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Solve sec(x)^2-2*tan(x)=4 for 0<=x<=360

We know sin(x)^2+cos(x)^2=1Dividing by cos(x)^2: tan(x)^2+1=sec(x)^2Substitute into the Equation and Rearrange to get: tan(x)^2-2*tan(x)-3=0Let y = tan(x): y^2-2y-3=0Factorising: (y-3)(y+1)=0so y = 3 and ...

Answered by Mukesh R. • Maths tutor

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The second and fourth term of a geometric series is 100 and 225 respectively. Find the common ratio and first term of the series. Round your answer to 2 d.p if necessary

Formula for a Geometric series for term n = arⁿ, where a = the first term and r = common ratio.Therefore, with the information given we can write that ar² = 100 and ar⁴ =...

Answered by • Maths tutor

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Find the area contained under the curve y =3x^2 - x^3 between 0 and 3

Equation of curve is: y = y =3x² - x³To find area need to integrate between 0 and 3So integrating each term gives x³ - x⁴/4 + ...

Answered by Juan R. • Maths tutor

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A projectile is thrown from the ground at 30 degrees from the horizontal direction with an initial speed of 20m/s. What is the horizontal distance travelled before it hits the ground? Take the acceleration due to gravity as 9.8m/s^2

Draw diagram outlining the symmetric parabolic shape of the projectile's motion. Find vertical component of the initial speed using SOH CAH TOA. sin(30) = opposite/hypotenuse = x/30therefore, x = 30sin(30...

Answered by Raphael D. • Maths tutor

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The curve C has equation y = 3x^4 – 8x^3 – 3 Find (i) dy/dx (ii) the co-ordinates of the stationary point(s)

i) dy/dx=12x^3-24x^2ii) the stationary points occur when dy/dx = 0 so we must find the solutions to 12x^3-24x^2=0.12x^3-24x^2= 12x^2(x-2)=0Therefore our stationary points are when 12x^2=0 ie x=0 and x-2=0...

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