第 12 类 RD Sharma 解决方案 – 第 17 章增加和减少函数 – 练习 17.2 |设置 2

问题 11. 证明 f(x) = cos ² x 是 (0, π/2) 上的减函数。

解决方案：

We have,

f(x) = cos² x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(cos^2 x)$

f'(x) = 2 cos x (– sin x)

f'(x) = – sin 2x

Now for 0 < x < π/2,

=> sin 2x > 0

=> – sin 2x < 0

=> f'(x) < 0

Thus, f(x) is decreasing on x ∈ (0, π/2).

Hence proved.

问题 12. 证明 f(x) = sin x 是 (–π/2, π/2) 上的递增函数。

解决方案：

We have,

f(x) = sin x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(sin x)$

f'(x) = cos x

Now for –π/2 < x < π/2,

=> cos x > 0

=> f'(x) > 0

Thus, f(x) is increasing on x ∈ (–π/2, π/2).

Hence proved.

问题 13. 证明 f(x) = cos x 是 (0, π) 上的减函数，在 (–π, 0) 上增加且在 (–π, π) 上既不增加也不减少。

解决方案：

We have,

f(x) = cos x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(cos x)$

f'(x) = – sin x

Now for 0 < x < π,

=> sin x > 0

=> – sin x < 0

=> f’(x) < 0

And for –π < x < 0,

=> sin x < 0

=> – sin x > 0

=> f’(x) > 0

Therefore, f(x) is decreasing in (0, π) and increasing in (–π, 0).

Hence f(x) is neither increasing nor decreasing in (–π, π).

 Hence proved.

问题 14. 证明 f(x) = tan x 是 (–π/2, π/2) 上的增函数。

解决方案：

We have,

f(x) = tan x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(tan x)$

f'(x) = sec² x

Now for –π/2 < x < π/2,

=> sec² x > 0

=> f’(x) > 0

Thus, f(x) is increasing on interval (–π/2, π/2).

Hence proved.

问题 15. 证明 f(x) = tan ^–1 (sin x + cos x) 是区间 (π/4, π /2) 上的减函数。

解决方案：

We have,

f(x) = tan^–1 (sin x + cos x)

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(tan^{-1} (sin x + cos x))$

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

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

f'(x) = $\frac{cosx-sinx}{1+sin^2x+cos^2+2sinxcosx}$

f'(x) = $\frac{cosx-sinx}{1+1+2sinxcosx}$

f'(x) = $\frac{cosx-sinx}{2+2sinxcosx}$

f'(x) = $\frac{cosx-sinx}{2(1+sinxcosx)}$

Now for π/4 < x < π/2,

=> $\frac{cosx-sinx}{2(1+sinxcosx)}$ < 0

=> f’(x) < 0

Thus, f(x) is decreasing on interval (π/4, π/2).

Hence proved.

问题 16. 证明函数f(x) = sin (2x + π/4) 在 (3π/8, 5π/8) 上递减。

解决方案：

We have,

f(x) = sin (2x + π/4)

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(sin (2x + π/4))$

f'(x) = 2 cos (2x + π/4) 

Now we have, 3π/8 < x < 5π/8

=> 3π/4 < 2x < 5π/4

=> 3π/4  + π/4 < 2x + π/4  < 5π/4 + π/4

=> π < 2x + π/4 + 3π/2

As, 2x + π/4 lies in 3rd quadrant, we get,

=> cos (2x + π/4) < 0

=> 2 cos (2x + π/4) < 0

=> f'(x) < 0

Thus, f(x) is decreasing on interval (3π/8, 5π/8).

Hence proved.

问题 17. 证明函数f(x) = cot ^–1 (sin x + cos x) 在 (0, π/4) 上增加并在 (π/4, π/2) 上减少。

解决方案：

We have,

f(x) = cot^–1 (sin x + cos x)

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(cot^{-1} (sin x + cos x))$

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

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

f'(x) = $\frac{cosx-sinx}{1+sin^2x+cos^2+2sinxcosx}$

f'(x) = $\frac{cosx-sinx}{1+1+2sinxcosx}$

f'(x) = $\frac{cosx-sinx}{2+2sinxcosx}$

f'(x) = $\frac{cosx-sinx}{2(1+sinxcosx)}$

Now for π/4 < x < π/2,

=> $\frac{cosx-sinx}{2(1+sinxcosx)}$ < 0

=> cos x – sin x < 0

=> f’(x) < 0

Also for 0 < x < π/4,

=> $\frac{cosx-sinx}{2(1+sinxcosx)}$ > 0

=> cos x – sin x > 0

=> f'(x) > 0

Thus, f(x) is increasing on interval (0, π/4) and decreasing on intervals (π/4, π/2).

Hence proved.

问题 18. 证明 f(x) = (x – 1) e ^x + 1 是所有 x > 0 的递增函数。

解决方案：

We have,

f(x) = (x – 1) e^x + 1

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}((x - 1) e^x + 1)$

f'(x) = e^x + (x – 1) e^x

f'(x) = e^x(1+ x – 1)

f'(x) = x e^x

Now for x > 0,

⇒ e^x > 0

⇒ x e^x > 0

⇒ f’(x) > 0

Thus f(x) is increasing on interval x > 0.

