CIE Binomial Theorem 2020

$\def\frac{\dfrac}$

 1. (CIE $0606 / 2018 / \mathrm{w} / 11 / \mathrm{q} 3)$

The coefficient of $x^{2}$ in the expansion of $(2-x)(3+k x)^{6}$ is equal to 972 . Find the possible values of the constant $k$.


2. (CIE 0606/2018/w/13/q1)

(a) In the expansion of $(2+p x)^{5}$ the coefficient of $x^{3}$ is equal to $-\frac{8}{25}$. Find the value of the constant $p$.

(b) Find the term independent of $x$ in the expansion of $\left(2 x^{2}+\frac{1}{4 x^{2}}\right)^{3}$


3. $(\mathrm{CIF} 0606 / 2019 / \mathrm{s} / 11 / \mathrm{q} 4)$

(i) The first 3 terms, in ascending powers of $x$, in the expansion of $(2+b x)^{8}$ can be written as $a+256 x+c x^{2}$. Find the valuc of each of the constants $a, b$ and $c$.

(ii) Using the values found in part (i), find the term independent of $x$ in the expansion of

$$(2+b x)^{8}\left(2 x-\frac{3}{x}\right)^{2}$$


4. $(\mathrm{CIE} 0606 / 2019 / \mathrm{m} / 12 / \mathrm{q} 3)$

(i) Find the first 3 terms in the expansion, in ascending powers of $x$, of $\left(3-\frac{x}{9}\right)^{6}$. Give the terms in their simplest form.

(ii) Hence find the term independent of $x$ in the expansion of $\left(3-\frac{x}{9}\right)^{6}\left(x-\frac{2}{x}\right)^{2}$.


5. (CIE $0606 / 2019 / \mathrm{w} / 12 / \mathrm{q} 3)$

The first three terms in the expansion of $\left(1-\frac{x}{7}\right)^{14}(1-2 x)^{4}$ can be written as $1+a x+b x^{2}$. Find the value of each of the constants $a$ and $b$.


6. $(\mathrm{CIE} 0606 / 2019 / \mathrm{w} / 21 / \mathrm{q} 10)$

(i) Expand $(3+x)^{4}$ evaluating each coefficient.

In the expansion of $\left(x-\frac{p}{x}\right)(3+x)^{4}$ the coefficient of $x$ is zero.

(ii) Find the value of the constant $p$. $[2]$

(iii) Hence find the term independent of $x$. [1]

(iv) Show that the coefficient of $x^{2}$ is 90 . $[2]$


7. $(\mathrm{CIE} 0606 / 2019 / \mathrm{s} / 22 / \mathrm{q} 8)$

(a) In the binomial expansion of $\left(a-\frac{x}{2}\right)^{6}$, the coefficient of $x^{3}$ is 120 times the coefficient of $x^{5}$. Find the possible values of the constant $a$.

(b) (i) Expand $(1+2 x)^{20}$ in ascending powers of $x$, as far as the term in $x^{3}$. Simplify each term.

(ii) Use your expansion to show that the value of $0.98^{20}$ is $0.67$ to 2 decimal places.


8. (CIE $0606 / 2019 / \mathrm{s} / 23 / \mathrm{q} 8)$

(a) (i) Given that $\left(x^{2}-\frac{1}{p x}\right)^{8}=x^{16}-4 x^{13}+q x^{10}+r x^{7}+\ldots .$ find the value of each of the constants $p$, $q$ and $r$

(ii) Explain why there is no term independent of $x$ in the binomial expansion of $\left(x^{2}-\frac{1}{p x}\right)^{3}$

(b) In the binomial expansion of $\left(1-\frac{\sqrt{x}}{2}\right)^{n}$, where $n$ is a positive integer, the coefficient of $x$ is 30 . Form an equation in $n$ and hence find the value of $n$.


9. $(\mathrm{CIE} 0606 / 2020 / \mathrm{m} / 12 / \mathrm{q} 3)$

The first 3 terms in the expansion of $(3-a x)^{5}$, in ascending powers of $x$, can be written in the form $b-81 x+c x^{2}$. Find the value of each of $a, b$ and $c$.


