Download IITian Academy App for accessing Online Mock Tests and More..
Question
Find the term \({x^3}\) in the expansion of \({\left( {\frac{2}{3}x – 3} \right)^8}\) .
Answer/Explanation
Markscheme
evidence of using binomial expansion (M1)
e.g. selecting correct term, \({a^8}{b^0} + \left( {\begin{array}{*{20}{c}}
8\\
1
\end{array}} \right){a^7}b + \left( {\begin{array}{*{20}{c}}
8\\
2
\end{array}} \right){a^6}{b^2} + \ldots \)
evidence of calculating the factors, in any order A1A1A1
e.g. 56 , \(\frac{{{2^2}}}{{{3^3}}}\) , \( – {3^5}\) , \(\left( {\begin{array}{*{20}{c}}
8\\
5
\end{array}} \right){\left( {\frac{2}{3}x} \right)^3}{( – 3)^5}\)
\( – 4032{x^3}\) (accept = \( – 4030{x^3}\) to 3 s.f.) A1 N2
[5 marks]
Question
Expand \({(x – 2)^4}\) and simplify your result.
Find the term in \({x^3}\) in \((3x + 4){(x – 2)^4}\) .
Answer/Explanation
Markscheme
evidence of expanding M1
e.g. \({(x – 2)^4} = {x^4} + 4{x^3}( – 2) + 6{x^2}{( – 2)^2} + 4x{( – 2)^3} + {( – 2)^4}\) A2 N2
\({(x – 2)^4} = {x^4} – 8{x^3} + 24{x^2} – 32x + 16\)
[3 marks]
finding coefficients, \(3 \times 24( = 72)\) , \(4 \times( – 8)( = – 32)\) (A1)(A1)
term is \(40{x^3}\) A1 N3
[3 marks]
Question
Let \(f(x) = {x^3} – 4x + 1\) .
Expand \({(x + h)^3}\) .
Use the formula \(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) – f(x)}}{h}\) to show that the derivative of \(f(x)\) is \(3{x^2} – 4\) .
The tangent to the curve of f at the point \({\text{P}}(1{\text{, }} – 2)\) is parallel to the tangent at a point Q. Find the coordinates of Q.
The graph of f is decreasing for \(p < x < q\) . Find the value of p and of q.
Write down the range of values for the gradient of \(f\) .
Answer/Explanation
Markscheme
attempt to expand (M1)
\({(x + h)^3} = {x^3} + 3{x^2}h + 3x{h^2} + {h^3}\) A1 N2
[2 marks]
evidence of substituting \(x + h\) (M1)
correct substitution A1
e.g. \(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{{(x + h)}^3} – 4(x + h) + 1 – ({x^3} – 4x + 1)}}{h}\)
simplifying A1
e.g. \(\frac{{({x^3} + 3{x^2}h + 3x{h^2} + {h^3} – 4x – 4h + 1 – {x^3} + 4x – 1)}}{h}\)
factoring out h A1
e.g. \(\frac{{h(3{x^2} + 3xh + {h^2} – 4)}}{h}\)
\(f'(x) = 3{x^2} – 4\) AG N0
[4 marks]
\(f'(1) = – 1\) (A1)
setting up an appropriate equation M1
e.g. \(3{x^2} – 4 = – 1\)
at Q, \(x = – 1,y = 4\) (Q is \(( – 1{\text{, }}4)\)) A1 A1
[4 marks]
recognizing that f is decreasing when \(f'(x) < 0\) R1
correct values for p and q (but do not accept \(p = 1.15{\text{, }}q = – 1.15\) ) A1A1 N1N1
e.g. \(p = – 1.15{\text{, }}q = 1.15\) ; \( \pm \frac{2}{{\sqrt 3 }}\) ; an interval such as \( – 1.15 \le x \le 1.15\)
[3 marks]
\(f'(x) \ge – 4\) , \(y \ge – 4\) , \(\left[ { – 4,\infty } \right[\) A2 N2
[2 marks]
Question
Find the term in \({x^4}\) in the expansion of \({\left( {3{x^2} – \frac{2}{x}} \right)^5}\) .
