MCQ
The solution of $\frac{{dy}}{{dx}} = {e^x}(\sin x + \cos x)$ is
- A$y = {e^x}(\sin x - \cos x) + c$
- B$y = {e^x}(\cos x - \sin x) + c$
- ✓$y = {e^x}\sin x + c$
- D$y = {e^x}\cos x + c$
==> $dy = {e^x}(\sin x + \cos x)dx$
On integrating, we get $y = {e^x}\sin x + c$.
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$f_1(x)=\left\{\begin{array}{lll}|x| & \text { if } & x<0, \\ e^x & \text { if } & x \geq 0 ;\end{array}\right.$
$f_2(x)=x^2$
$f_3(x)=\left\{\begin{array}{ccc}\sin x & \text { if } & x < 0, \\ x & \text { if } & x \geq 0\end{array}\right.$ and
$f_4(x)=\left\{\begin{array}{ccc}f_2\left(f_1(x)\right) & \text { if } & x < 0, \\ f_2\left(f_1(x)\right)-1 & \text { if } & x \geq 0\end{array}\right.$
| List $I$ | List $II$ |
| $P.$ $ f_4$ is | $1.$ onto but not one-one |
| $Q.$ $f_3$ is | $2.$ neither continuous nor one-one |
| $R.$ $f _2 \circ f _1$ is | $3.$ differentiable but not one-one |
| $S.$ $ f_2$ is | $4.$ continuous and one-one |
Codes: $ \quad P \quad Q \quad R \quad S $