[JEE Main 2005] source ocr uncertain if x is so small that x 3 and higher powers
[Source OCR-uncertain] If $x$ is so small that $x^3$ and higher powers are neglected, the approximate value of $\dfrac{(1+x)^2-\left(1+\frac x2\right)^3}{(1-x)^{1/2}}$ is
(a) $1-\dfrac{3x^2}8$
(b) $3x+\dfrac{3x^2}8$
(c) $-\dfrac{3x^2}8$
(d) $\dfrac x2-\dfrac{3x^2}8$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
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Correct answer: (d) $\dfrac x2-\dfrac{3x^2}8$
Numerator $\approx\frac x2+\frac{x^2}4$; dividing by $(1-x)^{1/2}\approx1-\frac x2-\frac{x^2}8$ leaves a leading $\frac x2$ term, matching option (d).
JEE Main 2005 · Binomial Theorem — verified solution by the Vidaara Team.
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