Question
Find the first three terms,in ascending powers of x,in the expansion of
(i)\(\left ( 1-x \right )^{6}\)
(ii)\(\left ( 1+2x \right )^{6}\)
(ii)Hence find the coefficient of \(x^{2}\) in the expansion of \(\left [ \left ( 1-x \right )\left ( 1+2x \right ) \right ]^{6}\)
▶️Answer/Explanation
(i) Here, \(a = 1\) and \(b = -x\), and \(n = 6\). We will find the terms up to \(x^2\):
The term with \(x^0\): \(\binom{6}{0} (1)^{6-0} (-x)^0 = 1\)
The term with \(x^1\): \(\binom{6}{1} (1)^{6-1} (-x)^1 = -6x\)
The term with \(x^2\): \(\binom{6}{2} (1)^{6-2} (-x)^2 = 15x^2\)
So the first three terms are \(1 – 6x + 15x^2\).
(ii) Here, \(a = 1\) and \(b = 2x\), and \(n = 6\). We will find the terms up to \(x^2\):
The term with \(x^0\): \(\binom{6}{0} (1)^{6-0} (2x)^0 = 1\)
The term with \(x^1\): \(\binom{6}{1} (1)^{6-1} (2x)^1 = 12x\)
The term with \(x^2\): \(\binom{6}{2} (1)^{6-2} (2x)^2 = 60x^2\)
So the first three terms are \(1 + 12x + 60x^2\).
(iii) The expansion of \([(1-x)(1+2x)]^6\) is equivalent to the product of the expansions of \((1-x)^6\) and \((1+2x)^6\).
The coefficient of \(x^2\) in the final expansion will come from the following products of terms from the expansions of \((1-x)^6\) and \((1+2x)^6\):
The \(x^0\) term from \((1-x)^6\) (which is \(1\)) multiplied by the \(x^2\) term from \((1+2x)^6\) (which is \(60x^2\)).
The \(x^1\) term from \((1-x)^6\) (which is \(-6x\)) multiplied by the \(x^1\) term from \((1+2x)^6\) (which is \(12x\)).
The \(x^2\) term from \((1-x)^6\) (which is \(15x^2\)) multiplied by the \(x^0\) term from \((1+2x)^6\) (which is \(1\)).
Now , coefficient of \(x^2\):
\( 1 \cdot 60x^2 + (-6x) \cdot 12x + 15x^2 \cdot 1 = 60x^2 – 72x^2 + 15x^2 = (60 – 72 + 15)x^2 = 3x^2 \)
Therefore, the coefficient of \(x^2\) in the expansion of \([(1-x)(1+2x)]^6\) is \(3\).
Question
Find the term independent of x in the expansion of \(\left ( 4x^{3} +\frac{1}{2x}\right )^{8}\)
▶️Answer/Explanation
Let’s denote the general term in the expansion as \( T_{k+1} \) where \( k \) is the index of the term:
\( T_{k+1} = \binom{8}{k} (4x^3)^{8-k} \left(\frac{1}{2x}\right)^k \)
\( 3(8-k) – k = 0 \)
\( 24 – 3k – k = 0 \)
\( 24 – 4k = 0 \)
\( 4k = 24 \)
\( k = 6 \)
Now we substitute \( k = 6 \) into the general term to find the term that is independent of \( x \):
\( T_{6+1} = \binom{8}{6} (4x^3)^{8-6} \left(\frac{1}{2x}\right)^6 \)
\( T_{7} = \binom{8}{6} (4x^3)^2 \left(\frac{1}{2x}\right)^6 \)
\( T_{7} = \binom{8}{6} (16x^6) \left(\frac{1}{64x^6}\right) \)
\( T_{7} = \binom{8}{6} \frac{16}{64} \)
\( T_{7} = 28 \cdot \frac{1}{4} \)
\( T_{7} = 7 \)
Thus, the term in the expansion of \( \left(4 x^3+\frac{1}{2 x}\right)^8 \) that is independent of \( x \) is 7.
Question
Find the coefficient of \(x^{6}\) in the expansion of \(\left ( 2x^{3} -\frac{1}{x^{2}}\right )^{7}\).
▶️Answer/Explanation
The binomial theorem states that for any real numbers \(a\) and \(b\), and any positive integer \(n\), the expansion of \((a + b)^n\) is given by:
\( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
In each term of the expansion, the powers of \(x\) from \(a\) and \(b\) must add up to 6.
