Free Objective Test 01 Practice Test - 11th and 12th 

$(x+a{)}^n$ + $(x-a{)}^n$ = 2$[x^n{+}^n{C}_2{x}^{n-2}{a}^2{+}^n{C}_4{x}^{n-4}{a}^4{+}^n{C}_6{x}^{n-6}{a}^6+.......]$

Here, n = 6, x =  $\sqrt{2}$, a = 1;

${}^6{C}_2$ = 15, ${}^6{C}_4$ = 15, ${}^6{C}_6$ = 1

∴ ($\sqrt{2}+1{)}^6$ + ($\sqrt{2}-1{)}^6$ = 2$\left[\right(\sqrt{2}{)}^6+15.(\sqrt{2}{)}^4.1+15(\sqrt{2}{)}^2.1+1.1]$

$=2[8+15\times 4+15\times 2+1]=198$

$(1+3x+3x^2+x^3{)}^6$ = $\left(\right(1+x{)}^3{)}^6$ = $(1+x{)}^{18}$

Hence the middle term is ${10}^{th}$.

Middle term of $(1+2x{)}^{2n}$ is $T_{n+1}$ = ${}^{2}n{C}_n$$x^n$

=  $\frac{\left(2n\right)!}{n!n!}$$x^n$ =  $\frac{\mathrm{1.3.5.......}(2n-1)}{n!}$ $2^n$$x^n$.

${}^{18}{C}_{2r+3}$ = ${}^{18}{C}_{r-3}$   ⇒ 2r + 3 + r - 3 = 18   ⇒ r = 6  

$T_{r+1}$ = ${}^5{C}_r$$(x^2{)}^{5-r}$($\frac{k}{x}{)}^r$

For coefficient of x, 10 - 2r - r = 1  ⇒ r = 3

Hence, $T_{3+1}$ = ${}^5{C}_3$$(x^2{)}^{5-3}$($\frac{k}{x}{)}^3$

According to question, 10$k^3$ = 270  ⇒ k = 3.

In the expansion of $(a{x}^2+\frac{1}{bx}{)}^{11}$, the general

term is

$T_{r+1}$ = ${}^{11}{C}_r$$(a{x}^2{)}^{11-r}$$(\frac{1}{bx}{)}^r$ =  ${}^{11}{C}_r$$(a{)}^{11-r}$$\left(\frac{1}{b^r}\right)x^{22-3r}$

For $x^7$, we must have 22 - 3r = 7  ⇒ r = 5, and

the coefficient of $x^7$ = ${}^{11}{C}_5$.$a^{11-5}\frac{1}{5}$) = ${}^{11}{C}_5$  $\frac{a^6}{b^5}$

Similarly, in the expansion of $(ax-\frac{1}{b{x}^2}{)}^{11}$, the

general term is $T_{r+1}$ =  ${}^{11}{C}_r(-1{)}^r$ $\frac{a^{11-r}}{b^r}$.$x^{11-3r}$

For $x^{-7}$ we must have, 11 - 3r = -7  ⇒ r = 6, and

the coefficient of $x^{-7}$ is ${}^{11}{C}_6$ $\frac{a^5}{b^6}$ =  ${}^{11}{C}_5$ $\frac{a^5}{b^6}$.

As given,  ${}^{11}{C}_5$ $\frac{a^6}{b^5}$ =  ${}^{11}{C}_5$ $\frac{a^5}{b^6}$ ⇒ ab = 1.

(x - 1)(x - 2)(x - 3)..............(x - 100)

Number of terms = 100;

∴ Coefficient of $x^{99}$

= (x - 1)(x - 2)(x - 3)........(x -100)

= (-1 - 2 - 3 - .......... - 100) = -(1 + 2 +.........+ 100)

= -  $\frac{100\times 101}{2}$ = - 5050.

Since $(n+2{)}^{th}$ term is the middle term in the 

expansion of $(1+x{)}^{2n+2}$, therefore p = ${}^{2n+2}{C}_{n+1}$.

Since $(n+1{)}^{th}$ and $(n+2{)}^{th}$ terms are middle terms in the expansion of $(1+x{)}^{2n+1}$,

therefore q = ${}^{2n+1}{C}_n$ and r =${}^{2n+1}{C}_{2n+1}$.

 But ,${}^{2n+1}{C}_n$ +${}^{2n+1}{C}_{n+1}$ = ${}^{2n+2}{C}_{n+1}$

∴ q + r = p

$T_3$ = ${}^5{C}_2$. $x^2$$(x^{lo{g}_{10}x}{)}^3$ = ${10}^6$

Put ${}^5{C}_2$ = 10 [ ∵$lo{g}_{10}10=1$].

