Maths

    \[\text { The binomial distribution whose mean is } 9 \text { and the variance is } 2.25 \text { is }\]

$$ \begin{array}{l} \text { Mean }=n p \\ \text { Variance }=n p(1-p) \end{array} $$ Therefore according to question $$ \begin{array}{l} 1-\mathrm{p}=\mathrm{q}=\frac{\text { Variance }}{\text {...

read more

    \[\text { If }\left(\tan ^{-1} x\right)^{2}+\left(\cot ^{-1} x\right)^{2}=\frac{5 \pi^{2}}{8}, \text { then } x \text { equals to }\]

$$ \begin{array}{l} \text { We have, }\left(\tan ^{-1} x\right)^{2}+\left(\cot ^{-1} x\right)^{2}=\frac{5 \pi^{2}}{8} \\ \Rightarrow\left(\tan ^{-1} x+\cot ^{-1} x\right)^{2}-2 \tan ^{-1} x \cdot...

read more

Evaluate

$$ \begin{array}{l} \int \frac{2 x^{3}-1}{x^{4}+x} d x \\ \Rightarrow \int \frac{\left(4 x^{3}+1\right)-\left(2 x^{3}+2\right)}{x^{4}+x} d x \\ \Rightarrow \int \frac{4 x^{3}+1}{x^{4}+x} d x-2 \int...

read more

    \[\text { Prove that: } 2 \tan ^{-1}\left(\frac{1}{5}\right)+\sec ^{-1}\left(\frac{5 \sqrt{2}}{7}\right)+2 \tan ^{-1}\left(\frac{1}{8}\right)=\frac{\pi}{4}\]

$$ \begin{array}{l} 2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8} \\ =2\left[\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right]+\sec ^{-1} \frac{5 \sqrt{2}}{7}...

read more

Solve \int_{0}^{x / 4} \tan ^{2} x d x

$\tan ^{2} \mathrm{x}=\sec ^{2} \mathrm{x}-1$ $\Rightarrow \int_{0}^{\pi / 4}-1 \mathrm{dx}+\int_{0}^{\pi / 4}\left(\sec ^{2} \mathrm{x}\right) \mathrm{dx}$ $=-\pi / 4+[\tan \mathrm{x}\}_{0}^{\pi /...

read more

Find the general solution of the differential equation
*** QuickLaTeX cannot compile formula:
\begin{array}{l} \frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0

*** Error message:
\begin{array} on input line 8 ended by \end{document}.
leading text: \end{document}
Improper \prevdepth.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Missing \cr inserted.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
You can't use `\end' in internal vertical mode.
leading text: \end{document}
\begin{array} on input line 8 ended by \end{document}.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Emergency stop.

$$ \begin{array}{l} \frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0 \\ \frac{d y}{d x}=\frac{-\sqrt{1-y^{2}}}{1+x^{2}} \\ \Rightarrow \int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{-d...

read more

\int \operatorname{cosec} x d x=?

$$ \begin{array}{l} \mathrm{I}=\int \csc x \mathrm{dx} \\ \Rightarrow \mathrm{I}=\int \frac{\csc x(\csc x-\cot x)}{(\csc x-\cot x)} \mathrm{dx} \\ \text { Put } \csc x-\cot x=t \\...

read more