The general term in the expansion is:
\( \binom{7}{k} (2x^3)^{7-k} \left(-\frac{1}{x^2}\right)^k \)
\( = \binom{7}{k} 2^{7-k} x^{21-3k} \cdot (-1)^k x^{-2k}\)
\(= \binom{7}{k} 2^{7-k} (-1)^k x^{21-5k} \)
\( 21 – 5k = 6 \)
\( 5k = 21 – 6 \)
\( 5k = 15 \)
\( k = 3 \)
Therefore, the term with \(x^6\) will come from the term in the expansion where \(k = 3\).
\( \binom{7}{3} 2^{7-3} (-1)^3 x^6 \)
\( = 35 \cdot 2^4 \cdot (-1) x^6 \)
\( = 35 \cdot 16 \cdot (-1) x^6 \)
\( = -560 x^6 \)
So, the coefficient of \(x^6\) in the expansion of \(\left(2 x^3-\frac{1}{x^2}\right)^7\) is \(-560\).
Question
Find the term independent of x in the expansion of \(\left ( 2x+\frac{1}{2x^{3}} \right )^{8}\).
▶️Answer/Explanation
The general term in the expansion of \( (a + b)^n \) is given by:
\( T_{k+1} = \binom{n}{k} \cdot a^{n-k} \cdot b^k \)
For the given binomial \( \left(2x + \frac{1}{2x^3}\right)^8 \), let’s denote \( a = 2x \) and \( b = \frac{1}{2x^3} \), and we have \( n = 8 \).
The term independent of \( x \) is where the powers of \( x \) cancel each other out. So, for the \( k \)-th term in the expansion, we need:
\( (2x)^{8-k} \cdot \left(\frac{1}{2x^3}\right)^k \)
This term will be independent of \( x \) if the exponent of \( x \) is zero:
\( 1 \cdot (8 – k) – 3 \cdot k = 0 \)
\( 8 – k – 3k = 0 \)
\( 8 – 4k = 0 \)
\( k = \frac{8}{4} \)
\( k = 2 \)
Nowe,substituting k=2
\( T_{2+1} = \binom{8}{2} \cdot (2x)^{8-2} \cdot \left(\frac{1}{2x^3}\right)^2 \)
\( T_{3} = \binom{8}{2} \cdot (2x)^6 \cdot \left(\frac{1}{4x^6}\right) \)
\( T_{3} = \binom{8}{2} \cdot \frac{(2^6 x^6)}{4x^6} \)
\( T_{3} = 28 \cdot \frac{64}{4} \)
\( T_{3} = 28 \cdot 16 \)
\( T_{3} = 448 \)
Thus, the term in the expansion of \( \left(2 x + \frac{1}{2 x^3}\right)^8 \) that is independent of \( x \) is \( 448 \).
Question
(i) Find the first three terms when \(\left ( 2+3x \right )^{6}\) is expanded in ascending powers of x.
(ii) In the expansion of \(\left ( 1+ax \right )\left ( 2+3x \right )^{6}\) , the coefficient of x is zero. Find the value of a.
▶️Answer/Explanation
(i)We’ll use the binomial theorem to expand \((2+3x)^6\). The binomial theorem states that:
\( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
Here, \(a = 2\) and \(b = 3x\), and we’re expanding to the power of 6. The first three terms correspond to \(k = 0, 1, 2\).
\(\Rightarrow \)For \(k = 0\):
\( \binom{6}{0} 2^6 (3x)^0 = 1 \cdot 64 \cdot 1 = 64 \)
\(\Rightarrow\) For \(k = 1\):
\( \binom{6}{1} 2^5 (3x)^1 = 6 \cdot 32 \cdot 3x = 576x \)
\(\Rightarrow \)For \(k = 2\):
\( \binom{6}{2} 2^4 (3x)^2 = 15 \cdot 16 \cdot 9x^2 = 2160x^2 \)
So, the first three terms are \(64 + 576x + 2160x^2\).
(ii)The expansion of \((2+3x)^6\) has already been calculated, and we are interested in the terms that will contribute to the \(x\) term when multiplied by \((1+ax)\).
So, the coefficient of \(x\) in the entire expansion is \(64a + 576\). Setting this equal to zero gives:
\( 64a + 576 = 0 \)
\( 64a = -576 \)
\( a = \frac{-576}{64} \)
\( a = -9 \)
Hence, the value of \(a\) is \(-9\).
