Question

Let $f\left(x\right)=x^2+b_{1}x+c_1,g\left(x\right)=x^2+b_{2}x+c_2$. Let the real roots of $f\left(x\right)=0$ be $\alpha ,\beta$ and real roots of $g\left(x\right)=0$ be $\alpha +h,\beta +h$. The least value of $f\left(x\right)$ is $-\frac{1}{4}$. The least value of $g\left(x\right)$ occurs at $x=-\frac{7}{2}$.

The least value of $g\left(x\right)$ is

• A. $-\frac{1}{4}$
• B. $-1$
• C. $-\frac{1}{3}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A)

$-\frac{1}{4}$

minimum value is nothing but $-\Delta /4a$
so for $f\left(x\right)$ ${\Delta }_1=4c_1-b_1^2$
so for $g\left(x\right)$ ${\Delta }_2=4c_2-b_2^2$
$(\alpha +h)(\beta +h)=c_2$
$\alpha \beta =c_1$
$\alpha +\beta =b_1$
$(\alpha +h)+(\beta +h)=b_2$
solving both we get
$2\alpha \beta -{\alpha }^2-{\beta }^2$=${\Delta }_1$$={\Delta }_2$
so both have same minimum value $-\frac{1}{4}$