Question

Let the parabolas $y=x^2+ax+b$ and $y=x(c-x)$ touch each other at the point $(1,0)$. Then

• A. $a=-3$
• B. $b=1$
• C. $c=2$
• D. $b+c=3$

Answer 1

Answer: (A), (D)

$a=-3$
$b+c=3$

As point $(1,0)$ lies on
$y=x^2+ax+b\Rightarrow a+b=-1$ ...(1)
And also lies on $y=x(c-x)\Rightarrow c=1$ ...(2)
They touch each other, so slope of tangents drawn from $(1,0)$ will be same, then
$\frac{dy}{dx}=2x+a=c-2x\Rightarrow a=c-4\Rightarrow a=-3$ ...(3)
Using (1) and (3) we get
$-3+b=1\Rightarrow b=2$ ...(4)
Now using (2) and (4) we get
$b+c=2+1=3$
Hence, options 'A' and 'D' are correct.