. \] \[ = 8^{\frac{2}{3}} \left( \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right)^{\frac{2}{3}}. \] Since \[ 8^{\frac{2}{3}}=4, \] we get \[ =4 \left( \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right)^{\frac{2}{3}}. \]
Step 2: Simplify the fraction inside the bracket. Using \[ \frac{1+a}{1+b}\approx 1+a-b, \] we obtain \[ \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \approx 1+\frac{3x}{64}-\frac{3x}{8}. \] \[ = 1-\frac{21x}{64}. \] Therefore, \[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} \approx 4\left(1-\frac{21x}{64}\right)^{\frac{2}{3}}. \]
Step 3: Apply the binomial approximation. Using \[ (1+u)^n\approx 1+nu, \] with \[ u=-\frac{21x}{64}, \qquad n=\frac23, \] we get \[ \left(1-\frac{21x}{64}\right)^{\frac23} \approx 1+\frac23\left(-\frac{21x}{64}\right). \] \[ = 1-\frac{7x}{32}. \] Hence, \[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} \approx 4\left(1-\frac{7x}{32}\right). \] \[ = 4-\frac{28x}{32}. \] \[ = 4-\frac{7x}{8}. \]
Step 4: Write the final answer. \[ \boxed{4-\frac{7x}{8}} \]
Step 1: Rewrite the given expression. \[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} = \left(\frac{64\left(1+\frac{3x}{64}\right)} {8\left(1+\frac{3x}{8}\right)}\right)^{\frac{2}{3}}. \] \[ = \left[ 8\cdot \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right]^{\frac{2}{3}}. \] \[ = 8^{\frac{2}{3}} \left( \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right)^{\frac{2}{3}}. \] Since \[ 8^{\frac{2}{3}}=4, \] we get \[ =4 \left( \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \right)^{\frac{2}{3}}. \]
Step 2: Simplify the fraction inside the bracket. Using \[ \frac{1+a}{1+b}\approx 1+a-b, \] we obtain \[ \frac{1+\frac{3x}{64}} {1+\frac{3x}{8}} \approx 1+\frac{3x}{64}-\frac{3x}{8}. \] \[ = 1-\frac{21x}{64}. \] Therefore, \[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} \approx 4\left(1-\frac{21x}{64}\right)^{\frac{2}{3}}. \]
Step 3: Apply the binomial approximation. Using \[ (1+u)^n\approx 1+nu, \] with \[ u=-\frac{21x}{64}, \qquad n=\frac23, \] we get \[ \left(1-\frac{21x}{64}\right)^{\frac23} \approx 1+\frac23\left(-\frac{21x}{64}\right). \] \[ = 1-\frac{7x}{32}. \] Hence, \[ \left(\frac{3x+64}{3x+8}\right)^{\frac{2}{3}} \approx 4\left(1-\frac{7x}{32}\right). \] \[ = 4-\frac{28x}{32}. \] \[ = 4-\frac{7x}{8}. \]
Step 4: Write the final answer. \[ \boxed{4-\frac{7x}{8}} \]
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