Question

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

• A. $35/16$
• B. $35/48$
• C. $7/16$
• D. $5/16$

Answer 1

The correct option is B $35/48$

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

At $x=1$ At $x=2$

$f\left(x\right)=6-6a$ $f\left(x\right)=18-24a$

$f\left(x\right)=3x^2-12ax$.

At $x=\frac{7}{4}$

$f\left(x\right)=3{\left(\frac{7}{4}\right)}^2-12a\left(\frac{7}{4}\right)+5$

So according to long range mean value theorem,

$\frac{1/-24a-(6-6a)}{2-1}=3{\left(\frac{7}{4}\right)}^3-12a\left(\frac{7}{4}\right)+5$

$12-18a=3{\left(\frac{7}{4}\right)}^2-12a\left(\frac{7}{4}\right)+5$

$3a=\frac{147}{16}-\frac{112}{16}=\frac{35}{16}$

$a=\frac{35}{48}$

$B$ is correct.