JEE Main 2022 · 30 Jun · Shift 1 · Mathematics Q6

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MathematicsSingle correct+4 / -1Question 6 of 90

For two positive real numbers a and b such that $\dfrac1{a^2}+\dfrac1{b^3}=4$ , the minimum value of the constant term in the expansion of $\left(ax^{\frac18}+bx^{-\frac1{12}}\right)^{10}$ is :

\dfrac{105}2

\dfrac{105}4

\dfrac{105}8

\dfrac{105}{16}

Answer and worked solution

Official answer: C

Constant term. $\frac{10-r}8=\frac r{12}$ gives $r=6$ : ${}^{10}C_6a^4b^6=210(a^2b^3)^2$ .

AM-GM. $4=\frac1{a^2}+\frac1{b^3}\ge\frac2{\sqrt{a^2b^3}}$ , so $a^2b^3\ge\frac14$ .

Minimum $\frac{210}{16}=\frac{105}8$ .

(C)

Topics:General and middle terms AM-GM-HM inequalities

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JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 6