A MATH HELP! - Singapore Forums by SGClub.com  Home Photos Member List Register Mark Forums Read  Why aren't you a member of SGClub.com yet?? » Join 130,000+ other members in chatting. » Make lots of new friends here. » Keep up-to-date with current events. » Participate in Club outings. » Download lots of Free Stuff! Registration just takes 2mins and is absolutely free so join our community today! I Want to Choose my Own Personal Nickname Now! 22nd June 2013, 11:23 AM #1 (permalink) New SGClubber Posts: 23 Join Date: Nov 2012 Likes: 17 Liked 3 Times in 2 Posts Gender: A MATH HELP! Tweet 1. Given that x^1/4 + x^-1/4 = 3, find the value of x + 1/x 2. The points (5, -1) and (5,15) are on the circumference of a circle whose centre, C, lies above the x-axis. The line y = -3 is a tangent to the circle. Find (i) the radius of the circle, (ii) the two possible x-coordinates of C, (iii) the two possible equations of the circle. 3. The figure shows part of the graphs of the curve y = e^2x+1 and y= 2cosx. Find the (a) coordinates of A and (b) the shaded area. THANKS! Sponsors: Last edited by snowmine; 22nd June 2013 at 11:23 AM.    22nd June 2013, 11:44 AM #2 (permalink) Experienced SGClubber    Posts: 2,404 Join Date: Jul 2009 Likes: 0 Liked 211 Times in 174 Posts Gender: Re: A MATH HELP! Tweet Worked out the first one for you: x^1/4 + x^-1/4 = 3 ------------(1) Taking fourth power on both sides for (1): we have x + 4x^(1/2) + 6 + 4x^(-1/2) + 1/x = 81 x +1/x + 4 [ x^(1/2) + 2 + x^(-1/2) ] = 75 ------------(2) Squaring both sides for (1): x^(1/2) + 2 + x^(-1/2) = 9 x^(1/2) + 2 + x^(-1/2) = 7 ----------------(3) Substituting (3) into (2): x +1/x + 4(7) = 75 Hence, x +1/x = 47 (shown) Hope this helps. Peace. __________________ White Group Mathematics Free A Level H2 Maths Resource Site Last edited by whitecorp; 22nd June 2013 at 11:44 AM.   Members who Liked this post by whitecorp: snowmine (22nd June 2013) 22nd June 2013, 12:03 PM #3 (permalink) Experienced SGClubber    Posts: 2,404 Join Date: Jul 2009 Likes: 0 Liked 211 Times in 174 Posts Gender: Re: A MATH HELP! Tweet Here's the solution for Q2: If the centre of the circle is denoted by (a, b), recognize that the point of tangency for the line y=-3 with the circle is actually ( a, -3) . (ie either the highest or the lowest point on the circle) Equation of circle is given by (x-a)^2 +(y-b)^2 = r^2 ----------(1) Now we know that the points (a ,-3) , (5, -1) and (5, 15) lie on the circle, substituting these into (1) gives (-3-b)^2 = r^2 ====> (3+b)^2 =r^2 ----------(2) (5-a)^2 +(-1-b)^2 = r^2 =======> (5-a)^2 +(1+b)^2 = r^2 ----------(3) (5-a)^2 +(15-b)^2 = r^2 ----------(4) (4)-(3): (15-b)^2 - (1+b)^2 =0 Expanding gives -32b + 224 =0, ie b=7 Substituting b=7 into (2) gives r=10 Hence radius of circle is 10 units. (shown) Substituting b=7 and r=10 into (3) gives (5-a)^2 +64 = 100 (5-a)^2 = 36 5-a =6 or 5-a =-6 therefore, the two possible x-coordinates of the circle are a=-1 or a= 11 (shown) Lastly, with all the information in place, the two possible equations of the circle are given by (x+1)^2 +(y-7)^2 = 100 or (x-11)^2 +(y-7)^2 = 100 (shown) Hope this helps. Peace. __________________ White Group Mathematics Free A Level H2 Maths Resource Site Last edited by whitecorp; 22nd June 2013 at 12:05 PM.   Members who Liked this post by whitecorp: snowmine (22nd June 2013) 22nd June 2013, 12:19 PM #4 (permalink) Addicted SGClubber  Posts: 874 Join Date: Sep 2010 Likes: 6 Liked 48 Times in 41 Posts Gender: Re: A MATH HELP! Tweet 1.Squaring both sides, (x^(1/4)+x^(-1/4))² =x^(1/2)+2x^(1/4)x^(-1/4)+x^(-1/2) =x^(1/2)+x^(-1/2)+2 =3²=9 -> x^(1/2)+x^(-1/2)=7 Squaring both sides again, x+2x^(1/2)x^(-1/2)+x^(-1)=49 x+x^(-1)+2=49 x+1/x=47 2. (i) The easiest way I see to get the radius of the circle is to see that since y=-3 is 2 units below (5,-1), then the higher horizontal tangent will be 2 units above (5,15), which is y=17. So the radius is (17+3)/2=10 units. (ii) The y-coordinate of the 2 centres is easily found by 17-10=7. There are multiple ways I see to get the x-coordinate, the fastest that I can see is this: Let the centre of C2 be (a,7). See that a-5 is the horizontal distance between C2 and the vertical line cutting through (5,-1) and (5,15). By Pythagoras theorem, (a-5)²+((15+1)/2)²=10² which gives a-5=6 and a=11. The x-coordinate of C1 is 5-6=-1. So the 2 centres are (-1,7) and (11,7) (iii)Should be fairly straightforward. 3. I think you're missing some info, can't get A. Edit: Would like to point out that my solution for Q2 lacks rigour, whitecorp's solution would be more 'complete'. __________________ Are you ready to get serious? Last edited by Betakuwe; 22nd June 2013 at 12:33 PM.   Members who Liked this post by Betakuwe: snowmine (22nd June 2013) 22nd June 2013, 10:15 PM #5 (permalink) New SGClubber Posts: 23 Join Date: Nov 2012 Likes: 17 Liked 3 Times in 2 Posts Gender: Re: A MATH HELP! Tweet Thanks whitecorp and betakuwe for your help!    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