Question

(a)show that the expression $\frac{p{x}^2+3x-4}{p+3x-4x^2}$ will be capable of all values when x is real, provided that p has any value between $1$ and $7$.
(b) Suppose $f\left(x\right)$ is such a quadratic expression that it is positive for all real x.
If $g\left(x\right)=f\left(x\right)+f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)$, then for any real x prove that $g\left(x\right)>0$.

Answer 1

(a) Let $\frac{p{x}^2+3x-4}{p+3x+4x^2}=y$.
Then $(4y-p)x^2+3x(1-y)-(4+yp)=0$
Now $b^2-4ac=9{(1-y)}^2+4(4y+p)(4+yp)$
$=(16p+9)y^2+2(2p^2+23)y+(16p+9)\ge 0$
for real values of x.
Arguing as in part (b), $B^2-4AC<0$
and the sign is same as of $A=16p+9$ which is to be $+ive$
$\therefore$ $4{(2p^2+23)}^2-4{(16p+9)}^2<0$
and $16p+9>0$
or $16{(p+4)}^2(p^2-8p+7)<0$
and $16p+9>0$
or ${(p+4)}^2(p-1)(p-7)<0$
and $16p+9>0$
Both these inequalities are satisfied it $1<p<7$.
(b) Let $f\left(x\right)=a{x}^2+bx+c$. Then
$g\left(x\right)=f\left(x\right)+f^{\prime }\left(x\right)+F^{\prime \prime }\left(x\right)$
$=a{x}^2+bx+c+2ax+b+2a$
$=a{x}^2+(b+2a)x+c+b+2a$
Since $f\left(x\right)$ is positive for all real values of x.
We must have $b^2-4ac<0$ and $a>0$.......(1)
Now discriminant of $g\left(x\right)$ is given by
$B^2-4AC={(b+2a)}^2-4a(c+b+2a)$
$=b^2+4ab+4a^2-4ac-4ab-8a^2$
$=b^2-4ab-4a^2<0$
[$\because b^2-4ac<0$ by (1) and $-4a^2<0$]
Hence sign of $g\left(x\right)$ is the same as that of first term i.e. of a, which is positive