The number of real roots of the equation $${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$ is :
Solution
$${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$<br><br>For equation to be defined,<br><br>x<sup>2</sup> + x $\ge$ 0<br><br>$\Rightarrow$ x<sup>2</sup> + x + 1 $\ge$ 1<br><br>$\therefore$ Only possibility that the equation is defined <br><br>x<sup>2</sup> + x = 0 $\Rightarrow$ x = 0; x = $-$1<br><br>None of these values satisfy<br><br>$\therefore$ No of roots = 0
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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