{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MA1505_08S2_Test_Outline_Solutions[1]

# MA1505_08S2_Test_Outline_Solutions[1] - MA 1505 Mathematics...

This preview shows pages 1–3. Sign up to view the full content.

MA 1505 Mathematics 1 2008/2009 Semester 2 Mid-term Test Outline Solutions Question 1. Routine key steps : dy dx = 4 x 3 - 8 x 3 . At x = 1, the slope of the tangent is - 4. Equation of the tangent is y = 9 - 4 x . The y -intercept is p = 9. Question 2. Similar to Tutorial 1, Question 4 ( c ): dy dx = dy dt dx dt = 1 + sin t cos t d dt dy dx = · · · = 1 + sin t cos 2 t . Thus, d 2 y dx 2 = 1 + sin t cos 2 t dx dt = 1 + sin t cos 3 t . Question 3. If x denotes the length of the side of the rectangular area parallel to the road, then the total length of fence required is f ( x ) = x + 2 5 , 000 x = x + 10 , 000 x , where x > 0 . Then f 0 ( x ) = 1 - 10 , 000 x 2 and f 00 ( x ) = 20 , 000 x 3 > 0 . Now, for x > 0 we have f 0 ( x ) = 0 ⇐⇒ x = 100 gives local (and absolute) minimum . Thus, f (100) = 200.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
MA1505 Mid-term Test Outline Solutions Question 4. Let y = x x - e 3 x . Then ln y = x ln x x - e 3 = ln x x - e 3 1 x . By L’Hopital’s rule, lim x + ln y = lim x + x - e 3 x · - e 3 ( x - e 3 ) 2 - 1 x 2 = lim x + xe 3 x - e 3 = e 3 . Thus, lim x + x x - e 3 x = e e 3 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

MA1505_08S2_Test_Outline_Solutions[1] - MA 1505 Mathematics...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online