CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

[Aarav]

Hello All,

Please carry the discussions for QQAD problems 51 to 100 on this thread.

Have a nice and fruitful time over here

[nbangalorekar]

Aarav hope to have a more exciting and mind blowing session with this.......

[sabsebadapaagal]

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Quantitative Question # 051
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Given p and q be positive such that 2 >= p-q, the min value of 2/(p+q) + q/2 is


(1) √2 - 1/2 (2) (√2 + 1)/2 (3) 1 (4) 1/√2 (5) none of these

[sabsebadapaagal]

Originally Posted by sabsebadapaagal

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Quantitative Question # 051
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Given p and q be positive such that 2 >= p-q, the min value of 2/(p+q) + q/2 is


(1) √2 - 1/2 (2) (√2 + 1)/2 (3) 1 (4) 1/√2 (5) none of these

min(2/(p+q) + q/2) >= 2(q/(p+q))^1/2 (AM > GM)
=> 2(q/(q+4))^1/2 ---------(A)

now substitute q=1 ... we get a value something less than 1 thus options 2,3 gets discarded.... now we are left with options 1, 4 and 5

on working further we find that (A) is minimum at q=4/7 which gives 2(q/(q+4))^1/2 = (1/2)^1/2

hence option 4

[srikar2097]

Originally Posted by sabsebadapaagal

min(2/(p+q) + q/2) >= 2(q/(p+q))^1/2 (AM > GM)
=> 2(q/(q+4))^1/2 ---------(A)

now substitute q=1 ... we get a value something less than 1 thus options 2,3 gets discarded.... now we are left with options 1, 4 and 5

on working further we find that (A) is minimum at q=4/7 which gives 2(q/(q+4))^1/2 = (1/2)^1/2

hence option 4

Didn't get how you applied AM ≥ GM here?

[tarunad]

guys the ans should be (a) , the exp is 1/(p-1) +(p-1)/2 -1/2 which is greater than (a)

[kamalesh123]

My approach.

Substitute q = p-2 + k where k >=0;

Applying Rolle's theorem, I got the answer as √2 - 1/2 . Option (1).

[sabsebadapaagal]

Originally Posted by srikar2097

Didn't get how you applied AM ≥ GM here?

AM >= GM can be applied cos both terms are positive ...... have I missed something ?

[tarunad]

hi kamalesh123 ,
Can you let us know how to use rolles theorm,
i heard that its a very useful one for quadratics

[rajeev_hts]

f=2/(p+q) +q/2

2>p-q => q>=p-2

f=1/(p-1) +p/2 -1

differentiating and putting eq to 0
p=1+2^1/2

Min(f)= 2^0.5 - 0.5

option(1)

I am feeling like question was designed to trap this approach and I have become victim of that.