Question
In the expansion of \(\left ( 1-\frac{2x}{a} \right )\left ( a+x \right )^{5}\), where a is a non-zero constant, show that the coefficient of \(x^{2}\) is zero.
▶️Answer/Explanation
The binomial theorem states that:
\( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
Here, \(a = a\) and \(b = x\), and we are expanding to the power of 5. We are interested in the terms up to \(x^2\), so we will consider \(k = 0, 1, 2\).
1. For \(k = 0\):
\( \binom{5}{0} a^5 x^0 = 1 \cdot a^5 = a^5 \)
2. For \(k = 1\):
\( \binom{5}{1} a^4 x^1 = 5a^4 x \)
3. For \(k = 2\):
\( \binom{5}{2} a^3 x^2 = 10a^3 x^2 \)
So, the expansion of \((a + x)^5\) up to the \(x^2\) term is \(a^5 + 5a^4 x + 10a^3 x^2\).
\( \left(1 – \frac{2x}{a}\right)(a^5 + 5a^4 x + 10a^3 x^2) \)
Expanding this,
1. \(1 \times 10a^3 x^2 = 10a^3 x^2\)
2. \(-\frac{2x}{a} \times 5a^4 x = -10a^3 x^2\)
Now, adding these
\( 10a^3 x^2 – 10a^3 x^2 = 0 \)
The \(x^2\) terms cancel each other out, thus proving that the coefficient of \(x^2\) in the expansion of \(\left(1-\frac{2x}{a}\right)(a+x)^5\) is zero.
Question
In the expansion of \(\left ( 2+ax \right )^{7}\), the coefficient of x is equal to the coefficient of \(x^{2}\). Find the value of the non-zero constant a.
▶️Answer/Explanation
The general term in the binomial expansion is given by:
\( T_{k+1} = \binom{7}{k} (2)^{7-k} (ax)^k \)
Coefficient of \( x \):
\( T_2 = \binom{7}{1} (2)^{6} (ax)^1 = 7 \cdot 64 \cdot a = 448a \)
Coefficient of \( x^2 \):
\( T_3 = \binom{7}{2} (2)^{5} (ax)^2 = 21 \cdot 32 \cdot a^2 = 672a^2 \)
For these two coefficients to be equal, we set them equal to each other.
Solving for \( a \):
\( 448a = 672a^2 \)
\( 1 = \frac{672}{448} a \)
\( 1 = \frac{3}{2} a \)
\( a = \frac{2}{3} \)
Thus, the value of the non-zero constant \( a \) that makes the coefficient of \( x \) equal to the coefficient of \( x^2 \) in the expansion of \( (2+ax)^7 \) is \( a = \frac{2}{3} \).
Question
The coefficients of \(x^2\) and \(x^3\) in the expansion of \((3 − 2x)^6\) are a and b respectively. Find the value of \(\frac{a}{b}\).
▶️Answer/Explanation
The coefficient of \(x^2\) arises when \(k = 2\) in the binomial expansion. The term in the expansion for \(k = 2\) is:
\( \binom{6}{2} (3)^{6-2} (-2x)^2 = \binom{6}{2} \cdot 3^4 \cdot 4x^2 \)
\( \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \)
So, the coefficient of \(x^2\) is
\( 15 \cdot 3^4 \cdot 4 = 15 \cdot 81 \cdot 4 = 4860 \)
Thus, \(a = 4860\).
Coefficient of \(x^3\)
The coefficient of \(x^3\) arises when \(k = 3\). The term in the expansion for \(k = 3\) is:
\( \binom{6}{3} (3)^{6-3} (-2x)^3 = \binom{6}{3} \cdot 3^3 \cdot (-8x^3) \)
\( \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)
So, the coefficient of \(x^3\) is
\( 20 \cdot 3^3 \cdot (-8) = 20 \cdot 27 \cdot (-8) = -4320 \)
Thus, \(b = -4320\).
Now,
\( \frac{a}{b} = \frac{4860}{-4320} = -\frac{9}{8} \)
Therefore, \(\frac{a}{b} = -\frac{9}{8}\).
Question
The function f is defined for x≥ 0 by \(f(x) =(4x + 1)^\frac{3}{2}\).
(i) Find f′(x) and f′′(x).
The first, second and third terms of a geometric progression are respectively f(2), f′(2) and kf′′(2).
(ii) Find the value of the constant k.