Answer/Explanation
Markscheme
evidence of substituting into binomial expansion (M1)
e.g. \({a^5} + \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right){a^4}b + \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right){a^3}{b^2} + \ldots \)
identifying correct term for \({x^4}\) (M1)
evidence of calculating the factors, in any order A1A1A1
e.g. \(\left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right),27{x^6},\frac{4}{{{x^2}}}\) ; \(10{(3{x^2})^3}{\left( {\frac{{ – 2}}{x}} \right)^2}\)
Note: Award A1 for each correct factor.
\({\rm{term}} = 1080{x^4}\) A1 N2
Note: Award M1M1A1A1A1A0 for 1080 with working shown.
[6 marks]
Question
Consider the expansion of \({(3{x^2} + 2)^9}\) .
Write down the number of terms in the expansion.
Find the term in \({x^4}\) .
Answer/Explanation
Markscheme
10 terms A1 N1
[1 mark]
evidence of binomial expansion (M1)
e.g. \({a^9}{b^0} + \left( \begin{array}{l}
9\\
1
\end{array} \right){a^8}b + \left( \begin{array}{l}
9\\
2
\end{array} \right){a^7}{b^2} + \ldots \), \(\left( \begin{array}{l}
9\\
r
\end{array} \right){(a)^{n – r}}{(b)^r}\) , Pascal’s triangle
evidence of correct term (A1)
e.g. 8th term, \(r = 7\) , \(\left( \begin{array}{l}
9\\
7
\end{array} \right)\) , \({(3{x^2})^2}{2^7}\)
correct expression of complete term (A1)
e.g. \(\left( \begin{array}{l}
9\\
7
\end{array} \right){(3{x^2})^2}{(2)^7}\) , \(_2^9C{(3{x^2})^2}{(2)^7}\) , \(36 \times 9 \times 128\)
\(41472{x^4}\) (accept \(41500{x^4}\) ) A1 N2
[4 marks]
Question
Consider the expansion of \({\left( {2{x^3} + \frac{b}{x}} \right)^8} = 256{x^{24}} + 3072{x^{20}} + \ldots + k{x^0} + \ldots \) .
Find b.
Find k.
Answer/Explanation
Markscheme
valid attempt to find term in \({x^{20}}\) (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
8\\
1
\end{array}} \right)({2^7})(b)\) , \({(2{x^3})^7}\left( {\frac{b}{x}} \right) = 3072\)
correct equation A1
e.g. \(\left( {\begin{array}{*{20}{c}}
8\\
1
\end{array}} \right)({2^7})(b) = 3072\)
\(b = 3\) A1 N2
[3 marks]
evidence of choosing correct term (M1)
e.g. 7th term, \(r = 6\)
correct expression A1
e.g. \(\left( {\begin{array}{*{20}{c}}
8\\
6
\end{array}} \right){(2{x^3})^2}{\left( {\frac{3}{x}} \right)^6}\)
\(k = 81648\) (accept \(81600\) ) A1 N2
[3 marks]
Question
The third term in the expansion of \({(2x + p)^6}\) is \(60{x^4}\) . Find the possible values of p .
Answer/Explanation
Markscheme
attempt to expand binomial (M1)
e.g. \({(2x)^6}{p^0} + \left( {\begin{array}{*{20}{c}}
6\\
1
\end{array}} \right){(2x)^5}{(p)^1} + \ldots \) , \(\left( {\begin{array}{*{20}{c}}
n\\
r
\end{array}} \right){(2x)^r}{(p)^{n – r}}\)
one correct calculation for term in \({x^4}\) in the expansion for power 6 (A1)
e.g. 15 , \(16{x^4}\)
correct expression for term in \({x^4}\) (A1)
e.g. \(\left( {\begin{array}{*{20}{c}}
6\\
2
\end{array}} \right){(2x)^4}{(p)^2}\) , \({15.2^4}{p^2}\)
Notes: Accept sloppy notation e.g. omission of brackets around \(2x\) .
