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)(a)\((1-x)^{6}=1-6x+15x^{2}\)

(b)\(\left ( 1+2x \right )^{6}=1+12x+60x^{2}\)

(ii) Product of (a) and (b)

\(\rightarrow 60-72+15=3\)

Question

Find the term independent of x in the expansion of \(\left ( 4x^{3} +\frac{1}{2x}\right )^{8}\)

Answer/Explanation

\(\left [ ^{8}C_{6} \right ]\times \left [ 16 \right ]\times \left ( x^{6} \right )\times \left [ \frac{1}{\left ( 2^{6} \right )\left ( x^{6} \right )} \right ]\)

7

Question

Find the coefficient of \(x^{6}\) in the expansion of \(\left ( 2x^{3} -\frac{1}{x^{2}}\right )^{7}\).

Answer/Explanation

\(\left [ ^{7} C_{3}\right ]\times \left [ \left ( 2x^{3} \right ) ^{4}\right ]\times \left [ \left ( -\frac{1}{x^{2}} \right ) ^{3}\right ]\)

\(35\times 2^{4}\times (-1)^{3}\) leading to their answer 

\(-560\left ( x^{6} \right )\) as answer

Question

 Find the term independent of x in the expansion of \(\left ( 2x+\frac{1}{2x^{3}} \right )^{8}\).

Answer/Explanation

\(^{8}C_{6}\left ( 2x \right )^{6}\left ( \frac{1}{2x^{3}} \right )^{2}\)

\(28\times 64\times \frac{1}{4}\) (powers and factorials evaluated)

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)\(64+576x+2160x^{2}\)

(ii)\(576a(x^{2})+216(x^{2})=0\)

\(a=-\frac{2160}{576}\)

\(\left ( -\frac{15}{4} \right )\) or -3.75

 

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

\((a+x)^{5}=a^{5}+^{5}C_{1}a^{4}x+^{5}C_{2}a^{3}x^{2}+….\)

\(\left ( -\frac{2}{a}\times (their 5a^{4})+(their 10a^{3}) \right )(x^{2})\)

0

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

\(^{7}C_{1}\times 2^{6}\times a=^{7}C^{2}\times 2^{5}\times a^{2}\)

a=\left ( \frac{7\times 2^{6}}{21\times 2^{5}} \right )=\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

\(\left ( 3-2x \right )^{6}\)

Coefficient of \( x^{2}=3^{4}+(-2)^{2}\times _{6}C_{2}=a\)

Coefficient of x^{3}=3^{3}+(-2)^{3}\times _{6}C_{3}=b

Question.

Find the term independent of x in the expansion of \((x –\frac{3}{2x})^2\).

Answer/Explanation

\(\left ( x-\frac{3}{2x} \right )^{6}\)

Term is \(^{6}C_{3}\times x^{3}\times \left ( \frac{-3}{2x} \right )3\)

\(\rightarrow -67.5\) oe

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) \(f{}’\left ( x \right )=\left [ \frac{3}{2}\left ( 4x+1 \right )^{\frac{1}{2}} \right ]\left [ 4 \right ]\)

 \(f{}”\left ( x \right )=6\times \frac{1}{2}\times \left ( 4x+1 \right )^{-\frac{1}{2}}\times 4\)

(ii) f(2),\(f{}’\left ( 2 \right )\),\(kf{}”\left ( 2 \right )=27\),4k OR 12

\(\frac{27}{18}=\frac{18}{4k}\) oe OR \(kf{}”\left ( 2 \right )=12\Rightarrow k=3\)

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

\(^{5}C_{2}\left ( \frac{1}{ax} \right )^{3}\left ( 2ax^{2} \right )^{2}\)

\(10\times \frac{1}{a^{3}}\times 4a^{2}=5\) soi 

a=8 cao 

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

Ans:(i) \(80(x^4)\), \(-32(x^5)\)

(ii) \((-32 +80p)(x^{5})=0\)

\(p=\frac{2}{5}\) or  \(\frac{32}{80}\)

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)  Ind term =\((2x) ^{3}\times (\frac{k}{x})^{3}\times _{6}C^{3}\textrm{}=540\rightarrow K=1^{\frac{1}{2}}\)

(ii) Term, in \(15\times 16\times k^{2}\)=\((or540x^{2})\)

Question 

The coefficient of \(x^{3}\) in the expansion of \( (1-px)^{5}\) is −2160. Find the value of the constant p.

Answer/Explanation

\(_{3}^{5}\textrm{C}\left [ (-) (px)^{3}\right ](-1)10p^{3}=-2160\)

then divide and take cube root,

\(p=6\)

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.                                                                                 [4]

     (b) Find the coefficient of x6 in the expansion of \((1-x^{3})(2x^{2}+\frac{a}{x})^{6}\)    [1]

Answer/Explanation

Ans

5 (a) \(6C2\times \left [ 2(x^{2}) \right ]^{4}\times \left [ \frac{a}{(x)} \right ]^{2}, \ 6C3\times \left [ 2(x^{2}) \right ]^{3}\times \left [ \frac{a}{(x)}^{3} \right ]\)

            \(15\times 2^{4}\times a^{2}=20\times 2^{3}\times a^{3}\)

            \(a=\frac{15\times 2^{4}}{20\times 2^{3}}=\frac{3}{2}\)

5 (b) 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

kx2 + 2x – k = kx – 2  leading to kx2 + (-k + 2) x – k + 2 [=0]

(-k + 2)2 – 4k( – k + 2)

5k2 – 12k + 4 or (-k + 2) (-k + 2 -4k)

(- k + 2) (-5k + 2)

\(\frac{2}{5} < k < 2\)

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

Ans:

(a) \(5C2[2(x)]^3[\frac{a}{(x^2)}]^2\)
\(10 \times 8 \times a^2 (\frac{x^3}{x^4})=720(\frac{1}{x})\)
a = ±3

(b) \(5C4[2(x)][\frac{their a}{(x^2)}]^4\)
810 identified

Question

 (a) Find the first three terms in the expansion of (3 − 2x)5 in ascending powers of x.     [3]

   (b) Hence find the coefficient of x2 in the expansion of (4 + x)2 (3 − 2x)5

Answer/Explanation

Ans

3 (a) 243 
         −810x 
         +1080x2 
3 (b) (4 +x)2 = 16 + 8x +x2 
          Coefficient of x2 is 16 × 1080 + 8 × (−810) + 243 
           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 x2 in the expansion of (1+x)5(1-2x)6.

Answer/Explanation

Ans:
a) 1-5x+10x2
b) 1-12x+60x2
c) (1+5x+10x2)(1-12x+60x2) leading to 60-60+10
     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) \(6_{C_{2}}\times \left ( 3x \right )^{4} \left ( \frac{2}{x^{2}} \right )^{2}\)

\(15 \times 3^{4}\times 2^{2}\)

4860

(b) Their 4860 and one other relevant term

Other term = \(6 C3 \left ( 3x \right )^{3}\left ( \frac{2}{x^{2}}\right )^{3} or 6C3\times 3 ^{3} \times 2^{3} or 4320\)

[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

\(\left ( kx+\frac{1}{x} \right )^{5}+\left ( 1-\frac{2}{x} \right )^{8}\)

    Coefficient in \(\left ( kx+\frac{1}{x} \right )^{5}=10\times k^{2}\)

    Coefficient in \(\left ( 1-\frac{2}{x} \right )^{8}=8\times -2\)

    10k² − 16 = 74→k = 3