▶️Answer/Explanation
(i) \( f(x) = (4x + 1)^{\frac{3}{2}} \):
Using chain rule,
\( f'(x) = \frac{3}{2}(4x + 1)^{\frac{1}{2}} \cdot 4 = 6(4x + 1)^{\frac{1}{2}} \)
Differentiating \( f'(x) \) again:
\( f”(x) = 6 \cdot \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 = 12(4x + 1)^{-\frac{1}{2}} \)
(ii) The first, second, and third terms of a geometric progression are given as \( f(2) \), \( f'(2) \), and \( kf”(2) \) respectively.
The common ratio \( r \) of a geometric progression can be found by dividing the second term by the first term. Thus,
\( r = \frac{f'(2)}{f(2)} \)
The third term can be expressed as the second term multiplied by the common ratio \( r \), so we have:
\( kf”(2) = f'(2) \cdot r \)
Substituting \( r \) and solving for \( k \):
\( kf”(2) = f'(2) \cdot \frac{f'(2)}{f(2)} \)
Now, substituting the values of \( f(2) \), \( f'(2) \), and \( f”(2) \):
\( f(2) = (4 \cdot 2 + 1)^{\frac{3}{2}} = 9^{\frac{3}{2}} = 27 \)
\( f'(2) = 6(4 \cdot 2 + 1)^{\frac{1}{2}} = 6 \cdot 3 = 18 \)
\( f”(2) = 12(4 \cdot 2 + 1)^{-\frac{1}{2}} = 12 \cdot \frac{1}{3} = 4 \)
Thus,
\( k \cdot 4 = 18 \cdot \frac{18}{27} \)
\( k = \frac{18}{4} \cdot \frac{18}{27} = \frac{9}{2} \cdot \frac{2}{3} = 3 \)
Therefore, the value of the constant \( k \) is 3.
Question
In the expansion of \(\left ( \frac{1}{ax}+2ax^{2} \right )^{5}\) the coefficient of x is 5. Find the value of the constant a.
▶️Answer/Explanation
The general term of the expansion is given by:
\( T_{k+1} = \binom{5}{k} \left(\frac{1}{ax}\right)^{5-k} (2ax^2)^k \)
\( -1 \cdot (5-k) + 2k = 1 \)
\( -5 + k + 2k = 1 \)
\( 3k = 6 \)
\( k = 2 \)
Now,
\( T_{2+1} = \binom{5}{2} \left(\frac{1}{ax}\right)^{5-2} (2ax^2)^2 \)
\( T_3 = 10 \cdot \left(\frac{1}{ax}\right)^3 \cdot (4a^2x^4) \)
\( T_3 = 10 \cdot \frac{4a^2x^4}{a^3x^3} \)
\( T_3 = 40 \cdot \frac{a^2x}{a^3} \)
\( T_3 = 40 \cdot \frac{x}{a} \)
The coefficient of \( x \) in this term is \( \frac{40}{a} \). We are given that the coefficient is 5, so we have:
\( \frac{40}{a} = 5 \)
\( a = \frac{40}{5} \)
\( a = 8 \)
So, the value of the constant \( a \) is 8.
Question
(i) Find the coefficients of \(x^{4}\) and \(x^{5}\) in the expansion of \((1 – 2x)^{5}\).
(ii) It is given that, when \((1 + px)(1 – 2x)^{5}\) is expanded, there is no term in \(x^{5}\). Find the value of the constant p.
▶️Answer/Explanation
(i)The binomial theorem states that for any real numbers \(a\) and \(b\), and any positive integer \(n\), the expansion of \((a + b)^n\) is given by:
\( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \)
In this case, \(a = 1\) and \(b = -2x\), and we’re expanding to the power of 5.
The general term in the expansion is given by:
\( \binom{5}{k} (1)^{5-k} (-2x)^k = \binom{5}{k} (-2)^k x^k \)
Coefficient of \(x^4\): This term arises when \(k = 4\).
\( \binom{5}{4} (-2)^4 x^4 = 5 \cdot 16 x^4 = 80x^4 \)
So, the coefficient of \(x^4\) is 80.
Coefficient of \(x^5\): This term arises when \(k = 5\).
\( \binom{5}{5} (-2)^5 x^5 = 1 \cdot -32 x^5 = -32x^5 \)
So, the coefficient of \(x^5\) is -32.
