Question

For all $x \in R$, if $m{x^2} - 9mx + (5m + 1) > 0$, then $m$ lies in the interval.
A) $\left( {0,\dfrac{4}{{61}}} \right)$
B) $\left[ {\dfrac{4}{{61}},\dfrac{4}{{61}}} \right)$
C) $\left( {\dfrac{{ - 61}}{4},0} \right]$
D) None of these

Answer 1

Since, the given quadratic equation is greater than $0$, therefore, the condition for $a{x^2} + bx + c > 0$ is $a > 0$ and $D < 0$, here $D$ is the discriminant of the quadratic equation and it can be evaluated by the formula ${b^2} - 4ac$.

Complete step-by-step answer:
We are given that for all $x \in R$, if $m{x^2} - 9mx + (5m + 1) > 0$.
We have to find the interval for $m$which satisfies the given condition.
First, we compare the given quadratic equation with the standard form of the quadratic equation $a{x^2} + bx + c$.
Therefore, $a = m,\,b = - 9m$and $c = 5m + 1$
Now, we know that the condition for $a{x^2} + bx + c > 0$ is $a > 0$ and $D < 0$, here $D$ is the discriminant of the quadratic equation and it can be evaluated by the formula ${b^2} - 4ac$.
It means $m > 0...(1)$
Now, we evaluate the discriminant which is ${b^2} - 4ac$.
$D = {( - 9m)^2} - 4m(5m + 1) \\\ \Rightarrow D = 81{m^2} - 20{m^2} - 4m \\\ \Rightarrow D = 61{m^2} - 4m \\\$
Now,
Therefore,
Solve the inequality and evaluate the interval for .

Since, and the above product is less than , therefore, to $D < 0$make the inequality true $61m - 4 < 0$
Solve the inequality.
$61m < 4 \\\ \Rightarrow m < \dfrac{4}{{61}}.....(2) \\\$
The interval for $m$ will be the combine solution of inequality $(1)$ and $(2)$
Since, the value of $m$is greater than $0$but less than $\dfrac{4}{{61}}$, therefore, the values of $m$ lies between $0$and $\dfrac{4}{{61}}$, both the values are not included because in our inequality there is not equality sign.
Therefore, the interval notation for the values of $m$will be:
$\left( {0,\dfrac{4}{{61}}} \right)$
Hence, option (A) is correct.

Note: When the inequality includes the symbol $\leqslant , \geqslant$then the interval notation for the values would be written in close bracket that is [;] and When the inequality includes the symbol <, > then the interval notation for the values would be written in open bracket that is (;).