Application of Derivatives Exercise: 16.4 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to study

1. Rolle's Theorem
2. Geometrical Interpretation of Rolle's Theorem
3. Lagrange's Mean Value Theorem

Exercise 16.4

1. Verify Rolle’s theorem for the following functions:

a. f(x) = 3x² – 4 in [-1,1]

Solution:

Since, f(x) = 3x² – 4, is a polynomial function, it is continuous in [-1,1].

And, f’(x) = 6x, exists for all x in (-1,1).

So, f(x) is differentiable in (-1,1).

Again, f(-1) = 3(-1)² – 4 = 3 – 4 = -1.

And f(1) = 3(1)² – 4 = 3 – 4 = -1

So, f(-1) = f(1)

Hence, f(x) satisfies all conditions of Rolle’s theorem so there exists a C in (-1,1) such that f’(C) = 0 → 6C → C = 0.

Hence, C = 0 ԑ (-1,1), and Rolle’s theorem is verified.

b. f(x) = 2x² – 3x + 1 in $\left[ {\frac{1}{2},1} \right]$

Solution:

Since, f(x) = 2x² – 3x + 1 is a polynomial function, so it is continuous in $\left[ {\frac{1}{2},1} \right]$.

And f’(x) = 4x – 3 exists for all x in $\left( {\frac{1}{2},1} \right)$.

Again,f$\left( {\frac{1}{2}} \right)$ = 2.${\left( {\frac{1}{2}} \right)^2}$ = 3.$\frac{1}{2}$ + 1 = $\frac{{1 - 3 + 2}}{2}$ = $\frac{0}{2}$ = 0.

And f(1) = 2.1² – 3.1 + 1 = 2 – 3 + 1 = 0.

So, ${\rm{f}}\left( {\frac{1}{2}} \right)$ = f(1).

Hence, f(x) satisfies the condition of Rolle’s theorem so there exists C in $\left( {\frac{1}{2},1} \right)$, such that f’(C) = 0.

Or, 4C – 3 = 0 → C = $\frac{3}{4}$ ԑ $\left( {\frac{1}{2},1} \right)$ so the Rolle’s theorem is verified.

c. f(x) = (x + 1)(x – 2) in [-1,2]

Solution:

Since, f(x) = (x + 1)(x – 2) = x² – x – 2 is a polynomial function, so it is continuous in $\left[ { - 1,2} \right]$.

And f’(x) = 2x – 1 exists for all x in $\left( { - 1,2} \right)$. Thus, f(x) is differentiable in (-1,2).

Again,f$\left( { - 1} \right)$ = (-1)² – (-1) – 2 = 0.

And f(2) = 2² – 2 – 2 = 0.

So, f(-1) = f(2).

Hence, f(x) satisfies the condition of Rolle’s theorem so there exists C in $\left( { - 1,2} \right)$, such that

f’(C) = 0 → 2C – 1 = 0.

So, C = $\frac{1}{2}$ ԑ (-1,2).

Hence, Rolle’s theorem is verified.

d. f(x) = x(x – 1)²  in [0,1]

Solution:

Since f(x) = x(x – 1)² is a polynomial function, so it is continuous in $\left[ {0,1} \right]$.

And f’(x) = 1.(x – 1)² + x.2(x – 1)

Or, f’(x) = x² – 2x + 1 + 2x² – 2x = 3x² – 4x + 1.

Thus, f(x) is differentiable in $\left({0,1} \right)$.

Again,f$\left( 0 \right)$ = 0.

And f(1) = 0

So, f(0) = f(1).

Hence, f(x) satisfies the condition of Rolle’s theorem so there exists C in $\left( {0,1} \right)$, such that

f’(C) = 0 → 3C² – 4C + 1 = 0

So, C = 1, $\frac{1}{3}$ where 1 ∉ (0,1) but $\frac{1}{3}$ԑ (0,1)

Hence, Rolle’s theorem is verified.

e. f(x) = (x – 1)(x – 2)(x – 3) in [1,3]

Solution:

Since, f(x) = (x – 1)(x – 2)(x – 3) = x³ – 3x² + 11x – 6is a polynomial function, so it is continuous in $\left[ {1,3} \right]$.

And f’(x) = 3x² – 12x + 11, exists for all x in (1,3).

Thus, f(x) is differentiable in $\left( {1,3} \right)$.

Again,f$\left( 1 \right)$ = 0.

And f(3) = 0

So, f(1) = f(3).

Hence, f(x) satisfies the condition of Rolle’s theorem so there exists C in $\left( {1,3} \right)$, such that

f’(C) = 0 → 3C² – 12C + 11 = 0

So, C = $\frac{{4 \pm \sqrt 3 }}{3}$ where C = $\frac{{4 - \sqrt 3 }}{3}$ does not belong to (1,3) but C = $\frac{{4 + \sqrt 3 }}{4}$ ԑ (1,3).

