Q117 / 180 · KCET 2015 · Mathematics
If \( \mathrm{P}=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right| \) and \( \mathrm{Q}=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right| \),then \( \frac{d Q}{d x}= \)
- A\( 3 P+1 \)
- B\( 1-3 P \)
- C\( -3 P \)
- Correct answerD\( 3 \mathrm{P} \)
Solution
Given that
\( P=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|, Q=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right| \)
So, \( P=x^{2}-1 \)
And \( Q=x\left(x^{2}-1\right)-(x-1)+(1-x) \)
\( =x^{3}-3 x+2 \)
Now, \( \frac{d Q}{d x}=3 x^{2}-3 \)
\( =3\left(x^{2}-1\right)=3 P \)
\( P=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|, Q=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right| \)
So, \( P=x^{2}-1 \)
And \( Q=x\left(x^{2}-1\right)-(x-1)+(1-x) \)
\( =x^{3}-3 x+2 \)
Now, \( \frac{d Q}{d x}=3 x^{2}-3 \)
\( =3\left(x^{2}-1\right)=3 P \)
Hence, the correct answer is option: 'D'.