Question:
Let \(x, y, z\) be positive real numbers such that \(x, y, z\) are in GP and \(\tan^{-1} x, \tan^{-1} y, \tan^{-1} z\) are in AP. Then which one of the following is correct?
मान लीजिए \(x, y, z\) धनात्मक वास्तविक संख्याएँ हैं जैसे कि \(x, y, z\) गुणोत्तर श्रेणी (GP) में हैं और \(\tan^{-1} x, \tan^{-1} y, \tan^{-1} z\) समांतर श्रेणी (AP) में हैं। तो निम्नलिखित में से कौन सा सही है?
Options:
- \(x = y = z\)
- \(xz = 1\)
- \(x \ne y\) and \(y = z\)
- \(x = y\) and \(y \ne z\)
- \(x = y = z\)
- \(xz = 1\)
- \(x \ne y\) तथा \(y = z\)
- \(x = y\) तथा \(y \ne z\)
Correct Answer: a)
Explanation:
Since \(x, y, z\) are in GP, \(y^2 = xz\).
Since \(\tan^{-1} x, \tan^{-1} y, \tan^{-1} z\) are in AP:
\[2\tan^{-1} y = \tan^{-1} x + \tan^{-1} z \implies \tan^{-1}\left(\frac{2y}{1 – y^2}\right) = \tan^{-1}\left(\frac{x + z}{1 – xz}\right)\]
\[\frac{2y}{1 – y^2} = \frac{x + z}{1 – xz}\]
Since \(xz = y^2\), the denominators are identical, which gives:
\[2y = x + z\]
Thus, \(x, y, z\) are also in AP.
Three positive numbers that are simultaneously in AP and GP must be equal: \(x = y = z\).
Hence, the correct option is (a).
चूँकि \(x, y, z\) GP में हैं: \(y^2 = xz\)।
\(\tan^{-1}\) पदों के AP में होने से:
\[\frac{2y}{1 – y^2} = \frac{x + z}{1 – xz}\]
चूँकि \(xz = y^2\), अतः \(2y = x + z\), जिसका अर्थ है कि \(x, y, z\) AP में भी हैं।
यदि संख्याएँ AP और GP दोनों में हों, तो वे समान होती हैं: \(x = y = z\)।
अतः सही विकल्प (a) है।