let \(a, b, c\) be real numbers such that \(2 a+3 b+6 c=0\) and \(g(x) \equiv a x^2+b x+c=0\) has atleast…
let \(a, b, c\) be real numbers such that \(2 a+3 b+6 c=0\) and \(g(x) \equiv a x^2+b x+c=0\) has atleast one root in the interval \((1,2)\). If a function \(f:[1,2] \rightarrow R\) for which Rolle's mean value theorem holds is such that \(f(x)\) is a primitive of \(g(x)\), then \(f(x)=\)
\(x^3-3 x^2+2 x\)
\(3 x^3-6 x^2+2 x\)
\(12 x^3-14 x^2+3 x\)
\(3 x^3-x\)
Solution
Given,
\(2 a+3 b+6 c=0\)...(i)
and \(g(x)=a x^2+b x+c=0\)
According to given information,
\(\begin{aligned}
& f(x)=\int g(x) d x=\int\left(a x^2+b x+c\right) d x \\
& f(x)=\frac{a}{3} x^3+\frac{b}{2} x^2+c x
\end{aligned}\)
\(\begin{aligned}
\text{Now, } f(2) & =\frac{a}{3}(2)^3+\frac{b}{2}(2)^2+c(2) \\
& =\frac{8}{3} a+2 b+2 c \quad \ldots (ii)
\end{aligned}\)
\(\begin{aligned}
\text{and } f(\mathrm{l}) & =\frac{a}{3}(\mathrm{l})^3+\frac{b}{2}(\mathrm{l})^2+c(\mathrm{l}) \\
& =\frac{a}{3}+\frac{b}{2}+c \quad \ldots (iii)
\end{aligned}\)
Here, \(f(2)=f(1)\)
\(\frac{8}{3} a+2 b+2 c=\frac{a}{3}+\frac{b}{2}+c\)
\(\Rightarrow \quad 14 a+9 b+6 c=0\)...(iv)
By solving Eqs. (i) and (iv), we get
\(\begin{aligned}
a & =3, b=-6 \text { and } c=2 \\
\therefore \quad f(x) & =\frac{3}{3} x^3-\frac{6}{2} x^2+2 x=x^3-3 x^2+2 x
\end{aligned}\)