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 $(4n+2),n\in N$ 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)

$\frac{2\cdot \left(\frac{x}{2}\right)+3\cdot \left(\frac{y}{3}\right)+5\left(\frac{z}{5}\right)}{10}\ge {\left(\frac{x^2}{2^2}\cdot \frac{y^3}{3^3}\cdot \frac{z^5}{5^5}\right)}^{\frac{1}{10}}$
$x^2{y}^3{z}^5\le 2^2\cdot 3^3\cdot 5^5$
$\therefore N=2^2\cdot 3^3\cdot 5^5$
(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 $(4n+2),n\in N$ i.e.,
$2(2n+1),n\in N$
$=(3+1)(5+1)-1=23$
(S) Product of all the divisors of $N=(N{)}^m=(N{)}^{36}$
$m=\frac{\text{ Total number of divisors }}{2}=\frac{72}{2}=36$