MCQ
$\int_{\,0}^{\,\pi } {{e^{{{\sin }^2}x}}{{\cos }^3}x\,dx} $ is equals to
- A$ - 1$
- ✓$0$
- C$1$
- D$\pi $
$ \Rightarrow I = \int_0^\pi {{e^{{{\sin }^2}(\pi - x)}}{{\cos }^3}(\pi - x)\,} dx$ ..$(i)$
$ \Rightarrow I = - \int_0^\pi {{e^{{{\sin }^2}x}}{{\cos }^3}x\,dx} $ ..$(ii)$
Adding $(i)$ and $(ii),$ we get
$2I = 0$ ==> $I = 0$.
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Then which of the following options is/are correct?
$(1)$ For $x =1$, there exists a unit vector $\alpha \hat{ i }+\beta \hat{ j }+\gamma \hat{ k }$ for which $R \left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
$(2)$ There exists a real number $x$ such that $P Q=Q P$
$(3)$ $\operatorname{det} R=\operatorname{det}\left[\begin{array}{lll}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{array}\right]+8$, for all $x \in R$
$(4)$ For $x=0$, if $R\left[\begin{array}{l}1 \\ a \\ b\end{array}\right]=6\left[\begin{array}{l}1 \\ a \\ b\end{array}\right]$, then $a+b=5$