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This is one of our many quizzes on Mean value theorem. Click on this to access more such competitive quizzes on Differentiation.

Mean value theorem Quiz 2

mathematics

 Choose the correct value for $ c$  for the function $ f(x) = a{x^3} + b{x^2} + cx$  in $ [0,1]$  using Lagrange's theorem. 

 Apply the formula  $ \dfrac{{f(b) - f(a)}}{{b - a}} = f'(c)$ and then use the formula $ \dfrac{{ - b\sqrt {{b^2} - 4ac} }}{{2a}}$ to get the value.  	 

 The function given is a polynomial function which is continuous for all values of $ x$ which exist in the interval $ [0,1]$. So, $ f(x)$ is derivable at $ (0,1)$.  $ \\ $ Differentiating the given function: $ \\ f'(x) = 3{x^2}a + 2bx + c \\ $ All the conditions of Lagrange's mean value theorem are satisfied. So there exist at least one value $ c$ in $ (0,1)$ such that $ \dfrac{{f(b) - f(a)}}{{b - a}} = f'(c)$  $ \\ $ The value of $ f(1)$ is: $ \\ f(1) = a + b + c \\ $  And the value of $ f(0) = a{(0)^3} + b{(0)^2} + c(0) = 0$. $ \\ $ On substituting we get, $ \\ \dfrac{{a + b + c - 0}}{{1 - 0}} = 3{c^2}a + 2cb + c \\ \Rightarrow 3{c^2}a + 2bc - (a + b) = 0 \\ $ Applying the formula $ \dfrac{{ - b\sqrt {{b^2} - 4ac} }}{{2a}}$ to get the value of $ c$  $ \\ c = \dfrac{{ - 2b\sqrt {4{b^2} + 4.3a.(a + b)} }}{{6a}} \\ c = \dfrac{{ - b\sqrt {3{a^2} + {b^2} + 3ab} }}{{3a}}$ 

 Mark the point on the parabola given $ y = {(x - 2)^2}$ where the tangent is parallel to the chord which is joining the points $ (3,0)$ and $ (4,1)$.

 Apply the formula  $ \dfrac{{f(b) - f(a)}}{{b - a}} = f'(c)$ and then solve. 

 Given function $ y = {(x - 2)^2}$ is  polynomial function which is continuous in the interval $ [3,4]$  $ \\ $ Differentiation of the function is $ f'(x) = 2(x - 2)$, so $ f(x)$ is differentiable in $ (3,4)$  $ \\ $ As both of the points of the Lagrange's mean value theorem get satisfied so there exists a point $ c \in (3,4)$ such that $ f'(c) = \dfrac{{f(4) - f(3)}}{{4 - 3}}$  $ \\ $ The value of $ f(4)$  is: $ \\  f(4) = {(4 - 2)^2} = {(2)^2} = 4 \\ $ The value of $ f(3)$ is: $ \\ f(3) = {(3 - 2)^2} = {(1)^2} = 1 \\ $ Substituting the values in the above formula we get, $ \\ 2(c - 2) = f'(c) = \dfrac{{4 - 1}}{1} \\ 2(c - 2) = 3 \\ \Rightarrow c - 2 = \dfrac{3}{2} \\ \Rightarrow c = \dfrac{7}{2} \in (3,4) \\ $ $\therefore x = \dfrac{7}{2}$ and $ y = {(x - 2)^2} = {\left(\dfrac{7}{2} - 2\right)^2} = \dfrac{9}{4}$ 

 The function given $ f(x) = {x^3} + a{x^2} + 2bx$ follows Rolle’s theorem, $ x \in [1,2]$ at the point $ x = \dfrac{4}{3}$. Tick the correct values for $ a$  and $ b$.

 Put $ f(1) = f(2)$ to get an equation. Then take $ f'(c) = 0$ at $ x = \dfrac{4}{3}$ and get another equation. Then solve these two equations to get the values. 

 Given function $ f(x) = {x^3} + a{x^2} + 2bx$ is a polynomial function which is continuous on $ [1,2]$ $ \\ $ Differentiating the given function we get, $ f'(x) = 3{x^2} + 2ax + 2b$ which exists uniquely on $ (1,2)$ $ \\ $ So $ f(x)$ is derivable on $ (1,2)$  $ \\ $ The value of $ f(1)$ is : $ \\ f(1) = {(1)^3} + a{(1)^2} + 2b(1) = 1 + a + 2b \\ $ The value of $ f(2)$ is : $ \\ f(2) = {(2)^3} + a{(2)^2} + 2b(2) = 8 + 4a + 4b \\ $ As the question says that the function follows Rolle’s theorem so $ f(1) = f(2)$. Applying this condition we get, $ \\  1 + a + 2b = 8 + 4a + 4b \\ \Rightarrow 3a + 2b + 7 = 0 \\  $  According to the question $ x = \dfrac{4}{3} \Rightarrow f'(c) = 0$  $ \\ $ So putting $ x = \dfrac{4}{3}$ in $ f'(x)$ we get, $ \\ 3{\left(\dfrac{4}{3}\right)^2} + 2a\left(\dfrac{4}{3}\right) + 2b \\ \Rightarrow 8a + 6b + 16 = 0 \\ $  Solving $ 3a + 2b + 7 = 0$ and $ 8a + 6b + 16 = 0$ simultaneously we get, $ \\ $ $ a = 5$ and $ b =  - 4$ 

