(a) Let px2+3x−4p+3x+4x2=y.
Then (4y−p)x2+3x(1−y)−(4+yp)=0
Now b2−4ac=9(1−y)2+4(4y+p)(4+yp)
=(16p+9)y2+2(2p2+23)y+(16p+9)≥0
for real values of x.
Arguing as in part (b), B2−4AC<0
and the sign is same as of A=16p+9 which is to be +ive
∴ 4(2p2+23)2−4(16p+9)2<0
and 16p+9>0
or 16(p+4)2(p2−8p+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(x)=ax2+bx+c. Then
g(x)=f(x)+f′(x)+F′′(x)
=ax2+bx+c+2ax+b+2a
=ax2+(b+2a)x+c+b+2a
Since f(x) is positive for all real values of x.
We must have b2−4ac<0 and a>0.......(1)
Now discriminant of g(x) is given by
B2−4AC=(b+2a)2−4a(c+b+2a)
=b2+4ab+4a2−4ac−4ab−8a2
=b2−4ab−4a2<0
[∵b2−4ac<0 by (1) and −4a2<0]
Hence sign of g(x) is the same as that of first term i.e. of a, which is positive