The gradient of a curve is given by $$$\frac{\mathrm{d} y}{\mathrm{~d} x}=6(3 x-5)^{3}-k x^{2}$$$, where $$$k$$$ is a constant. The curve has a stationary point at $$$(2,-3.5)$$$. Find the value of $$$k$$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_s21_qp_12 Year：2021 Question No：11(a)

Answer:

At stationary point $\frac{\mathrm{d} y}{\mathrm{~d} x}=0$ so $6(3 \times 2-5)^{3}-k \times 2^{2}=0$
$
[k=] \frac{3}{2}
$

Knowledge points:

1.7.4 locate stationary points and determine their nature, and use information about stationary points in sketching graphs. (Including use of the second derivative for identifying maxima and minima; alternatives may be used in questions where no method is specified.) (Knowledge of points of inflexion is not included.)

Answer:

Knowledge points:

Solution:

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