Theory of Equations 1 Question 5

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

(a) 6

(b) 8

(c) 7

(d) 3

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Answer:

Correct Answer: 6. (c)

Solution:

  1. The quadratic expression

$a x^{2}+b x+c, x \in R$ is always positive,

if $a>0$ and $D<0$.

So, the quadratic expression

$(1+2 m) x^{2}-2(1+3 m) x+4(1+m), x \in R$ will be

always positive, if $1+2 m>0$

and $D=4(1+3 m)^{2}-4(2 m+1) 4(1+m)<0$

From inequality Eq. (i), we get

$ m>-\frac{1}{2} $

From inequality Eq. (ii), we get

$ \begin{aligned} & 1+9 m^{2}+6 m-4\left(2 m^{2}+3 m+1\right)<0 \\ \Rightarrow \quad & m^{2}-6 m-3<0 \\ \Rightarrow \quad & {[m-(3+\sqrt{12})][m-(3-\sqrt{12})]<0 } \\ & \quad\left[\because m^{2}-6 m-3=0 \Rightarrow m=\frac{6 \pm \sqrt{36+12}}{2}=3 \pm \sqrt{12}\right] \end{aligned} $

$ \Rightarrow \quad 3-\sqrt{12}<m<3+\sqrt{12} $

From inequalities Eqs. (iii) and (iv), the integral values of $m$ are $0,1,2,3,4,5,6$

Hence, the number of integral values of $m$ is 7 .



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