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 \[x = \dfrac{7}{4}\] is parallel to the chord that joins the points of intersection of the curve with the ordinates \[x = 1\] and \[x = 2\]. Then find the value of \[a\].
A. \[\dfrac{{35}}{{16}}\]
B. \[\dfrac{{35}}{{48}}\]
C. \[\dfrac{7}{6}\]
D. \[\dfrac{5}{{16}}\]

Answer 1

Hint: First, differentiate the given function with respect to the variable \[x\]. Then apply the condition of Lagrange’s mean value theorem. Substitute the value of the first derivative, \[a = 1\] and \[b = 2\] in the condition to solve the condition of Lagrange’s mean value theorem. In the end, substitute \[x = \dfrac{7}{4}\] in the simplified equation to reach the required answer.

Formula used
Lagrange’s mean value theorem: Let a function \[f:\left[ {a,b} \right] \to R\] be a continuous function and differentiable on the open interval \[\left( {a,b} \right)\]. Then there exists some \[c \in \left( {a,b} \right)\] such that, \[f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\]
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step by step solution
Given:
The function \[f\left( x \right) = {x^3} - 6a{x^2} + 5x\] satisfies the conditions of Lagrange’s mean value theorem for the interval \[\left[ {1,2} \right]\].
The tangent to the curve \[y = f\left( x \right)\] at \[x = \dfrac{7}{4}\] is parallel to the chord that joins the points of intersection of the curve with the ordinates \[x = 1\] and \[x = 2\].

Let’s differentiate the given function with respect to the variable \[x\].
\[f'\left( x \right) = 3{x^2} - 12ax + 5\] \[.....\left( 1 \right)\]

Since the given function satisfies the conditions of Lagrange’s mean value theorem for the interval \[\left[ {1,2} \right]\].
We get,
\[f'\left( x \right) = \dfrac{{f\left( 2 \right) - f\left( 1 \right)}}{{2 - 1}}\]
Substitute the value of \[f'\left( x \right)\] from the equation \[\left( 1 \right)\] and the values in the above equation.
\[3{x^2} - 12ax + 5 = \dfrac{{\left[ {{{\left( 2 \right)}^3} - 6a{{\left( 2 \right)}^2} + 5\left( 2 \right)} \right] - \left[ {{{\left( 1 \right)}^3} - 6a{{\left( 1 \right)}^2} + 5\left( 1 \right)} \right]}}{{2 - 1}}\]
\[ \Rightarrow 3{x^2} - 12ax + 5 = \dfrac{{\left[ {8 - 24a + 10} \right] - \left[ {1 - 6a + 5} \right]}}{1}\]
\[ \Rightarrow 3{x^2} - 12ax + 5 = - 24a + 18 - 6 + 6a\]
\[ \Rightarrow 3{x^2} - 12ax + 5 = 12 - 18a\]

Now substitute \[x = \dfrac{7}{4}\] in the above equation.
\[3{\left( {\dfrac{7}{4}} \right)^2} - 12a\left( {\dfrac{7}{4}} \right) + 5 = 12 - 18a\]
\[ \Rightarrow 3\left( {\dfrac{{49}}{{16}}} \right) - 21a + 5 = 12 - 18a\]
\[ \Rightarrow 21a - 18a = \dfrac{{147}}{{16}} - 12 + 5\]
\[ \Rightarrow 3a = \dfrac{{147}}{{16}} - 7\]
\[ \Rightarrow 3a = \dfrac{{147 - 112}}{{16}}\]
\[ \Rightarrow 3a = \dfrac{{35}}{{16}}\]
Divide both sides by 3.
\[a = \dfrac{{35}}{{48}}\]

Hence option B is correct.

Note:Lagrange’s mean value theorem states that for a curve between two points there exists a point where the tangent is parallel to the secant line passing through these two points of the curve.