IIT JEE , AIEEE Forum - Mathematics , Algebra

ayush_2008

if x^-x-k<0 for at least one real value of x then k lies in the interval -----------

srujana

If x^2-x-k<0 for at least one real value of x then k lies in the interval ----

Let f(x) =x^2-x

The graph of f(x) is something like

$\Rightarrow$ f(x) attains minimum value at $x=\frac{1}{2}$ and $f \left(\frac{1}{2}\right)=-\frac{1}{4}$

For the function to attain -ve value for at least one value of x, the point of minimum, $\left(\frac{1}{2},-\frac{1}{4}\right)$ can at the most be shifted by $\frac{1}{4}$ units above X axis

$\Rightarrow$ $-k<\frac{1}{4}$

$\Rightarrow$ $k>- \frac{1}{4}$

Hence $k \in \left(-\frac{1}{4},\infty \right)$

Decoder

it is x^-x ... :D:D ... now solve :):)

rudra.panda

well the subject is quadratic :P

srujana

nice q decoder :D

now our $f(x)=x^{-x}$

we see that the max occurs at $x=\frac{1}{e}$

and as $x\rightarrow \infty...f(x)\rightarrow 0$

now even if the graph is pulled down by infinitesimally small value...the fn attains -ve value fr atleast one value of x

the ans should be $k\le \lim_{x\to\infty}x^{-x}$

Ask & Discuss Questions with Community & Experts