341det A = a3+b3+c3-3abc ≥ 0
We are given (det A)2=1 and so det A =1
That means a3+b3+c3=4
Q1. \textup{If A=}\begin{bmatrix} a & b & c\\ b & c & a\\ c & a & b \end{bmatrix}\textup{, where a,b,c}\; \epsilon \; R^{+}\textup{ and abc=1,}\; A^{T}A=I\textup{, then }a^3+b^3+c^3\textup{ is}
(a) a+b+c (b) a+b+c+3 (c) 3 (d) 4
Q2. MULTI ANSWER
\textup{Which of the following functions (does or) do not have a maximum or minimum point?}
(a) 25 - (x-9)13 (b) (x-8)12 (c) (x-7)24 (d) x9 + 3x8 + 11
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9 Answers
341
1\begin{bmatrix} \sum a^2 & \sum ab & \sum ab \\ \sum ab & \sum a^2 & \sum ab \\ \sum ab & \sum ab & \sum a^2 \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
\\\sum{a^2}=1\\\sum{ab}=0
a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)
=(a+b+c)(1-0)
so a3+b3+c3=a+b+c+3abc=a+b+c+3
1062. ans given AD
i think only A...
1 . @che... then shouldnt it be a multi anser..
also look at bhatt sir's answer abv..
(the question was a single option one)
341the problem is you have said a,b,c are +ve real numbers. In that case we cannot have \sum ab = 0
For the second one,they are talking about this concept: We have f'(x0) = 0. But f"(x0) = 0. What more do we need to do to establish whether x0 is an extremum point or not?
