Show that the temperature of the water when it is removed from the cooker is \({90^ \circ }{\text{C}}\).

The following table shows values for \(m\) and \(T(m)\).

(i)     Write down the value of \(s\).

(ii)    Draw the graph of \(T(m)\) for \(0 \leqslant m \leqslant 10\) . Use a scale of \(1{\text{ cm}}\) to represent \(1\) minute on the horizontal axis and a scale of \(1{\text{ cm}}\) to represent \({10^ \circ }{\text{C}}\) on the vertical axis.

(iii)   Use your graph to find how long it takes for the temperature to reach \({56^ \circ }{\text{C}}\). Show your method clearly.

(iv)   Write down the temperature approached by the water after a long time. Justify your answer.

Consider the function \(S(m) = 20m - 40\) for \(2 \leqslant m \leqslant 6\) .

The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.

Draw the graph of \(S(m)\) on the same set of axes used for part (b).

Consider the function \(S(m) = 20m - 40\) for \(2 \leqslant m \leqslant 6\) .

The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.

Comment on the meaning of the constant \(20\) in the formula for \(S(m)\) in relation to the temperature of the soup.

Consider the function \(S(m) = 20m - 40\) for \(2 \leqslant m \leqslant 6\) .

The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.

(i)     Use your graph to solve the equation \(S(m) = T(m)\) . Show your method clearly.

(ii)    Hence describe by using inequalities the set of values of \(m\) for which \(S(m) > T(m)\).

\(T(0) = 20 + 70 \times {2.72^{ - 0.4 \times 0}} = 90\)     (M1)(A1)(AG)

Note: (M1) for taking \(m = 0\) , (A1) for substituting \(0\) into the formula. For the A mark to be awarded \(90\) must be justified by correct method.

[2 marks]

(i)    21.3     (A1)

(ii)  

     (A4)(ft)

Note: Scales and labels (A1). Smooth curve (A1). All points correct including the \(y\)-intercept (A2), 1 point incorrect (A1), otherwise (A0). Follow through from their value of \(s\).

(iii)    \(m = 1.7{\text{ minutes}}\) (Accept \( \pm 0.2\) )     (A2)(ft)

Note: Follow through from candidate's graph. Accept answers in minutes and seconds if consistent with graph. If answer incorrect and correct line(s) seen on graph award (M1)(A0).

(iv)    \({20^ \circ }{\text{C}}\)     (A1)(ft)

The curve behaves asymptotically to the line \(y = 20\) or similar.     (A1)

OR

The room temperature is 20 or similar

OR

When \({\text{m}}\) is a very large number the term \(70 \times {2.72^{ - 0.4{\text{m}}}}\) tends to zero or similar.

Note: Follow through from their graph if appropriate.

[9 marks]

     (A1)(A1)

Notes: (A1) for correct line, (A1) for domain. If line not drawn on same set of axes award at most (A1)(A0).

[2 marks]

It indicates by how much the temperature increases per minute.     (A1)

[1 mark]

(i)     \(m = 3.8\) (Accept \( \pm 0.1\) )     (A2)(ft)

Note: Follow through from candidate's graph. Accept answers in minutes and seconds if consistent with graph. If answer incorrect and correct line(s) seen on graph award (M1)(A0).

(ii)    \(3.8 < m \leqslant 6\)     (A1)(A1)(ft)

Note: (A1) for \(m > 3.8\) and (A1) for \(m \leqslant 6\). Follow through from candidate's answer to part (e)(i). If candidate was already penalized in (c) for domain and does not state \(m \leqslant 6\) then award (A2)(ft).

Good marks were gained in this question, mainly from parts (a) to (d). Very few students answered the show that question using a backwards process. These students did not gain full marks. Labelled and neat exponential graphs were drawn. Marks were lost sometimes for starting the curve at point \((1{\text{, }}66.9)\) instead of at \((0{\text{, }}90)\). Also there were students using a ruler to help them joined up the points for which they lost one mark as the graph must be smooth. Candidates managed to find the time at which the temperature was \(56\). However, those students that gave their answer as a coordinate pair lost the answer mark. Some students justified the behaviour of the curve by mentioning the asymptote but the big majority said that the room temperature was \(20\) and it was awarded full marks. Most of the students drew the straight line correct and in the given domain. Those students that drew the line in a separate set of axes could not answer the last part of the question. This part question proved to be difficult for many students and so worked as a discriminating one. One of the most common errors seen in part (b)(iii) and (e)(i) was to give the answer as a point instead of giving just the first coordinate of that point.

