MCQ
If $\int_{}^{} {{e^x}\sin x\;dx = \frac{1}{2}{e^x}\;.\;a + c} $, then $a = $
- ✓$\sin x - \cos x$
- B$\cos x - \sin x$
- C$ - \cos x - \sin x$
- D$\cos x + \sin x$
Let $I = \int_{}^{} {{e^x}\sin x\,dx} = - {e^x}\cos x + \int_{}^{} {{e^x}\cos x\,dx + c} $
$ = - {e^x}\cos x + {e^x}\sin x - \int_{}^{} {{e^x}\sin x\,dx + c} $
$ \Rightarrow 2I = {e^x}( - \cos x + \sin x) + c$. Now from $(i),$
we get $\frac{1}{2}{e^x}a = \frac{1}{2}{e^x}(\sin x - \cos x) \Rightarrow a = \sin x - \cos x.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is