(ii) The \(x^5\) term in the expanded expression arises from two sources:
1. The \(x^4\) term from \((1 – 2x)^5\) (which is \(80x^4\)) multiplied by the \(px\) term from \((1 + px)\).
2. The \(x^5\) term from \((1 – 2x)^5\) (which is \(-32x^5\)) multiplied by the constant term from \((1 + px)\) (which is 1).
So, the coefficient of \(x^5\) in the entire expansion is the sum of these contributions, which should equal zero:
\( 80px^5 – 32x^5 = 0 \)
\( 80p – 32 =\)
\( 80p = 32 \)
\( p = \frac{32}{80} = \frac{2}{5} \)
Therefore, the value of the constant \(p\) is \(\frac{2}{5}\).
Question
The term independent of x in the expansion of \(\left ( 2x+\frac{k}{x} \right )^{6}\) ,, where k is a constant, is 540.
(i) Find the value of k.
(ii) For this value of k, find the coefficient of \( x^{2}\) in the expansion.15\times \(16\times k^{2}=(or540x^{2})\)
▶️Answer/Explanation
(i)The general term in a binomial expansion is given by:
\( T_{r+1} = \binom{n}{r} (a)^{n-r} (b)^r \)
For the given binomial \( (2x + \frac{k}{x})^6 \), we set \( a = 2x \) and \( b = \frac{k}{x} \), and \( n = 6 \).
\( (2x)^{6-r} \left(\frac{k}{x}\right)^r = x^{6-r-r} k^r = x^{6-2r} k^r \)
\( 6 – 2r = 0 \)
\( r = 3 \)
\( T_{3+1} = \binom{6}{3} (2x)^{6-3} \left(\frac{k}{x}\right)^3 \)
\( T_4 = \binom{6}{3} (2x)^3 \frac{k^3}{x^3} \)
\( T_4 = 20 \cdot 8k^3 \)
\( T_4 = 160k^3 \)
\( 160k^3 = 540 \)
\( k^3 = \frac{540}{160} \)
\( k^3 = \frac{27}{8} \)
\( k = \left(\frac{27}{8}\right)^{\frac{1}{3}} \)
\( k = \frac{3}{2} \)
So, the value of \( k \) is \( \frac{3}{2} \) or 1.5.
(ii) With \( k = \frac{3}{2} \), we look for the term in the expansion where the power of \( x \) is 2.
\( 6 – 2r = 2 \)
\( 6 – 2 = 2r \)
\( 4 = 2r \)
\( r = 2 \)
Using the general term formula:
\( T_{2+1} = \binom{6}{2} (2x)^{6-2} \left(\frac{3}{2x}\right)^2 \)
\( T_3 = 15 \cdot (2x)^4 \left(\frac{9}{4x^2}\right) \)
\( T_3 = 15 \cdot 16x^2 \frac{9}{4} \)
\( T_3 = 15 \cdot 4 \cdot 9x^2 \)
\( T_3 = 540x^2 \)
Thus, the coefficient of \( x^2 \) in the expansion is 540.
Question
The coefficient of \(x^{3}\) in the expansion of \( (1-px)^{5}\) is −2160. Find the value of the constant p.
▶️Answer/Explanation
The general term of the expansion is given by:
\( T_{k+1} = \binom{n}{k} \cdot (1)^{n-k} \cdot (-px)^k \)
For the term that contains \( x^3 \), we need to find \( T_4 \) where \( k = 3 \):
\( T_{3+1} = \binom{5}{3} \cdot (1)^{5-3} \cdot (-px)^3 \)
\( T_4 = 10 \cdot (-px)^3 \)
\( T_4 = 10 \cdot (-p)^3 \cdot x^3 \)
This gives us the coefficient of \( x^3 \) as \( -10p^3 \). We are given that this coefficient is -2160, so we set the expression equal to -2160 and solve for \( p \):
\( -10p^3 = -2160 \)
\( p^3 = \frac{2160}{10} \)
\( p^3 = 216 \)
\( p = \sqrt[3]{216} \)
\( p = 6 \)
So, the value of the constant \( p \) is 6.
Question
In the expansion of \((2x^{2}+\frac{1}{x})^{6}\), the coefficients of x6 and x3 are equal.
(a) Find the value of the non-zero constant a.
