第 12 课 NCERT 解决方案 - 数学第一部分 - 导数的应用 - 练习 6.2|设置 2

第 6 章导数的应用——练习 6.2|设置 1

问题 11. 证明 f(x) = x ² – x + 1 给出的函数f 在 (– 1, 1) 上既不严格增加也不严格减少。

解决方案：

Given: f(x) = x² – x + 1

f'(x) = 2x – 1

For strictly increasing, f'(x) > 0

2x – 1 > 0

x > 1/2 

So, f(x) function is increasing for x > 1/2 in the interval (1/2, 1)          -(Given interval is (-1, 1)

Similarly, for decreasing f'(x) < 0

2x – 1 < 0

x < 1/2  

So, f(x) function is increasing for x < 1/2 in the interval (-1, 1/2)          -(Given interval is (-1, 1)

Hence, the function f(x) = x² – x + 1 is neither strictly increasing nor decreasing. 

问题 12. 以下哪个函数在 (0, π/2) 上递减。

(A) cos x (B) cos 2x (C) cos 3x (D) tan x

解决方案：

(A) f(x) = cos x

f'(x) = -sin x

Now in (0, π/2) interval, sin x is positive(because it is second quadrant)

So, -sin x < 0

∴ f'(x) < 0

f(x) = cos x is strictly decreasing on(0, π/2).

(B) f(x) = cos 2x

f'(x) = -2 sin 2x  

Now in (0, π/2) interval, sin x is positive(because it is second quadrant)

-sin 2x < 0 

∴ f'(x) < 0, 

f(x) = cos 2x is strictly decreasing on(0, π/2).

(C) f(x) = cos 3x

f'(x) = -3sin 3x

Let 3x = t

So in sin 3x = sin t

When t ∈(0, π), sin t + >0 or 3x ∈ (0, π)

But when π/3 < x < π/2

π < 3x < 3π/2

Here sin 3x < 0

So, in x ∈ (0, π/3),

 f'(x) = -3sin 3x < 0 & in x∈(π/3, π/2), f'(x) = -3sin 3x > 0

f'(x) is changing signs, hence f(x) is not strictly decreasing.

(D) f(x) = tan x

f'(x) = sec²x

Now in x ∈ (0, π/2), sec²x > 0

Hence, f(x) is strictly increasing on(0, π/2).

So, option (A) and (B) are decreasing on (0, π/2).

问题 13. 由 f(x) = x ¹⁰⁰ + sin x – 1 给出的函数f 在以下哪个区间递减？

(A) (0, 1) (B) π/2, π (C) 0, π/2 (D) 这些都不是

解决方案：

f(x) = x¹⁰⁰ + sin x – 1

f'(x) = 100x⁹⁹ + cos x

(A) In (0, 1) interval, x > 0, so 100x⁹⁹ > 0

and for cos x: (0, 1°) = (0, 0.57°) > 0

Hence, f(x)is strictly increasing in interval(0, 1) 

(B) In (π/2, π) interval, 

For 100x⁹⁹: x ∈ (π/2, π) = (11/7, 22/7) = (1.5, 3.1) > 1

So, x⁹⁹ > 1. Hence 100x⁹⁹ > 100

For Cos x: (π/2, π) in second quadrant and in second quadrant cos x is negative, so the value is in be -1 and 0.

Hence, f(x)is strictly increasing in interval (π/2, π)

(C) In (0, π/2) interval, both cos x > 0 and 100x⁹⁹ > 0

So f'(x) > 0 

Hence, f(x)is strictly increasing in interval (0, π/2)

So, the correct option is (D).

问题 14. 对于由 f(x) = x ² + ax + 1 给出的函数f 的什么值在 (1, 2) 上增加？

解决方案：

Given: f(x) = x² + ax + 1

f'(x) = 2x + a

Now, x ∈ (1, 2), 2x ∈ (2, 4)

2x + a ∈ (2 + a, 4 + a)

For f(x) to be strictly increasing, f'(x) > 0

If the minimum value of f'(x) > 0 then 

f'(x) on its entire domain will be > 0.

f'(x)_min > 0

2 + a > 0

a > -2 

问题 15. 让 I 是与 [–1, 1] 不相交的任何区间。证明函数f 由$f(x)=x+\frac{1}{x}$在 I 上增加。

解决方案：

Clearly the maximum interval I is R-(-1,1)

Now, f(x) = $x+\frac{x}{1}$

f'(x) = $1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$

It is given that I be any interval disjoint from [–1, 1]

So, for every x ∈ I either x < -1 or x > 1

So, for x < -1, f'(x) is positive.

So, for x < 1, f'(x) is positive.

Hence, f'(x) > 0 ∀ x ∈ I, so, f(x) is strictly increasing on I.

问题 16. 证明 f(x) = log sin x 给出的函数f 在 (0, π/2) 上增加并在 (π/2, π) 上减少。

解决方案：

f(x) = log sin x

f'(x) = $\frac{1}{\sin x}.\cos x=\cot x$      

Interval (0, π/2), it is first quadrant, here cot x is positive. 

So, f'(x) = cot x is positive (i.e., cot x > 0)

Hence, f(x) is strictly increasing in interval (0, π/2)

Interval (π/2, π), it is second quadrant, here cot x is negative.

So, f'(x) = cot x is negative (i.e., cot x < 0)

Hence, f(x) is strictly decreasing in interval (π/2, π)

问题 17. 证明 f(x) 给出的函数f = log|cos x|在 (0, π/2) 上减少并在 (π/2, π) 上增加。

解决方案：

f(x) = log cos x

f'(x) = 1/cos x (-sin x) = -tan x          

Interval (0, π/2), it is first quadrant, here tan x is positive. 

So, f'(x) = -tan x is negative(i.e., tan x < 0)

Hence, f(x) is strictly decreasing in interval (0, π/2)

Interval (π/2, π), it is second quadrant, here tan x is negative.

So, f'(x) = -tan x is positive (i.e., tan x > 0)

Hence, f(x) is strictly increasing in interval (π/2, π)

问题 18. 证明 f(x) = x ³ – 3x ² + 3x – 100 给出的函数在 R 中是递增的。

解决方案：

f(x) = x³ – 3x² + 3x – 100

f'(x) = 3x² – 6x + 3

f'(x) = 3(x² – 2x + 1)

f'(x) = 3(x – 1)² ≥ 0 ∀ x in R

So f(x) is strictly increasing in R. 

问题 19. y = x ² e ^-x增加的区间是

(A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)

解决方案：

Given, f(x) = x²e^-x

f'(x) = x²(-e^-x) + e^-x.2x

f'(x) = e^-x(2x – x²)

f'(x) = e^-x.x(2 – x)

For f(x) to be increasing, f'(x) ≥ 0

So, f'(x) ≥ 0

e^-x.x.(2 – x) ≥ 0

x.(2 – x) ≥ 0

x(x – 2) ≥ 0

x ∈ [0, 2]

So, the f(x) is strictly increasing in interval (0, 2). Correct option in D.