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An object of mass 3kg is held at rest on a rough plane. The plane is inclined at 30º to the horizontal and has a coefficient of friction of 0.2. The object is released, what acceleration does the object move with?

We need to use Newtons law F=ma going down the slope.
We can see that the only forces acting in this direction are the component of the weight and friction, so we have that: F = Wsin30 - μR = 3a

Answered by Asha D. • Maths tutor

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Show that (sec(x))^2 /(sec(x)+1)(sec(x)-1) can be written as (cosec(x))^2.

( sec²(x))/((sec(x)+1)(sec(x)-1))Then, by the rule of 'difference of two squares', we know that this equals= (sec²(x))/(sec²(x)-1)= (sec²x/tan²x)s...

Answered by Rishi S. • Maths tutor

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Find the values of A between and including 0 and 360 degrees for tan(2A) = 3tan(A)

You cannot work with this equation in the current form so you must use identities to find an equivalent form that you can work with. It is known that tan(2A) = 2tan(A) / 1-tan²(A) so set this e...

Answered by Daniel M. • Maths tutor

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Express the equation cosecθ(3 cos 2θ+7)+11=0 in the form asin^2(θ) + bsin(θ) + c = 0, where a, b and c are constants.

We must first use the identity cosecθ = 1/sinθ. Now the equation becomes (1/sinθ)(3 cos 2θ+7)+11=0. Since we know that the question is asking for the answer in the form of asin²θ + bsinθ + c = ...

Answered by George L. • Maths tutor

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Explain why for any constant a, if y = a^x then dy/dx = a^x(ln(a))

So let's start with taking the natural log on both sides of y=a^x, giving us ln(y) = ln(a^x). Using the laws of logarithms we can write this as ln(y) = xln(a).Next, we differentiate bo...

Answered by James M. • Maths tutor

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