Question

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

• A. $\frac{35}{16}$
• B. $\frac{35}{48}$
• C. $\frac{7}{6}$
• D. $\frac{5}{16}$

Answer 1

The correct option is B

$\frac{35}{48}$

Explanation for correct option:

Step 1: Solve for the value of $f\text{'}\left(\frac{7}{4}\right)$

Given function, $f\left(x\right)=x^3-6a{x}^2+5x$

Upon differentiating, we get,

$\Rightarrow f\text{'}\left(x\right)=3x^2-12ax+5$

It is given that the function is parallel to the chord that joins the points of intersection of the curve with coordinates $x=1,2$ at $x=\frac{7}{4}$

So, substitute $x=\frac{7}{4}$ in $f\text{'}\left(x\right)=3x^2-12ax+5$

$$\begin{gathered} \Rightarrow f\text{'}\left(\frac{7}{4}\right)=3{\left(\frac{7}{4}\right)}^2-12a\left(\frac{7}{4}\right)+5 \\ \Rightarrow =3\times \frac{49}{16}-21a+5 \\ \Rightarrow =\frac{147}{16}-21a+5 \end{gathered}$$

Step 2: Solve for the value of $a$

Now, from Lagrange's mean value theorem we know that, $f\text{'}\left(x\right)=\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\right]$

Evaluate $f\left(a\right)$ and $f\left(b\right)$ by replacing values in $f\left(x\right)=x^3-6a{x}^2+5x$

$$\begin{gathered} \Rightarrow f\left(a\right)=f\left(1\right) \\ f\left(a\right)={\left(1\right)}^3-6\left(1\right)a+5\left(1\right) \\ f\left(a\right)=6-6a \end{gathered}$$

$$\begin{gathered} \Rightarrow f\left(b\right)=f\left(2\right) \\ f\left(b\right)={\left(2\right)}^3-6{\left(2\right)}^{2}a+5\left(2\right) \\ f\left(b\right)=8-24a+10 \\ f\left(b\right)=18-24a \end{gathered}$$

Substituting all the values in $f\text{'}\left(x\right)=\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\right]$ i.e. $f\text{'}\left(\frac{7}{4}\right)=\left[\frac{f\left(b\right)-f\left(a\right)}{b-a}\right]$, we get,

$$\begin{gathered} \Rightarrow \frac{147}{16}-21a+5=\left[\frac{18-24a-6+6a}{2-1}\right] \\ \Rightarrow \frac{147}{16}+5-12=3a \\ \Rightarrow a=\frac{35}{48} \end{gathered}$$

Hence, option (B), i.e. $\frac{35}{48}$ is the correct answer.