MCQ
Solution of the differential equation $\frac{{dy}}{{dx}}\tan y = \sin (x + y) + \sin (x - y)$ is
- ✓$\sec y + 2\cos x = c$
- B$\sec y - 2\cos x = c$
- C$\cos y - 2\sin x = c$
- D$\tan y - 2\sec y = c$
$\frac{{dy}}{{dx}}(\tan y) = 2\sin x\cos y$ ==> $\frac{{\sin y}}{{{{\cos }^2}y}}dy = 2\sin xdx$
==> $\int {\frac{{\sin y}}{{{{\cos }^2}y}}} dy = 2\int {\sin xdx} $ ==> $\frac{1}{{\cos y}} = - 2\cos x + c$
$\therefore$ $\sec y + 2\cos x = c$.
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Statement $1:$ $\mathop {\lim }\limits_{x \to {2^ - }} \,f(x)$ exists.
Statement $2:$ $f$ is continuous at $x = 2.$
Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
| $x=-1$ | $x=0$ | $x=2$ | |
| $f(x)$ | $3$ | $6$ | $0$ |
| $g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$