Hence proved.

问题 19. 证明函数x ² – x + 1 在 (0, 1) 上既不增加也不减少。

解决方案：

We have,

f(x) = x² – x + 1

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^2 - x + 1)$

f'(x) = 2x – 1 + 0

f'(x) = 2x – 1

Now for 0 < x < 1/2, we have

=> 2x – 1 < 0

=> f(x) < 0

Also for 1/2 < x < 1,

=> 2x – 1 > 0

=> f(x) > 0

Thus f(x) is increasing on interval (1/2, 1) and decreasing on interval (0, 1/2).

Hence, the function is neither increasing nor decreasing on (0, 1).

Hence proved.

问题 20. 证明 f(x) = x ⁹ + 4x ⁷ + 11 是所有 x ∈ R 的递增函数。

解决方案：

We have,

f(x) = x⁹ + 4x⁷ + 11

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^9 + 4x^7 + 11)$

f'(x) = 9x⁸ + 28x⁶ + 0

f'(x) = 9x⁸ + 28x⁶ 

f'(x) = x⁶ (9x² + 28) 

As it is given, x ∈ R, we get,

=> x⁶ > 0

Also, we can conclude that,

=> 9x² + 28 > 0

This gives us, f'(x) > 0.

Hence, the function is increasing on the interval x ∈ R.

Hence proved.

问题 21. 证明 f(x) = x ³ – 6x ² + 12x – 18 在 R 上增加。

解决方案：

We have,

f(x) = x³ – 6x² + 12x – 18

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^3 - 6x^2 + 12x - 18)$

f'(x) = 3x² – 12x + 12 –  0

f'(x) = 3x² – 12x + 12 

f'(x) = 3 (x² – 4x + 4)

f'(x) = 3 (x – 2)²

Now for x ∈ R, we get,

=> (x – 2)² > 0

=> 3 (x – 2)² > 0

=> f'(x) > 0

Hence, the function is increasing on the interval x ∈ R.

Hence proved.

问题 22. 说明函数f(x) 何时在区间 [a, b] 上增加。测试函数f(x) = x ² – 6x + 3 是否在区间 [4, 6] 上增加。

解决方案：

A function f(x) is said to be increasing on an interval [a, b] if f(x) > 0.

We have,

f(x) = x² – 6x + 3

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(x^2 - 6x + 3)$

f'(x) = 2x – 6 + 0

f'(x) = 2x – 6

f'(x) = 2(x – 3)

Now for x ∈ [4, 6], we get,

=> 4 ≤ x ≤ 6

=> 1 ≤ (x – 3) ≤ 3

=> x – 3 > 0

=> f'(x) > 0

Thus, the function is increasing on the interval [4, 6].

Hence proved.

问题 23. 证明 f(x) = sin x – cos x 是 (–π/4, π/4) 上的递增函数。

解决方案：

We have,

f(x) = sin x – cos x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(sin x - cos x)$

f'(x) = cos x + sin x

f'(x) = $\sqrt{2}[\frac{1}{\sqrt{2}}cosx+\frac{1}{\sqrt{2}}sinx]$

f'(x) = $\sqrt{2}[sin\frac{\pi}{4}cosx+cos\frac{\pi}{4}sinx]$

f'(x) = $\sqrt{2}sin(\frac{\pi}{4}+x)$

Now we have, x ∈ (–π/4, π/4)

=> –π/4 < x < π/4

=> 0 < (x + π/4) < π/2

=> sin 0 < sin (x + π/4) < sin π/2

=> 0 < sin (x + π/4) < 1

=> sin (x + π/4) > 0

=> √2 sin (x + π/4) > 0

=> f'(x) > 0

Thus, the function is increasing on the interval (–π/4, π/4).

Hence proved.

问题 24. 证明 f(x) = tan ^–1 x – x 是 R 上的递减函数。

解决方案：

We have,

f(x) = tan^–1 x – x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(tan^{-1} x - x)$

f'(x) = $\frac{1}{1+x^2}-1$

f'(x) = $\frac{1-1-x^2}{1+x^2}$

f'(x) = $\frac{-x^2}{1+x^2}$

Now for x ∈ R, we have,

=> x² > 0 and 1 + x² > 0

=> $\frac{x^2}{1+x^2}$ > 0

=> $\frac{-x^2}{1+x^2}$ < 0

=> f'(x) < 0

Thus, f(x) is a decreasing function on the interval x ∈ R.

Hence proved.

问题 25. 确定 f(x) = –x/2 + sin x 在 (–π/3, π/3) 上是增函数还是减函数。

解决方案：

We have,

f(x) = –x/2 + sin x

On differentiating both sides with respect to x, we get

f'(x) = $\frac{d}{dx}(\frac{-x}{2} + sin x)$

f'(x) = $\frac{-1}{2} + cosx$

Now, we have

=> x ∈ (–π/3, π/3)

=> –π/3 < x < π/3

=> cos (–π/3) < cos x < cos (π/3)

=> 1/2 < cos x < 1/2

=> $\frac{-1}{2} + cosx$ > 0

=> f'(x) > 0

Thus, f(x) is a increasing function on the interval x ∈ (–π/3, π/3).

Hence proved.