Accept absence of \(x\) in middle factor.
correct term (A1)
e.g. \(240{p^2}{x^4}\) (accept absence of \({x^4}\) )
setting up equation with their coefficient equal to 60 M1
e.g. \(\left( {\begin{array}{*{20}{c}}
6\\
2
\end{array}} \right){(2)^4}{(p)^2} = 60\) , \(240{p^2}{x^4} = 60{x^4}\) , \({p^2} = \frac{{60}}{{240}}\)
\(p = \pm \frac{1}{2}(p = \pm 0.5)\) A1A1 N3
[7 marks]
Question
In the expansion of \({(3x – 2)^{12}}\) , the term in \({x^5}\) can be expressed as \(\left( {\begin{array}{*{20}{c}}
{12}\\
r
\end{array}} \right) \times {(3x)^p} \times {( – 2)^q}\) .
(a) Write down the value of \(p\) , of \(q\) and of \(r\) .
(b) Find the coefficient of the term in \({x^5}\) .
Write down the value of \(p\) , of \(q\) and of \(r\) .
Find the coefficient of the term in \({x^5}\) .
Answer/Explanation
Markscheme
(a) \(p = 5\) , \(q = 7\) , \(r = 7\) (accept \(r = 5\)) A1A1A1 N3
[3 marks]
(b) correct working (A1)
eg \(\left( {\begin{array}{*{20}{c}}
{12}\\
7
\end{array}} \right) \times {(3x)^5} \times {( – 2)^7}\) , \(792\) , \(243\) , \( – {2^7}\) , \(24634368\)
coefficient of term in \({x^5}\) is \( – 24634368\) A1 N2
Note: Do not award the final A1 for an answer that contains \(x\).
[2 marks]
Total [5 marks]
\(p = 5\) , \(q = 7\) , \(r = 7\) (accept \(r = 5\)) A1A1A1 N3
[3 marks]
correct working (A1)
eg \(\left( {\begin{array}{*{20}{c}}
{12}\\
7
\end{array}} \right) \times {(3x)^5} \times {( – 2)^7}\) , \(792\) , \(243\) , \( – {2^7}\) , \(24634368\)
coefficient of term in \({x^5}\) is \( – 24634368\) A1 N2
Note: Do not award the final A1 for an answer that contains \(x\).
[2 marks]
Total [5 marks]
Question
The constant term in the expansion of \({\left( {\frac{x}{a} + \frac{{{a^2}}}{x}} \right)^6}\) , where \(a \in \mathbb{R}\) is \(1280\). Find \(a\) .
Answer/Explanation
Markscheme
evidence of binomial expansion (M1)
eg selecting correct term,\({\left( {\frac{x}{a}} \right)^6}{\left( {\frac{{{a^2}}}{x}} \right)^0} + \left( \begin{array}{l}
6\\
1
\end{array} \right){\left( {\frac{x}{a}} \right)^5}{\left( {\frac{{{a^2}}}{x}} \right)^1} + \ldots \)
evidence of identifying constant term in expansion for power \(6\) (A1)
eg \(r = 3\) , 4th term
evidence of correct term (may be seen in equation) A2
eg \(20\frac{{{a^6}}}{{{a^3}}}\) , \(\left( \begin{array}{l}
6\\
3
\end{array} \right){\left( {\frac{x}{a}} \right)^3}{\left( {\frac{{{a^2}}}{x}} \right)^3}\)
attempt to set up their equation (M1)
eg \(\left( \begin{array}{l}
6\\
3
\end{array} \right){\left( {\frac{x}{a}} \right)^3}{\left( {\frac{{{a^2}}}{x}} \right)^3} = 1280\), \({a^3} = 1280\)
correct equation in one variable \(a\) (A1)
eg \(20{a^3} = 1280\) , \({a^3} = 64\)
\(a = 4\) A1 N4
[7 marks]
Question
Consider the expansion of \({(x + 3)^{10}}\).
Write down the number of terms in this expansion.
Find the term containing \({x^3}\).
