Question

let a,b,c belongs to R such that 2a+3b+6c=0 and g(x) be the antiderivative of f(x)=ax^2+bx+c. if the slope of the tangent drawn to the curve y=g(x) at (1,g(1)) and (2,g(2)) are equal then
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Answer 1

$\textbf{Given:}$

$g(x)\,\text{be the anti-derivative of}\,f(x)=ax^2+bx+c\;\text{and}\;2a+3b+6c=0$

$\textbf{Solution:}$

$\text{Since g(x) is the anti-derivative of f(x), we have}$

$\bf\dfrac{dg}{dx}=f(x)$

$\dfrac{dg}{dx}=ax^2+bx+c$

$\text{But,}$

$\text{Slope of the tangent drawn to the curve y=g(x) at (1,g(1)) and (2,g(2)) are equal}$

$\dfrac{dg}{dx}|_(1,g(1))=\dfrac{dg}{dx}|_(2,g(2))$

$a(1)^2+b(1)+c=a(2)^2+b(2)+c$

$a+b+c=4a+2b+c$

$\implies\,b=-3a$

$\text{Now,}$

$2a+3b+6c=0$

$2a-9a+6c=0$

$-7a+6c=0$

$7a-6c=0$

$7a=6c$

$\dfrac{a}{c}=\dfrac{6}{7}$

$\implies\,a:c=6:7$

$\textbf{Answer:}$

$\bf\,a:c=6:7$