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 :

• A. \( {{105} \over 2} \)
• B. \( {{105} \over 4} \)
• C. \( {{105} \over 8} \)
• D. \( {{105} \over 16} \)

Hint:

Find r for which the power of x vanishes, then minimise \(a^4b^6=(a^2b^3)^2\) under \(1/a^2+1/b^3=4\) using AM-GM.

Answer 1

The Correct Option is C

Step 1: General term \(T_{r+1}=\binom{10}{r}a^{10-r}b^{r}x^{\frac{10-r}{8}-\frac{r}{12}}\). Constant term: \(\frac{10-r}{8}=\frac{r}{12}\Rightarrow r=6\).
Step 2: Constant term \(=\binom{10}{6}a^4b^6=210\,a^4b^6\).
Step 3: AM-GM on \(\frac1{a^2},\frac1{b^3}\): \(2=\dfrac{4}{2}\ge\sqrt{\dfrac{1}{a^2b^3}}\Rightarrow a^2b^3\ge\dfrac14\Rightarrow a^4b^6\ge\dfrac1{16}\) (equality when \(\frac1{a^2}=\frac1{b^3}=2\)).
Step 4: Minimum constant term \(=\dfrac{210}{16}=\dfrac{105}{8}\).
Answer: (C) \(\dfrac{105}{8}\)