JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 10)
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 :
1
2
4
0
Explanation
$${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$
For equation to be defined,
x2 + x $$\ge$$ 0
$$\Rightarrow$$ x2 + x + 1 $$\ge$$ 1
$$\therefore$$ Only possibility that the equation is defined
x2 + x = 0 $$\Rightarrow$$ x = 0; x = $$-$$1
None of these values satisfy
$$\therefore$$ No of roots = 0
For equation to be defined,
x2 + x $$\ge$$ 0
$$\Rightarrow$$ x2 + x + 1 $$\ge$$ 1
$$\therefore$$ Only possibility that the equation is defined
x2 + x = 0 $$\Rightarrow$$ x = 0; x = $$-$$1
None of these values satisfy
$$\therefore$$ No of roots = 0
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