Question

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