Eng Mathematics

Question

Let $f(x)=ax^2+bx+c;$ a, b, c, $\in$ R and $a \neq 0$. Suppose $f(x)>0$ for all $x \in R$. Let g(x)=f(x)+f ’(x)+f ’’(x)$. Then

• A. g(x)>0 for all $x \in R$
• B. g(x)<0 for all $x \in R$
• C. g(x)=0 has real roots
• D. None of the above

Answer 1

since f(x)>0 ==> b^2-4*a*c<0

also g(x)=ax^2+(2a+b)x+(2a+b+c) ==> its D=(2a+b)^2 - 4*a*(2a+b+c)

Now one can easily prove that D<0 even for g(x) (Note that you have to use b^2-4ac<0 to prove it)

Hence no real roots for g(x) ie g(x)>0 for all x.

 

Proof for D<0:--

as  D=(2a+b)^2 - 4*a*(2a+b+c)

==> D=(4a^2+4ab+b^2)-8a^2-4ab-4ac

==> D=-4a^2+b^2-4ac

==> since b^2-4ac<0 for all x  and -4a^2 is -ve for all a

==> D<0 for all x

Answer 2

$f(x)=ax^2+bx+c$

$f'(x)=2ax+b$

$f''(x)=2a$

$g(x)=2a+2ax+b+ax^2+bx+c$

Now, $a,b,c,x\epsilon R$

Range of $R =-\alpha to +\alpha$

a) Not possible if $a=3,x=-2,b=-21 and c=1$

b) Not possible if $a,b,c,x=+ve$

c) Not possible always

d) Correct Answer here

Comments

Brother, If possible please explain D<0 for g(x) ...

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as  D=(2a+b)^2 - 4*a*(2a+b+c)

==> D=(4a^2+4ab+b^2)-8a^2-4ab-4ac

==> D=-4a^2+b^2-4ac

==> since b^2-4ac<0 for all x  and -4a^2 is -ve for all a

==> D<0 for all x

 

I hope now your doubt is clear......

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Thanks brother, I was missing one term in g(x).