Matrices And Determinants Ques 68
The total number of distincts $x \in R$ for which $\left|\begin{array}{ccc}x & x^{2} & 1+x^{3} \\ 2 x & 4 x^{2} & 1+8 x^{3} \\ 3 x & 9 x^{2} & 1+27 x^{3}\end{array}\right|=10$ is
(2016 Adv.)
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Answer:
Correct Answer: 68.(2)
Solution:
Formula:
- Given, $\left|\begin{array}{ccc}x & x^{2} & 1+x^{3} \\ 2 x & 4 x^{2} & 1+8 x^{3} \\ 3 x & 9 x^{2} & 1+27 x^{3}\end{array}\right|=10$
$ \Rightarrow \quad x \cdot x^{2}\left|\begin{array}{ccc} 1 & 1 & 1+x^{3} \\ 2 & 4 & 1+8 x^{3} \\ 3 & 9 & 1+27 x^{3} \end{array}\right|=10 $
Apply $R_{2} \rightarrow R_{2}-2 R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1}$, we get
$ \begin{aligned} & x^{3}\left|\begin{array}{ccc} 1 & 1 & 1+x^{3} \\ 0 & 2 & -1+6 x^{3} \\ 0 & 6 & -2+24 x^{3} \end{array}\right|=10 \\ & \Rightarrow \quad x^{3} \cdot\left|\begin{array}{cc} 2 & 6 x^{3}-1 \\ 6 & 24 x^{3}-2 \end{array}\right|=10 \end{aligned} $
$ \begin{array}{rlrl} \Rightarrow & x^{3}\left(48 x^{3}-4-36 x^{3}+6\right) =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{array} $
Hence, the number of real solutions is $2$ .
BETA