Answer/Explanation
Markscheme
11 terms A1 N1
[1 mark]
evidence of binomial expansion (M1)
eg \(\left( \begin{array}{c}n\\r\end{array} \right)\) \({a^{n – r}}{b^r}\), attempt to expand
evidence of choosing correct term (A1)
eg \({8^{{\text{th}}}}{\text{ term, }}r = 7\), \(\left( \begin{array}{c}10\\7\end{array} \right)\), \({(x)^3}{(3)^7}\)
correct working (A1)
eg \(\left( \begin{array}{c}10\\7\end{array} \right)\) \({(x)^3}{(3)^7}\), \(\left( \begin{array}{c}10\\3\end{array} \right)\)\({(x)^3}{(3)^7}\),
\(262440{x^3}{\text{ (accept }}262000{x^3})\) A1 N3
[4 marks]
Question
Consider the expansion of \({x^2}{\left( {3{x^2} + \frac{k}{x}} \right)^8}\). The constant term is \({\text{16 128}}\).
Find \(k\).
Answer/Explanation
Markscheme
valid approach (M1)
eg \({\text{\(\left( \begin{array}{c}8\\r\end{array} \right)\)}}{\left( {3{x^2}} \right)^{8 – r}}{\left( {\frac{k}{x}} \right)^r}\),
\({\left( {3{x^2}} \right)^8} + {\text{\(\left( \begin{array}{c}8\\1\end{array} \right)\)}}{\left( {3{x^2}} \right)^7}\left( {\frac{k}{x}} \right) + {\text{\(\left( \begin{array}{c}8\\2\end{array} \right)\)}}{\left( {3{x^2}} \right)^6}{\left( {\frac{k}{x}} \right)^2} + \ldots \), Pascal’s triangle to \({9^{{\text{th}}}}\) line
attempt to find value of r which gives term in \({x^0}\) (M1)
eg exponent in binomial must give \({x^{ – 2}},{\text{ }}{x^2}{\left( {{x^2}} \right)^{8 – r}}{\left( {\frac{k}{x}} \right)^r} = {x^0}\)
correct working (A1)
eg \(2(8 – r) – r = – 2,{\text{ }}18 – 3r = 0,{\text{ }}2r + ( – 8 + r) = – 2\)
evidence of correct term (A1)
eg \({\text{\(\left( \begin{array}{c}8\\2\end{array} \right)\), \(\left( \begin{array}{c}8\\6\end{array} \right)\)}}{\left( {3{x^2}} \right)^2}{\left( {\frac{k}{x}} \right)^2},{\text{ }}r = 6,{\text{ }}r = 2\)
equating their term and 16128 to solve for \(k\) M1
eg \({x^2}{\text{\(\left( \begin{array}{c}8\\6\end{array} \right)\)}}{\left( {3{x^2}} \right)^2}{\left( {\frac{k}{x}} \right)^6} = 16128,{\text{ }}{k^6} = \frac{{16128}}{{28(9)}}\)
\(k = \pm 2\) A1A1 N2
Note: If no working shown, award N0 for \(k = 2\).
Total [7 marks]
Question
Consider the expansion of \({\left( {\frac{{{x^3}}}{2} + \frac{p}{x}} \right)^8}\). The constant term is \(5103\). Find the possible values of \(p\).
Answer/Explanation
Markscheme
valid approach to find the required term (M1)
eg\(\;\;\;\)\(\left( {\begin{array}{*{20}{c}} 8 \\ r \end{array}} \right){\left( {\frac{{{x^3}}}{2}} \right)^{8 – r}}{\left( {\frac{p}{x}} \right)^r},{\text{ }}{\left( {\frac{{{x^3}}}{2}} \right)^8}{\left( {\frac{p}{x}} \right)^0} + \left( {\begin{array}{*{20}{c}} 8 \\ 1 \end{array}} \right){\left( {\frac{{{x^3}}}{2}} \right)^7}{\left( {\frac{p}{x}} \right)^1} + \ldots \), Pascal’s triangle to required value
identifying constant term (may be indicated in expansion) (A1)
eg\(\;\;\;\)\({7^{{\text{th}}}}{\text{ term, }}r = 6,{\text{ }}{\left( {\frac{1}{2}} \right)^2},{\text{ }}\left( {\begin{array}{*{20}{c}} 8 \\ 6 \end{array}} \right),{\text{ }}{\left( {\frac{{{x^3}}}{2}} \right)^2}{\left( {\frac{p}{x}} \right)^6}\)
correct calculation (may be seen in expansion) (A1)
eg\(\;\;\;\)\(\left( {\begin{array}{*{20}{c}} 8 \\ 6 \end{array}} \right){\left( {\frac{{{x^3}}}{2}} \right)^2}{\left( {\frac{p}{x}} \right)^6},{\text{ }}\frac{{8 \times 7}}{2} \times \frac{{{p^6}}}{{{2^2}}}\)
setting up equation with their constant term equal to \(5103\) M1
eg\(\;\;\;\)\(\left( {\begin{array}{*{20}{c}} 8 \\ 6 \end{array}} \right){\left( {\frac{{{x^3}}}{2}} \right)^2}{\left( {\frac{p}{x}} \right)^6} = 5103,{\text{ }}{p^6} = \frac{{5103}}{7}\)
\(p = \pm 3\) A1A1 N3
[6 marks]
Question
Consider the expansion of \({(2x + 3)^8}\).
