SATHEE: Matrices and Determinants 2 Question 44

Matrices and Determinants 2 Question 44

48. The total number of distincts $x \in R$ for which $| \begin{array}{ccc} x & x^{2} & 1 + x^{3} \\ 2 x & 4 x^{2} & 1 + 8 x^{3} \\ 3 x & 9 x^{2} & 1 + 27 x^{3} \end{array} | = 10$ is

(2016 Adv.)

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Answer:

Correct Answer: 48. (2)

Solution:

Given, $| \begin{array}{ccc} x & x^{2} & 1 + x^{3} \\ 2 x & 4 x^{2} & 1 + 8 x^{3} \\ 3 x & 9 x^{2} & 1 + 27 x^{3} \end{array} | = 10$

$\Rightarrow x \cdot x^{2} | \begin{array}{ccc} 1 & 1 & 1 + x^{3} \\ 2 & 4 & 1 + 8 x^{3} \\ 3 & 9 & 1 + 27 x^{3} \end{array} | = 10$

Apply $R_{2} \to R_{2} - 2 R_{1}$ and $R_{3} \to R_{3} - 3 R_{1}$ , we get

$\begin{aligned} x^{3} | \begin{array}{ccc} 1 & 1 & 1 + x^{3} \\ 0 & 2 & - 1 + 6 x^{3} \\ 0 & 6 & - 2 + 24 x^{3} \end{array} | = 10 \\ \Rightarrow x^{3} \cdot | \begin{array}{cc} 2 & 6 x^{3} - 1 \\ 6 & 24 x^{3} - 2 \end{array} | = 10 \end{aligned}$

$\begin{aligned} \Rightarrow & x^{3} (48 x^{3} - 4 - 36 x^{3} + 6) = 10 \\ \Rightarrow & 12 x^{6} + 2 x^{3} = 10 \\ \Rightarrow & 6 x^{6} + x^{3} - 5 = 0 \\ \Rightarrow & x^{3} = \frac{5}{6}, - 1 \\ x = (\frac{5}{6})^{1 / 3}, - 1 \end{aligned}$

Hence, the number of real solutions is 2 .

Matrices and Determinants 2 Question 44

48. The total number of distincts $x \in R$ for which $| \begin{array}{ccc} x & x^{2} & 1 + x^{3} \\ 2 x & 4 x^{2} & 1 + 8 x^{3} \\ 3 x & 9 x^{2} & 1 + 27 x^{3} \end{array} | = 10$ is

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Matrices and Determinants 2 Question 44

48. The total number of distincts x∈R for which |xx21+x32x4x21+8x33x9x21+27x3|=10 is

Answer:

Solution:

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48. The total number of distincts $x \in R$ for which $| \begin{array}{ccc} x & x^{2} & 1 + x^{3} \\ 2 x & 4 x^{2} & 1 + 8 x^{3} \\ 3 x & 9 x^{2} & 1 + 27 x^{3} \end{array} | = 10$ is