MHT CET · Maths · Inverse Trigonometric Functions
Considering only the Principal values of inverse functions, the set
\(A=\left\{x \geq 0 \left\lvert\, \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}\right.\right\}\)
- A contains two elements.
- B contains more than two elements.
- C is an empty set.
- D is a singleton set.
Answer & Solution
Correct Answer
(D) is a singleton set.
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4} \\ \therefore \quad & \tan ^{-1}\left(\frac{2 x+3 x}{1-(2 x)(3 x)}\right)=\frac{\pi}{4}\end{aligned}\)
\(\begin{array}{ll}
\therefore & \frac{5 x}{1-6 x^2}=\tan \left(\frac{\pi}{4}\right) \\
\therefore & 5 x=1-6 x^2 \\
\therefore & 6 x^2+5 x-1=0 \\
\therefore & 6 x^2+6 x-x-1=0 \\
\therefore & (6 x-1)(x+1)=0 \\
\therefore & x=\frac{1}{6} \ldots[\because x \geq 0]
\end{array}\)
\(\therefore \quad\) set A is a singleton set.
\(\begin{array}{ll}
\therefore & \frac{5 x}{1-6 x^2}=\tan \left(\frac{\pi}{4}\right) \\
\therefore & 5 x=1-6 x^2 \\
\therefore & 6 x^2+5 x-1=0 \\
\therefore & 6 x^2+6 x-x-1=0 \\
\therefore & (6 x-1)(x+1)=0 \\
\therefore & x=\frac{1}{6} \ldots[\because x \geq 0]
\end{array}\)
\(\therefore \quad\) set A is a singleton set.
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