Find the number of real solutions - tan-1-x x-1 -sin-1-x2-x-1- -2

The correct option is D 2 Our requirement is x^ 2 +x≥ 0 Second requirement √( x ^ 2 +x+1 )≤ 1 ⇒ x ^ 2 +x≤ 0 Hence x=0,x= 1 ...

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Solving Simultaneous Trigonometric Equations

Find the numb...

Question

Find the number of real solutions :
${tan}^{- 1} \sqrt{x (x + 1)} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}$

1

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t a n^{- 1} (\sqrt{x (x + 1)} + s i n^{- 1} \sqrt{(x^{2} + x + 1)} = \frac{π}{2}

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

The number of real solutions of the equation $t a n^{- 1} \sqrt{x (x + 1)} + s i n^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}$ is

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

s i n^{- 1} [x - \frac{x^{2}}{2} + \frac{x^{3}}{4} . . . .] = π / 2 - c o s^{- 1} [x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} . . . .]

0 < | x | < \sqrt{2}

t a n^{- 1} \sqrt{x (x + 1)} + s i n^{- 1} \sqrt{x^{2} + x + 1} = π / 2

- \sqrt{2} < x < 0

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