MCQ
Write the function in the simplest form: $\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}, x \neq 0$
- A$\frac{1}{2} \sin ^{-1} x$
- B$ \tan ^{-1} x$
- C$\frac{1}{2} \cot ^{-1} x$
- ✓$\frac{1}{2} \tan ^{-1} x$
✓
Answer
Correct option: D.
$\frac{1}{2} \tan ^{-1} x$
d
$\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}$
$\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}$
Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$
$\therefore \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\tan ^{-1} \frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}$
$=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right)$
$\tan ^{-1}\left(\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)$
$=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Let $a =$$\mathop {Lim}\limits_{x \to 1} \,\,\frac{x}{{\ln \,x}}\; - \;\frac{1}{{x\,\ln \,x}}$ ; $b =$$\mathop {Lim}\limits_{x \to 0} \,\,\frac{{{x^3} - 16x}}{{4x + {x^2}}}$ ; $c =$$\mathop {Lim}\limits_{x \to 0} \,\,\frac{{\ln (1 + \sin x)}}{x}$ and $d =$$\mathop {Lim}\limits_{x \to - 1} \,\,\frac{{{{(x + 1)}^3}}}{{3\left( {\sin (x + 1) - (x + 1)} \right)}}$ , then the matrix $\left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]$ is
→For the function $f(x) = {e^x},a = 0,b = 1$, the value of $ c$ in mean value theorem will be
→If $3\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)-4\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)+2\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)=\frac{\pi}{3}$ is equal to:
→- $\frac{1}{\sqrt3}$
- $-\frac{1}{\sqrt3}$
- $\sqrt3$
- $-\frac{\sqrt3}{4}$
If $L.M.V.$ theorem is true for $f(x) = x(x-1)(x-2);\, x \in [0,\, 1/2]$ , then $C =$ ?
→If $\int \frac{\cos \theta}{5+7 \sin \theta-2 \cos ^{2} \theta} d \theta=A \log _{e}|B(\theta)|+C$ where $C$ is a constant of integration, then $\frac{ B (\theta)}{ A }$ can be
→$\int {\frac{{{{\sin }^3}2x}}{{{{\cos }^5}2x}}dx = } $
→The value of $k$ for which the set of equations $x + ky + 3z = 0,$ $3x + ky - 2z = 0,$ $2x + 3y - 4z = 0$ has a non trivial solution over the set of rationals is
→In figure, which of the following is not true?
→- $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec0$
-
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec0$
-
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec0$
-
$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec0$
Statement-$1$ : ${\cot ^{ - 1}}\left[ {\frac{{\log \left( {e/{x^2}} \right)}}{{\log \left( {e{x^2}} \right)}}} \right] + {\cot ^{ - 1}}\left[ {\frac{{\log (e{x^2})}}{{\log (e/{x^2})}}} \right]$ = $\frac {\pi}{2}$
→Statement-$2$ : ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right]$ = ${\tan ^{ - 1}}\,1 + \,{\tan ^{ - 1}}\left( {\log {x^2}} \right)$
Choose the correct answer from the given four options.
For the following probability distribution E(X) is equal to:
| X | -4 | -3 | -2 | -1 | 0 |
| P(X) | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |
- 0
- -1
- -2
- -1.8