1062) If n = 2m
then ans = 1 + 3 + 5 + .. + 2m-1
= m^2 = n^2/4
If n = 2m+1
then ans = 1 + 3 + 5 + ... + 2m+1
= (m+1)^2 = (n+1)^2/4
$(1) Solve the equation $x^3-x^2+9\lambda x-\lambda=0$. if It is known to have only\\\\ positive roots. and also find value of $\lambda$\\\\ (2) Determine The no. of local extreme of \\\\$f(x)=(x-1).(x-2)^2.(x-3)^3.(x-4)^4.................(x-n)^n$\\\\ Where $x\in R$ and $n$(integer) $>1.$
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2 Answers
11)
let the roots be a,b,c
a+b+c=1
ab+bc+ca=-9λ
abc=λ
since a,b,c are all positive so -9λ has to be positive
so λ<0
but abc=λ means λhas to be ≥0
hence λ=0
which gives x=0
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