Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

inverse trigonometric functions

Multiple Choice Questions

191.
If $2 \sinh^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}) = \log (\frac{1 + x}{1 - x})$ , then x is equal to

a

$\frac{1}{a}$

$\sqrt{1 - a^{2}}$

$\frac{1}{\sqrt{1 - a^{2}}}$

$Given that, 2 \sinh^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}) = \log (\frac{1 + x}{1 - x}) \Rightarrow 2 \sinh^{- 1} (x) = 2 \log (x + \sqrt{1 + x^{2}}) [∵ \sinh^{- 1} (x) = \log (x + \sqrt{1 + x^{2}})] = \log {(x + \sqrt{1 + x^{2}})}^{2} = \log (x^{2} + 1 + x^{2} + 2 x \sqrt{1 + x^{2}}) 2 \sinh^{- 1} (x) = \log (2 x^{2} + 1 + 2 x \sqrt{1 + x^{2}}) = \log [\frac{2 a}{1 - a^{2}} + 1 + 2 \times \frac{a}{\sqrt{1 - a^{2}}} \times \sqrt{1 + \frac{a^{2}}{1 - a^{2}}}] [∵ x = \frac{a}{\sqrt{1 - a^{2}}}] = \log [\frac{2 a}{1 - a^{2}} + 1 + \frac{2 a}{\sqrt{1 - a^{2}}} \times \frac{1}{\sqrt{1 - a^{2}}}] = \log [\frac{2 a}{1 - a^{2}} + 1 + \frac{2 a}{1 - a^{2}}] = \log [\frac{2 a^{2} + 1 - a^{2} + 2 a}{1 - a^{2}}] = \log [\frac{a^{2} + 2 a + 1}{1 - a^{2}}]$

$Put the value of 2 \sinh^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}), we get \log [\frac{a^{2} + 2 a + 1}{1 - a^{2}}] = \log [\frac{1 + x}{1 - x}] \Rightarrow \frac{a^{2} + 2 a + 1}{1 - a^{2}} = \frac{1 + x}{1 - x} \Rightarrow \frac{a^{2} + 2 a + 1 + 1 - a^{2}}{a^{2} + 2 a + 1 - 1 + a^{2}} = \frac{1 + x + 1 - x}{1 - x - 1 + x} \Rightarrow \frac{2 + 2 a}{2 a^{2} + 2 a} = \frac{1}{x} [∵ by componendo and divdendo rule] \Rightarrow \frac{2 (1 + a)}{2 a (1 + a)} = \frac{1}{x} \Rightarrow \frac{1}{a} = \frac{1}{x} \Rightarrow x = a$

192.

$If x = \sin (2 \tan^{- 1} (2)) and y = \sin (\frac{1}{2} \tan^{- 1} (\frac{4}{3})), then$

$x > y$
x = y
x = 0 = y
$x < y$

193.

$If \cosh (x) = \frac{5}{4}, then \cosh (3 x) = ?$

$\frac{61}{16}$
$\frac{63}{16}$
$\frac{65}{16}$
$\frac{61}{63}$

194.

$In a ∆ ABC, if < A = 90 °, then \cos^{- 1} (\frac{R}{r_{2} + r_{3}}) = ?$

$90 °$
$30 °$
$60 °$
$45 °$

195.

$If \cot (\cos^{- 1} (x)) = \sec \{\tan^{- 1} (\frac{a}{\sqrt{b^{2} - a^{2}}})\} : b > a, then x =$

$\frac{b}{\sqrt{2 b^{2} - a^{2}}}$
$\frac{\sqrt{b^{2} - a^{2}}}{ab}$
$\frac{a}{\sqrt{2 b^{2} - a^{2}}}$
$\frac{\sqrt{b^{2} - a^{2}}}{a}$

196.

$2 π - [\sin^{- 1} (\frac{4}{5}) + \sin^{- 1} (\frac{5}{13}) + \sin^{- 1} (\frac{16}{65})] = ?$

$\frac{π}{2}$
$π$
$\frac{3 π}{2}$
$- \frac{π}{2}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

191. If 2sinh-1a1 - a2 = log1 + x1 - x, then x is equal to a 1a 1 - a2 11 - a2

191.
If $2 \sinh^{- 1} (\frac{a}{\sqrt{1 - a^{2}}}) = \log (\frac{1 + x}{1 - x})$ , then x is equal to

a

$\frac{1}{a}$

$\sqrt{1 - a^{2}}$

$\frac{1}{\sqrt{1 - a^{2}}}$