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Given that x = 1/2 is a root of the equation 2x^3 – 9x^2 + kx – 13 = 0, find the value of k and the other roots of the equation.

Firstly, we note that a 'root' is simply a solution of the equation (at least, in this case). Let's start from what we've been told: x = 1/2 is a root of the equation. Since it's a solution, let's sub it in ...
EL
Answered by Eugene L. • Maths tutor
6120 Views

June 2008 C1 Paper Differentiation Question

a) ask student: purpose of differentiation (i), how to perform differentiation (ii) (i) to find the gradient of a curve (ii) multiply the coefficient of x by its indice then minus one from the indici perform...
NG
Answered by Nicola G. • Maths tutor
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Using the result: ∫(2xsin(x)cos(x))dx = -1⁄2[xcos(2x)-1⁄2sin(2x)] calculate ∫sin²(x) dx using integration by parts

Recall that ∫uv'=uv- ∫u'v Set u=sin²(x), v'=1 Therefore u'=2sin(x)cos(x) and v=x which gives us the following: ∫sin²(x)dx = xsin²(x) - ∫2xsin(x)cos(x)dx The second integral in the above expression is given i...
NM
Answered by Nick M. • Maths tutor
6023 Views

solve 2cos^2(x) - cos(x) = 0 on the interval 0<=x < 180

we start y factoring and solving for each equation: cos(x) (2cos(x) - 1) = 0 this means: cos(x) = 0 and cos(x) = 1/2 from the first equation we get: x = 90 and from the second equation using the known trigon...
DS
Answered by Dimitris S. • Maths tutor
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Express 2cos(x) + 5sin(x) in the form Rsin(x + a) where 0<a<90

Expanding Rsin(x + a): Rsin(x + a) = Rsin(x)cos(a) + Rcos(x)sin(a) Comparing coefficients of sin(x), cos(x) with first expression leads to: Rsin(a) = 2, Rcos(a) = 5 Dividing these equations gives: tan(a) = 2...
DH
Answered by Dan H. • Maths tutor
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