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

2024
Views:

465

Pages:

3
Author:

Shahbaz A.

6663-01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary Differentiation: shahbaz ahmed July 2024 Q 5. The curve C with equation y = f (x), x ̸= 0, passes through the point (3, 7 21 ) , Given that f ′ (x) = 2x + 3 x2 , (a) find f(x). (b) Verify that f(–2) = 5. (c) Find an equation for the tangent to C at the point (–2, 5), giving your answer in the form ax + by + c = 0, where a, b and c are integers. Solution (a) 1 (3, 7 12 ) = (3, 15 2 ) f ′ (x) = 2x + d(f (x)) dx 3 x2 = 2x + 3x−2 d(f (x)) = (2x + 3x−2 )dx d(f (x)) = 2x dx + 3x−2 dx R R d(f (x)) = [2x dx + 3x−2 dx] R R f (x) = 2 x dx + 3 x−2 dx −1 2 f (x) = 2[ x2 ] + 3[ x−1 ] + C f (x) = x2 − 3 x +C 3 3 +C Putting x=3 f (3) = 32 − 15 2 =9−1+C 15 2 =8+C 8+C = C= 15 2 15 2 −8 2 C= 15−16 2 = − 12 Since f (x) = x2 − 3 x +C Putting C = − 12 f (x) = x2 − 3 x − 1 2 3 x − 1 2 (b) Since f (x) = x2 − Putting x = −2 3 )− f (−2) = (−2)2 − ( −2 f (−2) = 4 + 3 2 − 1 2 1 2 f (−2) = 4 + 1 = 5 (c) Given that f ′ (x) = 2x + 3 x2 , Putting x=-2 Given that f ′ (−2) = 2(−2) + 3 (−2)2 , f ′ (−2) = −4 + 34 , f ′ (−2) = −16+3 4 = − 13 4 = m, Where m is the slope of the tangent through the point (−2, 5) Equation of the tangent through the point (−2, 5) with slope m = − 13 4 y − 5 = − 13 4 [x − (−2)] y − 5 = − 13 4 (x + 2) 4y − 20 = −13(x + 2) 4y − 20 = −13x − 26 13x + 4y − 20 + 26 = 0 13x + 4y + 6 = 0 3