Write down the number of terms in this expansion.
Find the term in \({x^3}\).
Answer/Explanation
Markscheme
9 terms A1 N1
[1 mark]
valid approach to find the required term (M1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 8 \\ r \end{array}} \right){(2x)^{8 – r}}{(3)^r},{\text{ }}{(2x)^8}{(3)^0} + {(2x)^7}{(3)^1} + \ldots \), Pascal’s triangle to \({{\text{8}}^{{\text{th}}}}\) row
identifying correct term (may be indicated in expansion) (A1)
eg\(\;\;\;{{\text{6}}^{{\text{th}}}}{\text{ term, }}r = 5,{\text{ }}\left( {\begin{array}{*{20}{c}} 8 \\ 5 \end{array}} \right),{\text{ (2x}}{{\text{)}}^3}{(3)^5}\)
correct working (may be seen in expansion) (A1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 8 \\ 5 \end{array}} \right){(2x)^3}{(3)^5},{\text{ }}56 \times {2^3} \times {3^5}\)
\(108864{x^3}\;\;\;\)(accept \(109000{x^3}\)) A1 N3
[4 marks]
Notes: Do not award any marks if there is clear evidence of adding instead of multiplying.
Do not award final A1 for a final answer of \(108864\), even if \(108864{x^3}\) is seen previously.
If no working shown award N2 for \(108864\).
Question
The third term in the expansion of \({(x + k)^8}\) is \(63{x^6}\). Find the possible values of \(k\).
Answer/Explanation
Markscheme
valid approach to find the required term (M1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 8 \\ r \end{array}} \right){x^{8 – r}}{k^r}\), Pascal’s triangle to \({{\text{8}}^{{\text{th}}}}\) row, \({x^8} + 8{x^7}k + 28{x^6}{k^2} + \ldots \)
identifying correct term (may be indicated in expansion) (A1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 8 \\ 2 \end{array}} \right){x^6}{k^2},{\text{ }}\left( {\begin{array}{*{20}{c}} 8 \\ 6 \end{array}} \right){x^6}{k^2},{\text{ }}r = 2\)
setting up equation in \(k\) with their coefficient/term (M1)
eg\(\;\;\;28{k^2}{x^6} = 63{x^6},{\text{ }}\left( {\begin{array}{*{20}{c}} 8 \\ 6 \end{array}} \right){k^2} = 63\)
\(k = \pm 1.5{\text{ (exact)}}\) A1A1 N3
[5 marks]
| Marks available | 4 |
| Reference code | 16M.2.sl.TZ1.4 |
Question
Find the term in \({x^6}\) in the expansion of \({(x + 2)^9}\).
Hence, find the term in \({x^7}\) in the expansion of \(5x{(x + 2)^9}\).
