CBSE (Science) Class 12CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution for Verify the Hypothesis and Conclusion of Lagrange'S Man Value Theorem for the Function F(X) = 1 4 X − 1 , 1≤ X ≤ 4 ? - CBSE (Science) Class 12 - Mathematics

Login
Create free account


      Forgot password?

Question

Verify the  hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = \[\frac{1}{4x - 1},\] 1≤ x ≤ 4 ?

 

Solution

The given function is \[f\left( x \right) = \frac{1}{4x - 1}\].

Since for each \[x \in \left[ 1, 4 \right]\] , vthe function attains a unique definite value, \[f\left( x \right)\]  is continuous on \[\left[ 1, 4 \right]\] .

Also,

\[f'\left( x \right) = \frac{- 4}{\left( 4x - 1 \right)^2}\] exists for all \[x \in \left[ 1, 4 \right]\]
Thus, both the conditions of Lagrange's mean value theorem are satisfied.
Consequently, there exists some 
\[c \in \left( 1, 4 \right)\] such that 
\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1} = \frac{f\left( 4 \right) - f\left( 1 \right)}{3}\]
Now, 
\[f\left( x \right) = \frac{1}{4x - 1}\]
\[\Rightarrow\] \[f'\left( x \right) = \frac{- 4}{\left( 4x - 1 \right)^2}\] ,
\[f\left( 4 \right) = \frac{1}{15}, f\left( 1 \right) = \frac{1}{3}\]
\[\therefore\] \[f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1}\]

\[\Rightarrow f'\left( x \right) = \frac{\frac{1}{15} - \frac{1}{3}}{4 - 1} = \frac{- 4}{45}\]

\[ \Rightarrow \frac{- 4}{\left( 4x - 1 \right)^2} = \frac{- 4}{45}\]

\[ \Rightarrow \left( 4x - 1 \right)^2 = 45\]

\[ \Rightarrow 16 x^2 - 8x - 44 = 0\]

\[ \Rightarrow 4 x^2 - 2x - 11 = 0\]

\[ \Rightarrow x = \frac{1}{4}\left( 1 \pm 3\sqrt{5} \right)\]

Thus, 

\[c = \frac{1}{4}\left( 1 + 3\sqrt{5} \right) \in \left( 1, 4 \right)\] such that 

\[f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1}\]. 

Hence, Lagrange's theorem is verified.

  Is there an error in this question or solution?
Solution Verify the Hypothesis and Conclusion of Lagrange'S Man Value Theorem for the Function F(X) = 1 4 X − 1 , 1≤ X ≤ 4 ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
S
View in app×