Hướng dẫn Giải bài 5 (trang 10, SGK 12 Giải Tích)

Hướng dẫn Giải Bài 5 (Trang 10, SGK Giải Tích 12)

Câu hỏi: Chứng minh các bất đẳng thức sau: a) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>tan</mi><mfenced><mi>x</mi></mfenced><mo> </mo><mo>></mo><mi>x</mi><mo> </mo><mfenced><mrow><mn>0</mn><mo><</mo><mi>x</mi><mo><</mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></mrow></mfenced></math> b) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>tan</mi><mfenced><mi>x</mi></mfenced><mo> </mo><mo>></mo><mo> </mo><mi>x</mi><mo> </mo><mo>+</mo><mo> </mo><mfrac><msup><mi>x</mi><mn>3</mn></msup><mn>3</mn></mfrac><mo> </mo><mfenced><mrow><mn>0</mn><mo><</mo><mi>x</mi><mo><</mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></mrow></mfenced></math>   Hướng dẫn Giải: a) Hàm số f(x) = tanx - x liên tục trên nữa khoảng <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>[</mo><mo> </mo><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>)</mo></math> và có đạo hàm  f'(x) = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mrow><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mi>x</mi></mfenced></mrow></mfrac><mo> </mo><mo>-</mo><mo> </mo><mn>1</mn><mo>></mo><mn>0</mn><mo> </mo><mo>∀</mo><mi>x</mi><mo> </mo><mo>∈</mo><mo> </mo><mfenced><mrow><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></mrow></mfenced></math> Do đó f(x) đồng biến trên <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>[</mo><mo> </mo><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>)</mo><mo> </mo></math> Với 0 < x < <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></math> ta có f(x) > f(0) = 0 <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo></math> tanx > x; <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∀</mo><mi>x</mi><mo> </mo><mo>∈</mo><mfenced><mrow><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></mrow></mfenced></math> b) Hàm số g(x) = tanx - x - <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><msup><mi>x</mi><mn>3</mn></msup><mn>3</mn></mfrac></math> liên tục trên <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>[</mo><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>)</mo></math> có đạo hàm g'(x) = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mrow><msup><mi>cos</mi><mn>2</mn></msup><mfenced><mi>x</mi></mfenced></mrow></mfrac><mo>-</mo><mn>1</mn><mo>-</mo><msup><mi>x</mi><mn>2</mn></msup><mo> </mo><mo>=</mo><mo> </mo><msup><mi>tan</mi><mn>2</mn></msup><mi>x</mi><mo> </mo><mo>-</mo><msup><mi>x</mi><mn>2</mn></msup><mo> </mo><mo>=</mo><mo> </mo><mfenced><mrow><mi>tan</mi><mi>x</mi><mo>-</mo><mi>x</mi></mrow></mfenced><mfenced><mrow><mi>tan</mi><mi>x</mi><mo>+</mo><mi>x</mi></mrow></mfenced><mo> </mo><mo>></mo><mn>0</mn><mo>,</mo><mo> </mo><mo>∀</mo><mi>x</mi><mo> </mo><mo>∈</mo><mfenced><mrow><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></mrow></mfenced></math> Do đó g đồng biến trên <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>[</mo><mo> </mo><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac><mo>)</mo></math> Với 0 < x <<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></math> ta có g(x) > g(0) = 0 <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>tan</mi><mi>x</mi><mo> </mo><mo>></mo><mo> </mo><mi>x</mi><mo> </mo><mo>+</mo><mo> </mo><mfrac><msup><mi>x</mi><mn>3</mn></msup><mn>3</mn></mfrac><mo>;</mo><mo> </mo><mo>∀</mo><mi>x</mi><mo> </mo><mo>∈</mo><mfenced><mrow><mo> </mo><mn>0</mn><mo>;</mo><mo> </mo><mfrac><mi mathvariant="normal">π</mi><mn>2</mn></mfrac></mrow></mfenced></math>