Question

If the function $f(x)=x^3-6a{x}^2+5x$ satisfies the condition of Lagrange's mean value theorem over the interval $[1,2]$ and the tangent to the curve $y=f(x)$ at $x=\frac{7}{4}$ is parallel to the chord that joins the points of intersection of the curve with ordinates $x=1$ and $x=2$, then the value of $a$ is

• A. $\frac{35}{16}$
• B. $\frac{35}{48}$
• C. $\frac{7}{16}$
• D. $\frac{5}{16}$

Answer 1

Answer: (B)

Given, $f(x)=x^3-6a{x}^2+5x$
Now, $f(2)=8-24a=10=18-24a$
and $f(1)=1-6a+5=6-6a$
Also, $f^{\prime }(x)=3x^2-12ax+5$
$\therefore$ From Lagrange's Mean Value Theorem, we have
$f^{\prime }(x)=\frac{f(2)-f(1)}{2-1}=\frac{(18-24a)-(6-6a)}{1}\Rightarrow f^{\prime }(x)=12-18a$
Also $f^{\prime }\left(\frac{7}{4}\right)=3\times \frac{49}{16}-12a\times \frac{7}{4}+5=12-18a$
$\Rightarrow 3a=\frac{147}{16}-7\Rightarrow 3a=\frac{35}{16}\Rightarrow a=\frac{35}{48}$