# Please someone could you mark and correct my solutions if I made a mistake (1 Viewer)

#### fan96

##### 617 pages
2a) You calculated the gradient of the curve only at the given point, and assumed it (the gradient) was the same for all other points on the curve.

The question asked you to find the function which has gradient function $\bg_white {dy}/{dx} = \cos 2x$ for all values of $\bg_white x$ (in the domain), and which also passes through $\bg_white (\pi/2,\, 1)$.

The linear equation you gave for your answer has gradient function $\bg_white {dy}/{dx} = \cos 2x$ at $\bg_white (\pi/2,\, 1)$ but not necessarily for any other point on the curve.

You should start by integrating the gradient function and then use the given point to find the value of the constant of integration. The answer should be $\bg_white y = 1/2 \,\sin 2x + 1$.

2c) Between your 7th and 8th lines, note that $\bg_white -1/2 \, \cos(240^\circ) \neq -1/2 \, \cos(60^\circ)$. Rather,$\bg_white -1/2 \, \cos(240^\circ) = 1/2 \, \cos(60^\circ)$ . As a result, the negative sign in your 9th line should be a positive, giving $\bg_white A = 9/4\, \mathrm{u}^2$

#### HeroWise

##### Active Member
oki imma do every single question and send u a cheat sheet so you can cgeck with the answers

#### kpad5991

##### Member
oki imma do every single question and send u a cheat sheet so you can cgeck with the answers
Thank you so so much