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Maths-

General

Easy

Question

If matrix A = $open square brackets table row a b c row b c a row c a b end table close square brackets$ where a, b, c are real positive numbers, abc = 1 and A^T A = 1, then the value of a³ + b³ + c³ is -

3
2
4
1

The correct answer is: 4

A^T A = I $rightwards double arrow$ |A^TA| = |I| $rightwards double arrow$ |A|² = 1
$rightwards double arrow$ (a³ + b³ + c³ – 3abc)² = 1
$rightwards double arrow$ a³ + b³ + c³ – 3abc = 1
(since a, b, c are positive real number $rightwards double arrow$ a³ + b³ + c³ $greater or equal than$ 3abc from AM $greater or equal than$ GM)
$rightwards double arrow$ a³ + b³ + c³ = 4.
Hence (C) is correct answer.

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Statement-I :If $x equals cos space theta left parenthesis 1 plus sin space theta right parenthesis plus sin space theta left parenthesis 1 minus cos space theta right parenthesis$ and $y equals cos space theta left parenthesis 1 plus cos space theta right parenthesis minus sin space theta left parenthesis 1 minus sin space theta right parenthesis$ then $x plus y less or equal than 3$
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Statement-II : $sum from straight r equals 1 to straight n minus 1 of cos squared space open parentheses fraction numerator straight r pi over denominator straight n end fraction close parentheses equals straight n over 2 minus 1$
Statement-III : $sin space open parentheses pi over n close parentheses plus sin space open parentheses fraction numerator 3 pi over denominator n end fraction close parentheses plus sin space open parentheses fraction numerator 5 pi over denominator n end fraction close parentheses plus midline horizontal ellipsis plus n text terms end text equals 0$

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