 Solve the question to get the value of $ c$ for the function $ f(x) = \dfrac{1}{{3x - 1}}$ in $ [1,4]$  using Lagrange's mean value theorem. 

 The given function $ f(x) = \dfrac{1}{{3x - 1}}$ is continuous in $ [1,4]$. $ \\ $ The differentiation of the function given is $ f'(x) = \dfrac{{ - 3}}{{{{(3x - 1)}^2}}}$ which exists in $ (1,4)$ $ \\ $ The value of $ f(1)$ is: $ \\ f(1) = \dfrac{1}{{3(1) - 1}} = \dfrac{1}{2} \\ $ The value of $ f(2)$ is: $ \\ f(4) = \dfrac{1}{{3(4) - 1}} = \dfrac{1}{{11}} \\ $ As both the conditions of the Lagran’e get satisfied so there must exist a real value of $ c$ in $ (1,4)$ such that $ \dfrac{{f(b) - f(a)}}{{b - a}} = f'(c)$  $  \\ \dfrac{{ - 3}}{{{{(3c - 1)}^2}}} = \dfrac{{\dfrac{1}{{11}} - \dfrac{1}{2}}}{3} \\   \Rightarrow \dfrac{{ - 3}}{{{{(3c - 1)}^2}}} = \dfrac{{\dfrac{{2 - 11}}{{22}}}}{3} \\   \Rightarrow \dfrac{{ - 3}}{{{{(3c - 1)}^2}}} = \dfrac{{ - 3}}{{22}} \\   \Rightarrow {(3c - 1)^2} = 22 \\   \Rightarrow 3c - 1 =  \pm \sqrt {22}  \\   \Rightarrow c = \dfrac{{1 \pm \sqrt {22} }}{3} \\ $  $ \\ $ $ c = \dfrac{{1 + \sqrt {22} }}{3} = 1.88 \in (1,4)$ and at $ c = \dfrac{{1 - \sqrt {22} }}{3} =  - 1.23$ does not belong to $ (1,4)$.

 Determine a point on the curve given$ y = \sqrt {{x^2} - 1} $  which is defined in$ [1,3]$  in which the tangent is parallel to the chord joining the endpoints present on the curve using the Lagrange mean value theorem. 

 Apply the formula $ \dfrac{{f(b) - f(a)}}{{b - a}} = f'(c)$  and then substitute the value you find in the function given to get $ y$ 

 The function given is $ y = {({x^2} - 1)^{\frac{1}{2}}}$ which exists $ \forall x \in [1,3]$ so the given function is continuous in interval$ [1,3]$. $ \\  f'(x) = \dfrac{1}{2}.\dfrac{{2x}}{{\sqrt {{x^2} - 1} }} = \dfrac{x}{{\sqrt {{x^2} - 1} }}$. This exists for all the values of the $ x$ except$  \in (1,3)$ so the condition of the Lagrange's mean value theorem satisfies.  $ \\ $ Now applying the formula $ \dfrac{{f(b) - f(a)}}{{b - a}} = f'(c)$  $ \\ $ The value of $ f(b) = f(3) = \sqrt {{{(3)}^2} - 1}  = \sqrt 8  = 2\sqrt 2 $ $ \\ $ The value of $ f(a) = f(1) = \sqrt {{{(1)}^2} - 1}  = 0$ $ \\ $ Substituting the values in formula we get, $ \\ \dfrac{{2\sqrt 2  - 0}}{{3 - 1}} = \dfrac{c}{{\sqrt {{c^2} - 1} }} \\ \Rightarrow \sqrt 2 \sqrt {{c^2} - 1}  = c \\  \Rightarrow {c^2} = 2{c^2} - 6 \\ $ $ \\ \Rightarrow c = \sqrt 6 $ which lies in the interval $ (2,4)$. $ \\ $ When $ x = c = \sqrt 6 $ then the value of $ y$ is $ y = \sqrt {{x^2} - 4}  = \sqrt {6 - 4}  = \sqrt 2 $ 