Q61 / 180 · KCET 2011 · Mathematics
If $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$, then the value of $x^{b+c} \cdot y^{c+a} \cdot z^{a+b}$ is
- Correct answerA1
- B2
- C0
- D$-1$
Solution
Given, $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (say)
$\ldots(i)$ $\Rightarrow \quad x=e^{k(b-c)}, y=e^{k(c-a)}, z=e^{k(a-b)}$
Now, $\quad x^{b+c} \cdot y^{c+a} \cdot z^{a+b}$ $=e^{k(b-c)(b+c)} \cdot e^{k(c+a)(c-a)} \cdot e^{k(a+b)(a-b)}$
$$
\begin{aligned}
&=e^{k\left(b^{2}-c^{2}\right)} \cdot e^{k\left(c^{2}-a^{2}\right)} \cdot e^{k\left(a^{2}-b^{2}\right)} \\
&=e^{k\left(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2}\right)}=e^{k \cdot 0}=1
\end{aligned}
$$
Hence, the correct answer is option: 'A'.