The smallest value of the constant m 0 for | vedclass

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

Question

(C) Given the inequality $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.Multiplying by $x$ (since $x > 0$),we get $9mx^2 - x + 1 \geq 0$.For a

Vedclass · Accepted Answer

$\frac{1}{36}$

Vedclass · Answer

(C) Given the inequality $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$.Multiplying by $x$ (since $x > 0$),we get $9mx^2 - x + 1 \geq 0$.For a quadratic expression $ax^2 + bx + c \geq 0$ to hold for all $x > 0$,we analyze the discriminant $D = b^2 - 4ac$.Here,$a = 9m$,$b = -1$,and $c = 1$.The discriminant $D = (-1)^2 - 4(9m)(1) = 1 - 36m$.For the quadratic to be non-negative for all $x$,we require $D \leq 0$.$1 - 36m \leq 0 \implies 36m \geq 1 \implies m \geq \frac{1}{36}$.Thus,the smallest value of $m$ is $\frac{1}{36}$.

The smallest value of the constant $m > 0$ for which $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$ is:

Similar Questions

The roots of the equation $({a^2} + {b^2}){t^2} - 2(ac + bd)t + ({c^2} + {d^2}) = 0$ are equal. Then:

The number of the real roots of the equation $(x+1)^{2}+|x-5|=\frac{27}{4}$ is ....... .

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3x^2-10x-24=0$,and $\alpha(\beta+\gamma), \beta(\gamma+\alpha), \gamma(\alpha+\beta)$ are the roots of the equation $x^3+px^2+qx+r=0$,then $q=$

The value of $k$ for which the quadratic equation $kx^2 + 1 = kx + 3x - 11x^2$ has real and equal roots is:

Let $\theta$,$0 < \theta < \pi / 2$,be an angle such that the equation $x^2 + 4x \cos \theta + \cot \theta = 0$ has equal roots for $x$. Then $\theta$ in radians is

Vedclass Test Series

Exam Paper Generator

Online Exam Module