第 12 课 NCERT 解决方案 - 数学第一部分 - 第 5 章连续性和可微性 - 练习 5.8

问题 1. 验证函数f(x) = x ² + 2x – 8, x ∈ [– 4, 2] 的罗尔定理。

解决方案：

Now f(x) = x² + 2x – 8 is a polynomial

So, f(x) is continuous in the interval [-4,2] and differentiable in the interval (- 4,2)

f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0

f(2) = 2² + 4 – 8 = 8 – 8 = 0

f(-4) = f(2)

As Conditions of Rolle’s theorem are satisfied.

Then there exists some c in (-4, 2) such that f′(c) = 0

f'(x) = 2x + 2

f’ (c) = 2c + 2 = 0

c = – 1, 

and -1 ∈ [-4,2]

Hence, f’ (c) = 0 at c = – 1.

问题 2. 检查罗尔定理是否适用于以下任何函数。你能从这些例子中谈谈罗尔定理的反面吗？

(i) f(x) = [x] 对于 x ∈ [5, 9]

解决方案：

In the interval [5, 9], 

Now, f (x) = [x] which is neither continuous nor derivable at Integers. 

f (x) is neither continuous nor derivable at x = 6,7,8 

Hence, Rolle’s theorem is NOT applicable

(ii) f(x) = [x] 对于 x ∈ [– 2, 2]

解决方案：

In the interval [– 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Rolle’s theorem is NOT applicable

(iii) f(x) = x ² – 1 对于 x ∈ [1, 2]

解决方案：

Now f(x) = x² – 1 is a polynomial

So, f(x) is continuous in the interval [1, 2] and differentiable in the interval (1,2)

f(1) = (1)²  – 1 = 0

f(2) = 2² – 1 = 3

f(-4) ≠ f(2)

As Conditions of Rolle’s theorem are NOT satisfied.

Hence, Rolle’s theorem is NOT applicable

问题 3. 如果 f : [– 5, 5] → R 是一个可微函数，并且如果 f′(x) 在任何地方都不为零，则证明 f(– 5) ≠ f(5)。

解决方案：

For Rolle’s theorem

f is continuous in [a, b] ………(1)

f is derivable in [a, b] ………(2)

f (a) = f (b)  ………(3)

then f’ (c)=0, c ∈ (a, b)

So as, f is continuous and derivable

but f ‘(c) ≠ 0 

It concludes, f(a) ≠ f(b) 

 f(-5) ≠ f(5)

问题 4. 验证均值定理，如果 f(x) = x ² – 4x – 3 在区间 [a, b] 中，其中 a = 1 和 b = 4。

解决方案：

Now f(x) = x² – 4x -3 is a polynomial

So, f(x) is continuous in the interval [1,4] and differentiable in the interval (1,4)

f(1) = (1)² – 4(1) – 3 = -6

f(4) = 4² – 4(4) – 3 = -3

f′(c) = 2c – 4

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,4) such that 

f′(c) = $\frac{f(b)-f(a)}{b-a}$

= $\frac{-3-(-6)}{4-1}$

= 1

2c – 4 = 1

c = 5/2

and c = 5/2 ∈ (1,4)

问题 5. 验证均值定理，如果 f(x) = x ³ – 5x ² – 3x 在区间 [a, b] 中，其中 a = 1 和 b = 3。找到所有 c ∈ (1, 3) f'(c) = 0。

解决方案：

Now f(x) = x³– 5x²– 3x is a polynomial

So, f(x) is continuous in the interval [1,3] and differentiable in the interval (1,3)

f(1) = (1)³– 5(1)²– 3(1) = -7

f(3) = 3³– 5(3)²– 3(3) = -27

f′(c) = 3c² – 5(2c) – 3

f′(c) = 3c² – 10c – 3

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,3) such that

f′(c) = $\frac{f(b)-f(a)}{b-a}$

= $\frac{f(3)-f(1)}{b-a}$

= $\frac{-27-(-7)}{3-1}$

= $\frac{-20}{2}$

3c² – 10c – 3 = -10

3c² – 10c + 7 = 0

3c2 – 3c – 7c + 7 = 0

3c (c-1) – 7(c -1) = 0

(3c -7) (c-1) = 0

c = 7/3 or c = 1

As, 1 ∉ (1,3)

So, c = 7/3 ∈ (1,3)

According to the Rolle’s Theorem

As, f(3) ≠ f(1), Then there does not exist some c ∈ (1,3) such that f′(c) = 0

问题 6. 检查均值定理对上述练习 2 中给出的所有三个函数的适用性。

(i) f(x) = [x] 对于 x ∈ [5, 9]

解决方案：

In the interval [5, 9],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = 6,7,8

Hence, Mean value theorem is NOT applicable

(ii) f(x) = [x] 对于 x ∈ [– 2, 2]

解决方案：

In the interval [– 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Mean value theorem is NOT applicable

(iii) f(x) = x ² – 1 对于 x ∈ [1, 2]

解决方案：

Now f(x) = x² – 1 is a polynomial

So, f(x) is continuous in the interval [1,2] and differentiable in the interval (1,2)

f(1) = (1)² – 1 = 0

f(2) = 2² -1 = 3

f′(c) = 2c

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,2) such that

f′(c) = $\frac{f(b)-f(a)}{b-a}$

= $\frac{f(2)-f(1)}{2-1}$

= $\frac{3-0}{2-1}$

= 3

2c = 3

c = 3/2

and c = 3/2 ∈ (1,4)