12th NCERT Continuity And Differentiability Exercise 5.8 Questions 6

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Question (1)

Verify Rolle's theorem for the function f(x) = xSolution

Since f(x) = xHere a = -4, b = 2.

\[f(a) = f( - 4) = {( - 4)^2} + 2( - 4) - 8 = 16 - 8 - 8 = 0\] \[f(b) = f(2) = {(2)^2} + 2(2) - 8 = 4 + 4 - 8 = 0\] \[f(a) = f(b)\] Rolle's theorem is applicable. There exists c ∈ (-4,2) such that f'(c) = 0.

\[{\rm{f(x) = }}{{\rm{x}}^2} + 2x - 8\] \[f'(x) = 2x + 2\] \[f'(c) = 2c + 2\] \[f'(c) = 0 \Rightarrow 2c + 2 = 0\] \[c = - 1 \in ( - 4,2)\] So c = -1.

Question (2)

Examine if Rolle's theorem is applicable to any of the following functions.Can you say some thing about the converse of Rolle's theorem from these example?(i) f(x) = [ x ] for x ∈ [ 5, 9]

Solution

Since f(x) = [ x ], = greatest integer not greater than x,for any value x = a. as x → a

for any value x = a. as x → a

So not continuous for any value of x.

So Rolle's theorem is not applicable.

(ii) f(x) = [ x ] for x ∈ [-2, 2]

Solution

Since f(x) = [ x ], = greatest integer not greater than x,for any value x = a. as x → a

for any value x = a. as x → a

So not continuous for any value of x.

So Rolle's theorem is not applicable.

(iii) f(x) = x

Solution

f(x) is a polynomial, so continuous in [1,2] and differentiable in (1, 2).\[f(a) = f(1) = {(1)^2} - 2 = 1 - 2 = - 1\] \[f(b) = f(2) = {(2)^2} - 2 = 4 - 2 = 2\] \[f(a) \ne f(b)\] So Rolle,s theorem is not applicable.

Question (3)

If f : [-5, 5] → R is a differentiable function and if f'(x) does not vanish any where, then prove that f(-5) ≠ f(5).Solution

As f(x) is differentiable in [ -5, 5], so it will be continuous in (-5, 5)Since f'(x) does not vanish , f'(c) ≠ 0.

By Mean Value Theorem, \[f'(c) \ne 0\] \[\frac{{f(5) - f( - 5)}}{{5 - ( - 5)}} \ne 0\] \[f(5) - f( - 5) \ne 0\] \[f(5) \ne f( - 5)\]

Question (4)

Verify Mean Value Theorem, if f(x) = xSolution

Since f(x) is a polynomial it is continuous in interval [ 1, 4] and differentiable in interval (1, 4).\[f(a) = f(1) = {(1)^2} - 4(1) - 3 = 1 - 4 - 3 = - 6\] \[f(b) = f(4) = {(4)^2} - 4(4) - 3 = 16 - 16 - 3 = - 3\] \[f(a) \ne f(b)\] So Mean Value Theorem is Applicable . So there exists c ∈ (1, 4) such that \[f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\] \[f(x) = {x^2} - 4x - 3\] \[f'(x) = 2x - 4 \Rightarrow f'(c) = 2c - 4\] \[f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\] \[2c - 4 = \frac{{ - 3 - ( - 6)}}{{4 - 1}}\] \[2c - 4 = 1 \Rightarrow c = \frac{5}{2}\] 5/2 ∈ (1, 4). So c = 5/2.

Question (5)

Verify Mean Value Theorem, if f(x) = xSolution

Since f(x) is a polynomial it is continuous in interval [ 1, 3] and differentiable in interval (1, 3).\[f(a) = f(1) = {(1)^3} - 5{(1)^2} - 3(1) = 1 - 5 - 3 = - 7\] \[f(b) = f(3) = {(3)^3} - 5{(3)^2} - 3(3) = 27 - 45 - 9 = - 27\] \[f(a) \ne f(b)\] So Mean Value Theorem is Applicable . There exists c ∈ (1,3) such that \[f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\] \[f(x) = {x^3} - 5{x^2} - 3x\] \[f'(x) = 3{x^2} - 10x - 3 \Rightarrow f'(c) = 3{c^2} - 10c - 3\] \[f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\] \[3{c^2} - 10c - 3 = \frac{{ - 27 - ( - 7)}}{{3 - 1}}\] \[3{c^2} - 10c - 3 = - 10\] \[3{c^2} - 10c + 7 = 0\] \[(c - 1)(3c - 7) = 0\] \[c = 1\;\;or\;\;c = \frac{7}{3}\] c = 1 ∉ (1, 3) so c≠ 1. c = 7/3 ∈ (1, 3).

So c = 7/3

If f'(c) = 0, \[3{c^2} - 10c - 3 = 0\] \[c = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] \[c = \frac{{10 \pm \sqrt {100 + 36} }}{{2(3)}}\] \[c = \frac{{10 \pm \sqrt {136} }}{6}\] \[c = \frac{{10 \pm 2\sqrt {34} }}{6} = \frac{5}{3} \pm \frac{{\sqrt {34} }}{3}\] \[\left( {\frac{5}{3} + \frac{{\sqrt {34} }}{3}} \right) \in (1,3)\; \quad \text{But} \quad \left( {\frac{5}{3} - \frac{{\sqrt {34} }}{3}} \right) \notin (1,3)\] \[c = \left( {\frac{5}{3} + \frac{{\sqrt {34} }}{3}} \right)\]

Question (6)

Examine the applicability of Mean Value Theorem for all three functions given in the above exercise2.(i) f(x) = [ x ] for x ∈ [ 5, 9]

Solution

Since f(x) = [ x ], = greatest integer not greater than x,for any value x = a. as x → a

for any value x = a. as x → a

So not continuous for any value of x.

So Mean Value Theorem is not applicable.

(ii) f(x) = [ x ] for x ∈ [-2, 2]

Solution

Since f(x) = [ x ], = greatest integer not greater than x,for any value x = a. as x → a

for any value x = a. as x → a

So not continuous for any value of x.

So Mean Value Theorem is not applicable.

(iii) f(x) = x

Solution

f(x) is a polynomial, so continuous in [1,2] and differentiable in (1, 2).\[f(a) = f(1) = {(1)^2} - 2 = 1 - 2 = - 1\] \[f(b) = f(2) = {(2)^2} - 2 = 4 - 2 = 2\] \[f(a) \ne f(b)\] So Mean Value Theorem is Applicable . So there exists c ∈ (1, 2) such that \[f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\] \[{\rm{f(x) = }}{{\rm{x}}^2} - 1,\] \[f'(x) = 2x \Rightarrow f'(c) = 2c\] \[f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\] \[2c = \frac{{2 - ( - 1)}}{{2 - 1}} = 3 \Rightarrow c = \frac{3}{2}\] 3/2 ∈ (1, 2), so c = 3/2.