6. maximum mark: 11\nthe line ( y = 7x - 2 ) is tangent to the curve ( y = ae^{x - 1}+bx^{2} ) at ( x = 1…

6. maximum mark: 11\nthe line ( y = 7x - 2 ) is tangent to the curve ( y = ae^{x - 1}+bx^{2} ) at ( x = 1 ).\n(a) use the fact that the tangent meets the curve to show that ( a + b = 5 ).\n(b) use the fact that the tangent has the same gradient as the curve to find another relationship between ( a ) and ( b ).\n(c) hence find the values of ( a ) and ( b ).\n(d) find the gradient of the curve when ( x = 0 ).

6. maximum mark: 11\nthe line ( y = 7x - 2 ) is tangent to the curve ( y = ae^{x - 1}+bx^{2} ) at ( x = 1 ).\n(a) use the fact that the tangent meets the curve to show that ( a + b = 5 ).\n(b) use the fact that the tangent has the same gradient as the curve to find another relationship between ( a ) and ( b ).\n(c) hence find the values of ( a ) and ( b ).\n(d) find the gradient of the curve when ( x = 0 ).

Answer

Explanation:

Step1: Substitute (x = 1) into the line and curve equations

For the line (y=7x - 2), when (x = 1), (y=7\times1-2=5). For the curve (y = ae^{x - 1}+bx^{2}), when (x = 1), (y=a\times e^{1 - 1}+b\times1^{2}=a + b). Since the tangent meets the curve at (x = 1), then (a + b=5).

Step2: Differentiate the curve equation

Differentiate (y = ae^{x - 1}+bx^{2}) with respect to (x). Using the chain - rule ((e^{u})^\prime=e^{u}\cdot u^\prime) (where (u=x - 1)) and ((x^{n})^\prime=nx^{n - 1}), we get (y^\prime=ae^{x - 1}+2bx). The slope of the line (y = 7x-2) is (m = 7). When (x = 1), (y^\prime|_{x = 1}=ae^{1 - 1}+2b\times1=a + 2b). Since the tangent has the same gradient as the curve at (x = 1), then (a+2b = 7).

Step3: Solve the system of equations

We have the system of equations (\begin{cases}a + b=5\a + 2b=7\end{cases}). Subtract the first equation from the second equation: ((a + 2b)-(a + b)=7 - 5). (a+2b - a - b=2), so (b = 2). Substitute (b = 2) into (a + b=5), we get (a=5 - b). Then (a=5-2 = 3).

Step4: Find the gradient of the curve at (x = 0)

We know (y^\prime=ae^{x - 1}+2bx), and (a = 3), (b = 2). Substitute (x = 0) into (y^\prime): (y^\prime|_{x = 0}=3e^{0 - 1}+2\times2\times0=\frac{3}{e}).

Answer:

(a) By substituting (x = 1) into (y=7x - 2) ((y = 5)) and (y=ae^{x - 1}+bx^{2}) ((y=a + b)), we show (a + b=5). (b) Another relationship is (a + 2b=7). (c) (a = 3), (b = 2). (d) The gradient of the curve when (x = 0) is (\frac{3}{e}).