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(C) For the quadratic expression $f(x) = ax^2 + bx + c$ to be always positive for all $x \in R$,we must have $a > 0$ and the discriminant $D < 0$.Step

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$7$

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(C) For the quadratic expression $f(x) = ax^2 + bx + c$ to be always positive for all $x \in R$,we must have $a > 0$ and the discriminant $D Step $1$: Condition $a > 0$$1 + 2m > 0 \Rightarrow 2m > -1 \Rightarrow m > -\frac{1}{2}$.Step $2$: Condition $D $D = [-2(1 + 3m)]^2 - 4(1 + 2m)(4(1 + m)) $4(1 + 6m + 9m^2) - 16(1 + 3m + 2m^2) Divide by $4$: $(1 + 6m + 9m^2) - 4(1 + 3m + 2m^2) $1 + 6m + 9m^2 - 4 - 12m - 8m^2 $m^2 - 6m - 3 Step $3$: Solve the inequality $m^2 - 6m - 3 The roots of $m^2 - 6m - 3 = 0$ are $m = \frac{6 \pm \sqrt{36 - 4(1)(-3)}}{2} = \frac{6 \pm \sqrt{48}}{2} = 3 \pm 2\sqrt{3}$.Since $\sqrt{3} \approx 1.732$,$2\sqrt{3} \approx 3.464$.So,$3 - 3.464 Step $4$: Intersection with $m > -0.5$The interval is $(-0.464, 6.464)$.The integral values of $m$ are ${0, 1, 2, 3, 4, 5, 6}$.There are $7$ such values.

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is:

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