Question

If $ac>b^2$ then the sum of the coefficients in the expansion of $(a{\alpha }^2{x}^2+2b\alpha x+c{)}^n,(a,b,c,\alpha \in R,n\in N)$ is

• A. Positive if $a>0$.
• B. Positive if $c>0$.
• C. Negative if $a<0,n$ is odd.
• D. Positive if $c<0,n$ is even.

Answer 1

Answer: (A), (C)

The correct options are
A Positive if $a>0$.
C Negative if $a<0,n$ is odd.

If $ac>b^2$, then $ac-b^2>0$ or $b^2-ac<0$

Now, ${(a{\alpha }^2{x}^2+2b\alpha x+c)}^n={(a{\left(\alpha x\right)}^2+2b\left(\alpha x\right)+c)}^n$

Here, $\Delta =4b^2-4ac=4(b^2-ac)<0$

Hence, it has no roots.

the point where slope is 0 in $a{x}^2+bx+c$ is

$y=\frac{-(b^2-4ac)}{4a}$

so for $a>0$, y=+ve as ($\Delta <0$)

$a<0$, y=-ve

hence sum of coefficient is +ve for $a>0$

So, correct option is $A$ and $C$.