MCQ
The solution of $\frac{{{d^2}y}}{{d{x^2}}} = \cos x - \sin x$is
- ✓$y = - \cos x + \sin x + {c_1}x + {c_2}$
- B$y = - \cos x - \sin x + {c_1}x + {c_2}$
- C$y = \cos x - \sin x + {c_1}{x^2} + {c_2}x$
- D$y = \cos x + \sin x + {c_1}{x^2} + {c_2}x$
$\frac{{dy}}{{dx}} = \sin x + \cos x + {c_1}$
Again $y = - \cos x + \sin x + {c_1}x + {c_2}$.
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$(i)$ $f _1(x)=\sin \left(\sqrt{1- e ^{-x^2}}\right)$
$(ii)$ $f_2(x)=\left\{\begin{array}{ll}\frac{|\sin x|}{\tan ^{-1} x} & \text { if } x \neq 0 \\ 1 & \text { if } x=0\end{array}\right.$, where the inverse trigonometric function of $\tan ^{-1} x$
assume values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,
$(iii)$ $f_3(x)=\left[\sin \left(\log _c(x+2)\right)\right]$, where, for $t \in R ,[t]$ denotes the greatest integer less than or equal to $t$,
(iv) $f_4(x)=\left\{\begin{array}{ll}x^2 \sin \left(\frac{1}{x}\right) & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{array}\right.$
| $LIST I$ | $LIST II$ |
| $P$ The function $f _1$ is | $1$ $NOT$ continuous at $x=0$ |
| $Q$ The function $f _2$ is | $2$ continuous at $x =0$ and $NOT$ differentiable at $x =0$ |
| $R$ The function $f_3$ is | $3$ differentiable at $x=0$ and its derivative is $NO$T continuous at $x =0$ |
| $S$ The function $f _4$ is | $4$ differentiable at $x =0$ and its derivative is continuous at $x =0$ |
The correct option is: