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(A)to(r);(B)to(s);(C)to(t);(D)to(p,q,r,s,t)
Solution :
(A):.AMgeGM <br> :.(ab^(2)+ac^(2)+bc^(2)+ba^(2)+ca^(2)+cb^(2))/(6)ge(ab^(2)*ac^(2)*bc^(2)*ba^(2)*ca^(2)*cb^(2))^((1)/(6))=abc <br> :.a(b^(2)+c^(2))+b(c^(2)+a^(2))+c(a^(2)+b^(2))ge6abc <br> implies lambda=6 <br> (B) :.AMgeGM <br> For b,c, we get ((b+c))/(2)gesqrt(bc) <br> implies (b+c)ge2sqrt(bc)" " "....." <br> For c,a, we get <br> ((c+a))/(2)ge sqrt(ca) <br> implies (c+a)ge 2sqrt(ca)" " ".....(ii)" <br> and a, b, we get <br> ((a+b))/(2)ge sqrt(ab) <br> implies (a+b)ge 2sqrt(ab)" " ".....(iii)" <br> On multiplying Eqs. (i), (ii) and (iii), we get <br> (b+c)(c+a)(a+b)ge8abc <br> implies (1-a)(1-b)(1-c)ge 8abc [:. a+b+c=1]<br> implies ((1)/(a)-1)((1)/(b)-1)((1)/(c)-1)ge 8 <br> :. lambda=8 <br> (C ) :.AMgeHM <br> ((s-a)+(s-b)+(s-c))/(3)ge(3)/((1)/(s-a)+(1)/(s-b)+(1)/(s-c)) <br> implies (3s-(a+b+c))/(3)ge(3)/((1)/(s-a)+(1)/(s-b)+(1)/(s-c)) <br> implies (3s-s)/(3)ge(3)/(((1)/(s-a)+(1)/(s-b)+(1)/(s-c))) <br> implies(2s)/(s-a)+(2s)/(s-b)+(2s)/(s-c)ge9 <br> Here, lambda-1=9 <br> :. lambda =10 <br> (D) If a, b, c are in GP. <br> a^(lambda),b^(lambda),c^(lambda) are also in GP. <br> Then, AMge GM <br> (a^(lambda)+c^(lambda))/(2)geb^(lambda) <br> impliesa^(lambda)+c^(lambda)ge2b^(lambda) <br> implies((a)/(b))^(lambda)+((c)/(b))^(lambda)ge2 <br> :. lambdainR <br> Hence, lambda=2,4,6,8,10 Transcript

TimeTranscript
00:00 - 00:59match the following column column 1 and column to column when we are having a part if a greater than zero value of so letter start with solution let's solve party and then will solve BC and so here we are having ABC is a positive and we have to find the minimum value of this one relation and that relation is a m is greater than GM is greater than equal to H M Vinod that I am is always greater than game and so will use this relation to simplify
01:00 - 01:59this one letter start with part 1 and part 1 is here and here we are having a greater than zero be greater than zero and C greater than zero and we have to find the minimum value of P B square + c square then b square + c square then see a square + b square simplify this one will get a b square + c square + b square + c square b + c square then + expression we are having a total six terms and will apply the concept that I am is greater than equal to GM reply this one will get
02:00 - 02:59EM of the 6th term will be a b square + AC square + b square a square b + b square + c square and 1 whole divided by 6 will be greater than equal to will be greater than equal to of all this term and all terms are E square into x square into x square into x square x square and square and whole power one upon 6 simplify this one and you get here will have a b square + c square + b square + a square b +
03:00 - 03:59c square + b square and whole divided by 6 8 and here will have a greater than your will have a t e square then a square b square then b b and b square power one upon 6 simplify this one and will get a square + c square + b square + c square + c square + b square is greater than equal to 6 ABC ok Subah comparing that one will have a Lambda ABC year so by comparing the cylinder ABC will get
04:00 - 04:59Lambda is equal to 6 year so for the first option we option A we're having a answer is that is are so for me correct Kolam correct match is are here and it is 6 let's do other another part That is b&b part we are having a if a b c are positive a + b + C is equal to one year and minimum value of 1 upon 3 minus one upon B - 11 - 1 is limited in Lambda is again we are to find the Lambda and a PCR positive so let's do this part B part part B and it again we are having a b c positive greater than zero and relation given that is a + b + C is equal
05:00 - 05:59to one year and we have to find minimum value of 1 upon 3 minus 11 upon B - 1 and 1 upon 1 minus 10 so from this expression a + b + C is equal to 1 by dividing both side by a will get one upon x is equal to 1 + b upon 3 + 3 upon ok so similarly will get one upon we also one upon B will be 1 + 1 upon B + c upon we will get one account so one account will be 1 + 1 + 2 upon 3 + 1 upon 2 will put this one in the expression and will get 1 upon 1 minus 10 will get
06:00 - 06:59b + c / then by putting one upon the will get / be here by coaching mein kaun si will get a + b / now we know that is greater than equal to GM so this apply a m greater than GM on Thambi + 23 + Europe + b o b + c is greater than equal to twice of 100 BC similarly will have a + c will be greater than twice of under root AC and similarly will have a + b greater than equal to twice of underwood ab so we'll put this one in the relationship will get b + c to b + c is greater than 2 under root BC
07:00 - 07:59/ then a + C + 200 AC / the and then to under root ab / 2 by simplifying this one will get this one is the minimum minimum side so will have a now b + c b + c / into a + b into a + b / will be greater than equal to twice twice into voice into twice and under root so under today and today is will become a year or we can say that under root of
