Question

Let $a,b,c\in R$ be such that $2a+3b+6c=0$ and $g(x)$ be the anti derivative of $f(x)=a{x}^2+bx+c.$ If the slopes of the tangents drawn to the curve $y=g(x)$ at $(1,g(l))$ and $(2,g(2))$ are equal, then

• A. $\frac{a}{3}=\frac{b}{-8}=\frac{c}{3}$
• B. $\frac{a}{6}=\frac{b}{-18}=\frac{c}{7}$
• C. $\frac{a}{3}=\frac{b}{-6}=\frac{c}{2}$
• D. $a=b=c=-1$

Answer 1

Answer: (B)

We have,
$2a+3b+6c=0$ ...(i)
$g(x)$ is anti derivative of $f(x)=a{x}^2+bx+c$
$\therefore g^{\prime }(x)=a{x}^2+bx+c$
Given, $g^{\prime }(1)=g^{\prime }(2)$
$\Rightarrow a+b+c=4a+2b+c$
$\Rightarrow 3a+b=0$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{a}{-6}=\frac{b}{18}=\frac{c}{-7}$
$\Rightarrow \frac{a}{6}=\frac{b}{-18}=\frac{c}{7}$