uestion 1:



(1+x)p <= (1 + xp)



where,



(a) p>1



(b) 0<=p<=1



(c) x>0



(d) x<0



 



plz solve with proper explanation

C0 + C1(X) + C2(X^2) + C3(C^3) + C4(C^4) + ... Cp(X^p)   <=  1 + X^p

C1(X) + C2(X^2) + C3(C^3) + C4(C^4) + ... C(p-1)[X^(p-1)]  <= 0

The above inequality holds true only when X<0, if its multiple answer type question, then p>1 also should be marked... or else it is generally understood

hope u got it...!!!

no dude u r totally wrong with the answer .....the answer is  b and c

dats k... i got a&d n b&c both answers rite... i didn't get the method for b&c...

i would b glad if u could rectify my mistake... thanq...

Expanding (1+x)^p and cancelling 1,px,we get,

p(p-1)/2!.x²+p(p-1)(p-2)/3!.x³+......If p>1 and an integer then

pc₂,pc₃,...,pc_p are +ve and though x<0,the expression can't be -ver,bcoz every alternate +ve term exceeds -ve term in magnitude.

So,0p1,then p(p-1)/2!,... are -ve and if x>0,the product is -ve and hence,the whole expression is -ve.Thus,the answers are b,c.

@allamraju

let me get cleared with d question...

the question is  (1+x)^p <= 1+x^p

but not  (1+x)^p <= 1+px

so if u take x= -1 & p=2... the inequality holds... => a&d is also correct...

let me know if i'm wrong...

Oh..I didn't understand whether it is x^p or xp.He even didn't use the symbol ^ in between.For (1+x)p,It could recognise bcoz if it is multiplication then there is no meaning to the question.Anyway,The same explanation works for that too.

The expression starts with p.x and in the middle there will be no x^p term.That's all.

x^p is d last term... so first n last term both get cancelled... n hence a,d & b,c both d solutions satisfy...

but one thing i didn't get... he included dis question in differential calculus... so is thr ny way of solving it by DC...???

guys this is a question of differential calculys itself....based on increasing decresing fncs....but the explanation i have in the book is VERY vague n concise...very ough to understand