MCQ
If $y = {x^{{x^{x......\infty }}}}$, then ${{dy} \over {dx}} = $
- A${{{y^2}} \over {x(1 + y\log x)}}$
- ✓${{{y^2}} \over {x(1 - y\log x)}}$
- C${y \over {x(1 + y\log x)}}$
- D${y \over {x(1 - y\log x)}}$
Therefore, on differentiating $\frac{{dy}}{{dx}} = \frac{{{y^2}}}{{x(1 - y\log x)}}$.
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Statement $-2$ : $f(x) = \frac{1}{{\sqrt {1 - {x^2}} }} + \left[ {\frac{{{x^2} + x + 1}}{4}} \right]$ , where $[.]$ is greatest integer function. Function $f(x)$ is even function