08:00 - 08:59a square then the square and square and / / ABC Sona final will have a b + c / into a + b into a + b / will be greater than equal to will have eat pear and ABC and under root a square b square c square bracket cancel out so this will be greater than 8 aur we can say that again will change in the starting starting expression and that was once Upon A -1 and 1 upon B - 1 and one upon x minus 1 is greater than it so here minimum value is IIT for part B part the minimum value is 80 part
09:00 - 09:59be correct matches so party correct match is now next is the part C now parts is greater than zero B greater than zero is greater than or equal to a + b + c and the minimum value of 2 + 2 + 2 upon x minus 3 is then Lambda is so party let's start party so part c a greater than 0.0 and later than zero here and S is given that is a + b + c year and we need to find the minimum value of 2 as divided by x minus plus us divided by x minus b + 2 x divided by x minus 2 find
10:00 - 10:59the minimum value of this and its minimum values given that is Lambda -1 ok so first find the minimum value of this one so we know that here I am is greater than hm you know that I am is always greater than hm so let's take the amorphous minus a minus b minus 3 x minus plus minus b + X - 3 and whole divided by 3 is the a.m. and it will be greater than equal to of hm hm hm will be 3 / inverse of the whole term and universe of the atom is one upon x minus 1 + 1 upon x minus b + 1 upon x minus 3 year will simplify this one and will get ss11 3 minus a +
11:00 - 11:59b + c divided by 3 and this one is greater than equal to 3 / 1 upon x minus 1 + 1 upon x minus b + 1 upon x minus 3 year will simplify and we know that s is equal to a + b + c so will have a Prius - / 3 and this one is greater than equal to 3 / 1 upon x minus 1 + 1 upon x minus b + 1 upon x minus 3 into x minus 2 is to 3 minus is to you know that to US and cross multiplying one upon x minus 1 upon x minus b will have one upon x minus y + 1 upon x minus b + 1 upon x minus 3 is greater than equal to cos x 33
12:00 - 12:59nine year and this one is the final answer that is the twice upon x minus 1 + 2 upon x minus b + 2 upon x minus 3 is greater than equal to 90 minimum value is 9 minimum value of this expression is 9 and given that minimum value is Lambda -1 -1 is nine year and that is why equal to 10 cm is equal to 10 for option for part c and aur part Lambda Eastern part C Allahabad correct Majesty now next and last part last part is D here and let's see what is the D in the depart if A is greater than 20 20
13:00 - 13:59is greater than zero and a b c are in GP given that a b c are in GP then the minimum value of a upon B Lambda + cm1b Lambda is 2 then Lambda is the part of the year and we are having a ABC that is greater than zero and a b c are in GP and we know that when a relation is b square is equal to AC AC is equal to b square now next next we have to find the minimum value of a and b h m and n + 1 upon the power Lambda and minimum values given minimum value of this one is and that is equal to 2 find minimum value of this expression so we know that I am is greater than g n m is greater than GM so let's take to travel that is power
14:00 - 14:59Lambda and power Lambda and just find the aim is greater than GM so I am is greater than equal to gmb know that and this students will use this relation to am will be a Lambda + c level are divided by 2 and GM will be under root of Lambda Lambda here and we know that the relation between relation between ABC is b square equal to AC so we can write a Lambda + Lambda divided by 2 is greater than equal to under root AC AC is b square b square Power Limited here and will simplify this one will get a Lambda + Lambda divided by 2 is greater than equal to real simplify this one so will get D square and
15:00 - 15:59Ruby Lambda and square and Android get cancer Sobi Laila year and now will have a Lambda + Lambda / bid is greater than equal to 2 year so we can see that whole Lambda + / b whole Lambda is greater than equal to 2 year so its minimum value does not depend upon the Lambda we have started from am is greater than GM and we have used the relation that this are in the GPS or b square equal to AC and we have used at relation here and we got this expression here again is not depending upon any value can be any value here so maybe belongs to a belongs to Arya it can
16:00 - 16:59take any value so that belongs to H all the option will be correct here so I will match with all p q r s t r i can take any value so 2 4 6 8 10 what is the correct and Devil correctly matched with all all values of the column so finally we are having answer and answer is that is measured with our then be is measured with as then she is measured with T and D is matched with p q r s t and u value

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