Inverse Circular Functions Ques 24
- Considering only the principal values of inverse functions, the set $A=x \geq 0: \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$
(2019 Main, 12 Jan I)
(a) is an empty set
(b) is a singleton
(c) contains more than two elements
(d) contains two elements
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Answer:
Correct Answer: 24.(b)
Solution:
Formula:
Identities of Addition and Substraction:
- Given equation is
$ \begin{aligned} & \tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}, x \geq 0 \\ & \Rightarrow \quad \tan ^{-1} \frac{5 x}{1-6 x^{2}}=\frac{\pi}{4}, 6 x^{2}<1 \\ & \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} (\frac{x+y}{1-x y}), x y<1\right] \\ & \Rightarrow \quad \frac{5 x}{1-6 x^{2}}=1, x^{2}<\frac{1}{6} \\ & \Rightarrow \quad 6 x^{2}+5 x-1=0,0 \leq x<\frac{1}{\sqrt{6}} \quad[\because x \geq 0] \\ & \Rightarrow 6 x^{2}+6 x-x-1=0, \quad 0 \leq x<\frac{1}{\sqrt{6}} \\ & \Rightarrow 6 x(x+1)-1(x+1)=0, \end{aligned} $
$ \begin{array}{lll} \Rightarrow & (6 x-1)(x+1)=0, & 0 \leq x<\frac{1}{\sqrt{6}} \\ \Rightarrow & x=\frac{1}{6},-1, & 0 \leq x<\frac{1}{\sqrt{6}} \\ \Rightarrow & x=\frac{1}{6}, & {\left[\because 0 \leq x<\frac{1}{\sqrt{6}}\right]} \end{array} $
So ’ $A$ ’ is a singleton set.
$\therefore$ The solution of given differential equation represents a circle with centre on the $X$-axis.
BETA
