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

[Piyush Padgil]

Originally Posted by mehrozkhn

sol:
let A = [2/(p+q) + q/2]
we have to find minimum value of A
now p-q <= 2
we can see
p+q = p - q + 2q
since q is positive
p + q >= p-q
=> p + q >= 2
=> 2/(p+q) >= 1 ........(1)
now minimum value of q is equal to zero
hence A(min) = 1 + 0
= 1
hence answer is (3)

Sorry buddy but I differ with you here....
p-q<=2 => p-q can be 1,0,-1.....
Now p+q >=p-q does not mean that p+q >=2... Consider p-q = 0 for understanding

[thebornattitude]

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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

Solution-

Let p=q+2-m.......(i)

so equation becomes ,

2/(2+2q-m) +q/2.....(ii)

now we have to find a way of writing q/2 in the form of denominator of 1'st term, then only we can apply any inequality,else by putting values we cant find, as no of dependents are higher..

so q/2 can be written as
(2+2q-m)/4 + (m-2)/4
put this is (ii) to get

2/(2+2q-m) + (2+2q-m)/4 + (m-2)/4.......(iii)

now take 1'st 2 terms,,,and apply AM>GM, we get

2/(2+2q-m) + (2+2q-m)/4 >=2/root(2)= root(2).....(iv)

so, we get
2/(2+2q-m) + (2+2q-m)/4 + (m-2)/4>= root(2) + (m-2)/4

this will be further minimized when m is taken to be minimum ie m=0....hence


2/(2+2q-m) + (2+2q-m)/4 + (m-2)/4>=root(2) -1/2.......

hence the answer- option (1)

[Aarav]

Originally Posted by arbit_rageur

but in more complicated cases, when having two variables and minimizing two parts of the expression separately ,it may become an issue. ....Your thoughts?

I know -> the Inequalities problems can be tricky at times, and hence suggested to pause for few seconds before marking the final answer.

Last year in QQAD Tests we had one simple problem on Inequality that stumped everyone due to issues you mentioned above.

[arbit_rageur]

A small mistake...The penultimate line should read 2/(2+2q)+q/2>=sqrt(2)-1/2.

Originally Posted by arbit_rageur

2>=p-q

So 2+q>=p. => 2/(p+q)+q/2>=2/(2+2q)+q/2>=sqrt(2)-1.

So (1).

[thebornattitude]

And yah just one very gud pt pointed out by @arbit ..just saw now,,,,,

We have to check for q to be postive here...else it doesnt hold true...

Checked just now, and found that

q= {m+2(root(2)-1)}/2....which is positive, hence holds true.

[Aarav]

QQAD 050 -> Case 3 was left for students to be solved. I hope the solution makes sense to all? You can pin-point me the issues if you find any since last day I was out of station and travelling and hence couldn't spend much time with this problem.

[neynetr]

QQAD: 51

p-q <= 2

let 2/(p+q) + q/2 be k... which is to be minimised.
the p = 4/(2k-q) - q
substitusting this value of p back:

4/(2k-q) - 2q <= 2
=> 2q^2 + q(2-4k) + (4-4k) = 0

as p and q are positive so by that condition (c/a >= 0 and -b/a < 0)
we get k =< 1 and k > 1/2

moving further for real values of q the discriminant should be real.

=> 16k^2 + 16k - 28 >= 0
=> k < -1/2 - sqrt(2)
or
k > -1/2 + sqrt(2)

so min value of k is sqrt(2) - 1/2
option (A)

[v-factor]

2 >= p-q

So p <= (q+2) max value of p = q+2

Replacing in the equation we get

2/(2q+2) + q/2 => 1/(q+1) + q/2

I'm stuck here.....now how to proceed...how to apply AM>=GM here....plz help

2nd attempt:
writing q in terms of denominator of 1st term
1/(q+1) + (q+1)/2 - 1/2
Applying AM>=GM for 1st 2 terms
1/(q+1) + (q+1)/2 - 1/2 >= (2)^1/2 - 1/2
Hence option 1

[slam]

Originally Posted by neynetr

QQAD: 51

p-q <= 2

let 2/(p+q) + q/2 be k... which is to be minimised.
the p = 4/(2k-q) - q
substitusting this value of p back:

4/(2k-q) - 2q <= 2
=> 2q^2 + q(2-4k) + (4-4k) = 0

as p and q are positive so by that condition (c/a >= 0 and -b/a < 0)
we get k =< 1 and k > 1/2

moving further for real values of q the discriminant should be real.

=> 16k^2 + 16k - 28 >= 0
=> k < -1/2 - sqrt(2)
or
k > -1/2 + sqrt(2)

so min value of k is sqrt(2) - 1/2
option (A)

Nice method! It helped me understand the problem.

However, I still don't get where my own method went wrong. I used p - q = 2 - k, substituted the value of q in the expression, and then differentiated to find the maxima/minima. Any ideas on why this shouldn't be used here?

[ayan gupto]

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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

ans:
min[2/(p+q) + q/2]......@
will be min. when (p+q) = max
from equation (p-q) <= 2
we have p <= 2+q
substituting in eq. @
=> min[1/(q+1) + q/2].........b

now lets assume q+1 = a
therefore q = a-1
substituting in eq.b
=>min[1/a + (a-1)/2]
=min[1/a + a/2 -1/2]
now use AM >= GM for the first 2 terms
=> {1/a+a/2}/2 >= 1/sqrt(2)
=>{1/a+a/2}>= 2/sqrt(2)

hence min[1/a + a/2 -1/2] = 2/sqrt(2) - 1/2
=sqrt(2)-1/2

hence option (1).