MCQ
If ${\tan ^{ - 1}}2x + {\tan ^{ - 1}}3x = \frac{\pi }{4}$, then $x =$
- A$-1$
- ✓$\frac{1}{6}$
- C$ - 1,\,\frac{1}{6}$
- DNone of these
$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{2x + 3x}}{{1 - (2x)\,(3x)}}} \right) = \frac{\pi }{4}\,$
$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{5x}}{{1 - 6{x^2}}}} \right)\, = {\tan ^{ - 1}}(1)$
$ \Rightarrow \,\,\frac{{5x}}{{1 - 6{x^2}}} = 1\,$
$ \Rightarrow \,\,1 - 6{x^2} = 5x$
$\, \Rightarrow \,\,6{x^2} + 5x - 1 = 0$
$ \Rightarrow \,\,(x + 1)\,\left( {x - \frac{1}{6}} \right) = 0\,$
$ \Rightarrow \,\,x = - 1,\,\,\frac{1}{6}$
But $-1$ does not hold.
Trick : Check with the options.
Obviously the equation holds for $x = \frac{1}{6}$, but not for $-1$.
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$(A)$ a unique point in the interval $\left(n, n+\frac{1}{2}\right)$
$(B)$ a unique point in the interval $\left(n+\frac{1}{2}, n+1\right)$
$(C)$ a unique point in the interval $( n , n +1$ )
$(D)$ two points in the interval $( n , n +1)$