Complex Numbers And Quadratic Equations question 28
Question: If $ a,b,c $ are real numbers such that $ a+b+c=0, $ then the quadratic equation $ 3ax^{2}+2bx+c=0 $ has [MNR 1992; DCE 1999]
Options:
A) At least one root in [0, 1]
B) At least one root in [1, 2]
C) At least one root in $ [-1,0] $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ {f}’(x) $ denotes the quadratic expression $ f’(x)\equiv 3ax^{2}+2bx+c $ , whose antiderivative be denoted by $ f(x)=ax^{3}+bx^{2}+cx $ Now $ f(x) $ being a polynomial in $ R,f(x) $ is continuous and differentiable on R. To apply Rolle’s theorem. We observe that $ f(0)=0 $ and $ f(1)=a+b+c=0, $ by hypothesis. So there must exist at least one value of x, say $ x=\alpha \in (0,1) $ such that $ {f}’(\alpha )=0\Leftrightarrow 3a{{\alpha }^{2}}+2b\alpha +c=0 $ That is, $ {f}’(x)=3ax^{2}+2bx+c=0 $ has at least one root in [0, 1].
BETA
