Theory Of Equations Ques 71
- Let $f(x)$ be a quadratic expression which is positive for all real values of $x$. If $g(x)=f(x)+f^{\prime}(x)+f^{\prime \prime}(x)$, then for any real $x$
$(1990,2 M)$
(a) $g(x)<0$
(b) $g(x)>0$
(c) $g(x)=0$
(d) $g(x) \geq 0$
Analytical & Descriptive Questions
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Answer:
Correct Answer: 71.(b)
Solution:
- Let $f(x)=a x^{2}+b x+c>0, \forall x \in R$
$ \begin{aligned} \Rightarrow & a & >0 \\ \text { and } & b^{2}-4 a c & <0 ……(i) \end{aligned} $
$ \begin{array}{ll} \therefore & g(x)=f(x)+f^{\prime}(x)+f^{\prime \prime}(x) \\ \Rightarrow & g(x)=a x^{2}+b x+c+2 a x+b+2 a \\ \Rightarrow & g(x)=a x^{2}+x(b+2 a)+(c+b+2 a) \end{array} $
whose discriminant
$ \begin{aligned} & =(b+2 a)^{2}-4 a(c+b+2 a) \\ & =b^{2}+4 a^{2}+4 a b-4 a c-4 a b-8 a^{2} \end{aligned} $
$ =b^{2}-4 a^{2}-4 a c=\left(b^{2}-4 a c\right)-4 a^{2}<0 \quad \text { [from Eq. (i)] } $
$\therefore g(x)>0 \forall x$, as $a>0$ and discriminant $<0$.
Thus, $g(x)>0, \forall x \in R$.
BETA
