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The number of real roots of the equation \tan ^{-1} \sqrt{\mathrm{x}(\mathrm{x}+1)}+\sin ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}+1}=\frac{\pi}{4} is :
 
Option: 1 1
Option: 2 2
Option: 3 4  
Option: 4 0

Answers (1)

best_answer

\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4} .

Domain

\begin{aligned} &1, \quad x(x+1) \geqslant 0 \\ &\Rightarrow \quad x \in(-\infty,-1] \cup[0, \infty) \\ &2 \cdot \quad x^{2}+x+1 \geqslant 0 \Rightarrow x \in R \end{aligned}

\begin{aligned} \\&\text { 3. } \sqrt{x^{2}+x+1} \leq 1 \\ \Rightarrow & x^{2}+x+1 \leq 1 \\ \Rightarrow & x^{2}+x \leq 0 \\ \Rightarrow & x(x+1) \leq 0 \\ \Rightarrow & x \in[-1,0] \end{aligned}

Overall domain is intersection of these 3

\Rightarrow \text { domain }=\{-1,0\} \text {. }

 

For x= -1

Equation is \tan ^{-1}(0)+\sin ^{-1}(1)=\frac{\pi}{4} \Rightarrow \frac{\pi}{2} = \frac{\pi}{4}

So x= -1 do not satisfy it

 

For x= 0

Equation is \tan ^{-1}(0)+\sin ^{-1}(1)=\frac{\pi}{4} \Rightarrow \frac{\pi}{2} = \frac{\pi}{4}

x= 0 also does not satisfy it

So, no solution

Hence, the correct answer is option (4)

Posted by

Kuldeep Maurya

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