@Brooklyn said: sin^3A+cos^3A find its maximum and min values?
solve this one too guys
did anyone receive admit card??????
@ang264 said: did anyone receive admit card??????
sabka aa gaya hai. agar nahi aaya to download kar lo.
a*sinx + b*cosx lies between -(a^2 + b^2)^1/2 and (a^2 + b^2)^1/2
Standard result :D
@JS-M said:1+ sinA/(1-sinA) = 4+ 2rt3solve it ......sinA = rt3/2A = 60
Bhai how did u get 1+ sinA/(1-sinA)=4+2rt3???
@JS-M said:let the distance by which faulty engine traveled is X km and speed S km/h{80/S+X-80/(3/5S)} - X/3/5S = 1 3/5...........(1)X/2 = X-80/1.45............(2)find X from (2) and put in (1)bt option se kare to in NO TIME....
Bhai what is 80 and 13/5 in the equation"{80/S+X-80/(3/5S)} - X/3/5S = 1 3/5...........(1)"
some important formulas
1. if α and β are the roots of equation ax2 + bx + c = 0, roots of cx” + bx + a = 0 are 1/α and 1/β.
if α and β are the roots of equation ax2 + bx + c = 0, roots of ax2 – bx + c = 0 are -α and -β.
2. n(n + l)(2n + 1) is always divisible by 6.
3. 3^2n leaves remainder = 1 when divided by 8.
4. n^3 + (n + 1 )^3 + (n + 2 )^3 is always divisible by 9.
5. 10^(2n + 1) + 1 is always divisible by 11.
6. n(n^2- 1) is always divisible by 6.
7. n^2+ n is always even.
8. 2^3n-1 is always divisible by 7.
9. n^3 + 2n is always divisible by 3.
@mickym said: From a square piece of cardboard measuring 2 cm on each side , a box with no top is to be formed by cutting out from each corner a square with side b and bending up the flaps . The value of b for which box has the greatest volume ?
1/3 ??
10.3^4n – 4^3n is always divisible by 17.
11.Minimum value of a^2.sec^2Ɵ + b^2.cosec^2Ɵ is (a + b)^2; (0°
for eg. minimum value of 49 sec^2Ɵ + 64.cosec^2Ɵ is (7 + 8)^2 = 225.
12.among all shapes with the same perimeter a circle has the largest area.
13.if one diagonal of a quadrilateral bisects the other, then it also bisects the quadrilateral.
14.sum of all the angles of a convex quadrilateral = (n – 2)180°
15.number of diagonals in a convex quadrilateral = 0.5n(n – 3)
16.let P, Q are the midpoints of the nonparallel sides BC and AD of a trapezium ABCD.Then, ΔAPD = ΔCQB.
17.|x| = x if x ≥ 0 and |x| = – x if x ≤ 0
18.(a – b)^2n = (b – a)^2n and (a – b)^2n+1 = – (b – a)^2n+1
wat's ur approach@007pagal said:maximise b(2-2b)(2-2b)
& for that how r u getting b= 1/3
@007pagal said:maximise b(2-2b)(2-2b)
Table of Trigonometric Identities is attached below...
@mickym said:& for that how r u getting b= 1/3
after differentiating once and equating to zero i am getting two values 1/3 and 1,
and 1/3 gives -ve value after second diff.
and 1/3 gives -ve value after second diff.
@sh.arora10 said:Let the actual speed of train be X miles/hour.Let the initial distance covered by train is d1.After fault its speed will be reduced to 3X/5 miles/hour.So the distance travelled is (3X/5) * 2 = 6X/5 = d2 ----> (i)So total distance = X + 6X/5 = 11X/5Now if the fault had occurred at d1 + 503X/5 * 5/4 = d2 - 50 -----> (ii)subtracting (ii) from (i)(i) - (ii) = (6X/5) - (3X/4) = 509X/20 = 50 or X = 1000/9So the distance d2 = 133.33 milesSo d1 = 11*1000/(9*5) - 133.33 = 111.11So total distance is 133.33+111.11 = 244.44 miles
Plz explain this equation:3X/5 * 5/4 = d2 - 50 -----> (ii)
@007pagal said:after differentiating once and equating to zero i am getting two values 1/3 and 1, and 1/3 gives -ve value after second diff.
@007pagal said:after differentiating once and equating to zero i am getting two values 1/3 and 1, and 1/3 gives -ve value after second diff.
CAn u write the exp u got after 1st diff & 2nd diff
@sh.arora10 said:Let the actual speed of train be X miles/hour.Let the initial distance covered by train is d1.After fault its speed will be reduced to 3X/5 miles/hour.So the distance travelled is (3X/5) * 2 = 6X/5 = d2 ----> (i)So total distance = X + 6X/5 = 11X/5Now if the fault had occurred at d1 + 503X/5 * 5/4 = d2 - 50 -----> (ii)subtracting (ii) from (i)(i) - (ii) = (6X/5) - (3X/4) = 509X/20 = 50 or X = 1000/9So the distance d2 = 133.33 milesSo d1 = 11*1000/(9*5) - 133.33 = 111.11So total distance is 133.33+111.11 = 244.44 miles
Also explain this equation"So d1 = 11*1000/(9*5) - 133.33 = 111.11"??
initially 4b^3 - 8b^
after 1st diff: 12b^2 - 16b + 4
after 2nd: 24b - 16
@JS-M said:yr easy way ..........16/2{50+ (16-1)*5} = 1000now v need >1000, then it will be (1105)answer me error ni hai.....
But bhai start toh 25 rs karna hai na???