For two positive real numbers a and b such that [1 a^2 + 1 b^3 = 4], then minimum value of…

Question

For two positive real numbers a and b such that $${1 \over {{a^2}}} + {1 \over {{b^3}}} = 4$$, then minimum value of the constant term in the expansion of $${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$$ is :

Accepted Answer

Given, Binomial expansion,

$${\left( {a{x^{{1 \over 8}}} + b{x^{ - {1 \over {12}}}}} \right)^{10}}$$

General term,

$${T_{r + 1}} = {}^{10}{C_r}\,.\,{\left( {a{x^{{1 \over 8}}}} \right)^{10 - r}}\,.\,{\left( {b{x^{ - {1 \over {12}}}}} \right)^r}$$

$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{10 - r} \over 8}}}\,.\,{x^{ - {r \over {12}}}}$$

$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{\left( {{{10 - r} \over 8} - {r \over {12}}} \right)}}$$

$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 3r - 2r} \over {24}}}}$$

$$ = {}^{10}{C_r}\,.\,{a^{10 - r}}\,.\,{b^r}\,.\,{x^{{{30 - 5r} \over {24}}}}$$

For constant term,

$${{30 - 5r} \over {24}} = 0$$

$$ \Rightarrow r = 6$$

$$\therefore$$ Constant term,

$${T_{r + 1}} = {T_{6 + 1}} = {}^{10}{C_6}\,.\,{a^4}\,.\,{b^6}$$

$$ = {{10!} \over {6!\,4!}}{a^4}\,.\,{b^6}$$

$$ = {{10 \times 9 \times 8 \times 7} \over {4 \times 3 \times 2 \times 1}}\,.\,{a^4}\,.\,{b^6}$$

$= 210{a^4}{b^6}$

We know, $$GM \ge HM$$

For terms a2 and b3,

$$\sqrt {{a^2}{b^3}}  \ge {2 \over {{1 \over {{a^2}}} + {1 \over {{b^3}}}}}$$

$$ \Rightarrow \sqrt {{a^2}{b^3}}  \ge {2 \over 4}$$

$$ \Rightarrow {a^2}{b^3} \ge {1 \over 4}$$

$$ \Rightarrow {({a^2}{b^3})^2} \ge {1 \over {16}}$$

$$\therefore$$ $${a^4}{b^6} \ge {1 \over {16}}$$

$$\therefore$$ Minimum value of $${a^4}{b^6} = {1 \over {16}}$$

$$\therefore$$ Minimum value of constant term

$${T_7} = 210 \times {a^4}{b^6}$$

$$ = 210 \times {1 \over {16}}$$

$$ = {{105} \over 8}$$

Solutions · 1