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Inverse Trigonometric Functions

Multiple Choice Questions

131.

$2 \cos^{- 1} (x) = \sin^{- 1} (2 x \sqrt{1 - x^{2}})$ is valid for all values of x satisfying

$- 1 \leq x \leq 1$
$0 \leq x \leq 1$
$\frac{1}{\sqrt{2}} \leq x \leq 1$
$0 \leq x \leq \frac{1}{\sqrt{2}}$

132.

If ${msin}^{- 1} (x) = \log_{e} (y), then (1 - x^{2}) y'' - xy'$ is equal to

m²y
- m²y
2y
- 2y

133.

If $\frac{3 x + 1}{(x - 1) (x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 3}$ , $\sin^{- 1} (\frac{A}{B})$ is equal to

$\frac{π}{2}$
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{4}$

134.

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

one
four
two
infinitely many

135.

If $\frac{{(x + 1)}^{2}}{x^{3} + x} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1}$ , then $\sin^{- 1} (A) + \tan^{- 1} (B) + \sec^{- 1} (C)$ is equal to

$\frac{π}{2}$
$\frac{π}{6}$
0
$\frac{5 π}{6}$

136.

$\cos (2 \cos^{- 1} (\frac{1}{5}) + \sin^{- 1} (\frac{1}{5}))$ is equal to

$\frac{1}{5}$
$\frac{- 2 \sqrt{6}}{5}$
$- \frac{1}{5}$
$\frac{\sqrt{3}}{2}$

137.

The value of $\tan^{- 1} (\frac{x}{y}) - \tan^{- 1} (\frac{x - y}{x + y})$ (where, x, y > 0) is

$\frac{π}{4}$
$- \frac{π}{4}$
$\frac{π}{2}$
$- \frac{π}{2}$

138.
If y = (tan^-1(x)), then (x² + 1)²y₂ + 2x(x² + 1)y₁ is equal to

4

0

2

1

$Given, y = {(\tan^{- 1} (x))}^{2} On differentlating both sides w . r . t x, we get \frac{dy}{dx} = 2 (\tan^{- 1} (x)) \times \frac{d}{dx} (\tan^{- 1} (x)) \Rightarrow \frac{dy}{dx} = \frac{2 \tan^{- 1} (x)}{1 + x^{2}} \Rightarrow (1 + x^{2}) \frac{dy}{dx} = 2 \tan^{- 1} (x) Again, dlfferenating both sides w . r . t . x, we get (1 + x^{2}) \frac{d}{dx} (\frac{dy}{dx}) + \frac{dy}{dx} \frac{d}{dx} (1 + x^{2}) = 2 \frac{d}{dx} (\tan^{- 1} (x)) \Rightarrow (1 + x^{2}) \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} (0 + 2 x) = \frac{2}{1 + x^{2}} \Rightarrow (1 + x^{2}) \frac{d^{2} y}{{dx}^{2}} + 2 x (1 + x^{2}) \frac{dy}{dx} = 2 or {(1 + x^{2})}^{2} y_{2} + 2 x (1 + x^{2}) y_{1} = 2$

139.

Given $0 \leq x \leq \frac{1}{2}$ , then the value of $\tan [\sin^{- 1} \{\frac{x}{\sqrt{2}} + \frac{\sqrt{1 - x^{2}}}{\sqrt{2}}\} - \sin^{- 1} (x)]$ is