(b) Find the coefficient of x6 in the expansion of \((1-x^{3})(2x^{2}+\frac{a}{x})^{6}\)
▶️Answer/Explanation
(a) For the coefficient of \( x^6 \), we set \( 4(2) – 2k = 6 \), which gives \( k = 2 \). This corresponds to the term:
\( 6C2 \times \left(2x^2\right)^4 \times \left(\frac{a}{x}\right)^2 \)
For the coefficient of \( x^3 \), we set \( 3(2) – 3k = 3 \), which gives \( k = 3 \). This corresponds to the term:
\( 6C3 \times \left(2x^2\right)^3 \times \left(\frac{a}{x}\right)^3 \)
\( 15 \times 2^4 \times a^2 = 20 \times 2^3 \times a^3 \)
\( a = \frac{15 \times 2^4}{20 \times 2^3} = \frac{3}{2} \)
So, \( a = \frac{3}{2} \).
(b) Finding the coefficient of \( x^6 \) in \( \left(1 – x^3\right) \left(2x^2 + \frac{3}{2x}\right)^6 \):
Since the coefficients of \( x^6 \) and \( x^3 \) in \( \left(2x^2 + \frac{3}{2x}\right)^6 \) are equal (as proven in part (a)), the \( x^6 \) term in the expansion of \( \left(1 – x^3\right) \left(2x^2 + \frac{3}{2x}\right)^6 \) will cancel out.
Thus, the coefficient of \( x^6 \) in the given expression is \( 0 \).
Question
A curve has equation y = kx2 + 2x − k and a line has equation \(y = kx − 2\), where k is a constant. Find the set of values of k for which the curve and line do not intersect.
▶️Answer/Explanation
To find the set of values of \( k \) for which the curve \( y = kx^2 + 2x – k \) and the line \( y = kx – 2 \) do not intersect, we should equate their equations and find the values of \( k \) for which the resulting quadratic equation has no real solutions. This means that the discriminant of the quadratic equation should be less than zero.
\( kx^2 + 2x – k = kx – 2 \)
\( kx^2 + (2 – k)x – k + 2 = 0 \)
\( kx^2 – kx + 2x – k + 2 = 0 \)
\( kx^2 – (k – 2)x – k + 2 = 0 \)
This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = k \), \( b = -k + 2 \), and \( c = -k + 2 \).
The discriminant \( \Delta \) of a quadratic equation is given by \( \Delta = b^2 – 4ac \).
For the quadratic equation to have no real solutions, \( \Delta < 0 \).
\( \Delta = (-k + 2)^2 – 4(k)(-k + 2) \)
\( \Delta = (k^2 – 4k + 4) – 4k(-k + 2) \)
\( \Delta = k^2 – 4k + 4 + 4k^2 – 8k \)
\( \Delta = 5k^2 – 12k + 4 \)
For no real solutions, we require \( \Delta < 0 \):
\( 5k^2 – 12k + 4 < 0 \)
Solving this quadratic inequality will give us the range of values for \( k \) where the curve and line do not intersect.
The roots of the equation \( 5k^2 – 12k + 4 = 0 \) can be found using the quadratic formula:
\( k = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
where \( a = 5 \), \( b = -12 \), and \( c = 4 \).
The roots are:
\( k = \frac{12 \pm \sqrt{(-12)^2 – 4 \cdot 5 \cdot 4}}{2 \cdot 5} \)
\( k = \frac{12 \pm \sqrt{144 – 80}}{10} \)
\( k = \frac{12 \pm \sqrt{64}}{10} \)
\( k = \frac{12 \pm 8}{10} \)
So the roots are:
\( k_1 = \frac{20}{10} = 2 \)
\( k_2 = \frac{4}{10} = \frac{2}{5} \)
The quadratic inequality \( 5k^2 – 12k + 4 < 0 \) is satisfied between these roots.
Therefore, the set of values of \( k \) for which the curve and line do not intersect is \( \frac{2}{5} < k < 2 \).
Question
The coefficient of \(\frac{1}{x}\) in the expansion of \((2x+\frac{a}{x^2})^5\) is 720.
(a) Find the possible values of the constant a.
(b) Hence find the coefficient of \(\frac{1}{x^7}\) in the expansion.
