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Mathematics: Post your doubts here!
I am getting the same answer as yours but the answer listed at the back of the book is [21(15)^(1/2)]/5] :confused: -
Mathematics: Post your doubts here!
I am getting the same answer i.e '5/13'. Answers written at the back of the book are incorrect sometimes. -
Mathematics: Post your doubts here!
AB = OB - OA AB = (2)(-2)(11) - (-1)(2)(5) AB = (3)(-4)(6) '(3)(-4)(6)' is the direction of the plane. 3x - 4y + 6z = k The plane passes through 'b' therefore you can find the value of 'k' using the coordinates of 'b'. 3x - 4y + 6z = k 3(2) - 4(-2) + 6(11) = k 80 = k Put back the value... -
Mathematics: Post your doubts here!
Q7(i). u = 2 + i u* = 2 - i u + u * = 2 + 1 + 2 - i = 4 + 0i Mark the points O (0,o), A (2,1), B (2,-1) and C (4,0) on an argand diagram and join all the points. The resulting shape will be Parallelogram with adjacent sides equal. Q7(ii). u/u* (2 + i)/(2 - i) [(2 + i)/(2 - i)] x [(2 +... -
Mathematics: Post your doubts here!
The attached file has a question from June 2010 while as, above, you had asked for a question from November 2009. :confused: -
Mathematics: Post your doubts here!
Q10(i). (dA/dt) = k [(4/3)πr^3] (dA/dt) = (dr/dt) x (dA/dr) A = 4πr^2 (dA/dr) = 8πr (dr/dt) = 2 (dA/dt) = (dr/dt) x (dA/dr) (dA/dt) = (2) x (8πr) (dA/dt) = 16πr (dA/dt) = k [(4/3)πr^3] (dr/dt) x (dA/dr) = k [(4/3)πr^3] 16πr = k [(4/3)πr^3] Substitute '5' in the place of 'r' in the... -
Mathematics: Post your doubts here!
Please view the attached .pdf file, the scanned solution of Q10's inside it. -
Mathematics: Post your doubts here!
r = (4)(2)(-1) + t(2)(-1)(-2) - (0)(2)(4) r = (4)(0)(-5) + t (2)(-1)(-2) r = (4+2t) (-t) (-5-2t) (4+2t) . (2) (-t) . (-1) = 0 (-5-2t) . (-2) 8 + 4t + t + 10 + 4t = 0 t = -2 r = (4+2t) (-t) (-5-2t) r = (0)(2)(-1) Finding the modulus. Modulus = [ (0)^2 + (2)^2 + (-1)^2 ]^(1/2)... -
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Mathematics: Post your doubts here!
Q7(i). (e^2t)(dx/dt) = cos^2 x [1/(cos^2 x)] dx = (e^-2t) dt (sec^2 x) dx = (e^-2t) dt Integrate both the sides. tan x = -(e^-2t)/2 + c Put 'x=0' and 't=0' in the above equation in order to find the value of the constant 'c'. tan x = -(e^-2t)/2 + c 0 = -1/2 + c 1/2 = c Put back the... -
Mathematics: Post your doubts here!
Take the cross product of the direction of the line and the direction of the plane and equate it to '0'. (2).(2) (-1).(b) = 0 (-2).(c) 4 - b- 2c = 0 4 = b + 2c Simultaneously solve both the obtained equations to find the value of 'b' and ''c'. 4 = b + 2c 4 - 2c = b c - 2b = 7 c -... -
Mathematics: Post your doubts here!
Q6, June 2008. xy(x+y) = 2a^3 x^2y + xy^2 = 2a^3 Differentiating. 2xy + (x^2)(dx/dy) + y^2 + (2xy)(dy/dx) = 0 (x^2 + 2xy)(dy/dx) = - y^2 - 2xy dy/dx = ( - y^2 - 2xy )/(x^2 + 2xy) As the tangent is parallel to the x-axis, dy/dx=0. dy/dx = ( - y^2 - 2xy )/(x^2 + 2xy) 0 = ( - y^2 - 2xy... -
Mathematics: Post your doubts here!
Q6(ii), June 2007 2 sin x = x 2 sin x - x Substitute 'π/2' in the place of 'x'. 2 sin x - x 2 sin π/4 - π/4 0.628 Substitute '2π/3' in the place of 'x'. 2 sin x - x 2 sin 2π/3 - 2π/3 -0.36 There's a simple rule of verifying. In iteration, you'll be given two values and a formula... -
Mathematics: Post your doubts here!
All what you'll need to do is differentiate the gradient of the normal to the curve and to find its maximum point, equate it to '0' and obtain two values of x. y = (1 + x)√(1 − x^2) (dy/dx) = (1+x)[(-2x)/(2√(1 − x^2)] + 1[√(1 − x^2] 0 = (-x)(1+x) + 1 - x^2 0 = -x - x^2 + 1 -x^2 0 -2x^2 - x +... -
Mathematics: Post your doubts here!
Q6(i), June 2007 Area of triangle AOB: A = (1/2) a b sin x A = (1/2) r^2 sin x Area of the sector: s = r x A = (1/2) r x s A = (1/2) r x rx A = (1/2) r^2 x Area of the Triangle = (1/2) Area of the sector (1/2) r^2 sin x = (1/4) r^2 x sin x = x/2 2 sin x = x Q6(ii), June 2007 2... -
Mathematics: Post your doubts here!
If you've done the first part of this question, you must have found out the direction vector of the plane ABC which comes out to be (4,2,1). We'll use the same method to find the direction vector of the plane OAB and then find the acute angle between these two planes. Plane OAB. AO = (-2, 0... -
Mathematics: Post your doubts here!
Q2: e^x + e^2x = e^3x Take 'y=e^x'. e^x + e^2x = e^3x y + y^2 - y^3 = 0 y ( 1 + y - y^2) = 0 - y^2 + y + 1 = 0 y^2 - y - 1 = 0 a=1, b=-1, c=-1 y = [-b+-√(b^2 - 4ac)]/2a y = [1+-√(5)]/2 y = 1.62 Put this value of 'y' back into the equation 'y = e^x' to calculate the value of 'x'... -
Mathematics: Post your doubts here!
This's the diagram with the correct shading: First of all, you'll construct a circle of radius '2' with centre at the origin (0,0). According to the inequality '|z|<2', area inside the circle will be shaded. Next, you'll construct a perpendicular bisector for the inequality '|z-u^2|<|z-u|'... -
Mathematics: Post your doubts here!
Q3, June 2007: y = x sin 2x x=π/4 Keep 'x=π/4' in the curve equation to find the value of 'y'. y = x sin 2x y = π/4 sin 2π/4 y = π/4 Differentiate the curve equation and then keep 'x=π/4' to find the value of the gradient y = x sin 2x (dy/dx) = sin 2x + 2x cos 2x (dy/dx) = sin 2π/4 +... -
Mathematics: Post your doubts here!
Simply plug in the values 'π/2' and '2π/3' in the place of 'x' in the equation 'x - 2 sin x'. One value will come out to be negative while as the other will come out to be positive if 'a' satisfies this equation. x - 2 sin x Put 'π/2' in the place of 'x'. π/2 - 2 sin π/2 - 0.43 Now put...