10. $(\mathrm{CIE} 0606 / 2020 / \mathrm{s} / 12 / \mathrm{q} 3)$

(a) Find the first 3 terms in the expansion of $\left(4-\frac{x}{16}\right)^{6}$ in ascending powers of $x$. Give cach term in its simplest form.

(b) Hence find the term independent of $x$ in the expansion of $\left(4-\frac{x}{16}\right)^{6}\left(x-\frac{1}{x}\right)^{2}$.


11. $(\mathrm{CIE} 0606 / 2020 / \mathrm{w} / 12 / \mathrm{q} 5)$

Find the coefficient of $x^{2}$ in the expansion of $\left(x-\frac{3}{x}\right)\left(x+\frac{2}{x}\right)^{5}$


12. (CIE $0606 / 2020 / \mathrm{w} / 13 / \mathrm{q} 5)$

Given that the coefficient of $x^{2}$ in the expansion of $(1+x)\left(1-\frac{x}{2}\right)^{n}$ is $\frac{25}{4}$, find the value of the positive integer $n$.


13. $(\mathrm{CIE} 0606 / 2020 / \mathrm{w} / 21 / \mathrm{q} 5)$

The first three terms in the expansion of $(a+b x)^{5}(1+x)$ are $32-208 x+c x^{2} .$ Find the value of each of the integers $a, b$ and $c$. [7]


14. $(\mathrm{CIE} 0606 / 2020 / \mathrm{w} / 21 / \mathrm{q} 7)$

DO NOT USE A CALCULATOR IN THIS QUESTION.

$$p(x)=2 x^{3}-3 x^{2}-23 x+12$$

(a) Find the value of $\mathrm{p}\left(\frac{1}{2}\right)$.

(b) Write $\mathrm{p}(x)$ as the product of three linear factors and hence solve $\mathrm{p}(x)=0$


15. $(\mathrm{CIE} 0606 / 2020 / \mathrm{s} / 21 / \mathrm{q} 8)$

(a) Expand $(2-x)^{5}$, simplifying each coefficient.

(b) Hence solve $\frac{\mathrm{e}^{(2-x)^{3}} \times e^{80 r}}{\mathrm{e}^{104^{4}+32}}=\mathrm{e}^{-x^{3}}$.


16. $(\mathrm{CIE} 0606 / 2020 / \mathrm{s} / 23 / \mathrm{q} 9)$

DO NOT USE A CALCULATOR IN THIS QUESTION.

(a) Find the term independent of $x$ in the binomial expansion of $\left(3 x-\frac{1}{x}\right)^{6}$.

(b) In the expansion of $\left(1+\frac{x}{2}\right)^{n}$ the coefficient of $x^{4}$ is half the coefficient of $x^{6}$. Find the value of the positive constant $n$.


Answer

1. $k=1, \frac{-2}{5}$

2. (a) $p=-\frac{1}{5}$ (b) $\frac{35}{8}$

3. (i) $a=256, b=\frac{1}{4}, c=112$

(ii) $-2064$

4. (i) $729-162 x+15 x^{2}$

(ii) $-2856$

5. $a=-10, b=\frac{293}{7}$

6. (i) $81+108 x+54 x^{2}+12 x^{3}+x^{4}$

(ii) $p=1.5$ (iii) $-162$ (iv) Show

7. (a) $a=\pm 3$

(b) (i) $1+40 x+760 x^{2}+9120 x^{3}$

(ii) Show

8. (a)(i) $\quad p=2, q=7, r=-7$

(ii) Valid explanation (b) $n=16$

9. $a=\frac{1}{5}, b=243, c=\frac{54}{5} 12345678910$

10. (a) $4096-384 x+15 x^{2}$

(b) $-8177$

11. 10

$12 . n=10$

13. $a=2, b=-3, c=480$

14. (a) $p\left(\frac{1}{2}\right)=0$,

(b) $p(x)=(2 x-1)(x-4)(x+3), x=\frac{1}{2}, 4,-3$

15. (a) $32-80 x+80 x^{2}-40 x^{3}+10 x^{4}-x^{5}$

(b) $x=0, x=2$

16. (a) $-540$ (b) $n=20$


Post a Comment

0 Comments