Good marks were gained in this question, mainly from parts (a) to (d). Very few students answered the show that question using a backwards process. These students did not gain full marks. Labelled and neat exponential graphs were drawn. Marks were lost sometimes for starting the curve at point \((1{\text{, }}66.9)\) instead of at \((0{\text{, }}90)\). Also there were students using a ruler to help them joined up the points for which they lost one mark as the graph must be smooth. Candidates managed to find the time at which the temperature was \(56\). However, those students that gave their answer as a coordinate pair lost the answer mark. Some students justified the behaviour of the curve by mentioning the asymptote but the big majority said that the room temperature was \(20\) and it was awarded full marks. Most of the students drew the straight line correct and in the given domain. Those students that drew the line in a separate set of axes could not answer the last part of the question. This part question proved to be difficult for many students and so worked as a discriminating one. One of the most common errors seen in part (b)(iii) and (e)(i) was to give the answer as a point instead of giving just the first coordinate of that point.

Good marks were gained in this question, mainly from parts (a) to (d). Very few students answered the show that question using a backwards process. These students did not gain full marks. Labelled and neat exponential graphs were drawn. Marks were lost sometimes for starting the curve at point \((1{\text{, }}66.9)\) instead of at \((0{\text{, }}90)\). Also there were students using a ruler to help them joined up the points for which they lost one mark as the graph must be smooth. Candidates managed to find the time at which the temperature was \(56\). However, those students that gave their answer as a coordinate pair lost the answer mark. Some students justified the behaviour of the curve by mentioning the asymptote but the big majority said that the room temperature was \(20\) and it was awarded full marks. Most of the students drew the straight line correct and in the given domain. Those students that drew the line in a separate set of axes could not answer the last part of the question. This part question proved to be difficult for many students and so worked as a discriminating one. One of the most common errors seen in part (b)(iii) and (e)(i) was to give the answer as a point instead of giving just the first coordinate of that point.

Good marks were gained in this question, mainly from parts (a) to (d). Very few students answered the show that question using a backwards process. These students did not gain full marks. Labelled and neat exponential graphs were drawn. Marks were lost sometimes for starting the curve at point \((1{\text{, }}66.9)\) instead of at \((0{\text{, }}90)\). Also there were students using a ruler to help them joined up the points for which they lost one mark as the graph must be smooth. Candidates managed to find the time at which the temperature was \(56\). However, those students that gave their answer as a coordinate pair lost the answer mark. Some students justified the behaviour of the curve by mentioning the asymptote but the big majority said that the room temperature was \(20\) and it was awarded full marks. Most of the students drew the straight line correct and in the given domain. Those students that drew the line in a separate set of axes could not answer the last part of the question. This part question proved to be difficult for many students and so worked as a discriminating one. One of the most common errors seen in part (b)(iii) and (e)(i) was to give the answer as a point instead of giving just the first coordinate of that point.

Good marks were gained in this question, mainly from parts (a) to (d). Very few students answered the show that question using a backwards process. These students did not gain full marks. Labelled and neat exponential graphs were drawn. Marks were lost sometimes for starting the curve at point \((1{\text{, }}66.9)\) instead of at \((0{\text{, }}90)\). Also there were students using a ruler to help them joined up the points for which they lost one mark as the graph must be smooth. Candidates managed to find the time at which the temperature was \(56\). However, those students that gave their answer as a coordinate pair lost the answer mark. Some students justified the behaviour of the curve by mentioning the asymptote but the big majority said that the room temperature was \(20\) and it was awarded full marks. Most of the students drew the straight line correct and in the given domain. Those students that drew the line in a separate set of axes could not answer the last part of the question. This part question proved to be difficult for many students and so worked as a discriminating one. One of the most common errors seen in part (b)(iii) and (e)(i) was to give the answer as a point instead of giving just the first coordinate of that point.