If x = 10, then ${10}^3$.${10}^{2.1}$ = ${10}^5$ is satisfied.

Hence x = 10.

$\frac{(1-3x{)}^{1/2}+(1-x{)}^{5/3}}{2[1-\frac{x}{4}{]}^{1/2}}\frac{1}{2}\left[\right(1-\frac{3}{2}x)+(1-\frac{5x}{3}\left)\right](1+\frac{x}{8})\frac{1}{2}[2-\frac{19}{6}x]](1+\frac{x}{8})=\frac{1}{2}[2-(-\frac{1}{4}+\frac{19}{6}\left)x\right]=1-\frac{35}{24}x\therefore (a,b)=(1,-\frac{35}{24})$

As given $(1+ax{)}^n$ = 1 + 8x + 24$x^2$ + .........

⇒ 1 +  $\frac{n}{1}$ax +  $\frac{n(n-1)}{1.2}$$a^2{x}^2$ + ......... = 1 + 8x + 24$x^2$ + ...........

⇒na = 8, $\frac{n(n-1)}{1.2}$$a^2$ = 24 ⇒ na(n-1)a = 48

⇒8(8 - a) = 48 ⇒ 8 - a = 6 ⇒ a = 2 ⇒ n = 4.

$n^p$ - n is divisible by p for any natural number

greater than 1. It is Fermet's theorem.

Trick: Let n = 4 and p = 2

Then $(4{)}^2$ - 4 = 16 - 4 = 12, it is divisible by 2.

So,it is true for any natural number greater than 1.

$(2+\sqrt{(}2){)}^4$ =  $(\sqrt{2}{)}^4$$(\sqrt{2}+1{)}^4$

 = 4[${}^4{C}_0$ + ${}^4{C}_1$($\sqrt{2})$ + ${}^4{C}_2$($\sqrt{2}{)}^2$ + ${}^4{C}_3$($\sqrt{2}{)}^3$ +${}^4{C}_4$ ($\sqrt{2}{)}^4$]

= 4 [1 + 4 $\sqrt{2}$ +  $\frac{4.3}{2}$.2 +  $\frac{\mathrm{4.3.2}}{\mathrm{1.2.3}}$.2 $\sqrt{2}$ + 4]

= 4[1 + 4 $\sqrt{2}$ + 12 + 8 $\sqrt{2}$ + 4] = 4[17 + 12 $\sqrt{2}$]

= 4[17 + 17]  = 4[34] = 136

$T_{r+1}$ = ${}^{642}{C}_r(5^{\frac{1}{2}}{)}^{642-r}.(7^{\frac{1}{6}}{)}^r$

Obviously, r should be a multiple of 6.

Total number of terms =  $\frac{642}{6}$ = 107,

but first term for r = 0 is also an integer. Hence total terms are 107 + 1 = 108.

Let $a_1,a_2,a_3,a_4$ be respectively the coefficients of $(r+1{)}^{\text{th}},(r+2{)}^{\text{th}},(r+3{)}^{\text{th}}\text{and}(r+4{)}^{\text{th}}$ terms in the expansion of $(1+x{)}^n$.
Then $a_1={{}^{n}C}_r,a_2={{}^{n}C}_{r+1},a_3={{}^{n}C}_{r+2},a_4={{}^{n}C}_{r+3}$
Now,
$\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{{{}^{n}C}_r}{{{}^{n}C}_r+{{}^{n}C}_{r+1}}+\frac{{{}^{n}C}_{r+2}}{{{}^{n}C}_{r+2}+{{}^{n}C}_{r+3}}=\frac{{{}^{n}C}_r}{{{}^{n+1}C}_{r+1}}+\frac{{{}^{n}C}_{r+2}}{{{}^{n+1}C}_{r+3}}=\frac{{{}^{n}C}_r}{\frac{n+1}{r+1}{{}^{n}C}_r}+\frac{{{}^{n}C}_r}{\frac{n+1}{r+3}{{}^{n}C}_{r+2}}=\frac{r+1}{n+1}+\frac{r+3}{n+3}=\frac{2(r+2)}{n+1}$
Also, solving  the R.H.S, we get
$\frac{2a_2}{a_2+a_3}=2\frac{{{}^{n}C}_{r+1}}{{{}^{n}C}_{r+1}+{{}^{n}C}_{r+2}}=2\frac{{{}^{n}C}_{r+1}}{{{}^{n+1}C}_{r+2}}=\frac{2(r+2)}{n+1}$