OCR A-Level Maths A Pure calculus overview quiz quiz

$\cos 4x$

$4\cos 4x$

$-4\cos 4x$

$\frac{1}{4}\cos 4x$

$e^x$

$xe^x$

$e^x(1 + x)$

$e^x(x - 1)$

minimum

point of inflection

cannot tell

maximum

$\ln|x| + c$

$-\frac{1}{x^2} + c$

$\frac{x^2}{2} + c$

$1 + c$

$6$

$8$

$9$

$4$

All of the ordinates, including the very first one and the very last one in the sum

Only the first and the last ordinates, with all of the interior ones left as they are

Only the interior ordinates

None of the ordinates at all, since every one of them is simply added once in the sum

is the factor that is easiest to integrate, so that finding $v$ takes the least work

is always the exponential factor in the product, whatever the other factor happens to be

has the highest power of $x$, so that differentiating it lowers the power each time

simplifies when differentiated

$\int u^3\,du$

$\int 2u^3\,du$

$\int u^3\,dx$

$\int x u^3\,du$

$P = kt + c$

$P = P_0 e^{kt}$

$P = P_0 + kt$

$P = e^{kt} + c$

It guarantees the function is differentiable, so its gradient exists everywhere on the interval

It makes the function increasing throughout the interval, so it can only cross zero once

It guarantees the function actually crosses zero between the values

It is not needed at all, since opposite signs alone are enough to guarantee a root exists

$f(x_n)$ is large, since a large function value pushes the next estimate too far away

the starting value is exact, because then the method has nothing left to do and stalls

the function is a polynomial, since the method only works on trigonometric functions

$f'(x_n)$ is zero or very small

$\int_0^2 (2x - x^2)\,dx$

$\int_0^2 (x^2 - 2x)\,dx$

$\int_0^2 x^2\,dx$

$\int_0^2 2x\,dx$