MCQ
Solution of $\cos x\frac{{dy}}{{dx}} + y\sin x = 1$ is
- A$y\sec x\tan x = c$
- ✓$y\sec x= \tan x c$
- C$y\tan x = \sec x + c$
- D$y\tan x = \sec x\tan x + c$
$\therefore $ $I.F.$ $ = {e^{\int_{}^{} {\tan xdx} }} = {e^{\log \sec x}} = \sec x$
Hence solution is $y\sec x = \int_{}^{} {{{\sec }^2}x + c} = \tan x + c$.
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$\begin{gathered}
f\left( x \right) = \left[ \begin{gathered}
{\cos ^{ - 1}}\left( \mu \right) + {x^2},0 < x < 1 \hfill \\
4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x \geqslant 1 \hfill \\
\end{gathered} \right.,f\left( x \right) \hfill \\
\hfill \\ \end{gathered}$ can have a local minimum at $x =$ $1$, if the value of $\mu$ lies in the interval