MCQ
Given function $f(x) = \left( {{{{e^{2x}} - 1} \over {{e^{2x}} + 1}}} \right)$ is
- ✓Increasing
- BDecreasing
- CEven
- DNone of these
==>$f( - x) = \frac{{{e^{ - 2x}} - 1}}{{{e^{ - 2x}} + 1}} = \frac{{1 - {e^{2x}}}}{{1 + {e^{2x}}}}$==>$f(x) = - \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} = - f(x)$
==> $f(x)$ is an odd function.
Again $f(x) = \frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}} \Rightarrow f'(x) = \frac{{4{e^{2x}}}}{{{{(1 + {e^{2x}})}^2}}} > 0\,\forall \,n \in R$
==> $f(x)$ is an increasing function.
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${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$ , ${{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a$
, ${{c}_{2}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{1}}}{|{{b}_{1}}{{|}^{2}}}{{b}_{1}}$
,${{c}_{3}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.{{b}_{2}}}{|{{b}_{2}}{{|}^{2}}}{{b}_{2}}$,
${{c}_{4}}=a-\frac{c.a}{|a{{|}^{2}}}a$.
Then which of the following is a set of mutually orthogonal vectors is
If $y(\pi)=\pi+2$, then the value of $y\left(\frac{\pi}{2}\right)$ is: