Question

The number of integral values of m for which the quadratic expression:
$\left( 1+2m \right){{x}^{2}}-2\left( 1+3m \right)x+4\left( 1+m \right):x\in R$ is always positive is
A. 8
B. 7
C. 6
D. 3

Answer 1

To solve this question, we will use the fact that, for any quadratic equation of the form $a{{x}^{2}}+bx+c$ to be positive we have $a>0$ and D discriminant which is ${{b}^{2}}-4ac<0$. We will compare this two values in our given equation and then try to find the value of m in some range. Also, we will use that if any equation $a{{x}^{2}}+bx+c$ has its root as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
A quadratic equation of the form $a{{x}^{2}}+bx+c$ is always positive if $a>0$ and its discriminant $D<0$
We have discriminant D of $a{{x}^{2}}+bx+c$ the form $D={{b}^{2}}-4ac$
So, a quadratic expression of the form $a{{x}^{2}}+bx+c$ is always positive if $a>0\text{ and }{{b}^{2}}-4ac<0$

Complete step-by-step solution:
We are given our quadratic expression as:
$\left( 1+2m \right){{x}^{2}}-2\left( 1+3m \right)x+4\left( 1+m \right)$
Comparing it from $a{{x}^{2}}+bx+c$ we have

& a=\left( 1+2m \right) \\\ & b=-2\left( 1+3m \right) \\\ & c=4\left( 1+m \right) \\\ \end{aligned}$$ From above theory we have the quadratic expression $$\left( 1+2m \right){{x}^{2}}-2\left( 1+3m \right)x+4\left( 1+m \right)$$ is always positive if $a>0$ is $$\left( 1+2m \right)>0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$$ and discriminant $$D={{b}^{2}}-4ac<0$$ that is $${{\left( -2\left( 1+3m \right) \right)}^{2}}-4\times \left( 1+2m \right)4\left( 1+m \right)<0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}$$ So, we have now to solve equation (i) and (ii) to get out integral values of m. From equation (i) we have $$1+2m>0\Rightarrow 2m>-1$$ Dividing by 2 both sides we get: $$m>\dfrac{-1}{2}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}$$ Consider equation (ii) now, we have $$\begin{aligned} & {{\left( -2\left( 1+3m \right) \right)}^{2}}-4\times \left( 1+2m \right)4\left( 1+m \right)<0 \\\ & \Rightarrow 4{{\left( 1+3m \right)}^{2}}-16\left( 1+2m \right)\left( 1+m \right)<0 \\\ \end{aligned}$$ Opening square using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ we get: $$\Rightarrow 4\left( 1+9{{m}^{2}}+6m \right)-16\left( 1+m \right)\left( 1+2m \right)<0$$ Multiplying the brackets and solving further we get: $$\begin{aligned} & 4+36{{m}^{2}}+24m-16\left( 1+2m+m+2{{m}^{2}} \right)<0 \\\ & \Rightarrow 4+36{{m}^{2}}+24m-16-48m-32{{m}^{2}}<0 \\\ \end{aligned}$$ Taking terms having ${{m}^{2}},m$ one side we have: $$\begin{aligned} & \left( 36-32 \right){{m}^{2}}+\left( 24-48 \right)m+4-16<0 \\\ & \Rightarrow 4{{m}^{2}}+\left( -24m \right)-12<0 \\\ \end{aligned}$$ Taking 4 common we get: $$4\left( {{m}^{2}}-6m-3 \right)<0$$ Now as $$4\ne 0\text{ and }4>0\Rightarrow {{m}^{2}}-6m-3<0$$ We have roots of quadratic equation $a{{x}^{2}}+bx+c$ as $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ Comparing this for above equation ${{m}^{2}}-6m-3$ we get roots as: $$\begin{aligned} & m=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 1\times(-3)}}{2\times 1} \\\ & \Rightarrow m=\dfrac{6\pm \sqrt{36+12}}{2} \\\ & \Rightarrow m=\dfrac{6\pm \sqrt{48}}{2} \\\ \end{aligned}$$ Taking LCM of 48 as: $$\begin{aligned} & \begin{matrix} 2 \\\ 2 \\\ 2 \\\ \end{matrix}\begin{matrix} \left| \underline {48} \right. \\\ \left| \underline {24} \right. \\\ \left| \underline {12} \right. \\\ \end{matrix} \\\ & \begin{matrix} 2 \\\ 2 \\\ \end{matrix}\begin{matrix} \left|\underline 6 \right. \\\ \left|\underline 3 \right. \\\ \end{matrix} \\\ & 1 \\\ \end{aligned}$$ We have, $$\begin{aligned} & 48=2\times 2\times 2\times 2\times 3 \\\ & \Rightarrow \sqrt{48}=2\times 2\sqrt{3}=4\sqrt{3} \\\ \end{aligned}$$ So, we have $$\sqrt{48}=4\sqrt{3}$$ So, we have the value of $m=\dfrac{6\pm 4\sqrt{3}}{2}$ Taking 2 common from numerator and denominator we get: $$\begin{aligned} & m=\dfrac{\left( 3\pm 2\sqrt{3} \right)}{{}} \\\ & m=3\pm 2\sqrt{3} \\\ \end{aligned}$$ As, we had ${{m}^{2}}-6m-3<0$ then this can be written as $$\left( m-\left( 3+2\sqrt{3} \right) \right)\left( m-\left( 3-2\sqrt{3} \right) \right)<0$$ Product of two terms is less than 0 then it implies that; If $$\left( x-a \right)\left( x-b \right)<0\Rightarrow a\text{ }<\text{ }x\text{ }<\text{ }b$$ So, using this in above we have: $$\begin{aligned} & \left( m-\left( 3+2\sqrt{3} \right) \right)\left( m-\left( 3-2\sqrt{3} \right) \right)\text{ }<\text{ }0 \\\ & \Rightarrow 3-2\sqrt{3}\text{ }<\text{ }m\text{ }<\text{ }3+2\sqrt{3} \\\ \end{aligned}$$ As here we had $3-2\sqrt{3}\text{ }<\text{ }m\text{ }<\text{ }3+2\sqrt{3}$ Hence, finally we have m as $$\Rightarrow 3-2\sqrt{3}\text{ }<\text{ }m\text{ }<\text{ }3+2\sqrt{3}$$ Opening $3-2\sqrt{3}$ and using $\sqrt{3}=1.732$ we get: $$\begin{aligned} & 3-2\times 1.732=3-3.464=-0.464 \\\ & \text{and }3+2\times 1.732=3+3.464=6.464 \\\ \end{aligned}$$ Then, value of m lies as: $$-0.464\text{ }<\text{ }m\text{ }<\text{ }6.464$$ All integral values of m lying in their range is $$m=0,1,2,3,4,5,6$$ So, we have 7 integral values of m possible such that, $$\left( 1+2m \right){{x}^{2}}-2\left( 1+3m \right)x+4\left( 1+m \right)\text{ is positive}$$ Also, from equation (iii) we had $$m\text{ }>\text{ }\dfrac{-1}{2}\Rightarrow m\text{ }>\text{ }-0.5$$ So, anyway above values are valid. **Therefore, there are 7 integral values of m possible. Option B is correct.** **Note:** The student can make mistake at the point where $$\left( m-\left( 3+2\sqrt{3} \right) \right)\left( m-\left( 3-2\sqrt{3} \right) \right)<0$$ is considered. This step does not imply that both $$\left( m-\left( 3+2\sqrt{3} \right) \right)\text{ and }\left( m-\left( 3-2\sqrt{3} \right) \right)$$ are negative. Because if product of two numbers is negative then there is a possibility of one of them to be positive. Therefore, we will not use both $$\left( m-\left( 3+2\sqrt{3} \right) \right)\left( m-\left( 3-2\sqrt{3} \right) \right)<0$$ to be negative. There is one more key point that, if the roots of quadratic equation are $\alpha \text{ and }\beta $ then the quadratic equation $a{{x}^{2}}+bx+c$ can be written as $\left( x-\alpha \right)\left( x-\beta \right)$