Theory Of Equations Ques 49
- 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: 49.(c)
Solution:
- 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 ……(i) $
and $D=4(1+3 m)^{2}-4(2 m+1) 4(1+m)<0 ……(ii) $
From inequality Eq. (i), we get
$ m>-\frac{1}{2} ……(iii) $
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} ……(iv) $
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 .
BETA
