Test Questions.

6) 11

6) it will be 13..

ambiguous cases will be - 9,2,2 and 6,6,1

2) simple maths- ans will be 22

5)after some logical guesswork - 95

@narayan - we have to check the cases where the sum of factors are same . (there is no other set of factors which sum up to 11 other than the one u hav provided)

it is not mentioned that two integers have to be same ?

Qn -5 , Solution:

Since n has 4 divisors it must be of the form pq or m³, where p,q,m are all primes.

Case 1. When n = m³, the given condition translates to 4(m+ m²)= m³+ 1. not possible since m does not divide 1.

Case 2. When n= pq. The given condition translates to 4(p+q)= pq+1, or (p-4)(q-4)= 15

We obtain systems p- 4 = 1, q- 4 = 15 , and p- 4 = 3 , q-4 = 5

this gives n = pq = 95

edit: din saw prophet sir and aditya's posts

Qn-1 (all real solution) -1

is the 7th answer 1√2-1 or √2+1?

is the 7th answer 32+√6..???

7) ans is coming to (3+√17)/2 ≈ 3.56

For the first one, adding the two equations and then adding 1 to both sides produces

(1+a)(1+b)(1+c) = 8044 = 2 \ \times \ 2 \ \times \ 2011

So a = b =1 and c=2010.

You can easily check that the equations are satisfied.

So a+b+c=2012

section B
2)
taking lcm- S= \sum_{1}^{2011}{\frac{a_{i}+b_{i}}{a_{i}b_{i}}}
S= \sum_{1}^{2011}{\frac{2}{-a-a^{2}}}S= -2\sum_{1}^{2011}{\frac{a+1 -a}{a(a+1)}}
S= -2\sum_{1}^{2011}{\frac{1}{a} -\frac{1}{a+1}}

S= -20111006

Please tell where do you get such questions?
For 9. if we take any particular point the region comes out to be a circle. Cant we simply find the area of circle?
But i also noticed the 'at least' part.
But i dont know what it signifies.

For 1. one real solution is -1 for sure. but i dont know if there are any other solutions too.