JEE MAIN - Mathematics (2005 - No. 42)
If $$x$$ is so small that $${x^3}$$ and higher powers of $$x$$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as
$$1 - {3 \over 8}{x^2}$$
$$3x + {3 \over 8}{x^2}$$
$$ - {3 \over 8}{x^2}$$
$${x \over 2} - {3 \over 8}{x^2}$$
Explanation
$${\left( {1 + x} \right)^{{3 \over 2}}}$$ = 1 + $${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$$
= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $$x$$ is so small, so $${x^3}$$ and higher powers of $$x$$ neglected)
$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$
= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$
= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$
= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$
= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$
= $$ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$
[ As x3 is so small we can ignore $$-{3 \over {16}}{x^3}$$]
= $$ - {3 \over 8}{x^2}$$
= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $$x$$ is so small, so $${x^3}$$ and higher powers of $$x$$ neglected)
$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$
= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$
= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$
= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$
= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$
= $$ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$
[ As x3 is so small we can ignore $$-{3 \over {16}}{x^3}$$]
= $$ - {3 \over 8}{x^2}$$
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