The number of real roots of the equation tan | vedclass

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(A) The given equation is $	an ^{-1} \sqrt{x^{2}+x}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$.For the equation to be defined,the arguments of the in

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$0$

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(A) The given equation is $	an ^{-1} \sqrt{x^{2}+x}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4}$.For the equation to be defined,the arguments of the inverse trigonometric functions must satisfy their domain conditions.$1$. For $	an ^{-1} \sqrt{x^{2}+x}$,we require $x^{2}+x \geq 0$.$2$. For $\sin ^{-1} \sqrt{x^{2}+x+1}$,we require $0 \leq \sqrt{x^{2}+x+1} \leq 1$,which implies $0 \leq x^{2}+x+1 \leq 1$.Since $x^{2}+x+1 = (x+\frac{1}{2})^{2} + \frac{3}{4}$,the minimum value of $x^{2}+x+1$ is $\frac{3}{4}$.Thus,the condition $x^{2}+x+1 \leq 1$ implies $x^{2}+x+1 \leq 1$,or $x^{2}+x \leq 0$.Combining $x^{2}+x \geq 0$ and $x^{2}+x \leq 0$,we must have $x^{2}+x = 0$.This gives $x(x+1) = 0$,so $x=0$ or $x=-1$.If $x=0$,the equation becomes $	an ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2} 
eq \frac{\pi}{4}$.If $x=-1$,the equation becomes $	an ^{-1} \sqrt{0} + \sin ^{-1} \sqrt{1} = 0 + \frac{\pi}{2} = \frac{\pi}{2} 
eq \frac{\pi}{4}$.Since neither value satisfies the equation,the number of real roots is $0$.

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

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