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