Hence, Rolle’s theorem is verified.

f. f(x) = sinx in $\left[ {0,{\rm{\pi }}} \right]$

Solution:

Since, f(x) = sinx is continuous in $\left[ {0,{\rm{\pi }}} \right]$.

And f’(x) = cosx, exists for all x ԑ [0,π]

Thus, f(x) is differentiable in $\left( {0,{\rm{\pi }}} \right)$.

Also, f(0) = sin 0 = 0

Again,f$\left( {\rm{\pi }} \right)$ = sinπ = 0.

So, f(0) = f(π).

Hence, f(x) satisfies the condition of Rolle’s theorem so there exists C in $\left( {0,{\rm{\pi }}} \right)$, such that

f’(C) = 0 → ,cosC = 0. → C = $\frac{{\rm{\pi }}}{2}$ ԑ (0,π).

Hence, Rolle’s theorem is verified.

g. f(x) = cos2x in $\left[ {--{\rm{\pi }},{\rm{\pi }}} \right]$

Solution:

Since, f(x) = cos2x is continuous in $\left[ {--{\rm{\pi }},{\rm{\pi }}} \right]$.

And f’(x) = -2sin2x, exists for all x in (-π,π).

Also,f$\left( { - {\rm{\pi }}} \right)$ = cos(-2π) = 1.

And f(π) = cos2π = 1.

So, f(2π) = f(-2π).

Hence, f(x) satisfies the condition of Rolle’s theorem so there exists C in $\left( {--{\rm{\pi }},{\rm{\pi }}} \right)$, such that

f’(C) = 0 →

So, -2sin2C = 0 → sin2C = 0 → C = 0.

Thus, C = 0 ԑ (-π,π)

Hence, Rolle’s theorem is verified.

h. f(x) = $\sqrt {25 - {{\rm{x}}^2}} $ in $\left[ {--5,5} \right]$

Solution:

Since, f(x) = $\sqrt {25 - {{\rm{x}}^2}} $is continuous in $\left[ {--5,5} \right]$.

And f’(x) = $ - \frac{{\rm{x}}}{{\sqrt {25 - {{\rm{x}}^2}} }}$, exists for all x in (-5,5).

Also,f$\left( { - 5} \right)$ = 0

And f(5) = 0

So, f(-5) = f(5).

Hence, f(x) satisfies the condition of Rolle’s theorem so there exists C in $\left( { - 5,5} \right)$, such that

f’(C) = 0 →

So, $ - \frac{{\rm{C}}}{{\sqrt {25 - {{\rm{C}}^2}} }}$ = 0 → C = 0 ԑ(-5,5).

Hence, Rolle’s theorem is verified.

2. Using Rolle’s theorem, find a point on each of the curves represented by the following function, where the tangent is parallel to the x-axis. 

a. f(x) = x² – 1 in [-2,2]

Solution:

Since f(x) = x² – 1 is continuous on [-2,2] and differential in (-2,2)

Also, f(-2) = f(2) = 3, so f(x) satisfies all conditions of Rolle’s theorem so there exists a C ԑ (-2,2) at which the tangent is parallel to x – axis, such that, f’(C) = 0 → 2C = 0 → C = 0 i.e. x = 0.

When x = 0 then y = f(0) = 0² – 1 = - 1.

Hence, the required point is (0,-1).

b. f(x) = 6x – x² in [0,6]

Since f(x) = 6x – x² is continuous on [0,6] and f’(x) = 6 – 2x exists in (0,6) and f(0) = f(6) = 0, so f(x) satisfies all conditions of Rolle’s theorem so there exists a x ԑ (0,6) at which the tangent is parallel to x-axis,

So, f’(x) = 0

→ 6 – 2x = 0 → x = 3 ԑ (0,6)

For x = 3, y = f(3) = 6.3 – 3² = 9.

So, the required point is (3,9).

c. f(x) = sinx x, ԑ (0,π)

Since f(x) = sinx is continuous on [0,π] and f’(x) exists in (0,π) and f(0) = f(π) = 0, so f(x) satisfies all conditions of Rolle’s theorem so there exists a x ԑ (0,π) at which the tangent is parallel to x-axis,

So, f’(x) = 0

→ cosx = 0 → x = $\frac{{\rm{\pi }}}{2}$ ԑ (0,π].

So, y = f$\left( {\frac{{\rm{\pi }}}{2}} \right)$ = sin $\frac{{\rm{\pi }}}{2}$ = 1.

So, required point is ($\frac{{\rm{\pi }}}{2}$,1).

