JEE Advanced 2011 Paper 1Question 1
The minimum value of the sum of real numbers $a^{-5}$, $a^{-4}$, $3a^{-3}$, $1$, $a^8$ and $a^{10}$ with $a>0$ is
Differentiation and its Application
Answer$8$
Solution
Concept. $S'(a)$ is increasing and vanishes at $a=1$, so $a=1$ gives the minimum. Formulas used. (1) If $S'$ is increasing and $S'(1)=0$, then $S$ has its least value at $1$Step 1 - Derivative. $$S'(a)=-5a^{-6}-4a^{-5}-9a^{-4}+8a^7+10a^9.$$ $S'(1)=-5-4-9+8+10=0$. Step 2 - $S'$ increasing. Each term increases with $a$ (negative powers carry a minus sign). So $S'<0$ for $a<1$ and $S'>0$ for $a>1$. Step 3 - Value. By (1), $S_{\min}=S(1)=1+1+3+1+1+1=8$. Answer: 8