MCQ
$\sin \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{4}{5}} \right) = $
- ✓$\frac{1}{{\sqrt {10} }}$
- B$ - \frac{1}{{\sqrt {10} }}$
- C$\frac{1}{{10}}$
- D$ - \frac{1}{{10}}$
Now $\sin \left( {\frac{1}{2}{{\cos }^{ - 1}}\frac{4}{5}} \right) = \sin \left( {\frac{x}{2}} \right)$ .....$(ii)$
From $(i),$ $\cos x = \frac{4}{5}$
==> $1 - 2{\sin ^2}\frac{x}{2} = \frac{4}{5}$
==> $2{\sin ^2}\frac{x}{2} = 1 - \frac{4}{5} = \frac{1}{5}$
$ \Rightarrow \sin \frac{x}{2} = \sqrt {\frac{1}{{10}}} $.
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$f(x)=\left\{\begin{array}{ccc}x^{5} \sin \left(\frac{1}{x}\right)+5 x^{2}& , & x<0 \\ 0 & , & x=0 \\ x^{5} \cos \left(\frac{1}{x}\right)+\lambda x^{2} & , & x>0\end{array} .\right.$
The value of $\lambda$ for which $f^{\prime \prime}(0)$ exists, is