MCQ
${\cot ^{ - 1}}3 + {\rm{cose}}{{\rm{c}}^{ - 1}}\sqrt 5 =$
- A$\frac{\pi }{3}$
- ✓$\frac{\pi }{4}$
- C$\frac{\pi }{6}$
- D$\frac{\pi }{2}$
$ = {\cot ^{ - 1}}\left( {\frac{{3 \times 2 - 1}}{{3 + 2}}} \right) = {\cot ^{ - 1}}(1) = \frac{\pi }{4}$.
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$f(x)=\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x$
$f(x)=\left\{\begin{array}{ll} \frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0 \end{array}\right.$
is continuous at $x=0,$ where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $X$.
Then :
$\left|\begin{array}{lll}(1+\alpha)^2 & (1+2 \alpha)^2 & (1+3 \alpha)^2 \\ (2+\alpha)^2 & (2+2 \alpha)^2 & (2+3 \alpha)^2 \\ (3+\alpha)^2 & (3+2 \alpha)^2 & (3+3 \alpha)^2\end{array}\right|=-648 \alpha$ ?
$(A)$ $-4$ $(B)$ $9$ $(C)$ $-9$ $(D)$ $4$