SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, 2ADV C3 2021 HSC 9 MC

Let  `h(x) = f(g(x))`  where the function `f(x)` is an odd function and the function `g(x)` is an even function.

The tangent to  `y = h(x)`  at  `x = k`, where  `k > 0`, has the equation  `y = mx + c`.

What is the equation of the tangent to  `y = h(x)`  at  `x = –k`?

  1. `y = mx + c`
  2. `y = -mx + c`
  3. `y = mx - c`
  4. `y = -mx - c`
Show Answers Only

`B`

Show Worked Solution

`h(x) = f(g(x))`

Mean mark 52%.

`h^{′}(x) = g^{′}(x) · f^{′}(g(x))`

`text(At)\ \ x = k,`

`h(k) = f(g(k))`

`h^{′}(k) = g^{′}(x) · f^{′}(g(x)) = m`
 

`text(At)\ \ x = –k,`

`h(–k)=f(g(–k))=f(g(k))=h(k)`

`-> c\ text(is the same)`

`h^{′}(–k)` `= g^{′}(–k) · f^{′}(g(–k))`
  `= -g^{′}(k) · f^{′}(g(k))\ \ \ \ (g(k)\ text{is even →}\ g^{′}(–k)=–g^{′}(k))`
  `= -m`

 
`:.\ text(Equation of tangent is)\ \ y = -mx + c`

`=> B`