▶️Answer/Explanation
(a)The general term in the expansion is:
\( \binom{5}{k} (2x)^{5-k} \left(\frac{a}{x^2}\right)^k = \binom{5}{k} 2^{5-k} a^k x^{5-k-2k} \)
We need to find the term where the exponent of \(x\) is -1 (for the coefficient of \(\frac{1}{x}\)). Setting the exponent equal to -1:
\( 5 – k – 2k = -1 \)
\( 5 – 3k = -1 \)
\( 3k = 6 \)
\( k = 2 \)
Substituting \(k = 2\) into the general term:
\( \binom{5}{2} 2^{5-2} a^2 x^{-1} = 10 \cdot 8 \cdot a^2 \cdot \frac{1}{x} = 80a^2 \cdot \frac{1}{x} \)
Since the coefficient of \(\frac{1}{x}\) is given as 720, we have:
\( 80a^2 = 720 \)
\( a^2 = \frac{720}{80} \)
\( a^2 = 9 \)
\( a = \pm 3 \)
(b) Now, we find the term in the expansion where the exponent of \(x\) is -7.
\( 5 – k – 2k = -7 \)
\( 5 – 3k = -7 \)
\( 3k = 12 \)
\( k = 4 \)
Substituting \(k = 4\) into the general term:
\( \binom{5}{4} 2^{5-4} a^4 x^{-7} = 5 \cdot 2 \cdot a^4 \cdot \frac{1}{x^7} = 10a^4 \cdot \frac{1}{x^7} \)
Using the value of \(a\) as \(\pm 3\), the coefficient of \(\frac{1}{x^7}\) is:
\( 10(\pm 3)^4 = 10 \cdot 81 = 810 \)
Therefore, the coefficient of \(\frac{1}{x^7}\) in the expansion is 810.
Question
(a) Find the first three terms in the expansion of (3 − 2x)5 in ascending powers of x.
(b) Hence find the coefficient of x2 in the expansion of (4 + x)2 (3 − 2x)5
▶️Answer/Explanation
(a) Expanding \((3 – 2x)^5\):
Using the binomial theorem, \((a – b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}(-b)^k\),
\(
\begin{aligned}
(3 – 2x)^5 &= \binom{5}{0}3^5(-2x)^0 + \binom{5}{1}3^4(-2x)^1 + \binom{5}{2}3^3(-2x)^2 + \ldots \\
&= 1 \cdot 3^5 \cdot 1 – 5 \cdot 3^4 \cdot 2x + 10 \cdot 3^3 \cdot (2x)^2 + \ldots \\
&= 243 – 5 \cdot 81 \cdot 2x + 10 \cdot 27 \cdot 4x^2 + \ldots \\
&= 243 – 810x + 1080x^2 + \ldots
\end{aligned}
\)
The first three terms are \(243\), \(-810x\), and \(1080x^2\).
(b) The expanded form of \((4 + x)^2\) is \( 16 + 8x + x^2 \).
From part (a), the first three terms of \((3 – 2x)^5\) are \( 243 \), \( -810x \), and \( 1080x^2 \).
To find the coefficient of \( x^2 \) in the product \((4 + x)^2(3 – 2x)^5\), we need to look at the combinations of terms from each expansion that will produce an \( x^2 \) term:
1. \( 16 \) (from \( (4 + x)^2 \)) times \( 1080x^2 \) (from \( (3 – 2x)^5 \)).
2. \( 8x \) (from \( (4 + x)^2 \)) times \( -810x \) (from \( (3 – 2x)^5 \)).
3. \( x^2 \) (from \( (4 + x)^2 \)) times \( 243 \) (from \( (3 – 2x)^5 \)).
Now,
1. \( 16 \times 1080x^2 = 17280x^2 \)
2. \( 8x \times -810x = -6480x^2 \)
3. \( x^2 \times 243 = 243x^2 \)
Adding these up gives the total coefficient of \( x^2 \) in the expansion:
\( 17280x^2 – 6480x^2 + 243x^2 = 10800x^2 + 243x^2 = 11043x^2 \)
So, the coefficient of \( x^2 \) in the expansion of \((4 + x)^2(3 – 2x)^5\) is \( 11043 \).
.
Question
a) Find the first three terms in the expansion, in ascending powers of x, of (1+x)5.
b) Find the first terms in the expansion, in ascending powers of x, (1-2x)6.
c) Hence find the coefficient of \(x^2\) in the expansion of (1+x)5(1-2x)6.
▶️Answer/Explanation
(a) The first three terms in the expansion of \( (1 + x)^5 \):
Using the binomial theorem, these terms are:
1. \( \binom{5}{0}x^0 = 1 \)
2. \( \binom{5}{1}x^1 = 5x \)
3. \( \binom{5}{2}x^2 = 10x^2 \)
So, the first three terms are \( 1 + 5x + 10x^2 \).
