Question

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$, where $a,b,c,d$ are real and $3b^2<c^2$, is an increasing function and $g\left(x\right)=a{f}^{\prime }\left(x\right)+b{f}^{\prime \prime }\left(x\right)+c^2$. lf $G\left(x\right)={\int }_{\alpha }^{x}g\left(t\right)dt,\alpha \in R$, then for $\alpha <x<\alpha +1$,

• A. G(x) is a decreasing function
• B. G(x) is an increasing function
• C. G(x) is neither increasing nor decreasing
• D. G(x) is a one-one function

Answer 1

Answer: (B), (D)

The correct options are
B G(x) is a one-one function
D G(x) is an increasing function

$f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$f^{\prime }\left(x\right)=3a{x}^2+2bx+c$
Since, $f\left(x\right)$ is increasing.

$\Rightarrow f^{\prime }\left(x\right)=3a{x}^2+2bx+c>0$
$\Rightarrow a>0$ and $4b^2-12ac<0$
$\Rightarrow a>0$ and $b^2<3ac$
Also,

$g\left(x\right)=a{f}^{\prime }\left(x\right)+b{f}^{\prime \prime }\left(x\right)+c^2$
$\Rightarrow g\left(x\right)=a(3a{x}^2+2bx+c)+b(6ax+2b)+c^2$
$\Rightarrow g\left(x\right)=3a^2{x}^2+8abx+2b^2+c^2+ac$

$D=64a^2{b}^2-4\left(3a^2\right)(2b^2+c^2+ac)$
$\Rightarrow D=4a^2(10b^2-3c^2-3ac)<4a^2(10b^2-3c^2-b^2)$
$\Rightarrow D<4a^2(9b^2-3c^2)$
$\Rightarrow D<12a^2(3b^2-c^2)$

Since $\Rightarrow 3b^2-c^2<0$,
$\Rightarrow D<0$
$\Rightarrow g\left(x\right)>0$ for all $x\in R$.

$G\left(x\right)={\int }_{\alpha }^{x}g\left(t\right)dt$ is a strictly increasing function.
Hence,

$G\left(x\right)$ is one-one in the given interval.