### 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 $$\left [ \left ( 1-x \right )\left ( 1+2x \right ) \right ]^{6}$$

(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 $$\left ( 4x^{3} +\frac{1}{2x}\right )^{8}$$

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 $$\left ( 2x^{3} -\frac{1}{x^{2}}\right )^{7}$$.

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 $$\left ( 2x+\frac{1}{2x^{3}} \right )^{8}$$.

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.

(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.

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. $$-\frac{2x}{a} \times 5a^4 x = -10a^3 x^2$$
$$10a^3 x^2 – 10a^3 x^2 = 0$$
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.

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 $$\frac{a}{b}$$.

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.

(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.

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.

(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 $$16\times k^{2}=(or540x^{2})$$

(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.

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 $$(1-x^{3})(2x^{2}+\frac{a}{x})^{6}$$

(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.

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.

(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

(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.

(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) $$\left ( 3x + \frac{2}{x^{2}} \right )^{6}$$ $$\left ( 1 – x^{3} \right )$$

(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.

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, $$k = 3$$.

Scroll to Top