(b) The first three terms in the expansion of \( (1 – 2x)^6 \):
Using the binomial theorem, these terms are:
1. \( \binom{6}{0}(-2x)^0 = 1 \)
2. \( \binom{6}{1}(-2x)^1 = -12x \)
3. \( \binom{6}{2}(-2x)^2 = 60x^2 \)
So, the first three terms are \( 1 – 12x + 60x^2 \).
(c) Finding the coefficient of \( x^2 \) in the expansion of \( (1 + x)^5(1 – 2x)^6 \):
The product is \( (1 + 5x + 10x^2)(1 – 12x + 60x^2) \).
1. The \( x^2 \) term from \( (1 + x)^5 \) (which is \( 10x^2 \)) and the constant term from \( (1 – 2x)^6 \) (which is \( 1 \)).
2. The \( x \) term from \( (1 + x)^5 \) (which is \( 5x \)) and the \( x \) term from \( (1 – 2x)^6 \) (which is \( -12x \)).
3. The constant term from \( (1 + x)^5 \) (which is \( 1 \)) and the \( x^2 \) term from \( (1 – 2x)^6 \) (which is \( 60x^2 \)).
The contributions to the \( x^2 \) term are:
1. \( 10x^2 \cdot 1 = 10x^2 \)
2. \( 5x \cdot -12x = -60x^2 \)
3. \( 1 \cdot 60x^2 = 60x^2 \)
Adding these contributions gives the coefficient of \( x^2 \) in the expansion:
\( 10x^2 – 60x^2 + 60x^2 = 10x^2 \)
Therefore, the coefficient of \( x^2 \) in the expansion of \( (1 + x)^5(1 – 2x)^6 \) is indeed \( 10 \).
Question
Find the term independent of x in each of the following expansions.
(a) \(\left ( 3x + \frac{2}{x^{2}} \right )^{6}\)
(b) \(\left ( 3x + \frac{2}{x^{2}} \right )^{6}\) \(\left ( 1 – x^{3} \right )\)
▶️Answer/Explanation
(a) We use the binomial theorem, which states that \( (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \). The term independent of \( x \) occurs when the powers of \( x \) in \( 3x \) and \( \frac{2}{x^2} \) cancel each other out.
\(\binom{6}{2} (3x)^4 \left(\frac{2}{x^2}\right)^2 \)
\( \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \)
\( 15 \times 3^4 \times 2^2 = 15 \times 81 \times 4 = 4860 \)
So, the term independent of \( x \) in \( \left(3x + \frac{2}{x^2}\right)^6 \) is 4860.
(b) Here, we already know the term independent of \( x \) in \( \left(3x + \frac{2}{x^2}\right)^6 \) is 4860. Now we need to consider how multiplying by \( (1 – x^3) \) affects this:
The term \( 4860 \) multiplied by \( 1 \) remains \( 4860 \).
This is the \( x^3 \) term from the expansion, which is:
\( \binom{6}{3} (3x)^3 \left(\frac{2}{x^2}\right)^3 \)
\( \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)
\( 20 \times 3^3 \times 2^3 = 20 \times 27 \times 8 = 4320 \)
The term independent of \( x \) in \( \left(3x + \frac{2}{x^2}\right)^6 (1 – x^3) \) is therefore:
\( 4860 – 4320 = 540 \)
Question
The coefficient of \(\frac{1}{x}\) in the expansion of \((kx+\frac{1}{x})^{5}+(1-\frac{2}{x})^{8}\) is 74. find the value of positive constant k.
▶️Answer/Explanation
The coefficient of \( \frac{1}{x} \) in this expansion is given by \( \binom{5}{3} k^2 = 10k^2 \).
For \( \left(1 – \frac{2}{x}\right)^8 \), the term containing \( \frac{1}{x} \) arises when \( r = 1 \), leading to \( \binom{8}{1} (-2)^1 = -16 \).
The total coefficient of \( \frac{1}{x} \) in the combined expression is the sum of these coefficients:
\( 10k^2 – 16 = 74 \)
Solving for \( k \), we have:
\( 10k^2 = 74 + 16 \)
\( 10k^2 = 90 \)
\( k^2 = \frac{90}{10} \)
\( k^2 = 9 \)
\( k = \pm3 \)
Since \( k \) is a positive constant, \( k = 3 \).