3. Verify Lagrange’s mean value theorem for the following functions:

a. f(x) = 3x² – 2x in [1,3]

Solution:

Since f(x) = 3x² – 2x is continuous [1,3] and f’(x) = 6x – 2 exists on (1,3) so there exists a C in (1,3) such that,

Or, f’(C) = $\frac{{{\rm{f}}\left( 3 \right) - {\rm{f}}\left( 1 \right)}}{{3 - 1}}$$\left[ {{\rm{f}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}} \right]$

Or, 6C – 2 = $\frac{{\left( {{{3.3}^2} - 2.3} \right) - \left( {{{3.1}^2} - 2.1} \right)}}{{3 - 1}}$   [a = 1,b = 3]

Or, 12C – 4 = 21 – 1 = 20.

So, C = $\frac{{24}}{{12}}$ = 2 ԑ (1,3).

Hence, the mean value theorem is verified.

b. f(x) = x² – 2x + 4 in [1,5]

Solution:

Since f(x) = x² – 2x + 4 is continuous [1,5] and f’(x) = 2x – 2 exists on (1,5) so there exists    $\left[ {{\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}} \right]$

Or, 2C – 2 = $\frac{{\left( {5 - 2.5 + 4} \right) - \left( {{1^2} - 2.1 + 4} \right)}}{{5 - 1}}$   [a = 1,b = 5]

Or, 8C – 8 = 19 – 3 = 16.

So, C – 1 = 2

So, C = 3 ԑ (1,5)

Hence, the mean value theorem is verified.

c. f(x) = 2x² – 10x + 29 in [2,7]

Solution:

Since f(x) = 2x² – 10x + 29 is continuous [2,7] and f’(x) = 4x – 10 exists on (2,7) so there exists a C in (2,7) such that,

$\left[ {{\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}} \right]$ = $\frac{{{\rm{f}}\left( 7 \right) - {\rm{f}}\left( 2 \right)}}{{7 - 2}}$

Or, 4C – 10 = $\frac{{({{2.7}^2} - 10.7 + 29( - \left( {{{2.2}^2} - 10.2 + 29} \right)}}{5}$

Or, 20C – 50 = 98 – 70 + 29 – 8 + 20 – 29

Or, 20C – 50 = 40

So, C = $\frac{9}{2}$ ԑ (2,7).

Hence, the mean value theorem is verified.

d. f(x) = x³ + x² – 6x  in [-1,4]

Solution:

Since f(x) = x³ + x² – 6x  is continuous [-1,4] and f’(x) = 3x² + 2x – 6  exists for all x in (-1,4) so there exists a C in (-1,4) such that,

$\left[ {{\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}} \right]$ = $\frac{{{\rm{f}}\left( 4 \right) - {\rm{f}}\left( 1 \right)}}{{4 - 1}}$

Or, 3C² + 2C – 6 = $\frac{{\left( {{4^3} + {4^2} - 6.4} \right) - \left\{ {{{\left( { - 1} \right)}^3} + {{\left( { - 1} \right)}^2} - 6\left( { - 1} \right)} \right\}}}{5}$.

Or, 3C² + 2C – 6 = $\frac{{56 - 6}}{5}$ = $\frac{{50}}{5}$ = 10.

Or, 3C² + 2C – 16 = 0

So, C = $\frac{{ - 2 \pm \sqrt {4 - 4.3\left( { - 16} \right)} }}{{2.3}}$ = $\frac{{ - 2 \pm \sqrt {196} }}{6}$.

So, C = 2 ԑ (-1,4).  [but $ - \frac{8}{3}$∉ (-1,4)]

Hence, the mean value theorem is verified for C = 2.

e. f(x) = x(x – 1)² = x³ – 2x² + x in [0,2]

Solution:

Since f(x) = x(x – 1)² = x³ – 2x² + x is continuous [0,2] and f’(x) = 3x² – 4x + 1  exists for all x in (0,2) so from the mean value theorem there exists a C in (0,2) such that,

$\left[ {{\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}} \right]$ = $\frac{{{\rm{f}}\left( 2 \right) - {\rm{f}}\left( 0 \right)}}{{2 - 0}}$

Or, 3C² – 4C + 1 = $\frac{{2.{{\left( {2 - 1} \right)}^2} - 0}}{2}$ = $\frac{2}{2}$ = 1.

Or, 3C² – 4C = 0.

