MCQ
If ${\cot ^{ - 1}}x + {\tan ^{ - 1}}3 = \frac{\pi }{2}$, then $x =$
- A$1/3$
- B$1/4$
- ✓$3$
- D$4$
$ \Rightarrow \,\,{\cot ^{ - 1}}x + {\tan ^{ - 1}}3 = \frac{\pi }{2}\,\, $
$\Rightarrow \,\,{\tan ^{ - 1}}\frac{1}{x} + {\tan ^{ - 1}}3 = \frac{\pi }{2}$
$ \Rightarrow \,\,{\tan ^{ - 1}}\left( {\frac{{\frac{1}{x} + 3}}{{1 - \frac{1}{x}.3}}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{0}} \right)$
$ \Rightarrow \,\,\frac{{3x + 1}}{{x - 3}} = \frac{1}{0}\,\, \Rightarrow \,\,x = 3$
Aliter : As we know that, ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2},$
therefore for the given question, $ x$ should be $3.$
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$(I)$ The curve $y=f(x)$ intersects the $x$-axis exactly at one point
$(II)$ The curve $y=f(x)$ intersects the $x$-axis at $\mathrm{x}=\cos \frac{\pi}{12}$
Then
$f(x)=x^2+\frac{5}{12}$ and $g(x)=\left\{\begin{array}{cc}2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4}\end{array}\right.$
If $\alpha$ is the area of the region
$\left\{( x , y ) \in R \times R :| x | \leq \frac{3}{4}, 0 \leq y \leq \min \{f( x ), g( x )\}\right\},$
then the value of $9 \alpha$ is. . . . . .
Statement $-1$ : The probability that system of equations has a solution is $1$ .
Statement $-2$ : The probability that the system of equations has a unique solution is $\frac {3}{8}$ .