Q197 / 270 · NDA 2017 (Phase 1) · Mathematics
Let $\mathrm{x}, \mathrm{y}, \mathrm{z}$ be positive real numbers such that $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are in GP and $\tan ^{-1} x, \tan ^{-1} y$ and $\tan ^{-1} z$ are in AP. Then which one of the following is correct?
- Correct answerA$\mathrm{x}=\mathrm{y}=\mathrm{z}$
- B$\mathrm{xz}=1$
- C$x \neq y$ and $y=z$
- D$x=y$ and $y \neq z$
Solution
$\begin{aligned}
& 2 \tan ^{-1} y=\tan ^{-1} x+\tan ^{-1} z \\
& \left(\therefore \tan ^{-1}, \tan ^{-1} y, \tan ^{-1} z \text { are in A.P }\right) \\
& \text { and } x, y, z \text { are in G.P } \quad \therefore \quad y^2=x z...(i)
\end{aligned}$
$\begin{aligned}
& \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-x z} \\
& \Rightarrow \frac{2 y}{1-y^2}=\frac{x+z}{1-y^2} \\
& \Rightarrow 2 y=x+z \\
& x, y, z \text { are in A.P. }
\end{aligned}$
(from (i))
Given $x, y, z$ are also in G.P
So, $x=y=z$
Hence, the correct answer is option: 'A'.