Question

Let $x , y , z$ be positive numbers such that $x + y + z =10$. If maximum value of $x ^2 y ^3 z ^5$ is $N$, then

List -I		List -II
P	number of odd divisors of $N$ is	1	16
Q	number of divisor of $N$ which have atleast two zeroes at the end is	2	23
R	number of divisors of $N$ which are of the form $(4 n +2), n \in N$ is	3	24
S	product of all the divisors of $N$ is $( N )^{ m }$ where $m$ is	4	36

• A. P= 3 ,Q= 2 ,R= 1 ,S= 4
• B. P=1 ,Q=2,R= 4 ,S=3
• C. P=3 ,Q=1,R= 2 ,S=4
• D. P= 1 ,Q=2,R= 3,S= 4

Answer 1

Answer: (C)

$\frac{2 \cdot\left(\frac{ x }{2}\right)+3 \cdot\left(\frac{ y }{3}\right)+5\left(\frac{ z }{5}\right)}{10} \geq\left(\frac{ x ^2}{2^2} \cdot \frac{ y ^3}{3^3} \cdot \frac{ z ^5}{5^5}\right)^{\frac{1}{10}}$
$x ^2 y ^3 z ^5 \leq 2^2 \cdot 3^3 \cdot 5^5 $
$\therefore N =2^2 \cdot 3^3 \cdot 5^5$
(P) Number of odd divisors of $N =(3+1)(5+1)=24$
(Q) Number of divisors of $N$ which have atleast two zeroes at the end is $=(3+1)(4)=16$
(R) Number of divisors of $N$ which are of the form $(4 n+2), n \in N$ i.e.,
$2(2 n+1), n \in N $
$=(3+1)(5+1)-1=23$
(S) Product of all the divisors of $N =( N )^{ m }=( N )^{36}$
$m =\frac{\text { Total number of divisors }}{2}=\frac{72}{2}=36$