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It is given that a, b are real and p, q are purely imaginary - HSC - SSCE Mathematics Extension 2 - Question 9 - 2018 - Paper 1

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It is given that a, b are real and p, q are purely imaginary. Which pair of inequalities must always be true? A. $a^2 p^2 + b^2 q^2 \leq 2abpq$, $a^2 b^2 + p^2 q^... show full transcript

Worked Solution & Example Answer:It is given that a, b are real and p, q are purely imaginary - HSC - SSCE Mathematics Extension 2 - Question 9 - 2018 - Paper 1

Step 1

A. $a^2 p^2 + b^2 q^2 \leq 2abpq$, $a^2 b^2 + p^2 q^2 \geq 2abpq$

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Answer

To determine whether this inequality holds, we can analyze the values of 'a', 'b', 'p', and 'q'. The first inequality is not necessarily always true because it contradicts the conditions set for real and imaginary numbers. Hence, it cannot be universally valid.

Step 2

B. $a^2 p^2 + b^2 q^2 \leq 2abpq$, $a^2 b^2 + p^2 q^2 \leq 2abpq$

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Answer

This pair of inequalities appears to hold under the Cauchy-Schwarz inequality, which states that for any real numbers and imaginary numbers as defined, these relations are valid. This is the correct answer.

Step 3

C. $a^2 p^2 + b^2 q^2 \geq 2abpq$, $a^2 b^2 + p^2 q^2 \leq 2abpq$

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Answer

The first inequality here is derived from the reverse Cauchy-Schwarz inequality which is not always true given the conditions of 'a', 'b', 'p', and 'q'. Therefore, this cannot be valid.

Step 4

D. $a^2 p^2 + b^2 q^2 \geq 2abpq$, $a^2 b^2 + p^2 q^2 \geq 2abpq$

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Answer

Similar to the analysis in option C, this set of inequalities cannot be universally verified under the given conditions. Thus, it is not a valid option.

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