NECO 2024 · Paper 2 · Q5

(a)
If $y = (3x^3 + 2x^2 + 1)(3x^2 + 4)$ , find $\dfrac{dy}{dx}$ .
(b)
Given that $\sin(x + 30)^\circ = \cos(2x + 33)^\circ$ , find the value of $x$ .

Worked solution (try it first)

(a)

Use the product rule: if $y = uv$ , then $\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$ .
Here $u = 3x^3 + 2x^2 + 1$ , so $\frac{du}{dx} = 9x^2 + 4x$ , and $v = 3x^2 + 4$ , so $\frac{dv}{dx} = 6x$ .
Then $\frac{dy}{dx} = (9x^2 + 4x)(3x^2 + 4) + (3x^3 + 2x^2 + 1)(6x)$ .
Expand: $(27x^4 + 12x^3 + 36x^2 + 16x) + (18x^4 + 12x^3 + 6x)$ .
Collect like terms: $\frac{dy}{dx} = 45x^4 + 24x^3 + 36x^2 + 22x$ .

(b)

The sine of an angle equals the cosine of its complement, so when $\sin A = \cos B$ (both acute), $A + B = 90^\circ$ .
So $(x + 30) + (2x + 33) = 90$ .
Then $3x + 63 = 90$ , $3x = 27$ and $x = 9$ .

Report a problem with this question