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Mathematical Methods — Question 5

QCAA 2020, Paper 1 · 1 mark

Q5 · 2020 · Technology-freeSimple familiar1 mark

QUESTION 5

The equation of the tangent to the curve

f(t)=te^t

at

t=1

is

(A)

y=et

(B)

y=2et-e

(C)

y=et-e^2+1

(D)

y=2et-2e^2+1

ABCD

Question linkOriginal paper

WORKED SOLUTION

Answer B

1 mark

ANSWER

B

Worked solution

f'(t)=e^t+te^t=e^t(1+t)

, so the gradient at

t=1

is

2e

and the point is

(1,e)

. Thus

y-e=2e(t-1)

, or

y=2et-e

. Hence B.

Selects option B.

[1 mark]

Worked explanation by QCE Vault; correct option verified against the QCAA answer key.

View the full QCAA marking guide

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