So, C = 0∉ (0,2).  [but ${\rm{C}} = \frac{4}{3}$ ԑ (0,2)]

Hence, the mean value theorem is verified for C = $\frac{4}{3}$.

f. f(x) = $\sqrt {{{\rm{x}}^2} + 4} $, x ԑ [2,4]

Solution:

Since f(x) = $\sqrt {{{\rm{x}}^2} + 4} $ is continuous [2,4] and f’(x) = $\frac{{\rm{x}}}{{\sqrt {{{\rm{x}}^2} - 4} }}$ exists for all x in (2,4) so from the mean value theorem there exists a C in (2,4) such that,

$\left[ {{\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}} \right]$ = $\frac{{{\rm{f}}\left( 4 \right) - {\rm{f}}\left( 2 \right)}}{{4 - 2}}$

Or,$\frac{{\rm{C}}}{{\sqrt {{{\rm{C}}^2} - 4} }}$ = $\frac{{\sqrt {16 - 4}  - \sqrt {4 - 4} }}{{4 - 2}}$ = $\frac{{2\sqrt 3 }}{2}$ = $\sqrt 3 $.

Or, C = $\sqrt {3\left( {{{\rm{C}}^2} - 4} \right)} $.

Or, C² = 3C² – 12

Or, C ^{= ±}$\sqrt 6 $^.

So, C = $\sqrt 6 $ ԑ (2,4). 

Hence, the mean value theorem is verified.

g. f(x) = ${{\rm{e}}^{\rm{x}}}$, x ԑ [0,1]

Solution:

Since f(x) = ${{\rm{e}}^{\rm{x}}}$ is continuous on [0,1] and f’(x) = ${{\rm{e}}^{\rm{x}}}$ exists for all x in (0,1) so that there exists a C in (0,1) such that,

$\left[ {{\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}} \right]$ = $\frac{{{\rm{f}}\left( 1 \right) - {\rm{f}}\left( 0 \right)}}{{1 - 0}}$

Or, e^C = $\frac{{{\rm{e}} - 1}}{1}$→ e^C = e – 1.

So, C = log(e – 1) ԑ (0,1)

Hence, the mean value theorem is verified. 

4. a) Using Lagrange’s Mean Value theorem, find the point on the curve f(x) = x(x – 2), the tangents at which is parallel to the chord joining the points (1,-1) and (4,8).

Solution:

Since x ranges from 1 to 4 so f(x) = x(x – 2) = x² – 2x, x ԑ [1,4] is continuous in [1,4] and f’(x) = 2x – 2 exists for all x in (1,4) so there exists a C in (1,4) such that,

Or, ${\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}$ = $\frac{{{\rm{f}}\left( 4 \right) - {\rm{f}}\left( 1 \right)}}{{4 - 1}}$.

Or, 2C – 2 = $\frac{{4\left( {4 - 2} \right) - 1\left( {1 - 2} \right)}}{3}$ = $\frac{{8 + 1}}{3}$ = 3.

So, C’ = $\frac{5}{2}$ ԑ(1,4).

Thus, C = $\frac{5}{2}$ is the abscissa of the point at which the tangent drawn is parallel to the chord joining the point (1,-1) and (4,8). Not put x = $\frac{5}{2}$ in y = f(x) = x(x – 2).

We have, y = $\frac{5}{2}\left( {\frac{5}{2} - 2} \right)$ = $\frac{5}{2}.\frac{1}{2}$ = $\frac{5}{4}$.

So, the required point is $\left( {\frac{5}{2},\frac{5}{4}} \right)$.

b. Examine whether the function f(x) = x² – 6x + 1 satisfies Lagrange’s mean value theorem. If it satisfies find the coordinates of the points at which the tangent is parallel to the chord joining points A91,-4) and B(3,-8).

Solution:

Since x ranges from 1 to 3 so f(x) = x² – 6x + 1, satisfies all the conditions of the mean value theorem  in [1,3],so there exists a C in (1,3) such that,

Or, ${\rm{f'}}\left( {\rm{C}} \right) = \frac{{{\rm{f}}\left( {\rm{b}} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{b}} - {\rm{a}}}}$ = $\frac{{{\rm{f}}\left( 3 \right) - {\rm{f}}\left( 1 \right)}}{{3 - 1}}$.

Or, 2C – 6 = $\frac{{\left( {{3^2} - 6.3 + 1} \right) - \left( {{1^2} - 6.1 + 1} \right)}}{2}$ = $\frac{{ - 8 + 4}}{2}$ = -2.

So, C = 2 ԑ (1,3).

Thus, C = 2 is the abscissa of the point at which the tangent drawn is parallel to the chord joining the point to the chord joining the points A(1,-4) and B(4,8).

 Thus, putting x = 2 in y = x² – 6x + 1.

We have, y = 2² – 6.2 + 1 = -7.

So, the required point is (2,-7).

 

5. Show that Rolle’s theorem can not be applied to the function.

Solution:

f(x) = 1 – x^2/3→ f(x) = 1 – $\frac{2}{3}$.$\frac{1}{{{{\rm{x}}^{\frac{1}{3}}}}}$.

Which does not exist at x = 0 ԑ (-1,1)

So, Rolle’s theorem cannot be applied.