KVPY Sample Paper KVPY Stream-SX Model Paper-15

If \[a,\text{ }b,\text{ }c,\text{ }d~\in R\] such that \[\frac{a+2c}{b+3d}+\frac{4}{3}=0,\] then the equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] has-

A) At least one root in \[(-\,1,\,\,0)\]

B) At least one root in (0, 1)

C) No root in \[(-\,1,\,\,1)\]

D) No root in (0, 2)

Correct Answer: B

Solution :

\[\frac{a+2c}{b+3d}+\frac{4}{3}=0\]

\[\Leftrightarrow 3a+4b+6c+12d=0\]

\[\Leftrightarrow \frac{1}{4}a+\frac{b}{3}+\frac{c}{2}+d=0\]

Consider \[f(x)=\frac{a{{x}^{4}}}{4}+\frac{b{{x}^{3}}}{3}+\frac{c{{x}^{2}}}{2}+dx\]

Then, \[f(0)=0=f(1)\]

\[\therefore f(x)\] satisfies the conditions of Rolle?s theorem in [0,1].

Hence, f?(x) = 0 has at least one solution in (0,1).