MCQ
The solution set of the equation ${\sin ^{ - 1}}x = 2{\tan ^{ - 1}}x$ is
  • A
    $\{1, 2\}$
  • B
    $\{-1, 2\}$
  • $\{-1,1, 0\}$
  • D
    $\{1, \frac{1}{2} , 0\}$

Answer

Correct option: C.
$\{-1,1, 0\}$
c
(c) ${\sin ^{ - 1}}x = 2{\tan ^{ - 1}}x$

==> ${\sin ^{ - 1}}x = {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}}$

$ \Rightarrow \frac{{2x}}{{1 + {x^2}}} = x$ ==> ${x^3} - x = 0$

==> $x(x + 1)(x - 1) = 0$ ==> $x = \left\{ { - 1,\,\,1,\,\,0} \right\}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If a system of the equation ${(\alpha + 1)^3}x + {(\alpha + 2)^3}y - {(\alpha + 3)^3} = 0$ and $(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0,x + y - 1 = 0$ is constant. what is the value of $\alpha $
If $y \frac{d y}{d x}=x\left[\frac{y^{2}}{x^{2}}+\frac{\phi\left(\frac{y^{2}}{x^{2}}\right)}{\phi^{\prime}\left(\frac{y^{2}}{x^{2}}\right)}\right], x>0, \phi>0$, and $y(1)=-1$ then $\phi\left(\frac{\mathrm{y}^{2}}{4}\right)$ is equal to :
The solution of $\frac{{dy}}{{dx}} = \frac{{{e^x}({{\sin }^2}x + \sin 2x)}}{{y(2\log y + 1)}}$ is
For $3 \times 3$ matrices $M$ and $N$, which of the following statement$(s)$ is (are) $NOT$ correct?

$(A)$ $N ^{\top} M N$ is symmetric or skew symmetric, according as $M$ is symmetric or skew symmetric

$(B)$ $M N-N M$ is skew symmetric for all symmetric matrices $M$ and $N$

$(C)$ $M N$ is symetric for all symmetric matrices $M$ and $N$

$(D)$ $(\operatorname{adj} M)(\operatorname{adj} N)=\operatorname{adj}(M N)$ for all invertible matrices $M$ and $N$

$\int_{a}^{b}\text{x}^2\text{dx}=$
If the area of the region $\left\{( x , y ): x ^{\frac{2}{3}}+ y ^{\frac{2}{3}} \leq 1 x + y \geq 0, y \geq 0\right\}$ is $A$, then $\frac{256 A }{\pi}$
The area bounded by the curves $y = x (1 - \ln x) ; \,\,x = e^{-1}$ and positive $X-$ axis between $x = e^{-1}$ and $x = e$ is :
The coordinates of the midpoints of the line segment joining the points $(2, 3, 4)$ and $(8, -3, 8)$ are:
The solution of the differential equation ${(x + y)^2}\frac{{dy}}{{dx}} = {a^2}$ is
Area of the region bounded by the curve $\text{y}=\sqrt{49-\text{x}^2}$ and the $x -$ axis is: