Q46 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
Let $a, b, c, d \in \mathbf{R}$ be such that $a d-b c \neq 0$ and $e$ be a positive number other than 1 . If $x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|$ and $\Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$, then the values of $x$ and $y$ are respectively.
- Correct answerA$e^{\frac{\Delta_1}{\Delta_3}}, e^{\frac{\Delta_2}{\Delta_3}}$
- B$e^{\frac{\Delta_3}{\Delta_2}}, e^{\frac{\Delta_1}{\Delta_2}}$
- C$e^{\frac{-\Delta_1}{\Delta_3}}, e^{\frac{-\Delta_2}{\Delta_3}}$
- D$e^{\frac{\Delta_2}{\Delta_1}}, e^{\frac{\Delta_3}{\Delta_1}}$
Solution
We have,
$\begin{aligned}
\Delta_1 & =\left|\begin{array}{ll}
m & b \\
n & d
\end{array}\right|=m d-b n \\
\Delta_2 & =\left|\begin{array}{ll}
a & m \\
c & n
\end{array}\right|=a n-m c \\
\Delta_3 & =\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|=a d-b c \\
x^a y^b & =e^m, x^c y^d=e^n \\
y & =\left(\frac{e^m}{x^a}\right)^{1 / b} \text { and } y=\left(\frac{e^n}{x^c}\right)^{1 / d}
\end{aligned}$
$\begin{array}{ll}
\therefore \quad & \frac{e^{m / b}}{x^{a / b}}=\frac{e^{n / d}}{x^{c / d}} \Rightarrow x^{c / d-a / b}=e^{n / d-m / b} \\
\Rightarrow \quad x & =\left(e^{n / d-m / b}\right)^{\frac{1}{c}-\frac{a}{b}} \Rightarrow x=e^{\left(\frac{b n-m d}{b d} \times \frac{b d}{b c-a d}\right)} \\
\Rightarrow & x=e^{\frac{b n-m d}{b c-a d}} \Rightarrow x=e^{\frac{m d-b n}{a d-b c}}=e^{\frac{\Delta_1}{\Delta_3}}
\end{array}$
Putting the value of $x$ in $y=\frac{e^{m / b}}{x^{a / b}}$ we get
$y=e^{\frac{a n-m c}{a d-b c}}=e^{\frac{\Delta_2}{\Delta_3}}$
Hence, the correct answer is option: 'A'.