MHT CET · Maths · Inverse Trigonometric Functions
Considering only the principal values of an inverse function, the set
\(\mathrm{A}=\left\{x \geq 0 / \tan ^{-1} x+\tan ^{-1} 6 x=\frac{\pi}{4}\right\}\)
- 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
Consider, \(\tan ^{-1} x+\tan ^{-1} 6 x=\frac{\pi}{4}\)
\(\begin{aligned}
& \Rightarrow \tan ^{-1}\left(\frac{7 x}{1-6 x^2}\right)=\frac{\pi}{4} \\
& \quad \quad \ldots\left[\tan ^{-1}(x)+\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \\
& \Rightarrow \frac{7 x}{1-6 x^2}=1 \\
& \Rightarrow 7 x=1-6 x^2 \\
& \Rightarrow 6 x^2+7 x-1=0 \\
& \Rightarrow x=\frac{-7 \pm \sqrt{73}}{12} \\
& \quad \text { Since } x \geq 0 \\
& \therefore \quad x=\frac{-7+\sqrt{73}}{12}
\end{aligned}\)
\(\therefore \quad\) A is a singleton set.
\(\begin{aligned}
& \Rightarrow \tan ^{-1}\left(\frac{7 x}{1-6 x^2}\right)=\frac{\pi}{4} \\
& \quad \quad \ldots\left[\tan ^{-1}(x)+\tan ^{-1}(y)=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right] \\
& \Rightarrow \frac{7 x}{1-6 x^2}=1 \\
& \Rightarrow 7 x=1-6 x^2 \\
& \Rightarrow 6 x^2+7 x-1=0 \\
& \Rightarrow x=\frac{-7 \pm \sqrt{73}}{12} \\
& \quad \text { Since } x \geq 0 \\
& \therefore \quad x=\frac{-7+\sqrt{73}}{12}
\end{aligned}\)
\(\therefore \quad\) A is a singleton set.
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