Answer/Explanation
Markscheme
valid approach to find the required term (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 9 \\ r \end{array}} \right){(x)^{9 – r}}{(2)^r},{\text{ }}{x^9} + 9{x^8}(2) + \left( {\begin{array}{*{20}{c}} 9 \\ 2 \end{array}} \right){x^7}{(2)^2} + \ldots \), Pascal’s triangle to the 9th row
identifying correct term (may be indicated in expansion) (A1)
eg\(\,\,\,\,\,\)4th term, \(r = 6,{\text{ }}\left( {\begin{array}{*{20}{c}} 9 \\ 3 \end{array}} \right),{\text{ }}{(x)^6}{(2)^3}\)
correct calculation (may be seen in expansion) (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 9 \\ 3 \end{array}} \right){(x)^6}{(2)^3},{\text{ }}84 \times {2^3}\)
672\({x^6}\) A1 N3
[4 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)recognizing \({x^7}\) is found when multiplying \(5x \times 672{x^6}\)
\(3360{x^7}\) A1 N2
[2 marks]
Question
Consider the expansion of \({\left( {{x^2} + \frac{2}{x}} \right)^{10}}\).
Write down the number of terms of this expansion.
Find the coefficient of \({x^8}\).
Answer/Explanation
Markscheme
11 terms A1 N1
[1 mark]
valid approach (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {10} \\ r \end{array}} \right){({x^2})^{10 – r}}{\left( {\frac{2}{x}} \right)^r},{\text{ }}{a^{10}}{b^0} + \left( {\begin{array}{*{20}{c}} {10} \\ 1 \end{array}} \right){a^9}{b^1}\left( {\begin{array}{*{20}{c}} {10} \\ 2 \end{array}} \right){a^8}{b^2} + \ldots \)
Pascal’s triangle to \({11^{th}}\) row
valid attempt to find value of \(r\) which gives term in \({x^8}\) (M1)
eg\(\,\,\,\,\,\)\({({x^2})^{10 – r}}\left( {\frac{1}{{{x^r}}}} \right) = {x^8},{\text{ }}{x^{2r}}{\left( {\frac{2}{x}} \right)^{10 – r}} = {x^8}\)
identifying required term (may be indicated in expansion) (A1)
eg\(\,\,\,\,\,\)\(r = 6,{\text{ }}{{\text{5}}^{{\text{th}}}}{\text{ term, }}{{\text{7}}^{{\text{th}}}}{\text{ term}}\)
correct working (may be seen in expansion) (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {10} \\ 6 \end{array}} \right){({x^2})^6}{\left( {\frac{2}{x}} \right)^4},{\text{ }}210 \times 16\)
3360 A1 N3
[5 marks]
Question
In the expansion of \(a{x^3}{(2 + ax)^{11}}\), the coefficient of the term in \({x^5}\) is 11880. Find the value of \(a\).
Answer/Explanation
Markscheme
valid approach for expansion (must have correct substitution for parameters, but accept an incorrect value for \(r\)) (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {11} \\ r \end{array}} \right){(2)^{11 – r}}a{x^r},{\text{ }}\left( {\begin{array}{*{20}{c}} {11} \\ 3 \end{array}} \right){(2)^8}{(ax)^3},{\text{ }}{2^{11}} + \left( {\begin{array}{*{20}{c}} {11} \\ 1 \end{array}} \right){(2)^{10}}{(ax)^1} + \left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){(2)^9}(ax) + \ldots \)
recognizing need to find term in \({x^2}\) in binomial expansion (A1)
eg\(\,\,\,\,\,\)\(r = 2,{\text{ }}{(ax)^2}\)
correct term or coefficient in binomial expansion (may be seen in equation) (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){(ax)^2}{(2)^9},{\text{ }}55({a^2}{x^2})(512),{\text{ }}28160{a^2}\)
setting up equation in \({x^5}\) with their coefficient/term (do not accept other powers of \(x\)) (M1)
eg\(\,\,\,\,\,\)\(a{x^3}\left( {\begin{array}{*{20}{c}} {11} \\ 2 \end{array}} \right){(ax)^2}{(2)^9} = 11880{x^5}\)
correct equation (A1)
eg\(\,\,\,\,\,\)\(28160{a^3} = 11880\)
\(a = \frac{3}{4}\) A1 N3
[6 marks]
Question
Consider the expansion of \({\left( {2x + \frac{k}{x}} \right)^9}\), where k > 0 . The coefficient of the term in x3 is equal to the coefficient of the term in x5. Find k.
Markscheme
valid approach to find one of the required terms (must have correct substitution for parameters but accept “r” or an incorrect value for r) (M1)
