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 ).\nfind 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 ).\nfind 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 rules (\frac{d}{dx}(e^{u})=e^{u}\cdot u') (where (u=x - 1), (u'=1)) and (\frac{d}{dx}(x^{n})=nx^{n - 1}).
  • (y'=ae^{x - 1}+2bx).
  • The slope of the line (y = 7x-2) is (m = 7). When (x = 1), the slope of the curve (y'=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 (\begin{cases}a + b=5\a+2b=7\end{cases}).
  • Subtract the first equation from the second: ((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=5-2 = 3).

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

  • The derivative of the curve (y'=ae^{x - 1}+2bx). Substitute (a = 3), (b = 2) into (y').
  • (y'=3e^{x - 1}+4x).
  • When (x = 0), (y'=3e^{-1}+4\times0=\frac{3}{e}).

Answer:

(a = 3), (b = 2), and the gradient of the curve when (x = 0) is (\frac{3}{e})