The number of real roots of the equation tan-1-xx-1-sin-1-x2-x-1-4 is

Tan^ 1√(x(x+1))+sin^ 1√(x^2+x+1)=π4 Finding domain x(x + 1) ≥ nbsp;0 …(1) and nbsp;0≤ x^2 + x + 1 ≤ 1 ⇒ 1≤ x^2 + x ≤ 0 …(2) From (1) and (2), we have ...

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Byju's Answer

Standard XII

Mathematics

De-Moivre's Theorem

The number of...

Question

The number of real roots of the equation ${tan}^{- 1} \sqrt{x (x + 1)} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{4}$ is

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Solution

The correct option is D $0$
${tan}^{- 1} \sqrt{x (x + 1)} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{4}$
Finding domain
$x (x + 1) \geq 0 \dots (1)$
and $0 \leq x^{2} + x + 1 \leq 1$
$\Rightarrow - 1 \leq x^{2} + x \leq 0 \dots (2)$
from $(1)$ and $(2)$ we have $x^{2} + x = 0 \Rightarrow x = 0, - 1$
When, $x = 0$ or $- 1$
$L H S = \frac{π}{2} \neq \frac{π}{4} \Rightarrow$ No solution

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The number of real roots of the equation tan−1√x(x+1)+sin−1√x2+x+1=π4 is

The correct option is D 0tan−1√x(x+1)+sin−1√x2+x+1=π4 Finding domain x(x+1)≥0 …(1) and 0≤x2+x+1≤1 ⇒−1≤x2+x≤0 …(2) from (1) and (2) we have x2+x=0⇒x=0,–1 When, x=0 or –1 LHS=π2≠π4⇒ No solution

The number of real roots of the equation ${tan}^{- 1} \sqrt{x (x + 1)} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{4}$ is