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# If the function $f(x)={{x}^{3}}-6a{{x}^{2}}+5x$ satisfies the conditions of Lagrange’s mean theorem for the interval $\left[ 1,2 \right]$ and the tangent to the curve $y=f(x)$ at $x=\dfrac{7}{4}$ is parallel to the chord joining the points of intersection of the curve with the coordinates $x=1$ and $x=2$ . Then the value of a is? \begin{align} & \text{A}\text{. }\dfrac{35}{16} \\ & \text{B}\text{. }\dfrac{35}{48} \\ & \text{C}\text{. }\dfrac{7}{16} \\ & \text{D}\text{. }\dfrac{5}{16} \\ \end{align}

Last updated date: 13th Jun 2024
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Hint: We will use Lagrange’s mean theorem to find the value of $a$ . The Lagrange’s mean theorem is given as $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ .
Here, $a,b$ are the points of the function and $c$ is the point of the curve. We put the values in the equation and obtain a desired answer.

We have given the function $f(x)={{x}^{3}}-6a{{x}^{2}}+5x$ satisfies the conditions of Lagrange’s mean theorem for the interval $\left[ 1,2 \right]$ and the tangent to the curve $y=f(x)$ at $x=\dfrac{7}{4}$ .
We have to find the value of $a$ .
Now, as given in the question the function $f(x)={{x}^{3}}-6a{{x}^{2}}+5x$ satisfies the conditions of Lagrange’s mean theorem for the interval [1,2].
So, according to Lagrange’s mean theorem if the function is continuous and differentiable on points $\left[ 1,2 \right]$ , there must exists a real number $c\in \left( 1,2 \right)$ such as
$f'(c)=\dfrac{f(b)-f(a)}{b-a}$
$\Rightarrow f'(c)=\dfrac{f(2)-f(1)}{2-1}$
We have $f(x)={{x}^{3}}-6a{{x}^{2}}+5x$
So, first we put $x=1$ , we get
\begin{align} & f(1)={{1}^{3}}-6a\times {{1}^{2}}+5\times 1 \\ & f(1)=1-6a+5 \\ & f(1)=6-6a \\ \end{align}
Now, we put $x=2$ , we get
\begin{align} & f(2)={{2}^{3}}-6a\times {{2}^{2}}+5\times 2 \\ & f(2)=8-24a+10 \\ & f(2)=18-24a \\ \end{align}
Now, we have to calculate $f'(c)$ .
We have $f(x)={{x}^{3}}-6a{{x}^{2}}+5x$
So, $f'(x)=3{{x}^{2}}-12ax+5\text{ }\left[ As\text{ }{{x}^{a}}=a{{x}^{a-1}} \right]$
Now, we have given that the curve $y=f(x)$ at $x=\dfrac{7}{4}$ is parallel to the chord joining the points of intersection of the curve with the coordinates $x=1$ and $x=2$ .
$f'(c)=3{{\left( \dfrac{7}{4} \right)}^{2}}-12a\left( \dfrac{7}{4} \right)+5$
Now, the Lagrange’s mean theorem will be
$f'(c)=\dfrac{f(b)-f(a)}{b-a}$
Substituting the values in above equation, we get
\begin{align} & f'(c)=\dfrac{f(2)-f(1)}{2-1} \\ & 3{{\left( \dfrac{7}{4} \right)}^{2}}-12a\left( \dfrac{7}{4} \right)+5=\dfrac{\left( 18-24a \right)-\left( 6-6a \right)}{1} \\ \end{align}
\begin{align} & 3\left( \dfrac{49}{16} \right)-21a+5=\left( 18-24a \right)-\left( 6-6a \right) \\ & \dfrac{147}{16}-21a+5=18-24a-6+6a \\ & \dfrac{147}{16}-21a+5=12-18a \\ & \dfrac{147}{16}+5-12=21a-18a \\ & \dfrac{147+80-192}{16}=3a \\ & \dfrac{35}{16}=3a \\ & a=\dfrac{35}{16\times 3} \\ & a=\dfrac{35}{48} \\ \end{align}
So, the value of $a$ is $\dfrac{35}{48}$ .
So, the correct answer is “Option B”.

Note: In this particular question it is given that the function $f(x)={{x}^{3}}-6a{{x}^{2}}+5x$ satisfies the conditions of Lagrange’s mean theorem. If not given we need to check that the function satisfied all the necessary conditions such as the function is continuous and differentiable on the given points. If function satisfies all conditions then we apply Lagrange’s mean theorem.