The number of real values of x satisfying the equation tan-1x1-x2-tan-11x3-3-4- is-

The correct option is B 1 tan ^ 1x^4+1 x^2x^3 x^5 x = 3π4 x^4+1 x^2x^3 x^5 x = 1 x^4 + 1 x^2 + x^3 x^5 x = 0 Real roots = 1 ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

The number of...

Question

The number of real values of x satisfying the equation ${tan}^{- 1} (\frac{x}{1 - x^{2}}) + {tan}^{- 1} (\frac{1}{x^{3}}) = \frac{3 π}{4}$ , is?

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Solution

The correct option is B $1$
${tan}^{- 1} \frac{x^{4} + 1 - x^{2}}{x^{3} - x^{5} - x} = \frac{3 π}{4}$
$\frac{x^{4} + 1 - x^{2}}{x^{3} - x^{5} - x} = - 1$
$x^{4} + 1 - x^{2} + x^{3} - x^{5} - x = 0$
Real roots = $1$

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The number of real values of x satisfying the equation tan−1(x1−x2)+tan−1(1x3)=3π4, is?

The correct option is B 1tan−1x4+1−x2x3−x5−x=3π4x4+1−x2x3−x5−x=−1x4+1−x2+x3−x5−x=0Real roots = 1

The number of real values of x satisfying the equation ${tan}^{- 1} (\frac{x}{1 - x^{2}}) + {tan}^{- 1} (\frac{1}{x^{3}}) = \frac{3 π}{4}$ , is?

The correct option is B $1$
${tan}^{- 1} \frac{x^{4} + 1 - x^{2}}{x^{3} - x^{5} - x} = \frac{3 π}{4}$
$\frac{x^{4} + 1 - x^{2}}{x^{3} - x^{5} - x} = - 1$
$x^{4} + 1 - x^{2} + x^{3} - x^{5} - x = 0$
Real roots = $1$