The diagram shows the curve with parametric equations $$\[ x=\ln (2 t+3), \quad y=\frac{2 t-3}{2 t+3} . \]$$ The curve crosses the $$\(y\)$$-axis at the point $$\(A\)$$ and the $$\(x\)$$-axis at the point $$\(B\)$$. Find the gradient of the curve at $$\(B\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No:9709_w21_qp_21 Year:2021 Question No:5(c)
Answer:
Attempt to find value of \(t\) corresponding to \(y=0\)
Obtain \(t=\frac{3}{2}\) and hence gradient is 1
Obtain \(t=\frac{3}{2}\) and hence gradient is 1
Knowledge points:
1.7.1 understand the gradient of a curve at a point as the limit of the gradients of a suitable sequence of chords, and use the notations for first and second derivatives (Only an informal understanding of the idea of a limit is expected.)
2.2.1 understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base)
Solution:
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