MCQ
Solution of the differential equation $\sin \frac{{dy}}{{dx}} = a$ with $y(0) = 1$ is
- A${\sin ^{ - 1}}\frac{{(y - 1)}}{x} = a$
- ✓$\sin \frac{{(y - 1)}}{x} = a$
- C$\sin \frac{{(1 - y)}}{{(1 + x)}} = a$
- D$\sin \frac{y}{{(x + 1)}} = a$
Integrating both sides,$\int_{}^{} {dy} = \int_{}^{} {{{\sin }^{ - 1}}a\,dx} $
$y = x{\sin ^{ - 1}}a + c$ and $y(0) = 0 + c = 1$, $\therefore c = 1$
$\therefore y = x{\sin ^{ - 1}}a + 1$ ==> $a = \sin \frac{{y - 1}}{x}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$
$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$
$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$
has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
Statement $-2$ : The equation in $\alpha $
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
$f(x)=\frac{2 x}{\sqrt{1+9 x^2}}$. If the composition of $f, \underbrace{(f \circ f \circ f \circ \ldots \circ f)}_{10 \text { times }}(x)=\frac{2^{10} x}{\sqrt{1+9 \alpha x^2}}$, then the value of $\sqrt{3 \alpha+1}$ is equal to....................
| $X$ | $0$ | $2$ | $4$ | $6$ | $8$ |
| $P(X)$ | $a$ | $2a$ | $a+b$ | $2b$ | $3b$ |
is $ \frac{46}{9}$ , then the variance of the distribution is