University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

125

Pages:

4
Author:

Shahbaz A.

Cambridge International AS and A Level 9709-22 Paper 2 Pure Mathematics 2 February-March 2023 shahbaz ahmed July 2024 Q6 Cambridge International AS and A Level The diagram shows the curve with equation y = 4e2x +9 ex +2 . point M and crosses the y-axis at the point P. (a) Find the exact value of the gradient of the curve at P. (b) Find the exact coordinates of M. Solution 1 The curve has a minimum y= 4e2x + 9 ex + 2 Let f (x) = 4e2x + 9 d[f (x)] dx = d(4e2x +9) dx d[f (x)] dx = d(4e2x ) dx d[f (x)] dx = 4 d(edx d[f (x)] dx = 4e2x d(2x) dx d[f (x)] dx dx = 4 × 2e2x dx d[f (x)] dx = 8e2x 2x + ) d dx (9) +0 Similarly Let g(x)=ex + 2 d[g(x)] dx = d(ex +2) dx 2 d[g(x)] dx = d(ex ) dx + d[g(x)] dx = d(ex ) dx +0 d[g(x)] dx = ex d(2) dx Putting in the formula d d f (x) − f (x) dx g(x) g(x) dx dy = dx [g(x)]2 (ex + 2)(8e2x ) − (4e2x + 9)ex dy = dx [e2 + 2]2 At x=0 e0 = 1 Hence dy dx = (1+2)(8)−(4+9) (1+2)2 dy dx = 24−13 9 Since dy dx = 11 9 at point P is gradient at the same Gradient at P= 11 9 (b) Since M is the minimum point dy (ex + 2)(8e2x ) − (4e2x + 9)ex = =0 dx [e2 + 2]2 =⇒ (ex + 2)(8e2x ) − (4e2x + 9)ex = 0 ex [(ex + 2)(8ex ) − (4e2x + 9)] = 0 ex = 0 =⇒ x −→ −∞ [(ex + 2)(8ex ) − (4e2x + 9)] = 0 8e2x + 16ex − 4e2x − 9 = 0 4e2x + 16ex − 9 = 0 4e2x + (18 − 2)ex − 9 = 0 3 4e2x + 18ex − 2ex − 9 = 0 4e2x + 18ex − 2ex − 9 = 0 2ex (2ex + 9) − 1(2ex + 9) = 0 (2ex + 9)(2ex − 1) = 0 2ex + 9 = 0 has no solution 2ex − 1 = 0 1 2 ex = e2x = 1 4 Putting in y= y= 4× 14 +9 1 2 +2 y= 10 4e2x + 9 ex + 2 5 2 y = 10 × 2 5 =4 Also ex = 1 2 ln ex = ln 12 x ln e = ln 21 Since ln e = 1 x = ln 12 Coordinates of the point M are x = ln 12 y=4 4