If the expression left( {mx - 1 + frac{1}{x}} | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) We know that $a{x^2} + bx + c \ge 0$ if $a > 0$

and ${b^2} - 4ac \le 0$.

? $mx - 1 + \frac{1}{x} = 0\,\,\, \Rightarrow \,\,\frac{{m{x^2}

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(c) We know that $a{x^2} + bx + c \ge 0$ if $a > 0$

and ${b^2} - 4ac \le 0$.

? $mx - 1 + \frac{1}{x} = 0\,\,\, \Rightarrow \,\,\frac{{m{x^2} - x + 1}}{x} \ge 0$

==> $m{x^2} - x + 1 \ge 0$ and $x > 0$

Now $m{x^2} - x + 1 \ge 0$ if $m > 0$ and $1 - 4m \le 0$

or if $m > 0$ and $m \ge \frac{1}{4}$

Thus the minimum value of m is $\frac{1}{4}$.

If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be

Similar Questions

The number of integers $k$ for which the equation $x^3-27 x+k=0$ has at least two distinct integer roots is

If $a \in R$ and the equation $ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$ (where $[x]$ denotes the greatest integer $\leq\,x$)has no integral solution ,then all possible values of $a$ lie in the interval

If $a, b, c$ are real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1$, then $(3 a+5 b-8 c)^2+(-8 a+3 b+5 c)^2$ $+(5 a-8 b+3 c)^2$ is equal to

The number of real solutions of the equation $e ^{4 x }+4 e ^{3 x }-58 e ^{2 x }+4 e ^{ x }+1=0$ is..........

Sum of the solutions of the equation $\left[ {{x^2}} \right] - 2x + 1 = 0$ is (where $[.]$ denotes greatest integer function)